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1
.gitignore vendored
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out out
*.pdf

8
1/10-mechanical-similarity.md
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@ -82,3 +82,11 @@ which express $\bar{U}$ and $\bar{T}$ in terms of the total energy of the system
In particular, for small oscillations ($k=2$) we have $\bar{T} = \bar{U}$, i.e. the mean values of the kinetic and potential energies are equal. For a Newtonian interaction $(k = - 1)$, $2\bar{T} = - \bar{U}$, and $E=-\bar{T}$, in accordance with the fact that, in such an interaction, the motion takes place in a finite region of space only if the total energy is negative (see `1/15`). In particular, for small oscillations ($k=2$) we have $\bar{T} = \bar{U}$, i.e. the mean values of the kinetic and potential energies are equal. For a Newtonian interaction $(k = - 1)$, $2\bar{T} = - \bar{U}$, and $E=-\bar{T}$, in accordance with the fact that, in such an interaction, the motion takes place in a finite region of space only if the total energy is negative (see `1/15`).
[^1]: The expression on the right of `1/10.5` is sometimes called the virial of the system. [^1]: The expression on the right of `1/10.5` is sometimes called the virial of the system.
<!-- PROBLEMS -->
<!-- PROBLEM 1. Find the ratio of the times in the same path for particles having different -->
<!-- masses but the same potential energy. -->
<!-- SOLUTION. t'/t = (m'/m). -->
<!-- PROBLEM 2. Find the ratio of the times in the same path for particles having the same mass -->
<!-- but potential energies differing by a constant factor. -->
<!-- SOLUTION. I'/t -->

10
1/10_problems.md
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@ -1,10 +0,0 @@
---
---
PROBLEMS
PROBLEM 1. Find the ratio of the times in the same path for particles having different
masses but the same potential energy.
SOLUTION. t'/t = (m'/m).
PROBLEM 2. Find the ratio of the times in the same path for particles having the same mass
but potential energies differing by a constant factor.
SOLUTION. I'/t

35
1/11-motion-in-one-dimension.md
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@ -22,7 +22,8 @@ The equations of motion corresponding to these Lagrangians can be integrated in
``` ```
The two arbitrary constants in the solution of the equations of motion are here represented by the total energy E and the constant of integration. The two arbitrary constants in the solution of the equations of motion are here represented by the total energy E and the constant of integration.
Since the kinetic energy is essentially positive, the total energy always exceeds the potential energy, i.e. the motion can take place only in those regions of space where U(x) < E. For example, let the function U(x) be of the form shown in 'TODO Fig. 6 (p. 26)'. If we draw in the figure a horizontal line corresponding to a given value of the total energy, we immediately find the possible regions of motion. In the example of Fig. 6, the motion can occur only in the range AB or in the range to the right of C.
Since the kinetic energy is essentially positive, the total energy always exceeds the potential energy, i.e. the motion can take place only in those regions of space where $U(x) \lt E$. For example, let the function U(x) be of the form shown in 'TODO Fig. 6 (p. 26)'. If we draw in the figure a horizontal line corresponding to a given value of the total energy, we immediately find the possible regions of motion. In the example of Fig. 6, the motion can occur only in the range AB or in the range to the right of C.
The points at which the potential energy equals the total energy, The points at which the potential energy equals the total energy,
```load ```load
@ -40,3 +41,35 @@ from X1 to X2 (because of the reversibility property, `1/5`) or, by `1/11.3`),
``` ```
where $x_1$ and $x_2$ are roots of equation `1/11.4` for the given value of $E$. This formula gives the period of the motion as a function of the total energy of the particle. where $x_1$ and $x_2$ are roots of equation `1/11.4` for the given value of $E$. This formula gives the period of the motion as a function of the total energy of the particle.
<!-- PROBLEMS -->
<!-- PROBLEM 1. Determine the period of oscillations of a simple pendulum (a particle of mass -->
<!-- m suspended by a string of length l in a gravitational field) as a function of the amplitude of -->
<!-- the oscillations. -->
<!-- SOLUTION. The energy of the pendulum is E = 1ml2j2-mgl cos = -mgl cos to, where -->
<!-- o is the angle between the string and the vertical, and to the maximum value of . Calculating -->
<!-- the period as the time required to go from = 0 to = Do, multiplied by four, we find -->
<!-- -cos -->
<!-- po) -->
<!-- The substitution sin $ = sin 10/sin 100 converts this to T = /(l/g)K(sin 100), where -->
<!-- 1/75 -->
<!-- - -->
<!-- is the complete elliptic integral of the first kind. For sin 100 22 100 < 1 (small oscillations), -->
<!-- an expansion of the function K gives -->
<!-- T = -->
<!-- §12 -->
<!-- Determination of the potential energy -->
<!-- 27 -->
<!-- The first term corresponds to the familiar formula. -->
<!-- PROBLEM 2. Determine the period of oscillation, as a function of the energy, when a -->
<!-- particle of mass m moves in fields for which the potential energy is -->
<!-- (a) U = Alx -->
<!-- (b) U = Uo/cosh2ax, -U0 0, (c) U = Uotan2ax. -->
<!-- SOLUTION. (a): -->
<!-- T = -->
<!-- By the substitution yn = u the integral is reduced to a beta function, which can be expressed -->
<!-- in terms of gamma functions: -->
<!-- The dependence of T on E is in accordance with the law of mechanical similarity (10.2), -->
<!-- (10.3). -->
<!-- (b) T = (7/a)V(2m/E). -->
<!-- (c) T =(t/a)v[2m/(E+U0)] -->

31
1/11_problems.md
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@ -1,31 +0,0 @@
PROBLEMS
PROBLEM 1. Determine the period of oscillations of a simple pendulum (a particle of mass
m suspended by a string of length l in a gravitational field) as a function of the amplitude of
the oscillations.
SOLUTION. The energy of the pendulum is E = 1ml2j2-mgl cos = -mgl cos to, where
o is the angle between the string and the vertical, and to the maximum value of . Calculating
the period as the time required to go from = 0 to = Do, multiplied by four, we find
-cos
po)
The substitution sin $ = sin 10/sin 100 converts this to T = /(l/g)K(sin 100), where
1/75
-
is the complete elliptic integral of the first kind. For sin 100 22 100 < 1 (small oscillations),
an expansion of the function K gives
T =
§12
Determination of the potential energy
27
The first term corresponds to the familiar formula.
PROBLEM 2. Determine the period of oscillation, as a function of the energy, when a
particle of mass m moves in fields for which the potential energy is
(a) U = Alx
(b) U = Uo/cosh2ax, -U0 0, (c) U = Uotan2ax.
SOLUTION. (a):
T =
By the substitution yn = u the integral is reduced to a beta function, which can be expressed
in terms of gamma functions:
The dependence of T on E is in accordance with the law of mechanical similarity (10.2),
(10.3).
(b) T = (7/a)V(2m/E).
(c) T =(t/a)v[2m/(E+U0)]

12
1/13-the-reduced-mass.md
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@ -34,4 +34,14 @@ where
is called the *reduced mass*. The function `1/13.3` is formally identical with the Lagrangian of a particle of mass $m$ moving in an external field $U(\v{r})$ which is symmetrical about a fixed origin. is called the *reduced mass*. The function `1/13.3` is formally identical with the Lagrangian of a particle of mass $m$ moving in an external field $U(\v{r})$ which is symmetrical about a fixed origin.
Thus the problem of the motion of two interacting particles is equivalent to that of the motion of one particle in a given external field $U(\v{r})$. From the solution $\v{r} = \v{r}(t)$ of this problem, the paths $\v{r}_1 = \v{r}_1(t)$ and $\v{r}_2 = \v{r}_2(t)$ of the two particles separately, relative to their common centre of mass, are obtained by means of formulae `1/13.2`. <!-- PROBLEM -->
<!-- A system consists of one particle of mass M and n particles with equal masses m. Eliminate -->
<!-- the motion of the centre of mass and so reduce the problem to one involving n particles. -->
<!-- SOLUTION. Let R be the radius vector of the particle of mass M, and Ra (a = 1, 2, ..., n) -->
<!-- those of the particles of mass m. We put ra = Ra-R and take the origin to be at the centre -->
<!-- of mass: MR+mER = 0. Hence where =M + nm; Ra = R + ra. -->
<!-- Substitution in the Lagrangian L = gives -->
<!-- ra. -->
<!-- The potential energy depends only on the distances between the particles, and so can be -->
<!-- written as a function of the ra. -->
<!-- Thus the problem of the motion of two interacting particles is equivalent to that of the motion of one particle in a given external field $U(\v{r})$. From the solution $\v{r} = \v{r}(t)$ of this problem, the paths $\v{r}_1 = \v{r}_1(t)$ and $\v{r}_2 = \v{r}_2(t)$ of the two particles separately, relative to their common centre of mass, are obtained by means of formulae `1/13.2`. -->

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1/13_problem.md
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@ -1,11 +0,0 @@
PROBLEM
A system consists of one particle of mass M and n particles with equal masses m. Eliminate
the motion of the centre of mass and so reduce the problem to one involving n particles.
SOLUTION. Let R be the radius vector of the particle of mass M, and Ra (a = 1, 2, ..., n)
those of the particles of mass m. We put ra = Ra-R and take the origin to be at the centre
of mass: MR+mER = 0. Hence where =M + nm; Ra = R + ra.
Substitution in the Lagrangian L = gives
ra.
The potential energy depends only on the distances between the particles, and so can be
written as a function of the ra.

159
1/14-motion-in-a-central-field.md
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@ -21,3 +21,162 @@ In the present case, the generalised momentum $p_\phi=mr^2\dot{\phi}$ is the sam
``` ```
This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The expression $\mfrac{1}{2}r\cdot r\dd{\phi}$ is the area of the sector bounded by two neighbouring radius vectors and an element of the path `1/fig8` This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The expression $\mfrac{1}{2}r\cdot r\dd{\phi}$ is the area of the sector bounded by two neighbouring radius vectors and an element of the path `1/fig8`
§14
Motion in a central field
31
(Fig. 8). Calling this area df, we can write the angular momentum of the par-
ticle as
M = 2mf,
(14.3)
where the derivative f is called the sectorial velocity. Hence the conservation
of angular momentum implies the constancy of the sectorial velocity: in equal
times the radius vector of the particle sweeps out equal areas (Kepler's second
law).t
rdd
dd
0
FIG. 8
The complete solution of the problem of the motion of a particle in a central
field is most simply obtained by starting from the laws of conservation of
energy and angular momentum, without writing out the equations of motion
themselves. Expressing in terms of M from (14.2) and substituting in the
expression for the energy, we obtain
E = =
(14.4)
Hence
(14.5)
or, integrating,
constant.
(14.6)
Writing (14.2) as do = M dt/mr2, substituting dt from (14.5) and integrating,
we find
constant.
(14.7)
Formulae (14.6) and (14.7) give the general solution of the problem. The
latter formula gives the relation between r and , i.e. the equation of the path.
Formula (14.6) gives the distance r from the centre as an implicit function of
time. The angle o, it should be noted, always varies monotonically with time,
since (14.2) shows that & can never change sign.
t The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.
32
Integration of the Equations of Motion
§14
The expression (14.4) shows that the radial part of the motion can be re-
garded as taking place in one dimension in a field where the "effective poten-
tial energy" is
(14.8)
The quantity M2/2mr2 is called the centrifugal energy. The values of r for which
U(r)
(14.9)
determine the limits of the motion as regards distance from the centre.
When equation (14.9) is satisfied, the radial velocity j is zero. This does not
mean that the particle comes to rest as in true one-dimensional motion, since
the angular velocity o is not zero. The value j = 0 indicates a turning point
of the path, where r(t) begins to decrease instead of increasing, or vice versa.
If the range in which r may vary is limited only by the condition r > rmin,
the motion is infinite: the particle comes from, and returns to, infinity.
If the range of r has two limits rmin and rmax, the motion is finite and the
path lies entirely within the annulus bounded by the circles r = rmax and
r = rmin- This does not mean, however, that the path must be a closed curve.
During the time in which r varies from rmax to rmin and back, the radius
vector turns through an angle Ao which, according to (14.7), is given by
Mdr/r2
(14.10)
The condition for the path to be closed is that this angle should be a rational
fraction of 2n, i.e. that Ao = 2nm/n, where m and n are integers. In that case,
after n periods, the radius vector of the particle will have made m complete
revolutions and will occupy its original position, so that the path is closed.
Such cases are exceptional, however, and when the form of U(r) is arbitrary
the angle is not a rational fraction of 2nr. In general, therefore, the path
of a particle executing a finite motion is not closed. It passes through the
minimum and maximum distances an infinity of times, and after infinite time
it covers the entire annulus between the two bounding circles. The path
shown in Fig. 9 is an example.
There are only two types of central field in which all finite motions take
place in closed paths. They are those in which the potential energy of the
particle varies as 1/r or as r2. The former case is discussed in §15; the latter
is that of the space oscillator (see §23, Problem 3).
At a turning point the square root in (14.5), and therefore the integrands
in (14.6) and (14.7), change sign. If the angle is measured from the direc-
tion of the radius vector to the turning point, the parts of the path on each
side of that point differ only in the sign of for each value of r, i.e. the path
is symmetrical about the line = 0. Starting, say, from a point where = rmax
the particle traverses a segment of the path as far as a point with r rmin,
§14
Motion in a central field
33
then follows a symmetrically placed segment to the next point where r = rmax,
and so on. Thus the entire path is obtained by repeating identical segments
forwards and backwards. This applies also to infinite paths, which consist of
two symmetrical branches extending from the turning point (r = rmin) to
infinity.
'max
min
so
FIG. 9
The presence of the centrifugal energy when M # 0, which becomes
infinite as 1/22 when r -> 0, generally renders it impossible for the particle to
reach the centre of the field, even if the field is an attractive one. A "fall" of
the particle to the centre is possible only if the potential energy tends suffi-
ciently rapidly to -00 as r 0. From the inequality
1mr2 = E- U(r) - M2/2mr2
or r2U(Y)+M2/2m < Er2, it follows that r can take values tending to zero
only if
(14.11)
i.e. U(r) must tend to - 8 either as - a/r2 with a > M2/2m, or proportionally
to - 1/rn with n > 2.
PROBLEMS
PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass
m moving on the surface of a sphere of radius l in a gravitational field).
SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the
polar axis vertically downwards, the Lagrangian of the pendulum is
1ml2(02 + 62 sin20) +mgl cos 0.
2*
34
Integration of the Equations of Motion
§14
The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the
z-component of angular momentum, is conserved:
(1)
The energy is
E = cos 0
(2)
= 0.
Hence
(3)
where the "effective potential energy" is
Ueff(0) = COS 0.
For the angle o we find, using (1),
do
(4)
The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively.
The range of 0 in which the motion takes place is that where E > Ueff, and its limits
are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots
between -1 and +1; these define two circles of latitude on the sphere, between which the
path lies.
PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a
cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational
field.
SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the
polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co-
ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is
= a.
By the same method as in Problem 1, we find
==
The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots;
these define two horizontal circles on the cone, between which the path lies.
PROBLEM 3. Integrate the equations of motion for a pendulum of mass M2, with a mass M1
at the point of support which can move on a horizontal line lying in the plane in which m2
moves (Fig. 2, §5).
SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The
generalised momentum Px, which is the horizontal component of the total momentum of the
system, is therefore conserved
Px = cos = constant.
(1)
The system may always be taken to be at rest as a whole. Then the constant in (1) is zero
and integration gives
(m1+m2)x+m2) sin = constant,
(2)
which expresses the fact that the centre of mass of the system does not move horizontally.

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1/15-keplers-problem.md Normal file
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@ -0,0 +1,216 @@
§15
Kepler's problem
35
Using (1), we find the energy in the form
E
(3)
Hence
Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
dulum, which moves in an arc of a circle.
$15. Kepler's problem
An important class of central fields is formed by those in which the poten-
tial energy is inversely proportional to r, and the force accordingly inversely
proportional to r2. They include the fields of Newtonian gravitational attrac-
tion and of Coulomb electrostatic interaction; the latter may be either attrac-
tive or repulsive.
Let us first consider an attractive field, where
U=-a/r
(15.1)
with a a positive constant. The "effective" potential energy
(15.2)
is
of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
r
8
it tends to zero from negative values ; for r = M2/ma it has a minimum value
Ueff, min = -mx2/2M2.
(15.3)
Ueff
FIG. 10
It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
for E > 0.
36
Integration of the Equations of Motion
§15
The shape of the path is obtained from the general formula (14.7). Substi-
tuting there U = - a/r and effecting the elementary integration, we have
o =
- constant.
Taking the origin of such that the constant is zero, and putting
P = M2/ma, e= [1 1+(2EM2/mo2)]
(15.4)
we can write the equation of the path as
p/r = 1+e coso.
(15.5)
This is the equation of a conic section with one focus at the origin; 2p is called
the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
is seen from (15.5) to be such that the point where = 0 is the point nearest
to the origin (called the perihelion).
In the equivalent problem of two particles interacting according to the law
(15.1), the orbit of each particle is a conic section, with one focus at the centre
of mass of the two particles.
It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
has been said earlier in this section. According to the formulae of analytical
geometry, the major and minor semi-axes of the ellipse are
a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
(15.6)
y
X
2b
ae
2a
FIG. 11
The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
becomes a circle. It may be noted that the major axis of the ellipse depends
only on the energy of the particle, and not on its angular momentum. The
least and greatest distances from the centre of the field (the focus of the
ellipse) are
rmin = =p/(1+e)=a(1-e),
max=p(1-e)=a(1+e). = = (15.7)
These expressions, with a and e given by (15.6) and (15.4), can, of course,
also be obtained directly as the roots of the equation Ueff(r) = E.
§15
Kepler's problem
37
The period T of revolution in an elliptical orbit is conveniently found by
using the law of conservation of angular momentum in the form of the area
integral (14.3). Integrating this equation with respect to time from zero to
T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
f = nab, and by using the formulae (15.6) we find
T = 2ma3/2-(m/a)
= ma((m2E3).
(15.8)
The proportionality between the square of the period and the cube of the
linear dimension of the orbit has already been demonstrated in §10. It may
also be noted that the period depends only on the energy of the particle.
For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
tance of the perihelion from the focus is
rmin ==pl(e+1)=a(e-1), = =
(15.9)
where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
y
p
ale-1)
FIG. 12
If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
perihelion distance rmin = 1p. This case occurs if the particle starts from rest
at infinity.
The co-ordinates of the particle as functions of time in the orbit may be
found by means of the general formula (14.6). They may be represented in a
convenient parametric form as follows.
Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
we can write the integral (14.6) for the time as
t
=
=
38
Integration of the Equations of Motion
§15
The obvious substitution r-a = - ae cos $ converts the integral to
sioant
If time is measured in such a way that the constant is zero, we have the
following parametric dependence of r on t:
r = a(1-e cos ), t =
(15.10)
the particle being at perihelion at t = 0. The Cartesian co-ordinates
x = r cos o, y = r sin (the x and y axes being respectively parallel to the
major and minor axes of the ellipse) can likewise be expressed in terms of
the parameter $. From (15.5) and (15.10) we have
ex = = =
y is equal to W(r2-x2). Thus
x = a(cos & - e),
y = =av(1-e2) $.
(15.11)
A complete passage round the ellipse corresponds to an increase of $ from 0
to 2nr.
Entirely similar calculations for the hyperbolic orbits give
r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
(15.12)
x = a(e-cosh ) y = a1/(e2-1)sinh &
where the parameter $ varies from - 00 to + 00.
Let us now consider motion in a repulsive field, where
U
(a>0).
(15.13)
Here the effective potential energy is
Utt
and decreases monotonically from + 00 to zero as r varies from zero to
infinity. The energy of the particle must be positive, and the motion is always
infinite. The calculations are exactly similar to those for the attractive field.
The path is a hyperbola (or, if E = 0, a parabola):
pr r = =1-e coso, =
(15.14)
where P and e are again given by (15.4). The path passes the centre of the
field in the manner shown in Fig. 13. The perihelion distance is
rmin =p(e-1)=a(e+1). =
(15.15)
The time dependence is given by the parametric equations
= =(ma3/a)(esinh+) =
(15.16)
x
= a(cosh & e ,
y = av((e2-1) sinh &
§15
Kepler's problem
39
To conclude this section, we shall show that there is an integral of the mo-
tion which exists only in fields U = a/r (with either sign of a). It is easy to
verify by direct calculation that the quantity
vxM+ar/r
(15.17)
is constant. For its total time derivative is v
since M = mr xv,
Putting mv = ar/r3 from the equation of motion, we find that this expression
vanishes.
y
0
(I+e)
FIG. 13
The direction of the conserved vector (15.17) is along the major axis from
the focus to the perihelion, and its magnitude is ae. This is most simply
seen by considering its value at perihelion.
It should be emphasised that the integral (15.17) of the motion, like M and
E, is a one-valued function of the state (position and velocity) of the particle.
We shall see in §50 that the existence of such a further one-valued integral
is due to the degeneracy of the motion.
PROBLEMS
PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
moving in a parabola in a field U = -a/r.
SOLUTION. In the integral
we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
the required dependence:
r=1p(1+n2),
t=
y=pn.
40
Integration of the Equations of Motion
§15
The parameter n varies from - 00 to +00.
PROBLEM 2. Integrate the equations of motion for a particle in a central field
U = - a/r2 (a > 0).
SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
(a) for E > andM 0 and
(b) for E>0 0nd and M 2/2m a,
(c) for E <0 and Ms1
In all three cases
In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
origin as
00. The fall from a given value of r takes place in a finite time, namely
PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
(14.10), which we write as
in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
the first-order term gives the required change so:
(1)
where we have changed from the integration over r to one over , along the path of the "un-
perturbed" motion.
In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.

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§16. Disintegration of particles
IN many cases the laws of conservation of momentum and energy alone can
be used to obtain important results concerning the properties of various mech-
anical processes. It should be noted that these properties are independent of
the particular type of interaction between the particles involved.
Let us consider a "spontaneous" disintegration (that is, one not due to
external forces) of a particle into two "constituent parts", i.e. into two other
particles which move independently after the disintegration.
This process is most simply described in a frame of reference in which the
particle is at rest before the disintegration. The law of conservation of momen-
tum shows that the sum of the momenta of the two particles formed in the
disintegration is then zero; that is, the particles move apart with equal and
opposite momenta. The magnitude Po of either momentum is given by the
law of conservation of energy:
here M1 and m2 are the masses of the particles, E1t and E2i their internal
energies, and E the internal energy of the original particle. If € is the "dis-
integration energy", i.e. the difference
E= E:-E11-E2i,
(16.1)
which must obviously be positive, then
(16.2)
which determines Po; here m is the reduced mass of the two particles. The
velocities are V10 = Po/m1, V20 = Po/m2.
Let us now change to a frame of reference in which the primary particle
moves with velocity V before the break-up. This frame is usually called the
laboratory system, or L system, in contradistinction to the centre-of-mass
system, or C system, in which the total momentum is zero. Let us consider
one of the resulting particles, and let V and V0 be its velocities in the L and
the C system respectively. Evidently V = V+vo, or V -V = V0, and SO
(16.3)
where 0 is the angle at which this particle moves relative to the direction of
the velocity V. This equation gives the velocity of the particle as a function
41
42
Collisions Between Particles
§16
of its direction of motion in the L system. In Fig. 14 the velocity V is repre-
sented by a vector drawn to any point on a circle+ of radius vo from a point
A at a distance V from the centre. The cases V < vo and V>00 are shown
in Figs. 14a, b respectively. In the former case 0 can have any value, but in
the latter case the particle can move only forwards, at an angle 0 which does
not exceed Omax, given by
(16.4)
this is the direction of the tangent from the point A to the circle.
C
V
V
VO
VO
max
oo
oo
A
V
A
V
(a) V<VO
)V>Vo
FIG. 14
The relation between the angles 0 and Oo in the L and C systems is evi-
dently (Fig. 14)
tan
(16.5)
If this equation is solved for cos Oo, we obtain
V
0
(16.6)
For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus
sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however,
the relation is not one-to-one: for each value of 0 there are two values of Oo,
which correspond to vectors V0 drawn from the centre of the circle to the
points B and C (Fig. 14b), and are given by the two signs in (16.6).
In physical applications we are usually concerned with the disintegration
of not one but many similar particles, and this raises the problem of the
distribution of the resulting particles in direction, energy, etc. We shall
assume that the primary particles are randomly oriented in space, i.e. iso-
tropically on average.
t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral
section.
§16
Disintegration of particles
43
In the C system, this problem is very easily solved: every resulting particle
(of a given kind) has the same energy, and their directions of motion are
isotropically distributed. The latter fact depends on the assumption that the
primary particles are randomly oriented, and can be expressed by saying
that the fraction of particles entering a solid angle element doo is proportional
to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is
obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is
(16.7)
The corresponding distributions in the L system are obtained by an
appropriate transformation. For example, let us calculate the kinetic energy
distribution in the L system. Squaring the equation V = V0 + V, we have
2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the
kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of
particle is under consideration, and substituting in (16.7), we find the re-
quired distribution:
(1/2mvov) dT.
(16.8)
The kinetic energy can take values between Tmin = 3m(e0-V)2 and
Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed
uniformly over this range.
When a particle disintegrates into more than two parts, the laws of con-
servation of energy and momentum naturally allow considerably more free-
dom as regards the velocities and directions of motion of the resulting particles.
In particular, the energies of these particles in the C system do not have
determinate values. There is, however, an upper limit to the kinetic energy
of any one of the resulting particles. To determine the limit, we consider
the system formed by all these particles except the one concerned (whose
mass is M1, say), and denote the "internal energy" of that system by Ei'.
Then the kinetic energy of the particle M1 is, by (16.1) and (16.2),
T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the
primary particle. It is evident that T10 has its greatest possible value
when E/' is least. For this to be so, all the resulting particles except M1
must be moving with the same velocity. Then Ei is simply the sum of their
internal energies, and the difference E;-E-E; is the disintegration
energy E. Thus
T10,max = (M - M1) E M.
(16.9)
PROBLEMS
PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra-
tion into two particles.
SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling
010 simply Oo and using formula (16.5) for each of the two particles, we can put
7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two
equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then
44
Collisions Between Particles
§17
form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using
(16.2),
(m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6
PROBLEM 2. Find the angular distribution of the resulting particles in the L system.
SOLUTION. When vo > V, we substitute (16.6), with the plus sign of the radical, in (16.7),
obtaining
(0 .
When vo < V, both possible relations between Oo and 0 must be taken into account. Since,
when 0 increases, one value of Oo increases and the other decreases, the difference (not the
sum) of the expressions d cos Oo with the two signs of the radical in (16.6) must be taken.
The result is
(0 max).
PROBLEM 3. Determine the range of possible values of the angle 0 between the directions
of motion of the two resulting particles in the L system.
SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula (16.5)
(see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema
of the resulting expression gives the following ranges of 0, depending on the relative magni-
tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10
< 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by
sin =
§17. Elastic collisions
A collision between two particles is said to be elastic if it involves no change
in their internal state. Accordingly, when the law of conservation of energy
is applied to such a collision, the internal energy of the particles may be
neglected.
The collision is most simply described in a frame of reference in which the
centre of mass of the two particles is at rest (the C system). As in $16, we
distinguish by the suffix 0 the values of quantities in that system. The velo-
cities of the particles before the collision are related to their velocities V1 and
V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
where V = V1-V2; see (13.2).
Because of the law of conservation of momentum, the momenta of the two
particles remain equal and opposite after the collision, and are also unchanged
in magnitude, by the law of conservation of energy. Thus, in the C system
the collision simply rotates the velocities, which remain opposite in direction
and unchanged in magnitude. If we denote by no a unit vector in the direc-
tion of the velocity of the particle M1 after the collision, then the velocities
of the two particles after the collision (distinguished by primes) are
V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
(17.1)
§17
Elastic collisions
45
In order to return to the L system, we must add to these expressions the
velocity V of the centre of mass. The velocities in the L system after the
collision are therefore
V1' =
(17.2)
V2' =
No further information about the collision can be obtained from the laws
of conservation of momentum and energy. The direction of the vector no
depends on the law of interaction of the particles and on their relative position
during the collision.
The results obtained above may be interpreted geometrically. Here it is
more convenient to use momenta instead of velocities. Multiplying equations
(17.2) by M1 and M2 respectively, we obtain
(17.3)
P2' muno+m2(p1+p2)/(m1+m2)
where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
mv and use the construction shown in Fig. 15. If the unit vector no is along
OC, the vectors AC and CB give the momenta P1' and P2' respectively.
When p1 and P2 are given, the radius of the circle and the points A and B
are fixed, but the point C may be anywhere on the circle.
C
p'
no
P'2
B
A
FIG. 15
Let us consider in more detail the case where one of the particles (m2, say) is
at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
momentum P1 of the particle M1 before the collision. The point A lies inside
or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
46
Collisions Between Particles
§17
diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
are the angles between the directions of motion after the collision and the
direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
gives the direction of no, is the angle through which the direction of motion
of m1 is turned in the C system. It is evident from the figure that 01 and O2
can be expressed in terms of X by
(17.4)
C
p'
P2
pi
P2
0
max
10,
X
O2
O2
B
B
A
0
A
Q
0
(a) m < m2
(b) m, m m m
AB=p : AO/OB= m/m2
FIG. 16
We may give also the formulae for the magnitudes of the velocities of the
two particles after the collision, likewise expressed in terms of X:
ib
(17.5)
The sum A1 + O2 is the angle between the directions of motion of the
particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
if M1 > M2.
When the two particles are moving afterwards in the same or in opposite
directions (head-on collision), we have X=TT, i.e. the point C lies on the
diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
In this case the velocities after the collision are
(17.6)
This value of V2' has the greatest possible magnitude, and the maximum
§17
Elastic collisions
47
energy which can be acquired in the collision by a particle originally at rest
is therefore
(17.7)
where E1 = 1M1U12 is the initial energy of the incident particle.
If M1 < M2, the velocity of M1 after the collision can have any direction.
If M1 > M2, however, this particle can be deflected only through an angle
not exceeding Omax from its original direction; this maximum value of A1
corresponds to the position of C for which AC is a tangent to the circle
(Fig. 16b). Evidently
sin Omax = OC|OA = M2/M1.
(17.8)
The collision of two particles of equal mass, of which one is initially at
rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
C
p'
P2
Q2
B
A
0
FIG. 17
Then
01=1x,
A2 = 1(-x),
(17.9)
12
=
(17.10)
After the collision the particles move at right angles to each other.
PROBLEM
Express the velocity of each particle after a collision between a moving particle (m1) and
another at rest (m2) in terms of their directions of motion in the L system.
SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
Hence
for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.
48
Collisions Between Particles
§18
§18. Scattering
As already mentioned in §17, a complete calculation of the result of a
collision between two particles (i.e. the determination of the angle x) requires
the solution of the equations of motion for the particular law of interaction
involved.
We shall first consider the equivalent problem of the deflection of a single
particle of mass m moving in a field U(r) whose centre is at rest (and is at
the centre of mass of the two particles in the original problem).
As has been shown in $14, the path of a particle in a central field is sym-
metrical about a line from the centre to the nearest point in the orbit (OA
in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
say) with this line. The angle X through which the particle is deflected as it
passes the centre is seen from Fig. 18 to be
X = -200.
(18.1)
A
X
to
FIG. 18
The angle do itself is given, according to (14.7), by
(M/r2) dr
(18.2)
taken between the nearest approach to the centre and infinity. It should be
recalled that rmin is a zero of the radicand.
For an infinite motion, such as that considered here, it is convenient to
use instead of the constants E and M the velocity Voo of the particle at infinity
and the impact parameter p. The latter is the length of the perpendicular
from the centre O to the direction of Voo, i.e. the distance at which the particle
would pass the centre if there were no field of force (Fig. 18). The energy
and the angular momentum are given in terms of these quantities by
E = 1mvoo²,
M = mpVoo,
(18.3)
§18
Scattering
49
and formula (18.2) becomes
dr
(18.4)
Together with (18.1), this gives X as a function of p.
In physical applications we are usually concerned not with the deflection
of a single particle but with the scattering of a beam of identical particles
incident with uniform velocity Voo on the scattering centre. The different
particles in the beam have different impact parameters and are therefore
scattered through different angles X. Let dN be the number of particles
scattered per unit time through angles between X and X + dx. This number
itself is not suitable for describing the scattering process, since it is propor-
tional to the density of the incident beam. We therefore use the ratio
do = dN/n,
(18.5)
where n is the number of particles passing in unit time through unit area of
the beam cross-section (the beam being assumed uniform over its cross-
section). This ratio has the dimensions of area and is called the effective
scattering cross-section. It is entirely determined by the form of the scattering
field and is the most important characteristic of the scattering process.
We shall suppose that the relation between X and P is one-to-one; this is
so if the angle of scattering is a monotonically decreasing function of the
impact parameter. In that case, only those particles whose impact parameters
lie between p(x) and p(x) + dp(x) are scattered at angles between X and
+ dx. The number of such particles is equal to the product of n and the
area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
effective cross-section is therefore
do = 2mp dp.
(18.6)
In order to find the dependence of do on the angle of scattering, we need
only rewrite (18.6) as
do = 2(x)|dp(x)/dx|dx
(18.7)
Here we use the modulus of the derivative dp/dx, since the derivative may
be (and usually is) negative. t Often do is referred to the solid angle element
do instead of the plane angle element dx. The solid angle between cones
with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
(18.7)
do.
(18.8)
t If the function p(x) is many-valued, we must obviously take the sum of such expressions
as (18.7) over all the branches of this function.
50
Collisions Between Particles
§18
Returning now to the problem of the scattering of a beam of particles, not
by a fixed centre of force, but by other particles initially at rest, we can say
that (18.7) gives the effective cross-section as a function of the angle of
scattering in the centre-of-mass system. To find the corresponding expression
as a function of the scattering angle 0 in the laboratory system, we must
express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
expressions for both the scattering cross-section for the incident beam of
particles (x in terms of 01) and that for the particles initially at rest (x in terms
of O2).
PROBLEMS
PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
for r>a).
SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
the path consists of two straight lines symmetrical about the radius to the point where the
particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
a sin to = a sin 1(-x) = a cos 1x.
A
to
p
&
FIG. 19
Substituting in (18.7) or (18.8), we have
do = 1ma2 sin X do,
(1)
i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
the total cross-section o = na2, in accordance with the fact that the "impact area" which the
particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
and m2 that of the sphere) we have
do1,
where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
stituting X = 201 from (17.9) in (1).

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§17. Elastic collisions
A collision between two particles is said to be elastic if it involves no change
in their internal state. Accordingly, when the law of conservation of energy
is applied to such a collision, the internal energy of the particles may be
neglected.
The collision is most simply described in a frame of reference in which the
centre of mass of the two particles is at rest (the C system). As in $16, we
distinguish by the suffix 0 the values of quantities in that system. The velo-
cities of the particles before the collision are related to their velocities V1 and
V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
where V = V1-V2; see (13.2).
Because of the law of conservation of momentum, the momenta of the two
particles remain equal and opposite after the collision, and are also unchanged
in magnitude, by the law of conservation of energy. Thus, in the C system
the collision simply rotates the velocities, which remain opposite in direction
and unchanged in magnitude. If we denote by no a unit vector in the direc-
tion of the velocity of the particle M1 after the collision, then the velocities
of the two particles after the collision (distinguished by primes) are
V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
(17.1)
§17
Elastic collisions
45
In order to return to the L system, we must add to these expressions the
velocity V of the centre of mass. The velocities in the L system after the
collision are therefore
V1' =
(17.2)
V2' =
No further information about the collision can be obtained from the laws
of conservation of momentum and energy. The direction of the vector no
depends on the law of interaction of the particles and on their relative position
during the collision.
The results obtained above may be interpreted geometrically. Here it is
more convenient to use momenta instead of velocities. Multiplying equations
(17.2) by M1 and M2 respectively, we obtain
(17.3)
P2' muno+m2(p1+p2)/(m1+m2)
where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
mv and use the construction shown in Fig. 15. If the unit vector no is along
OC, the vectors AC and CB give the momenta P1' and P2' respectively.
When p1 and P2 are given, the radius of the circle and the points A and B
are fixed, but the point C may be anywhere on the circle.
C
p'
no
P'2
B
A
FIG. 15
Let us consider in more detail the case where one of the particles (m2, say) is
at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
momentum P1 of the particle M1 before the collision. The point A lies inside
or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
46
Collisions Between Particles
§17
diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
are the angles between the directions of motion after the collision and the
direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
gives the direction of no, is the angle through which the direction of motion
of m1 is turned in the C system. It is evident from the figure that 01 and O2
can be expressed in terms of X by
(17.4)
C
p'
P2
pi
P2
0
max
10,
X
O2
O2
B
B
A
0
A
Q
0
(a) m < m2
(b) m, m m m
AB=p : AO/OB= m/m2
FIG. 16
We may give also the formulae for the magnitudes of the velocities of the
two particles after the collision, likewise expressed in terms of X:
ib
(17.5)
The sum A1 + O2 is the angle between the directions of motion of the
particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
if M1 > M2.
When the two particles are moving afterwards in the same or in opposite
directions (head-on collision), we have X=TT, i.e. the point C lies on the
diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
In this case the velocities after the collision are
(17.6)
This value of V2' has the greatest possible magnitude, and the maximum
§17
Elastic collisions
47
energy which can be acquired in the collision by a particle originally at rest
is therefore
(17.7)
where E1 = 1M1U12 is the initial energy of the incident particle.
If M1 < M2, the velocity of M1 after the collision can have any direction.
If M1 > M2, however, this particle can be deflected only through an angle
not exceeding Omax from its original direction; this maximum value of A1
corresponds to the position of C for which AC is a tangent to the circle
(Fig. 16b). Evidently
sin Omax = OC|OA = M2/M1.
(17.8)
The collision of two particles of equal mass, of which one is initially at
rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
C
p'
P2
Q2
B
A
0
FIG. 17
Then
01=1x,
A2 = 1(-x),
(17.9)
12
=
(17.10)
After the collision the particles move at right angles to each other.
PROBLEM
Express the velocity of each particle after a collision between a moving particle (m1) and
another at rest (m2) in terms of their directions of motion in the L system.
SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
Hence
for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.

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§18
§18. Scattering
As already mentioned in §17, a complete calculation of the result of a
collision between two particles (i.e. the determination of the angle x) requires
the solution of the equations of motion for the particular law of interaction
involved.
We shall first consider the equivalent problem of the deflection of a single
particle of mass m moving in a field U(r) whose centre is at rest (and is at
the centre of mass of the two particles in the original problem).
As has been shown in $14, the path of a particle in a central field is sym-
metrical about a line from the centre to the nearest point in the orbit (OA
in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
say) with this line. The angle X through which the particle is deflected as it
passes the centre is seen from Fig. 18 to be
X = -200.
(18.1)
A
X
to
FIG. 18
The angle do itself is given, according to (14.7), by
(M/r2) dr
(18.2)
taken between the nearest approach to the centre and infinity. It should be
recalled that rmin is a zero of the radicand.
For an infinite motion, such as that considered here, it is convenient to
use instead of the constants E and M the velocity Voo of the particle at infinity
and the impact parameter p. The latter is the length of the perpendicular
from the centre O to the direction of Voo, i.e. the distance at which the particle
would pass the centre if there were no field of force (Fig. 18). The energy
and the angular momentum are given in terms of these quantities by
E = 1mvoo²,
M = mpVoo,
(18.3)
§18
Scattering
49
and formula (18.2) becomes
dr
(18.4)
Together with (18.1), this gives X as a function of p.
In physical applications we are usually concerned not with the deflection
of a single particle but with the scattering of a beam of identical particles
incident with uniform velocity Voo on the scattering centre. The different
particles in the beam have different impact parameters and are therefore
scattered through different angles X. Let dN be the number of particles
scattered per unit time through angles between X and X + dx. This number
itself is not suitable for describing the scattering process, since it is propor-
tional to the density of the incident beam. We therefore use the ratio
do = dN/n,
(18.5)
where n is the number of particles passing in unit time through unit area of
the beam cross-section (the beam being assumed uniform over its cross-
section). This ratio has the dimensions of area and is called the effective
scattering cross-section. It is entirely determined by the form of the scattering
field and is the most important characteristic of the scattering process.
We shall suppose that the relation between X and P is one-to-one; this is
so if the angle of scattering is a monotonically decreasing function of the
impact parameter. In that case, only those particles whose impact parameters
lie between p(x) and p(x) + dp(x) are scattered at angles between X and
+ dx. The number of such particles is equal to the product of n and the
area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
effective cross-section is therefore
do = 2mp dp.
(18.6)
In order to find the dependence of do on the angle of scattering, we need
only rewrite (18.6) as
do = 2(x)|dp(x)/dx|dx
(18.7)
Here we use the modulus of the derivative dp/dx, since the derivative may
be (and usually is) negative. t Often do is referred to the solid angle element
do instead of the plane angle element dx. The solid angle between cones
with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
(18.7)
do.
(18.8)
t If the function p(x) is many-valued, we must obviously take the sum of such expressions
as (18.7) over all the branches of this function.
50
Collisions Between Particles
§18
Returning now to the problem of the scattering of a beam of particles, not
by a fixed centre of force, but by other particles initially at rest, we can say
that (18.7) gives the effective cross-section as a function of the angle of
scattering in the centre-of-mass system. To find the corresponding expression
as a function of the scattering angle 0 in the laboratory system, we must
express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
expressions for both the scattering cross-section for the incident beam of
particles (x in terms of 01) and that for the particles initially at rest (x in terms
of O2).
PROBLEMS
PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
for r>a).
SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
the path consists of two straight lines symmetrical about the radius to the point where the
particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
a sin to = a sin 1(-x) = a cos 1x.
A
to
p
&
FIG. 19
Substituting in (18.7) or (18.8), we have
do = 1ma2 sin X do,
(1)
i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
the total cross-section o = na2, in accordance with the fact that the "impact area" which the
particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
and m2 that of the sphere) we have
do1,
where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
stituting X = 201 from (17.9) in (1).

0
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76
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@ -86,11 +86,79 @@ is said to be uniform. The potential energy in such a field is evidently
To conclude this section, we may make the following remarks concerning the application of Lagrange's equations to various problems. It is often necessary to deal with mechanical systems in which the interaction between different bodies (or particles) takes the form of constraints, i.e. restrictions on their relative position. In practice, such constraints are effected by means of rods, strings, hinges and so on. This introduces a new factor into the problem, in that the motion of the bodies results in friction at their points of contact, and the problem in general ceases to be one of pure mechanics `1/25`. In many cases, however, the friction in the system is so slight that its effect on the motion is entirely negligible. If the masses of the constraining elements of the system are also negligible, the effect of the constraints is simply to reduce the number of degrees of freedom $S$ of the system to a value less than $3N$. To determine the motion of the system, the Lagrangian `1/5.5` can again be used, with a set of independent generalised co-ordinates equal in number to the actual degrees of freedom. To conclude this section, we may make the following remarks concerning the application of Lagrange's equations to various problems. It is often necessary to deal with mechanical systems in which the interaction between different bodies (or particles) takes the form of constraints, i.e. restrictions on their relative position. In practice, such constraints are effected by means of rods, strings, hinges and so on. This introduces a new factor into the problem, in that the motion of the bodies results in friction at their points of contact, and the problem in general ceases to be one of pure mechanics `1/25`. In many cases, however, the friction in the system is so slight that its effect on the motion is entirely negligible. If the masses of the constraining elements of the system are also negligible, the effect of the constraints is simply to reduce the number of degrees of freedom $S$ of the system to a value less than $3N$. To determine the motion of the system, the Lagrangian `1/5.5` can again be used, with a set of independent generalised co-ordinates equal in number to the actual degrees of freedom.
## Problems <!-- ## Problems -->
Find the Lagrangian for each of the following systems when placed in a uniform gravitational field (acceleration $g$). <!-- Find the Lagrangian for each of the following systems when placed in a uniform gravitational field (acceleration $g$). -->
#### PROBLEM 1. A coplanar double pendulum <!-- #### PROBLEM 1. A coplanar double pendulum -->
`1/fig1` <!-- `1/fig1` -->
<!-- Find the Lagrangian for each of the following systems when placed in a uniform gravitational field (acceleration g). -->
<!-- TODO figures -->
<!-- §5 -->
<!-- The Lagrangian for a system of particles -->
<!-- 11 -->
<!-- PROBLEM 1. A coplanar double pendulum (Fig. 1). -->
<!-- '!!!!!!! -->
<!-- pl4 -->
<!-- m, -->
<!-- 2 -->
<!-- L2 -->
<!-- M2 -->
<!-- FIG. 1 -->
<!-- SOLUTION. We take as co-ordinates the angles 01 and O2 which the strings l1 and l2 make -->
<!-- with the vertical. Then we have, for the particle M1, T1 = U = -migh cos 01. In -->
<!-- order to find the kinetic energy of the second particle, we express its Cartesian co-ordinates -->
<!-- x2, y2 (with the origin at the point of support and the y-axis vertically downwards) in terms -->
<!-- of the angles 01 and O2: X2 = l1 sin 01 sin O2, y2 = l1 cos 1+12 cos O2. Then we find -->
<!-- T2 = tm2(x22++22) -->
<!-- = -->
<!-- Finally -->
<!-- L = +m2) m2/1/20102 cos(01-02)+(m1+m2)gl cos 01 + m2gl2 cos O2. -->
<!-- PROBLEM 2. A simple pendulum of mass M2, with a mass M1 at the point of support which -->
<!-- can move on a horizontal line lying in the plane in which m2 moves (Fig. 2). -->
<!-- x -->
<!-- m, -->
<!-- l -->
<!-- M2 -->
<!-- FIG. 2 -->
<!-- SOLUTION. Using the co-ordinate x of M1 and the angle between the string and the -->
<!-- vertical, we have -->
<!-- L = cos b) +m2gl cos p. -->
<!-- PROBLEM 3. A simple pendulum of mass m whose point of support (a) moves uniformly -->
<!-- on a vertical circle with constant frequency y (Fig. 3), (b) oscillates horizontally in the plane -->
<!-- of motion of the pendulum according to the law x = a cos yt, (c) oscillates vertically accord- -->
<!-- ing to the law y = a cos yt. -->
<!-- SOLUTION. (a) The co-ordinates of m are x = a cos rt+l sin , y = -a sin rt+l cos o. -->
<!-- The Lagrangian is -->
<!-- L = sin(-t) +mgl cos ; -->
<!-- here terms depending only on time have been omitted, together with the total time derivative -->
<!-- of mlay cos(-t). -->
<!-- 12 -->
<!-- The Equations of Motion -->
<!-- §5 -->
<!-- (b) The co-ordinates of m are x = a cos yt+l sin , y = l cos p. The Lagrangian is (omit- -->
<!-- ting total derivatives) -->
<!-- L = st sin +mgl cos . -->
<!-- (c) Similarly -->
<!-- L = cos st cos +mgl cos . -->
<!-- ly -->
<!-- m -->
<!-- FIG. 3 -->
<!-- PROBLEM 4. The system shown in Fig. 4. The particle M2 moves on a vertical axis and the -->
<!-- whole system rotates about this axis with a constant angular velocity S2. -->
<!-- A -->
<!-- a -->
<!-- m -->
<!-- m, -->
<!-- a -->
<!-- m2 -->
<!-- FIG. 4 -->
<!-- SOLUTION. Let 0 be the angle between one of the segments a and the vertical, and $ the -->
<!-- angle of rotation of the system about the axis; ; b = S. For each particle M1, the infinitesimal -->
<!-- displacement is given by dl-2 = a2 sin² 0 do2. The distance of M2 from the point -->
<!-- of support A is 2a cos 0, and so dl2 = -2a sin 0 d0. The Lagrangian is -->
<!-- L = m1a2(j2 + Q2 sin20) +2m2a202 sin20+2(m1 +m2)ga cos 0. -->

72
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@ -1,72 +0,0 @@
---
section: The Equations of Motion
---
Find the Lagrangian for each of the following systems when placed in a uniform gravitational field (acceleration g).
TODO figures
§5
The Lagrangian for a system of particles
11
PROBLEM 1. A coplanar double pendulum (Fig. 1).
'!!!!!!!
pl4
m,
2
L2
M2
FIG. 1
SOLUTION. We take as co-ordinates the angles 01 and O2 which the strings l1 and l2 make
with the vertical. Then we have, for the particle M1, T1 = U = -migh cos 01. In
order to find the kinetic energy of the second particle, we express its Cartesian co-ordinates
x2, y2 (with the origin at the point of support and the y-axis vertically downwards) in terms
of the angles 01 and O2: X2 = l1 sin 01 sin O2, y2 = l1 cos 1+12 cos O2. Then we find
T2 = tm2(x22++22)
=
Finally
L = +m2) m2/1/20102 cos(01-02)+(m1+m2)gl cos 01 + m2gl2 cos O2.
PROBLEM 2. A simple pendulum of mass M2, with a mass M1 at the point of support which
can move on a horizontal line lying in the plane in which m2 moves (Fig. 2).
x
m,
l
M2
FIG. 2
SOLUTION. Using the co-ordinate x of M1 and the angle between the string and the
vertical, we have
L = cos b) +m2gl cos p.
PROBLEM 3. A simple pendulum of mass m whose point of support (a) moves uniformly
on a vertical circle with constant frequency y (Fig. 3), (b) oscillates horizontally in the plane
of motion of the pendulum according to the law x = a cos yt, (c) oscillates vertically accord-
ing to the law y = a cos yt.
SOLUTION. (a) The co-ordinates of m are x = a cos rt+l sin , y = -a sin rt+l cos o.
The Lagrangian is
L = sin(-t) +mgl cos ;
here terms depending only on time have been omitted, together with the total time derivative
of mlay cos(-t).
12
The Equations of Motion
§5
(b) The co-ordinates of m are x = a cos yt+l sin , y = l cos p. The Lagrangian is (omit-
ting total derivatives)
L = st sin +mgl cos .
(c) Similarly
L = cos st cos +mgl cos .
ly
m
FIG. 3
PROBLEM 4. The system shown in Fig. 4. The particle M2 moves on a vertical axis and the
whole system rotates about this axis with a constant angular velocity S2.
A
a
m
m,
a
m2
FIG. 4
SOLUTION. Let 0 be the angle between one of the segments a and the vertical, and $ the
angle of rotation of the system about the axis; ; b = S. For each particle M1, the infinitesimal
displacement is given by dl-2 = a2 sin² 0 do2. The distance of M2 from the point
of support A is 2a cos 0, and so dl2 = -2a sin 0 d0. The Lagrangian is
L = m1a2(j2 + Q2 sin20) +2m2a202 sin20+2(m1 +m2)ga cos 0.

12
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@ -73,3 +73,15 @@ are called generalised forces. In this notation, Lagrange's equations are
``` ```
In Cartesian co-ordinates the generalised momenta are the components of the vectors $\v{p}_a$. In general, however, the $p_i$ are linear homogeneous functions of the generalised velocities $\dot{q}_i$, and do not reduce to products of mass and velocity. In Cartesian co-ordinates the generalised momenta are the components of the vectors $\v{p}_a$. In general, however, the $p_i$ are linear homogeneous functions of the generalised velocities $\dot{q}_i$, and do not reduce to products of mass and velocity.
<!-- PROBLEM -->
<!-- A particle of mass m moving with velocity V1 leaves a half-space in which its potential energy -->
<!-- is a constant U1 and enters another in which its potential energy is a different constant U2. -->
<!-- Determine the change in the direction of motion of the particle. -->
<!-- SOLUTION. The potential energy is independent of the co-ordinates whose axes are parallel -->
<!-- to the plane separating the half-spaces. The component of momentum in that plane is -->
<!-- therefore conserved. Denoting by 01 and O2 the angles between the normal to the plane and -->
<!-- the velocities V1 and V2 of the particle before and after passing the plane, we have V1 sin 01 -->
<!-- = U2 sin O2. The relation between V1 and V2 is given by the law of conservation of energy, -->
<!-- and the result is -->
<!-- sin -->

15
1/7_problem.md
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@ -1,15 +0,0 @@
---
section: Conservation laws
---
PROBLEM
A particle of mass m moving with velocity V1 leaves a half-space in which its potential energy
is a constant U1 and enters another in which its potential energy is a different constant U2.
Determine the change in the direction of motion of the particle.
SOLUTION. The potential energy is independent of the co-ordinates whose axes are parallel
to the plane separating the half-spaces. The component of momentum in that plane is
therefore conserved. Denoting by 01 and O2 the angles between the normal to the plane and
the velocities V1 and V2 of the particle before and after passing the plane, we have V1 sin 01
= U2 sin O2. The relation between V1 and V2 is given by the law of conservation of energy,
and the result is
sin

9
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@ -62,3 +62,12 @@ E
``` ```
This formula gives the law of transformation of energy from one frame to another, corresponding to formula `1/8.1` for momentum. If the centre of mass is at rest in $K'$, then $\v{P}' = 0$, $E' = E_i$, and we have `1/8.4`. This formula gives the law of transformation of energy from one frame to another, corresponding to formula `1/8.1` for momentum. If the centre of mass is at rest in $K'$, then $\v{P}' = 0$, $E' = E_i$, and we have `1/8.4`.
<!-- PROBLEM -->
<!-- Find the law of transformation of the action S from one inertial frame to another. -->
<!-- SOLUTION. The Lagrangian is equal to the difference of the kinetic and potential energies, -->
<!-- and is evidently transformed in accordance with a formula analogous to (8.5): -->
<!-- Integrating this with respect to time, we obtain the required law of transformation of the -->
<!-- action: -->
<!-- S=S'tuV.R'+1uV2t, -->
<!-- where R' is the radius vector of the centre of mass in the frame K'. -->

11
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@ -1,11 +0,0 @@
---
---
PROBLEM
Find the law of transformation of the action S from one inertial frame to another.
SOLUTION. The Lagrangian is equal to the difference of the kinetic and potential energies,
and is evidently transformed in accordance with a formula analogous to (8.5):
Integrating this with respect to time, we obtain the required law of transformation of the
action:
S=S'tuV.R'+1uV2t,
where R' is the radius vector of the centre of mass in the frame K'.

28
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@ -128,3 +128,31 @@ $$
The Lagrangian is, in terms of these co-ordinates, The Lagrangian is, in terms of these co-ordinates,
and substitution of this in `1/9.7` gives `1/9.8`. and substitution of this in `1/9.7` gives `1/9.8`.
<!-- PROBLEMS -->
<!-- PROBLEM 1. Obtain expressions for the Cartesian components and the magnitude of the -->
<!-- angular momentum of a particle in cylindrical co-ordinates r, , Z. -->
<!-- SOLUTION. Mx = m(rz-zi) sin - -mrzo cos , -->
<!-- My = -m(rz-zi) cos -mrzo sin , -->
<!-- Mz = mr2 -->
<!-- M2 = -->
<!-- PROBLEM 2. The same as Problem 1, but in spherical co-ordinates r, 0, o. -->
<!-- SOLUTION. Mx = -mr2(8 sin + sin 0 cos 0 cos ), -->
<!-- My = mr2(j cos - sin 0 cos 0 sin b), -->
<!-- Mz = mr2sin20, -->
<!-- M2 = -->
<!-- PROBLEM 3. Which components of momentum P and angular momentum M are conserved -->
<!-- in motion in the following fields? -->
<!-- (a) the field of an infinite homogeneous plane, (b) that of an infinite homogeneous cylinder, -->
<!-- (c) that of an infinite homogeneous prism, (d) that of two points, (e) that of an infinite homo- -->
<!-- geneous half-plane, (f) that of a homogeneous cone, (g) that of a homogeneous circular torus, -->
<!-- (h) that of an infinite homogeneous cylindrical helix. -->
<!-- SOLUTION. (a) Px, Py, Mz (if the plane is the xy-plane), (b) M, Pz (if the axis of the -->
<!-- cylinder is the z-axis), (c) P (if the edges of the prism are parallel to the z-axis), -->
<!-- (d) Mz (if the line joining the points is the z-axis), (e) Py (if the edge of the half- -->
<!-- plane is the y-axis), (f) Mz (if the axis of the cone is the z-axis), (g) Mz (if the axis -->
<!-- of the torus is the z-axis), (h) the Lagrangian is unchanged by a rotation through an angle -->
<!-- so about the axis of the helix (let this be the z-axis) together with a translation through a -->
<!-- distance h86/2m along the axis (h being the pitch of the helix). Hence SL = 8z aL/dz+ -->
<!-- +80 0L/26 = = 0, so that I+hPz/2n = constant. -->

30
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@ -1,30 +0,0 @@
---
---
PROBLEMS
PROBLEM 1. Obtain expressions for the Cartesian components and the magnitude of the
angular momentum of a particle in cylindrical co-ordinates r, , Z.
SOLUTION. Mx = m(rz-zi) sin - -mrzo cos ,
My = -m(rz-zi) cos -mrzo sin ,
Mz = mr2
M2 =
PROBLEM 2. The same as Problem 1, but in spherical co-ordinates r, 0, o.
SOLUTION. Mx = -mr2(8 sin + sin 0 cos 0 cos ),
My = mr2(j cos - sin 0 cos 0 sin b),
Mz = mr2sin20,
M2 =
PROBLEM 3. Which components of momentum P and angular momentum M are conserved
in motion in the following fields?
(a) the field of an infinite homogeneous plane, (b) that of an infinite homogeneous cylinder,
(c) that of an infinite homogeneous prism, (d) that of two points, (e) that of an infinite homo-
geneous half-plane, (f) that of a homogeneous cone, (g) that of a homogeneous circular torus,
(h) that of an infinite homogeneous cylindrical helix.
SOLUTION. (a) Px, Py, Mz (if the plane is the xy-plane), (b) M, Pz (if the axis of the
cylinder is the z-axis), (c) P (if the edges of the prism are parallel to the z-axis),
(d) Mz (if the line joining the points is the z-axis), (e) Py (if the edge of the half-
plane is the y-axis), (f) Mz (if the axis of the cone is the z-axis), (g) Mz (if the axis
of the torus is the z-axis), (h) the Lagrangian is unchanged by a rotation through an angle
so about the axis of the helix (let this be the z-axis) together with a translation through a
distance h86/2m along the axis (h being the pitch of the helix). Hence SL = 8z aL/dz+
+80 0L/26 = = 0, so that I+hPz/2n = constant.

72
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@ -28,43 +28,43 @@ III. INTEGRATION OF THE EQUATIONS OF MOTION
</span> </span>
14. [Motion in a central field](14-motion-in-a-central-field.html) 14. [Motion in a central field](14-motion-in-a-central-field.html)
15. [Kepler's problem]() 15. [Kepler's problem](15-keplers-problem.html)
IV. COLLISION BETWEEN PARTICLES IV. COLLISION BETWEEN PARTICLES
16. Disintegration of particles 16. [Disintegration of particles](16-disintegration-of-particles.html)
17. Elastic collisions 17. [Elastic collisions](17-elastic-collisions.html)
18. Scattering 18. [Scattering](18-scattering.html)
19. Rutherford's formula 19. [Rutherford's formula](19-rutherfords-formula.html)
20. Small-angle scattering 20. [Small-angle scattering](20-small-angle-scattering.html)
V. SMALL OSCILLATIONS V. SMALL OSCILLATIONS
21. Free oscillations in one dimension 21. [Free oscillations in one dimension](21-free-oscillations-in-one-dimension.html)
22. Forced oscillations 22. [Forced oscillations](22-forced-oscillations.html)
23. Oscillations of systems with more than one degree of freedom 23. [Oscillations of systems with more than one degree of freedom](23-oscillations-of-systems-with-more-than-one-degree-of-freedom.html)
24. Vibrations of molecules 24. [Vibrations of molecules](24-vibrations-of-molecules.html)
25. Damped oscillations 25. [Damped oscillations](25-damped-oscillations.html)
26. Forced oscillations under friction 26. [Forced oscillations under friction](26-forced-oscillations-under-friction.html)
27. Parametric resonance 27. [Parametric resonance](27-parametric-resonance.html)
28. Anharmonic oscillations 28. [Anharmonic oscillations](28-anharmonic-oscillations.html)
29. Resonance in non-linear oscillations 29. [Resonance in non-linear oscillations](29-resonance-in-non-linear-oscillations.html)
30. Motion in a rapidly oscillating field 30. [Motion in a rapidly oscillating field](30-motion-in-a-rapidly-oscillating-field.html)
VI. MOTION OF A RIGID BODY VI. MOTION OF A RIGID BODY
31. Angular velocity 31. [Angular velocity](31-angular-velocity.html)
32. The inertia tensor 32. [The inertia tensor](32-the-inertia-tensor.html)
33. Angular momentum of a rigid body 33. [Angular momentum of a rigid body](33-angular-momentum-of-a-rigid-body.html)
34. The equations of motion of a rigid body 34. [The equations of motion of a rigid body](34-the-equations-of-motion-of-a-rigid-body.html)
35. Eulerian angles 35. [Eulerian angles](35-eulerian-angles.html)
36. Euler's equations 36. [Euler's equations](36-eulers-equations.html)
37. The asymmetrical top 37. [The asymmetrical top](37-the-asymmetrical-top.html)
38. Rigid bodies in contact 38. [Rigid bodies in contact](38-rigid-bodies-in-contact.html)
39. Motion in a non-inertial frame of reference 39. [Motion in a non-inertial frame of reference](39-motion-in-a-non-inertial-frame-of-reference.html)
VII. THE CANONICAL EQUATIONS VII. THE CANONICAL EQUATIONS
40. Hamilton's equations 40. [Hamilton's equations](40-hamiltons-equations.html)
41. The Routhian 41. [The Routhian](41-the-routhian.html)
42. Poisson brackets 42. [Poisson brackets](42-poisson-brackets.html)
43. The action as a function of the co-ordinates 43. [The action as a function of the co-ordinates](43-the-actions-as-a-function-of-the-co-ordinates.html)
44. Maupertuis' principle 44. [Maupertuis' principle](44-maupertuis-principle.html)
45. Canonical transformations 45. [Canonical transformations](45-canonical-transformations.html)
46. Liouville's theorem 46. [Liouville's theorem](46-louivilles-theorem.html)
47. The Hamilton-Jacobi equation 47. [The Hamilton-Jacobi equation](47-the-hamilton-jacobi-equations.html)
48. Separation of the variables 48. [Separation of the variables](48-separation-of-the-variables.html)
49. Adiabatic invariants 49. [Adiabatic invariants](49-adiabatic-invariants.html)
50. General properties of motion in `s` dimensions 50. [General properties of motion in `s` dimensions](50-general-properties-of-motion-in-s-dimensions.html)

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@ -0,0 +1,591 @@
Kepler's problem
35
Using (1), we find the energy in the form
E
(3)
Hence
Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
dulum, which moves in an arc of a circle.
$15. Kepler's problem
An important class of central fields is formed by those in which the poten-
tial energy is inversely proportional to r, and the force accordingly inversely
proportional to r2. They include the fields of Newtonian gravitational attrac-
tion and of Coulomb electrostatic interaction; the latter may be either attrac-
tive or repulsive.
Let us first consider an attractive field, where
U=-a/r
(15.1)
with a a positive constant. The "effective" potential energy
(15.2)
is
of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
r
8
it tends to zero from negative values ; for r = M2/ma it has a minimum value
Ueff, min = -mx2/2M2.
(15.3)
Ueff
FIG. 10
It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
for E > 0.
36
Integration of the Equations of Motion
§15
The shape of the path is obtained from the general formula (14.7). Substi-
tuting there U = - a/r and effecting the elementary integration, we have
o =
- constant.
Taking the origin of such that the constant is zero, and putting
P = M2/ma, e= [1 1+(2EM2/mo2)]
(15.4)
we can write the equation of the path as
p/r = 1+e coso.
(15.5)
This is the equation of a conic section with one focus at the origin; 2p is called
the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
is seen from (15.5) to be such that the point where = 0 is the point nearest
to the origin (called the perihelion).
In the equivalent problem of two particles interacting according to the law
(15.1), the orbit of each particle is a conic section, with one focus at the centre
of mass of the two particles.
It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
has been said earlier in this section. According to the formulae of analytical
geometry, the major and minor semi-axes of the ellipse are
a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
(15.6)
y
X
2b
ae
2a
FIG. 11
The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
becomes a circle. It may be noted that the major axis of the ellipse depends
only on the energy of the particle, and not on its angular momentum. The
least and greatest distances from the centre of the field (the focus of the
ellipse) are
rmin = =p/(1+e)=a(1-e),
max=p(1-e)=a(1+e). = = (15.7)
These expressions, with a and e given by (15.6) and (15.4), can, of course,
also be obtained directly as the roots of the equation Ueff(r) = E.
§15
Kepler's problem
37
The period T of revolution in an elliptical orbit is conveniently found by
using the law of conservation of angular momentum in the form of the area
integral (14.3). Integrating this equation with respect to time from zero to
T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
f = nab, and by using the formulae (15.6) we find
T = 2ma3/2-(m/a)
= ma((m2E3).
(15.8)
The proportionality between the square of the period and the cube of the
linear dimension of the orbit has already been demonstrated in §10. It may
also be noted that the period depends only on the energy of the particle.
For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
tance of the perihelion from the focus is
rmin ==pl(e+1)=a(e-1), = =
(15.9)
where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
y
p
ale-1)
FIG. 12
If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
perihelion distance rmin = 1p. This case occurs if the particle starts from rest
at infinity.
The co-ordinates of the particle as functions of time in the orbit may be
found by means of the general formula (14.6). They may be represented in a
convenient parametric form as follows.
Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
we can write the integral (14.6) for the time as
t
=
=
38
Integration of the Equations of Motion
§15
The obvious substitution r-a = - ae cos $ converts the integral to
sioant
If time is measured in such a way that the constant is zero, we have the
following parametric dependence of r on t:
r = a(1-e cos ), t =
(15.10)
the particle being at perihelion at t = 0. The Cartesian co-ordinates
x = r cos o, y = r sin (the x and y axes being respectively parallel to the
major and minor axes of the ellipse) can likewise be expressed in terms of
the parameter $. From (15.5) and (15.10) we have
ex = = =
y is equal to W(r2-x2). Thus
x = a(cos & - e),
y = =av(1-e2) $.
(15.11)
A complete passage round the ellipse corresponds to an increase of $ from 0
to 2nr.
Entirely similar calculations for the hyperbolic orbits give
r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
(15.12)
x = a(e-cosh ) y = a1/(e2-1)sinh &
where the parameter $ varies from - 00 to + 00.
Let us now consider motion in a repulsive field, where
U
(a>0).
(15.13)
Here the effective potential energy is
Utt
and decreases monotonically from + 00 to zero as r varies from zero to
infinity. The energy of the particle must be positive, and the motion is always
infinite. The calculations are exactly similar to those for the attractive field.
The path is a hyperbola (or, if E = 0, a parabola):
pr r = =1-e coso, =
(15.14)
where P and e are again given by (15.4). The path passes the centre of the
field in the manner shown in Fig. 13. The perihelion distance is
rmin =p(e-1)=a(e+1). =
(15.15)
The time dependence is given by the parametric equations
= =(ma3/a)(esinh+) =
(15.16)
x
= a(cosh & e ,
y = av((e2-1) sinh &
§15
Kepler's problem
39
To conclude this section, we shall show that there is an integral of the mo-
tion which exists only in fields U = a/r (with either sign of a). It is easy to
verify by direct calculation that the quantity
vxM+ar/r
(15.17)
is constant. For its total time derivative is v
since M = mr xv,
Putting mv = ar/r3 from the equation of motion, we find that this expression
vanishes.
y
0
(I+e)
FIG. 13
The direction of the conserved vector (15.17) is along the major axis from
the focus to the perihelion, and its magnitude is ae. This is most simply
seen by considering its value at perihelion.
It should be emphasised that the integral (15.17) of the motion, like M and
E, is a one-valued function of the state (position and velocity) of the particle.
We shall see in §50 that the existence of such a further one-valued integral
is due to the degeneracy of the motion.
PROBLEMS
PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
moving in a parabola in a field U = -a/r.
SOLUTION. In the integral
we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
the required dependence:
r=1p(1+n2),
t=
y=pn.
40
Integration of the Equations of Motion
§15
The parameter n varies from - 00 to +00.
PROBLEM 2. Integrate the equations of motion for a particle in a central field
U = - a/r2 (a > 0).
SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
(a) for E > andM 0 and
(b) for E>0 0nd and M 2/2m a,
(c) for E <0 and Ms1
In all three cases
In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
origin as
00. The fall from a given value of r takes place in a finite time, namely
PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
(14.10), which we write as
in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
the first-order term gives the required change so:
(1)
where we have changed from the integration over r to one over , along the path of the "un-
perturbed" motion.
In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.
§14
Motion in a central field
31
(Fig. 8). Calling this area df, we can write the angular momentum of the par-
ticle as
M = 2mf,
(14.3)
where the derivative f is called the sectorial velocity. Hence the conservation
of angular momentum implies the constancy of the sectorial velocity: in equal
times the radius vector of the particle sweeps out equal areas (Kepler's second
law).t
rdd
dd
0
FIG. 8
The complete solution of the problem of the motion of a particle in a central
field is most simply obtained by starting from the laws of conservation of
energy and angular momentum, without writing out the equations of motion
themselves. Expressing in terms of M from (14.2) and substituting in the
expression for the energy, we obtain
E = =
(14.4)
Hence
(14.5)
or, integrating,
constant.
(14.6)
Writing (14.2) as do = M dt/mr2, substituting dt from (14.5) and integrating,
we find
constant.
(14.7)
Formulae (14.6) and (14.7) give the general solution of the problem. The
latter formula gives the relation between r and , i.e. the equation of the path.
Formula (14.6) gives the distance r from the centre as an implicit function of
time. The angle o, it should be noted, always varies monotonically with time,
since (14.2) shows that & can never change sign.
t The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.
32
Integration of the Equations of Motion
§14
The expression (14.4) shows that the radial part of the motion can be re-
garded as taking place in one dimension in a field where the "effective poten-
tial energy" is
(14.8)
The quantity M2/2mr2 is called the centrifugal energy. The values of r for which
U(r)
(14.9)
determine the limits of the motion as regards distance from the centre.
When equation (14.9) is satisfied, the radial velocity j is zero. This does not
mean that the particle comes to rest as in true one-dimensional motion, since
the angular velocity o is not zero. The value j = 0 indicates a turning point
of the path, where r(t) begins to decrease instead of increasing, or vice versa.
If the range in which r may vary is limited only by the condition r > rmin,
the motion is infinite: the particle comes from, and returns to, infinity.
If the range of r has two limits rmin and rmax, the motion is finite and the
path lies entirely within the annulus bounded by the circles r = rmax and
r = rmin- This does not mean, however, that the path must be a closed curve.
During the time in which r varies from rmax to rmin and back, the radius
vector turns through an angle Ao which, according to (14.7), is given by
Mdr/r2
(14.10)
The condition for the path to be closed is that this angle should be a rational
fraction of 2n, i.e. that Ao = 2nm/n, where m and n are integers. In that case,
after n periods, the radius vector of the particle will have made m complete
revolutions and will occupy its original position, so that the path is closed.
Such cases are exceptional, however, and when the form of U(r) is arbitrary
the angle is not a rational fraction of 2nr. In general, therefore, the path
of a particle executing a finite motion is not closed. It passes through the
minimum and maximum distances an infinity of times, and after infinite time
it covers the entire annulus between the two bounding circles. The path
shown in Fig. 9 is an example.
There are only two types of central field in which all finite motions take
place in closed paths. They are those in which the potential energy of the
particle varies as 1/r or as r2. The former case is discussed in §15; the latter
is that of the space oscillator (see §23, Problem 3).
At a turning point the square root in (14.5), and therefore the integrands
in (14.6) and (14.7), change sign. If the angle is measured from the direc-
tion of the radius vector to the turning point, the parts of the path on each
side of that point differ only in the sign of for each value of r, i.e. the path
is symmetrical about the line = 0. Starting, say, from a point where = rmax
the particle traverses a segment of the path as far as a point with r rmin,
§14
Motion in a central field
33
then follows a symmetrically placed segment to the next point where r = rmax,
and so on. Thus the entire path is obtained by repeating identical segments
forwards and backwards. This applies also to infinite paths, which consist of
two symmetrical branches extending from the turning point (r = rmin) to
infinity.
'max
min
so
FIG. 9
The presence of the centrifugal energy when M # 0, which becomes
infinite as 1/22 when r -> 0, generally renders it impossible for the particle to
reach the centre of the field, even if the field is an attractive one. A "fall" of
the particle to the centre is possible only if the potential energy tends suffi-
ciently rapidly to -00 as r 0. From the inequality
1mr2 = E- U(r) - M2/2mr2
or r2U(Y)+M2/2m < Er2, it follows that r can take values tending to zero
only if
(14.11)
i.e. U(r) must tend to - 8 either as - a/r2 with a > M2/2m, or proportionally
to - 1/rn with n > 2.
PROBLEMS
PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass
m moving on the surface of a sphere of radius l in a gravitational field).
SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the
polar axis vertically downwards, the Lagrangian of the pendulum is
1ml2(02 + 62 sin20) +mgl cos 0.
2*
34
Integration of the Equations of Motion
§14
The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the
z-component of angular momentum, is conserved:
(1)
The energy is
E = cos 0
(2)
= 0.
Hence
(3)
where the "effective potential energy" is
Ueff(0) = COS 0.
For the angle o we find, using (1),
do
(4)
The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively.
The range of 0 in which the motion takes place is that where E > Ueff, and its limits
are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots
between -1 and +1; these define two circles of latitude on the sphere, between which the
path lies.
PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a
cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational
field.
SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the
polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co-
ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is
= a.
By the same method as in Problem 1, we find
==
The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots;
these define two horizontal circles on the cone, between which the path lies.
PROBLEM 3. Integrate the equations of motion for a pendulum of mass M2, with a mass M1
at the point of support which can move on a horizontal line lying in the plane in which m2
moves (Fig. 2, §5).
SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The
generalised momentum Px, which is the horizontal component of the total momentum of the
system, is therefore conserved
Px = cos = constant.
(1)
The system may always be taken to be at rest as a whole. Then the constant in (1) is zero
and integration gives
(m1+m2)x+m2) sin = constant,
(2)
which expresses the fact that the centre of mass of the system does not move horizontally.
§15
Kepler's problem
35
Using (1), we find the energy in the form
E
(3)
Hence
Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
dulum, which moves in an arc of a circle.
$15. Kepler's problem
An important class of central fields is formed by those in which the poten-
tial energy is inversely proportional to r, and the force accordingly inversely
proportional to r2. They include the fields of Newtonian gravitational attrac-
tion and of Coulomb electrostatic interaction; the latter may be either attrac-
tive or repulsive.
Let us first consider an attractive field, where
U=-a/r
(15.1)
with a a positive constant. The "effective" potential energy
(15.2)
is
of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
r
8
it tends to zero from negative values ; for r = M2/ma it has a minimum value
Ueff, min = -mx2/2M2.
(15.3)
Ueff
FIG. 10
It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
for E > 0.
36
Integration of the Equations of Motion
§15
The shape of the path is obtained from the general formula (14.7). Substi-
tuting there U = - a/r and effecting the elementary integration, we have
o =
- constant.
Taking the origin of such that the constant is zero, and putting
P = M2/ma, e= [1 1+(2EM2/mo2)]
(15.4)
we can write the equation of the path as
p/r = 1+e coso.
(15.5)
This is the equation of a conic section with one focus at the origin; 2p is called
the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
is seen from (15.5) to be such that the point where = 0 is the point nearest
to the origin (called the perihelion).
In the equivalent problem of two particles interacting according to the law
(15.1), the orbit of each particle is a conic section, with one focus at the centre
of mass of the two particles.
It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
has been said earlier in this section. According to the formulae of analytical
geometry, the major and minor semi-axes of the ellipse are
a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
(15.6)
y
X
2b
ae
2a
FIG. 11
The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
becomes a circle. It may be noted that the major axis of the ellipse depends
only on the energy of the particle, and not on its angular momentum. The
least and greatest distances from the centre of the field (the focus of the
ellipse) are
rmin = =p/(1+e)=a(1-e),
max=p(1-e)=a(1+e). = = (15.7)
These expressions, with a and e given by (15.6) and (15.4), can, of course,
also be obtained directly as the roots of the equation Ueff(r) = E.
§15
Kepler's problem
37
The period T of revolution in an elliptical orbit is conveniently found by
using the law of conservation of angular momentum in the form of the area
integral (14.3). Integrating this equation with respect to time from zero to
T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
f = nab, and by using the formulae (15.6) we find
T = 2ma3/2-(m/a)
= ma((m2E3).
(15.8)
The proportionality between the square of the period and the cube of the
linear dimension of the orbit has already been demonstrated in §10. It may
also be noted that the period depends only on the energy of the particle.
For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
tance of the perihelion from the focus is
rmin ==pl(e+1)=a(e-1), = =
(15.9)
where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
y
p
ale-1)
FIG. 12
If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
perihelion distance rmin = 1p. This case occurs if the particle starts from rest
at infinity.
The co-ordinates of the particle as functions of time in the orbit may be
found by means of the general formula (14.6). They may be represented in a
convenient parametric form as follows.
Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
we can write the integral (14.6) for the time as
t
=
=
38
Integration of the Equations of Motion
§15
The obvious substitution r-a = - ae cos $ converts the integral to
sioant
If time is measured in such a way that the constant is zero, we have the
following parametric dependence of r on t:
r = a(1-e cos ), t =
(15.10)
the particle being at perihelion at t = 0. The Cartesian co-ordinates
x = r cos o, y = r sin (the x and y axes being respectively parallel to the
major and minor axes of the ellipse) can likewise be expressed in terms of
the parameter $. From (15.5) and (15.10) we have
ex = = =
y is equal to W(r2-x2). Thus
x = a(cos & - e),
y = =av(1-e2) $.
(15.11)
A complete passage round the ellipse corresponds to an increase of $ from 0
to 2nr.
Entirely similar calculations for the hyperbolic orbits give
r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
(15.12)
x = a(e-cosh ) y = a1/(e2-1)sinh &
where the parameter $ varies from - 00 to + 00.
Let us now consider motion in a repulsive field, where
U
(a>0).
(15.13)
Here the effective potential energy is
Utt
and decreases monotonically from + 00 to zero as r varies from zero to
infinity. The energy of the particle must be positive, and the motion is always
infinite. The calculations are exactly similar to those for the attractive field.
The path is a hyperbola (or, if E = 0, a parabola):
pr r = =1-e coso, =
(15.14)
where P and e are again given by (15.4). The path passes the centre of the
field in the manner shown in Fig. 13. The perihelion distance is
rmin =p(e-1)=a(e+1). =
(15.15)
The time dependence is given by the parametric equations
= =(ma3/a)(esinh+) =
(15.16)
x
= a(cosh & e ,
y = av((e2-1) sinh &
§15
Kepler's problem
39
To conclude this section, we shall show that there is an integral of the mo-
tion which exists only in fields U = a/r (with either sign of a). It is easy to
verify by direct calculation that the quantity
vxM+ar/r
(15.17)
is constant. For its total time derivative is v
since M = mr xv,
Putting mv = ar/r3 from the equation of motion, we find that this expression
vanishes.
y
0
(I+e)
FIG. 13
The direction of the conserved vector (15.17) is along the major axis from
the focus to the perihelion, and its magnitude is ae. This is most simply
seen by considering its value at perihelion.
It should be emphasised that the integral (15.17) of the motion, like M and
E, is a one-valued function of the state (position and velocity) of the particle.
We shall see in §50 that the existence of such a further one-valued integral
is due to the degeneracy of the motion.
PROBLEMS
PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
moving in a parabola in a field U = -a/r.
SOLUTION. In the integral
we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
the required dependence:
r=1p(1+n2),
t=
y=pn.
40
Integration of the Equations of Motion
§15
The parameter n varies from - 00 to +00.
PROBLEM 2. Integrate the equations of motion for a particle in a central field
U = - a/r2 (a > 0).
SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
(a) for E > andM 0 and
(b) for E>0 0nd and M 2/2m a,
(c) for E <0 and Ms1
In all three cases
In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
origin as
00. The fall from a given value of r takes place in a finite time, namely
PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
(14.10), which we write as
in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
the first-order term gives the required change so:
(1)
where we have changed from the integration over r to one over , along the path of the "un-
perturbed" motion.
In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.
CHAPTER IV
COLLISIONS BETWEEN PARTICLES

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IN many cases the laws of conservation of momentum and energy alone can
be used to obtain important results concerning the properties of various mech-
anical processes. It should be noted that these properties are independent of
the particular type of interaction between the particles involved.
Let us consider a "spontaneous" disintegration (that is, one not due to
external forces) of a particle into two "constituent parts", i.e. into two other
particles which move independently after the disintegration.
This process is most simply described in a frame of reference in which the
particle is at rest before the disintegration. The law of conservation of momen-
tum shows that the sum of the momenta of the two particles formed in the
disintegration is then zero; that is, the particles move apart with equal and
opposite momenta. The magnitude Po of either momentum is given by the
law of conservation of energy:
here M1 and m2 are the masses of the particles, E1t and E2i their internal
energies, and E the internal energy of the original particle. If € is the "dis-
integration energy", i.e. the difference
E= E:-E11-E2i,
(16.1)
which must obviously be positive, then
(16.2)
which determines Po; here m is the reduced mass of the two particles. The
velocities are V10 = Po/m1, V20 = Po/m2.
Let us now change to a frame of reference in which the primary particle
moves with velocity V before the break-up. This frame is usually called the
laboratory system, or L system, in contradistinction to the centre-of-mass
system, or C system, in which the total momentum is zero. Let us consider
one of the resulting particles, and let V and V0 be its velocities in the L and
the C system respectively. Evidently V = V+vo, or V -V = V0, and SO
(16.3)
where 0 is the angle at which this particle moves relative to the direction of
the velocity V. This equation gives the velocity of the particle as a function
41
42
Collisions Between Particles
§16
of its direction of motion in the L system. In Fig. 14 the velocity V is repre-
sented by a vector drawn to any point on a circle+ of radius vo from a point
A at a distance V from the centre. The cases V < vo and V>00 are shown
in Figs. 14a, b respectively. In the former case 0 can have any value, but in
the latter case the particle can move only forwards, at an angle 0 which does
not exceed Omax, given by
(16.4)
this is the direction of the tangent from the point A to the circle.
C
V
V
VO
VO
max
oo
oo
A
V
A
V
(a) V<VO
)V>Vo
FIG. 14
The relation between the angles 0 and Oo in the L and C systems is evi-
dently (Fig. 14)
tan
(16.5)
If this equation is solved for cos Oo, we obtain
V
0
(16.6)
For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus
sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however,
the relation is not one-to-one: for each value of 0 there are two values of Oo,
which correspond to vectors V0 drawn from the centre of the circle to the
points B and C (Fig. 14b), and are given by the two signs in (16.6).
In physical applications we are usually concerned with the disintegration
of not one but many similar particles, and this raises the problem of the
distribution of the resulting particles in direction, energy, etc. We shall
assume that the primary particles are randomly oriented in space, i.e. iso-
tropically on average.
t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral
section.
§16
Disintegration of particles
43
In the C system, this problem is very easily solved: every resulting particle
(of a given kind) has the same energy, and their directions of motion are
isotropically distributed. The latter fact depends on the assumption that the
primary particles are randomly oriented, and can be expressed by saying
that the fraction of particles entering a solid angle element doo is proportional
to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is
obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is
(16.7)
The corresponding distributions in the L system are obtained by an
appropriate transformation. For example, let us calculate the kinetic energy
distribution in the L system. Squaring the equation V = V0 + V, we have
2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the
kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of
particle is under consideration, and substituting in (16.7), we find the re-
quired distribution:
(1/2mvov) dT.
(16.8)
The kinetic energy can take values between Tmin = 3m(e0-V)2 and
Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed
uniformly over this range.
When a particle disintegrates into more than two parts, the laws of con-
servation of energy and momentum naturally allow considerably more free-
dom as regards the velocities and directions of motion of the resulting particles.
In particular, the energies of these particles in the C system do not have
determinate values. There is, however, an upper limit to the kinetic energy
of any one of the resulting particles. To determine the limit, we consider
the system formed by all these particles except the one concerned (whose
mass is M1, say), and denote the "internal energy" of that system by Ei'.
Then the kinetic energy of the particle M1 is, by (16.1) and (16.2),
T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the
primary particle. It is evident that T10 has its greatest possible value
when E/' is least. For this to be so, all the resulting particles except M1
must be moving with the same velocity. Then Ei is simply the sum of their
internal energies, and the difference E;-E-E; is the disintegration
energy E. Thus
T10,max = (M - M1) E M.
(16.9)
PROBLEMS
PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra-
tion into two particles.
SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling
010 simply Oo and using formula (16.5) for each of the two particles, we can put
7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two
equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then
44
Collisions Between Particles

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form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using
(16.2),
(m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6
PROBLEM 2. Find the angular distribution of the resulting particles in the L system.
SOLUTION. When vo > V, we substitute (16.6), with the plus sign of the radical, in (16.7),
obtaining
(0 .
When vo < V, both possible relations between Oo and 0 must be taken into account. Since,
when 0 increases, one value of Oo increases and the other decreases, the difference (not the
sum) of the expressions d cos Oo with the two signs of the radical in (16.6) must be taken.
The result is
(0 max).
PROBLEM 3. Determine the range of possible values of the angle 0 between the directions
of motion of the two resulting particles in the L system.
SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula (16.5)
(see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema
of the resulting expression gives the following ranges of 0, depending on the relative magni-
tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10
< 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by
sin =
§17. Elastic collisions
A collision between two particles is said to be elastic if it involves no change
in their internal state. Accordingly, when the law of conservation of energy
is applied to such a collision, the internal energy of the particles may be
neglected.
The collision is most simply described in a frame of reference in which the
centre of mass of the two particles is at rest (the C system). As in $16, we
distinguish by the suffix 0 the values of quantities in that system. The velo-
cities of the particles before the collision are related to their velocities V1 and
V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
where V = V1-V2; see (13.2).
Because of the law of conservation of momentum, the momenta of the two
particles remain equal and opposite after the collision, and are also unchanged
in magnitude, by the law of conservation of energy. Thus, in the C system
the collision simply rotates the velocities, which remain opposite in direction
and unchanged in magnitude. If we denote by no a unit vector in the direc-
tion of the velocity of the particle M1 after the collision, then the velocities
of the two particles after the collision (distinguished by primes) are
V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
(17.1)
§17
Elastic collisions
45
In order to return to the L system, we must add to these expressions the
velocity V of the centre of mass. The velocities in the L system after the
collision are therefore
V1' =
(17.2)
V2' =
No further information about the collision can be obtained from the laws
of conservation of momentum and energy. The direction of the vector no
depends on the law of interaction of the particles and on their relative position
during the collision.
The results obtained above may be interpreted geometrically. Here it is
more convenient to use momenta instead of velocities. Multiplying equations
(17.2) by M1 and M2 respectively, we obtain
(17.3)
P2' muno+m2(p1+p2)/(m1+m2)
where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
mv and use the construction shown in Fig. 15. If the unit vector no is along
OC, the vectors AC and CB give the momenta P1' and P2' respectively.
When p1 and P2 are given, the radius of the circle and the points A and B
are fixed, but the point C may be anywhere on the circle.
C
p'
no
P'2
B
A
FIG. 15
Let us consider in more detail the case where one of the particles (m2, say) is
at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
momentum P1 of the particle M1 before the collision. The point A lies inside
or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
46
Collisions Between Particles
§17
diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
are the angles between the directions of motion after the collision and the
direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
gives the direction of no, is the angle through which the direction of motion
of m1 is turned in the C system. It is evident from the figure that 01 and O2
can be expressed in terms of X by
(17.4)
C
p'
P2
pi
P2
0
max
10,
X
O2
O2
B
B
A
0
A
Q
0
(a) m < m2
(b) m, m m m
AB=p : AO/OB= m/m2
FIG. 16
We may give also the formulae for the magnitudes of the velocities of the
two particles after the collision, likewise expressed in terms of X:
ib
(17.5)
The sum A1 + O2 is the angle between the directions of motion of the
particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
if M1 > M2.
When the two particles are moving afterwards in the same or in opposite
directions (head-on collision), we have X=TT, i.e. the point C lies on the
diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
In this case the velocities after the collision are
(17.6)
This value of V2' has the greatest possible magnitude, and the maximum
§17
Elastic collisions
47
energy which can be acquired in the collision by a particle originally at rest
is therefore
(17.7)
where E1 = 1M1U12 is the initial energy of the incident particle.
If M1 < M2, the velocity of M1 after the collision can have any direction.
If M1 > M2, however, this particle can be deflected only through an angle
not exceeding Omax from its original direction; this maximum value of A1
corresponds to the position of C for which AC is a tangent to the circle
(Fig. 16b). Evidently
sin Omax = OC|OA = M2/M1.
(17.8)
The collision of two particles of equal mass, of which one is initially at
rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
C
p'
P2
Q2
B
A
0
FIG. 17
Then
01=1x,
A2 = 1(-x),
(17.9)
12
=
(17.10)
After the collision the particles move at right angles to each other.
PROBLEM
Express the velocity of each particle after a collision between a moving particle (m1) and
another at rest (m2) in terms of their directions of motion in the L system.
SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
Hence
for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.
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§18. Scattering
As already mentioned in §17, a complete calculation of the result of a
collision between two particles (i.e. the determination of the angle x) requires
the solution of the equations of motion for the particular law of interaction
involved.
We shall first consider the equivalent problem of the deflection of a single
particle of mass m moving in a field U(r) whose centre is at rest (and is at
the centre of mass of the two particles in the original problem).
As has been shown in $14, the path of a particle in a central field is sym-
metrical about a line from the centre to the nearest point in the orbit (OA
in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
say) with this line. The angle X through which the particle is deflected as it
passes the centre is seen from Fig. 18 to be
X = -200.
(18.1)
A
X
to
FIG. 18
The angle do itself is given, according to (14.7), by
(M/r2) dr
(18.2)
taken between the nearest approach to the centre and infinity. It should be
recalled that rmin is a zero of the radicand.
For an infinite motion, such as that considered here, it is convenient to
use instead of the constants E and M the velocity Voo of the particle at infinity
and the impact parameter p. The latter is the length of the perpendicular
from the centre O to the direction of Voo, i.e. the distance at which the particle
would pass the centre if there were no field of force (Fig. 18). The energy
and the angular momentum are given in terms of these quantities by
E = 1mvoo²,
M = mpVoo,
(18.3)
§18
Scattering
49
and formula (18.2) becomes
dr
(18.4)
Together with (18.1), this gives X as a function of p.
In physical applications we are usually concerned not with the deflection
of a single particle but with the scattering of a beam of identical particles
incident with uniform velocity Voo on the scattering centre. The different
particles in the beam have different impact parameters and are therefore
scattered through different angles X. Let dN be the number of particles
scattered per unit time through angles between X and X + dx. This number
itself is not suitable for describing the scattering process, since it is propor-
tional to the density of the incident beam. We therefore use the ratio
do = dN/n,
(18.5)
where n is the number of particles passing in unit time through unit area of
the beam cross-section (the beam being assumed uniform over its cross-
section). This ratio has the dimensions of area and is called the effective
scattering cross-section. It is entirely determined by the form of the scattering
field and is the most important characteristic of the scattering process.
We shall suppose that the relation between X and P is one-to-one; this is
so if the angle of scattering is a monotonically decreasing function of the
impact parameter. In that case, only those particles whose impact parameters
lie between p(x) and p(x) + dp(x) are scattered at angles between X and
+ dx. The number of such particles is equal to the product of n and the
area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
effective cross-section is therefore
do = 2mp dp.
(18.6)
In order to find the dependence of do on the angle of scattering, we need
only rewrite (18.6) as
do = 2(x)|dp(x)/dx|dx
(18.7)
Here we use the modulus of the derivative dp/dx, since the derivative may
be (and usually is) negative. t Often do is referred to the solid angle element
do instead of the plane angle element dx. The solid angle between cones
with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
(18.7)
do.
(18.8)
t If the function p(x) is many-valued, we must obviously take the sum of such expressions
as (18.7) over all the branches of this function.
50
Collisions Between Particles
§18
Returning now to the problem of the scattering of a beam of particles, not
by a fixed centre of force, but by other particles initially at rest, we can say
that (18.7) gives the effective cross-section as a function of the angle of
scattering in the centre-of-mass system. To find the corresponding expression
as a function of the scattering angle 0 in the laboratory system, we must
express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
expressions for both the scattering cross-section for the incident beam of
particles (x in terms of 01) and that for the particles initially at rest (x in terms
of O2).
PROBLEMS
PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
for r>a).
SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
the path consists of two straight lines symmetrical about the radius to the point where the
particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
a sin to = a sin 1(-x) = a cos 1x.
A
to
p
&
FIG. 19
Substituting in (18.7) or (18.8), we have
do = 1ma2 sin X do,
(1)
i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
the total cross-section o = na2, in accordance with the fact that the "impact area" which the
particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
and m2 that of the sphere) we have
do1,
where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
stituting X = 201 from (17.9) in (1).
§18
Scattering
51
For a sphere originally at rest, X = n-202 in all cases, and substitution in (1) gives
do2 = a2|cos 02 do2.
PROBLEM 2. Express the effective cross-section (Problem 1) as a function of the energy E
lost by a scattered particle.
SOLUTION. The energy lost by a particle of mass M1 is equal to that gained by the sphere of
mass M2. From (17.5) and (17.7), € = E2' = [2m22m2/(m1+m2)2] voo2 sin21x = Emax sin21x,
whence de = 1 mex sin X dx; substituting in (1), Problem 1, we have do = na2 de/emax. The
scattered particles are uniformly distributed with respect to € in the range from zero to
Emax.
PROBLEM 3. Find the effective cross-section as a function of the velocity Uoo for particles
scattered in a field U -rrn.
SOLUTION. According to (10.3), if the potential energy is a homogeneous function of order
k = -n, then similar paths are such that p ~v-2/n, or p = Voo-2/nf(x), the angles of deflec-
tion X being equal for similar paths. Substitution in (18.6) gives do ~ Voo-4/n do.
PROBLEM 4. Determine the effective cross-section for a particle to "fall" to the centre of
a field U = -a/r2.
SOLUTION. The particles which "fall" to the centre are those for which 2a > mp20002 (see
(14.11)), i.e. for which the impact parameter does not exceed Pmax = (2a/mvoo). The
effective cross-section is therefore o = Pmax2 = 2na/mvo².
PROBLEM 5. The same as Problem 4, but for a field U = -a/rn (n 2, a > 0).
SOLUTION. The effective potential energy Ueff = depends on r in the
manner shown in Fig. 20. Its maximum value is
Ueff
U0
FIG. 20
The particles which "fall" to the centre are those for which U0 < E. The condition U0 = E
gives Pmax, whence
=
PROBLEM 6. Determine the effective cross-section for particles of mass m1 to strike a sphere
of mass m2 and radius R to which they are attracted in accordance with Newton's law.
SOLUTION. The condition for a particle to reach the sphere is that rmin < R, where r'min
is the point on the path which is nearest to the centre of the sphere. The greatest possible
value of P is given by rmin = R; this is equivalent to Ueff(R) = E or
= , where a = ymim2 (2 being the gravitational constant) and we have put m 22 M1 on
the assumption that m2 > M1. Solving for pmax2, we finally obtain o = mR2(1+2ym2/Rvv3).
52
Collisions Between Particles
§18
When
Voo
8 the effective cross-section tends, of course, to the geometrical cross-section
of the sphere.
PROBLEM 7. Deduce the form of a scattering field U(r), given the effective cross-section
as a function of the angle of scattering for a given energy E. It is assumed that U(r) decreases
monotonically with r (a repulsive field), with U(0) > E and U(00) = 0 (O. B. FIRSOV 1953).
SOLUTION. Integration of do with respect to the scattering angle gives, according to the
formula
(1)
the square of the impact parameter, so that p(x) (and therefore x(p)) is known.
We put
s=1/r,
=1/p2,
[[1-(U|E)]
(2)
Then formulae (18.1), (18.2) become
1/
(3)
where so(x) is the root of the equation xw2(so)-so2 = 0.
Equation (3) is an integral equation for the function w(s), and may be solved by a method
similar to that used in $12. Dividing both sides of (3) by (a-x) and integrating with respect
to x from zero to a, we find
so(a)
dx ds
so(a)
ds
or, integrating by parts on the left-hand side,
This relation is differentiated with respect to a, and then so(a) is replaced by s simply;
accordingly a is replaced by s2/w2, and the result is, in differential form,
=
(11/20)
or
dx
This equation can be integrated immediately if the order of integration on the right-hand
side is inverted. Since for S = 0 (i.e. r oo) we must have W = 1 (i.e. U = 0), we have,

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Rutherford's formula
53
on returning to the original variables r and P, the following two equivalent forms of the final
result:
=====
(dx/dp)
(4)
This formula determines implicitly the function w(r) (and therefore U(r)) for all r > rmin,
i.e. in the range of r which can be reached by a scattered particle of given energy E.
§19. Rutherford's formula
One of the most important applications of the formulae derived above is
to the scattering of charged particles in a Coulomb field. Putting in (18.4)
U = a/r and effecting the elementary integration, we obtain
whence p2 = (a2/m2voo4) tan2oo, or, putting 00 = 1(-) from (18.1),
p2 =
(19.1)
Differentiating this expression with respect to X and substituting in (18.7)
or (18.8) gives
do = (a/v2cosxdx/sin31
(19.2)
or
do =
(19.3)
This is Rutherford's formula. It may be noted that the effective cross-section
is independent of the sign of a, so that the result is equally valid for repulsive
and attractive Coulomb fields.
Formula (19.3) gives the effective cross-section in the frame of reference
in which the centre of mass of the colliding particles is at rest. The trans-
formation to the laboratory system is effected by means of formulae (17.4).
For particles initially at rest we substitute X = 77 - 202 in (19.2) and obtain
do2 = 2n(a/mvoo2)2 sin de2/cos302
=
(19.4)
The same transformation for the incident particles leads, in general, to a very
complex formula, and we shall merely note two particular cases.
If the mass M2 of the scattering particle is large compared with the mass
M1 of the scattered particle, then X 2 O1 and m 22 M1, SO that
do1 = = (a/4E1)2do1/sin4301
(19.5)
where E1 = 1M1U..02 is the energy of the incident particle.
54
Collisions Between Particles
§ 19
If the masses of the two particles are equal (m1 = M2, m = 1M1), then by
(17.9) X = 201, and substitution in (19.2) gives
do1 = 2(/E1)2 cos 01 d01/sin³01
=
(19.6)
If the particles are entirely identical, that which was initially at rest cannot
be distinguished after the collision. The total effective cross-section for all
particles is obtained by adding do1 and do2, and replacing A1 and O2 by their
common value 0:
do
=
do.
(19.7)
Let us return to the general formula (19.2) and use it to determine the
distribution of the scattered particles with respect to the energy lost in the
collision. When the masses of the scattered (m1) and scattering (m2) particles
are arbitrary, the velocity acquired by the latter is given in terms of the angle
of scattering in the C system by V2' = [2m1/(m1+m2)]%"00 sin 1x; see (17.5).
The energy acquired by M2 and lost by M1 is therefore E = 1M2U2'2
= (2m2/m2)/0002 sin21x. Expressing sin 1x in terms of E and substituting
in (19.2), we obtain
do = de/e2.
(19.8)
This is the required formula: it gives the effective cross-section as a function
of the energy loss E, which takes values from zero to Emax = 2m2voo2/m2.
PROBLEMS
PROBLEM 1. Find the effective cross-section for scattering in a field U = a/r2 (a > 0).
SOLUTION. The angle of deflection is
The effective cross-section is
do
sin
PROBLEM 2. Find the effective cross-section for scattering by a spherical "potential well"
of radius a and "depth" U0 (i.e. a field with U = 0 for r > a and U = - U0 for r < a).
SOLUTION. The particle moves in a straight line which is "refracted" on entering and leav- -
ing the well. According to §7, Problem, the angle of incidence a and the angle of refraction
B (Fig. 21) are such that sin x/sin B = n, where n = W(1+2U0/mVo2). The angle of deflection
is X = 2(a-B). Hence
=
Eliminating a from this equation and the relation a sin a p, which is evident from the
diagram, we find the relation between P and X:
cos
1x

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Small-angle scattering
55
Finally, differentiating, we have the effective cross-section
cos
do.
The angle X varies from zero (for p=0) to Xmax (for P = a), where cos 1xmax = 1/n.
The total effective cross-section, obtained by integrating do over all angles within the cone
Xmax, is, of course, equal to the geometrical cross-section 2
a
to
a
FIG. 21
§20. Small-angle scattering
The calculation of the effective cross-section is much simplified if only
those collisions are considered for which the impact parameter is large, so
that the field U is weak and the angles of deflection are small. The calculation
can be carried out in the laboratory system, and the centre-of-mass system
need not be used.
We take the x-axis in the direction of the initial momentum of the scattered
particle M1, and the xy-plane in the plane of scattering. Denoting by P1' the
momentum of the particle after scattering, we evidently have sin 01 = P1y'/P1'.
For small deflections, sin 01 may be approximately replaced by 01, and P1' in
the denominator by the initial momentum P1 = MIUoo:
(20.1)
Next, since Py = Fy, the total increment of momentum in the y-direction is
(20.2)
The
force Fy = - JULY (dU/dr)dr/dy = (dU/dr)y/r.
Since the integral (20.2) already contains the small quantity U, it can be
calculated, in the same approximation, by assuming that the particle is not
deflected at all from its initial path, i.e. that it moves in a straight line y = p
with uniform velocity Voo. Thus we put in (20.2) Fy = - (dU/dr)p/r,
dt = dx/voo. The result is
56
Collisions Between Particles
§20
Finally, we change the integration over x to one over r. Since, for a straight
path, r2 = x2+p2, when x varies from - 00 to + 80, r varies from 8 to P
and back. The integral over x therefore becomes twice the integral over r
from p to 80, and dx = r dr/v(r2-p2). The angle of scattering O1 is thus
given byt
(20.3)
and this is the form of the function 01(p) for small deflections. The effective
cross-section for scattering (in the L system) is obtained from (18.8) with 01
instead of X, where sin 01 may now be replaced by A1:
(20.4)
PROBLEMS
PROBLEM 1. Derive formula (20.3) from (18.4).
SOLUTION. In order to avoid spurious divergences, we write (18.4) in the form
PO
and take as the upper limit some large finite quantity R, afterwards taking the value as R
00.
Since U is small, we expand the square root in powers of U, and approximately replace
rmin by p:
dr
The first integral tends to 1/11 as R 00. The second integral is integrated by parts, giving
=
This is equivalent to (20.3).
PROBLEM 2. Determine the effective cross-section for small-angle scattering in a field
U=a/m(n) 0).
t If the above derivation is applied in the C system, the expression obtained for X is the
same with m in place of M1, in accordance with the fact that the small angles 01 and X are
related by (see (17.4)) 01 = m2x/(m1 +m2).
§20
Small-angle scattering
57
SOLUTION. From (20.3) we have
dr
The substitution p2/r2 = U converts the integral to a beta function, which can be expressed
in terms of gamma functions:
Expressing P in terms of 01 and substituting in (20.4), we obtain
do1.
3
CHAPTER V
SMALL OSCILLATIONS
$21. Free oscillations in one dimension
A VERY common form of motion of mechanical systems is what are called
small oscillations of a system about a position of stable equilibrium. We shall
consider first of all the simplest case, that of a system with only one degree
of freedom.
Stable equilibrium corresponds to a position of the system in which its
potential energy U(q) is a minimum. A movement away from this position
results in the setting up of a force - dU/dq which tends to return the system
to equilibrium. Let the equilibrium value of the generalised co-ordinate
q be 90. For small deviations from the equilibrium position, it is sufficient
to retain the first non-vanishing term in the expansion of the difference
U(q) - U(90) in powers of q-qo. In general this is the second-order term:
U(q) - U(q0) 112 1k(q-90)2, where k is a positive coefficient, the value of the
second derivative U"(q) for q = 90. We shall measure the potential energy
from its minimum value, i.e. put U(qo) = 0, and use the symbol
x = q-90
(21.1)
for the deviation of the co-ordinate from its equilibrium value. Thus
U(x) = .
(21.2)
The kinetic energy of a system with one degree of freedom is in general
of the form 1a(q)q2 = 1a(q)x2. In the same approximation, it is sufficient to
replace the function a(q) by its value at q = qo. Putting for brevity a(go) = m,
we have the following expression for the Lagrangian of a system executing
small oscillations in one dimension:
L = 1mx2-1kx2.
(21.3)
The corresponding equation of motion is
m+kx=0,
(21.4)
or
w2x=0,
(21.5)
where
w= ((k/m).
(21.6)
+ It should be noticed that m is the mass only if x is the Cartesian co-ordinate.
+ Such a system is often called a one-dimensional oscillator.
58

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Free oscillations in one dimension
59
Two independent solutions of the linear differential equation (21.5) are
cos wt and sin wt, and its general solution is therefore
COS wt +C2 sin wt.
(21.7)
This expression can also be written
x = a cos(wt + a).
(21.8)
Since cos(wt+a) = cos wt cos a - sin wt sin a, a comparison with (21.7)
shows that the arbitrary constants a and a are related to C1 and C2 by
tan a = - C2/C1.
(21.9)
Thus, near a position of stable equilibrium, a system executes harmonic
oscillations. The coefficient a of the periodic factor in (21.8) is called the
amplitude of the oscillations, and the argument of the cosine is their phase;
a is the initial value of the phase, and evidently depends on the choice of
the origin of time. The quantity w is called the angular frequency of the oscil-
lations; in theoretical physics, however, it is usually called simply the fre-
quency, and we shall use this name henceforward.
The frequency is a fundamental characteristic of the oscillations, and is
independent of the initial conditions of the motion. According to formula
(21.6) it is entirely determined by the properties of the mechanical system
itself. It should be emphasised, however, that this property of the frequency
depends on the assumption that the oscillations are small, and ceases to hold
in higher approximations. Mathematically, it depends on the fact that the
potential energy is a quadratic function of the co-ordinate.
The energy of a system executing small oscillations is E =
= 1m(x2+w2x2) or, substituting (21.8),
E =
(21.10)
It is proportional to the square of the amplitude.
The time dependence of the co-ordinate of an oscillating system is often
conveniently represented as the real part of a complex expression:
x = re[A exp(iwt)],
(21.11)
where A is a complex constant; putting
A = a exp(ix),
(21.12)
we return to the expression (21.8). The constant A is called the complex
amplitude; its modulus is the ordinary amplitude, and its argument is the
initial phase.
The use of exponential factors is mathematically simpler than that of
trigonometrical ones because they are unchanged in form by differentiation.
t It therefore does not hold good if the function U(x) has at x = 0 a minimum of
higher order, i.e. U ~ xn with n > 2; see §11, Problem 2(a).
60
Small Oscillations
§21
So long as all the operations concerned are linear (addition, multiplication
by constants, differentiation, integration), we may omit the sign re through-
out and take the real part of the final result.
PROBLEMS
PROBLEM 1. Express the amplitude and initial phase of the oscillations in terms of the
initial co-ordinate xo and velocity vo.
SOLUTION. a = (xx2+002/w2), tan a = -vo/wxo.
PROBLEM 2. Find the ratio of frequencies w and w' of the oscillations of two diatomic
molecules consisting of atoms of different isotopes, the masses of the atoms being M1, m2 and
'M1', m2'.
SOLUTION. Since the atoms of the isotopes interact in the same way, we have k = k'.
The coefficients m in the kinetic energies of the molecules are their reduced masses. Accord-
ing to (21.6) we therefore have
PROBLEM 3. Find the frequency of oscillations of a particle of mass m which is free to
move along a line and is attached to a spring whose other end is fixed at a point A (Fig. 22)
at a distance l from the line. A force F is required to extend the spring to length l.
A
X
FIG. 22
SOLUTION. The potential energy of the spring is (to within higher-order terms) equal to
the force F multiplied by the extension Sl of the spring. For x < l we have 81 = (12++2) -
=
x2/21, so that U = Fx2/21. Since the kinetic energy is 1mx2, we have = V(F/ml).
PROBLEM 4. The same as Problem 3, but for a particle of mass m moving on a circle of
radius r (Fig. 23).
m
&
FIG. 23

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Forced oscillations
61
SOLUTION. In this case the extension of the spring is (if
= cos
The kinetic energy is T = 1mr2o2, and the frequency is therefore w = V[F(++1)/mrl].
PROBLEM 5. Find the frequency of oscillations of the pendulum shown in Fig. 2 (§5),
whose point of support carries a mass M1 and is free to move horizontally.
SOLUTION. For < 1 the formula derived in $14, Problem 3 gives
T
Hence
PROBLEM 6. Determine the form of a curve such that the frequency of oscillations of a
particle on it under the force of gravity is independent of the amplitude.
SOLUTION. The curve satisfying the given condition is one for which the potential energy
of a particle moving on it is U = 1ks2, where s is the length of the arc from the position of
equilibrium. The kinetic energy T = 1ms2, where m is the mass of the particle, and the fre-
quency is then w = (k/m) whatever the initial value of S.
In a gravitational field U = mgy, where y is the vertical co-ordinate. Hence we have
1ks2 = mgy or y = w2s2/2g. But ds2 = dx2+dy², whence
dy = SV[(g/2w2y)-1] dy.
The integration is conveniently effected by means of the substitution y = g(1-cos E)/4ww,
which yields x = g(s+sin 5)/4w2. These two equations give, in parametric form, the equation
of the required curve, which is a cycloid.
$22. Forced oscillations
Let us now consider oscillations of a system on which a variable external
force acts. These are called forced oscillations, whereas those discussed in
§21 are free oscillations. Since the oscillations are again supposed small, it
is implied that the external field is weak, because otherwise it could cause the
displacement x to take too large values.
The system now has, besides the potential energy 1kx2, the additional
potential energy Ue(x, t) resulting from the external field. Expanding this
additional term as a series of powers of the small quantity x, we have
Ue(x, t) 12 x[dUe/dx]x_0. The first term is a function of time only,
and may therefore be omitted from the Lagrangian, as being the total time
derivative of another function of time. In the second term - [dUe/dx]x_0 is
the external "force" acting on the system in the equilibrium position, and
is a given function of time, which we denote by F(t). Thus the potential
energy involves a further term -xF(t), and the Lagrangian of the system
is
L
=
(22.1)
The corresponding equation of motion is m+kx = F(t) or
(22.2)
where we have again introduced the frequency w of the free oscillations.
The general solution of this inhomogeneous linear differential equation
with constant coefficients is x = xo+x1, where xo is the general solution of
62
Small Oscillations
§22
the corresponding homogeneous equation and X1 is a particular integral of
the inhomogeneous equation. In the present case xo represents the free
oscillations discussed in $21.
Let us consider a case of especial interest, where the external force is itself
a simple periodic function of time, of some frequency y:
F(t) = f cos(yt+)).
(22.3)
We seek a particular integral of equation (22.2) in the form X1 = b cos(yt+B),
with the same periodic factor. Substitution in that equation gives
b = f/m(w2-r2); adding the solution of the homogeneous equation, we
obtain the general integral in the form
(22.4)
The arbitrary constants a and a are found from the initial conditions.
Thus a system under the action of a periodic force executes a motion which
is a combination of two oscillations, one with the intrinsic frequency w of
the system and one with the frequency y of the force.
The solution (22.4) is not valid when resonance occurs, i.e. when the fre-
quency y of the external force is equal to the intrinsic frequency w of the
system. To find the general solution of the equation of motion in this case,
we rewrite (22.4) as
x
=
a
where a now has a different value. Asy->w, the second term is indetermin-
ate, of the form 0/0. Resolving the indeterminacy by L'Hospital's rule, we
have
x = acos(wt+a)+(f/2mw)tsin(wt+B). =
(22.5)
Thus the amplitude of oscillations in resonance increases linearly with the
time (until the oscillations are no longer small and the whole theory given
above becomes invalid).
Let us also ascertain the nature of small oscillations near resonance, when
y w+E with E a small quantity. We put the general solution in the com-
plex form
= A exp(iwt) exp[i(w+t)) = [A+B exp(iet)]exp(ist)
(22.6)
Since the quantity A+B exp(iet) varies only slightly over the period 2n/w
of the factor exp(iwt), the motion near resonance may be regarded as small
oscillations of variable amplitude.t Denoting this amplitude by C, we have
= A B exp(iet)|. Writing A and B in the form a exp(ix) and b exp(iB)
respectively, we obtain
(22.7)
t The "constant" term in the phase of the oscillation also varies.
§22
Forced oscillations
63
Thus the amplitude varies periodically with frequency E between the limits
|a-b a+b. This phenomenon is called beats.
The equation of motion (22.2) can be integrated in a general form for an
arbitrary external force F(t). This is easily done by rewriting the equation
as
or
=
(22.8)
where
s=xtiwx
(22.9)
is a complex quantity. Equation (22.8) is of the first order. Its solution when
the right-hand side is replaced by zero is $ = A exp(iwt) with constant A.
As before, we seek a solution of the inhomogeneous equation in the form
$ = A(t) exp(iwt), obtaining for the function A(t) the equation À(t)
= F(t) exp(-iwt)/m. Integration gives the solution of (22.9):
& = -
(22.10)
where the constant of integration so is the value of $ at the instant t = 0.
This is the required general solution; the function x(t) is given by the imagin-
ary part of (22.10), divided by w.t
The energy of a system executing forced oscillations is naturally not con-
served, since the system gains energy from the source of the external field.
Let us determine the total energy transmitted to the system during all time,
assuming its initial energy to be zero. According to formula (22.10), with
the lower limit of integration - 00 instead of zero and with ( - 00) = 0,
we have for t
00
exp(-iwt)dt|
The energy of the system is
E = 1m(x2+w2x2)= = 1ME2.
(22.11)
Substituting we obtain the energy transferred
(22.12)
t The force F(t) must, of course, be written in real form.
64
Small Oscillations
§22
it is determined by the squared modulus of the Fourier component of the
force F(t) whose frequency is the intrinsic frequency of the system.
In particular, if the external force acts only during a time short in com-
parison with 1/w, we can put exp(-iwt) Ill 1. Then
This result is obvious: it expresses the fact that a force of short duration
gives the system a momentum I F dt without bringing about a perceptible
displacement.
PROBLEMS
PROBLEM 1. Determine the forced oscillations of a system under a force F(t) of the follow-
ing forms, if at time t = 0 the system is at rest in equilibrium (x = x = 0): (a) F = Fo,
a constant, (b) F = at, (c) F = Fo exp(-at), (d) F = Fo exp(-at) cos Bt.
SOLUTION. (a) x = (Fo/mw2)(1-cos wt). The action of the constant force results in a dis-
placement of the position of equilibrium about which the oscillations take place.
(b) x = (a/mw3)(wt-sin wt).
(c) x = - cos wt +(a/w) sin wt].
(d) x = wt + sin wt +
+exp(-at)[(wpta2-B2) cos Bt-2aB sin
This last case is conveniently treated by writing the force in the complex form
F=Foexp(-ati)t].
PROBLEM 2. Determine the final amplitude for the oscillations of a system under a force
which is zero for t<0, Fot/T for 0 <t<<, and Fo for t > T (Fig. 24), if up to time
t = 0 the system is at rest in equilibrium.
F
Fo
,
T
FIG. 24
SOLUTION. During the interval 0<+<T the oscillations are determined by the initial
condition as x = (Fo/mTw3)(wt-sin wt). For t > T we seek a solution in the form
=c1w(t-T)+c2 sin w(t - T)+Fo/mw2
The continuity of x and x at t = T gives C1 = -(Fo/mTw3) sin wT, C2 = (Fo/mTw3 X
X (1 - cos wT). The amplitude is a = = (2Fo/mTw3) sin twT. This is the smaller,
the more slowly the force Fo is applied (i.e. the greater T).
PROBLEM 3. The same as Problem 2, but for a constant force Fo which acts for a finite
time T (Fig. 25).

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Oscillations of systems with more than one degree of freedom
65
SOLUTION. As in Problem 2, or more simply by using formula (22.10). For t > T we have
free oscillations about x =0, and
dt
FO
f
T
FIG. 25
The squared modulus of & gives the amplitude from the relation = The result is
a = (2Fo/mw2) sin twT.
PROBLEM 4. The same as Problem 2, but for a force Fot/T which acts between t = 0 and
t = T (Fig. 26).
F
FO
,
T
FIG. 26
SOLUTION. By the same method we obtain
a = (Fo/Tmw3)/[wT2-2wT sin wT+2(1-cos - wT)].
PROBLEM 5. The same as Problem 2, but for a force Fo sin wt which acts between t = 0
and t = T = 2n/w (Fig. 27).
F
T
FIG. 27
SOLUTION. Substituting in (22.10) F(t) = Fo sin wt = Fo[exp(iwt)-exp(-iwt)]/2i and
integrating from 0 to T, we obtain a = Fon/mw2.
$23. Oscillations of systems with more than one degree of freedom
The theory of free oscillations of systems with S degrees of freedom is
analogous to that given in §21 for the case S = 1.
3*
66
Small Oscillations
§23
Let the potential energy of the system U as a function of the generalised
co-ordinates qi (i = 1, 2, ..., s) have a minimum for qi = qio. Putting
Xi=qi-qio
(23.1)
for the small displacements from equilibrium and expanding U as a function
of the xi as far as the quadratic terms, we obtain the potential energy as a
positive definite quadratic form
(23.2)
where we again take the minimum value of the potential energy as zero.
Since the coefficients kik and kki in (23.2) multiply the same quantity XiXK,
it is clear that they may always be considered equal: kik = kki.
In the kinetic energy, which has the general form () (see (5.5)),
we put qi = qio in the coefficients aik and, denoting aik(90) by Mik, obtain
the kinetic energy as a positive definite quadratic form
Emission
(23.3)
The coefficients Mik also may always be regarded as symmetrical: Mik=Mki.
Thus the Lagrangian of a system executing small free oscillations is
(23.4)
i,k
Let us now derive the equations of motion. To determine the derivatives
involved, we write the total differential of the Lagrangian:
- kikxi dxk - kikxxdxi).
i,k
Since the value of the sum is obviously independent of the naming of the
suffixes, we can interchange i and k in the first and third terms in the paren-
theses. Using the symmetry of Mik and kik, we have
dL =
Hence
k
Lagrange's equations are therefore
(i=1,2,...,s);
(23.5)
they form a set of S linear homogeneous differential equations with constant
coefficients.
As usual, we seek the S unknown functions xx(t) in the form
xx = Ak explicut),
(23.6)
where Ak are some constants to be determined. Substituting (23.6) in the
§23
Oscillations of systems with more than one degree of freedom
67
equations (23.5) and cancelling exp(iwt), we obtain a set of linear homo-
geneous algebraic equations to be satisfied by the Ak:
(23.7)
If this system has non-zero solutions, the determinant of the coefficients
must vanish:
(23.8)
This is the characteristic equation and is of degree S in w2. In general, it has
S different real positive roots W&2 (a = 1,2,...,s); in particular cases, some of
these roots may coincide. The quantities Wa thus determined are the charac-
teristic frequencies or eigenfrequencies of the system.
It is evident from physical arguments that the roots of equation (23.8) are
real and positive. For the existence of an imaginary part of w would mean
the presence, in the time dependence of the co-ordinates XK (23.6), and SO
of the velocities XK, of an exponentially decreasing or increasing factor. Such
a factor is inadmissible, since it would lead to a time variation of the total
energy E = U+: T of the system, which would therefore not be conserved.
The same result may also be derived mathematically. Multiplying equation
(23.7) by Ai* and summing over i, we have = 0,
whence w2 = . The quadratic forms in the numerator
and denominator of this expression are real, since the coefficients kik and
Mik are real and symmetrical: (kA*Ak)* = kikAAk* = k
= kikAkAi*. They are also positive, and therefore w2 is positive.t
The frequencies Wa having been found, we substitute each of them in
equations (23.7) and find the corresponding coefficients Ak. If all the roots
Wa of the characteristic equation are different, the coefficients Ak are pro-
portional to the minors of the determinant (23.8) with w = Wa. Let these
minors be . A particular solution of the differential equations (23.5) is
therefore X1c = Ca exp(iwat), where Ca is an arbitrary complex constant.
The general solution is the sum of S particular solutions. Taking the real
part, we write
III
(23.9)
where
(23.10)
Thus the time variation of each co-ordinate of the system is a super-
position of S simple periodic oscillations O1, O2, ..., Os with arbitrary ampli-
tudes and phases but definite frequencies.
t The fact that a quadratic form with the coefficients kik is positive definite is seen from
their definition (23.2) for real values of the variables. If the complex quantities Ak are written
explicitly as ak +ibk, we have, again using the symmetry of kik, kikAi* Ak = kik(ai-ibi)
X
= kikaiak kikbibk, which is the sum of two positive definite forms.
68
Small Oscillations
§23
The question naturally arises whether the generalised co-ordinates can be
chosen in such a way that each of them executes only one simple oscillation.
The form of the general integral (23.9) points to the answer. For, regarding
the S equations (23.9) as a set of equations for S unknowns Oa, as we can
express O1, O2, ..., Os in terms of the co-ordinates X1, X2, ..., Xs. The
quantities Oa may therefore be regarded as new generalised co-ordinates,
called normal co-ordinates, and they execute simple periodic oscillations,
called normal oscillations of the system.
The normal co-ordinates Oa are seen from their definition to satisfy the
equations
Oatwaia = 0.
(23.11)
This means that in normal co-ordinates the equations of motion become S
independent equations. The acceleration in each normal co-ordinate depends
only on the value of that co-ordinate, and its time dependence is entirely
determined by the initial values of the co-ordinate and of the corresponding
velocity. In other words, the normal oscillations of the system are completely
independent.
It is evident that the Lagrangian expressed in terms of normal co-ordinates
is a sum of expressions each of which corresponds to oscillation in one dimen-
sion with one of the frequencies was i.e. it is of the form
(23.12)
where the Ma are positive constants. Mathematically, this means that the
transformation (23.9) simultaneously puts both quadratic forms-the kinetic
energy (23.3) and the potential energy (23.2)-in diagonal form.
The normal co-ordinates are usually chosen so as to make the coefficients
of the squared velocities in the Lagrangian equal to one-half. This can be
achieved by simply defining new normal co-ordinates Qx by
Qa = VMaOa.
(23.13)
Then
The above discussion needs little alteration when some roots of the charac-
teristic equation coincide. The general form (23.9), (23.10) of the integral of
the equations of motion remains unchanged, with the same number S of
terms, and the only difference is that the coefficients corresponding to
multiple roots are not the minors of the determinant, which in this case
vanish.
t The impossibility of terms in the general integral which contain powers of the time as
well as the exponential factors is seen from the same argument as that which shows that the
frequencies are real: such terms would violate the law of conservation of energy
§23
Oscillations of systems with more than one degree of freedom
69
Each multiple (or, as we say, degenerate) frequency corresponds to a number
of normal co-ordinates equal to its multiplicity, but the choice of these co-
ordinates is not unique. The normal co-ordinates with equal Wa enter the
kinetic and potential energies as sums Q and Qa2 which are transformed
in the same way, and they can be linearly transformed in any manner which
does not alter these sums of squares.
The normal co-ordinates are very easily found for three-dimensional oscil-
lations of a single particle in a constant external field. Taking the origin of
Cartesian co-ordinates at the point where the potential energy U(x,y,2) is
a minimum, we obtain this energy as a quadratic form in the variables x, y, Z,
and the kinetic energy T = m(x2+yj++2) (where m is the mass of the
particle) does not depend on the orientation of the co-ordinate axes. We
therefore have only to reduce the potential energy to diagonal form by an
appropriate choice of axes. Then
L =
(23.14)
and the normal oscillations take place in the x,y and 2 directions with fre-
quencies = (k1/m), w2=1/(k2/m), w3=1/(k3/m). In the particular
case of a central field (k1 =k2=kg III three frequencies
are equal (see Problem 3).
The use of normal co-ordinates makes possible the reduction of a problem
of forced oscillations of a system with more than one degree of freedom to a
series of problems of forced oscillation in one dimension. The Lagrangian of
the system, including the variable external forces, is
(23.15)
where L is the Lagrangian for free oscillations. Replacing the co-ordinates
X1c by normal co-ordinates, we have
(23.16)
where we have put
The corresponding equations of motion
(23.17)
each involve only one unknown function Qa(t).
PROBLEMS
PROBLEM 1. Determine the oscillations of a system with two degrees of freedom whose
Lagrangian is L = (two identical one-dimensional systems of
eigenfrequency wo coupled by an interaction - axy).
70
Small Oscillations

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SOLUTION. The equations of motion are x+ wo2x = ay, j + wo2y = ax. The substitution
(23.6) gives
Ax(wo2-w2) = aAy,
(1)
The characteristic equation is (wo2-w2)2= a2, whence w12 = wo2-a, w22 = wo2-+x. For
w = W1, the equations (1) give Ax = Ay, and for w = w2, Ax = -Ay. Hence x =
(Q1+Q2)/V2, y = (Q1-Q2)/V2, the coefficients 1/V2 resulting from the normalisation
of the normal co-ordinates as in equation (23.13).
For a < wo2 (weak coupling) we have W1 all wo-1x, W2 ill wotla. The variation of x
and y is in this case a superposition of two oscillations with almost equal frequencies, i.e.
beats of frequency W2-W1 = a (see $22). The amplitude of y is a minimum when that of x
is a maximum, and vice versa.
PROBLEM 2. Determine the small oscillations of a coplanar double pendulum (Fig. 1, $5).
SOLUTION. For small oscillations (01 < 1, 02 < 1), the Lagrangian derived in §5, Problem
1, becomes
L =
The equations of motion are
= 0, lio +1202+802
=
0.
Substitution of (23.6) gives
41(m1+m2)(g-h1w2)-A2w2m2l2 0, = 0.
The roots of the characteristic equation are
((ma3
As m1
8 the frequencies tend to the values (g/l1) and /(g/l2), corresponding to indepen-
dent oscillations of the two pendulums.
PROBLEM 3. Find the path of a particle in a central field U = 1kr2 (called a space oscillator).
SOLUTION. As in any central field, the path lies in a plane, which we take as the xy-plane.
The variation of each co-ordinate x,y is a simple oscillation with the same frequency
= v(k/m): x = a cos(wt+a), y=b cos(wt+), or x = a cos , y = b cos(+8)
= b cos 8 cos -b sin 8 sin , where = wt +a, 8 = B-a. Solving for cos o and sin o and
equating the sum of their squares to unity, we find the equation of the path:
This is an ellipse with its centre at the origin.t When 8 = 0 or IT, the path degenerates to a
segment of a straight line.
$24. Vibrations of molecules
If we have a system of interacting particles not in an external field, not all
of its degrees of freedom relate to oscillations. A typical example is that of
molecules. Besides motions in which the atoms oscillate about their positions
of equilibrium in the molecule, the whole molecule can execute translational
and rotational motions.
Three degrees of freedom correspond to translational motion, and in general
the same number to rotation, so that, of the 3n degrees of freedom of a mole-
cule containing n atoms, 3n-6 - correspond to vibration. An exception is formed
t The fact that the path in a field with potential energy U = 1kr2 is a closed curve has
already been mentioned in $14.
§24
Vibrations of molecules
71
by molecules in which the atoms are collinear, for which there are only two
rotational degrees of freedom (since rotation about the line of atoms is of no
significance), and therefore 3n-5 vibrational degrees of freedom.
In solving a mechanical problem of molecular oscillations, it is convenient
to eliminate immediately the translational and rotational degrees of freedom.
The former can be removed by equating to zero the total momentum of the
molecule. Since this condition implies that the centre of mass of the molecule
is at rest, it can be expressed by saying that the three co-ordinates of the
centre of mass are constant. Putting ra = rao+Ua, where ra0 is the radius
vector of the equilibrium position of the ath atom, and Ua its deviation from
this position, we have the condition = constant = or
= 0.
(24.1)
To eliminate the rotation of the molecule, its total angular momentum
must be equated to zero. Since the angular momentum is not the total time
derivative of a function of the co-ordinates, the condition that it is zero can-
not in general be expressed by saying that some such function is zero. For
small oscillations, however, this can in fact be done. Putting again
ra = rao+ua and neglecting small quantities of the second order in the
displacements Ua, we can write the angular momentum of the molecule as
= .
The condition for this to be zero is therefore, in the same approximation,
0,
(24.2)
in which the origin may be chosen arbitrarily.
The normal vibrations of the molecule may be classified according to the
corresponding motion of the atoms on the basis of a consideration of the sym-
metry of the equilibrium positions of the atoms in the molecule. There is
a general method of doing so, based on the use of group theory, which we
discuss elsewhere. Here we shall consider only some elementary examples.
If all n atoms in a molecule lie in one plane, we can distinguish normal
vibrations in which the atoms remain in that plane from those where they
do not. The number of each kind is readily determined. Since, for motion
in a plane, there are 2n degrees of freedom, of which two are translational
and one rotational, the number of normal vibrations which leave the atoms
in the plane is 2n-3. The remaining (3n-6)-(2n-3) = n-3 vibrational
degrees of freedom correspond to vibrations in which the atoms move out
of the plane.
For a linear molecule we can distinguish longitudinal vibrations, which
maintain the linear form, from vibrations which bring the atoms out of line.
Since a motion of n particles in a line corresponds to n degrees of freedom,
of which one is translational, the number of vibrations which leave the atoms
t See Quantum Mechanics, $100, Pergamon Press, Oxford 1965.
72
Small Oscillations
§24
in line is n - 1. Since the total number of vibrational degrees of freedom of a
linear molecule is 3n - 5, there are 2n-4 which bring the atoms out of line.
These 2n-4 vibrations, however, correspond to only n-2 different fre-
quencies, since each such vibration can occur in two mutually perpendicular
planes through the axis of the molecule. It is evident from symmetry that
each such pair of normal vibrations have equal frequencies.
PROBLEMS
PROBLEM 1. Determine the frequencies of vibrations of a symmetrical linear triatomic
molecule ABA (Fig. 28). It is assumed that the potential energy of the molecule depends
only on the distances AB and BA and the angle ABA.
3
B
A
(o)
(b)
(c)
FIG. 28
SOLUTION. The longitudinal displacements X1, X2, X3 of the atoms are related, according
to (24.1), by MA(X1+x3) +mBX2 = 0. Using this, we eliminate X2 from the Lagrangian of the
longitudinal motion
L =
and use new co-ordinatesQa=x1tx,Qx1-x3.The result
where u = 2mA+mB is the mass of the molecule. Hence we see that Qa and Qs are normal
co-ordinates (not yet normalised). The co-ordinate Qa corresponds to a vibration anti-
symmetrical about the centre of the molecule (x1 = x3; Fig. 28a), with frequency
wa = (k1u/mAmB). The co-ordinate Q8 corresponds to a symmetrical vibration (x1 = -x3;
Fig. 28b), with frequency
The transverse displacements y1, y2, y3 of the atoms are, according to (24.1) and (24.2),
related by mA(y1+y2) +mBy2 = 0, y1 = y3 (a symmetrical bending of the molecule; Fig. 28c).
The potential energy of this vibration can be written as 1/22/282, where 8 is the deviation of the
angle ABA from the value IT, given in terms of the displacements by 8 = [(y1-yy)+(ys-y2)]/L.
Expressing y1,y2, y3 in terms of 8, we obtain the Lagrangian of the transverse motion:
L =
whence the frequency is = V(2k2u/mAmB).
t Calculations of the vibrations of more complex molecules are given by M. V. VOL'KENSH-
TEIN, M. A. EL'YASHEVICH and B. I. STEPANOV, Molecular Vibrations (Kolebaniya molekul),
Moscow 1949; G. HERZBERG, Molecular Spectra and Molecular Structure: Infra-red and
Raman Spectra of Polyatomic Molecules, Van Nostrand, New York 1945,
§24
Vibrations of molecules
73
PROBLEM 2. The same as Problem 1, but for a triangular molecule ABA (Fig. 29).
y
A
A
3
2a
I
2
B
(a)
(b)
(c)
FIG. 29
SOLUTION. By (24.1) and (24.2) the x and y components of the displacements u of the
atoms are related by
0,
0,
(y1-y3) sin x-(x1+x3) cos a = 0.
The changes 8l1 and Sl2 in the distances AB and BA are obtained by taking the components
along these lines of the vectors U1-U2 and U3-U2:
8l1 = (x1-x2) sin cos a,
8l2 = -(x3-x2) sin at(y3-y2) cos a.
The change in the angle ABA is obtained by taking the components of those vectors per-
pendicular to AB and BA:
sin sin
a].
The Lagrangian of the molecule is
L
We use the new co-ordinates Qa = 1+x3, q81 = X1-x3, q82 = y1+y3. The components
of the vectors u are given in terms of these co-ordinates by X1 = 1(Qa+q x3 = 1(Qa-981),
X2
= -MAQa/MB, = -MAQ82/MB. The
Lagrangian becomes
L
=
qksici +
sin
a
cos
a.
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Small Oscillations

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Hence we see that the co-ordinate Qa corresponds to a normal vibration antisymmetrical
about the y-axis (x1 = x3, y1 = -y3; Fig. 29a) with frequency
The co-ordinates qs1, qs2 together correspond to two vibrations symmetrical about the
y-axis (x1 = -X3, y1 y3; Fig. 29b, c), whose frequencies Ws1, W82 are given by the roots
of the quadratic (in w2) characteristic equation
1
When 2 x = 75, all three frequencies become equal to those derived in Problem 1.
PROBLEM 3. The same as Problem 1, but for an unsymmetrical linear molecule ABC
(Fig. 30).
A
FIG. 30
SOLUTION. The longitudinal (x) and transverse (y) displacements of the atoms are related
by
mAX1+mBX2+mcx3 = 0, mAy1tmBy2+mcy3= 0,
MAhy1 = mcl2y3.
The potential energy of stretching and bending can be written
where 2l = li+l2. Calculations similar to those in Problem 1 give
for the transverse vibrations and the quadratic (in w2) equation
=
for the frequencies wil, W12 of the longitudinal vibrations.
$25. Damped oscillations
So far we have implied that all motion takes place in a vacuum, or else that
the effect of the surrounding medium on the motion may be neglected. In
reality, when a body moves in a medium, the latter exerts a resistance which
tends to retard the motion. The energy of the moving body is finally dissipated
by being converted into heat.
Motion under these conditions is no longer a purely mechanical process,
and allowance must be made for the motion of the medium itself and for the
internal thermal state of both the medium and the body. In particular, we
cannot in general assert that the acceleration of a moving body is a function
only of its co-ordinates and velocity at the instant considered; that is, there
are no equations of motion in the mechanical sense. Thus the problem of the
motion of a body in a medium is not one of mechanics.
There exists, however, a class of cases where motion in a medium can be
approximately described by including certain additional terms in the
§25
Damped oscillations
75
mechanical equations of motion. Such cases include oscillations with fre-
quencies small compared with those of the dissipative processes in the
medium. When this condition is fulfilled we may regard the body as being
acted on by a force of friction which depends (for a given homogeneous
medium) only on its velocity.
If, in addition, this velocity is sufficiently small, then the frictional force
can be expanded in powers of the velocity. The zero-order term in the expan-
sion is zero, since no friction acts on a body at rest, and so the first non-
vanishing term is proportional to the velocity. Thus the generalised frictional
force fir acting on a system executing small oscillations in one dimension
(co-ordinate x) may be written fir = - ax, where a is a positive coefficient
and the minus sign indicates that the force acts in the direction opposite to
that of the velocity. Adding this force on the right-hand side of the equation
of motion, we obtain (see (21.4))
mx = -kx-ax.
(25.1)
We divide this by m and put
k/m= wo2, a/m=2x; =
(25.2)
wo is the frequency of free oscillations of the system in the absence of friction,
and A is called the damping coefficient or damping decrement.
Thus the equation is
(25.3)
We again seek a solution x = exp(rt) and obtain r for the characteristic
equation r2+2xr + wo2 = 0, whence ¥1,2 = The general
solution of equation (25.3) is
c1exp(rit)+c2 exp(r2t).
Two cases must be distinguished. If wo, we have two complex con-
jugate values of r. The general solution of the equation of motion can then
be written as
where A is an arbitrary complex constant, or as
= aexp(-Xt)cos(wta),
(25.4)
with w = V(w02-2) and a and a real constants. The motion described by
these formulae consists of damped oscillations. It may be regarded as being
harmonic oscillations of exponentially decreasing amplitude. The rate of
decrease of the amplitude is given by the exponent X, and the "frequency"
w is less than that of free oscillations in the absence of friction. For 1 wo,
the difference between w and wo is of the second order of smallness. The
decrease in frequency as a result of friction is to be expected, since friction
retards motion.
t The dimensionless product XT (where T = 2n/w is the period) is called the logarithmic
damping decrement.
76
Small Oscillations
§25
If A < wo, the amplitude of the damped oscillation is almost unchanged
during the period 2n/w. It is then meaningful to consider the mean values
(over the period) of the squared co-ordinates and velocities, neglecting the
change in exp( - At) when taking the mean. These mean squares are evidently
proportional to exp(-2xt). Hence the mean energy of the system decreases
as
(25.5)
where E0 is the initial value of the energy.
Next, let A > wo. Then the values of r are both real and negative. The
general form of the solution is
-
(25.6)
We see that in this case, which occurs when the friction is sufficiently strong,
the motion consists of a decrease in /x/, i.e. an asymptotic approach (as t ->
00)
to the equilibrium position. This type of motion is called aperiodic damping.
Finally, in the special case where A = wo, the characteristic equation has
the double root r = - 1. The general solution of the differential equation is
then
(25.7)
This is a special case of aperiodic damping.
For a system with more than one degree of freedom, the generalised
frictional forces corresponding to the co-ordinates Xi are linear functions of
the velocities, of the form
=
(25.8)
From purely mechanical arguments we can draw no conclusions concerning
the symmetry properties of the coefficients aik as regards the suffixes i and
k, but the methods of statistical physics make it possible to demonstrate
that in all cases
aki.
(25.9)
Hence the expressions (25.8) can be written as the derivatives
=
(25.10)
of the quadratic form
(25.11)
which is called the dissipative function.
The forces (25.10) must be added to the right-hand side of Lagrange's
equations:
(25.12)
t See Statistical Physics, $123, Pergamon Press, Oxford 1969.

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Forced oscillations under friction
77
The dissipative function itself has an important physical significance: it
gives the rate of dissipation of energy in the system. This is easily seen by
calculating the time derivative of the mechanical energy of the system. We
have
aL
=
Since F is a quadratic function of the velocities, Euler's theorem on homo-
geneous functions shows that the sum on the right-hand side is equal to 2F.
Thus
dE/dt==2-2F,
(25.13)
i.e. the rate of change of the energy of the system is twice the dissipative
function. Since dissipative processes lead to loss of energy, it follows that
F > 0, i.e. the quadratic form (25.11) is positive definite.
The equations of small oscillations under friction are obtained by adding
the forces (25.8) to the right-hand sides of equations (23.5):
=
(25.14)
Putting in these equations XK = Ak exp(rt), we obtain, on cancelling exp(rt),
a set of linear algebraic equations for the constants Ak:
(25.15)
Equating to zero their determinant, we find the characteristic equation, which
determines the possible values of r:
(25.16)
This is an equation in r of degree 2s. Since all the coefficients are real,
its roots are either real, or complex conjugate pairs. The real roots must be
negative, and the complex roots must have negative real parts, since other-
wise the co-ordinates, velocities and energy of the system would increase
exponentially with time, whereas dissipative forces must lead to a decrease
of the energy.
§26. Forced oscillations under friction
The theory of forced oscillations under friction is entirely analogous to
that given in §22 for oscillations without friction. Here we shall consider
in detail the case of a periodic external force, which is of considerable interest.
78
Small Oscillations
§26
Adding to the right-hand side of equation (25.1) an external force f cos st
and dividing by m, we obtain the equation of motion:
+2*+wox=(fm)cos = yt.
(26.1)
The solution of this equation is more conveniently found in complex form,
and so we replace cos st on the right by exp(iyt):
exp(iyt).
We seek a particular integral in the form x = B exp(iyt), obtaining for B
the value
(26.2)
Writing B = exp(i8), we have
b tan 8 = 2xy/(y2-wo2).
(26.3)
Finally, taking the real part of the expression B exp(iyt) = b exp[i(yt+8)],
we find the particular integral of equation (26.1); adding to this the general
solution of that equation with zero on the right-hand side (and taking for
definiteness the case wo > 1), we have
x = a exp( - At) cos(wtta)+bcos(yt+8)
(26.4)
The first term decreases exponentially with time, so that, after a sufficient
time, only the second term remains:
x = b cos(yt+8).
(26.5)
The expression (26.3) for the amplitude b of the forced oscillation increases
as y approaches wo, but does not become infinite as it does in resonance
without friction. For a given amplitude f of the force, the amplitude of the
oscillations is greatest when y = V(w02-2)2); for A < wo, this differs from
wo only by a quantity of the second order of smallness.
Let us consider the range near resonance, putting y = wote with E small,
and suppose also that A < wo. Then we can approximately put, in (26.2),
22 =(y+wo)(y-wo) 22 2woe, 2ixy 22 2ixwo, SO that
B = -f/2m(e-ii))wo
(26.6)
or
b f/2mw01/(22+12),
tan 8 = N/E.
(26.7)
A property of the phase difference 8 between the oscillation and the external
force is that it is always negative, i.e. the oscillation "lags behind" the force.
Far from resonance on the side < wo, 8 0; on the side y > wo, 8
-77.
The change of 8 from zero to - II takes place in a frequency range near wo
which is narrow (of the order of A in width); 8 passes through - 1/2 when
y = wo. In the absence of friction, the phase of the forced oscillation changes
discontinuously by TT at y = wo (the second term in (22.4) changes sign);
when friction is allowed for, this discontinuity is smoothed out.
§26
Forced oscillations under friction
79
In steady motion, when the system executes the forced oscillations given
by (26.5), its energy remains unchanged. Energy is continually absorbed by
the system from the source of the external force and dissipated by friction.
Let I(y) be the mean amount of energy absorbed per unit time, which depends
on the frequency of the external force. By (25.13) we have I(y) = 2F, where
F is the average value (over the period of oscillation) of the dissipative func-
tion. For motion in one dimension, the expression (25.11) for the dissipative
function becomes F = 1ax2 = Amx2. Substituting (26.5), we have
F = mb22 sin2(yt+8).
The time average of the squared sine is 1/2 so that
I(y) = Mmb2y2. =
(26.8)
Near resonance we have, on substituting the amplitude of the oscillation
from (26.7),
I(e) =
(26.9)
This is called a dispersion-type frequency dependence of the absorption.
The half-width of the resonance curve (Fig. 31) is the value of E for which
I(e) is half its maximum value (E = 0). It is evident from (26.9) that in the
present case the half-width is just the damping coefficient A. The height of
the maximum is I(0) = f2/4mx, and is inversely proportional to . Thus,
I/I(O)
/2
€
-1
a
FIG. 31
when the damping coefficient decreases, the resonance curve becomes more
peaked. The area under the curve, however, remains unchanged. This area
is given by the integral
[ ((7) dy = [ I(e) de.
Since I(e) diminishes rapidly with increasing E, the region where |el is
large is of no importance, and the lower limit may be replaced by - 80, and
I(e) taken to have the form given by (26.9). Then we have
"
(26.10)
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Small Oscillations

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PROBLEM
Determine the forced oscillations due to an external force f = fo exp(at) COS st in the
presence of friction.
SOLUTION. We solve the complex equation of motion
2+wo2x = (fo/m) exp(at+iyt)
and then take the real part. The result is a forced oscillation of the form
x=bexp(at)cos(yt+8),
where
b =
tan s =
§27. Parametric resonance
There exist oscillatory systems which are not closed, but in which the
external action amounts only to a time variation of the parameters.t
The parameters of a one-dimensional system are the coefficients m and k
in the Lagrangian (21.3). If these are functions of time, the equation of
motion is
(27.1)
We introduce instead of t a new independent variable T such that
dr = dt/m(t); this reduces the equation to
d2x/d-2+mkx=0.
There is therefore no loss of generality in considering an equation of motion
of the form
(27.2)
obtained from (27.1) if m = constant.
The form of the function w(t) is given by the conditions of the problem.
Let us assume that this function is periodic with some frequency y and period
T = 2n/y. This means that w(t+T) = w(t), and so the equation (27.2) is
invariant under the transformation t t+ T. Hence, if x(t) is a solution of
the equation, so is x(t+T). That is, if x1(t) and x2(t) are two independent
integrals of equation (27.2), they must be transformed into linear combina-
tions of themselves when t is replaced by t + T. It is possible to choose X1
and X2 in such a way that, when t t+T, they are simply multiplied by
t A simple example is that of a pendulum whose point of support executes a given periodic
motion in a vertical direction (see Problem 3).
+ This choice is equivalent to reducing to diagonal form the matrix of the linear trans-
formation of x1(t) and x2(t), which involves the solution of the corresponding quadratic
secular equation. We shall suppose here that the roots of this equation do not coincide.
§27
Parametric resonance
81
constants: x1(t+T) = 1x1(t), x2(t+T) = u2x2(t). The most general functions
having this property are
(t) = 111t/TII1(t), x2(t) = M2t/T112(t),
(27.3)
where II1(t), II2(t) are purely periodic functions of time with period T.
The constants 1 and 2 in these functions must be related in a certain way.
Multiplying the equations +2(t)x1 = 0, 2+w2(t)x2 = 0 by X2 and X1
respectively and subtracting, we = = 0, or
X1X2-XIX2 = constant.
(27.4)
For any functions x1(t), x2(t) of the form (27.3), the expression on the left-
hand side of (27.4) is multiplied by H1U2 when t is replaced by t + T. Hence
it is clear that, if equation (27.4) is to hold, we must have
M1M2=1.
(27.5)
Further information about the constants M1, 2 can be obtained from the
fact that the coefficients in equation (27.2) are real. If x(t) is any integral of
such an equation, then the complex conjugate function x* (t) must also be
an integral. Hence it follows that U1, 2 must be the same as M1*, M2*, i.e.
either 1 = M2* or 1 and 2 are both real. In the former case, (27.5) gives
M1 = 1/1*, i.e. /1112 = 1/22/2 = 1: the constants M1 and 2 are of modulus
unity.
In the other case, two independent integrals of equation (27.2) are
x2(t) = -/I2(t),
(27.6)
with a positive or negative real value of u (Iu/ # 1). One of these functions
(x1 or X2 according as /x/ > 1 or /u/ <1) increases exponentially with time.
This means that the system at rest in equilibrium (x = 0) is unstable: any
deviation from this state, however small, is sufficient to lead to a rapidly
increasing displacement X. This is called parametric resonance.
It should be noticed that, when the initial values of x and x are exactly
zero, they remain zero, unlike what happens in ordinary resonance (§22),
in which the displacement increases with time (proportionally to t) even from
initial values of zero.
Let us determine the conditions for parametric resonance to occur in the
important case where the function w(t) differs only slightly from a constant
value wo and is a simple periodic function:
w2(1) = con2(1+h cosyt)
(27.7)
where
the
constant h 1; we shall suppose h positive, as may always be
done by suitably choosing the origin of time. As we shall see below, para-
metric resonance is strongest if the frequency of the function w(t) is nearly
twice wo. Hence we put y = 2wo+e, where E < wo.
82
Small Oscillations
§27
The solution of equation of motion+
+wo2[1+hcos(2wot)t]x
(27.8)
may be sought in the form
(27.9)
where a(t) and b(t) are functions of time which vary slowly in comparison
with the trigonometrical factors. This form of solution is, of course, not
exact. In reality, the function x(t) also involves terms with frequencies which
differ from wother by integral multiples of 2wo+e; these terms are, how-
ever, of a higher order of smallness with respect to h, and may be neglected
in a first approximation (see Problem 1).
We substitute (27.9) in (27.8) and retain only terms of the first order in
€, assuming that à ea, b ~ eb; the correctness of this assumption under
resonance conditions is confirmed by the result. The products of trigono-
metrical functions may be replaced by sums:
cos(wot1e)t.cos(2wote)t =
etc., and in accordance with what was said above we omit terms with fre-
quency 3(wo+1e). The result is
= 0.
If this equation is to be justified, the coefficients of the sine and cosine must
both be zero. This gives two linear differential equations for the functions
a(t) and b(t). As usual, we seek solutions proportional to exp(st). Then
= 0, 1(e-thwo)a- - sb = 0, and the compatibility condition
for these two algebraic equations gives
(27.10)
The condition for parametric resonance is that S is real, i.e. s2 > 0.1 Thus
parametric resonance occurs in the range
(27.11)
on either side of the frequency 2wo.ll The width of this range is proportional
to h, and the values of the amplification coefficient S of the oscillations in the
range are of the order of h also.
Parametric resonance also occurs when the frequency y with which the
parameter varies is close to any value 2wo/n with n integral. The width of the
t An equation of this form (with arbitrary y and h) is called in mathematical physics
Mathieu's equation.
+ The constant u in (27.6) is related to s by u = - exp(sn/wo); when t is replaced by
t+2n/2wo, the sine and cosine in (27.9) change sign.
II If we are interested only in the range of resonance, and not in the values of S in that
range, the calculations may be simplified by noting that S = 0 at the ends of the range, i.e.
the coefficients a and b in (27.9) are constants. This gives immediately € = thwo as in
(27.11).
§27
Parametric resonance
83
resonance range (region of instability) decreases rapidly with increasing N,
however, namely as hn (see Problem 2, footnote). The amplification co-
efficient of the oscillations also decreases.
The phenomenon of parametric resonance is maintained in the presence
of slight friction, but the region of instability becomes somewhat narrower.
As we have seen in §25, friction results in a damping of the amplitude of
oscillations as exp(- - At). Hence the amplification of the oscillations in para-
metric resonance is as exp[(s-1)t] with the positive S given by the solution
for the frictionless case, and the limit of the region of instability is given by
the equation - X = 0. Thus, with S given by (27.10), we have for the resonance
range, instead of (27.11),
(27.12)
It should be noticed that resonance is now possible not for arbitrarily
small amplitudes h, but only when h exceeds a "threshold" value hk. When
(27.12) holds, hk = 4X/wo. It can be shown that, for resonance near the fre-
quency 2wo/n, the threshold hk is proportional to X1/n, i.e. it increases with n.
PROBLEMS
PROBLEM 1. Obtain an expression correct as far as the term in h2 for the limits of the region
of instability for resonance near 2 = 2wo.
SOLUTION. We seek the solution of equation (27.8) in the form
x = ao cos(wo+1e)t +bo (wo+le)t +a1 cos 3( (wo+le)t +b1 sin 3(wo+le)t,
which includes terms of one higher order in h than (27.9). Since only the limits of the region
of instability are required, we treat the coefficients ao, bo, a1, b1 as constants in accordance
with the last footnote. Substituting in (27.8), we convert the products of trigonometrical
functions into sums and omit the terms of frequency 5(wo+1) in this approximation. The
result is
[
- cos(wo+l)
cos 3(wo+1e)tt
sin 3(wo+1e)t = 0.
In the terms of frequency wothe we retain terms of the second order of smallness, but in
those of frequency 3( (wo+1) only the first-order terms. Each of the expressions in brackets
must separately vanish. The last two give a1 = hao/16, b1 = hbo/16, and then the first two
give woe +thwo2+1e2-h2wo2/32 = 0.
Solving this as far as terms of order h2, we obtain the required limits of E:
= theo-h20003.
PROBLEM 2. Determine the limits of the region of instability in resonance near y = wo.
SOLUTION. Putting y = wote, we obtain the equation of motion
0.
Since the required limiting values of ~~h2, we seek a solution in the form
ao cos(wote)t sin(wote)t cos 2(wo+e)t +b1 sin 2(wo+e)t- +C1,
84
Small Oscillations

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which includes terms of the first two orders. To determine the limits of instability, we again
treat the coefficients as constants, obtaining
cos(wote)t-
+[-2woebo+thwo861] sin(wo+e)t.
+[-30002a1+thanoPao] cos 2(wote)t+
sin 2(wote)t+[c1wo+thwo2ao] 0.
Hence a1 = hao/6, b1 = hbo/6, C1 = -thao, and the limits aret € = -5h2wo/24, € = h2wo/24.
PROBLEM 3. Find the conditions for parametric resonance in small oscillations of a simple
pendulum whose point of support oscillates vertically.
SOLUTION. The Lagrangian derived in §5, Problem 3(c), gives for small oscillations
( < 1) the equation of motion + wo2[1+(4a/1) cos(2wo+t)) = 0, where wo2 = g/l.
Hence we see that the parameter h is here represented by 4all. The condition (27.11), for
example, becomes |
§28. Anharmonic oscillations
The whole of the theory of small oscillations discussed above is based on
the expansion of the potential and kinetic energies of the system in terms of
the co-ordinates and velocities, retaining only the second-order terms. The
equations of motion are then linear, and in this approximation we speak of
linear oscillations. Although such an expansion is entirely legitimate when
the amplitude of the oscillations is sufficiently small, in higher approxima-
tions (called anharmonic or non-linear oscillations) some minor but qualitatively
different properties of the motion appear.
Let us consider the expansion of the Lagrangian as far as the third-order
terms. In the potential energy there appear terms of degree three in the co-
ordinates Xi, and in the kinetic energy terms containing products of velocities
and co-ordinates, of the form XEXKXI. This difference from the previous
expression (23.3) is due to the retention of terms linear in x in the expansion
of the functions aik(q). Thus the Lagrangian is of the form
(28.1)
where Nikl, liki are further constant coefficients.
If we change from arbitrary co-ordinates Xi to the normal co-ordinates Qx
of the linear approximation, then, because this transformation is linear, the
third and fourth sums in (28.1) become similar sums with Qx and Qa in place
t
Generally, the width AE of the region of instability in resonance near the frequency
2wo/n is given by
AE =
a result due to M. BELL (Proceedings of the Glasgow Mathematical Association 3, 132, 1957).
§28
Anharmonic oscillations
85
of the co-ordinates Xi and the velocities Xr. Denoting the coefficients in these
new sums by dapy and Hapy's we have the Lagrangian in the form
(28.2)
a
a,B,Y
We shall not pause to write out in their entirety the equations of motion
derived from this Lagrangian. The important feature of these equations is
that they are of the form
(28.3)
where fa are homogeneous functions, of degree two, of the co-ordinates Q
and their time derivatives.
Using the method of successive approximations, we seek a solution of
these equations in the form
(28.4)
where Qa2, and the Qx(1) satisfy the "unperturbed" equations
i.e. they are ordinary harmonic oscillations:
(28.5)
Retaining only the second-order terms on the right-hand side of (28.3) in
the next approximation, we have for the Qx(2) the equations
(28.6)
where (28.5) is to be substituted on the right. This gives a set of inhomo-
geneous linear differential equations, in which the right-hand sides can be
represented as sums of simple periodic functions. For example,
cos(wpt + ag)
Thus the right-hand sides of equations (28.6) contain terms corresponding
to oscillations whose frequencies are the sums and differences of the eigen-
frequencies of the system. The solution of these equations must be sought
in a form involving similar periodic factors, and so we conclude that, in the
second approximation, additional oscillations with frequencies
wa+w
(28.7)
including the double frequencies 2wa and the frequency zero (corresponding
to a constant displacement), are superposed on the normal oscillations of the
system. These are called combination frequencies. The corresponding ampli-
tudes are proportional to the products Axap (or the squares aa2) of the cor-
responding normal amplitudes.
In higher approximations, when further terms are included in the expan-
sion of the Lagrangian, combination frequencies occur which are the sums
and differences of more than two Wa; and a further phenomenon also appears.
86
Small Oscillations
§28
In the third approximation, the combination frequencies include some which
coincide with the original frequencies W Wa+wp-wp). When the method
described above is used, the right-hand sides of the equations of motion there-
fore include resonance terms, which lead to terms in the solution whose
amplitude increases with time. It is physically evident, however, that the
magnitude of the oscillations cannot increase of itself in a closed system
with no external source of energy.
In reality, the fundamental frequencies Wa in higher approximations are
not equal to their "unperturbed" values wa(0) which appear in the quadratic
expression for the potential energy. The increasing terms in the solution
arise from an expansion of the type
which is obviously not legitimate when t is sufficiently large.
In going to higher approximations, therefore, the method of successive
approximations must be modified so that the periodic factors in the solution
shall contain the exact and not approximate values of the frequencies. The
necessary changes in the frequencies are found by solving the equations and
requiring that resonance terms should not in fact appear.
We may illustrate this method by taking the example of anharmonic oscil-
lations in one dimension, and writing the Lagrangian in the form
L =
(28.8)
The corresponding equation of motion is
(28.9)
We shall seek the solution as a series of successive approximations:
where
x(1) = a cos wt,
(28.10)
with the exact value of w, which in turn we express as w=wotw1)+w(2)+....
(The initial phase in x(1) can always be made zero by a suitable choice of the
origin of time.) The form (28.9) of the equation of motion is not the most
convenient, since, when (28.10) is substituted in (28.9), the left-hand side is
not exactly zero. We therefore rewrite it as
(28.11)
Putting x(1)+x(2), w wotwi and omitting terms of above the
second order of smallness, we obtain for x(2) the equation
= aa2 cos2wt+2wowlda cos wt
= 1xa2-1xa2 cos 2wt + 2wow1)a cos wt.
The condition for the resonance term to be absent from the right-hand side
is simply w(1) = 0, in agreement with the second approximation discussed

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Resonance in non-linear oscillations
87
at the beginning of this section. Solving the inhomogeneous linear equation
in the usual way, we have
(28.12)
Putting in (28.11) X wo+w(2), we obtain the equa-
tion for x(3)
= -
or, substituting on the right-hand side (28.10) and (28.12) and effecting
simple transformation,
wt.
Equating to zero the coefficient of the resonance term cos wt, we find the
correction to the fundamental frequency, which is proportional to the squared
amplitude of the oscillations:
(28.13)
The combination oscillation of the third order is
(28.14)
$29. Resonance in non-linear oscillations
When the anharmonic terms in forced oscillations of a system are taken
into account, the phenomena of resonance acquire new properties.
Adding to the right-hand side of equation (28.9) an external periodic force
of frequency y, we have
+2x+wo2x=(fm)cos = yt - ax2-Bx3;
(29.1)
here the frictional force, with damping coefficient A (assumed small) has also
been included. Strictly speaking, when non-linear terms are included in the
equation of free oscillations, the terms of higher order in the amplitude of
the external force (such as occur if it depends on the displacement x) should
also be included. We shall omit these terms merely to simplify the formulae;
they do not affect the qualitative results.
Let y = wote with E small, i.e. y be near the resonance value. To ascertain
the resulting type of motion, it is not necessary to consider equation (29.1)
if we argue as follows. In the linear approximation, the amplitude b is given
88
Small Oscillations
§29
near resonance, as a function of the amplitude f and frequency r of the
external force, by formula (26.7), which we write as
(29.2)
The non-linearity of the oscillations results in the appearance of an ampli-
tude dependence of the eigenfrequency, which we write as
wo+kb2,
(29.3)
the constant K being a definite function of the anharmonic coefficients (see
(28.13)). Accordingly, we replace wo by wo + kb2 in formula (29.2) (or, more
precisely, in the small difference y-wo). With y-wo=e, the resulting
equation is
=
(29.4)
or
Equation (29.4) is a cubic equation in b2, and its real roots give the ampli-
tude of the forced oscillations. Let us consider how this amplitude depends
on the frequency of the external force for a given amplitude f of that force.
When f is sufficiently small, the amplitude b is also small, so that powers
of b above the second may be neglected in (29.4), and we return to the form
of b(e) given by (29.2), represented by a symmetrical curve with a maximum
at the point E = 0 (Fig. 32a). As f increases, the curve changes its shape,
though at first it retains its single maximum, which moves to positive E if
K > 0 (Fig. 32b). At this stage only one of the three roots of equation (29.4)
is real.
When f reaches a certain value f k (to be determined below), however, the
nature of the curve changes. For all f > fk there is a range of frequencies in
which equation (29.4) has three real roots, corresponding to the portion
BCDE in Fig. 32c.
The limits of this range are determined by the condition db/de = 8 which
holds at the points D and C. Differentiating equation (29.4) with respect to
€, we have
db/de =
Hence the points D and C are determined by the simultaneous solution of
the equations
2-4kb2e+3k264+2 0
(29.5)
and (29.4). The corresponding values of E are both positive. The greatest
amplitude is reached where db/de = 0. This gives E = kb2, and from (29.4)
we have
bmax = f/2mwod;
(29.6)
this is the same as the maximum value given by (29.2).
§29
Resonance in non-linear oscillations
89
It may be shown (though we shall not pause to do so heret) that, of the
three real roots of equation (29.4), the middle one (represented by the dotted
part CD of the curve in Fig. 32c) corresponds to unstable oscillations of the
system: any action, no matter how slight, on a system in such a state causes
it to oscillate in a manner corresponding to the largest or smallest root (BC
or DE). Thus only the branches ABC and DEF correspond to actual oscil-
lations of the system. A remarkable feature here is the existence of a range of
frequencies in which two different amplitudes of oscillation are possible. For
example, as the frequency of the external force gradually increases, the ampli-
tude of the forced oscillations increases along ABC. At C there is a dis-
continuity of the amplitude, which falls abruptly to the value corresponding
to E, afterwards decreasing along the curve EF as the frequency increases
further. If the frequency is now diminished, the amplitude of the forced
oscillations varies along FD, afterwards increasing discontinuously from D
to B and then decreasing along BA.
b
(a)
to
b
(b)
f<f
b
(c)
f>tp
B
C
Di
A
E
F
€
FIG. 32
To calculate the value of fk, we notice that it is the value of f for which
the two roots of the quadratic equation in b2 (29.5) coincide; for f = f16, the
section CD reduces to a point of inflection. Equating to zero the discriminant
t The proof is given by, for example, N.N. BOGOLIUBOV and Y.A. MITROPOLSKY, Asymp-
totic Methods in the Theory of Non-Linear Oscillations, Hindustan Publishing Corporation,
Delhi 1961.
4
90
Small Oscillations
§29
of (29.5), we find E2 = 3X², and the corresponding double root is kb2 = 2e/3.
Substitution of these values of b and E in (29.4) gives
32m2wo2x3/31/3k.
(29.7)
Besides the change in the nature of the phenomena of resonance at fre-
quencies y 22 wo, the non-linearity of the oscillations leads also to new
resonances in which oscillations of frequency close to wo are excited by an
external force of frequency considerably different from wo.
Let the frequency of the external force y 22 two, i.e. y = two+e. In the
first (linear) approximation, it causes oscillations of the system with the same
frequency and with amplitude proportional to that of the force:
x(1)= (4f/3mwo2) cos(two+e)t
(see (22.4)). When the non-linear terms are included (second approximation),
these oscillations give rise to terms of frequency 2y 22 wo on the right-hand
side of the equation of motion (29.1). Substituting x(1) in the equation
= -
using the cosine of the double angle and retaining only the resonance term
on the right-hand side, we have
= - (8xf2/9m2w04) cos(wo+2e)t.
(29.8)
This equation differs from (29.1) only in that the amplitude f of the force is
replaced by an expression proportional to f2. This means that the resulting
resonance is of the same type as that considered above for frequencies
y 22 wo, but is less strong. The function b(e) is obtained by replacing f by
- 8xf2/9mwo4, and E by 2e, in (29.4):
62[(2e-kb2)2+12] = 16x2f4/81m4w010.
(29.9)
Next, let the frequency of the external force be 2= 2wote In the first
approximation, we have x(1) = - (f/3mwo2) cos(2wo+e)t. On substituting
in equation (29.1), we do not obtain terms representing an
external force in resonance such as occurred in the previous case. There is,
however, a parametric resonance resulting from the third-order term pro-
portional to the product x(1)x(2). If only this is retained out of the non-linear
terms, the equation for x(2) is
=
or
(29.10)
i.e. an equation of the type (27.8) (including friction), which leads, as we
have seen, to an instability of the oscillations in a certain range of frequencies.
§29
Resonance in non-linear oscillations
91
This equation, however, does not suffice to determine the resulting ampli-
tude of the oscillations. The attainment of a finite amplitude involves non-
linear effects, and to include these in the equation of motion we must retain
also the terms non-linear in x(2):
= cos(2wo+e)t. (29.11)
The problem can be considerably simplified by virtue of the following fact.
Putting on the right-hand side of (29.11) x(2) = b cos[(wo++)+8], where
b is the required amplitude of the resonance oscillations and 8 a constant
phase difference which is of no importance in what follows, and writing the
product of cosines as a sum, we obtain a term (afb/3mwo2)
of the ordinary resonance type (with respect to the eigenfrequency wo of the
system). The problem thus reduces to that considered at the beginning of
this section, namely ordinary resonance in a non-linear system, the only
differences being that the amplitude of the external force is here represented
by afb/3wo2, and E is replaced by 1/6. Making this change in equation (29.4),
we have
Solving for b, we find the possible values of the amplitude:
b=0,
(29.12)
(29.13)
1
(29.14)
Figure 33 shows the resulting dependence of b on € for K > 0; for K < 0
the curves are the reflections (in the b-axis) of those shown. The points B
and C correspond to the values E = To the left of
B, only the value b = 0 is possible, i.e. there is no resonance, and oscillations
of frequency near wo are not excited. Between B and C there are two roots,
b = 0(BC) and (29.13) (BE). Finally, to the right of C there are three roots
(29.12)-(29.14). Not all these, however, correspond to stable oscillations.
The value b = 0 is unstable on BC, and it can also be shown that the middle
root (29.14) always gives instability. The unstable values of b are shown in
Fig. 33 by dashed lines.
Let us examine, for example, the behaviour of a system initially "at rest"
as the frequency of the external force is gradually diminished. Until the point
t This segment corresponds to the region of parametric resonance (27.12), and a com-
parison of (29.10) and (27.8) gives 1h = 2af/3mwo4. The condition 12af/3mwo3 > 4X for
which the phenomenon can exist corresponds to h > hk.
+ It should be recalled that only resonance phenomena are under consideration. If these
phenomena are absent, the system is not literally at rest, but executes small forced oscillations
of frequency y.
92
Small Oscillations
§29
C is reached, b = 0, but at C the state of the system passes discontinuously
to the branch EB. As € decreases further, the amplitude of the oscillations
decreases to zero at B. When the frequency increases again, the amplitude
increases along BE.-
b
E
E
A
B
C D
FIG. 33
The cases of resonance discussed above are the principal ones which may
occur in a non-linear oscillating system. In higher approximations, resonances
appear at other frequencies also. Strictly speaking, a resonance must occur
at every frequency y for which ny + mwo = wo with n and m integers, i.e. for
every y = pwo/q with P and q integers. As the degree of approximation
increases, however, the strength of the resonances, and the widths of the
frequency ranges in which they occur, decrease so rapidly that in practice
only the resonances at frequencies y 2 pwo/q with small P and q can be ob-
served.
PROBLEM
Determine the function b(e) for resonance at frequencies y 22 3 wo.
SOLUTION. In the first approximation, x(1) = -(f/8mwo2) cos(3wo+t) For the second
approximation x(2) we have from (29.1) the equation
= -3,8x(1)x(2)2,
where only the term which gives the required resonance has been retained on the right-hand
side. Putting x(2) = b cos[(wo+)+8] and taking the resonance term out of the product
of three cosines, we obtain on the right-hand side the expression
(3,3b2f(32mwo2) cos[(wotle)t-28].
Hence it is evident that b(e) is obtained by replacing f by 3,8b2f/32wo², and E by JE, in
(29.4):
Ab4.
The roots of this equation are
b=0,
Fig. 34 shows a graph of the function b(e) for k>0. Only the value b=0 (the e-axis) and
the branch AB corresponds to stability. The point A corresponds to EK = 3(4x2)2-A3)/4kA,
t It must be noticed, however, that all the formulae derived here are valid only when the
amplitude b (and also E) is sufficiently small. In reality, the curves BE and CF meet, and at
their point of intersection the oscillation ceases; thereafter, b = 0.

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Motion in a rapidly oscillating field
93
bk2 = Oscillations exist only for € > Ek, and then b > bk. Since the state
b = 0 is always stable, an initial "push" is necessary in order to excite oscillations.
The formulae given above are valid only for small E. This condition is satisfied if 1 is small
and also the amplitude of the force is such that 2/wo < A KWO.
b
B
A
€
FIG. 34
§30. Motion in a rapidly oscillating field
Let us consider the motion of a particle subject both to a time-independent
field of potential U and to a force
f=f1coswt+fasin.ou
(30.1)
which varies in time with a high frequency w (f1, f2 being functions of the
co-ordinates only). By a "high" frequency we mean one such that w > 1/T,
where T is the order of magnitude of the period of the motion which the
particle would execute in the field U alone. The magnitude of f is not assumed
small in comparison with the forces due to the field U, but we shall assume
that the oscillation (denoted below by $) of the particle as a result of this
force is small.
To simplify the calculations, let us first consider motion in one dimension
in a field depending only on the space co-ordinate X. Then the equation of
motion of the particle ist
mx = -dU/dx+f.
(30.2)
It is evident, from the nature of the field in which the particle moves, that
it will traverse a smooth path and at the same time execute small oscillations
of frequency w about that path. Accordingly, we represent the function x(t)
as a sum:
(30.3)
where (t) corresponds to these small oscillations.
The mean value of the function (t) over its period 2n/w is zero, and the
function X(t) changes only slightly in that time. Denoting this average by a
bar, we therefore have x = X(t), i.e. X(t) describes the "smooth" motion of
t The co-ordinate x need not be Cartesian, and the coefficient m is therefore not neces-
sarily the mass of the particle, nor need it be constant as has been assumed in (30.2). This
assumption, however, does not affect the final result (see the last footnote to this section).
94
Small Oscillations
§30
the particle averaged over the rapid oscillations. We shall derive an equation
which determines the function X(t).t
Substituting (30.3) in (30.2) and expanding in powers of & as far as the
first-order terms, we obtain
(30.4)
This equation involves both oscillatory and "smooth" terms, which must
evidently be separately equal. For the oscillating terms we can put simply
mg = f(X, t);
(30.5)
the other terms contain the small factor & and are therefore of a higher order
of smallness (but the derivative sur is proportional to the large quantity w2
and so is not small). Integrating equation (30.5) with the function f given by
(30.1) (regarding X as a constant), we have
& = -f/mw2.
(30.6)
Next, we average equation (30.4) with respect to time (in the sense discussed
above). Since the mean values of the first powers of f and $ are zero, the result
is
dX
which involves only the function X(t). This equation can be written
mX = dUeff/dX,
(30.7)
where the "effective potential energy" is defined ast
Ueff = U+f2/2mw2
=
(30.8)
Comparing this expression with (30.6), we easily see that the term added to
U is just the mean kinetic energy of the oscillatory motion:
Ueff= U+1mg2
(30.9)
Thus the motion of the particle averaged over the oscillations is the same
as if the constant potential U were augmented by a constant quantity pro-
portional to the squared amplitude of the variable field.
t The principle of this derivation is due to P. L. KAPITZA (1951).
++ By means of somewhat more lengthy calculations it is easy to show that formulae (30.7)
and (30.8) remain valid even if m is a function of X.
§30
Motion in a rapidly oscillating field
95
The result can easily be generalised to the case of a system with any number
of degrees of freedom, described by generalised co-ordinates qi. The effective
potential energy is then given not by (30.8), but by
Unt = Ut
= U+ ,
(30.10)
where the quantities a-1ik, which are in general functions of the co-ordinates,
are the elements of the matrix inverse to the matrix of the coefficients aik in
the kinetic energy (5.5) of the system.
PROBLEMS
PROBLEM 1. Determine the positions of stable equilibrium of a pendulum whose point of
support oscillates vertically with a high frequency y
(g/l)).
SOLUTION. From the Lagrangian derived in §5, Problem 3(c), we see that in this case the
variable force is f = -mlay2 cos yt sin (the quantity x being here represented by the angle
b). The "effective potential energy" is therefore Ueff = mgl[-cos - & st(a2y2/4gl) sin2]. The
positions of stable equilibrium correspond to the minima of this function. The vertically
downward position ( = 0) is always stable. If the condition a2y2 > 2gl holds, the vertically
upward position ( = ) is also stable.
PROBLEM 2. The same as Problem 1, but for a pendulum whose point of support oscillates
horizontally.
SOLUTION. From the Lagrangian derived in §5, Problem 3(b), we find f = mlay2 cos yt
cos and Uell = mgl[-cos 3+(a2y2/4gl) cos2]. If a2y2 < 2gl, the position = 0 is stable.
If a2y2 > 2gl, on the other hand, the stable equilibrium position is given by cos = 2gl/a22.
CHAPTER VI
MOTION OF A RIGID BODY
$31. Angular velocity
A rigid body may be defined in mechanics as a system of particles such that
the distances between the particles do not vary. This condition can, of course,
be satisfied only approximately by systems which actually exist in nature.
The majority of solid bodies, however, change so little in shape and size
under ordinary conditions that these changes may be entirely neglected in
considering the laws of motion of the body as a whole.
In what follows, we shall often simplify the derivations by regarding a
rigid body as a discrete set of particles, but this in no way invalidates the
assertion that solid bodies may usually be regarded in mechanics as continu-
ous, and their internal structure disregarded. The passage from the formulae
which involve a summation over discrete particles to those for a continuous
body is effected by simply replacing the mass of each particle by the mass
P dV contained in a volume element dV (p being the density) and the sum-
mation by an integration over the volume of the body.
To describe the motion of a rigid body, we use two systems of co-ordinates:
a "fixed" (i.e. inertial) system XYZ, and a moving system X1 = x, X2 = y,
X3 = 2 which is supposed to be rigidly fixed in the body and to participate
in its motion. The origin of the moving system may conveniently be taken
to coincide with the centre of mass of the body.
The position of the body with respect to the fixed system of co-ordinates
is completely determined if the position of the moving system is specified.
Let the origin O of the moving system have the radius vector R (Fig. 35).
The orientation of the axes of that system relative to the fixed system is given
by three independent angles, which together with the three components of
the vector R make six co-ordinates. Thus a rigid body is a mechanical system
with six degrees of freedom.
Let us consider an arbitrary infinitesimal displacement of a rigid body.
It can be represented as the sum of two parts. One of these is an infinitesimal
translation of the body, whereby the centre of mass moves to its final position,
but the orientation of the axes of the moving system of co-ordinates is un-
changed. The other is an infinitesimal rotation about the centre of mass,
whereby the remainder of the body moves to its final position.
Let r be the radius vector of an arbitrary point P in a rigid body in the
moving system, and r the radius vector of the same point in the fixed system
(Fig. 35). Then the infinitesimal displacement dr of P consists of a displace-
ment dR, equal to that of the centre of mass, and a displacement doxr
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Angular velocity
97
relative to the centre of mass resulting from a rotation through an infinitesimal
angle do (see (9.1)): dr = dR + do xr. Dividing this equation by the time
dt during which the displacement occurs, and putting
dr/dt = V,
dR/dt =
do/dt = La
(31.1)
we obtain the relation
V = V+Sxr.
(31.2)
Z
X3
P
X2
r
o
R
X1
Y
X
FIG. 35
The vector V is the velocity of the centre of mass of the body, and is also
the translational velocity of the body. The vector S is called the angular
velocity of the rotation of the body; its direction, like that of do, is along the
axis of rotation. Thus the velocity V of any point in the body relative to the
fixed system of co-ordinates can be expressed in terms of the translational
velocity of the body and its angular velocity of rotation.
It should be emphasised that, in deriving formula (31.2), no use has been
made of the fact that the origin is located at the centre of mass. The advan-
tages of this choice of origin will become evident when we come to calculate
the energy of the moving body.
Let us now assume that the system of co-ordinates fixed in the body is
such that its origin is not at the centre of mass O, but at some point O' at
a distance a from O. Let the velocity of O' be V', and the angular velocity
of the new system of co-ordinates be S'. We again consider some point P
in the body, and denote by r' its radius vector with respect to O'. Then
= r'+a, and substitution in (31.2) gives V = V+2xa+2xr'. The
definition of V' and S' shows that V = Hence it follows that
(31.3)
The second of these equations is very important. We see that the angular
velocity of rotation, at any instant, of a system of co-ordinates fixed in
the body is independent of the particular system chosen. All such systems
t
To avoid any misunderstanding, it should be noted that this way of expressing the angular
velocity is somewhat arbitrary: the vector so exists only for an infinitesimal rotation, and not
for all finite rotations.
4*
98
Motion of a Rigid Body

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rotate with angular velocities S which are equal in magnitude and parallel
in direction. This enables us to call S the angular velocity of the body. The
velocity of the translational motion, however, does not have this "absolute"
property.
It is seen from the first formula (31.3) that, if V and S are, at any given
instant, perpendicular for some choice of the origin O, then V' and SS are
perpendicular for any other origin O'. Formula (31.2) shows that in this case
the velocities V of all points in the body are perpendicular to S. It is then
always possible+ to choose an origin O' whose velocity V' is zero, SO that the
motion of the body at the instant considered is a pure rotation about an axis
through O'. This axis is called the instantaneous axis of rotation.t
In what follows we shall always suppose that the origin of the moving
system is taken to be at the centre of mass of the body, and so the axis of
rotation passes through the centre of mass. In general both the magnitude
and the direction of S vary during the motion.
$32. The inertia tensor
To calculate the kinetic energy of a rigid body, we may consider it as a
discrete system of particles and put T = mv2, where the summation is
taken over all the particles in the body. Here, and in what follows, we simplify
the notation by omitting the suffix which denumerates the particles.
Substitution of (31.2) gives
T = Sxx+
The velocities V and S are the same for every point in the body. In the first
term, therefore, V2 can be taken outside the summation sign, and Em is
just the mass of the body, which we denote by u. In the second term we put
EmV Sxr = Emr VxS = VxS Emr. Since we take the origin of the
moving system to be at the centre of mass, this term is zero, because Emr = 0.
Finally, in the third term we expand the squared vector product. The result
is
(32.1)
Thus the kinetic energy of a rigid body can be written as the sum of two
parts. The first term in (32.1) is the kinetic energy of the translational motion,
and is of the same form as if the whole mass of the body were concentrated
at the centre of mass. The second term is the kinetic energy of the rotation
with angular velocity S about an axis passing through the centre of mass.
It should be emphasised that this division of the kinetic energy into two parts
is possible only because the origin of the co-ordinate system fixed in the
body has been taken to be at its centre of mass.
t O' may, of course, lie outside the body.
+ In the general case where V and SC are not perpendicular, the origin may be chosen so
as to make V and S parallel, i.e. so that the motion consists (at the instant in question) of a
rotation about some axis together with a translation along that axis.
§32
The inertia tensor
99
We may rewrite the kinetic energy of rotation in tensor form, i.e. in terms
of the components Xi and O of the vectors r and L. We have
Here we have used the identity Oi = SikOk, where dik is the unit tensor,
whose components are unity for i = k and zero for i # k. In terms of the
tensor
(32.2)
we have finally the following expression for the kinetic energy of a rigid
body:
T =
(32.3)
The Lagrangian for a rigid body is obtained from (32.3) by subtracting
the potential energy:
L =
(32.4)
The potential energy is in general a function of the six variables which define
the position of the rigid body, e.g. the three co-ordinates X, Y, Z of the
centre of mass and the three angles which specify the relative orientation of
the moving and fixed co-ordinate axes.
The tensor Iik is called the inertia tensor of the body. It is symmetrical,
i.e.
Ik=Iki
(32.5)
as is evident from the definition (32.2). For clarity, we may give its com-
ponents explicitly:
TEST
(32.6)
m(x2+y2)
The components Ixx, Iyy, Izz are called the moments of inertia about the
corresponding axes.
The inertia tensor is evidently additive: the moments of inertia of a body
are the sums of those of its parts.
t In this chapter, the letters i, k, l are tensor suffixes and take the values 1, 2, 3. The
summation rule will always be used, i.e. summation signs are omitted, but summation over
the values 1, 2, 3 is implied whenever a suffix occurs twice in any expression. Such a suffix is
called a dummy suffix. For example, AiBi = A . B, Ai2 = AiA1 = A², etc. It is obvious that
dummy suffixes can be replaced by any other like suffixes, except ones which already appear
elsewhere in the expression concerned.
100
Motion of a Rigid Body
§32
If the body is regarded as continuous, the sum in the definition (32.2)
becomes an integral over the volume of the body:
(32.7)
Like any symmetrical tensor of rank two, the inertia tensor can be reduced
to diagonal form by an appropriate choice of the directions of the axes
X1, x2, X3. These directions are called the principal axes of inertia, and the
corresponding values of the diagonal components of the tensor are called the
principal moments of inertia; we shall denote them by I, I2, I3. When the
axes X1, X2, X3 are so chosen, the kinetic energy of rotation takes the very
simple form
=
(32.8)
None of the three principal moments of inertia can exceed the sum of the
other two. For instance,
m(x12+x22) = I3.
(32.9)
A body whose three principal moments of inertia are all different is called
an asymmetrical top. If two are equal (I1 = I2 # I3), we have a symmetrical
top. In this case the direction of one of the principal axes in the x1x2-plane
may be chosen arbitrarily. If all three principal moments of inertia are equal,
the body is called a spherical top, and the three axes of inertia may be chosen
arbitrarily as any three mutually perpendicular axes.
The determination of the principal axes of inertia is much simplified if
the body is symmetrical, for it is clear that the position of the centre of mass
and the directions of the principal axes must have the same symmetry as
the body. For example, if the body has a plane of symmetry, the centre of
mass must lie in that plane, which also contains two of the principal axes of
inertia, while the third is perpendicular to the plane. An obvious case of this
kind is a coplanar system of particles. Here there is a simple relation between
the three principal moments of inertia. If the plane of the system is taken as
the x1x2-plane, then X3 = 0 for every particle, and so I = mx22, I2 = 12,
I3 = (12+x2)2, whence
(32.10)
If a body has an axis of symmetry of any order, the centre of mass must lie
on that axis, which is also one of the principal axes of inertia, while the other
two are perpendicular to it. If the axis is of order higher than the second,
the body is a symmetrical top. For any principal axis perpendicular to the
axis of symmetry can be turned through an angle different from 180° about the
latter, i.e. the choice of the perpendicular axes is not unique, and this can
happen only if the body is a symmetrical top.
A particular case here is a collinear system of particles. If the line of the
system is taken as the x3-axis, then X1 = X2 = 0 for every particle, and so
§32
The inertia tensor
101
two of the principal moments of inertia are equal and the third is zero:
I3 = 0.
(32.11)
Such a system is called a rotator. The characteristic property which distin-
guishes a rotator from other bodies is that it has only two, not three, rotational
degrees of freedom, corresponding to rotations about the X1 and X2 axes: it
is clearly meaningless to speak of the rotation of a straight line about itself.
Finally, we may note one further result concerning the calculation of the
inertia tensor. Although this tensor has been defined with respect to a system
of co-ordinates whose origin is at the centre of mass (as is necessary if the
fundamental formula (32.3) is to be valid), it may sometimes be more con-
veniently found by first calculating a similar tensor I' =
defined with respect to some other origin O'. If the distance OO' is repre-
sented by a vector a, then r = r'+a, Xi = x'i+ai; since, by the definition
of O, Emr = 0, we have
I'ikIk(a2ik-aiak).
(32.12)
Using this formula, we can easily calculate Iik if I'ik is known.
PROBLEMS
PROBLEM 1. Determine the principal moments of inertia for the following types of mole-
cule, regarded as systems of particles at fixed distances apart: (a) a molecule of collinear
atoms, (b) a triatomic molecule which is an isosceles triangle (Fig. 36), (c) a tetratomic
molecule which is an equilateral-based tetrahedron (Fig. 37).
m2
X2
m2
h
x
m
a
a
m
a
m
m
a
FIG. 36
FIG. 37
SOLUTION. (a)
Is = 0,
where Ma is the mass of the ath atom, lao the distance between the ath and bth atoms, and
the summation includes one term for every pair of atoms in the molecule.
For a diatomic molecule there is only one term in the sum, and the result is obvious it is
the product of the reduced mass of the two atoms and the square of the distance between
them: I1 = I2 = m1m2l2((m1+m2).
(b) The centre of mass is on the axis of symmetry of the triangle, at a distance X2 = mzh/u
from its base (h being the height of the triangle). The moments of inertia are I1 = 2m1m2h2/u,
I2 = 1m1a2, I3 = I+I2.
(c) The centre of mass is on the axis of symmetry of the tetrahedron, at a distance
X3 = mgh/u from its base (h being the height of the tetrahedron). The moments of inertia
102
Motion of a Rigid Body
§32
are I1 = = I3 = mia². If M1 = M2, h = (2/3)a, the molecule is a
regular tetrahedron and I1=I2 = I3 = mia2.
PROBLEM 2. Determine the principal moments of inertia for the following homogeneous
bodies: (a) a thin rod of length l, (b) a sphere of radius R, (c) a circular cylinder of radius R
and height h, (d) a rectangular parallelepiped of sides a, b, and c, (e) a circular cone of height
h and base radius R, (f) an ellipsoid of semiaxes a, b, c.
SOLUTION. (a) I1 = I2 = Trul2, I3 = 0 (we neglect the thickness of the rod).
(b) I1 = I2 = I3 = zuR2 (found by calculating the sum I+I+I3 = 2p dV).
(c) I1 = I2 = tu(R2+th2), I3 = tuR2 (where the x3-axis is along the axis of the cylinder).
(d) I1 = (2+c2), I2 = (a2+cc), I3 = 121(a2++b) (where the axes X1, x2, X3 are
along the sides a, b, c respectively).
(e) We first calculate the tensor I'ik with respect to axes whose origin is at the vertex of
the cone (Fig. 38). The calculation is simple if cylindrical co-ordinates are used, and the result
is I'1 = I'2 = I'3 = 2 The centre of mass is easily shown to be on the
axis of the cone and at a distance a = 3h from the vertex. Formula (32.12) therefore gives
I1 = I2 = I'1-2 = I3 = I'3 = TouR2.
X3,X3'
X2
xi
x2
FIG. 38
(f) The centre of mass is at the centre of the ellipsoid, and the principal axes of inertia are
along the axes of the ellipsoid. The integration over the volume of the ellipsoid can be reduced
to one over a sphere by the transformation x = as,y = bn, 2 = c5, which converts the equa-
tion of the surface of the ellipsoid 1 into that of the unit sphere
st+24's = 1.
For example, the moment of inertia about the x-axis is
dz
= tabcI'(b2 tc2),
where I' is the moment of inertia of a sphere of unit radius. Since the volume of the ellipsoid
is 4nabc/3, we find the moments of inertia I = tu(b2+c2), I2 = tu(a2+c2), I3 = tu(a2+b2).
PROBLEM 3. Determine the frequency of small oscillations of a compound pendulum (a
rigid body swinging about a fixed horizontal axis in a gravitational field).
SOLUTION. Let l be the distance between the centre of mass of the pendulum and the axis
about which it rotates, and a, B, y the angles between the principal axes of inertia and the
axis of rotation. We take as the variable co-ordinate the angle between the vertical
and a line through the centre of mass perpendicular to the axis of rotation. The velocity of
the centre of mass is V = 10, and the components of the angular velocity along the principal
§32
The inertia tensor
103
axes of inertia are o cos a, b cos B, b cos y. Assuming the angle to be small, we find the
potential energy U = ugl(1-cos 9 22 12. The Lagrangian is therefore
=
The frequency of the oscillations is consequently
w2 = cos2y).
PROBLEM 4. Find the kinetic energy of the system shown in Fig. 39: OA and AB are thin
uniform rods of length l hinged together at A. The rod OA rotates (in the plane of the diagram)
about O, while the end B of the rod AB slides along Ox.
A
l
l
x
B
FIG. 39
SOLUTION. The velocity of the centre of mass of the rod OA (which is at the middle of
the rod) is 110, where is the angle AOB. The kinetic energy of the rod OA is therefore
T1 = where u is the mass of each rod.
The Cartesian co-ordinates of the centre of mass of the rod AB are X = sl cos o, Y
= 1/ sin b. Since the angular velocity of rotation of this rod is also b, its kinetic energy is
T2 = = tul2(1- +8 sin2o)62 +1162. The total kinetic energy of this
system is therefore = I = Tzul2 (see Problem 2(a)).
PROBLEM 5. Find the kinetic energy of a cylinder of radius R rolling on a plane, if the mass
of the cylinder is so distributed that one of the principal axes of inertia is parallel to the axis
of the cylinder and at a distance a from it, and the moment of inertia about that principal
axis is I.
SOLUTION. Let be the angle between the vertical and a line from the centre of mass
perpendicular to the axis of the cylinder (Fig. 40). The motion of the cylinder at any instant
R
FIG. 40
may be regarded as a pure rotation about an instantaneous axis which coincides with the
line where the cylinder touches the plane. The angular velocity of this rotation is o, since
the angular velocity of rotation about all parallel axes is the same. The centre of mass is at a
distance V(a2+R2-2aR cos ) from the instantaneous axis, and its velocity is therefore
V = bv /(a2+R2-2aR cos ). The total kinetic energy is
T = cos
104
Motion of a Rigid Body
§32
PROBLEM 6. Find the kinetic energy of a homogeneous cylinder of radius a rolling inside
a cylindrical surface of radius R (Fig. 41).
R
FIG. 41
SOLUTION. We use the angle between the vertical and the line joining the centres of the
cylinders. The centre of mass of the rolling cylinder is on the axis, and its velocity is V =
o(R-a). We can calculate the angular velocity as that of a pure rotation about an instantaneous
axis which coincides with the line of contact of the cylinders it is Q = V/a = p(R-a)/a.
If I3 is the moment of inertia about the axis of the cylinder, then
T =
I3 being given by Problem 2(c).
PROBLEM 7. Find the kinetic energy of a homogeneous cone rolling on a plane.
SOLUTION. We denote by 0 the angle between the line OA in which the cone touches the
plane and some fixed direction in the plane (Fig. 42). The centre of mass is on the axis of the
cone, and its velocity V = a0 cos a, where 2a is the vertical angle of the cone and a the
Z
Y
A
FIG. 42
distance of the centre of mass from the vertex. The angular velocity can be calculated as
that of a pure rotation about the instantaneous axis OA: S2 = V/a sin a = é cot a. One of
the principal axes of inertia (x3) is along the axis of the cone, and we take another (x2) perpen-
dicular to the axis of the cone and to the line OA. Then the components of the vector S
(which is parallel to OA) along the principal axes of inertia are O sin a, 0, O cos a. The kinetic
energy is thus
=
= 3uh202(1+5 cos2x)/40,
where h is the height of the cone, and I1, I3 and a have been given in Problem 2(e).
PROBLEM 8. Find the kinetic energy of a homogeneous cone whose base rolls on a plane
and whose vertex is fixed at a height above the plane equal to the radius of the base, so that
the axis of the cone is parallel to the plane.
SOLUTION. We use the angle 0 between a fixed direction in the plane and the projection
of the axis of the cone on the plane (Fig. 43). Then the velocity of the centre of mass is V = aß,

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Angular momentum of a rigid body
105
the notation being as in Problem 7. The instantaneous axis of rotation is the generator OA
which passes through the point where the cone touches the plane. The centre of mass is at a
distance a sin a from this axis, and so l = V/a sin a = O/sin a. The components of the
vector Ca along the principal axes of inertia are, if the x2-axis is taken perpendicular to the
axis of the cone and to the line OA, O sin a = 6, 0, N cos a = è cot a. The kinetic energy
is therefore
T cot2a
= 314h282(sec2x+5)/40.
Z
0
Y
A
FIG. 43
PROBLEM 9. Find the kinetic energy of a homogeneous ellipsoid which rotates about one
of its axes (AB in Fig. 44) while that axis itself rotates about a line CD perpendicular to it
and passing through the centre of the ellipsoid.
SOLUTION. Let the angle of rotation about CD be 0, and that about AB (i.e. the angle
between CD and the x1-axis of inertia, which is perpendicular to AB) be . Then the com-
ponents of S along the axes of inertia are é cos , 0 sin , b, if the x3-axis is AB. Since the
centre of mass, at the centre of the ellipsoid, is at rest, the kinetic energy is
=
D
B
D
A
Do
A
1a
C
of
8
FIG. 44
FIG. 45
PROBLEM 10. The same as Problem 9, but for the case where the axis AB is not perpendicu-
lar to CD and is an axis of symmetry of the ellipsoid (Fig. 45).
SOLUTION. The components of Ca along the axis AB and the other two principal axes of
inertia, which are perpendicular to AB but otherwise arbitrary, are è cos a cos , è cos a X
sin , o to sin a. The kinetic energy is T = 11102 a)2.
$33. Angular momentum of a rigid body
The value of the angular momentum of a system depends, as we know, on
the point with respect to which it is defined. In the mechanics of a rigid body,
106
Motion of a Rigid Body
§33
the most appropriate point to choose for this purpose is the origin of the
moving system of co-ordinates, i.e. the centre of mass of the body, and in
what follows we shall denote by M the angular momentum SO defined.
According to formula (9.6), when the origin is taken at the centre of mass
of the body, the angular momentum M is equal to the "intrinsic" angular
momentum resulting from the motion relative to the centre of mass. In the
definition M = Emrxv we therefore replace V by Sxr:
M = =
or, in tensor notation,
Mi = OK
Finally, using the definition (32.2) of the inertia tensor, we have
(33.1)
If the axes X1, X2, X3 are the same as the principal axes of inertia, formula
(33.1) gives
M2 = I2DQ,
M3 = I303. =
(33.2)
In particular, for a spherical top, where all three principal moments of inertia
are equal, we have simply
M = IS,
(33.3)
i.e. the angular momentum vector is proportional to, and in the same direc-
tion as, the angular velocity vector. For an arbitrary body, however, the
vector M is not in general in the same direction as S; this happens only
when the body is rotating about one of its principal axes of inertia.
Let us consider a rigid body moving freely, i.e. not subject to any external
forces. We suppose that any uniform translational motion, which is of no
interest, is removed, leaving a free rotation of the body.
As in any closed system, the angular momentum of the freely rotating body
is constant. For a spherical top the condition M = constant gives C = con-
stant; that is, the most general free rotation of a spherical top is a uniform
rotation about an axis fixed in space.
The case of a rotator is equally simple. Here also M = IS, and the vector
S is perpendicular to the axis of the rotator. Hence a free rotation of a rotator
is a uniform rotation in one plane about an axis perpendicular to that plane.
The law of conservation of angular momentum also suffices to determine
the more complex free rotation of a symmetrical top. Using the fact that the
principal axes of inertia X1, X2 (perpendicular to the axis of symmetry (x3)
of the top) may be chosen arbitrarily, we take the x2-axis perpendicular to
the plane containing the constant vector M and the instantaneous position
of the x3-axis. Then M2 = 0, and formulae (33.2) show that Q2 = 0. This
means that the directions of M, St and the axis of the top are at every instant
in one plane (Fig. 46). Hence, in turn, it follows that the velocity V = Sxr
of every point on the axis of the top is at every instant perpendicular to that

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The equations of motion of a rigid body
107
plane. That is, the axis of the top rotates uniformly (see below) about the
direction of M, describing a circular cone. This is called regular precession
of the top. At the same time the top rotates uniformly about its own axis.
M
n
x3
22pr
x1
FIG. 46
The angular velocities of these two rotations can easily be expressed in
terms of the given angular momentum M and the angle 0 between the axis
of the top and the direction of M. The angular velocity of the top about its
own axis is just the component S3 of the vector S along the axis:
Q3 = M3/I3 = (M/I3) cos 0.
(33.4)
To determine the rate of precession Spr, the vector S must be resolved into
components along X3 and along M. The first of these gives no displacement
of the axis of the top, and the second component is therefore the required
angular velocity of precession. Fig. 46 shows that Spr sin 0 = Q1, and, since
S21 = M1/I1 = (M/I1) sin 0, we have
Spr r=M/I1.
(33.5)
$34. The equations of motion of a rigid body
Since a rigid body has, in general, six degrees of freedom, the general
equations of motion must be six in number. They can be put in a form which
gives the time derivatives of two vectors, the momentum and the angular
momentum of the body.
The first equation is obtained by simply summing the equations p = f
for each particle in the body, p being the momentum of the particle and f the
108
Motion of a Rigid Body
§34
force acting on it. In terms of the total momentum of the body P =
and total force acting on it F = Ef, we have
dP/dt = F.
(34.1)
Although F has been defined as the sum of all the forces f acting on the
various particles, including the forces due to other particles, F actually
includes only external forces: the forces of interaction between the particles
composing the body must cancel out, since if there are no external forces
the momentum of the body, like that of any closed system, must be conserved,
i.e. we must have F = 0.
If U is the potential energy of a rigid body in an external field, the force
F is obtained by differentiating U with respect to the co-ordinates of the
centre of mass of the body:
F = JUIR.
(34.2)
For, when the body undergoes a translation through a distance SR, the radius
vector r of every point in the body changes by SR, and so the change in the
potential energy is
SU = (U/dr) Sr = RR Couldr = SR SR.
It may be noted that equation (34.1) can also be obtained as Lagrange's
equation for the co-ordinates of the centre of mass, (d/dt)àL/JV = aL/JR,
with the Lagrangian (32.4), for which
OL/OV=,MV=P, 0L/JR = JU/OR = F.
Let us now derive the second equation of motion, which gives the time
derivative of the angular momentum M. To simplify the derivation, it is
convenient to choose the "fixed" (inertial) frame of reference in such a way
that the centre of mass is at rest in that frame at the instant considered.
We have M = (d/dt) Erxp = Eixp+. Our choice of the frame of
reference (with V = 0) means that the value of i at the instant considered is
the same as V = i. Since the vectors V and p = mv are parallel, ixp = 0.
Replacing p by the force f, we have finally
dM/dt = K,
(34.3)
where
K = .
(34.4)
Since M has been defined as the angular momentum about the centre of
mass (see the beginning of $33), it is unchanged when we go from one inertial
frame to another. This is seen from formula (9.5) with R = 0. We can there-
fore deduce that the equation of motion (34.3), though derived for a particular
frame of reference, is valid in any other inertial frame, by Galileo's relativity
principle.
The vector rxf is called the moment of the force f, and so K is the total
torque, i.e. the sum of the moments of all the forces acting on the body. Like
§34
The equations of motion of a rigid body
109
the total force F, the sum (34.4) need include only the external forces: by
the law of conservation of angular momentum, the sum of the moments of
the internal forces in a closed system must be zero.
The moment of a force, like the angular momentum, in general depends on
the choice of the origin about which it is defined. In (34.3) and (34.4) the
moments are defined with respect to the centre of mass of the body.
When the origin is moved a distance a, the new radius vector r' of each
point in the body is equal to r-a. Hence K = Erxf = Er'xf+ Eaxf or
K = K'+axF.
(34.5)
Hence we see, in particular, that the value of the torque is independent of
the choice of origin if the total force F = 0. In this case the body is said to
be acted on by a couple.
Equation (34.3) may be regarded as Lagrange's equation (d/dt) OL/OS
= 0L/dd for the "rotational co-ordinates". Differentiating the Lagrangian
(32.4) with respect to the components of the vector S2, we obtain
= IikOk = Mi. The change in the potential energy resulting from an
infinitesimal rotation SO of the body is SU = - Ef.Sr = -
= So. Erxf = -K.SO, whence
K =-20/00, =
(34.6)
so that aL/dd = 00/08 = K.
Let us assume that the vectors F and K are perpendicular. Then a vector a
can always be found such that K' given by formula (34.5) is zero and
K a x F.
(34.7)
The choice of a is not unique, since the addition to a of any vector parallel
to F does not affect equation (34.7). The condition K' = 0 thus gives a straight
line, not a point, in the moving system of co-ordinates. When K is perpendi-
cular to F, the effect of all the applied forces can therefore be reduced to that
of a single force F acting along this line.
Such a case is that of a uniform field of force, in which the force on a particle
is f = eE, with E a constant vector characterising the field and e characterising
the properties of a particle with respect to the field. Then F = Ee,
K = erxE. Assuming that # 0, we define a radius vector ro such that
(34.8)
Then the total torque is simply
=roxF
(34.9)
Thus, when a rigid body moves in a uniform field, the effect of the field
reduces to the action of a single force F applied at the point whose radius
vector is (34.8). The position of this point is entirely determined by the
t For example, in a uniform electric field E is the field strength and e the charge; in a
uniform gravitational field E is the acceleration g due to gravity and e is the mass m.
110
Motion of a Rigid Body

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properties of the body itself. In a gravitational field, for example, it is the
centre of mass.
$35. Eulerian angles
As has already been mentioned, the motion of a rigid body can be described
by means of the three co-ordinates of its centre of mass and any three angles
which determine the orientation of the axes X1, X2, X3 in the moving system of
co-ordinates relative to the fixed system X, Y, Z. These angles may often be
conveniently taken as what are called Eulerian angles.
Z
X2
Y
FIG. 47
Since we are here interested only in the angles between the co-ordinate
axes, we may take the origins of the two systems to coincide (Fig. 47). The
moving x1x2-plane intersects the fixed XY-plane in some line ON, called the
line of nodes. This line is evidently perpendicular to both the Z-axis and the
x3-axis; we take its positive direction as that of the vector product ZXX3
(where Z and X3 are unit vectors along the Z and X3 axes).
We take, as the quantities defining the position of the axes x1, X2, X3
relative to the axes X, Y, Z the angle 0 between the Z and X3 axes, the angle
between the X-axis and ON, and the angle as between the x1-axis and ON.
The angles and 4 are measured round the Z and X3 axes respectively in the
direction given by the corkscrew rule. The angle 0 takes values from 0 to TT,
and and 4 from 0 to 2n.t
t The angles 0 and - are respectively the polar angle and azimuth of the direction
X3 with respect to the axes X, Y, Z. The angles 0 and 12-- are respectively the polar angle
and azimuth of the direction Z with respect to the axes X1, X2, X3.
§35
Eulerian angles
111
Let us now express the components of the angular velocity vector S along
the moving axes X1, X2, X3 in terms of the Eulerian angles and their derivatives.
To do this, we must find the components along those axes of the angular
velocities 6, b, 4. The angular velocity è is along the line of nodes ON, and
its components are 1 = O cos 4/5, = - O sin 4/5, = 0. The angular velo-
city is along the Z-axis; its component along the x3-axis is 03 = cos 0, and
in the x1x2-plane sin A. Resolving the latter along the X1 and X2 axes, we
have 01 = sin 0 sin 4/s, O2 = sin 0 cos 4. Finally, the angular velocity is
is along the x3-axis.
Collecting the components along each axis, we have
S21 = 0 COS 4,
Q2 = sin 0 cosy-osiny,
(35.1)
S23 = o cos0+4. =
If the axes X1, X2, X3 are taken to be the principal axes of inertia of the body,
the rotational kinetic energy in terms of the Eulerian angles is obtained by
substituting (35.1) in (32.8).
For a symmetrical top (I1 = I2 # I3), a simple reduction gives
Trot =
(35.2)
This expression can also be more simply obtained by using the fact that the
choice of directions of the principal axes X1, X2 is arbitrary for a symmetrical
top. If the X1 axis is taken along the line of nodes ON, i.e. 4 = 0, the compo-
nents of the angular velocity are simply
O2 = o sin A,
(35.3)
As a simple example of the use of the Eulerian angles, we shall use them
to determine the free motion of a symmetrical top, already found in $33.
We take the Z-axis of the fixed system of co-ordinates in the direction of the
constant angular momentum M of the top. The x3-axis of the moving system
is along the axis of the top; let the x1-axis coincide with the line of nodes at
the instant considered. Then the components of the vector M are, by
formulae (35.3), M1 = I1 = I, M2 = IS2 = sin 0, M3 = I3Q3
= I3( cos 0+4). Since the x1-axis is perpendicular to the Z-axis, we have
M1 = 0, M2 = M sin 0, M3 = M cos 0. Comparison gives
0=0,
I = M,
=
(35.4)
The first of these equations gives 0 = constant, i.e. the angle between the
axis of the top and the direction of M is constant. The second equation gives
the angular velocity of precession = M/I1, in agreement with (33.5).
Finally, the third equation gives the angular velocity with which the top
rotates about its own axis: S3 = (M/I3) cos 0.
112
Motion of a Rigid Body
§35
PROBLEMS
PROBLEM 1. Reduce to quadratures the problem of the motion of a heavy symmetrical
top whose lowest point is fixed (Fig. 48).
SOLUTION. We take the common origin of the moving and fixed systems of co-ordinates
at the fixed point O of the top, and the Z-axis vertical. The Lagrangian of the top in a gravita-
tional field is L = (02 +02 sin ²0 + 1/3(1- cos 0)2-ugl - cos 0, where u is the mass
of the top and l the distance from its fixed point to the centre of mass.
Z
X3
x2
a
ug
Y
x1
N
FIG. 48
The co-ordinates 4 and are cyclic. Hence we have two integrals of the motion:
P4 = = cos 0) = constant = M3
(1)
= = cos 0 = constant III M2,
(2)
where I'1 = I1+ul2; the quantities P4 and Po are the components of the rotational angular
momentum about O along the X3 and Z axes respectively. The energy
E = cos 0
(3)
is also conserved.
From equations (1) and (2) we find
=
0)/I'1
sin 20,
(4)
(5)
Eliminating b and of from the energy (3) by means of equations (4) and (5), we obtain
E' =
where
E'
=
(6)
§35
Eulerian angles
113
Thus we have
t=
(7)
this is an elliptic integral. The angles 4 and are then expressed in terms of 0 by means of
integrals obtained from equations (4) and (5).
The range of variation of 0 during the motion is determined by the condition E' Ueff(0).
The function Uett(8) tends to infinity (if M3 # M2) when 0 tends to 0 or II, and has a minimum
between these values. Hence the equation E' = Ueff(0) has two roots, which determine the
limiting values 01 and O2 of the inclination of the axis of the top to the vertical.
When 0 varies from 01 to O2, the derivative o changes sign if and only if the difference
M-M3 cos 0 changes sign in that range of 0. If it does not change sign, the axis of the top
precesses monotonically about the vertical, at the same time oscillating up and down. The
latter oscillation is called nutation; see Fig. 49a, where the curve shows the track of the axis
on the surface of a sphere whose centre is at the fixed point of the top. If does change sign,
the direction of precession is opposite on the two limiting circles, and so the axis of the top
describes loops as it moves round the vertical (Fig. 49b). Finally, if one of 01, O2 is a zero of
M2-M3 cos 0, of and è vanish together on the corresponding limiting circle, and the path
of the axis is of the kind shown in Fig. 49c.
O2
O2
O2
(a)
(b)
(c)
FIG. 49
PROBLEM 2. Find the condition for the rotation of a top about a vertical axis to be stable.
SOLUTION. For 0 = 0, the X3 and Z axes coincide, so that M3 = Mz, E' = 0. Rotation
about this axis is stable if 0 = 0 is a minimum of the function Ueff(9). For small 0 we have
Ueff 22 whence the condition for stability is M32 > 41'1ugl or S232
> 41'1ugl/I32.
PROBLEM 3. Determine the motion of a top when the kinetic energy of its rotation about
its axis is large compared with its energy in the gravitational field (called a "fast" top).
SOLUTION. In a first approximation, neglecting gravity, there is a free precession of the
axis of the top about the direction of the angular momentum M, corresponding in this case
to the nutation of the top; according to (33.5), the angular velocity of this precession is
Sunu = M/I' 1.
(1)
In the next approximation, there is a slow precession of the angular momentum M about
the vertical (Fig. 50). To determine the rate of this precession, we average the exact equation
of motion (34.3) dM/dt = K over the nutation period. The moment of the force of gravity
on the top is K=uln3xg, where n3 is a unit vector along the axis of the top. It is evident
from symmetry that the result of averaging K over the "nutation cone" is to replace n3 by
its component (M/M) cos a in the direction of M, where a is the angle between M and the
axis of the top. Thus we have dM/dt = -(ul/M)gxM cos a. This shows that the vector M
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Motion of a Rigid Body

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precesses about the direction of g (i.e. the vertical) with a mean angular velocity
Spr (ul/M)g cos a
(2)
which is small compared with Senu
Spr
in
no
a
FIG. 50
In this approximation the quantities M and cos a in formulae (1) and (2) are constants,
although they are not exact integrals of the motion. To the same accuracy they are related
to the strictly conserved quantities E and M3 by M3 = M cos a,
§36. Euler's equations
The equations of motion given in §34 relate to the fixed system of co-
ordinates: the derivatives dP/dt and dM/dt in equations (34.1) and (34.3)
are the rates of change of the vectors P and M with respect to that system.
The simplest relation between the components of the rotational angular
momentum M of a rigid body and the components of the angular velocity
occurs, however, in the moving system of co-ordinates whose axes are the
principal axes of inertia. In order to use this relation, we must first transform
the equations of motion to the moving co-ordinates X1, X2, X3.
Let dA/dt be the rate of change of any vector A with respect to the fixed
system of co-ordinates. If the vector A does not change in the moving system,
its rate of change in the fixed system is due only to the rotation, so that
dA/dt = SxA; see §9, where it has been pointed out that formulae such as
(9.1) and (9.2) are valid for any vector. In the general case, the right-hand
side includes also the rate of change of the vector A with respect to the moving
system. Denoting this rate of change by d'A/dt, we obtain
dAdd
(36.1)
§36
Euler's equations
115
Using this general formula, we can immediately write equations (34.1) and
(34.3) in the form
=
K.
(36.2)
Since the differentiation with respect to time is here performed in the moving
system of co-ordinates, we can take the components of equations (36.2) along
the axes of that system, putting (d'P/dt)1 = dP1/dt, ..., (d'M/dt)1 = dM1/dt,
..., where the suffixes 1, 2, 3 denote the components along the axes x1, x2, X3.
In the first equation we replace P by V, obtaining
(36.3)
=
If the axes X1, X2, X3 are the principal axes of inertia, we can put M1 = I,
etc., in the second equation (36.2), obtaining
=
I2 = K2,
}
(36.4)
I3 = K3.
These are Euler's equations.
In free rotation, K = 0, so that Euler's equations become
= 0,
}
(36.5)
= 0.
As an example, let us apply these equations to the free rotation of a sym-
metrical top, which has already been discussed. Putting I1 = I2, we find from
the third equation SQ3 = 0, i.e. S3 = constant. We then write the first two
equations as O = -wS2, Q2 = wS1, where
=
(36.6)
is a constant. Multiplying the second equation by i and adding, we have
= so that S1+iD2 = A exp(iwt), where A is a
constant, which may be made real by a suitable choice of the origin of time.
Thus
S1 = A cos wt
Q2 = A sin wt.
(36.7)
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This result shows that the component of the angular velocity perpendicular
to the axis of the top rotates with an angular velocity w, remaining of constant
magnitude A = Since the component S3 along the axis of the
top is also constant, we conclude that the vector S rotates uniformly with
angular velocity w about the axis of the top, remaining unchanged in magni-
tude. On account of the relations M1 = , M2 = I2O2, M3 = I3O3 be-
tween the components of S and M, the angular momentum vector M evidently
executes a similar motion with respect to the axis of the top.
This description is naturally only a different view of the motion already
discussed in §33 and §35, where it was referred to the fixed system of co-
ordinates. In particular, the angular velocity of the vector M (the Z-axis in
Fig. 48, $35) about the x3-axis is, in terms of Eulerian angles, the same as
the angular velocity - 4. Using equations (35.4), we have
cos
or - is = I23(I3-I1)/I1, in agreement with (36.6).
§37. The asymmetrical top
We shall now apply Euler's equations to the still more complex problem
of the free rotation of an asymmetrical top, for which all three moments of
inertia are different. We assume for definiteness that
I3 > I2 I.
(37.1)
Two integrals of Euler's equations are known already from the laws of
conservation of energy and angular momentum:
= 2E,
(37.2)
= M2,
where the energy E and the magnitude M of the angular momentum are given
constants. These two equations, written in terms of the components of the
vector M, are
(37.3)
M2.
(37.4)
From these equations we can already draw some conclusions concerning
the nature of the motion. To do so, we notice that equations (37.3) and (37.4),
regarded as involving co-ordinates M1, M2, M3, are respectively the equation
of an ellipsoid with semiaxes (2EI1), (2EI2), (2EI3) and that of a sphere
of radius M. When the vector M moves relative to the axes of inertia of the
top, its terminus moves along the line of intersection of these two surfaces.
Fig. 51 shows a number of such lines of intersection of an ellipsoid with
§37
The asymmetrical top
117
spheres of various radii. The existence of an intersection is ensured by the
obviously valid inequalities
2EI1 < M2 < 2EI3,
(37.5)
which signify that the radius of the sphere (37.4) lies between the least and
greatest semiaxes of the ellipsoid (37.3).
x1
X2
FIG. 51
Let us examine the way in which these "paths"t of the terminus of the
vector M change as M varies (for a given value of E). When M2 is only slightly
greater than 2EI1, the sphere intersects the ellipsoid in two small closed curves
round the x1-axis near the corresponding poles of the ellipsoid; as M2 2EI1,
these curves shrink to points at the poles. When M2 increases, the curves
become larger, and for M2 = 2EI2 they become two plane curves (ellipses)
which intersect at the poles of the ellipsoid on the x2-axis. When M2 increases
further, two separate closed paths again appear, but now round the poles on
the
x3-axis; as M2 2EI3 they shrink to points at these poles.
First of all, we may note that, since the paths are closed, the motion of the
vector M relative to the top must be periodic; during one period the vector
M describes some conical surface and returns to its original position.
Next, an essential difference in the nature of the paths near the various
poles of the ellipsoid should be noted. Near the x1 and X3 axes, the paths lie
entirely in the neighbourhood of the corresponding poles, but the paths which
pass near the poles on the x2-axis go elsewhere to great distances from those
poles. This difference corresponds to a difference in the stability of the rota-
tion of the top about its three axes of inertia. Rotation about the x1 and X3
axes (corresponding to the least and greatest of the three moments of inertia)
t The corresponding curves described by the terminus of the vector Ca are called polhodes.
118
Motion of a Rigid Body
§37
is stable, in the sense that, if the top is made to deviate slightly from such a
state, the resulting motion is close to the original one. A rotation about the
x2-axis, however, is unstable: a small deviation is sufficient to give rise to a
motion which takes the top to positions far from its original one.
To determine the time dependence of the components of S (or of the com-
ponents of M, which are proportional to those of (2) we use Euler's equations
(36.5). We express S1 and S3 in terms of S2 by means of equations (37.2)
and (37.3):
S21 =
(37.6)
Q32 =
and substitute in the second equation (36.5), obtaining
dSQ2/dt (I3-I1)21-23/I2
= V{[(2EI3-M2-I2(I3-I2)22]
(37.7)
Integration of this equation gives the function t(S22) as an elliptic integral.
In reducing it to a standard form we shall suppose for definiteness that
M2 > 2EI2; if this inequality is reversed, the suffixes 1 and 3 are interchanged
in the following formulae. Using instead of t and S2 the new variables
(37.8)
S = S2V[I2(I3-I2)/(2EI3-M2)],
and defining a positive parameter k2 < 1 by
(37.9)
we obtain
ds
the origin of time being taken at an instant when S2 = 0. When this integral
is inverted we have a Jacobian elliptic function S = sn T, and this gives O2
as a function of time; S-1(t) and (33(t) are algebraic functions of 22(t) given
by (37.6). Using the definitions cn T = V(1-sn2r), dn T =
we find
Superscript(2) = [(2EI3-M2/I1(I3-I1)] CNT,
O2 =
(37.10)
O3 = dn T.
These are periodic functions, and their period in the variable T is 4K,
where K is a complete elliptic integral of the first kind:
=
(37.11)
§37
The asymmetrical top
119
The period in t is therefore
T =
(37.12)
After a time T the vector S returns to its original position relative to the
axes of the top. The top itself, however, does not return to its original position
relative to the fixed system of co-ordinates; see below.
For I = I2, of course, formulae (37.10) reduce to those obtained in §36
for a symmetrical top: as I I2, the parameter k2 0, and the elliptic
functions degenerate to circular functions: sn -> sin T, cn T cos
T,
dn T -> 1, and we return to formulae (36.7).
When M2 = 2EI3 we have Superscript(1) = S2 = 0, S3 = constant, i.e. the vector S
is always parallel to the x3-axis. This case corresponds to uniform rotation of
the top about the x3-axis. Similarly, for M2 = 2EI1 (when T III 0) we have
uniform rotation about the x1-axis.
Let us now determine the absolute motion of the top in space (i.e. its
motion relative to the fixed system of co-ordinates X, Y, Z). To do so, we
use the Eulerian angles 2/5, o, 0, between the axes X1, X2, X3 of the top and the
axes X, Y, Z, taking the fixed Z-axis in the direction of the constant vector M.
Since the polar angle and azimuth of the Z-axis with respect to the axes
x1, X2, X3 are respectively 0 and 1/77 - is (see the footnote to $35), we obtain on
taking the components of M along the axes X1, X2, X3
M sin 0 sin y = M1 = ,
M sin A cos is = M2 = I2O2,
(37.13)
M cos 0 = M3 = I3S23.
Hence
cos 0 = I3S3/M,
tan / =
(37.14)
and from formulae (37.10)
COS 0 = dn T,
(37.15)
tan 4 = cn r/snt,
which give the angles 0 and is as functions of time; like the components of the
vector S, they are periodic functions, with period (37.12).
The angle does not appear in formulae (37.13), and to calculate it we
must return to formulae (35.1), which express the components of S in terms
of the time derivatives of the Eulerian angles. Eliminating O from the equa-
tions S1 = sin 0 sin 4 + O cos 2/5, S2 = sin 0 cos 4-0 - sin 2/5, we obtain
& = (Superscript(2) sin 4+S2 cos 4)/sin 0, and then, using formulae (37.13),
do/dt =
(37.16)
The function (t) is obtained by integration, but the integrand involves
elliptic functions in a complicated way. By means of some fairly complex
120
Motion of a Rigid Body
§37
transformations, the integral can be expressed in terms of theta functions;
we shall not give the calculations, but only the final result.
The function (t) can be represented (apart from an arbitrary additive
constant) as a sum of two terms:
$(t) = (11(t)++2(t),
(37.17)
one of which is given by
(37.18)
where D01 is a theta function and a a real constant such that
sn(2ixK) = iv[I3(M2-2I1)/I1(2EI3-M2]
(37.19)
K and Tare given by (37.11) and (37.12). The function on the right-hand side
of (37.18) is periodic, with period 1T, so that 01(t) varies by 2n during a time
T. The second term in (37.17) is given by
(37.20)
This function increases by 2nr during a time T'. Thus the motion in is a
combination of two periodic motions, one of the periods (T) being the same
as the period of variation of the angles 4 and 0, while the other (T') is incom-
mensurable with T. This incommensurability has the result that the top does
not at any time return exactly to its original position.
PROBLEMS
PROBLEM 1. Determine the free rotation of a top about an axis near the x3-axis or the
x1-axis.
SOLUTION. Let the x3-axis be near the direction of M. Then the components M1 and M2
are small quantities, and the component M3 = M (apart from quantities of the second and
higher orders of smallness). To the same accuracy the first two Euler's equations (36.5) can
be written dM1/dt = DoM2(1-I3/I2), dM2/dt = QOM1(I3/I1-1), where So = M/I3. As
usual we seek solutions for M1 and M2 proportional to exp(iwt), obtaining for the frequency w
(1)
The values of M1 and M2 are
cos wt, sin wt,
(2)
where a is an arbitrary small constant. These formulae give the motion of the vector M
relative to the top. In Fig. 51, the terminus of the vector M describes, with frequency w,
a small ellipse about the pole on the x3-axis.
To determine the absolute motion of the top in space, we calculate its Eulerian angles.
In the present case the angle 0 between the x3-axis and the Z-axis (direction of M) is small,
t These are given by E. T. WHITTAKER, A Treatise on the Analytical Dynamics of Particles
and Rigid Bodies, 4th ed., Chapter VI, Dover, New York 1944.
§37
The asymmetrical top
121
and by formulae (37.14) tan of = M1/M2, cos 0) 2(1 (M3/M) 22
substituting (2), we obtain
tan 4 = V[I(I3-I2)/I2(I3-I1)] cot wt,
(3)
To find , we note that, by the third formula (35.1), we have, for 0 1,
Hence
= lot
(4)
omitting an arbitrary constant of integration.
A clearer idea of the nature of the motion of the top is obtained if we consider the change
in direction of the three axes of inertia. Let n1, n2, n3 be unit vectors along these axes. The
vectors n1 and n2 rotate uniformly in the XY-plane with frequency So, and at the same time
execute small transverse oscillations with frequency w. These oscillations are given by the
Z-components of the vectors:
22 M1/M = av(I3/I2-1) cos wt,
N2Z 22 M2/M = av(I3/I1-1) sin wt.
For the vector n3 we have, to the same accuracy, N3x 22 0 sin , N3y 22 -0 cos , n3z 1.
(The polar angle and azimuth of n3 with respect to the axes X, Y, Z are 0 and -; see
the footnote to 35.) We also write, using formulae (37.13),
naz=0sin(Qot-4)
= Asin Sot cos 4-0 cos lot sin 4
= (M 2/M) sin Dot-(M1/M) cos Sot
sin Sot sin N/1-1) cos Not cos wt
cos(so
Similarly
From this we see that the motion of n3 is a superposition of two rotations about the Z-axis
with frequencies So + w.
PROBLEM 2. Determine the free rotation of a top for which M2 = 2EI2.
SOLUTION. This case corresponds to the movement of the terminus of M along a curve
through the pole on the x2-axis (Fig. 51). Equation (37.7) becomes ds/dr = 1-s2,
= S = I2/20, where So = M/I2 = 2E|M. Integration of
this equation and the use of formulae (37.6) gives
sech T,
}
(1)
sech T.
To describe the absolute motion of the top, we use Eulerian angles, defining 0 as the angle
between the Z-axis (direction of M) and the x2-axis (not the x3-axis as previously). In formulae
(37.14) and (37.16), which relate the components of the vector CA to the Eulerian angles, we
5
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must cyclically permute the suffixes 1, 2, 3 to 3, 1, 2. Substitution of (1) in these formulae
then gives cos 0 = tanh T, = lot + constant, tan =
It is seen from these formulae that, as t 8, the vector SC asymptotically approaches the
x2-axis, which itself asymptotically approaches the Z-axis.
$38. Rigid bodies in contact
The equations of motion (34.1) and (34.3) show that the conditions of
equilibrium for a rigid body can be written as the vanishing of the total force
and total torque on the body:
F = f = 0 ,
K ==~rxf=0. =
(38.1)
Here the summation is over all the external forces acting on the body, and r
is the radius vector of the "point of application"; the origin with respect to
which the torque is defined may be chosen arbitrarily, since if F = 0 the
value of K does not depend on this choice (see (34.5)).
If we have a system of rigid bodies in contact, the conditions (38.1) for
each body separately must hold in equilibrium. The forces considered must
include those exerted on each body by those with which it is in contact. These
forces at the points of contact are called reactions. It is obvious that the mutual
reactions of any two bodies are equal in magnitude and opposite in direction.
In general, both the magnitudes and the directions of the reactions are
found by solving simultaneously the equations of equilibrium (38.1) for all the
bodies. In some cases, however, their directions are given by the conditions
of the problem. For example, if two bodies can slide freely on each other, the
reaction between them is normal to the surface.
If two bodies in contact are in relative motion, dissipative forces of friction
arise, in addition to the reaction.
There are two possible types of motion of bodies in contact-sliding and
rolling. In sliding, the reaction is perpendicular to the surfaces in contact,
and the friction is tangential. Pure rolling, on the other hand, is characterised
by the fact that there is no relative motion of the bodies at the point of
contact; that is, a rolling body is at every instant as it were fixed to the point
of contact. The reaction may be in any direction, i.e. it need not be normal
to the surfaces in contact. The friction in rolling appears as an additional
torque which opposes rolling.
If the friction in sliding is negligibly small, the surfaces concerned are
said to be perfectly smooth. If, on the other hand, only pure rolling without
sliding is possible, and the friction in rolling can be neglected, the surfaces
are said to be perfectly rough.
In both these cases the frictional forces do not appear explicitly in the pro-
blem, which is therefore purely one of mechanics. If, on the other hand, the
properties of the friction play an essential part in determining the motion,
then the latter is not a purely mechanical process (cf. $25).
Contact between two bodies reduces the number of their degrees of freedom
as compared with the case of free motion. Hitherto, in discussing such
§38
Rigid bodies in contact
123
problems, we have taken this reduction into account by using co-ordinates
which correspond directly to the actual number of degrees of freedom. In
rolling, however, such a choice of co-ordinates may be impossible.
The condition imposed on the motion of rolling bodies is that the velocities
of the points in contact should be equal; for example, when a body rolls on a
fixed surface, the velocity of the point of contact must be zero. In the general
case, this condition is expressed by the equations of constraint, of the form
E caide = 0,
(38.2)
where the Cai are functions of the co-ordinates only, and the suffix a denumer-
ates the equations. If the left-hand sides of these equations are not the total
time derivatives of some functions of the co-ordinates, the equations cannot
be integrated. In other words, they cannot be reduced to relations between the
co-ordinates only, which could be used to express the position of the bodies
in terms of fewer co-ordinates, corresponding to the actual number of degrees
of freedom. Such constraints are said to be non-holonomic, as opposed to
holonomic constraints, which impose relations between the co-ordinates only.
Let us consider, for example, the rolling of a sphere on a plane. As usual,
we denote by V the translational velocity (the velocity of the centre of the
sphere), and by Sa the angular velocity of rotation. The velocity of the point
of contact with the plane is found by putting r = - an in the general formula
V = +SXR; a is the radius of the sphere and n a unit vector along the
normal to the plane. The required condition is that there should be no sliding
at the point of contact, i.e.
V-aSxxn = 0.
(38.3)
This cannot be integrated: although the velocity V is the total time derivative
of the radius vector of the centre of the sphere, the angular velocity is not in
general the total time derivative of any co-ordinate. The constraint (38.3) is
therefore non-holonomic.t
Since the equations of non-holonomic constraints cannot be used to reduce
the number of co-ordinates, when such constraints are present it is necessary
to use co-ordinates which are not all independent. To derive the correspond-
ing Lagrange's equations, we return to the principle of least action.
The existence of the constraints (38.2) places certain restrictions on the
possible values of the variations of the co-ordinates: multiplying equations
(38.2) by St, we find that the variations dqi are not independent, but are
related by
(38.4)
t It may be noted that the similar constraint in the rolling of a cylinder is holonomic. In
that case the axis of rotation has a fixed direction in space, and hence la = do/dt is the total
derivative of the angle of rotation of the cylinder about its axis. The condition (38.3) can
therefore be integrated, and gives a relation between the angle and the co-ordinate of the
centre of mass.
124
Motion of a Rigid Body
§38
This must be taken into account in varying the action. According to
Lagrange's method of finding conditional extrema, we must add to the inte-
grand in the variation of the action
=
the left-hand sides of equations (38.4) multiplied by undetermined coeffici-
ents da (functions of the co-ordinates), and then equate the integral to zero.
In SO doing the variations dqi are regarded as entirely independent, and the
result is
(38.5)
These equations, together with the constraint equations (38.2), form a com-
plete set of equations for the unknowns qi and da.
The reaction forces do not appear in this treatment, and the contact of
the bodies is fully allowed for by means of the constraint equations. There
is, however, another method of deriving the equations of motion for bodies in
contact, in which the reactions are introduced explicitly. The essential feature
of this method, which is sometimes called d'Alembert's principle, is to write
for each of the bodies in contact the equations.
dP/dt==f,
(38.6)
wherein the forces f acting on each body include the reactions. The latter
are initially unknown and are determined, together with the motion of the
body, by solving the equations. This method is equally applicable for both
holonomic and non-holonomic constraints.
PROBLEMS
PROBLEM 1. Using d'Alembert's principle, find the equations of motion of a homogeneous
sphere rolling on a plane under an external force F and torque K.
SOLUTION. The constraint equation is (38.3). Denoting the reaction force at the point of
contact between the sphere and the plane by R, we have equations (38.6) in the form
u dV/dt = F+R,
(1)
dSu/dt = K-an xR,
(2)
where we have used the facts that P = V and, for a spherical top, M = ISE. Differentiating
the constraint equation (38.3) with respect to time, we have V = aS2xn. Substituting in
equation (1) and eliminating S by means of (2), we obtain (I/au)(F+R) = Kxn-aR+
+an(n . R), which relates R, F and K. Writing this equation in components and substitut-
ing I = zua2 (§32, Problem 2(b)), we have
R2 = -F2,
where the plane is taken as the xy-plane. Finally, substituting these expressions in (1), we
§38
Rigid bodies in contact
125
obtain the equations of motion involving only the given external force and torque:
dVx dt 7u 5 Ky
dt
The components Ox, Q2 y of the angular velocity are given in terms of Vx, Vy by the constraint
equation (38.3); for S2 we have the equation 2 dQ2/dt = K2, the z-component of equa-
tion (2).
PROBLEM 2. A uniform rod BD of weight P and length l rests against a wall as shown in
Fig. 52 and its lower end B is held by a string AB. Find the reaction of the wall and the ten-
sion in the string.
Rc
h
P
RB
T
A
B
FIG. 52
SOLUTION. The weight of the rod can be represented by a force P vertically downwards,
applied at its midpoint. The reactions RB and Rc are respectively vertically upwards and
perpendicular to the rod; the tension T in the string is directed from B to A. The solution
of the equations of equilibrium gives Rc = (Pl/4h) sin 2a, RB = P-Rcsin x, T = Rc cos a.
PROBLEM 3. A rod of weight P has one end A on a vertical plane and the other end B on
a horizontal plane (Fig. 53), and is held in position by two horizontal strings AD and BC,
RB
TA
A
RA
C
FIG. 53
126
Motion of a Rigid Body

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the latter being in the same vertical plane as AB. Determine the reactions of the planes and
the tensions in the strings.
SOLUTION. The tensions TA and TB are from A to D and from B to C respectively. The
reactions RA and RB are perpendicular to the corresponding planes. The solution of the
equations of equilibrium gives RB = P, TB = 1P cot a, RA = TB sin B, TA = TB cos B.
PROBLEM 4. Two rods of length l and negligible weight are hinged together, and their ends
are connected by a string AB (Fig. 54). They stand on a plane, and a force F is applied
at the midpoint of one rod. Determine the reactions.
RC
C
PA
F
1
RB
T
T
A
B
FIG. 54
SOLUTION. The tension T acts at A from A to B, and at B from B to A. The reactions RA
and RB at A and B are perpendicular to the plane. Let Rc be the reaction on the rod AC at
the hinge; then a reaction - -Rc acts on the rod BC. The condition that the sum of the moments
of the forces RB, T and - -Rc acting on the rod BC should be zero shows that Rc acts along
BC. The remaining conditions of equilibrium (for the two rods separately) give RA = 1F,
RB = 1F, Rc = 1F cosec a, T = 1F cot a, where a is the angle CAB.
§39. Motion in a non-inertial frame of reference
Up to this point we have always used inertial frames of reference in discuss-
ing the motion of mechanical systems. For example, the Lagrangian
L = 1mvo2- U,
(39.1)
and the corresponding equation of motion m dvo/dt = - au/dr, for a single
particle in an external field are valid only in an inertial frame. (In this section
the suffix 0 denotes quantities pertaining to an inertial frame.)
Let us now consider what the equations of motion will be in a non-inertial
frame of reference. The basis of the solution of this problem is again the
principle of least action, whose validity does not depend on the frame of
reference chosen. Lagrange's equations
(39.2)
are likewise valid, but the Lagrangian is no longer of the form (39.1), and to
derive it we must carry out the necessary transformation of the function Lo.
§39
Motion in a non-inertial frame of reference
127
This transformation is done in two steps. Let us first consider a frame of
reference K' which moves with a translational velocity V(t) relative to the
inertial frame K0. The velocities V0 and v' of a particle in the frames Ko and
K' respectively are related by
vo = v'+ V(t).
(39.3)
Substitution of this in (39.1) gives the Lagrangian in K':
L' = 1mv2+mv.+1mV2-U
Now V2(t) is a given function of time, and can be written as the total deriva-
tive with respect to t of some other function; the third term in L' can there-
fore be omitted. Next, v' = dr'/dt, where r' is the radius vector of the par-
ticle in the frame K'. Hence
mV(t)+v'= mV.dr/'dt = d(mV.r')/dt-mr'.dV/dt.
Substituting in the Lagrangian and again omitting the total time derivative,
we have finally
L' =
(39.4)
where W = dV/dt is the translational acceleration of the frame K'.
The Lagrange's equation derived from (39.4) is
(39.5)
Thus an accelerated translational motion of a frame of reference is equivalent,
as regards its effect on the equations of motion of a particle, to the application
of a uniform field of force equal to the mass of the particle multiplied by the
acceleration W, in the direction opposite to this acceleration.
Let us now bring in a further frame of reference K, whose origin coincides
with that of K', but which rotates relative to K' with angular velocity Su(t).
Thus K executes both a translational and a rotational motion relative to the
inertial frame Ko.
The velocity v' of the particle relative to K' is composed of its velocity
V
relative to K and the velocity Sxr of its rotation with K: v' = Lxr
(since the radius vectors r and r' in the frames K and K' coincide). Substitut-
ing this in the Lagrangian (39.4), we obtain
L = +mv.Sx+1m(xr)2-mW.r-
(39.6)
This is the general form of the Lagrangian of a particle in an arbitrary, not
necessarily inertial, frame of reference. The rotation of the frame leads to the
appearance in the Lagrangian of a term linear in the velocity of the particle.
To calculate the derivatives appearing in Lagrange's equation, we write
128
Motion of a Rigid Body
§39
the total differential
dL = mv.dv+mdv.Sxr+mv.Sxdr+
=
v.dv+mdv.xr+mdr.vxR+
The terms in dv and dr give
0L/dr X Q-mW-dU/0r. - -
Substitution of these expressions in (39.2) gives the required equation of
motion:
mdv/dt = (39.7)
We see that the "inertia forces" due to the rotation of the frame consist
of three terms. The force mrxo is due to the non-uniformity of the rotation,
but the other two terms appear even if the rotation is uniform. The force
2mvxs is called the Coriolis force; unlike any other (non-dissipative) force
hitherto considered, it depends on the velocity of the particle. The force
mSX(rxS) is called the centrifugal force. It lies in the plane through r and
S, is perpendicular to the axis of rotation (i.e. to S2), and is directed away
from the axis. The magnitude of this force is mpO2, where P is the distance
of the particle from the axis of rotation.
Let us now consider the particular case of a uniformly rotating frame with
no translational acceleration. Putting in (39.6) and (39.7) S = constant,
W = 0, we obtain the Lagrangian
L
=
(39.8)
and the equation of motion
mdv/dt = -
(39.9)
The energy of the particle in this case is obtained by substituting
p =
(39.10)
in E = p.v-L, which gives
E =
(39.11)
It should be noticed that the energy contains no term linear in the velocity.
The rotation of the frame simply adds to the energy a term depending only
on the co-ordinates of the particle and proportional to the square of the
angular velocity. This additional term - 1m(Sxr)2 is called the centrifugal
potential energy.
The velocity V of the particle relative to the uniformly rotating frame of
reference is related to its velocity V0 relative to the inertial frame Ko by
(39.12)
§39
Motion in a non-inertial frame of reference
129
The momentum p (39.10) of the particle in the frame K is therefore the same
as its momentum Po = MVO in the frame K0. The angular momenta
M = rxpo and M = rxp are likewise equal. The energies of the particle
in the two frames are not the same, however. Substituting V from (39.12) in
(39.11), we obtain E = 1mv02-mvo Sxr+U = 1mvo2 + mrxvo S.
The first two terms are the energy E0 in the frame K0. Using the angular
momentum M, we have
E = E0 n-M.S.
(39.13)
This formula gives the law of transformation of energy when we change to a
uniformly rotating frame. Although it has been derived for a single particle,
the derivation can evidently be generalised immediately to any system of
particles, and the same formula (39.13) is obtained.
PROBLEMS
PROBLEM 1. Find the deflection of a freely falling body from the vertical caused by the
Earth's rotation, assuming the angular velocity of this rotation to be small.
SOLUTION. In a gravitational field U = -mg. r, where g is the gravity acceleration
vector; neglecting the centrifugal force in equation (39.9) as containing the square of S, we
have the equation of motion
v = 2vxSu+g.
(1)
This equation may be solved by successive approximations. To do so, we put V = V1+V2,
where V1 is the solution of the equation V1 = g, i.e. V1 = gt+ (Vo being the initial velocity).
Substituting V = V1+v2in (1) and retaining only V1 on the right, we have for V2 the equation
V2 = 2v1xSc = 2tgxSt+2voxS. Integration gives
(2)
where h is the initial radius vector of the particle.
Let the z-axis be vertically upwards, and the x-axis towards the pole; then gx = gy = 0,
n sin 1, where A is the latitude (which for definite-
ness we take to be north). Putting V0 = 0 in (2), we find x = 0, =-1t300 cos A. Substitu-
tion of the time of fall t 22 (2h/g) gives finally x = 0,3 = - 1(2h/g)3/2 cos A, the negative
value indicating an eastward deflection.
PROBLEM 2. Determine the deflection from coplanarity of the path of a particle thrown
from the Earth's surface with velocity Vo.
SOLUTION. Let the xx-plane be such as to contain the velocity Vo. The initial altitude
h = 0. The lateral deviation is given by (2), Problem 1: y =
or, substituting the time of flight t 22 2voz/g, y =
PROBLEM 3. Determine the effect of the Earth's rotation on small oscillations of a pendulum
(the problem of Foucault's pendulum).
SOLUTION. Neglecting the vertical displacement of the pendulum, as being a quantity
of the second order of smallness, we can regard the motion as taking place in the horizontal
xy-plane. Omitting terms in N°, we have the equations of motion x+w2x = 20zy, j+w2y
= -20zx, where w is the frequency of oscillation of the pendulum if the Earth's rotation is
neglected. Multiplying the second equation by i and adding, we obtain a single equation
130
Motion of a Rigid Body
§39
+2i02s+w28 = 0 for the complex quantity $ = xtiy. For I2<<, the solution of this
equation is
$ = exp(-is2t) [A1 exp(iwt) +A2 exp(-iwt)]
or
xtiy = (xo+iyo) exp(-is2zt),
where the functions xo(t), yo(t) give the path of the pendulum when the Earth's rotation is
neglected. The effect of this rotation is therefore to turn the path about the vertical with
angular velocity Qz.
CHAPTER VII
THE CANONICAL EQUATIONS

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THE formulation of the laws of mechanics in terms of the Lagrangian, and
of Lagrange's equations derived from it, presupposes that the mechanical
state of a system is described by specifying its generalised co-ordinates and
velocities. This is not the only possible mode of description, however. A
number of advantages, especially in the study of certain general problems of
mechanics, attach to a description in terms of the generalised co-ordinates
and momenta of the system. The question therefore arises of the form of
the equations of motion corresponding to that formulation of mechanics.
The passage from one set of independent variables to another can be
effected by means of what is called in mathematics Legendre's transformation.
In the present case this transformation is as follows. The total differential
of the Lagrangian as a function of co-ordinates and velocities is
dL =
This expression may be written
(40.1)
since the derivatives aL/dqi are, by definition, the generalised momenta, and
aL/dqi = pi by Lagrange's equations. Writing the second term in (40.1) as
= - Eqi dpi, taking the differential d(piqi) to the left-hand
side, and reversing the signs, we obtain from (40.1)
The argument of the differential is the energy of the system (cf. §6);
expressed in terms of co-ordinates and momenta, it is called the Hamilton's
function or Hamiltonian of the system:
(40.2)
t The reader may find useful the following table showing certain differences between the
nomenclature used in this book and that which is generally used in the English literature.
Here
Elsewhere
Principle of least action
Hamilton's principle
Maupertuis' principle
Principle of least action
Maupertuis' principle
Action
Hamilton's principal function
Abbreviated action
Action
- -Translators.
131
132
The Canonical Equations
§40
From the equation in differentials
dH =
(40.3)
in which the independent variables are the co-ordinates and momenta, we
have the equations
=
(40.4)
These are the required equations of motion in the variables P and q, and
are called Hamilton's equations. They form a set of 2s first-order differential
equations for the 2s unknown functions Pi(t) and qi(t), replacing the S second-
order equations in the Lagrangian treatment. Because of their simplicity and
symmetry of form, they are also called canonical equations.
The total time derivative of the Hamiltonian is
Substitution of qi and pi from equations (40.4) shows that the last two terms
cancel, and so
dH/dt==Hoo.
(40.5)
In particular, if the Hamiltonian does not depend explicitly on time, then
dH/dt = 0, and we have the law of conservation of energy.
As well as the dynamical variables q, q or q, P, the Lagrangian and the
Hamiltonian involve various parameters which relate to the properties of the
mechanical system itself, or to the external forces on it. Let A be one such
parameter. Regarding it as a variable, we have instead of (40.1)
dL
and (40.3) becomes
dH =
Hence
(40.6)
which relates the derivatives of the Lagrangian and the Hamiltonian with
respect to the parameter A. The suffixes to the derivatives show the quantities
which are to be kept constant in the differentiation.
This result can be put in another way. Let the Lagrangian be of the form
L = Lo + L', where L' is a small correction to the function Lo. Then the
corresponding addition H' in the Hamiltonian H = H + H' is related to L'
by
(H')p,a - (L')
(40.7)
It may be noticed that, in transforming (40.1) into (40.3), we did not
include a term in dt to take account of a possible explicit time-dependence

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The Routhian
133
of the Lagrangian, since the time would there be only a parameter which
would not be involved in the transformation. Analogously to formula (40.6),
the partial time derivatives of L and H are related by
(40.8)
PROBLEMS
PROBLEM 1. Find the Hamiltonian for a single particle in Cartesian, cylindrical and
spherical co-ordinates.
SOLUTION. In Cartesian co-ordinates x, y, 2,
in cylindrical co-ordinates r, , z,
in spherical co-ordinates r, 0, ,
PROBLEM 2. Find the Hamiltonian for a particle in a uniformly rotating frame of reference.
SOLUTION. Expressing the velocity V in the energy (39.11) in terms of the momentum p
by (39.10), we have H = p2/2m-S rxp+U.
PROBLEM 3. Find the Hamiltonian for a system comprising one particle of mass M and n
particles each of mass m, excluding the motion of the centre of mass (see §13, Problem).
SOLUTION. The energy E is obtained from the Lagrangian found in §13, Problem, by
changing the sign of U. The generalised momenta are
Pa = OL/OV
Hence
-
= (mM/14)
=
=
Substitution in E gives
41. The Routhian
In some cases it is convenient, in changing to new variables, to replace
only some, and not all, of the generalised velocities by momenta. The trans-
formation is entirely similar to that given in 40.
To simplify the formulae, let us at first suppose that there are only two
co-ordinates q and E, say, and transform from the variables q, $, q, $ to
q, $, p, & where P is the generalised momentum corresponding to the co-
ordinate q.
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The Canonical Equations

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The differential of the Lagrangian L(q, $, q, §) is
dL = dq + (al/dg) ds (0L/as) d
d,
whence
= (0L/d) d.
If we define the Routhian as
= pq-L,
(41.1)
in which the velocity q is expressed in terms of the momentum P by means
of the equation P = 0L/dq, then its differential is
dR = - ds - (aL/a)
(41.2)
Hence
DRIP, p = OR/dq,
(41.3)
(41.4)
Substituting these equations in the Lagrangian for the co-ordinate $, we have
(41.5)
Thus the Routhian is a Hamiltonian with respect to the co-ordinate q
(equations (41.3)) and a Lagrangian with respect to the co-ordinate $ (equation
(41.5)).
According to the general definition the energy of the system is
E -p-L =
In terms of the Routhian it is
E=R-R,
(41.6)
as we find by substituting (41.1) and (41.4).
The generalisation of the above formulae to the case of several co-ordinates
q and & is evident.
The use of the Routhian may be convenient, in particular, when some of
the co-ordinates are cyclic. If the co-ordinates q are cyclic, they do not appear
in the Lagrangian, nor therefore in the Routhian, so that the latter is a func-
tion of P, $ and $. The momenta P corresponding to cyclic co-ordinates are
constant, as follows also from the second equation (41.3), which in this sense
contains no new information. When the momenta P are replaced by their
given constant values, equations (41.5) (d/dt) JR(p, $, 5)108 = JR(P, &, §) 128
become equations containing only the co-ordinates $, so that the cyclic co-
ordinates are entirely eliminated. If these equations are solved for the func-
tions (t), substitution of the latter on the right-hand sides of the equations
q = JR(p, $, E) gives the functions q(t) by direct integration.
PROBLEM
Find the Routhian for a symmetrical top in an external field U(, 0), eliminating the cyclic
co-ordinate 4 (where 4, , 0 are Eulerian angles).
§42
Poisson brackets
135
SOLUTION. The Lagrangian is = see
§35, Problem 1. The Routhian is
R = cos 0);
the first term is a constant and may be omitted.
42. Poisson brackets
Let f (p, q, t) be some function of co-ordinates, momenta and time. Its
total time derivative is
df
Substitution of the values of and Pk given by Hamilton's equations (40.4)
leads to the expression
(42.1)
where
(42.2)
dqk
This expression is called the Poisson bracket of the quantities H and f.
Those functions of the dynamical variables which remain constant during
the motion of the system are, as we know, called integrals of the motion.
We see from (42.1) that the condition for the quantity f to be an integral of
the motion (df/dt = 0) can be written
af(dt+[H,f]=0
(42.3)
If the integral of the motion is not explicitly dependent on the time, then
[H,f] = 0,
(42.4)
i.e. the Poisson bracket of the integral and the Hamiltonian must be zero.
For any two quantities f and g, the Poisson bracket is defined analogously
to (42.2):
(42.5)
The Poisson bracket has the following properties, which are easily derived
from its definition.
If the two functions are interchanged, the bracket changes sign; if one of
the functions is a constant c, the bracket is zero:
(42.6)
[f,c]=0.
(42.7)
Also
[f1+f2,g]=[f1,g)+[f2,g]
(42.8)
[f1f2,g] ]=fi[fa,8]+f2[f1,8] =
(42.9)
Taking the partial derivative of (42.5) with respect to time, we obtain
(42.10)
136
The Canonical Equations
§42
If one of the functions f and g is one of the momenta or co-ordinates, the
Poisson bracket reduces to a partial derivative:
(42.11)
(42.12)
Formula (42.11), for example, may be obtained by putting g = qk in (42.5);
the sum reduces to a single term, since dqk/dqi = 8kl and dqk/dpi = 0. Put-
ting in (42.11) and (42.12) the function f equal to qi and Pi we have, in parti-
cular,
[qi,qk] = [Pi, Pk] =0, [Pi, 9k] = Sik.
(42.13)
The relation
[f,[g,h]]+[g,[h,f]]+[h,[f,g]] = 0,
(42.14)
known as Jacobi's identity, holds between the Poisson brackets formed from
three functions f, g and h. To prove it, we first note the following result.
According to the definition (42.5), the Poisson bracket [f,g] is a bilinear
homogeneous function of the first derivatives of f and g. Hence the bracket
[h,[f,g]], for example, is a linear homogeneous function of the second
derivatives of f and g. The left-hand side of equation (42.14) is therefore a
linear homogeneous function of the second derivatives of all three functions
f, g and h. Let us collect the terms involving the second derivatives of f.
The first bracket contains no such terms, since it involves only the first
derivatives of f. The sum of the second and third brackets may be symboli-
cally written in terms of the linear differential operators D1 and D2, defined by
D1() = [g, ], D2(b) = [h, ]. Then
3,[h,f]]+[h,[f,g]] = [g, [h,f]]-[h,[g,f]
= D1[D2(f)]-D2[D1(f)]
= (D1D2-D2D1)f.
It is easy to see that this combination of linear differential operators cannot
involve the second derivatives of f. The general form of the linear differential
operators is
where & and Nk are arbitrary functions of the variables .... Then
and the difference of these,
§42
Poisson brackets
137
is again an operator involving only single differentiations. Thus the terms in
the second derivatives of f on the left-hand side of equation (42.14) cancel
and, since the same is of course true of g and h, the whole expression is identi-
cally zero.
An important property of the Poisson bracket is that, if f and g are two
integrals of the motion, their Poisson bracket is likewise an integral of the
motion:
[f,g] = constant. =
(42.15)
This is Poisson's theorem. The proof is very simple if f and g do not depend
explicitly on the time. Putting h = H in Jacobi's identity, we obtain
[H,[f,g]]+[f,[g,H]]+[g,[H,fl]=0.
Hence, if [H, g] =0 and [H,f] = 0, then [H,[f,g]] = 0, which is the
required result.
If the integrals f and g of the motion are explicitly time-dependent, we
put, from (42.1),
Using formula (42.10) and expressing the bracket [H, [f,g]] in terms of two
others by means of Jacobi's identity, we find
d
[
(42.16)
which evidently proves Poisson's theorem.
Of course, Poisson's theorem does not always supply further integrals of
the motion, since there are only 2s-1 - of these (s being the number of degrees
of freedom). In some cases the result is trivial, the Poisson bracket being a
constant. In other cases the integral obtained is simply a function of the ori-
ginal integrals f and g. If neither of these two possibilities occurs, however,
then the Poisson bracket is a further integral of the motion.
PROBLEMS
PROBLEM 1. Determine the Poisson brackets formed from the Cartesian components of
the momentum p and the angular momentum M = rxp of a particle.
SOLUTION. Formula (42.12) gives [Mx, Py] = -MM/Dy = -d(yp:-2py)/dy
=
-Pz,
and similarly [Mx, Px] = 0, [Mx, P2] = Py. The remaining brackets are obtained by cyclically
permuting the suffixes x, y, Z.
6
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PROBLEM 2. Determine the Poisson brackets formed from the components of M.
SOLUTION. A direct calculation from formula (42.5) gives [Mx, My] = -M2, [My, M]
= -Mx, [Mz, Mx] = -My.
Since the momenta and co-ordinates of different particles are mutually independent variables,
it is easy to see that the formulae derived in Problems 1 and 2 are valid also for the total
momentum and angular momentum of any system of particles.
PROBLEM 3. Show that [, M2] = 0, where is any function, spherically symmetrical
about the origin, of the co-ordinates and momentum of a particle.
SOLUTION. Such a function can depend on the components of the vectors r and p only
through the combinations r2, p2, r. p. Hence
and similarly for The required relation may be verified by direct calculation from
formula (42.5), using these formulae for the partial derivatives.
PROBLEM 4. Show that [f, M] = n xf, where f is a vector function of the co-ordinates
and momentum of a particle, and n is a unit vector parallel to the z-axis.
SOLUTION. An arbitrary vector f(r,p) may be written as f = where
01, O2, 03 are scalar functions. The required relation may be verified by direct calculation
from formulae (42.9), (42.11), (42.12) and the formula of Problem 3.
$43. The action as a function of the co-ordinates
In formulating the principle of least action, we have considered the integral
(43.1)
taken along a path between two given positions q(1) and q(2) which the system
occupies at given instants t1 and t2. In varying the action, we compared the
values of this integral for neighbouring paths with the same values of q(t1)
and q(t2). Only one of these paths corresponds to the actual motion, namely
the path for which the integral S has its minimum value.
Let us now consider another aspect of the concept of action, regarding S
as a quantity characterising the motion along the actual path, and compare
the values of S for paths having a common beginning at q(t1) = q(1), but
passing through different points at time t2. In other words, we consider the
action integral for the true path as a function of the co-ordinates at the upper
limit of integration.
The change in the action from one path to a neighbouring path is given
(if there is one degree of freedom) by the expression (2.5):
8S =
Since the paths of actual motion satisfy Lagrange's equations, the integral
in 8S is zero. In the first term we put Sq(t1) = 0, and denote the value of
§43
The action as a function of the co-ordinates
139
8q(t2) by 8q simply. Replacing 0L/dq by p, we have finally 8S = pdq or, in
the general case of any number of degrees of freedom,
ES==Pisqu-
(43.2)
From this relation it follows that the partial derivatives of the action with
respect to the co-ordinates are equal to the corresponding momenta:
=
(43.3)
The action may similarly be regarded as an explicit function of time, by
considering paths starting at a given instant t1 and at a given point q(1), and
ending at a given point q(2) at various times t2 = t. The partial derivative
asiat thus obtained may be found by an appropriate variation of the integral.
It is simpler, however, to use formula (43.3), proceeding as follows.
From the definition of the action, its total time derivative along the path is
dS/dt = L.
(43.4)
Next, regarding S as a function of co-ordinates and time, in the sense des-
cribed above, and using formula (43.3), we have
dS
A comparison gives asid = L- or
(43.5)
Formulae (43.3) and (43.5) may be represented by the expression
(43.6)
for the total differential of the action as a function of co-ordinates and time
at the upper limit of integration in (43.1). Let us now suppose that the co-
ordinates (and time) at the beginning of the motion, as well as at the end,
are variable. It is evident that the corresponding change in S will be given
by the difference of the expressions (43.6) for the beginning and end of the
path, i.e.
dsp
(43.7)
This relation shows that, whatever the external forces on the system during
its motion, its final state cannot be an arbitrary function of its initial state;
only those motions are possible for which the expression on the right-hand
side of equation (43.7) is a perfect differential. Thus the existence of the
principle of least action, quite apart from any particular form of the Lagran-
gian, imposes certain restrictions on the range of possible motions. In parti-
cular, it is possible to derive a number of general properties, independent
of the external fields, for beams of particles diverging from given points in
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space. The study of these properties forms a part of the subject of geometrical
optics.+
It is of interest to note that Hamilton's equations can be formally derived
from the condition of minimum action in the form
(43.8)
which follows from (43.6), if the co-ordinates and momenta are varied inde-
pendently. Again assuming for simplicity that there is only one co-ordinate
and momentum, we write the variation of the action as
= dt - (OH/dp)8p dt].
An integration by parts in the second term gives
At the limits of integration we must put 8q = 0, so that the integrated term
is zero. The remaining expression can be zero only if the two integrands
vanish separately, since the variations Sp and 8q are independent and arbitrary
dq = (OH/OP) dt, dp = - (dH/dq) dt, which, after division by dt, are
Hamilton's equations.
$44. Maupertuis' principle
The motion of a mechanical system is entirely determined by the principle
of least action: by solving the equations of motion which follow from that
principle, we can find both the form of the path and the position on the path
as a function of time.
If the problem is the more restricted one of determining only the path,
without reference to time, a simplified form of the principle of least action
may be used. We assume that the Lagrangian, and therefore the Hamilton-
ian, do not involve the time explicitly, SO that the energy of the system is
conserved: H(p, q) = E = constant. According to the principle of least action,
the variation of the action, for given initial and final co-ordinates and times
(to and t, say), is zero. If, however, we allow a variation of the final time t,
the initial and final co-ordinates remaining fixed, we have (cf.(43.7))
8S = -Hot.
(44.1)
We now compare, not all virtual motions of the system, but only those
which satisfy the law of conservation of energy. For such paths we can
replace H in (44.1) by a constant E, which gives
SS+Est=0.
(44.2)
t See The Classical Theory of Fields, Chapter 7, Pergamon Press, Oxford 1962.
§44
Maupertuis' principle
141
Writing the action in the form (43.8) and again replacing H by E, we have
(44.3)
The first term in this expression,
(44.4)
is sometimes called the abbreviated action.
Substituting (44.3) in (44.2), we find that
8S0=0.
(44.5)
Thus the abbreviated action has a minimum with respect to all paths which
satisfy the law of conservation of energy and pass through the final point
at any instant. In order to use such a variational principle, the momenta
(and so the whole integrand in (44.4)) must be expressed in terms of the
co-ordinates q and their differentials dq. To do this, we use the definition of
momentum:
(44.6)
and the law of conservation of energy:
E(g)
(44.7)
Expressing the differential dt in terms of the co-ordinates q and their differen-
tials dq by means of (44.7) and substituting in (44.6), we have the momenta
in terms of q and dq, with the energy E as a parameter. The variational prin-
ciple so obtained determines the path of the system, and is usually called
Maupertuis' principle, although its precise formulation is due to EULER and
LAGRANGE.
The above calculations may be carried out explicitly when the Lagrangian
takes its usual form (5.5) as the difference of the kinetic and potential energies:
The momenta are
and the energy is
The last equation gives
dt
(44.8)
142
The Canonical Equations
§44
substituting this in
Epides
we find the abbreviated action:
(44.9)
In particular, for a single particle the kinetic energy is T = 1/2 m(dl/dt)2,
where m is the mass of the particle and dl an element of its path; the variational
principle which determines the path is
${/[2m(B-U)]dl=0
(44.10)
where the integral is taken between two given points in space. This form is
due to JACOBI.
In free motion of the particle, U = 0, and (44.10) gives the trivial result
8 I dl = 0, i.e. the particle moves along the shortest path between the two
given points, i.e. in a straight line.
Let us return now to the expression (44.3) for the action and vary it with
respect to the parameter E. We have
substituting in (44.2), we obtain
(44.11)
When the abbreviated action has the form (44.9), this gives
=
(44.12)
which is just the integral of equation (44.8). Together with the equation of
the path, it entirely determines the motion.
PROBLEM
Derive the differential equation of the path from the variational principle (44.10).
SOLUTION. Effecting the variation, we have
f
In the second term we have used the fact that dl2 = dr2 and therefore dl d8l = dr. d&r.
Integrating this term by parts and then equating to zero the coefficient of Sr in the integrand,
we obtain the differential equation of the path:

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Canonical transformations
143
Expanding the derivative on the left-hand side and putting the force F = - auld gives
d2r/dl2=[F-(F.t)t]/2(E-U),
where t = dr/dl is a unit vector tangential to the path. The difference F-(F. t)t is the com-
ponent Fn of the force normal to the path. The derivative d2r/dl2 = dt/dl is known from
differential geometry to be n/R, where R is the radius of curvature of the path and n the unit
vector along the principal normal. Replacing E-U by 1mv2, we have (mv2/R)n = Fn, in
agreement with the familar expression for the normal acceleration in motion in a curved
path.
$45. Canonical transformations
The choice of the generalised co-ordinates q is subject to no restriction;
they may be any S quantities which uniquely define the position of the system
in space. The formal appearance of Lagrange's equations (2.6) does not
depend on this choice, and in that sense the equations may be said to be
invariant with respect to a transformation from the co-ordinates q1, q2,
to any other independent quantities Q1, Q2, The new co-ordinates Q are
functions of q, and we shall assume that they may explicitly depend on the
time, i.e. that the transformation is of the form
Qi=Qi(q,t)
(45.1)
(sometimes called a point transformation).
Since Lagrange's equations are unchanged by the transformation (45.1),
Hamilton's equations (40.4) are also unchanged. The latter equations, how-
ever, in fact allow a much wider range of transformations. This is, of course,
because in the Hamiltonian treatment the momenta P are variables inde-
pendent of and on an equal footing with the co-ordinates q. Hence the trans-
formation may be extended to include all the 2s independent variables P
and q:
Qt=Qi(p,q,t),
Pi = Pi(p, q,t).
(45.2)
This enlargement of the class of possible transformations is one of the im-
portant advantages of the Hamiltonian treatment.
The equations of motion do not, however, retain their canonical form
under all transformations of the form (45.2). Let us derive the conditions
which must be satisfied if the equations of motion in the new variables P, Q
are to be of the form
(45.3)
with some Hamiltonian H'(P,Q). When this happens the transformation is
said to be canonical.
The formulae for canonical transformations can be obtained as follows. It
has been shown at the end of §43 that Hamilton's equations can be derived
from the principle of least action in the form
(45.4)
144
The Canonical Equations
§45
in which the variation is applied to all the co-ordinates and momenta inde-
pendently. If the new variables P and Q also satisfy Hamilton's equations,
the principle of least action
0
(45.5)
must hold. The two forms (45.4) and (45.5) are equivalent only if their inte-
grands are the same apart from the total differential of some function F of
co-ordinates, momenta and time.t The difference between the two integrals
is then a constant, namely the difference of the values of F at the limits of
integration, which does not affect the variation. Thus we must have
=
Each canonical transformation is characterised by a particular function F,
called the generating function of the transformation.
Writing this relation as
(45.6)
we see that
Pi = 0F/dqi, =-0F/JQi H' = H+0F/dt;
(45.7)
here it is assumed that the generating function is given as a function of the
old and new co-ordinates and the time: F = F(q, Q, t). When F is known,
formulae (45.7) give the relation between p, q and P, Q as well as the new
Hamiltonian.
It may be convenient to express the generating function not in terms of the
variables q and Q but in terms of the old co-ordinates q and the new momenta
P. To derive the formulae for canonical transformations in this case, we must
effect the appropriate Legendre's transformation in (45.6), rewriting it as
=
The argument of the differential on the left-hand side, expressed in terms of
the variables q and P, is a new generating function (q, P, t), say. Thent
= Qi = ID/OPi, H' = H+d
(45.8)
We can similarly obtain the formulae for canonical transformations in-
volving generating functions which depend on the variables P and Q, or
p and P.
t We do not consider such trivial transformations as Pi = api, Qi = qt,H' = aH, with a an
arbitrary constant, whereby the integrands in (45.4) and (45.5) differ only by a constant
factor.
+ If the generating function is = fi(q, t)Pi, where the ft are arbitrary functions, we
obtain a transformation in which the new co-ordinates are Q = fi(q, t), i.e. are expressed
in terms of the old co-ordinates only (and not the momenta). This is a point transformation,
and is of course a particular canonical transformation.
§45
Canonical transformations
145
The relation between the two Hamiltonians is always of the same form:
the difference H' - H is the partial derivative of the generating function with
respect to time. In particular, if the generating function is independent of
time, then H' = H, i.e. the new Hamiltonian is obtained by simply substitut-
ing for P, q in H their values in terms of the new variables P, Q.
The wide range of the canonical transformations in the Hamiltonian treat-
ment deprives the generalised co-ordinates and momenta of a considerable
part of their original meaning. Since the transformations (45.2) relate each
of the quantities P, Q to both the co-ordinates q and the momenta P, the
variables Q are no longer purely spatial co-ordinates, and the distinction
between Q and P becomes essentially one of nomenclature. This is very
clearly seen, for example, from the transformation Q = Pi, Pi = -qi,
which obviously does not affect the canonical form of the equations and
amounts simply to calling the co-ordinates momenta and vice versa.
On account of this arbitrariness of nomenclature, the variables P and q in
the Hamiltonian treatment are often called simply canonically conjugate
quantities. The conditions relating such quantities can be expressed in terms
of Poisson brackets. To do this, we shall first prove a general theorem on the
invariance of Poisson brackets with respect to canonical transformations.
Let [f,g]p,a be the Poisson bracket, for two quantities f and g, in which
the differentiation is with respect to the variables P and q, and [f,g]p,Q that
in which the differentiation is with respect to P and Q. Then
(45.9)
The truth of this statement can be seen by direct calculation, using the for-
mulae of the canonical transformation. It can also be demonstrated by the
following argument.
First of all, it may be noticed that the time appears as a parameter in the
canonical transformations (45.7) and (45.8). It is therefore sufficient to prove
(45.9) for quantities which do not depend explicitly on time. Let us now
formally regard g as the Hamiltonian of some fictitious system. Then, by
formula (42.1), [f,g]p,a = df/dt. The derivative df/dt can depend only on
the properties of the motion of the fictitious system, and not on the particular
choice of variables. Hence the Poisson bracket [f,g] is unaltered by the
passage from one set of canonical variables to another.
Formulae (42.13) and (45.9) give
[Qi, Qk]p,a = 0, [Pi,Pk]p,a = 0,
(45.10)
These are the conditions, written in terms of Poisson brackets, which must
be satisfied by the new variables if the transformation P, q P, Q is canonical.
It is of interest to observe that the change in the quantities P, q during the
motion may itself be regarded as a series of canonical transformations. The
meaning of this statement is as follows. Let qt, Pt be the values of the canonical
t Whose generating function is
6*
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The Canonical Equations

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variables at time t, and qt+r, Pt+r their values at another time t +T. The latter
are some functions of the former (and involve T as a parameter):
If these formulae are regarded as a transformation from the variables Qt, Pt
to qt+r, Pttr, then this transformation is canonical. This is evident from the
expression ds = for the differential of the action S(qt++,
qt) taken along the true path, passing through the points qt and qt++ at given
times t and t + T (cf. (43.7)). A comparison of this formula with (45.6) shows
that - S is the generating function of the transformation.
46. Liouville's theorem
For the geometrical interpretation of mechanical phenomena, use is often
made of phase space. This is a space of 2s dimensions, whose co-ordinate axes
correspond to the S generalised co-ordinates and S momenta of the system
concerned. Each point in phase space corresponds to a definite state of the
system. When the system moves, the point representing it describes a curve
called the phase path.
The product of differentials dT = dq1 ... dqsdp1 dps may be regarded
as an element of volume in phase space. Let us now consider the integral
I dT taken over some region of phase space, and representing the volume of
that region. We shall show that this integral is invariant with respect to
canonical transformations; that is, if the variables P, q are replaced by
P, Q by a canonical transformation, then the volumes of the corresponding
regions of the spaces of P, and P, Q are equal:
...dqsdp1...dps =
(46.1)
The transformation of variables in a multiple integral is effected by the
formula I .jdQ1...dQsdP1...dPz = S... I Ddq1 dp1...dps,
where
(46.2)
is the Jacobian of the transformation. The proof of (46.1) therefore amounts
to proving that the Jacobian of every canonical transformation is unity:
D=1.
(46.3)
We shall use a well-known property of Jacobians whereby they can be
treated somewhat like fractions. "Dividing numerator and denominator" by
0(91, ..., qs, P1, Ps), we obtain
Another property of Jacobians is that, when the same quantities appear in
both the partial differentials, the Jacobian reduces to one in fewer variables,

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The Hamilton-Jacobi equation
147
in which these repeated quantities are regarded as constant in carrying out
the differentiations. Hence
(46.4)
P=constant
q=constant
The Jacobian in the numerator is, by definition, a determinant of order s
whose element in the ith row and kth column is Representing the
canonical transformation in terms of the generating function (q, P) as in
(45.8), we have = In the same way we find that the
ik-element of the determinant in the denominator of (46.4) is
This means that the two determinants differ only by the interchange of rows
and columns; they are therefore equal, so that the ratio (46.4) is equal to
unity. This completes the proof.
Let us now suppose that each point in the region of phase space considered
moves in the course of time in accordance with the equations of motion of the
mechanical system. The region as a whole therefore moves also, but its volume
remains unchanged:
f dr = constant.
(46.5)
This result, known as Liouville's theorem, follows at once from the invariance
of the volume in phase space under canonical transformations and from the
fact that the change in p and q during the motion may, as we showed at the end
of §45, be regarded as a canonical transformation.
In an entirely similar manner the integrals
11 2 dae dph
,
in which the integration is over manifolds of two, four, etc. dimensions in
phase space, may be shown to be invariant.
47. The Hamilton-Jacobi equation
In §43 the action has been considered as a function of co-ordinates and
time, and it has been shown that the partial derivative with respect to time
of this function S(q, t) is related to the Hamiltonian by
and its partial derivatives with respect to the co-ordinates are the momenta.
Accordingly replacing the momenta P in the Hamiltonian by the derivatives
as/aq, we have the equation
(47.1)
which must be satisfied by the function S(q, t). This first-order partial
differential equation is called the Hamilton-Jacobi equation.
148
The Canonical Equations
§47
Like Lagrange's equations and the canonical equations, the Hamilton-
Jacobi equation is the basis of a general method of integrating the equations
of motion.
Before describing this method, we should recall the fact that every first-
order partial differential equation has a solution depending on an arbitrary
function; such a solution is called the general integral of the equation. In
mechanical applications, the general integral of the Hamilton-Jacobi equation
is less important than a complete integral, which contains as many independent
arbitrary constants as there are independent variables.
The independent variables in the Hamilton-Jacobi equation are the time
and the co-ordinates. For a system with s degrees of freedom, therefore, a
complete integral of this equation must contain s+1 arbitrary constants.
Since the function S enters the equation only through its derivatives, one
of these constants is additive, so that a complete integral of the Hamilton-
Jacobi equation is
Sft,q,saas)+
(47.2)
where X1, ..., as and A are arbitrary constants.
Let us now ascertain the relation between a complete integral of the
Hamilton-Jacobi equation and the solution of the equations of motion which
is of interest. To do this, we effect a canonical transformation from the
variables q, P to new variables, taking the function f (t, q; a) as the
generating function, and the quantities a1, A2, ..., as as the new momenta.
Let the new co-ordinates be B1, B2, ..., Bs. Since the generating function
depends on the old co-ordinates and the new momenta, we use formulae
(45.8): Pi = af/dqi, Bi = af/dar, H' = H+dfdd. But since the function f
satisfies the Hamilton-Jacobi equation, we see that the new Hamiltonian is
zero: H' = H+af/dt = H+as/t = 0. Hence the canonical equations in
the new variables are di = 0, Bi = 0, whence
ay=constant,
Bi = constant.
(47.3)
By means of the S equations af/dai = Bi, the S co-ordinates q can be expressed
in terms of the time and the 2s constants a and B. This gives the general
integral of the equations of motion.
t Although the general integral of the Hamilton-Jacobi equation is not needed here, we
may show how it can be found from a complete integral. To do this, we regard A as an arbi-
trary function of the remaining constants: S = f(t, q1, ..., q8; a1, as) +A(a1, as). Re-
placing the Ai by functions of co-ordinates and time given by the S conditions asidar = 0,
we obtain the general integral in terms of the arbitrary function A(a1,..., as). For, when the
function S is obtained in this manner, we have
as
The quantities (as/dqs)a satisfy the Hamilton-Jacobi equation, since the function S(t, q; a)
is assumed to be a complete integral of that equation. The quantities asida therefore satisfy
the same equation.

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Separation of the variables
149
Thus the solution of the problem of the motion of a mechanical system by
the Hamilton-Jacobi method proceeds as follows. From the Hamiltonian,
we form the Hamilton-Jacobi equation, and find its complete integral (47.2).
Differentiating this with respect to the arbitrary constants a and equating
the derivatives to new constants B, we obtain S algebraic equations
asidar=Bt,
(47.4)
whose solution gives the co-ordinates q as functions of time and of the 2s
arbitrary constants. The momenta as functions of time may then be found
from the equations Pi = aslaqi.
If we have an incomplete integral of the Hamilton-Jacobi equation, depend-
ing on fewer than S arbitrary constants, it cannot give the general integral
of the equations of motion, but it can be used to simplify the finding of the
general integral. For example, if a function S involving one arbitrary con-
stant a is known, the relation asida = constant gives one equation between
q1, ..., qs and t.
The Hamilton-Jacobi equation takes a somewhat simpler form if the func-
tion H does not involve the time explicitly, i.e. if the system is conservative.
The time-dependence of the action is given by a term -Et:
S = So(g)-Et
(47.5)
(see 44), and substitution in (47.1) gives for the abbreviated action So(q)
the Hamilton-Jacobi equation in the form
(47.6)
$48. Separation of the variables
In a number of important cases, a complete integral of the Hamilton-
Jacobi equation can be found by "separating the variables", a name given to
the following method.
Let us assume that some co-ordinate, q1 say, and the corresponding
derivative asia appear in the Hamilton-Jacobi equation only in some
combination (q1, which does not involve the other co-ordinates, time,
or derivatives, i.e. the equation is of the form
(48.1)
where qi denotes all the co-ordinates except q1.
We seek a solution in the form of a sum:
(48.2)
150
The Canonical Equations
§48
substituting this in equation (48.1), we obtain
(48.3)
Let us suppose that the solution (48.2) has been found. Then, when it is
substituted in equation (48.3), the latter must become an identity, valid (in
particular) for any value of the co-ordinate q1. When q1 changes, only the
function is affected, and so, if equation (48.3) is an identity, must be a
constant. Thus equation (48.3) gives the two equations
(48.4)
= 0,
(48.5)
where a1 is an arbitrary constant. The first of these is an ordinary differential
equation, and the function S1(q1) is obtained from it by simple integration.
The remaining partial differential equation (48.5) involves fewer independent
variables.
If we can successively separate in this way all the S co-ordinates and the
time, the finding of a complete integral of the Hamilton-Jacobi equation is
reduced to quadratures. For a conservative system we have in practice to
separate only S variables (the co-ordinates) in equation (47.6), and when this
separation is complete the required integral is
(48.6)
where each of the functions Sk depends on only one co-ordinate; the energy
E, as a function of the arbitrary constants A1, As, is obtained by substituting
So = in equation (47.6).
A particular case is the separation of a cyclic variable. A cyclic co-ordinate
q1 does not appear explicitly in the Hamiltonian, nor therefore in the Hamilton-
Jacobi equation. The function (91, reduces to as/da simply, and
from equation (48.4) we have simply S1 = x1q1, so that
(48.7)
The constant a1 is just the constant value of the momentum P1 = asida
corresponding to the cyclic co-ordinate.
The appearance of the time in the term - Et for a conservative system
corresponds to the separation of the "cyclic variable" t.
Thus all the cases previously considered of the simplification of the integra-
tion of the equations of motion by the use of cyclic variables are embraced
by the method of separating the variables in the Hamilton-Jacobi equation.
To those cases are added others in which the variables can be separated even
though they are not cyclic. The Hamilton-Jacobi treatment is consequently
the most powerful method of finding the general integral of the equations of
motion.
§48
Separation of the variables
151
To make the variables separable in the Hamilton-Jacobi equation the
co-ordinates must be appropriately chosen. We shall consider some examples
of separating the variables in different co-ordinates, which may be of
physical interest in connection with problems of the motion of a particle in
various external fields.
(1) Spherical co-ordinates. In these co-ordinates (r, 0, ), the Hamiltonian is
and the variables can be separated if
U
=
where a(r), b(a), c(b) are arbitrary functions. The last term in this expression
for U is unlikely to be of physical interest, and we shall therefore take
U = a(r)+b(8)/r2.
(48.8)
In this case the Hamilton-Jacobi equation for the function So is
1
Since the co-ordinate is cyclic, we seek a solution in the form So
Pot+S1(T)+S2(9), obtaining for the functions S1(r) andS 2(0) the equations
(day)
=
E.
Integration gives finally
S = -
(48.9)
The arbitrary constants in (48.9) are Pp, B and E; on differentiating with
respect to these and equating the results to other constants, we have the
general solution of the equations of motion.
(2) Parabolic co-ordinates. The passage from cylindrical co-ordinates
(here denoted by p, o, 2) to parabolic co-ordinates E, N, o is effected by the
formulae
1(-n),pv(En).
(48.10)
The co-ordinates & and n take values from 0 to 00; the surfaces of constant
$ and n are easily seen to be two families of paraboloids of revolution, with
152
The Canonical Equations
§48
the z-axis as the axis of symmetry. The equations (48.10) can also be written,
in terms of
r = =
(48.11)
(i.e. the radius in spherical co-ordinates), as
$ = r++,
= r Z.
(48.12)
Let us now derive the Lagrangian of a particle in the co-ordinates $, n, o.
Differentiating the expressions (48.10) with respect to time and substituting
in the Lagrangian in cylindrical co-ordinates
L =
we obtain
L
=
(48.13)
The = and
the Hamiltonian is
(48.14)
The physically interesting cases of separable variables in these co-ordinates
correspond to a potential energy of the form
(48.15)
The equation for So is
2
=
E.
The cyclic co-ordinate can be separated as a term PoO. Multiplying the equa-
tion by m(s+n) and rearranging, we then have
Putting So = P&O + S2(n), we obtain the two equations
-B,
§48
Separation of the variables
153
integration of which gives finally
S
=
dn.
(48.16)
Here the arbitrary constants are Ps, B and E.
(3) Elliptic co-ordinates. These are E, n, o, defined by
(48.17)
The constant o is a parameter of the transformation. The co-ordinate $ takes
values from 1 to 80, and n from - 1 to + 1. The definitions which are geo-
metrically clearest+ are obtained in terms of the distances r1 and r2 to points
A1 and A2 on the z-axis for which 2 = to: r1 = V[(2-0)2+p2],
r2 = Substitution of (48.17) gives
= o(s-n), r2 = o(+n),
(48.18)
& = (r2+r1)/2o, n = (r2-r1)/2o. =
Transforming the Lagrangian from cylindrical to elliptic co-ordinates, we
find
L
=
(48.19)
The Hamiltonian is therefore
H
=
(48.20)
The physically interesting cases of separable variables correspond to a
potential energy
(48.21)
where a() and b(n) are arbitrary functions. The result of separating the
variables in the Hamilton-Jacobi equation is
S
=
1-n2
t The surfaces of constant $ are the ellipsoids = 1, of which A1 and
A2 are the foci; the surfaces of constant n are the hyperboloids 22/02/2-22/02(1-n2 = 1,
also with foci A1 and A2.
154
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PROBLEMS
PROBLEM 1. Find a complete integral of the Hamilton-Jacobi equation for motion of a
particle in a field U = a/r-Fz (a combination of a uniform field and a Coulomb field).
SOLUTION. The field is of the type (48.15), with a(f)=a1F,b(n)a+Fn2 Formula
(48.16) gives
S
=
with arbitrary constants Po, E,B. The constant B has in this case the significance that the one-
valued function of the co-ordinates and momenta of the particle
B
is conserved. The expression in the brackets is an integral of the motion for a pure Coulomb
field (see $15).
PROBLEM 2. The same as Problem 1, but for a field U = ai/r +az/r2 (the Coulomb field
of two fixed points at a distance 2a apart).
SOLUTION. This field is of the type (48.21), with a($) = (a1+az) /o, = (a1-az)n/o.
From formula (48.22) we find
S
=
The constant B here expresses the conservation of the quantity
B = cos 01+ cos 02),
where M is the total angular momentum of the particle, and 01 and O2 are the angles shown in
Fig. 55.
12
r
The
20
a
FIG. 55
$49. Adiabatic invariants
Let us consider a mechanical system executing a finite motion in one dimen-
sion and characterised by some parameter A which specifies the properties of
the system or of the external field in which it is placed, and let us suppose that
1 varies slowly (adiabatically) with time as the result of some external action;
by a "slow" variation we mean one in which A varies only slightly during the
period T of the motion:
di/dt < A.
(49.1)
§49
Adiabatic invariants
155
Such a system is not closed, and its energy E is not conserved. However, since
A varies only slowly, the rate of change E of the energy is proportional to the
rate of change 1 of the parameter. This means that the energy of the system
behaves as some function of A when the latter varies. In other words, there
is some combination of E and A which remains constant during the motion.
This quantity is called an adiabatic invariant.
Let H(p, q; A) be the Hamiltonian of the system, which depends on the
parameter A. According to formula (40.5), the total time derivative of the
energy of the system is dE/dt = OH/dt = (aH/dx)(d)/dt). In averaging this
equation over the period of the motion, we need not average the second
factor, since A (and therefore i) varies only slowly: dE/dt = (d)/dt)
and in the averaged function 01/01 we can regard only P and q, and not A, as
variable. That is, the averaging is taken over the motion which would occur
if A remained constant.
The averaging may be explicitly written
dE dt
According to Hamilton's equation q = OHOP, or dt = dq - (CH/OP). The
integration with respect to time can therefore be replaced by one with respect
to the co-ordinate, with the period T written as
here the $ sign denotes an integration over the complete range of variation
("there and back") of the co-ordinate during the period. Thus
dq/(HHap)
(49.2)
dt $ dq/(HHdp)
As has already been mentioned, the integrations in this formula must be
taken over the path for a given constant value of A. Along such a path the
Hamiltonian has a constant value E, and the momentum is a definite function
of the variable co-ordinate q and of the two independent constant parameters
E and A. Putting therefore P = p(q; E, 1) and differentiating with respect
to A the equation H(p, q; X) )=E, we have = 0, or
OH/OP ax ap
t If the motion of the system is a rotation, and the co-ordinate q is an angle of rotation ,
the integration with respect to must be taken over a "complete rotation", i.e. from 0 to 2nr.
156
The Canonical Equations
§49
Substituting this in the numerator of (49.2) and writing the integrand in the
denominator as ap/dE, we obtain
dt
(49.3)
dq
or
dt
Finally, this may be written as
dI/dt 0,
(49.4)
where
(49.5)
the integral being taken over the path for given E and A. This shows that, in
the approximation here considered, I remains constant when the parameter A
varies, i.e. I is an adiabatic invariant.
The quantity I is a function of the energy of the system (and of the para-
meter A). The partial derivative with respect to energy is given by 2m DI/DE
= $ (ap/dE) dq (i.e. the integral in the denominator in (49.3)) and is, apart from
a factor 2n, the period of the motion:
(49.6)
The integral (49.5) has a geometrical significance in terms of the phase
path of the system. In the case considered (one degree of freedom), the phase
space reduces to a two-dimensional space (i.e. a plane) with co-ordinates
P, q, and the phase path of a system executing a periodic motion is a closed
curve in the plane. The integral (49.5) taken round this curve is the area
enclosed. It can evidently be written equally well as the line integral
I = - $ q dp/2m and as the area integral I = II dp dq/2m.
As an example, let us determine the adiabatic invariant for a one-dimen-
sional oscillator. The Hamiltonian is H = where w is the
frequency of the oscillator. The equation of the phase path is given by the
law of conservation of energy H(p, q) = E. The path is an ellipse with semi-
axes (2mE) and V(2E/mw2), and its area, divided by 2nr, is
I=E/w.
(49.7)
t It can be shown that, if the function X(t) has no singularities, the difference of I from a
constant value is exponentially small.
§49
Adiabatic invariants
157
The adiabatic invariance of I signifies that, when the parameters of the
oscillator vary slowly, the energy is proportional to the frequency.
The equations of motion of a closed system with constant parameters
may be reformulated in terms of I. Let us effect a canonical transformation
of the variables P and q, taking I as the new "momentum". The generating
function is the abbreviated action So, expressed as a function of q and I. For
So is defined for a given energy of the system; in a closed system, I is a func-
tion of the energy alone, and so So can equally well be written as a function
So(q, I). The partial derivative (So/dq)E is the same as the derivative
( for constant I. Hence
(49.8)
corresponding to the first of the formulae (45.8) for a canonical trans-
formation. The second of these formulae gives the new "co-ordinate",
which we denote by W:
W = aso(q,I)/aI.
(49.9)
The variables I and W are called canonical variables; I is called the action
variable and W the angle variable.
Since the generating function So(q, I) does not depend explicitly on time,
the new Hamiltonian H' is just H expressed in terms of the new variables.
In other words, H' is the energy E(I), expressed as a function of the action
variable. Accordingly, Hamilton's equations in canonical variables are
i = 0,
w = dE(I)/dI.
(49.10)
The first of these shows that I is constant, as it should be; the energy is
constant, and I is so too. From the second equation we see that the angle
variable is a linear function of time:
W = (dE/dI)t + constant.
(49.11)
The action So(q, I) is a many-valued function of the co-ordinate. During
each period this function increases by
(49.12)
as is evident from the formula So = Spdq and the definition (49.5). During
the same time the angle variable therefore increases by
Aw = (S/I) =
(49.13)
t The exactness with which the adiabatic invariant (49.7) is conserved can be determined by
establishing the relation between the coefficients C in the asymptotic (t + 00) expressions
q = re[c exp(iw+t)] for the solution of the oscillator equation of motion q + w2(t) q = 0.
Here the frequency w is a slowly varying function of time, tending to constant limits w as
t
+ 00. The limiting values of I are given in terms of these coefficients by I = tw+/c+l2.
The solution is known from quantum mechanics, on account of the formal resemblance
between the above equation of motion and SCHRODINGER'S equation 4" + k2(x) 4 = 0 for
one-dimensional motion of a particle above a slowly varying (quasi-classical) "potential
barrier". The problem of finding the relation between the asymptotic (x + 00)
expressions
for & is equivalent to that of finding the "reflection coefficient" of the potential barrier; see
Quantum Mechanics, $52, Pergamon Press, Oxford 1965.
This method of determining the exactness of conservation of the adiabatic invariant for an
oscillator is due to L. P. PITAEVSKII. The relevant calculations are given by A. M. DYKHNE,
Soviet Physics JETP 11, 411, 1960. The analysis for the general case of an arbitrary finite
motion in one dimension is given by A.A. SLUTSKIN, Soviet Physics JETP 18, 676, 1964.
158
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as may also be seen directly from formula (49.11) and the expression (49.6)
for the period.
Conversely, if we express q and P, or any one-valued function F(p, q) of
them, in terms of the canonical variables, then they remain unchanged when
W increases by 2nd (with I constant). That is, any one-valued function F(p, q),
when expressed in terms of the canonical variables, is a periodic function of W
with period 2.
$50. General properties of motion in S dimensions
Let us consider a system with any number of degrees of freedom, executing
a motion finite in all the co-ordinates, and assume that the variables can be
completely separated in the Hamilton-Jacobi treatment. This means that,
when the co-ordinates are appropriately chosen, the abbreviated action
can be written in the form
(50.1)
as a sum of functions each depending on only one co-ordinate.
Since the generalised momenta are Pi = aso/dqi = dSi/dqi, each function
Si can be written
(50.2)
These are many-valued functions. Since the motion is finite, each co-ordinate
can take values only in a finite range. When qi varies "there and back" in this
range, the action increases by
(50.3)
where
(50.4)
the integral being taken over the variation of qi just mentioned.
Let us now effect a canonical transformation similar to that used in 49,
for the case of a single degree of freedom. The new variables are "action vari-
ables" Ii and "angle variables"
w(a(q
(50.5)
+ It should be emphasised, however, that this refers to the formal variation of the co-
ordinate qi over the whole possible range of values, not to its variation during the period of
the actual motion as in the case of motion in one dimension. An actual finite motion of a
system with several degrees of freedom not only is not in general periodic as a whole, but
does not even involve a periodic time variation of each co-ordinate separately (see below).
§50
General properties of motion in S dimensions
159
where the generating function is again the action expressed as a function of
the co-ordinates and the Ii. The equations of motion in these variables are
Ii = 0, w = de(I)/I, which give
I=constant,
(50.6)
+ constant.
(50.7)
We also find, analogously to (49.13), that a variation "there and back" of
the co-ordinate qi corresponds to a change of 2n in Wi:
Awi==2m
(50.8)
In other words, the quantities Wi(q, I) are many-valued functions of the co-
ordinates: when the latter vary and return to their original values, the Wi
may vary by any integral multiple of 2. This property may also be formulated
as a property of the function Wi(P, q), expressed in terms of the co-ordinates
and momenta, in the phase space of the system. Since the Ii, expressed in
terms of P and q, are one-valued functions, substitution of Ii(p, q) in wi(q, I)
gives a function wilp, q) which may vary by any integral multiple of 2n
(including zero) on passing round any closed path in phase space.
Hence it follows that any one-valued function F(P, q) of the state of the
system, if expressed in terms of the canonical variables, is a periodic function
of the angle variables, and its period in each variable is 2nr. It can be expanded
as a multiple Fourier series:
(50.9)
ls==
where l1, l2, ls are integers. Substituting the angle variables as functions
of time, we find that the time dependence of F is given by a sum of the form
(50.10)
lg==
Each term in this sum is a periodic function of time, with frequency
(50.11)
Since these frequencies are not in general commensurable, the sum itself is
not a periodic function, nor, in particular, are the co-ordinates q and
momenta P of the system.
Thus the motion of the system is in general not strictly periodic either as a
whole or in any co-ordinate. This means that, having passed through a given
state, the system does not return to that state in a finite time. We can say,
t Rotational co-ordinates (see the first footnote to 49) are not in one-to-one relation
with the state of the system, since the position of the latter is the same for all values of
differing by an integral multiple of 2nr. If the co-ordinates q include such angles, therefore,
these can appear in the function F(P, q) only in such expressions as cos and sin , which
are in one-to-one relation with the state of the system.
160
The Canonical Equations
§50
however, that in the course of a sufficient time the system passes arbitrarily
close to the given state. For this reason such a motion is said to be conditionally
periodic.
In certain particular cases, two or more of the fundamental frequencies
Wi = DE/DI are commensurable for arbitrary values of the Ii. This is called
degeneracy, and if all S frequencies are commensurable, the motion of the
system is said to be completely degenerate. In the latter case the motion is
evidently periodic, and the path of every particle is closed.
The existence of degeneracy leads, first of all, to a reduction in the number
of independent quantities Ii on which the energy of the system depends.
If two frequencies W1 and W2 are such that
(50.12)
where N1 and N2 are integers, then it follows that I1 and I2 appear in the energy
only as the sum n2I1+n1I2.
A very important property of degenerate motion is the increase in the
number of one-valued integrals of the motion over their number for a general
non-degenerate system with the same number of degrees of freedom. In the
latter case, of the 2s-1 integrals of the motion, only s functions of the state
of the system are one-valued; these may be, for example, the S quantities I
The remaining S - 1 integrals may be written as differences
(50.13)
The constancy of these quantities follows immediately from formula (50.7),
but they are not one-valued functions of the state of the system, because the
angle variables are not one-valued.
When there is degeneracy, the situation is different. For example, the rela-
tion (50.12) shows that, although the integral
WIN1-W2N2
(50.14)
is not one-valued, it is so except for the addition of an arbitrary integral
multiple of 2nr. Hence we need only take a trigonometrical function of this
quantity to obtain a further one-valued integral of the motion.
An example of degeneracy is motion in a field U = -a/r (see Problem).
There is consequently a further one-valued integral of the motion (15.17)
peculiar to this field, besides the two (since the motion is two-dimensional)
ordinary one-valued integrals, the angular momentum M and the energy E,
which hold for motion in any central field.
It may also be noted that the existence of further one-valued integrals
leads in turn to another property of degenerate motions: they allow a complete
separation of the variables for several (and not only one+) choices of the co-
t We ignore such trivial changes in the co-ordinates as q1' = q1'(q1), q2' = 92'(92).
§50
General properties of motion in S dimensions
161
ordinates. For the quantities Ii are one-valued integrals of the motion in
co-ordinates which allow separation of the variables. When degeneracy occurs,
the number of one-valued integrals exceeds S, and so the choice of those
which are the desired I is no longer unique.
As an example, we may again mention Keplerian motion, which allows
separation of the variables in both spherical and parabolic co-ordinates.
In §49 it has been shown that, for finite motion in one dimension, the
action variable is an adiabatic invariant. This statement holds also for systems
with more than one degree of freedom. Here we shall give a proof valid
for the general case.
Let X(t) be again a slowly varying parameter of the system. In the canonical
transformation from the variables P, q to I, W, the generating function is, as we
know, the action So(q, I). This depends on A as a parameter and, if A is a func-
tion of time, the function So(q, I; X(t)) depends explicitly on time. In such a
case the new Hamiltonian H' is not the same as H, i.e. the energy E(I), and
by the general formulae (45.8) for the canonical transformation we have
H' E(I)+asoldt = E(I)+A, where A III (aso/ad)r. Hamilton's equations
give
ig = -
(50.15)
We average this equation over a time large compared with the fundamental
periods of the system but small compared with the time during which the
parameter A varies appreciably. Because of the latter condition we need not
average 1 on the right-hand side, and in averaging the quantities we
may regard the motion of the system as taking place at a constant value of A
and therefore as having the properties of conditionally periodic motion
described above.
The action So is not a one-valued function of the co-ordinates: when q
returns to its initial value, So increases by an integral multiple of 2I. The
derivative A = (aso/ax), is a one-valued function, since the differentiation
is effected for constant Ii, and there is therefore no increase in So. Hence A,
expressed as a function of the angle variables Wr, is periodic. The mean value
of the derivatives of such a function is zero, and therefore by (50.15)
we have also
which shows that the quantities Ii are adiabatic invariants.
Finally, we may briefly discuss the properties of finite motion of closed
systems with S degrees of freedom in the most general case, where the vari-
ables in the Hamilton-Jacobi equation are not assumed to be separable.
The fundamental property of systems with separable variables is that the
integrals of the motion Ii, whose number is equal to the number of degrees
+ To simplify the formulae we assume that there is only one such parameter, but the proof
is valid for any number.
162
The Canonical Equations
§50
of freedom, are one-valued. In the general case where the variables are not
separable, however, the one-valued integrals of the motion include only
those whose constancy is derived from the homogeneity and isotropy of space
and time, namely energy, momentum and angular momentum.
The phase path of the system traverses those regions of phase space which
are defined by the given constant values of the one-valued integrals of the
motion. For a system with separable variables and S one-valued integrals,
these conditions define an s-dimensional manifold (hypersurface) in phase
space. During a sufficient time, the path of the system passes arbitrarily close
to every point on this hypersurface.
In a system where the variables are not separable, however, the number
of one-valued integrals is less than S, and the phase path occupies, completely
or partly, a manifold of more than S dimensions in phase space.
In degenerate systems, on the other hand, which have more than S integrals
of the motion, the phase path occupies a manifold of fewer than S dimensions.
If the Hamiltonian of the system differs only by small terms from one which
allows separation of the variables, then the properties of the motion are close
to those of a conditionally periodic motion, and the difference between the
two is of a much higher order of smallness than that of the additional terms in
the Hamiltonian.
PROBLEM
Calculate the action variables for elliptic motion in a field U = -a/r.
SOLUTION. In polar co-ordinates r, in the plane of the motion we have
'max
= 1+av(m2)E)
Hence the energy, expressed in terms of the action variables, is E = It
depends only on the sum Ir+I, and the motion is therefore degenerate; the two funda-
mental frequencies (in r and in b) coincide.
The parameters P and e of the orbit (see (15.4)) are related to Ir and I by
p=
Since Ir and I are adiabatic invariants, when the coefficient a or the mass m varies slowly
the eccentricity of the orbit remains unchanged, while its dimensions vary in inverse propor-
tion to a and to m.
INDEX
Acceleration, 1
Coriolis force, 128
Action, 2, 138ff.
Couple, 109
abbreviated, 141
Cross-section, effective, for scattering,
variable, 157
49ff.
Additivity of
C system, 41
angular momentum, 19
Cyclic co-ordinates, 30
energy, 14
integrals of the motion, 13
d'Alembert's principle, 124
Lagrangians, 4
Damped oscillations, 74ff.
mass, 17
Damping
momentum, 15
aperiodic, 76
Adiabatic invariants, 155, 161
coefficient, 75
Amplitude, 59
decrement, 75
complex, 59
Degeneracy, 39, 69, 160f.
Angle variable, 157
complete, 160
Angular momentum, 19ff.
Degrees of freedom, 1
of rigid body, 105ff.
Disintegration of particles, 41ff.
Angular velocity, 97f.
Dispersion-type absorption, 79
Area integral, 31n.
Dissipative function, 76f.
Dummy suffix, 99n.
Beats, 63
Brackets, Poisson, 135ff.
Eccentricity, 36
Eigenfrequencies, 67
Canonical equations (VII), 131ff.
Elastic collision, 44
Canonical transformation, 143ff.
Elliptic functions, 118f.
Canonical variables, 157
Elliptic integrals, 26, 118
Canonically conjugate quantities, 145
Energy, 14, 25f.
Central field, 21, 30
centrifugal, 32, 128
motion in, 30ff.
internal, 17
Centrally symmetric field, 21
kinetic, see Kinetic energy
Centre of field, 21
potential, see Potential energy
Centre of mass, 17
Equations of motion (I), 1ff.
system, 41
canonical (VII), 131ff.
Centrifugal force, 128
integration of (III), 25ff.
Centrifugal potential, 32, 128
of rigid body, 107ff.
Characteristic equation, 67
Eulerian angles, 110ff.
Characteristic frequencies, 67
Euler's equations, 115, 119
Closed system, 8
Collisions between particles (IV), 41ff.
Finite motion, 25
elastic, 44ff.
Force, 9
Combination frequencies, 85
generalised, 16
Complete integral, 148
Foucault's pendulum, 129f.
Conditionally periodic motion, 160
Frame of reference, 4
Conservation laws (II), 13ff.
inertial, 5f.
Conservative systems, 14
non-inertial, 126ff.
Conserved quantities, 13
Freedom, degrees of, 1
Constraints, 10
Frequency, 59
equations of, 123
circular, 59
holonomic, 123
combination, 85
Co-ordinates, 1
Friction, 75, 122
cyclic, 30
generalised, 1ff.
Galilean transformation, 6
normal, 68f.
Galileo's relativity principle, 6
163
164
Index
General integral, 148
Mechanical similarity, 22ff.
Generalised
Molecules, vibrations of, 70ff.
co-ordinates, 1ff.
Moment
forces, 16
of force, 108
momenta, 16
of inertia, 99ff.
velocities, 1ff.
principal, 100ff.
Generating function, 144
Momentum, 15f.
angular, see Angular momentum
Half-width, 79
generalised, 16
Hamiltonian, 131f.
moment of, see Angular momentum
Hamilton-Jacobi equation, 147ff.
Multi-dimensional motion, 158ff.
Hamilton's equations, 132
Hamilton's function, 131
Hamilton's principle, 2ff.
Newton's equations, 9
Holonomic constraint, 123
Newton's third law, 16
Nodes, line of, 110
Impact parameter, 48
Non-holonomic constraint, 123
Inertia
Normal co-ordinates, 68f.
law of, 5
Normal oscillations, 68
moments of, 99ff.
Nutation, 113
principal, 100ff.
principal axes of, 100
One-dimensional motion, 25ff., 58ff.
tensor, 99
Oscillations, see Small oscillations
Inertial frames, 5f.
Oscillator
Infinite motion, 25
one-dimensional, 58n.
Instantaneous axis, 98
space, 32, 70
Integrals of the motion, 13, 135
Jacobi's identity, 136
Particle, 1
Pendulums, 11f., 26, 33ff., 61, 70, 95,
Kepler's problem, 35ff.
102f., 129f.
Kepler's second law, 31
compound, 102f.
Kepler's third law, 23
conical, 34
Kinetic energy, 8, 15
Foucault's, 129f.
of rigid body, 98f.
spherical, 33f.
Perihelion, 36
Laboratory system, 41
movement of, 40
Lagrange's equations, 3f.
Phase, 59
Lagrangian, 2ff.
path, 146
for free motion, 5
space, 146
of free particle, 6ff.
Point transformation, 143
in non-inertial frame, 127
Poisson brackets, 135ff.
for one-dimensional motion, 25, 58
Poisson's theorem, 137
of rigid body, 99
Polhodes, 117n.
for small oscillations, 58, 61, 66, 69, 84
Potential energy, 8, 15
of system of particles, 8ff.
centrifugal, 32, 128
of two bodies, 29
effective, 32, 94
Latus rectum, 36
from period of oscillation, 27ff.
Least action, principle of, 2ff.
Potential well, 26, 54f.
Legendre's transformation, 131
Precession, regular, 107
Liouville's theorem, 147
L system, 41
Rapidly oscillating field, motion in, 93ff.
Reactions, 122
Mass, 7
Reduced mass, 29
additivity of, 17
Resonance, 62, 79
centre of, 17
in non-linear oscillations, 87ff.
reduced, 29
parametric, 80ff.
Mathieu's equation, 82n.
Rest, system at, 17
Maupertuis' principle, 141
Reversibility of motion, 9
Index
165
Rigid bodies, 96
Space
angular momentum of, 105ff.
homogeneity of, 5, 15
in contact, 122ff.
isotropy of, 5, 18
equations of motion of, 107ff.
Space oscillator, 32, 70
motion of (VI), 96ff.
Rolling, 122
Time
Rotator, 101, 106
homogeneity of, 5, 13ff.
Rough surface, 122
isotropy of, 8f.
Routhian, 134f.
Top
Rutherford's formula, 53f.
asymmetrical, 100, 116ff.
"fast", 113f.
spherical, 100, 106
Scattering, 48ff.
symmetrical, 100, 106f., 111f.
cross-section, effective, 49ff.
Torque, 108
Rutherford's formula for, 53f.
Turning points, 25, 32
small-angle, 55ff.
Two-body problem, 29
Sectorial velocity, 31
Separation of variables, 149ff.
Uniform field, 10
Similarity, mechanical, 22ff.
Sliding, 122
Variation, 2, 3
Small oscillations, 22, (V) 58ff.
first, 3
anharmonic, 84ff.
Velocity, 1
damped, 74ff.
angular, 97f.
forced, 61ff., 77ff.
sectorial, 31
free, 58ff., 65ff.
translational, 97
linear, 84
Virial, 23n.
non-linear, 84ff.
theorem, 23f.
normal, 68
Smooth surface, 122
Well, potential, 26, 54f.
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§14
Motion in a central field
31
(Fig. 8). Calling this area df, we can write the angular momentum of the par-
ticle as
M = 2mf,
(14.3)
where the derivative f is called the sectorial velocity. Hence the conservation
of angular momentum implies the constancy of the sectorial velocity: in equal
times the radius vector of the particle sweeps out equal areas (Kepler's second
law).t
rdd
dd
0
FIG. 8
The complete solution of the problem of the motion of a particle in a central
field is most simply obtained by starting from the laws of conservation of
energy and angular momentum, without writing out the equations of motion
themselves. Expressing in terms of M from (14.2) and substituting in the
expression for the energy, we obtain
E = =
(14.4)
Hence
(14.5)
or, integrating,
constant.
(14.6)
Writing (14.2) as do = M dt/mr2, substituting dt from (14.5) and integrating,
we find
constant.
(14.7)
Formulae (14.6) and (14.7) give the general solution of the problem. The
latter formula gives the relation between r and , i.e. the equation of the path.
Formula (14.6) gives the distance r from the centre as an implicit function of
time. The angle o, it should be noted, always varies monotonically with time,
since (14.2) shows that & can never change sign.
t The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.
32
Integration of the Equations of Motion
§14
The expression (14.4) shows that the radial part of the motion can be re-
garded as taking place in one dimension in a field where the "effective poten-
tial energy" is
(14.8)
The quantity M2/2mr2 is called the centrifugal energy. The values of r for which
U(r)
(14.9)
determine the limits of the motion as regards distance from the centre.
When equation (14.9) is satisfied, the radial velocity j is zero. This does not
mean that the particle comes to rest as in true one-dimensional motion, since
the angular velocity o is not zero. The value j = 0 indicates a turning point
of the path, where r(t) begins to decrease instead of increasing, or vice versa.
If the range in which r may vary is limited only by the condition r > rmin,
the motion is infinite: the particle comes from, and returns to, infinity.
If the range of r has two limits rmin and rmax, the motion is finite and the
path lies entirely within the annulus bounded by the circles r = rmax and
r = rmin- This does not mean, however, that the path must be a closed curve.
During the time in which r varies from rmax to rmin and back, the radius
vector turns through an angle Ao which, according to (14.7), is given by
Mdr/r2
(14.10)
The condition for the path to be closed is that this angle should be a rational
fraction of 2n, i.e. that Ao = 2nm/n, where m and n are integers. In that case,
after n periods, the radius vector of the particle will have made m complete
revolutions and will occupy its original position, so that the path is closed.
Such cases are exceptional, however, and when the form of U(r) is arbitrary
the angle is not a rational fraction of 2nr. In general, therefore, the path
of a particle executing a finite motion is not closed. It passes through the
minimum and maximum distances an infinity of times, and after infinite time
it covers the entire annulus between the two bounding circles. The path
shown in Fig. 9 is an example.
There are only two types of central field in which all finite motions take
place in closed paths. They are those in which the potential energy of the
particle varies as 1/r or as r2. The former case is discussed in §15; the latter
is that of the space oscillator (see §23, Problem 3).
At a turning point the square root in (14.5), and therefore the integrands
in (14.6) and (14.7), change sign. If the angle is measured from the direc-
tion of the radius vector to the turning point, the parts of the path on each
side of that point differ only in the sign of for each value of r, i.e. the path
is symmetrical about the line = 0. Starting, say, from a point where = rmax
the particle traverses a segment of the path as far as a point with r rmin,
§14
Motion in a central field
33
then follows a symmetrically placed segment to the next point where r = rmax,
and so on. Thus the entire path is obtained by repeating identical segments
forwards and backwards. This applies also to infinite paths, which consist of
two symmetrical branches extending from the turning point (r = rmin) to
infinity.
'max
min
so
FIG. 9
The presence of the centrifugal energy when M # 0, which becomes
infinite as 1/22 when r -> 0, generally renders it impossible for the particle to
reach the centre of the field, even if the field is an attractive one. A "fall" of
the particle to the centre is possible only if the potential energy tends suffi-
ciently rapidly to -00 as r 0. From the inequality
1mr2 = E- U(r) - M2/2mr2
or r2U(Y)+M2/2m < Er2, it follows that r can take values tending to zero
only if
(14.11)
i.e. U(r) must tend to - 8 either as - a/r2 with a > M2/2m, or proportionally
to - 1/rn with n > 2.
PROBLEMS
PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass
m moving on the surface of a sphere of radius l in a gravitational field).
SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the
polar axis vertically downwards, the Lagrangian of the pendulum is
1ml2(02 + 62 sin20) +mgl cos 0.
2*
34
Integration of the Equations of Motion
§14
The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the
z-component of angular momentum, is conserved:
(1)
The energy is
E = cos 0
(2)
= 0.
Hence
(3)
where the "effective potential energy" is
Ueff(0) = COS 0.
For the angle o we find, using (1),
do
(4)
The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively.
The range of 0 in which the motion takes place is that where E > Ueff, and its limits
are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots
between -1 and +1; these define two circles of latitude on the sphere, between which the
path lies.
PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a
cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational
field.
SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the
polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co-
ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is
= a.
By the same method as in Problem 1, we find
==
The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots;
these define two horizontal circles on the cone, between which the path lies.
PROBLEM 3. Integrate the equations of motion for a pendulum of mass M2, with a mass M1
at the point of support which can move on a horizontal line lying in the plane in which m2
moves (Fig. 2, §5).
SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The
generalised momentum Px, which is the horizontal component of the total momentum of the
system, is therefore conserved
Px = cos = constant.
(1)
The system may always be taken to be at rest as a whole. Then the constant in (1) is zero
and integration gives
(m1+m2)x+m2) sin = constant,
(2)
which expresses the fact that the centre of mass of the system does not move horizontally.
§15
Kepler's problem
35
Using (1), we find the energy in the form
E
(3)
Hence
Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
dulum, which moves in an arc of a circle.
$15. Kepler's problem
An important class of central fields is formed by those in which the poten-
tial energy is inversely proportional to r, and the force accordingly inversely
proportional to r2. They include the fields of Newtonian gravitational attrac-
tion and of Coulomb electrostatic interaction; the latter may be either attrac-
tive or repulsive.
Let us first consider an attractive field, where
U=-a/r
(15.1)
with a a positive constant. The "effective" potential energy
(15.2)
is
of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
r
8
it tends to zero from negative values ; for r = M2/ma it has a minimum value
Ueff, min = -mx2/2M2.
(15.3)
Ueff
FIG. 10
It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
for E > 0.
36
Integration of the Equations of Motion
§15
The shape of the path is obtained from the general formula (14.7). Substi-
tuting there U = - a/r and effecting the elementary integration, we have
o =
- constant.
Taking the origin of such that the constant is zero, and putting
P = M2/ma, e= [1 1+(2EM2/mo2)]
(15.4)
we can write the equation of the path as
p/r = 1+e coso.
(15.5)
This is the equation of a conic section with one focus at the origin; 2p is called
the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
is seen from (15.5) to be such that the point where = 0 is the point nearest
to the origin (called the perihelion).
In the equivalent problem of two particles interacting according to the law
(15.1), the orbit of each particle is a conic section, with one focus at the centre
of mass of the two particles.
It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
has been said earlier in this section. According to the formulae of analytical
geometry, the major and minor semi-axes of the ellipse are
a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
(15.6)
y
X
2b
ae
2a
FIG. 11
The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
becomes a circle. It may be noted that the major axis of the ellipse depends
only on the energy of the particle, and not on its angular momentum. The
least and greatest distances from the centre of the field (the focus of the
ellipse) are
rmin = =p/(1+e)=a(1-e),
max=p(1-e)=a(1+e). = = (15.7)
These expressions, with a and e given by (15.6) and (15.4), can, of course,
also be obtained directly as the roots of the equation Ueff(r) = E.
§15
Kepler's problem
37
The period T of revolution in an elliptical orbit is conveniently found by
using the law of conservation of angular momentum in the form of the area
integral (14.3). Integrating this equation with respect to time from zero to
T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
f = nab, and by using the formulae (15.6) we find
T = 2ma3/2-(m/a)
= ma((m2E3).
(15.8)
The proportionality between the square of the period and the cube of the
linear dimension of the orbit has already been demonstrated in §10. It may
also be noted that the period depends only on the energy of the particle.
For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
tance of the perihelion from the focus is
rmin ==pl(e+1)=a(e-1), = =
(15.9)
where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
y
p
ale-1)
FIG. 12
If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
perihelion distance rmin = 1p. This case occurs if the particle starts from rest
at infinity.
The co-ordinates of the particle as functions of time in the orbit may be
found by means of the general formula (14.6). They may be represented in a
convenient parametric form as follows.
Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
we can write the integral (14.6) for the time as
t
=
=
38
Integration of the Equations of Motion
§15
The obvious substitution r-a = - ae cos $ converts the integral to
sioant
If time is measured in such a way that the constant is zero, we have the
following parametric dependence of r on t:
r = a(1-e cos ), t =
(15.10)
the particle being at perihelion at t = 0. The Cartesian co-ordinates
x = r cos o, y = r sin (the x and y axes being respectively parallel to the
major and minor axes of the ellipse) can likewise be expressed in terms of
the parameter $. From (15.5) and (15.10) we have
ex = = =
y is equal to W(r2-x2). Thus
x = a(cos & - e),
y = =av(1-e2) $.
(15.11)
A complete passage round the ellipse corresponds to an increase of $ from 0
to 2nr.
Entirely similar calculations for the hyperbolic orbits give
r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
(15.12)
x = a(e-cosh ) y = a1/(e2-1)sinh &
where the parameter $ varies from - 00 to + 00.
Let us now consider motion in a repulsive field, where
U
(a>0).
(15.13)
Here the effective potential energy is
Utt
and decreases monotonically from + 00 to zero as r varies from zero to
infinity. The energy of the particle must be positive, and the motion is always
infinite. The calculations are exactly similar to those for the attractive field.
The path is a hyperbola (or, if E = 0, a parabola):
pr r = =1-e coso, =
(15.14)
where P and e are again given by (15.4). The path passes the centre of the
field in the manner shown in Fig. 13. The perihelion distance is
rmin =p(e-1)=a(e+1). =
(15.15)
The time dependence is given by the parametric equations
= =(ma3/a)(esinh+) =
(15.16)
x
= a(cosh & e ,
y = av((e2-1) sinh &
§15
Kepler's problem
39
To conclude this section, we shall show that there is an integral of the mo-
tion which exists only in fields U = a/r (with either sign of a). It is easy to
verify by direct calculation that the quantity
vxM+ar/r
(15.17)
is constant. For its total time derivative is v
since M = mr xv,
Putting mv = ar/r3 from the equation of motion, we find that this expression
vanishes.
y
0
(I+e)
FIG. 13
The direction of the conserved vector (15.17) is along the major axis from
the focus to the perihelion, and its magnitude is ae. This is most simply
seen by considering its value at perihelion.
It should be emphasised that the integral (15.17) of the motion, like M and
E, is a one-valued function of the state (position and velocity) of the particle.
We shall see in §50 that the existence of such a further one-valued integral
is due to the degeneracy of the motion.
PROBLEMS
PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
moving in a parabola in a field U = -a/r.
SOLUTION. In the integral
we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
the required dependence:
r=1p(1+n2),
t=
y=pn.
40
Integration of the Equations of Motion
§15
The parameter n varies from - 00 to +00.
PROBLEM 2. Integrate the equations of motion for a particle in a central field
U = - a/r2 (a > 0).
SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
(a) for E > andM 0 and
(b) for E>0 0nd and M 2/2m a,
(c) for E <0 and Ms1
In all three cases
In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
origin as
00. The fall from a given value of r takes place in a finite time, namely
PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
(14.10), which we write as
in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
the first-order term gives the required change so:
(1)
where we have changed from the integration over r to one over , along the path of the "un-
perturbed" motion.
In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.

413
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§29
Resonance in non-linear oscillations
91
This equation, however, does not suffice to determine the resulting ampli-
tude of the oscillations. The attainment of a finite amplitude involves non-
linear effects, and to include these in the equation of motion we must retain
also the terms non-linear in x(2):
= cos(2wo+e)t. (29.11)
The problem can be considerably simplified by virtue of the following fact.
Putting on the right-hand side of (29.11) x(2) = b cos[(wo++)+8], where
b is the required amplitude of the resonance oscillations and 8 a constant
phase difference which is of no importance in what follows, and writing the
product of cosines as a sum, we obtain a term (afb/3mwo2)
of the ordinary resonance type (with respect to the eigenfrequency wo of the
system). The problem thus reduces to that considered at the beginning of
this section, namely ordinary resonance in a non-linear system, the only
differences being that the amplitude of the external force is here represented
by afb/3wo2, and E is replaced by 1/6. Making this change in equation (29.4),
we have
Solving for b, we find the possible values of the amplitude:
b=0,
(29.12)
(29.13)
1
(29.14)
Figure 33 shows the resulting dependence of b on € for K > 0; for K < 0
the curves are the reflections (in the b-axis) of those shown. The points B
and C correspond to the values E = To the left of
B, only the value b = 0 is possible, i.e. there is no resonance, and oscillations
of frequency near wo are not excited. Between B and C there are two roots,
b = 0(BC) and (29.13) (BE). Finally, to the right of C there are three roots
(29.12)-(29.14). Not all these, however, correspond to stable oscillations.
The value b = 0 is unstable on BC, and it can also be shown that the middle
root (29.14) always gives instability. The unstable values of b are shown in
Fig. 33 by dashed lines.
Let us examine, for example, the behaviour of a system initially "at rest"
as the frequency of the external force is gradually diminished. Until the point
t This segment corresponds to the region of parametric resonance (27.12), and a com-
parison of (29.10) and (27.8) gives 1h = 2af/3mwo4. The condition 12af/3mwo3 > 4X for
which the phenomenon can exist corresponds to h > hk.
+ It should be recalled that only resonance phenomena are under consideration. If these
phenomena are absent, the system is not literally at rest, but executes small forced oscillations
of frequency y.
92
Small Oscillations
§29
C is reached, b = 0, but at C the state of the system passes discontinuously
to the branch EB. As € decreases further, the amplitude of the oscillations
decreases to zero at B. When the frequency increases again, the amplitude
increases along BE.-
b
E
E
A
B
C D
FIG. 33
The cases of resonance discussed above are the principal ones which may
occur in a non-linear oscillating system. In higher approximations, resonances
appear at other frequencies also. Strictly speaking, a resonance must occur
at every frequency y for which ny + mwo = wo with n and m integers, i.e. for
every y = pwo/q with P and q integers. As the degree of approximation
increases, however, the strength of the resonances, and the widths of the
frequency ranges in which they occur, decrease so rapidly that in practice
only the resonances at frequencies y 2 pwo/q with small P and q can be ob-
served.
PROBLEM
Determine the function b(e) for resonance at frequencies y 22 3 wo.
SOLUTION. In the first approximation, x(1) = -(f/8mwo2) cos(3wo+t) For the second
approximation x(2) we have from (29.1) the equation
= -3,8x(1)x(2)2,
where only the term which gives the required resonance has been retained on the right-hand
side. Putting x(2) = b cos[(wo+)+8] and taking the resonance term out of the product
of three cosines, we obtain on the right-hand side the expression
(3,3b2f(32mwo2) cos[(wotle)t-28].
Hence it is evident that b(e) is obtained by replacing f by 3,8b2f/32wo², and E by JE, in
(29.4):
Ab4.
The roots of this equation are
b=0,
Fig. 34 shows a graph of the function b(e) for k>0. Only the value b=0 (the e-axis) and
the branch AB corresponds to stability. The point A corresponds to EK = 3(4x2)2-A3)/4kA,
t It must be noticed, however, that all the formulae derived here are valid only when the
amplitude b (and also E) is sufficiently small. In reality, the curves BE and CF meet, and at
their point of intersection the oscillation ceases; thereafter, b = 0.
§30
Motion in a rapidly oscillating field
93
bk2 = Oscillations exist only for € > Ek, and then b > bk. Since the state
b = 0 is always stable, an initial "push" is necessary in order to excite oscillations.
The formulae given above are valid only for small E. This condition is satisfied if 1 is small
and also the amplitude of the force is such that 2/wo < A KWO.
b
B
A
€
FIG. 34
§30. Motion in a rapidly oscillating field
Let us consider the motion of a particle subject both to a time-independent
field of potential U and to a force
f=f1coswt+fasin.ou
(30.1)
which varies in time with a high frequency w (f1, f2 being functions of the
co-ordinates only). By a "high" frequency we mean one such that w > 1/T,
where T is the order of magnitude of the period of the motion which the
particle would execute in the field U alone. The magnitude of f is not assumed
small in comparison with the forces due to the field U, but we shall assume
that the oscillation (denoted below by $) of the particle as a result of this
force is small.
To simplify the calculations, let us first consider motion in one dimension
in a field depending only on the space co-ordinate X. Then the equation of
motion of the particle ist
mx = -dU/dx+f.
(30.2)
It is evident, from the nature of the field in which the particle moves, that
it will traverse a smooth path and at the same time execute small oscillations
of frequency w about that path. Accordingly, we represent the function x(t)
as a sum:
(30.3)
where (t) corresponds to these small oscillations.
The mean value of the function (t) over its period 2n/w is zero, and the
function X(t) changes only slightly in that time. Denoting this average by a
bar, we therefore have x = X(t), i.e. X(t) describes the "smooth" motion of
t The co-ordinate x need not be Cartesian, and the coefficient m is therefore not neces-
sarily the mass of the particle, nor need it be constant as has been assumed in (30.2). This
assumption, however, does not affect the final result (see the last footnote to this section).
94
Small Oscillations
§30
the particle averaged over the rapid oscillations. We shall derive an equation
which determines the function X(t).t
Substituting (30.3) in (30.2) and expanding in powers of & as far as the
first-order terms, we obtain
(30.4)
This equation involves both oscillatory and "smooth" terms, which must
evidently be separately equal. For the oscillating terms we can put simply
mg = f(X, t);
(30.5)
the other terms contain the small factor & and are therefore of a higher order
of smallness (but the derivative sur is proportional to the large quantity w2
and so is not small). Integrating equation (30.5) with the function f given by
(30.1) (regarding X as a constant), we have
& = -f/mw2.
(30.6)
Next, we average equation (30.4) with respect to time (in the sense discussed
above). Since the mean values of the first powers of f and $ are zero, the result
is
dX
which involves only the function X(t). This equation can be written
mX = dUeff/dX,
(30.7)
where the "effective potential energy" is defined ast
Ueff = U+f2/2mw2
=
(30.8)
Comparing this expression with (30.6), we easily see that the term added to
U is just the mean kinetic energy of the oscillatory motion:
Ueff= U+1mg2
(30.9)
Thus the motion of the particle averaged over the oscillations is the same
as if the constant potential U were augmented by a constant quantity pro-
portional to the squared amplitude of the variable field.
t The principle of this derivation is due to P. L. KAPITZA (1951).
++ By means of somewhat more lengthy calculations it is easy to show that formulae (30.7)
and (30.8) remain valid even if m is a function of X.
§30
Motion in a rapidly oscillating field
95
The result can easily be generalised to the case of a system with any number
of degrees of freedom, described by generalised co-ordinates qi. The effective
potential energy is then given not by (30.8), but by
Unt = Ut
= U+ ,
(30.10)
where the quantities a-1ik, which are in general functions of the co-ordinates,
are the elements of the matrix inverse to the matrix of the coefficients aik in
the kinetic energy (5.5) of the system.
PROBLEMS
PROBLEM 1. Determine the positions of stable equilibrium of a pendulum whose point of
support oscillates vertically with a high frequency y
(g/l)).
SOLUTION. From the Lagrangian derived in §5, Problem 3(c), we see that in this case the
variable force is f = -mlay2 cos yt sin (the quantity x being here represented by the angle
b). The "effective potential energy" is therefore Ueff = mgl[-cos - & st(a2y2/4gl) sin2]. The
positions of stable equilibrium correspond to the minima of this function. The vertically
downward position ( = 0) is always stable. If the condition a2y2 > 2gl holds, the vertically
upward position ( = ) is also stable.
PROBLEM 2. The same as Problem 1, but for a pendulum whose point of support oscillates
horizontally.
SOLUTION. From the Lagrangian derived in §5, Problem 3(b), we find f = mlay2 cos yt
cos and Uell = mgl[-cos 3+(a2y2/4gl) cos2]. If a2y2 < 2gl, the position = 0 is stable.
If a2y2 > 2gl, on the other hand, the stable equilibrium position is given by cos = 2gl/a22.
CHAPTER VI
MOTION OF A RIGID BODY
$31. Angular velocity
A rigid body may be defined in mechanics as a system of particles such that
the distances between the particles do not vary. This condition can, of course,
be satisfied only approximately by systems which actually exist in nature.
The majority of solid bodies, however, change so little in shape and size
under ordinary conditions that these changes may be entirely neglected in
considering the laws of motion of the body as a whole.
In what follows, we shall often simplify the derivations by regarding a
rigid body as a discrete set of particles, but this in no way invalidates the
assertion that solid bodies may usually be regarded in mechanics as continu-
ous, and their internal structure disregarded. The passage from the formulae
which involve a summation over discrete particles to those for a continuous
body is effected by simply replacing the mass of each particle by the mass
P dV contained in a volume element dV (p being the density) and the sum-
mation by an integration over the volume of the body.
To describe the motion of a rigid body, we use two systems of co-ordinates:
a "fixed" (i.e. inertial) system XYZ, and a moving system X1 = x, X2 = y,
X3 = 2 which is supposed to be rigidly fixed in the body and to participate
in its motion. The origin of the moving system may conveniently be taken
to coincide with the centre of mass of the body.
The position of the body with respect to the fixed system of co-ordinates
is completely determined if the position of the moving system is specified.
Let the origin O of the moving system have the radius vector R (Fig. 35).
The orientation of the axes of that system relative to the fixed system is given
by three independent angles, which together with the three components of
the vector R make six co-ordinates. Thus a rigid body is a mechanical system
with six degrees of freedom.
Let us consider an arbitrary infinitesimal displacement of a rigid body.
It can be represented as the sum of two parts. One of these is an infinitesimal
translation of the body, whereby the centre of mass moves to its final position,
but the orientation of the axes of the moving system of co-ordinates is un-
changed. The other is an infinitesimal rotation about the centre of mass,
whereby the remainder of the body moves to its final position.
Let r be the radius vector of an arbitrary point P in a rigid body in the
moving system, and r the radius vector of the same point in the fixed system
(Fig. 35). Then the infinitesimal displacement dr of P consists of a displace-
ment dR, equal to that of the centre of mass, and a displacement doxr
96
§31
Angular velocity
97
relative to the centre of mass resulting from a rotation through an infinitesimal
angle do (see (9.1)): dr = dR + do xr. Dividing this equation by the time
dt during which the displacement occurs, and putting
dr/dt = V,
dR/dt =
do/dt = La
(31.1)
we obtain the relation
V = V+Sxr.
(31.2)
Z
X3
P
X2
r
o
R
X1
Y
X
FIG. 35
The vector V is the velocity of the centre of mass of the body, and is also
the translational velocity of the body. The vector S is called the angular
velocity of the rotation of the body; its direction, like that of do, is along the
axis of rotation. Thus the velocity V of any point in the body relative to the
fixed system of co-ordinates can be expressed in terms of the translational
velocity of the body and its angular velocity of rotation.
It should be emphasised that, in deriving formula (31.2), no use has been
made of the fact that the origin is located at the centre of mass. The advan-
tages of this choice of origin will become evident when we come to calculate
the energy of the moving body.
Let us now assume that the system of co-ordinates fixed in the body is
such that its origin is not at the centre of mass O, but at some point O' at
a distance a from O. Let the velocity of O' be V', and the angular velocity
of the new system of co-ordinates be S'. We again consider some point P
in the body, and denote by r' its radius vector with respect to O'. Then
= r'+a, and substitution in (31.2) gives V = V+2xa+2xr'. The
definition of V' and S' shows that V = Hence it follows that
(31.3)
The second of these equations is very important. We see that the angular
velocity of rotation, at any instant, of a system of co-ordinates fixed in
the body is independent of the particular system chosen. All such systems
t
To avoid any misunderstanding, it should be noted that this way of expressing the angular
velocity is somewhat arbitrary: the vector so exists only for an infinitesimal rotation, and not
for all finite rotations.
4*
98
Motion of a Rigid Body
§32
rotate with angular velocities S which are equal in magnitude and parallel
in direction. This enables us to call S the angular velocity of the body. The
velocity of the translational motion, however, does not have this "absolute"
property.
It is seen from the first formula (31.3) that, if V and S are, at any given
instant, perpendicular for some choice of the origin O, then V' and SS are
perpendicular for any other origin O'. Formula (31.2) shows that in this case
the velocities V of all points in the body are perpendicular to S. It is then
always possible+ to choose an origin O' whose velocity V' is zero, SO that the
motion of the body at the instant considered is a pure rotation about an axis
through O'. This axis is called the instantaneous axis of rotation.t
In what follows we shall always suppose that the origin of the moving
system is taken to be at the centre of mass of the body, and so the axis of
rotation passes through the centre of mass. In general both the magnitude
and the direction of S vary during the motion.
$32. The inertia tensor
To calculate the kinetic energy of a rigid body, we may consider it as a
discrete system of particles and put T = mv2, where the summation is
taken over all the particles in the body. Here, and in what follows, we simplify
the notation by omitting the suffix which denumerates the particles.
Substitution of (31.2) gives
T = Sxx+
The velocities V and S are the same for every point in the body. In the first
term, therefore, V2 can be taken outside the summation sign, and Em is
just the mass of the body, which we denote by u. In the second term we put
EmV Sxr = Emr VxS = VxS Emr. Since we take the origin of the
moving system to be at the centre of mass, this term is zero, because Emr = 0.
Finally, in the third term we expand the squared vector product. The result
is
(32.1)
Thus the kinetic energy of a rigid body can be written as the sum of two
parts. The first term in (32.1) is the kinetic energy of the translational motion,
and is of the same form as if the whole mass of the body were concentrated
at the centre of mass. The second term is the kinetic energy of the rotation
with angular velocity S about an axis passing through the centre of mass.
It should be emphasised that this division of the kinetic energy into two parts
is possible only because the origin of the co-ordinate system fixed in the
body has been taken to be at its centre of mass.
t O' may, of course, lie outside the body.
+ In the general case where V and SC are not perpendicular, the origin may be chosen so
as to make V and S parallel, i.e. so that the motion consists (at the instant in question) of a
rotation about some axis together with a translation along that axis.
§32
The inertia tensor
99
We may rewrite the kinetic energy of rotation in tensor form, i.e. in terms
of the components Xi and O of the vectors r and L. We have
Here we have used the identity Oi = SikOk, where dik is the unit tensor,
whose components are unity for i = k and zero for i # k. In terms of the
tensor
(32.2)
we have finally the following expression for the kinetic energy of a rigid
body:
T =
(32.3)
The Lagrangian for a rigid body is obtained from (32.3) by subtracting
the potential energy:
L =
(32.4)
The potential energy is in general a function of the six variables which define
the position of the rigid body, e.g. the three co-ordinates X, Y, Z of the
centre of mass and the three angles which specify the relative orientation of
the moving and fixed co-ordinate axes.
The tensor Iik is called the inertia tensor of the body. It is symmetrical,
i.e.
Ik=Iki
(32.5)
as is evident from the definition (32.2). For clarity, we may give its com-
ponents explicitly:
TEST
(32.6)
m(x2+y2)
The components Ixx, Iyy, Izz are called the moments of inertia about the
corresponding axes.
The inertia tensor is evidently additive: the moments of inertia of a body
are the sums of those of its parts.
t In this chapter, the letters i, k, l are tensor suffixes and take the values 1, 2, 3. The
summation rule will always be used, i.e. summation signs are omitted, but summation over
the values 1, 2, 3 is implied whenever a suffix occurs twice in any expression. Such a suffix is
called a dummy suffix. For example, AiBi = A . B, Ai2 = AiA1 = A², etc. It is obvious that
dummy suffixes can be replaced by any other like suffixes, except ones which already appear
elsewhere in the expression concerned.
100
Motion of a Rigid Body
§32
If the body is regarded as continuous, the sum in the definition (32.2)
becomes an integral over the volume of the body:
(32.7)
Like any symmetrical tensor of rank two, the inertia tensor can be reduced
to diagonal form by an appropriate choice of the directions of the axes
X1, x2, X3. These directions are called the principal axes of inertia, and the
corresponding values of the diagonal components of the tensor are called the
principal moments of inertia; we shall denote them by I, I2, I3. When the
axes X1, X2, X3 are so chosen, the kinetic energy of rotation takes the very
simple form
=
(32.8)
None of the three principal moments of inertia can exceed the sum of the
other two. For instance,
m(x12+x22) = I3.
(32.9)
A body whose three principal moments of inertia are all different is called
an asymmetrical top. If two are equal (I1 = I2 # I3), we have a symmetrical
top. In this case the direction of one of the principal axes in the x1x2-plane
may be chosen arbitrarily. If all three principal moments of inertia are equal,
the body is called a spherical top, and the three axes of inertia may be chosen
arbitrarily as any three mutually perpendicular axes.
The determination of the principal axes of inertia is much simplified if
the body is symmetrical, for it is clear that the position of the centre of mass
and the directions of the principal axes must have the same symmetry as
the body. For example, if the body has a plane of symmetry, the centre of
mass must lie in that plane, which also contains two of the principal axes of
inertia, while the third is perpendicular to the plane. An obvious case of this
kind is a coplanar system of particles. Here there is a simple relation between
the three principal moments of inertia. If the plane of the system is taken as
the x1x2-plane, then X3 = 0 for every particle, and so I = mx22, I2 = 12,
I3 = (12+x2)2, whence
(32.10)
If a body has an axis of symmetry of any order, the centre of mass must lie
on that axis, which is also one of the principal axes of inertia, while the other
two are perpendicular to it. If the axis is of order higher than the second,
the body is a symmetrical top. For any principal axis perpendicular to the
axis of symmetry can be turned through an angle different from 180° about the
latter, i.e. the choice of the perpendicular axes is not unique, and this can
happen only if the body is a symmetrical top.
A particular case here is a collinear system of particles. If the line of the
system is taken as the x3-axis, then X1 = X2 = 0 for every particle, and so

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§32
The inertia tensor
101
two of the principal moments of inertia are equal and the third is zero:
I3 = 0.
(32.11)
Such a system is called a rotator. The characteristic property which distin-
guishes a rotator from other bodies is that it has only two, not three, rotational
degrees of freedom, corresponding to rotations about the X1 and X2 axes: it
is clearly meaningless to speak of the rotation of a straight line about itself.
Finally, we may note one further result concerning the calculation of the
inertia tensor. Although this tensor has been defined with respect to a system
of co-ordinates whose origin is at the centre of mass (as is necessary if the
fundamental formula (32.3) is to be valid), it may sometimes be more con-
veniently found by first calculating a similar tensor I' =
defined with respect to some other origin O'. If the distance OO' is repre-
sented by a vector a, then r = r'+a, Xi = x'i+ai; since, by the definition
of O, Emr = 0, we have
I'ikIk(a2ik-aiak).
(32.12)
Using this formula, we can easily calculate Iik if I'ik is known.
PROBLEMS
PROBLEM 1. Determine the principal moments of inertia for the following types of mole-
cule, regarded as systems of particles at fixed distances apart: (a) a molecule of collinear
atoms, (b) a triatomic molecule which is an isosceles triangle (Fig. 36), (c) a tetratomic
molecule which is an equilateral-based tetrahedron (Fig. 37).
m2
X2
m2
h
x
m
a
a
m
a
m
m
a
FIG. 36
FIG. 37
SOLUTION. (a)
Is = 0,
where Ma is the mass of the ath atom, lao the distance between the ath and bth atoms, and
the summation includes one term for every pair of atoms in the molecule.
For a diatomic molecule there is only one term in the sum, and the result is obvious it is
the product of the reduced mass of the two atoms and the square of the distance between
them: I1 = I2 = m1m2l2((m1+m2).
(b) The centre of mass is on the axis of symmetry of the triangle, at a distance X2 = mzh/u
from its base (h being the height of the triangle). The moments of inertia are I1 = 2m1m2h2/u,
I2 = 1m1a2, I3 = I+I2.
(c) The centre of mass is on the axis of symmetry of the tetrahedron, at a distance
X3 = mgh/u from its base (h being the height of the tetrahedron). The moments of inertia
102
Motion of a Rigid Body
§32
are I1 = = I3 = mia². If M1 = M2, h = (2/3)a, the molecule is a
regular tetrahedron and I1=I2 = I3 = mia2.
PROBLEM 2. Determine the principal moments of inertia for the following homogeneous
bodies: (a) a thin rod of length l, (b) a sphere of radius R, (c) a circular cylinder of radius R
and height h, (d) a rectangular parallelepiped of sides a, b, and c, (e) a circular cone of height
h and base radius R, (f) an ellipsoid of semiaxes a, b, c.
SOLUTION. (a) I1 = I2 = Trul2, I3 = 0 (we neglect the thickness of the rod).
(b) I1 = I2 = I3 = zuR2 (found by calculating the sum I+I+I3 = 2p dV).
(c) I1 = I2 = tu(R2+th2), I3 = tuR2 (where the x3-axis is along the axis of the cylinder).
(d) I1 = (2+c2), I2 = (a2+cc), I3 = 121(a2++b) (where the axes X1, x2, X3 are
along the sides a, b, c respectively).
(e) We first calculate the tensor I'ik with respect to axes whose origin is at the vertex of
the cone (Fig. 38). The calculation is simple if cylindrical co-ordinates are used, and the result
is I'1 = I'2 = I'3 = 2 The centre of mass is easily shown to be on the
axis of the cone and at a distance a = 3h from the vertex. Formula (32.12) therefore gives
I1 = I2 = I'1-2 = I3 = I'3 = TouR2.
X3,X3'
X2
xi
x2
FIG. 38
(f) The centre of mass is at the centre of the ellipsoid, and the principal axes of inertia are
along the axes of the ellipsoid. The integration over the volume of the ellipsoid can be reduced
to one over a sphere by the transformation x = as,y = bn, 2 = c5, which converts the equa-
tion of the surface of the ellipsoid 1 into that of the unit sphere
st+24's = 1.
For example, the moment of inertia about the x-axis is
dz
= tabcI'(b2 tc2),
where I' is the moment of inertia of a sphere of unit radius. Since the volume of the ellipsoid
is 4nabc/3, we find the moments of inertia I = tu(b2+c2), I2 = tu(a2+c2), I3 = tu(a2+b2).
PROBLEM 3. Determine the frequency of small oscillations of a compound pendulum (a
rigid body swinging about a fixed horizontal axis in a gravitational field).
SOLUTION. Let l be the distance between the centre of mass of the pendulum and the axis
about which it rotates, and a, B, y the angles between the principal axes of inertia and the
axis of rotation. We take as the variable co-ordinate the angle between the vertical
and a line through the centre of mass perpendicular to the axis of rotation. The velocity of
the centre of mass is V = 10, and the components of the angular velocity along the principal
§32
The inertia tensor
103
axes of inertia are o cos a, b cos B, b cos y. Assuming the angle to be small, we find the
potential energy U = ugl(1-cos 9 22 12. The Lagrangian is therefore
=
The frequency of the oscillations is consequently
w2 = cos2y).
PROBLEM 4. Find the kinetic energy of the system shown in Fig. 39: OA and AB are thin
uniform rods of length l hinged together at A. The rod OA rotates (in the plane of the diagram)
about O, while the end B of the rod AB slides along Ox.
A
l
l
x
B
FIG. 39
SOLUTION. The velocity of the centre of mass of the rod OA (which is at the middle of
the rod) is 110, where is the angle AOB. The kinetic energy of the rod OA is therefore
T1 = where u is the mass of each rod.
The Cartesian co-ordinates of the centre of mass of the rod AB are X = sl cos o, Y
= 1/ sin b. Since the angular velocity of rotation of this rod is also b, its kinetic energy is
T2 = = tul2(1- +8 sin2o)62 +1162. The total kinetic energy of this
system is therefore = I = Tzul2 (see Problem 2(a)).
PROBLEM 5. Find the kinetic energy of a cylinder of radius R rolling on a plane, if the mass
of the cylinder is so distributed that one of the principal axes of inertia is parallel to the axis
of the cylinder and at a distance a from it, and the moment of inertia about that principal
axis is I.
SOLUTION. Let be the angle between the vertical and a line from the centre of mass
perpendicular to the axis of the cylinder (Fig. 40). The motion of the cylinder at any instant
R
FIG. 40
may be regarded as a pure rotation about an instantaneous axis which coincides with the
line where the cylinder touches the plane. The angular velocity of this rotation is o, since
the angular velocity of rotation about all parallel axes is the same. The centre of mass is at a
distance V(a2+R2-2aR cos ) from the instantaneous axis, and its velocity is therefore
V = bv /(a2+R2-2aR cos ). The total kinetic energy is
T = cos
104
Motion of a Rigid Body
§32
PROBLEM 6. Find the kinetic energy of a homogeneous cylinder of radius a rolling inside
a cylindrical surface of radius R (Fig. 41).
R
FIG. 41
SOLUTION. We use the angle between the vertical and the line joining the centres of the
cylinders. The centre of mass of the rolling cylinder is on the axis, and its velocity is V =
o(R-a). We can calculate the angular velocity as that of a pure rotation about an instantaneous
axis which coincides with the line of contact of the cylinders it is Q = V/a = p(R-a)/a.
If I3 is the moment of inertia about the axis of the cylinder, then
T =
I3 being given by Problem 2(c).
PROBLEM 7. Find the kinetic energy of a homogeneous cone rolling on a plane.
SOLUTION. We denote by 0 the angle between the line OA in which the cone touches the
plane and some fixed direction in the plane (Fig. 42). The centre of mass is on the axis of the
cone, and its velocity V = a0 cos a, where 2a is the vertical angle of the cone and a the
Z
Y
A
FIG. 42
distance of the centre of mass from the vertex. The angular velocity can be calculated as
that of a pure rotation about the instantaneous axis OA: S2 = V/a sin a = é cot a. One of
the principal axes of inertia (x3) is along the axis of the cone, and we take another (x2) perpen-
dicular to the axis of the cone and to the line OA. Then the components of the vector S
(which is parallel to OA) along the principal axes of inertia are O sin a, 0, O cos a. The kinetic
energy is thus
=
= 3uh202(1+5 cos2x)/40,
where h is the height of the cone, and I1, I3 and a have been given in Problem 2(e).
PROBLEM 8. Find the kinetic energy of a homogeneous cone whose base rolls on a plane
and whose vertex is fixed at a height above the plane equal to the radius of the base, so that
the axis of the cone is parallel to the plane.
SOLUTION. We use the angle 0 between a fixed direction in the plane and the projection
of the axis of the cone on the plane (Fig. 43). Then the velocity of the centre of mass is V = aß,
§33
Angular momentum of a rigid body
105
the notation being as in Problem 7. The instantaneous axis of rotation is the generator OA
which passes through the point where the cone touches the plane. The centre of mass is at a
distance a sin a from this axis, and so l = V/a sin a = O/sin a. The components of the
vector Ca along the principal axes of inertia are, if the x2-axis is taken perpendicular to the
axis of the cone and to the line OA, O sin a = 6, 0, N cos a = è cot a. The kinetic energy
is therefore
T cot2a
= 314h282(sec2x+5)/40.
Z
0
Y
A
FIG. 43
PROBLEM 9. Find the kinetic energy of a homogeneous ellipsoid which rotates about one
of its axes (AB in Fig. 44) while that axis itself rotates about a line CD perpendicular to it
and passing through the centre of the ellipsoid.
SOLUTION. Let the angle of rotation about CD be 0, and that about AB (i.e. the angle
between CD and the x1-axis of inertia, which is perpendicular to AB) be . Then the com-
ponents of S along the axes of inertia are é cos , 0 sin , b, if the x3-axis is AB. Since the
centre of mass, at the centre of the ellipsoid, is at rest, the kinetic energy is
=
D
B
D
A
Do
A
1a
C
of
8
FIG. 44
FIG. 45
PROBLEM 10. The same as Problem 9, but for the case where the axis AB is not perpendicu-
lar to CD and is an axis of symmetry of the ellipsoid (Fig. 45).
SOLUTION. The components of Ca along the axis AB and the other two principal axes of
inertia, which are perpendicular to AB but otherwise arbitrary, are è cos a cos , è cos a X
sin , o to sin a. The kinetic energy is T = 11102 a)2.
$33. Angular momentum of a rigid body
The value of the angular momentum of a system depends, as we know, on
the point with respect to which it is defined. In the mechanics of a rigid body,
106
Motion of a Rigid Body
§33
the most appropriate point to choose for this purpose is the origin of the
moving system of co-ordinates, i.e. the centre of mass of the body, and in
what follows we shall denote by M the angular momentum SO defined.
According to formula (9.6), when the origin is taken at the centre of mass
of the body, the angular momentum M is equal to the "intrinsic" angular
momentum resulting from the motion relative to the centre of mass. In the
definition M = Emrxv we therefore replace V by Sxr:
M = =
or, in tensor notation,
Mi = OK
Finally, using the definition (32.2) of the inertia tensor, we have
(33.1)
If the axes X1, X2, X3 are the same as the principal axes of inertia, formula
(33.1) gives
M2 = I2DQ,
M3 = I303. =
(33.2)
In particular, for a spherical top, where all three principal moments of inertia
are equal, we have simply
M = IS,
(33.3)
i.e. the angular momentum vector is proportional to, and in the same direc-
tion as, the angular velocity vector. For an arbitrary body, however, the
vector M is not in general in the same direction as S; this happens only
when the body is rotating about one of its principal axes of inertia.
Let us consider a rigid body moving freely, i.e. not subject to any external
forces. We suppose that any uniform translational motion, which is of no
interest, is removed, leaving a free rotation of the body.
As in any closed system, the angular momentum of the freely rotating body
is constant. For a spherical top the condition M = constant gives C = con-
stant; that is, the most general free rotation of a spherical top is a uniform
rotation about an axis fixed in space.
The case of a rotator is equally simple. Here also M = IS, and the vector
S is perpendicular to the axis of the rotator. Hence a free rotation of a rotator
is a uniform rotation in one plane about an axis perpendicular to that plane.
The law of conservation of angular momentum also suffices to determine
the more complex free rotation of a symmetrical top. Using the fact that the
principal axes of inertia X1, X2 (perpendicular to the axis of symmetry (x3)
of the top) may be chosen arbitrarily, we take the x2-axis perpendicular to
the plane containing the constant vector M and the instantaneous position
of the x3-axis. Then M2 = 0, and formulae (33.2) show that Q2 = 0. This
means that the directions of M, St and the axis of the top are at every instant
in one plane (Fig. 46). Hence, in turn, it follows that the velocity V = Sxr
of every point on the axis of the top is at every instant perpendicular to that
§34
The equations of motion of a rigid body
107
plane. That is, the axis of the top rotates uniformly (see below) about the
direction of M, describing a circular cone. This is called regular precession
of the top. At the same time the top rotates uniformly about its own axis.
M
n
x3
22pr
x1
FIG. 46
The angular velocities of these two rotations can easily be expressed in
terms of the given angular momentum M and the angle 0 between the axis
of the top and the direction of M. The angular velocity of the top about its
own axis is just the component S3 of the vector S along the axis:
Q3 = M3/I3 = (M/I3) cos 0.
(33.4)
To determine the rate of precession Spr, the vector S must be resolved into
components along X3 and along M. The first of these gives no displacement
of the axis of the top, and the second component is therefore the required
angular velocity of precession. Fig. 46 shows that Spr sin 0 = Q1, and, since
S21 = M1/I1 = (M/I1) sin 0, we have
Spr r=M/I1.
(33.5)
$34. The equations of motion of a rigid body
Since a rigid body has, in general, six degrees of freedom, the general
equations of motion must be six in number. They can be put in a form which
gives the time derivatives of two vectors, the momentum and the angular
momentum of the body.
The first equation is obtained by simply summing the equations p = f
for each particle in the body, p being the momentum of the particle and f the
108
Motion of a Rigid Body
§34
force acting on it. In terms of the total momentum of the body P =
and total force acting on it F = Ef, we have
dP/dt = F.
(34.1)
Although F has been defined as the sum of all the forces f acting on the
various particles, including the forces due to other particles, F actually
includes only external forces: the forces of interaction between the particles
composing the body must cancel out, since if there are no external forces
the momentum of the body, like that of any closed system, must be conserved,
i.e. we must have F = 0.
If U is the potential energy of a rigid body in an external field, the force
F is obtained by differentiating U with respect to the co-ordinates of the
centre of mass of the body:
F = JUIR.
(34.2)
For, when the body undergoes a translation through a distance SR, the radius
vector r of every point in the body changes by SR, and so the change in the
potential energy is
SU = (U/dr) Sr = RR Couldr = SR SR.
It may be noted that equation (34.1) can also be obtained as Lagrange's
equation for the co-ordinates of the centre of mass, (d/dt)àL/JV = aL/JR,
with the Lagrangian (32.4), for which
OL/OV=,MV=P, 0L/JR = JU/OR = F.
Let us now derive the second equation of motion, which gives the time
derivative of the angular momentum M. To simplify the derivation, it is
convenient to choose the "fixed" (inertial) frame of reference in such a way
that the centre of mass is at rest in that frame at the instant considered.
We have M = (d/dt) Erxp = Eixp+. Our choice of the frame of
reference (with V = 0) means that the value of i at the instant considered is
the same as V = i. Since the vectors V and p = mv are parallel, ixp = 0.
Replacing p by the force f, we have finally
dM/dt = K,
(34.3)
where
K = .
(34.4)
Since M has been defined as the angular momentum about the centre of
mass (see the beginning of $33), it is unchanged when we go from one inertial
frame to another. This is seen from formula (9.5) with R = 0. We can there-
fore deduce that the equation of motion (34.3), though derived for a particular
frame of reference, is valid in any other inertial frame, by Galileo's relativity
principle.
The vector rxf is called the moment of the force f, and so K is the total
torque, i.e. the sum of the moments of all the forces acting on the body. Like
§34
The equations of motion of a rigid body
109
the total force F, the sum (34.4) need include only the external forces: by
the law of conservation of angular momentum, the sum of the moments of
the internal forces in a closed system must be zero.
The moment of a force, like the angular momentum, in general depends on
the choice of the origin about which it is defined. In (34.3) and (34.4) the
moments are defined with respect to the centre of mass of the body.
When the origin is moved a distance a, the new radius vector r' of each
point in the body is equal to r-a. Hence K = Erxf = Er'xf+ Eaxf or
K = K'+axF.
(34.5)
Hence we see, in particular, that the value of the torque is independent of
the choice of origin if the total force F = 0. In this case the body is said to
be acted on by a couple.
Equation (34.3) may be regarded as Lagrange's equation (d/dt) OL/OS
= 0L/dd for the "rotational co-ordinates". Differentiating the Lagrangian
(32.4) with respect to the components of the vector S2, we obtain
= IikOk = Mi. The change in the potential energy resulting from an
infinitesimal rotation SO of the body is SU = - Ef.Sr = -
= So. Erxf = -K.SO, whence
K =-20/00, =
(34.6)
so that aL/dd = 00/08 = K.
Let us assume that the vectors F and K are perpendicular. Then a vector a
can always be found such that K' given by formula (34.5) is zero and
K a x F.
(34.7)
The choice of a is not unique, since the addition to a of any vector parallel
to F does not affect equation (34.7). The condition K' = 0 thus gives a straight
line, not a point, in the moving system of co-ordinates. When K is perpendi-
cular to F, the effect of all the applied forces can therefore be reduced to that
of a single force F acting along this line.
Such a case is that of a uniform field of force, in which the force on a particle
is f = eE, with E a constant vector characterising the field and e characterising
the properties of a particle with respect to the field. Then F = Ee,
K = erxE. Assuming that # 0, we define a radius vector ro such that
(34.8)
Then the total torque is simply
=roxF
(34.9)
Thus, when a rigid body moves in a uniform field, the effect of the field
reduces to the action of a single force F applied at the point whose radius
vector is (34.8). The position of this point is entirely determined by the
t For example, in a uniform electric field E is the field strength and e the charge; in a
uniform gravitational field E is the acceleration g due to gravity and e is the mass m.
110
Motion of a Rigid Body
§35
properties of the body itself. In a gravitational field, for example, it is the
centre of mass.
$35. Eulerian angles
As has already been mentioned, the motion of a rigid body can be described
by means of the three co-ordinates of its centre of mass and any three angles
which determine the orientation of the axes X1, X2, X3 in the moving system of
co-ordinates relative to the fixed system X, Y, Z. These angles may often be
conveniently taken as what are called Eulerian angles.
Z
X2
Y
FIG. 47
Since we are here interested only in the angles between the co-ordinate
axes, we may take the origins of the two systems to coincide (Fig. 47). The
moving x1x2-plane intersects the fixed XY-plane in some line ON, called the
line of nodes. This line is evidently perpendicular to both the Z-axis and the
x3-axis; we take its positive direction as that of the vector product ZXX3
(where Z and X3 are unit vectors along the Z and X3 axes).
We take, as the quantities defining the position of the axes x1, X2, X3
relative to the axes X, Y, Z the angle 0 between the Z and X3 axes, the angle
between the X-axis and ON, and the angle as between the x1-axis and ON.
The angles and 4 are measured round the Z and X3 axes respectively in the
direction given by the corkscrew rule. The angle 0 takes values from 0 to TT,
and and 4 from 0 to 2n.t
t The angles 0 and - are respectively the polar angle and azimuth of the direction
X3 with respect to the axes X, Y, Z. The angles 0 and 12-- are respectively the polar angle
and azimuth of the direction Z with respect to the axes X1, X2, X3.

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§35
Eulerian angles
111
Let us now express the components of the angular velocity vector S along
the moving axes X1, X2, X3 in terms of the Eulerian angles and their derivatives.
To do this, we must find the components along those axes of the angular
velocities 6, b, 4. The angular velocity è is along the line of nodes ON, and
its components are 1 = O cos 4/5, = - O sin 4/5, = 0. The angular velo-
city is along the Z-axis; its component along the x3-axis is 03 = cos 0, and
in the x1x2-plane sin A. Resolving the latter along the X1 and X2 axes, we
have 01 = sin 0 sin 4/s, O2 = sin 0 cos 4. Finally, the angular velocity is
is along the x3-axis.
Collecting the components along each axis, we have
S21 = 0 COS 4,
Q2 = sin 0 cosy-osiny,
(35.1)
S23 = o cos0+4. =
If the axes X1, X2, X3 are taken to be the principal axes of inertia of the body,
the rotational kinetic energy in terms of the Eulerian angles is obtained by
substituting (35.1) in (32.8).
For a symmetrical top (I1 = I2 # I3), a simple reduction gives
Trot =
(35.2)
This expression can also be more simply obtained by using the fact that the
choice of directions of the principal axes X1, X2 is arbitrary for a symmetrical
top. If the X1 axis is taken along the line of nodes ON, i.e. 4 = 0, the compo-
nents of the angular velocity are simply
O2 = o sin A,
(35.3)
As a simple example of the use of the Eulerian angles, we shall use them
to determine the free motion of a symmetrical top, already found in $33.
We take the Z-axis of the fixed system of co-ordinates in the direction of the
constant angular momentum M of the top. The x3-axis of the moving system
is along the axis of the top; let the x1-axis coincide with the line of nodes at
the instant considered. Then the components of the vector M are, by
formulae (35.3), M1 = I1 = I, M2 = IS2 = sin 0, M3 = I3Q3
= I3( cos 0+4). Since the x1-axis is perpendicular to the Z-axis, we have
M1 = 0, M2 = M sin 0, M3 = M cos 0. Comparison gives
0=0,
I = M,
=
(35.4)
The first of these equations gives 0 = constant, i.e. the angle between the
axis of the top and the direction of M is constant. The second equation gives
the angular velocity of precession = M/I1, in agreement with (33.5).
Finally, the third equation gives the angular velocity with which the top
rotates about its own axis: S3 = (M/I3) cos 0.
112
Motion of a Rigid Body
§35
PROBLEMS
PROBLEM 1. Reduce to quadratures the problem of the motion of a heavy symmetrical
top whose lowest point is fixed (Fig. 48).
SOLUTION. We take the common origin of the moving and fixed systems of co-ordinates
at the fixed point O of the top, and the Z-axis vertical. The Lagrangian of the top in a gravita-
tional field is L = (02 +02 sin ²0 + 1/3(1- cos 0)2-ugl - cos 0, where u is the mass
of the top and l the distance from its fixed point to the centre of mass.
Z
X3
x2
a
ug
Y
x1
N
FIG. 48
The co-ordinates 4 and are cyclic. Hence we have two integrals of the motion:
P4 = = cos 0) = constant = M3
(1)
= = cos 0 = constant III M2,
(2)
where I'1 = I1+ul2; the quantities P4 and Po are the components of the rotational angular
momentum about O along the X3 and Z axes respectively. The energy
E = cos 0
(3)
is also conserved.
From equations (1) and (2) we find
=
0)/I'1
sin 20,
(4)
(5)
Eliminating b and of from the energy (3) by means of equations (4) and (5), we obtain
E' =
where
E'
=
(6)
§35
Eulerian angles
113
Thus we have
t=
(7)
this is an elliptic integral. The angles 4 and are then expressed in terms of 0 by means of
integrals obtained from equations (4) and (5).
The range of variation of 0 during the motion is determined by the condition E' Ueff(0).
The function Uett(8) tends to infinity (if M3 # M2) when 0 tends to 0 or II, and has a minimum
between these values. Hence the equation E' = Ueff(0) has two roots, which determine the
limiting values 01 and O2 of the inclination of the axis of the top to the vertical.
When 0 varies from 01 to O2, the derivative o changes sign if and only if the difference
M-M3 cos 0 changes sign in that range of 0. If it does not change sign, the axis of the top
precesses monotonically about the vertical, at the same time oscillating up and down. The
latter oscillation is called nutation; see Fig. 49a, where the curve shows the track of the axis
on the surface of a sphere whose centre is at the fixed point of the top. If does change sign,
the direction of precession is opposite on the two limiting circles, and so the axis of the top
describes loops as it moves round the vertical (Fig. 49b). Finally, if one of 01, O2 is a zero of
M2-M3 cos 0, of and è vanish together on the corresponding limiting circle, and the path
of the axis is of the kind shown in Fig. 49c.
O2
O2
O2
(a)
(b)
(c)
FIG. 49
PROBLEM 2. Find the condition for the rotation of a top about a vertical axis to be stable.
SOLUTION. For 0 = 0, the X3 and Z axes coincide, so that M3 = Mz, E' = 0. Rotation
about this axis is stable if 0 = 0 is a minimum of the function Ueff(9). For small 0 we have
Ueff 22 whence the condition for stability is M32 > 41'1ugl or S232
> 41'1ugl/I32.
PROBLEM 3. Determine the motion of a top when the kinetic energy of its rotation about
its axis is large compared with its energy in the gravitational field (called a "fast" top).
SOLUTION. In a first approximation, neglecting gravity, there is a free precession of the
axis of the top about the direction of the angular momentum M, corresponding in this case
to the nutation of the top; according to (33.5), the angular velocity of this precession is
Sunu = M/I' 1.
(1)
In the next approximation, there is a slow precession of the angular momentum M about
the vertical (Fig. 50). To determine the rate of this precession, we average the exact equation
of motion (34.3) dM/dt = K over the nutation period. The moment of the force of gravity
on the top is K=uln3xg, where n3 is a unit vector along the axis of the top. It is evident
from symmetry that the result of averaging K over the "nutation cone" is to replace n3 by
its component (M/M) cos a in the direction of M, where a is the angle between M and the
axis of the top. Thus we have dM/dt = -(ul/M)gxM cos a. This shows that the vector M
114
Motion of a Rigid Body
§36
precesses about the direction of g (i.e. the vertical) with a mean angular velocity
Spr (ul/M)g cos a
(2)
which is small compared with Senu
Spr
in
no
a
FIG. 50
In this approximation the quantities M and cos a in formulae (1) and (2) are constants,
although they are not exact integrals of the motion. To the same accuracy they are related
to the strictly conserved quantities E and M3 by M3 = M cos a,
§36. Euler's equations
The equations of motion given in §34 relate to the fixed system of co-
ordinates: the derivatives dP/dt and dM/dt in equations (34.1) and (34.3)
are the rates of change of the vectors P and M with respect to that system.
The simplest relation between the components of the rotational angular
momentum M of a rigid body and the components of the angular velocity
occurs, however, in the moving system of co-ordinates whose axes are the
principal axes of inertia. In order to use this relation, we must first transform
the equations of motion to the moving co-ordinates X1, X2, X3.
Let dA/dt be the rate of change of any vector A with respect to the fixed
system of co-ordinates. If the vector A does not change in the moving system,
its rate of change in the fixed system is due only to the rotation, so that
dA/dt = SxA; see §9, where it has been pointed out that formulae such as
(9.1) and (9.2) are valid for any vector. In the general case, the right-hand
side includes also the rate of change of the vector A with respect to the moving
system. Denoting this rate of change by d'A/dt, we obtain
dAdd
(36.1)
§36
Euler's equations
115
Using this general formula, we can immediately write equations (34.1) and
(34.3) in the form
=
K.
(36.2)
Since the differentiation with respect to time is here performed in the moving
system of co-ordinates, we can take the components of equations (36.2) along
the axes of that system, putting (d'P/dt)1 = dP1/dt, ..., (d'M/dt)1 = dM1/dt,
..., where the suffixes 1, 2, 3 denote the components along the axes x1, x2, X3.
In the first equation we replace P by V, obtaining
(36.3)
=
If the axes X1, X2, X3 are the principal axes of inertia, we can put M1 = I,
etc., in the second equation (36.2), obtaining
=
I2 = K2,
}
(36.4)
I3 = K3.
These are Euler's equations.
In free rotation, K = 0, so that Euler's equations become
= 0,
}
(36.5)
= 0.
As an example, let us apply these equations to the free rotation of a sym-
metrical top, which has already been discussed. Putting I1 = I2, we find from
the third equation SQ3 = 0, i.e. S3 = constant. We then write the first two
equations as O = -wS2, Q2 = wS1, where
=
(36.6)
is a constant. Multiplying the second equation by i and adding, we have
= so that S1+iD2 = A exp(iwt), where A is a
constant, which may be made real by a suitable choice of the origin of time.
Thus
S1 = A cos wt
Q2 = A sin wt.
(36.7)
116
Motion of a Rigid Body
§37
This result shows that the component of the angular velocity perpendicular
to the axis of the top rotates with an angular velocity w, remaining of constant
magnitude A = Since the component S3 along the axis of the
top is also constant, we conclude that the vector S rotates uniformly with
angular velocity w about the axis of the top, remaining unchanged in magni-
tude. On account of the relations M1 = , M2 = I2O2, M3 = I3O3 be-
tween the components of S and M, the angular momentum vector M evidently
executes a similar motion with respect to the axis of the top.
This description is naturally only a different view of the motion already
discussed in §33 and §35, where it was referred to the fixed system of co-
ordinates. In particular, the angular velocity of the vector M (the Z-axis in
Fig. 48, $35) about the x3-axis is, in terms of Eulerian angles, the same as
the angular velocity - 4. Using equations (35.4), we have
cos
or - is = I23(I3-I1)/I1, in agreement with (36.6).
§37. The asymmetrical top
We shall now apply Euler's equations to the still more complex problem
of the free rotation of an asymmetrical top, for which all three moments of
inertia are different. We assume for definiteness that
I3 > I2 I.
(37.1)
Two integrals of Euler's equations are known already from the laws of
conservation of energy and angular momentum:
= 2E,
(37.2)
= M2,
where the energy E and the magnitude M of the angular momentum are given
constants. These two equations, written in terms of the components of the
vector M, are
(37.3)
M2.
(37.4)
From these equations we can already draw some conclusions concerning
the nature of the motion. To do so, we notice that equations (37.3) and (37.4),
regarded as involving co-ordinates M1, M2, M3, are respectively the equation
of an ellipsoid with semiaxes (2EI1), (2EI2), (2EI3) and that of a sphere
of radius M. When the vector M moves relative to the axes of inertia of the
top, its terminus moves along the line of intersection of these two surfaces.
Fig. 51 shows a number of such lines of intersection of an ellipsoid with
§37
The asymmetrical top
117
spheres of various radii. The existence of an intersection is ensured by the
obviously valid inequalities
2EI1 < M2 < 2EI3,
(37.5)
which signify that the radius of the sphere (37.4) lies between the least and
greatest semiaxes of the ellipsoid (37.3).
x1
X2
FIG. 51
Let us examine the way in which these "paths"t of the terminus of the
vector M change as M varies (for a given value of E). When M2 is only slightly
greater than 2EI1, the sphere intersects the ellipsoid in two small closed curves
round the x1-axis near the corresponding poles of the ellipsoid; as M2 2EI1,
these curves shrink to points at the poles. When M2 increases, the curves
become larger, and for M2 = 2EI2 they become two plane curves (ellipses)
which intersect at the poles of the ellipsoid on the x2-axis. When M2 increases
further, two separate closed paths again appear, but now round the poles on
the
x3-axis; as M2 2EI3 they shrink to points at these poles.
First of all, we may note that, since the paths are closed, the motion of the
vector M relative to the top must be periodic; during one period the vector
M describes some conical surface and returns to its original position.
Next, an essential difference in the nature of the paths near the various
poles of the ellipsoid should be noted. Near the x1 and X3 axes, the paths lie
entirely in the neighbourhood of the corresponding poles, but the paths which
pass near the poles on the x2-axis go elsewhere to great distances from those
poles. This difference corresponds to a difference in the stability of the rota-
tion of the top about its three axes of inertia. Rotation about the x1 and X3
axes (corresponding to the least and greatest of the three moments of inertia)
t The corresponding curves described by the terminus of the vector Ca are called polhodes.
118
Motion of a Rigid Body
§37
is stable, in the sense that, if the top is made to deviate slightly from such a
state, the resulting motion is close to the original one. A rotation about the
x2-axis, however, is unstable: a small deviation is sufficient to give rise to a
motion which takes the top to positions far from its original one.
To determine the time dependence of the components of S (or of the com-
ponents of M, which are proportional to those of (2) we use Euler's equations
(36.5). We express S1 and S3 in terms of S2 by means of equations (37.2)
and (37.3):
S21 =
(37.6)
Q32 =
and substitute in the second equation (36.5), obtaining
dSQ2/dt (I3-I1)21-23/I2
= V{[(2EI3-M2-I2(I3-I2)22]
(37.7)
Integration of this equation gives the function t(S22) as an elliptic integral.
In reducing it to a standard form we shall suppose for definiteness that
M2 > 2EI2; if this inequality is reversed, the suffixes 1 and 3 are interchanged
in the following formulae. Using instead of t and S2 the new variables
(37.8)
S = S2V[I2(I3-I2)/(2EI3-M2)],
and defining a positive parameter k2 < 1 by
(37.9)
we obtain
ds
the origin of time being taken at an instant when S2 = 0. When this integral
is inverted we have a Jacobian elliptic function S = sn T, and this gives O2
as a function of time; S-1(t) and (33(t) are algebraic functions of 22(t) given
by (37.6). Using the definitions cn T = V(1-sn2r), dn T =
we find
Superscript(2) = [(2EI3-M2/I1(I3-I1)] CNT,
O2 =
(37.10)
O3 = dn T.
These are periodic functions, and their period in the variable T is 4K,
where K is a complete elliptic integral of the first kind:
=
(37.11)
§37
The asymmetrical top
119
The period in t is therefore
T =
(37.12)
After a time T the vector S returns to its original position relative to the
axes of the top. The top itself, however, does not return to its original position
relative to the fixed system of co-ordinates; see below.
For I = I2, of course, formulae (37.10) reduce to those obtained in §36
for a symmetrical top: as I I2, the parameter k2 0, and the elliptic
functions degenerate to circular functions: sn -> sin T, cn T cos
T,
dn T -> 1, and we return to formulae (36.7).
When M2 = 2EI3 we have Superscript(1) = S2 = 0, S3 = constant, i.e. the vector S
is always parallel to the x3-axis. This case corresponds to uniform rotation of
the top about the x3-axis. Similarly, for M2 = 2EI1 (when T III 0) we have
uniform rotation about the x1-axis.
Let us now determine the absolute motion of the top in space (i.e. its
motion relative to the fixed system of co-ordinates X, Y, Z). To do so, we
use the Eulerian angles 2/5, o, 0, between the axes X1, X2, X3 of the top and the
axes X, Y, Z, taking the fixed Z-axis in the direction of the constant vector M.
Since the polar angle and azimuth of the Z-axis with respect to the axes
x1, X2, X3 are respectively 0 and 1/77 - is (see the footnote to $35), we obtain on
taking the components of M along the axes X1, X2, X3
M sin 0 sin y = M1 = ,
M sin A cos is = M2 = I2O2,
(37.13)
M cos 0 = M3 = I3S23.
Hence
cos 0 = I3S3/M,
tan / =
(37.14)
and from formulae (37.10)
COS 0 = dn T,
(37.15)
tan 4 = cn r/snt,
which give the angles 0 and is as functions of time; like the components of the
vector S, they are periodic functions, with period (37.12).
The angle does not appear in formulae (37.13), and to calculate it we
must return to formulae (35.1), which express the components of S in terms
of the time derivatives of the Eulerian angles. Eliminating O from the equa-
tions S1 = sin 0 sin 4 + O cos 2/5, S2 = sin 0 cos 4-0 - sin 2/5, we obtain
& = (Superscript(2) sin 4+S2 cos 4)/sin 0, and then, using formulae (37.13),
do/dt =
(37.16)
The function (t) is obtained by integration, but the integrand involves
elliptic functions in a complicated way. By means of some fairly complex
120
Motion of a Rigid Body
§37
transformations, the integral can be expressed in terms of theta functions;
we shall not give the calculations, but only the final result.
The function (t) can be represented (apart from an arbitrary additive
constant) as a sum of two terms:
$(t) = (11(t)++2(t),
(37.17)
one of which is given by
(37.18)
where D01 is a theta function and a a real constant such that
sn(2ixK) = iv[I3(M2-2I1)/I1(2EI3-M2]
(37.19)
K and Tare given by (37.11) and (37.12). The function on the right-hand side
of (37.18) is periodic, with period 1T, so that 01(t) varies by 2n during a time
T. The second term in (37.17) is given by
(37.20)
This function increases by 2nr during a time T'. Thus the motion in is a
combination of two periodic motions, one of the periods (T) being the same
as the period of variation of the angles 4 and 0, while the other (T') is incom-
mensurable with T. This incommensurability has the result that the top does
not at any time return exactly to its original position.
PROBLEMS
PROBLEM 1. Determine the free rotation of a top about an axis near the x3-axis or the
x1-axis.
SOLUTION. Let the x3-axis be near the direction of M. Then the components M1 and M2
are small quantities, and the component M3 = M (apart from quantities of the second and
higher orders of smallness). To the same accuracy the first two Euler's equations (36.5) can
be written dM1/dt = DoM2(1-I3/I2), dM2/dt = QOM1(I3/I1-1), where So = M/I3. As
usual we seek solutions for M1 and M2 proportional to exp(iwt), obtaining for the frequency w
(1)
The values of M1 and M2 are
cos wt, sin wt,
(2)
where a is an arbitrary small constant. These formulae give the motion of the vector M
relative to the top. In Fig. 51, the terminus of the vector M describes, with frequency w,
a small ellipse about the pole on the x3-axis.
To determine the absolute motion of the top in space, we calculate its Eulerian angles.
In the present case the angle 0 between the x3-axis and the Z-axis (direction of M) is small,
t These are given by E. T. WHITTAKER, A Treatise on the Analytical Dynamics of Particles
and Rigid Bodies, 4th ed., Chapter VI, Dover, New York 1944.

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§37
The asymmetrical top
121
and by formulae (37.14) tan of = M1/M2, cos 0) 2(1 (M3/M) 22
substituting (2), we obtain
tan 4 = V[I(I3-I2)/I2(I3-I1)] cot wt,
(3)
To find , we note that, by the third formula (35.1), we have, for 0 1,
Hence
= lot
(4)
omitting an arbitrary constant of integration.
A clearer idea of the nature of the motion of the top is obtained if we consider the change
in direction of the three axes of inertia. Let n1, n2, n3 be unit vectors along these axes. The
vectors n1 and n2 rotate uniformly in the XY-plane with frequency So, and at the same time
execute small transverse oscillations with frequency w. These oscillations are given by the
Z-components of the vectors:
22 M1/M = av(I3/I2-1) cos wt,
N2Z 22 M2/M = av(I3/I1-1) sin wt.
For the vector n3 we have, to the same accuracy, N3x 22 0 sin , N3y 22 -0 cos , n3z 1.
(The polar angle and azimuth of n3 with respect to the axes X, Y, Z are 0 and -; see
the footnote to 35.) We also write, using formulae (37.13),
naz=0sin(Qot-4)
= Asin Sot cos 4-0 cos lot sin 4
= (M 2/M) sin Dot-(M1/M) cos Sot
sin Sot sin N/1-1) cos Not cos wt
cos(so
Similarly
From this we see that the motion of n3 is a superposition of two rotations about the Z-axis
with frequencies So + w.
PROBLEM 2. Determine the free rotation of a top for which M2 = 2EI2.
SOLUTION. This case corresponds to the movement of the terminus of M along a curve
through the pole on the x2-axis (Fig. 51). Equation (37.7) becomes ds/dr = 1-s2,
= S = I2/20, where So = M/I2 = 2E|M. Integration of
this equation and the use of formulae (37.6) gives
sech T,
}
(1)
sech T.
To describe the absolute motion of the top, we use Eulerian angles, defining 0 as the angle
between the Z-axis (direction of M) and the x2-axis (not the x3-axis as previously). In formulae
(37.14) and (37.16), which relate the components of the vector CA to the Eulerian angles, we
5
122
Motion of a Rigid Body
§38
must cyclically permute the suffixes 1, 2, 3 to 3, 1, 2. Substitution of (1) in these formulae
then gives cos 0 = tanh T, = lot + constant, tan =
It is seen from these formulae that, as t 8, the vector SC asymptotically approaches the
x2-axis, which itself asymptotically approaches the Z-axis.
$38. Rigid bodies in contact
The equations of motion (34.1) and (34.3) show that the conditions of
equilibrium for a rigid body can be written as the vanishing of the total force
and total torque on the body:
F = f = 0 ,
K ==~rxf=0. =
(38.1)
Here the summation is over all the external forces acting on the body, and r
is the radius vector of the "point of application"; the origin with respect to
which the torque is defined may be chosen arbitrarily, since if F = 0 the
value of K does not depend on this choice (see (34.5)).
If we have a system of rigid bodies in contact, the conditions (38.1) for
each body separately must hold in equilibrium. The forces considered must
include those exerted on each body by those with which it is in contact. These
forces at the points of contact are called reactions. It is obvious that the mutual
reactions of any two bodies are equal in magnitude and opposite in direction.
In general, both the magnitudes and the directions of the reactions are
found by solving simultaneously the equations of equilibrium (38.1) for all the
bodies. In some cases, however, their directions are given by the conditions
of the problem. For example, if two bodies can slide freely on each other, the
reaction between them is normal to the surface.
If two bodies in contact are in relative motion, dissipative forces of friction
arise, in addition to the reaction.
There are two possible types of motion of bodies in contact-sliding and
rolling. In sliding, the reaction is perpendicular to the surfaces in contact,
and the friction is tangential. Pure rolling, on the other hand, is characterised
by the fact that there is no relative motion of the bodies at the point of
contact; that is, a rolling body is at every instant as it were fixed to the point
of contact. The reaction may be in any direction, i.e. it need not be normal
to the surfaces in contact. The friction in rolling appears as an additional
torque which opposes rolling.
If the friction in sliding is negligibly small, the surfaces concerned are
said to be perfectly smooth. If, on the other hand, only pure rolling without
sliding is possible, and the friction in rolling can be neglected, the surfaces
are said to be perfectly rough.
In both these cases the frictional forces do not appear explicitly in the pro-
blem, which is therefore purely one of mechanics. If, on the other hand, the
properties of the friction play an essential part in determining the motion,
then the latter is not a purely mechanical process (cf. $25).
Contact between two bodies reduces the number of their degrees of freedom
as compared with the case of free motion. Hitherto, in discussing such
§38
Rigid bodies in contact
123
problems, we have taken this reduction into account by using co-ordinates
which correspond directly to the actual number of degrees of freedom. In
rolling, however, such a choice of co-ordinates may be impossible.
The condition imposed on the motion of rolling bodies is that the velocities
of the points in contact should be equal; for example, when a body rolls on a
fixed surface, the velocity of the point of contact must be zero. In the general
case, this condition is expressed by the equations of constraint, of the form
E caide = 0,
(38.2)
where the Cai are functions of the co-ordinates only, and the suffix a denumer-
ates the equations. If the left-hand sides of these equations are not the total
time derivatives of some functions of the co-ordinates, the equations cannot
be integrated. In other words, they cannot be reduced to relations between the
co-ordinates only, which could be used to express the position of the bodies
in terms of fewer co-ordinates, corresponding to the actual number of degrees
of freedom. Such constraints are said to be non-holonomic, as opposed to
holonomic constraints, which impose relations between the co-ordinates only.
Let us consider, for example, the rolling of a sphere on a plane. As usual,
we denote by V the translational velocity (the velocity of the centre of the
sphere), and by Sa the angular velocity of rotation. The velocity of the point
of contact with the plane is found by putting r = - an in the general formula
V = +SXR; a is the radius of the sphere and n a unit vector along the
normal to the plane. The required condition is that there should be no sliding
at the point of contact, i.e.
V-aSxxn = 0.
(38.3)
This cannot be integrated: although the velocity V is the total time derivative
of the radius vector of the centre of the sphere, the angular velocity is not in
general the total time derivative of any co-ordinate. The constraint (38.3) is
therefore non-holonomic.t
Since the equations of non-holonomic constraints cannot be used to reduce
the number of co-ordinates, when such constraints are present it is necessary
to use co-ordinates which are not all independent. To derive the correspond-
ing Lagrange's equations, we return to the principle of least action.
The existence of the constraints (38.2) places certain restrictions on the
possible values of the variations of the co-ordinates: multiplying equations
(38.2) by St, we find that the variations dqi are not independent, but are
related by
(38.4)
t It may be noted that the similar constraint in the rolling of a cylinder is holonomic. In
that case the axis of rotation has a fixed direction in space, and hence la = do/dt is the total
derivative of the angle of rotation of the cylinder about its axis. The condition (38.3) can
therefore be integrated, and gives a relation between the angle and the co-ordinate of the
centre of mass.
124
Motion of a Rigid Body
§38
This must be taken into account in varying the action. According to
Lagrange's method of finding conditional extrema, we must add to the inte-
grand in the variation of the action
=
the left-hand sides of equations (38.4) multiplied by undetermined coeffici-
ents da (functions of the co-ordinates), and then equate the integral to zero.
In SO doing the variations dqi are regarded as entirely independent, and the
result is
(38.5)
These equations, together with the constraint equations (38.2), form a com-
plete set of equations for the unknowns qi and da.
The reaction forces do not appear in this treatment, and the contact of
the bodies is fully allowed for by means of the constraint equations. There
is, however, another method of deriving the equations of motion for bodies in
contact, in which the reactions are introduced explicitly. The essential feature
of this method, which is sometimes called d'Alembert's principle, is to write
for each of the bodies in contact the equations.
dP/dt==f,
(38.6)
wherein the forces f acting on each body include the reactions. The latter
are initially unknown and are determined, together with the motion of the
body, by solving the equations. This method is equally applicable for both
holonomic and non-holonomic constraints.
PROBLEMS
PROBLEM 1. Using d'Alembert's principle, find the equations of motion of a homogeneous
sphere rolling on a plane under an external force F and torque K.
SOLUTION. The constraint equation is (38.3). Denoting the reaction force at the point of
contact between the sphere and the plane by R, we have equations (38.6) in the form
u dV/dt = F+R,
(1)
dSu/dt = K-an xR,
(2)
where we have used the facts that P = V and, for a spherical top, M = ISE. Differentiating
the constraint equation (38.3) with respect to time, we have V = aS2xn. Substituting in
equation (1) and eliminating S by means of (2), we obtain (I/au)(F+R) = Kxn-aR+
+an(n . R), which relates R, F and K. Writing this equation in components and substitut-
ing I = zua2 (§32, Problem 2(b)), we have
R2 = -F2,
where the plane is taken as the xy-plane. Finally, substituting these expressions in (1), we
§38
Rigid bodies in contact
125
obtain the equations of motion involving only the given external force and torque:
dVx dt 7u 5 Ky
dt
The components Ox, Q2 y of the angular velocity are given in terms of Vx, Vy by the constraint
equation (38.3); for S2 we have the equation 2 dQ2/dt = K2, the z-component of equa-
tion (2).
PROBLEM 2. A uniform rod BD of weight P and length l rests against a wall as shown in
Fig. 52 and its lower end B is held by a string AB. Find the reaction of the wall and the ten-
sion in the string.
Rc
h
P
RB
T
A
B
FIG. 52
SOLUTION. The weight of the rod can be represented by a force P vertically downwards,
applied at its midpoint. The reactions RB and Rc are respectively vertically upwards and
perpendicular to the rod; the tension T in the string is directed from B to A. The solution
of the equations of equilibrium gives Rc = (Pl/4h) sin 2a, RB = P-Rcsin x, T = Rc cos a.
PROBLEM 3. A rod of weight P has one end A on a vertical plane and the other end B on
a horizontal plane (Fig. 53), and is held in position by two horizontal strings AD and BC,
RB
TA
A
RA
C
FIG. 53
126
Motion of a Rigid Body
§39
the latter being in the same vertical plane as AB. Determine the reactions of the planes and
the tensions in the strings.
SOLUTION. The tensions TA and TB are from A to D and from B to C respectively. The
reactions RA and RB are perpendicular to the corresponding planes. The solution of the
equations of equilibrium gives RB = P, TB = 1P cot a, RA = TB sin B, TA = TB cos B.
PROBLEM 4. Two rods of length l and negligible weight are hinged together, and their ends
are connected by a string AB (Fig. 54). They stand on a plane, and a force F is applied
at the midpoint of one rod. Determine the reactions.
RC
C
PA
F
1
RB
T
T
A
B
FIG. 54
SOLUTION. The tension T acts at A from A to B, and at B from B to A. The reactions RA
and RB at A and B are perpendicular to the plane. Let Rc be the reaction on the rod AC at
the hinge; then a reaction - -Rc acts on the rod BC. The condition that the sum of the moments
of the forces RB, T and - -Rc acting on the rod BC should be zero shows that Rc acts along
BC. The remaining conditions of equilibrium (for the two rods separately) give RA = 1F,
RB = 1F, Rc = 1F cosec a, T = 1F cot a, where a is the angle CAB.
§39. Motion in a non-inertial frame of reference
Up to this point we have always used inertial frames of reference in discuss-
ing the motion of mechanical systems. For example, the Lagrangian
L = 1mvo2- U,
(39.1)
and the corresponding equation of motion m dvo/dt = - au/dr, for a single
particle in an external field are valid only in an inertial frame. (In this section
the suffix 0 denotes quantities pertaining to an inertial frame.)
Let us now consider what the equations of motion will be in a non-inertial
frame of reference. The basis of the solution of this problem is again the
principle of least action, whose validity does not depend on the frame of
reference chosen. Lagrange's equations
(39.2)
are likewise valid, but the Lagrangian is no longer of the form (39.1), and to
derive it we must carry out the necessary transformation of the function Lo.
§39
Motion in a non-inertial frame of reference
127
This transformation is done in two steps. Let us first consider a frame of
reference K' which moves with a translational velocity V(t) relative to the
inertial frame K0. The velocities V0 and v' of a particle in the frames Ko and
K' respectively are related by
vo = v'+ V(t).
(39.3)
Substitution of this in (39.1) gives the Lagrangian in K':
L' = 1mv2+mv.+1mV2-U
Now V2(t) is a given function of time, and can be written as the total deriva-
tive with respect to t of some other function; the third term in L' can there-
fore be omitted. Next, v' = dr'/dt, where r' is the radius vector of the par-
ticle in the frame K'. Hence
mV(t)+v'= mV.dr/'dt = d(mV.r')/dt-mr'.dV/dt.
Substituting in the Lagrangian and again omitting the total time derivative,
we have finally
L' =
(39.4)
where W = dV/dt is the translational acceleration of the frame K'.
The Lagrange's equation derived from (39.4) is
(39.5)
Thus an accelerated translational motion of a frame of reference is equivalent,
as regards its effect on the equations of motion of a particle, to the application
of a uniform field of force equal to the mass of the particle multiplied by the
acceleration W, in the direction opposite to this acceleration.
Let us now bring in a further frame of reference K, whose origin coincides
with that of K', but which rotates relative to K' with angular velocity Su(t).
Thus K executes both a translational and a rotational motion relative to the
inertial frame Ko.
The velocity v' of the particle relative to K' is composed of its velocity
V
relative to K and the velocity Sxr of its rotation with K: v' = Lxr
(since the radius vectors r and r' in the frames K and K' coincide). Substitut-
ing this in the Lagrangian (39.4), we obtain
L = +mv.Sx+1m(xr)2-mW.r-
(39.6)
This is the general form of the Lagrangian of a particle in an arbitrary, not
necessarily inertial, frame of reference. The rotation of the frame leads to the
appearance in the Lagrangian of a term linear in the velocity of the particle.
To calculate the derivatives appearing in Lagrange's equation, we write
128
Motion of a Rigid Body
§39
the total differential
dL = mv.dv+mdv.Sxr+mv.Sxdr+
=
v.dv+mdv.xr+mdr.vxR+
The terms in dv and dr give
0L/dr X Q-mW-dU/0r. - -
Substitution of these expressions in (39.2) gives the required equation of
motion:
mdv/dt = (39.7)
We see that the "inertia forces" due to the rotation of the frame consist
of three terms. The force mrxo is due to the non-uniformity of the rotation,
but the other two terms appear even if the rotation is uniform. The force
2mvxs is called the Coriolis force; unlike any other (non-dissipative) force
hitherto considered, it depends on the velocity of the particle. The force
mSX(rxS) is called the centrifugal force. It lies in the plane through r and
S, is perpendicular to the axis of rotation (i.e. to S2), and is directed away
from the axis. The magnitude of this force is mpO2, where P is the distance
of the particle from the axis of rotation.
Let us now consider the particular case of a uniformly rotating frame with
no translational acceleration. Putting in (39.6) and (39.7) S = constant,
W = 0, we obtain the Lagrangian
L
=
(39.8)
and the equation of motion
mdv/dt = -
(39.9)
The energy of the particle in this case is obtained by substituting
p =
(39.10)
in E = p.v-L, which gives
E =
(39.11)
It should be noticed that the energy contains no term linear in the velocity.
The rotation of the frame simply adds to the energy a term depending only
on the co-ordinates of the particle and proportional to the square of the
angular velocity. This additional term - 1m(Sxr)2 is called the centrifugal
potential energy.
The velocity V of the particle relative to the uniformly rotating frame of
reference is related to its velocity V0 relative to the inertial frame Ko by
(39.12)
§39
Motion in a non-inertial frame of reference
129
The momentum p (39.10) of the particle in the frame K is therefore the same
as its momentum Po = MVO in the frame K0. The angular momenta
M = rxpo and M = rxp are likewise equal. The energies of the particle
in the two frames are not the same, however. Substituting V from (39.12) in
(39.11), we obtain E = 1mv02-mvo Sxr+U = 1mvo2 + mrxvo S.
The first two terms are the energy E0 in the frame K0. Using the angular
momentum M, we have
E = E0 n-M.S.
(39.13)
This formula gives the law of transformation of energy when we change to a
uniformly rotating frame. Although it has been derived for a single particle,
the derivation can evidently be generalised immediately to any system of
particles, and the same formula (39.13) is obtained.
PROBLEMS
PROBLEM 1. Find the deflection of a freely falling body from the vertical caused by the
Earth's rotation, assuming the angular velocity of this rotation to be small.
SOLUTION. In a gravitational field U = -mg. r, where g is the gravity acceleration
vector; neglecting the centrifugal force in equation (39.9) as containing the square of S, we
have the equation of motion
v = 2vxSu+g.
(1)
This equation may be solved by successive approximations. To do so, we put V = V1+V2,
where V1 is the solution of the equation V1 = g, i.e. V1 = gt+ (Vo being the initial velocity).
Substituting V = V1+v2in (1) and retaining only V1 on the right, we have for V2 the equation
V2 = 2v1xSc = 2tgxSt+2voxS. Integration gives
(2)
where h is the initial radius vector of the particle.
Let the z-axis be vertically upwards, and the x-axis towards the pole; then gx = gy = 0,
n sin 1, where A is the latitude (which for definite-
ness we take to be north). Putting V0 = 0 in (2), we find x = 0, =-1t300 cos A. Substitu-
tion of the time of fall t 22 (2h/g) gives finally x = 0,3 = - 1(2h/g)3/2 cos A, the negative
value indicating an eastward deflection.
PROBLEM 2. Determine the deflection from coplanarity of the path of a particle thrown
from the Earth's surface with velocity Vo.
SOLUTION. Let the xx-plane be such as to contain the velocity Vo. The initial altitude
h = 0. The lateral deviation is given by (2), Problem 1: y =
or, substituting the time of flight t 22 2voz/g, y =
PROBLEM 3. Determine the effect of the Earth's rotation on small oscillations of a pendulum
(the problem of Foucault's pendulum).
SOLUTION. Neglecting the vertical displacement of the pendulum, as being a quantity
of the second order of smallness, we can regard the motion as taking place in the horizontal
xy-plane. Omitting terms in N°, we have the equations of motion x+w2x = 20zy, j+w2y
= -20zx, where w is the frequency of oscillation of the pendulum if the Earth's rotation is
neglected. Multiplying the second equation by i and adding, we obtain a single equation
130
Motion of a Rigid Body
§39
+2i02s+w28 = 0 for the complex quantity $ = xtiy. For I2<<, the solution of this
equation is
$ = exp(-is2t) [A1 exp(iwt) +A2 exp(-iwt)]
or
xtiy = (xo+iyo) exp(-is2zt),
where the functions xo(t), yo(t) give the path of the pendulum when the Earth's rotation is
neglected. The effect of this rotation is therefore to turn the path about the vertical with
angular velocity Qz.

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CHAPTER VII
THE CANONICAL EQUATIONS
§40. Hamilton's equations
THE formulation of the laws of mechanics in terms of the Lagrangian, and
of Lagrange's equations derived from it, presupposes that the mechanical
state of a system is described by specifying its generalised co-ordinates and
velocities. This is not the only possible mode of description, however. A
number of advantages, especially in the study of certain general problems of
mechanics, attach to a description in terms of the generalised co-ordinates
and momenta of the system. The question therefore arises of the form of
the equations of motion corresponding to that formulation of mechanics.
The passage from one set of independent variables to another can be
effected by means of what is called in mathematics Legendre's transformation.
In the present case this transformation is as follows. The total differential
of the Lagrangian as a function of co-ordinates and velocities is
dL =
This expression may be written
(40.1)
since the derivatives aL/dqi are, by definition, the generalised momenta, and
aL/dqi = pi by Lagrange's equations. Writing the second term in (40.1) as
= - Eqi dpi, taking the differential d(piqi) to the left-hand
side, and reversing the signs, we obtain from (40.1)
The argument of the differential is the energy of the system (cf. §6);
expressed in terms of co-ordinates and momenta, it is called the Hamilton's
function or Hamiltonian of the system:
(40.2)
t The reader may find useful the following table showing certain differences between the
nomenclature used in this book and that which is generally used in the English literature.
Here
Elsewhere
Principle of least action
Hamilton's principle
Maupertuis' principle
Principle of least action
Maupertuis' principle
Action
Hamilton's principal function
Abbreviated action
Action
- -Translators.
131
132
The Canonical Equations
§40
From the equation in differentials
dH =
(40.3)
in which the independent variables are the co-ordinates and momenta, we
have the equations
=
(40.4)
These are the required equations of motion in the variables P and q, and
are called Hamilton's equations. They form a set of 2s first-order differential
equations for the 2s unknown functions Pi(t) and qi(t), replacing the S second-
order equations in the Lagrangian treatment. Because of their simplicity and
symmetry of form, they are also called canonical equations.
The total time derivative of the Hamiltonian is
Substitution of qi and pi from equations (40.4) shows that the last two terms
cancel, and so
dH/dt==Hoo.
(40.5)
In particular, if the Hamiltonian does not depend explicitly on time, then
dH/dt = 0, and we have the law of conservation of energy.
As well as the dynamical variables q, q or q, P, the Lagrangian and the
Hamiltonian involve various parameters which relate to the properties of the
mechanical system itself, or to the external forces on it. Let A be one such
parameter. Regarding it as a variable, we have instead of (40.1)
dL
and (40.3) becomes
dH =
Hence
(40.6)
which relates the derivatives of the Lagrangian and the Hamiltonian with
respect to the parameter A. The suffixes to the derivatives show the quantities
which are to be kept constant in the differentiation.
This result can be put in another way. Let the Lagrangian be of the form
L = Lo + L', where L' is a small correction to the function Lo. Then the
corresponding addition H' in the Hamiltonian H = H + H' is related to L'
by
(H')p,a - (L')
(40.7)
It may be noticed that, in transforming (40.1) into (40.3), we did not
include a term in dt to take account of a possible explicit time-dependence
§41
The Routhian
133
of the Lagrangian, since the time would there be only a parameter which
would not be involved in the transformation. Analogously to formula (40.6),
the partial time derivatives of L and H are related by
(40.8)
PROBLEMS
PROBLEM 1. Find the Hamiltonian for a single particle in Cartesian, cylindrical and
spherical co-ordinates.
SOLUTION. In Cartesian co-ordinates x, y, 2,
in cylindrical co-ordinates r, , z,
in spherical co-ordinates r, 0, ,
PROBLEM 2. Find the Hamiltonian for a particle in a uniformly rotating frame of reference.
SOLUTION. Expressing the velocity V in the energy (39.11) in terms of the momentum p
by (39.10), we have H = p2/2m-S rxp+U.
PROBLEM 3. Find the Hamiltonian for a system comprising one particle of mass M and n
particles each of mass m, excluding the motion of the centre of mass (see §13, Problem).
SOLUTION. The energy E is obtained from the Lagrangian found in §13, Problem, by
changing the sign of U. The generalised momenta are
Pa = OL/OV
Hence
-
= (mM/14)
=
=
Substitution in E gives
41. The Routhian
In some cases it is convenient, in changing to new variables, to replace
only some, and not all, of the generalised velocities by momenta. The trans-
formation is entirely similar to that given in 40.
To simplify the formulae, let us at first suppose that there are only two
co-ordinates q and E, say, and transform from the variables q, $, q, $ to
q, $, p, & where P is the generalised momentum corresponding to the co-
ordinate q.
134
The Canonical Equations
§42
The differential of the Lagrangian L(q, $, q, §) is
dL = dq + (al/dg) ds (0L/as) d
d,
whence
= (0L/d) d.
If we define the Routhian as
= pq-L,
(41.1)
in which the velocity q is expressed in terms of the momentum P by means
of the equation P = 0L/dq, then its differential is
dR = - ds - (aL/a)
(41.2)
Hence
DRIP, p = OR/dq,
(41.3)
(41.4)
Substituting these equations in the Lagrangian for the co-ordinate $, we have
(41.5)
Thus the Routhian is a Hamiltonian with respect to the co-ordinate q
(equations (41.3)) and a Lagrangian with respect to the co-ordinate $ (equation
(41.5)).
According to the general definition the energy of the system is
E -p-L =
In terms of the Routhian it is
E=R-R,
(41.6)
as we find by substituting (41.1) and (41.4).
The generalisation of the above formulae to the case of several co-ordinates
q and & is evident.
The use of the Routhian may be convenient, in particular, when some of
the co-ordinates are cyclic. If the co-ordinates q are cyclic, they do not appear
in the Lagrangian, nor therefore in the Routhian, so that the latter is a func-
tion of P, $ and $. The momenta P corresponding to cyclic co-ordinates are
constant, as follows also from the second equation (41.3), which in this sense
contains no new information. When the momenta P are replaced by their
given constant values, equations (41.5) (d/dt) JR(p, $, 5)108 = JR(P, &, §) 128
become equations containing only the co-ordinates $, so that the cyclic co-
ordinates are entirely eliminated. If these equations are solved for the func-
tions (t), substitution of the latter on the right-hand sides of the equations
q = JR(p, $, E) gives the functions q(t) by direct integration.
PROBLEM
Find the Routhian for a symmetrical top in an external field U(, 0), eliminating the cyclic
co-ordinate 4 (where 4, , 0 are Eulerian angles).
§42
Poisson brackets
135
SOLUTION. The Lagrangian is = see
§35, Problem 1. The Routhian is
R = cos 0);
the first term is a constant and may be omitted.
42. Poisson brackets
Let f (p, q, t) be some function of co-ordinates, momenta and time. Its
total time derivative is
df
Substitution of the values of and Pk given by Hamilton's equations (40.4)
leads to the expression
(42.1)
where
(42.2)
dqk
This expression is called the Poisson bracket of the quantities H and f.
Those functions of the dynamical variables which remain constant during
the motion of the system are, as we know, called integrals of the motion.
We see from (42.1) that the condition for the quantity f to be an integral of
the motion (df/dt = 0) can be written
af(dt+[H,f]=0
(42.3)
If the integral of the motion is not explicitly dependent on the time, then
[H,f] = 0,
(42.4)
i.e. the Poisson bracket of the integral and the Hamiltonian must be zero.
For any two quantities f and g, the Poisson bracket is defined analogously
to (42.2):
(42.5)
The Poisson bracket has the following properties, which are easily derived
from its definition.
If the two functions are interchanged, the bracket changes sign; if one of
the functions is a constant c, the bracket is zero:
(42.6)
[f,c]=0.
(42.7)
Also
[f1+f2,g]=[f1,g)+[f2,g]
(42.8)
[f1f2,g] ]=fi[fa,8]+f2[f1,8] =
(42.9)
Taking the partial derivative of (42.5) with respect to time, we obtain
(42.10)
136
The Canonical Equations
§42
If one of the functions f and g is one of the momenta or co-ordinates, the
Poisson bracket reduces to a partial derivative:
(42.11)
(42.12)
Formula (42.11), for example, may be obtained by putting g = qk in (42.5);
the sum reduces to a single term, since dqk/dqi = 8kl and dqk/dpi = 0. Put-
ting in (42.11) and (42.12) the function f equal to qi and Pi we have, in parti-
cular,
[qi,qk] = [Pi, Pk] =0, [Pi, 9k] = Sik.
(42.13)
The relation
[f,[g,h]]+[g,[h,f]]+[h,[f,g]] = 0,
(42.14)
known as Jacobi's identity, holds between the Poisson brackets formed from
three functions f, g and h. To prove it, we first note the following result.
According to the definition (42.5), the Poisson bracket [f,g] is a bilinear
homogeneous function of the first derivatives of f and g. Hence the bracket
[h,[f,g]], for example, is a linear homogeneous function of the second
derivatives of f and g. The left-hand side of equation (42.14) is therefore a
linear homogeneous function of the second derivatives of all three functions
f, g and h. Let us collect the terms involving the second derivatives of f.
The first bracket contains no such terms, since it involves only the first
derivatives of f. The sum of the second and third brackets may be symboli-
cally written in terms of the linear differential operators D1 and D2, defined by
D1() = [g, ], D2(b) = [h, ]. Then
3,[h,f]]+[h,[f,g]] = [g, [h,f]]-[h,[g,f]
= D1[D2(f)]-D2[D1(f)]
= (D1D2-D2D1)f.
It is easy to see that this combination of linear differential operators cannot
involve the second derivatives of f. The general form of the linear differential
operators is
where & and Nk are arbitrary functions of the variables .... Then
and the difference of these,
§42
Poisson brackets
137
is again an operator involving only single differentiations. Thus the terms in
the second derivatives of f on the left-hand side of equation (42.14) cancel
and, since the same is of course true of g and h, the whole expression is identi-
cally zero.
An important property of the Poisson bracket is that, if f and g are two
integrals of the motion, their Poisson bracket is likewise an integral of the
motion:
[f,g] = constant. =
(42.15)
This is Poisson's theorem. The proof is very simple if f and g do not depend
explicitly on the time. Putting h = H in Jacobi's identity, we obtain
[H,[f,g]]+[f,[g,H]]+[g,[H,fl]=0.
Hence, if [H, g] =0 and [H,f] = 0, then [H,[f,g]] = 0, which is the
required result.
If the integrals f and g of the motion are explicitly time-dependent, we
put, from (42.1),
Using formula (42.10) and expressing the bracket [H, [f,g]] in terms of two
others by means of Jacobi's identity, we find
d
[
(42.16)
which evidently proves Poisson's theorem.
Of course, Poisson's theorem does not always supply further integrals of
the motion, since there are only 2s-1 - of these (s being the number of degrees
of freedom). In some cases the result is trivial, the Poisson bracket being a
constant. In other cases the integral obtained is simply a function of the ori-
ginal integrals f and g. If neither of these two possibilities occurs, however,
then the Poisson bracket is a further integral of the motion.
PROBLEMS
PROBLEM 1. Determine the Poisson brackets formed from the Cartesian components of
the momentum p and the angular momentum M = rxp of a particle.
SOLUTION. Formula (42.12) gives [Mx, Py] = -MM/Dy = -d(yp:-2py)/dy
=
-Pz,
and similarly [Mx, Px] = 0, [Mx, P2] = Py. The remaining brackets are obtained by cyclically
permuting the suffixes x, y, Z.
6
138
The Canonical Equations
§43
PROBLEM 2. Determine the Poisson brackets formed from the components of M.
SOLUTION. A direct calculation from formula (42.5) gives [Mx, My] = -M2, [My, M]
= -Mx, [Mz, Mx] = -My.
Since the momenta and co-ordinates of different particles are mutually independent variables,
it is easy to see that the formulae derived in Problems 1 and 2 are valid also for the total
momentum and angular momentum of any system of particles.
PROBLEM 3. Show that [, M2] = 0, where is any function, spherically symmetrical
about the origin, of the co-ordinates and momentum of a particle.
SOLUTION. Such a function can depend on the components of the vectors r and p only
through the combinations r2, p2, r. p. Hence
and similarly for The required relation may be verified by direct calculation from
formula (42.5), using these formulae for the partial derivatives.
PROBLEM 4. Show that [f, M] = n xf, where f is a vector function of the co-ordinates
and momentum of a particle, and n is a unit vector parallel to the z-axis.
SOLUTION. An arbitrary vector f(r,p) may be written as f = where
01, O2, 03 are scalar functions. The required relation may be verified by direct calculation
from formulae (42.9), (42.11), (42.12) and the formula of Problem 3.
$43. The action as a function of the co-ordinates
In formulating the principle of least action, we have considered the integral
(43.1)
taken along a path between two given positions q(1) and q(2) which the system
occupies at given instants t1 and t2. In varying the action, we compared the
values of this integral for neighbouring paths with the same values of q(t1)
and q(t2). Only one of these paths corresponds to the actual motion, namely
the path for which the integral S has its minimum value.
Let us now consider another aspect of the concept of action, regarding S
as a quantity characterising the motion along the actual path, and compare
the values of S for paths having a common beginning at q(t1) = q(1), but
passing through different points at time t2. In other words, we consider the
action integral for the true path as a function of the co-ordinates at the upper
limit of integration.
The change in the action from one path to a neighbouring path is given
(if there is one degree of freedom) by the expression (2.5):
8S =
Since the paths of actual motion satisfy Lagrange's equations, the integral
in 8S is zero. In the first term we put Sq(t1) = 0, and denote the value of
§43
The action as a function of the co-ordinates
139
8q(t2) by 8q simply. Replacing 0L/dq by p, we have finally 8S = pdq or, in
the general case of any number of degrees of freedom,
ES==Pisqu-
(43.2)
From this relation it follows that the partial derivatives of the action with
respect to the co-ordinates are equal to the corresponding momenta:
=
(43.3)
The action may similarly be regarded as an explicit function of time, by
considering paths starting at a given instant t1 and at a given point q(1), and
ending at a given point q(2) at various times t2 = t. The partial derivative
asiat thus obtained may be found by an appropriate variation of the integral.
It is simpler, however, to use formula (43.3), proceeding as follows.
From the definition of the action, its total time derivative along the path is
dS/dt = L.
(43.4)
Next, regarding S as a function of co-ordinates and time, in the sense des-
cribed above, and using formula (43.3), we have
dS
A comparison gives asid = L- or
(43.5)
Formulae (43.3) and (43.5) may be represented by the expression
(43.6)
for the total differential of the action as a function of co-ordinates and time
at the upper limit of integration in (43.1). Let us now suppose that the co-
ordinates (and time) at the beginning of the motion, as well as at the end,
are variable. It is evident that the corresponding change in S will be given
by the difference of the expressions (43.6) for the beginning and end of the
path, i.e.
dsp
(43.7)
This relation shows that, whatever the external forces on the system during
its motion, its final state cannot be an arbitrary function of its initial state;
only those motions are possible for which the expression on the right-hand
side of equation (43.7) is a perfect differential. Thus the existence of the
principle of least action, quite apart from any particular form of the Lagran-
gian, imposes certain restrictions on the range of possible motions. In parti-
cular, it is possible to derive a number of general properties, independent
of the external fields, for beams of particles diverging from given points in
140
The Canonical Equations
§44
space. The study of these properties forms a part of the subject of geometrical
optics.+
It is of interest to note that Hamilton's equations can be formally derived
from the condition of minimum action in the form
(43.8)
which follows from (43.6), if the co-ordinates and momenta are varied inde-
pendently. Again assuming for simplicity that there is only one co-ordinate
and momentum, we write the variation of the action as
= dt - (OH/dp)8p dt].
An integration by parts in the second term gives
At the limits of integration we must put 8q = 0, so that the integrated term
is zero. The remaining expression can be zero only if the two integrands
vanish separately, since the variations Sp and 8q are independent and arbitrary
dq = (OH/OP) dt, dp = - (dH/dq) dt, which, after division by dt, are
Hamilton's equations.
$44. Maupertuis' principle
The motion of a mechanical system is entirely determined by the principle
of least action: by solving the equations of motion which follow from that
principle, we can find both the form of the path and the position on the path
as a function of time.
If the problem is the more restricted one of determining only the path,
without reference to time, a simplified form of the principle of least action
may be used. We assume that the Lagrangian, and therefore the Hamilton-
ian, do not involve the time explicitly, SO that the energy of the system is
conserved: H(p, q) = E = constant. According to the principle of least action,
the variation of the action, for given initial and final co-ordinates and times
(to and t, say), is zero. If, however, we allow a variation of the final time t,
the initial and final co-ordinates remaining fixed, we have (cf.(43.7))
8S = -Hot.
(44.1)
We now compare, not all virtual motions of the system, but only those
which satisfy the law of conservation of energy. For such paths we can
replace H in (44.1) by a constant E, which gives
SS+Est=0.
(44.2)
t See The Classical Theory of Fields, Chapter 7, Pergamon Press, Oxford 1962.

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§44
Maupertuis' principle
141
Writing the action in the form (43.8) and again replacing H by E, we have
(44.3)
The first term in this expression,
(44.4)
is sometimes called the abbreviated action.
Substituting (44.3) in (44.2), we find that
8S0=0.
(44.5)
Thus the abbreviated action has a minimum with respect to all paths which
satisfy the law of conservation of energy and pass through the final point
at any instant. In order to use such a variational principle, the momenta
(and so the whole integrand in (44.4)) must be expressed in terms of the
co-ordinates q and their differentials dq. To do this, we use the definition of
momentum:
(44.6)
and the law of conservation of energy:
E(g)
(44.7)
Expressing the differential dt in terms of the co-ordinates q and their differen-
tials dq by means of (44.7) and substituting in (44.6), we have the momenta
in terms of q and dq, with the energy E as a parameter. The variational prin-
ciple so obtained determines the path of the system, and is usually called
Maupertuis' principle, although its precise formulation is due to EULER and
LAGRANGE.
The above calculations may be carried out explicitly when the Lagrangian
takes its usual form (5.5) as the difference of the kinetic and potential energies:
The momenta are
and the energy is
The last equation gives
dt
(44.8)
142
The Canonical Equations
§44
substituting this in
Epides
we find the abbreviated action:
(44.9)
In particular, for a single particle the kinetic energy is T = 1/2 m(dl/dt)2,
where m is the mass of the particle and dl an element of its path; the variational
principle which determines the path is
${/[2m(B-U)]dl=0
(44.10)
where the integral is taken between two given points in space. This form is
due to JACOBI.
In free motion of the particle, U = 0, and (44.10) gives the trivial result
8 I dl = 0, i.e. the particle moves along the shortest path between the two
given points, i.e. in a straight line.
Let us return now to the expression (44.3) for the action and vary it with
respect to the parameter E. We have
substituting in (44.2), we obtain
(44.11)
When the abbreviated action has the form (44.9), this gives
=
(44.12)
which is just the integral of equation (44.8). Together with the equation of
the path, it entirely determines the motion.
PROBLEM
Derive the differential equation of the path from the variational principle (44.10).
SOLUTION. Effecting the variation, we have
f
In the second term we have used the fact that dl2 = dr2 and therefore dl d8l = dr. d&r.
Integrating this term by parts and then equating to zero the coefficient of Sr in the integrand,
we obtain the differential equation of the path:
§45
Canonical transformations
143
Expanding the derivative on the left-hand side and putting the force F = - auld gives
d2r/dl2=[F-(F.t)t]/2(E-U),
where t = dr/dl is a unit vector tangential to the path. The difference F-(F. t)t is the com-
ponent Fn of the force normal to the path. The derivative d2r/dl2 = dt/dl is known from
differential geometry to be n/R, where R is the radius of curvature of the path and n the unit
vector along the principal normal. Replacing E-U by 1mv2, we have (mv2/R)n = Fn, in
agreement with the familar expression for the normal acceleration in motion in a curved
path.
$45. Canonical transformations
The choice of the generalised co-ordinates q is subject to no restriction;
they may be any S quantities which uniquely define the position of the system
in space. The formal appearance of Lagrange's equations (2.6) does not
depend on this choice, and in that sense the equations may be said to be
invariant with respect to a transformation from the co-ordinates q1, q2,
to any other independent quantities Q1, Q2, The new co-ordinates Q are
functions of q, and we shall assume that they may explicitly depend on the
time, i.e. that the transformation is of the form
Qi=Qi(q,t)
(45.1)
(sometimes called a point transformation).
Since Lagrange's equations are unchanged by the transformation (45.1),
Hamilton's equations (40.4) are also unchanged. The latter equations, how-
ever, in fact allow a much wider range of transformations. This is, of course,
because in the Hamiltonian treatment the momenta P are variables inde-
pendent of and on an equal footing with the co-ordinates q. Hence the trans-
formation may be extended to include all the 2s independent variables P
and q:
Qt=Qi(p,q,t),
Pi = Pi(p, q,t).
(45.2)
This enlargement of the class of possible transformations is one of the im-
portant advantages of the Hamiltonian treatment.
The equations of motion do not, however, retain their canonical form
under all transformations of the form (45.2). Let us derive the conditions
which must be satisfied if the equations of motion in the new variables P, Q
are to be of the form
(45.3)
with some Hamiltonian H'(P,Q). When this happens the transformation is
said to be canonical.
The formulae for canonical transformations can be obtained as follows. It
has been shown at the end of §43 that Hamilton's equations can be derived
from the principle of least action in the form
(45.4)
144
The Canonical Equations
§45
in which the variation is applied to all the co-ordinates and momenta inde-
pendently. If the new variables P and Q also satisfy Hamilton's equations,
the principle of least action
0
(45.5)
must hold. The two forms (45.4) and (45.5) are equivalent only if their inte-
grands are the same apart from the total differential of some function F of
co-ordinates, momenta and time.t The difference between the two integrals
is then a constant, namely the difference of the values of F at the limits of
integration, which does not affect the variation. Thus we must have
=
Each canonical transformation is characterised by a particular function F,
called the generating function of the transformation.
Writing this relation as
(45.6)
we see that
Pi = 0F/dqi, =-0F/JQi H' = H+0F/dt;
(45.7)
here it is assumed that the generating function is given as a function of the
old and new co-ordinates and the time: F = F(q, Q, t). When F is known,
formulae (45.7) give the relation between p, q and P, Q as well as the new
Hamiltonian.
It may be convenient to express the generating function not in terms of the
variables q and Q but in terms of the old co-ordinates q and the new momenta
P. To derive the formulae for canonical transformations in this case, we must
effect the appropriate Legendre's transformation in (45.6), rewriting it as
=
The argument of the differential on the left-hand side, expressed in terms of
the variables q and P, is a new generating function (q, P, t), say. Thent
= Qi = ID/OPi, H' = H+d
(45.8)
We can similarly obtain the formulae for canonical transformations in-
volving generating functions which depend on the variables P and Q, or
p and P.
t We do not consider such trivial transformations as Pi = api, Qi = qt,H' = aH, with a an
arbitrary constant, whereby the integrands in (45.4) and (45.5) differ only by a constant
factor.
+ If the generating function is = fi(q, t)Pi, where the ft are arbitrary functions, we
obtain a transformation in which the new co-ordinates are Q = fi(q, t), i.e. are expressed
in terms of the old co-ordinates only (and not the momenta). This is a point transformation,
and is of course a particular canonical transformation.
§45
Canonical transformations
145
The relation between the two Hamiltonians is always of the same form:
the difference H' - H is the partial derivative of the generating function with
respect to time. In particular, if the generating function is independent of
time, then H' = H, i.e. the new Hamiltonian is obtained by simply substitut-
ing for P, q in H their values in terms of the new variables P, Q.
The wide range of the canonical transformations in the Hamiltonian treat-
ment deprives the generalised co-ordinates and momenta of a considerable
part of their original meaning. Since the transformations (45.2) relate each
of the quantities P, Q to both the co-ordinates q and the momenta P, the
variables Q are no longer purely spatial co-ordinates, and the distinction
between Q and P becomes essentially one of nomenclature. This is very
clearly seen, for example, from the transformation Q = Pi, Pi = -qi,
which obviously does not affect the canonical form of the equations and
amounts simply to calling the co-ordinates momenta and vice versa.
On account of this arbitrariness of nomenclature, the variables P and q in
the Hamiltonian treatment are often called simply canonically conjugate
quantities. The conditions relating such quantities can be expressed in terms
of Poisson brackets. To do this, we shall first prove a general theorem on the
invariance of Poisson brackets with respect to canonical transformations.
Let [f,g]p,a be the Poisson bracket, for two quantities f and g, in which
the differentiation is with respect to the variables P and q, and [f,g]p,Q that
in which the differentiation is with respect to P and Q. Then
(45.9)
The truth of this statement can be seen by direct calculation, using the for-
mulae of the canonical transformation. It can also be demonstrated by the
following argument.
First of all, it may be noticed that the time appears as a parameter in the
canonical transformations (45.7) and (45.8). It is therefore sufficient to prove
(45.9) for quantities which do not depend explicitly on time. Let us now
formally regard g as the Hamiltonian of some fictitious system. Then, by
formula (42.1), [f,g]p,a = df/dt. The derivative df/dt can depend only on
the properties of the motion of the fictitious system, and not on the particular
choice of variables. Hence the Poisson bracket [f,g] is unaltered by the
passage from one set of canonical variables to another.
Formulae (42.13) and (45.9) give
[Qi, Qk]p,a = 0, [Pi,Pk]p,a = 0,
(45.10)
These are the conditions, written in terms of Poisson brackets, which must
be satisfied by the new variables if the transformation P, q P, Q is canonical.
It is of interest to observe that the change in the quantities P, q during the
motion may itself be regarded as a series of canonical transformations. The
meaning of this statement is as follows. Let qt, Pt be the values of the canonical
t Whose generating function is
6*
146
The Canonical Equations
§46
variables at time t, and qt+r, Pt+r their values at another time t +T. The latter
are some functions of the former (and involve T as a parameter):
If these formulae are regarded as a transformation from the variables Qt, Pt
to qt+r, Pttr, then this transformation is canonical. This is evident from the
expression ds = for the differential of the action S(qt++,
qt) taken along the true path, passing through the points qt and qt++ at given
times t and t + T (cf. (43.7)). A comparison of this formula with (45.6) shows
that - S is the generating function of the transformation.
46. Liouville's theorem
For the geometrical interpretation of mechanical phenomena, use is often
made of phase space. This is a space of 2s dimensions, whose co-ordinate axes
correspond to the S generalised co-ordinates and S momenta of the system
concerned. Each point in phase space corresponds to a definite state of the
system. When the system moves, the point representing it describes a curve
called the phase path.
The product of differentials dT = dq1 ... dqsdp1 dps may be regarded
as an element of volume in phase space. Let us now consider the integral
I dT taken over some region of phase space, and representing the volume of
that region. We shall show that this integral is invariant with respect to
canonical transformations; that is, if the variables P, q are replaced by
P, Q by a canonical transformation, then the volumes of the corresponding
regions of the spaces of P, and P, Q are equal:
...dqsdp1...dps =
(46.1)
The transformation of variables in a multiple integral is effected by the
formula I .jdQ1...dQsdP1...dPz = S... I Ddq1 dp1...dps,
where
(46.2)
is the Jacobian of the transformation. The proof of (46.1) therefore amounts
to proving that the Jacobian of every canonical transformation is unity:
D=1.
(46.3)
We shall use a well-known property of Jacobians whereby they can be
treated somewhat like fractions. "Dividing numerator and denominator" by
0(91, ..., qs, P1, Ps), we obtain
Another property of Jacobians is that, when the same quantities appear in
both the partial differentials, the Jacobian reduces to one in fewer variables,
§47
The Hamilton-Jacobi equation
147
in which these repeated quantities are regarded as constant in carrying out
the differentiations. Hence
(46.4)
P=constant
q=constant
The Jacobian in the numerator is, by definition, a determinant of order s
whose element in the ith row and kth column is Representing the
canonical transformation in terms of the generating function (q, P) as in
(45.8), we have = In the same way we find that the
ik-element of the determinant in the denominator of (46.4) is
This means that the two determinants differ only by the interchange of rows
and columns; they are therefore equal, so that the ratio (46.4) is equal to
unity. This completes the proof.
Let us now suppose that each point in the region of phase space considered
moves in the course of time in accordance with the equations of motion of the
mechanical system. The region as a whole therefore moves also, but its volume
remains unchanged:
f dr = constant.
(46.5)
This result, known as Liouville's theorem, follows at once from the invariance
of the volume in phase space under canonical transformations and from the
fact that the change in p and q during the motion may, as we showed at the end
of §45, be regarded as a canonical transformation.
In an entirely similar manner the integrals
11 2 dae dph
,
in which the integration is over manifolds of two, four, etc. dimensions in
phase space, may be shown to be invariant.
47. The Hamilton-Jacobi equation
In §43 the action has been considered as a function of co-ordinates and
time, and it has been shown that the partial derivative with respect to time
of this function S(q, t) is related to the Hamiltonian by
and its partial derivatives with respect to the co-ordinates are the momenta.
Accordingly replacing the momenta P in the Hamiltonian by the derivatives
as/aq, we have the equation
(47.1)
which must be satisfied by the function S(q, t). This first-order partial
differential equation is called the Hamilton-Jacobi equation.
148
The Canonical Equations
§47
Like Lagrange's equations and the canonical equations, the Hamilton-
Jacobi equation is the basis of a general method of integrating the equations
of motion.
Before describing this method, we should recall the fact that every first-
order partial differential equation has a solution depending on an arbitrary
function; such a solution is called the general integral of the equation. In
mechanical applications, the general integral of the Hamilton-Jacobi equation
is less important than a complete integral, which contains as many independent
arbitrary constants as there are independent variables.
The independent variables in the Hamilton-Jacobi equation are the time
and the co-ordinates. For a system with s degrees of freedom, therefore, a
complete integral of this equation must contain s+1 arbitrary constants.
Since the function S enters the equation only through its derivatives, one
of these constants is additive, so that a complete integral of the Hamilton-
Jacobi equation is
Sft,q,saas)+
(47.2)
where X1, ..., as and A are arbitrary constants.
Let us now ascertain the relation between a complete integral of the
Hamilton-Jacobi equation and the solution of the equations of motion which
is of interest. To do this, we effect a canonical transformation from the
variables q, P to new variables, taking the function f (t, q; a) as the
generating function, and the quantities a1, A2, ..., as as the new momenta.
Let the new co-ordinates be B1, B2, ..., Bs. Since the generating function
depends on the old co-ordinates and the new momenta, we use formulae
(45.8): Pi = af/dqi, Bi = af/dar, H' = H+dfdd. But since the function f
satisfies the Hamilton-Jacobi equation, we see that the new Hamiltonian is
zero: H' = H+af/dt = H+as/t = 0. Hence the canonical equations in
the new variables are di = 0, Bi = 0, whence
ay=constant,
Bi = constant.
(47.3)
By means of the S equations af/dai = Bi, the S co-ordinates q can be expressed
in terms of the time and the 2s constants a and B. This gives the general
integral of the equations of motion.
t Although the general integral of the Hamilton-Jacobi equation is not needed here, we
may show how it can be found from a complete integral. To do this, we regard A as an arbi-
trary function of the remaining constants: S = f(t, q1, ..., q8; a1, as) +A(a1, as). Re-
placing the Ai by functions of co-ordinates and time given by the S conditions asidar = 0,
we obtain the general integral in terms of the arbitrary function A(a1,..., as). For, when the
function S is obtained in this manner, we have
as
The quantities (as/dqs)a satisfy the Hamilton-Jacobi equation, since the function S(t, q; a)
is assumed to be a complete integral of that equation. The quantities asida therefore satisfy
the same equation.
§48
Separation of the variables
149
Thus the solution of the problem of the motion of a mechanical system by
the Hamilton-Jacobi method proceeds as follows. From the Hamiltonian,
we form the Hamilton-Jacobi equation, and find its complete integral (47.2).
Differentiating this with respect to the arbitrary constants a and equating
the derivatives to new constants B, we obtain S algebraic equations
asidar=Bt,
(47.4)
whose solution gives the co-ordinates q as functions of time and of the 2s
arbitrary constants. The momenta as functions of time may then be found
from the equations Pi = aslaqi.
If we have an incomplete integral of the Hamilton-Jacobi equation, depend-
ing on fewer than S arbitrary constants, it cannot give the general integral
of the equations of motion, but it can be used to simplify the finding of the
general integral. For example, if a function S involving one arbitrary con-
stant a is known, the relation asida = constant gives one equation between
q1, ..., qs and t.
The Hamilton-Jacobi equation takes a somewhat simpler form if the func-
tion H does not involve the time explicitly, i.e. if the system is conservative.
The time-dependence of the action is given by a term -Et:
S = So(g)-Et
(47.5)
(see 44), and substitution in (47.1) gives for the abbreviated action So(q)
the Hamilton-Jacobi equation in the form
(47.6)
$48. Separation of the variables
In a number of important cases, a complete integral of the Hamilton-
Jacobi equation can be found by "separating the variables", a name given to
the following method.
Let us assume that some co-ordinate, q1 say, and the corresponding
derivative asia appear in the Hamilton-Jacobi equation only in some
combination (q1, which does not involve the other co-ordinates, time,
or derivatives, i.e. the equation is of the form
(48.1)
where qi denotes all the co-ordinates except q1.
We seek a solution in the form of a sum:
(48.2)
150
The Canonical Equations
§48
substituting this in equation (48.1), we obtain
(48.3)
Let us suppose that the solution (48.2) has been found. Then, when it is
substituted in equation (48.3), the latter must become an identity, valid (in
particular) for any value of the co-ordinate q1. When q1 changes, only the
function is affected, and so, if equation (48.3) is an identity, must be a
constant. Thus equation (48.3) gives the two equations
(48.4)
= 0,
(48.5)
where a1 is an arbitrary constant. The first of these is an ordinary differential
equation, and the function S1(q1) is obtained from it by simple integration.
The remaining partial differential equation (48.5) involves fewer independent
variables.
If we can successively separate in this way all the S co-ordinates and the
time, the finding of a complete integral of the Hamilton-Jacobi equation is
reduced to quadratures. For a conservative system we have in practice to
separate only S variables (the co-ordinates) in equation (47.6), and when this
separation is complete the required integral is
(48.6)
where each of the functions Sk depends on only one co-ordinate; the energy
E, as a function of the arbitrary constants A1, As, is obtained by substituting
So = in equation (47.6).
A particular case is the separation of a cyclic variable. A cyclic co-ordinate
q1 does not appear explicitly in the Hamiltonian, nor therefore in the Hamilton-
Jacobi equation. The function (91, reduces to as/da simply, and
from equation (48.4) we have simply S1 = x1q1, so that
(48.7)
The constant a1 is just the constant value of the momentum P1 = asida
corresponding to the cyclic co-ordinate.
The appearance of the time in the term - Et for a conservative system
corresponds to the separation of the "cyclic variable" t.
Thus all the cases previously considered of the simplification of the integra-
tion of the equations of motion by the use of cyclic variables are embraced
by the method of separating the variables in the Hamilton-Jacobi equation.
To those cases are added others in which the variables can be separated even
though they are not cyclic. The Hamilton-Jacobi treatment is consequently
the most powerful method of finding the general integral of the equations of
motion.

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§48
Separation of the variables
151
To make the variables separable in the Hamilton-Jacobi equation the
co-ordinates must be appropriately chosen. We shall consider some examples
of separating the variables in different co-ordinates, which may be of
physical interest in connection with problems of the motion of a particle in
various external fields.
(1) Spherical co-ordinates. In these co-ordinates (r, 0, ), the Hamiltonian is
and the variables can be separated if
U
=
where a(r), b(a), c(b) are arbitrary functions. The last term in this expression
for U is unlikely to be of physical interest, and we shall therefore take
U = a(r)+b(8)/r2.
(48.8)
In this case the Hamilton-Jacobi equation for the function So is
1
Since the co-ordinate is cyclic, we seek a solution in the form So
Pot+S1(T)+S2(9), obtaining for the functions S1(r) andS 2(0) the equations
(day)
=
E.
Integration gives finally
S = -
(48.9)
The arbitrary constants in (48.9) are Pp, B and E; on differentiating with
respect to these and equating the results to other constants, we have the
general solution of the equations of motion.
(2) Parabolic co-ordinates. The passage from cylindrical co-ordinates
(here denoted by p, o, 2) to parabolic co-ordinates E, N, o is effected by the
formulae
1(-n),pv(En).
(48.10)
The co-ordinates & and n take values from 0 to 00; the surfaces of constant
$ and n are easily seen to be two families of paraboloids of revolution, with
152
The Canonical Equations
§48
the z-axis as the axis of symmetry. The equations (48.10) can also be written,
in terms of
r = =
(48.11)
(i.e. the radius in spherical co-ordinates), as
$ = r++,
= r Z.
(48.12)
Let us now derive the Lagrangian of a particle in the co-ordinates $, n, o.
Differentiating the expressions (48.10) with respect to time and substituting
in the Lagrangian in cylindrical co-ordinates
L =
we obtain
L
=
(48.13)
The = and
the Hamiltonian is
(48.14)
The physically interesting cases of separable variables in these co-ordinates
correspond to a potential energy of the form
(48.15)
The equation for So is
2
=
E.
The cyclic co-ordinate can be separated as a term PoO. Multiplying the equa-
tion by m(s+n) and rearranging, we then have
Putting So = P&O + S2(n), we obtain the two equations
-B,
§48
Separation of the variables
153
integration of which gives finally
S
=
dn.
(48.16)
Here the arbitrary constants are Ps, B and E.
(3) Elliptic co-ordinates. These are E, n, o, defined by
(48.17)
The constant o is a parameter of the transformation. The co-ordinate $ takes
values from 1 to 80, and n from - 1 to + 1. The definitions which are geo-
metrically clearest+ are obtained in terms of the distances r1 and r2 to points
A1 and A2 on the z-axis for which 2 = to: r1 = V[(2-0)2+p2],
r2 = Substitution of (48.17) gives
= o(s-n), r2 = o(+n),
(48.18)
& = (r2+r1)/2o, n = (r2-r1)/2o. =
Transforming the Lagrangian from cylindrical to elliptic co-ordinates, we
find
L
=
(48.19)
The Hamiltonian is therefore
H
=
(48.20)
The physically interesting cases of separable variables correspond to a
potential energy
(48.21)
where a() and b(n) are arbitrary functions. The result of separating the
variables in the Hamilton-Jacobi equation is
S
=
1-n2
t The surfaces of constant $ are the ellipsoids = 1, of which A1 and
A2 are the foci; the surfaces of constant n are the hyperboloids 22/02/2-22/02(1-n2 = 1,
also with foci A1 and A2.
154
The Canonical Equations
§49
PROBLEMS
PROBLEM 1. Find a complete integral of the Hamilton-Jacobi equation for motion of a
particle in a field U = a/r-Fz (a combination of a uniform field and a Coulomb field).
SOLUTION. The field is of the type (48.15), with a(f)=a1F,b(n)a+Fn2 Formula
(48.16) gives
S
=
with arbitrary constants Po, E,B. The constant B has in this case the significance that the one-
valued function of the co-ordinates and momenta of the particle
B
is conserved. The expression in the brackets is an integral of the motion for a pure Coulomb
field (see $15).
PROBLEM 2. The same as Problem 1, but for a field U = ai/r +az/r2 (the Coulomb field
of two fixed points at a distance 2a apart).
SOLUTION. This field is of the type (48.21), with a($) = (a1+az) /o, = (a1-az)n/o.
From formula (48.22) we find
S
=
The constant B here expresses the conservation of the quantity
B = cos 01+ cos 02),
where M is the total angular momentum of the particle, and 01 and O2 are the angles shown in
Fig. 55.
12
r
The
20
a
FIG. 55
$49. Adiabatic invariants
Let us consider a mechanical system executing a finite motion in one dimen-
sion and characterised by some parameter A which specifies the properties of
the system or of the external field in which it is placed, and let us suppose that
1 varies slowly (adiabatically) with time as the result of some external action;
by a "slow" variation we mean one in which A varies only slightly during the
period T of the motion:
di/dt < A.
(49.1)
§49
Adiabatic invariants
155
Such a system is not closed, and its energy E is not conserved. However, since
A varies only slowly, the rate of change E of the energy is proportional to the
rate of change 1 of the parameter. This means that the energy of the system
behaves as some function of A when the latter varies. In other words, there
is some combination of E and A which remains constant during the motion.
This quantity is called an adiabatic invariant.
Let H(p, q; A) be the Hamiltonian of the system, which depends on the
parameter A. According to formula (40.5), the total time derivative of the
energy of the system is dE/dt = OH/dt = (aH/dx)(d)/dt). In averaging this
equation over the period of the motion, we need not average the second
factor, since A (and therefore i) varies only slowly: dE/dt = (d)/dt)
and in the averaged function 01/01 we can regard only P and q, and not A, as
variable. That is, the averaging is taken over the motion which would occur
if A remained constant.
The averaging may be explicitly written
dE dt
According to Hamilton's equation q = OHOP, or dt = dq - (CH/OP). The
integration with respect to time can therefore be replaced by one with respect
to the co-ordinate, with the period T written as
here the $ sign denotes an integration over the complete range of variation
("there and back") of the co-ordinate during the period. Thus
dq/(HHap)
(49.2)
dt $ dq/(HHdp)
As has already been mentioned, the integrations in this formula must be
taken over the path for a given constant value of A. Along such a path the
Hamiltonian has a constant value E, and the momentum is a definite function
of the variable co-ordinate q and of the two independent constant parameters
E and A. Putting therefore P = p(q; E, 1) and differentiating with respect
to A the equation H(p, q; X) )=E, we have = 0, or
OH/OP ax ap
t If the motion of the system is a rotation, and the co-ordinate q is an angle of rotation ,
the integration with respect to must be taken over a "complete rotation", i.e. from 0 to 2nr.
156
The Canonical Equations
§49
Substituting this in the numerator of (49.2) and writing the integrand in the
denominator as ap/dE, we obtain
dt
(49.3)
dq
or
dt
Finally, this may be written as
dI/dt 0,
(49.4)
where
(49.5)
the integral being taken over the path for given E and A. This shows that, in
the approximation here considered, I remains constant when the parameter A
varies, i.e. I is an adiabatic invariant.
The quantity I is a function of the energy of the system (and of the para-
meter A). The partial derivative with respect to energy is given by 2m DI/DE
= $ (ap/dE) dq (i.e. the integral in the denominator in (49.3)) and is, apart from
a factor 2n, the period of the motion:
(49.6)
The integral (49.5) has a geometrical significance in terms of the phase
path of the system. In the case considered (one degree of freedom), the phase
space reduces to a two-dimensional space (i.e. a plane) with co-ordinates
P, q, and the phase path of a system executing a periodic motion is a closed
curve in the plane. The integral (49.5) taken round this curve is the area
enclosed. It can evidently be written equally well as the line integral
I = - $ q dp/2m and as the area integral I = II dp dq/2m.
As an example, let us determine the adiabatic invariant for a one-dimen-
sional oscillator. The Hamiltonian is H = where w is the
frequency of the oscillator. The equation of the phase path is given by the
law of conservation of energy H(p, q) = E. The path is an ellipse with semi-
axes (2mE) and V(2E/mw2), and its area, divided by 2nr, is
I=E/w.
(49.7)
t It can be shown that, if the function X(t) has no singularities, the difference of I from a
constant value is exponentially small.
§49
Adiabatic invariants
157
The adiabatic invariance of I signifies that, when the parameters of the
oscillator vary slowly, the energy is proportional to the frequency.
The equations of motion of a closed system with constant parameters
may be reformulated in terms of I. Let us effect a canonical transformation
of the variables P and q, taking I as the new "momentum". The generating
function is the abbreviated action So, expressed as a function of q and I. For
So is defined for a given energy of the system; in a closed system, I is a func-
tion of the energy alone, and so So can equally well be written as a function
So(q, I). The partial derivative (So/dq)E is the same as the derivative
( for constant I. Hence
(49.8)
corresponding to the first of the formulae (45.8) for a canonical trans-
formation. The second of these formulae gives the new "co-ordinate",
which we denote by W:
W = aso(q,I)/aI.
(49.9)
The variables I and W are called canonical variables; I is called the action
variable and W the angle variable.
Since the generating function So(q, I) does not depend explicitly on time,
the new Hamiltonian H' is just H expressed in terms of the new variables.
In other words, H' is the energy E(I), expressed as a function of the action
variable. Accordingly, Hamilton's equations in canonical variables are
i = 0,
w = dE(I)/dI.
(49.10)
The first of these shows that I is constant, as it should be; the energy is
constant, and I is so too. From the second equation we see that the angle
variable is a linear function of time:
W = (dE/dI)t + constant.
(49.11)
The action So(q, I) is a many-valued function of the co-ordinate. During
each period this function increases by
(49.12)
as is evident from the formula So = Spdq and the definition (49.5). During
the same time the angle variable therefore increases by
Aw = (S/I) =
(49.13)
t The exactness with which the adiabatic invariant (49.7) is conserved can be determined by
establishing the relation between the coefficients C in the asymptotic (t + 00) expressions
q = re[c exp(iw+t)] for the solution of the oscillator equation of motion q + w2(t) q = 0.
Here the frequency w is a slowly varying function of time, tending to constant limits w as
t
+ 00. The limiting values of I are given in terms of these coefficients by I = tw+/c+l2.
The solution is known from quantum mechanics, on account of the formal resemblance
between the above equation of motion and SCHRODINGER'S equation 4" + k2(x) 4 = 0 for
one-dimensional motion of a particle above a slowly varying (quasi-classical) "potential
barrier". The problem of finding the relation between the asymptotic (x + 00)
expressions
for & is equivalent to that of finding the "reflection coefficient" of the potential barrier; see
Quantum Mechanics, $52, Pergamon Press, Oxford 1965.
This method of determining the exactness of conservation of the adiabatic invariant for an
oscillator is due to L. P. PITAEVSKII. The relevant calculations are given by A. M. DYKHNE,
Soviet Physics JETP 11, 411, 1960. The analysis for the general case of an arbitrary finite
motion in one dimension is given by A.A. SLUTSKIN, Soviet Physics JETP 18, 676, 1964.
158
The Canonical Equations
§50
as may also be seen directly from formula (49.11) and the expression (49.6)
for the period.
Conversely, if we express q and P, or any one-valued function F(p, q) of
them, in terms of the canonical variables, then they remain unchanged when
W increases by 2nd (with I constant). That is, any one-valued function F(p, q),
when expressed in terms of the canonical variables, is a periodic function of W
with period 2.
$50. General properties of motion in S dimensions
Let us consider a system with any number of degrees of freedom, executing
a motion finite in all the co-ordinates, and assume that the variables can be
completely separated in the Hamilton-Jacobi treatment. This means that,
when the co-ordinates are appropriately chosen, the abbreviated action
can be written in the form
(50.1)
as a sum of functions each depending on only one co-ordinate.
Since the generalised momenta are Pi = aso/dqi = dSi/dqi, each function
Si can be written
(50.2)
These are many-valued functions. Since the motion is finite, each co-ordinate
can take values only in a finite range. When qi varies "there and back" in this
range, the action increases by
(50.3)
where
(50.4)
the integral being taken over the variation of qi just mentioned.
Let us now effect a canonical transformation similar to that used in 49,
for the case of a single degree of freedom. The new variables are "action vari-
ables" Ii and "angle variables"
w(a(q
(50.5)
+ It should be emphasised, however, that this refers to the formal variation of the co-
ordinate qi over the whole possible range of values, not to its variation during the period of
the actual motion as in the case of motion in one dimension. An actual finite motion of a
system with several degrees of freedom not only is not in general periodic as a whole, but
does not even involve a periodic time variation of each co-ordinate separately (see below).
§50
General properties of motion in S dimensions
159
where the generating function is again the action expressed as a function of
the co-ordinates and the Ii. The equations of motion in these variables are
Ii = 0, w = de(I)/I, which give
I=constant,
(50.6)
+ constant.
(50.7)
We also find, analogously to (49.13), that a variation "there and back" of
the co-ordinate qi corresponds to a change of 2n in Wi:
Awi==2m
(50.8)
In other words, the quantities Wi(q, I) are many-valued functions of the co-
ordinates: when the latter vary and return to their original values, the Wi
may vary by any integral multiple of 2. This property may also be formulated
as a property of the function Wi(P, q), expressed in terms of the co-ordinates
and momenta, in the phase space of the system. Since the Ii, expressed in
terms of P and q, are one-valued functions, substitution of Ii(p, q) in wi(q, I)
gives a function wilp, q) which may vary by any integral multiple of 2n
(including zero) on passing round any closed path in phase space.
Hence it follows that any one-valued function F(P, q) of the state of the
system, if expressed in terms of the canonical variables, is a periodic function
of the angle variables, and its period in each variable is 2nr. It can be expanded
as a multiple Fourier series:
(50.9)
ls==
where l1, l2, ls are integers. Substituting the angle variables as functions
of time, we find that the time dependence of F is given by a sum of the form
(50.10)
lg==
Each term in this sum is a periodic function of time, with frequency
(50.11)
Since these frequencies are not in general commensurable, the sum itself is
not a periodic function, nor, in particular, are the co-ordinates q and
momenta P of the system.
Thus the motion of the system is in general not strictly periodic either as a
whole or in any co-ordinate. This means that, having passed through a given
state, the system does not return to that state in a finite time. We can say,
t Rotational co-ordinates (see the first footnote to 49) are not in one-to-one relation
with the state of the system, since the position of the latter is the same for all values of
differing by an integral multiple of 2nr. If the co-ordinates q include such angles, therefore,
these can appear in the function F(P, q) only in such expressions as cos and sin , which
are in one-to-one relation with the state of the system.
160
The Canonical Equations
§50
however, that in the course of a sufficient time the system passes arbitrarily
close to the given state. For this reason such a motion is said to be conditionally
periodic.
In certain particular cases, two or more of the fundamental frequencies
Wi = DE/DI are commensurable for arbitrary values of the Ii. This is called
degeneracy, and if all S frequencies are commensurable, the motion of the
system is said to be completely degenerate. In the latter case the motion is
evidently periodic, and the path of every particle is closed.
The existence of degeneracy leads, first of all, to a reduction in the number
of independent quantities Ii on which the energy of the system depends.
If two frequencies W1 and W2 are such that
(50.12)
where N1 and N2 are integers, then it follows that I1 and I2 appear in the energy
only as the sum n2I1+n1I2.
A very important property of degenerate motion is the increase in the
number of one-valued integrals of the motion over their number for a general
non-degenerate system with the same number of degrees of freedom. In the
latter case, of the 2s-1 integrals of the motion, only s functions of the state
of the system are one-valued; these may be, for example, the S quantities I
The remaining S - 1 integrals may be written as differences
(50.13)
The constancy of these quantities follows immediately from formula (50.7),
but they are not one-valued functions of the state of the system, because the
angle variables are not one-valued.
When there is degeneracy, the situation is different. For example, the rela-
tion (50.12) shows that, although the integral
WIN1-W2N2
(50.14)
is not one-valued, it is so except for the addition of an arbitrary integral
multiple of 2nr. Hence we need only take a trigonometrical function of this
quantity to obtain a further one-valued integral of the motion.
An example of degeneracy is motion in a field U = -a/r (see Problem).
There is consequently a further one-valued integral of the motion (15.17)
peculiar to this field, besides the two (since the motion is two-dimensional)
ordinary one-valued integrals, the angular momentum M and the energy E,
which hold for motion in any central field.
It may also be noted that the existence of further one-valued integrals
leads in turn to another property of degenerate motions: they allow a complete
separation of the variables for several (and not only one+) choices of the co-
t We ignore such trivial changes in the co-ordinates as q1' = q1'(q1), q2' = 92'(92).

446
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§50
General properties of motion in S dimensions
161
ordinates. For the quantities Ii are one-valued integrals of the motion in
co-ordinates which allow separation of the variables. When degeneracy occurs,
the number of one-valued integrals exceeds S, and so the choice of those
which are the desired I is no longer unique.
As an example, we may again mention Keplerian motion, which allows
separation of the variables in both spherical and parabolic co-ordinates.
In §49 it has been shown that, for finite motion in one dimension, the
action variable is an adiabatic invariant. This statement holds also for systems
with more than one degree of freedom. Here we shall give a proof valid
for the general case.
Let X(t) be again a slowly varying parameter of the system. In the canonical
transformation from the variables P, q to I, W, the generating function is, as we
know, the action So(q, I). This depends on A as a parameter and, if A is a func-
tion of time, the function So(q, I; X(t)) depends explicitly on time. In such a
case the new Hamiltonian H' is not the same as H, i.e. the energy E(I), and
by the general formulae (45.8) for the canonical transformation we have
H' E(I)+asoldt = E(I)+A, where A III (aso/ad)r. Hamilton's equations
give
ig = -
(50.15)
We average this equation over a time large compared with the fundamental
periods of the system but small compared with the time during which the
parameter A varies appreciably. Because of the latter condition we need not
average 1 on the right-hand side, and in averaging the quantities we
may regard the motion of the system as taking place at a constant value of A
and therefore as having the properties of conditionally periodic motion
described above.
The action So is not a one-valued function of the co-ordinates: when q
returns to its initial value, So increases by an integral multiple of 2I. The
derivative A = (aso/ax), is a one-valued function, since the differentiation
is effected for constant Ii, and there is therefore no increase in So. Hence A,
expressed as a function of the angle variables Wr, is periodic. The mean value
of the derivatives of such a function is zero, and therefore by (50.15)
we have also
which shows that the quantities Ii are adiabatic invariants.
Finally, we may briefly discuss the properties of finite motion of closed
systems with S degrees of freedom in the most general case, where the vari-
ables in the Hamilton-Jacobi equation are not assumed to be separable.
The fundamental property of systems with separable variables is that the
integrals of the motion Ii, whose number is equal to the number of degrees
+ To simplify the formulae we assume that there is only one such parameter, but the proof
is valid for any number.
162
The Canonical Equations
§50
of freedom, are one-valued. In the general case where the variables are not
separable, however, the one-valued integrals of the motion include only
those whose constancy is derived from the homogeneity and isotropy of space
and time, namely energy, momentum and angular momentum.
The phase path of the system traverses those regions of phase space which
are defined by the given constant values of the one-valued integrals of the
motion. For a system with separable variables and S one-valued integrals,
these conditions define an s-dimensional manifold (hypersurface) in phase
space. During a sufficient time, the path of the system passes arbitrarily close
to every point on this hypersurface.
In a system where the variables are not separable, however, the number
of one-valued integrals is less than S, and the phase path occupies, completely
or partly, a manifold of more than S dimensions in phase space.
In degenerate systems, on the other hand, which have more than S integrals
of the motion, the phase path occupies a manifold of fewer than S dimensions.
If the Hamiltonian of the system differs only by small terms from one which
allows separation of the variables, then the properties of the motion are close
to those of a conditionally periodic motion, and the difference between the
two is of a much higher order of smallness than that of the additional terms in
the Hamiltonian.
PROBLEM
Calculate the action variables for elliptic motion in a field U = -a/r.
SOLUTION. In polar co-ordinates r, in the plane of the motion we have
'max
= 1+av(m2)E)
Hence the energy, expressed in terms of the action variables, is E = It
depends only on the sum Ir+I, and the motion is therefore degenerate; the two funda-
mental frequencies (in r and in b) coincide.
The parameters P and e of the orbit (see (15.4)) are related to Ir and I by
p=
Since Ir and I are adiabatic invariants, when the coefficient a or the mass m varies slowly
the eccentricity of the orbit remains unchanged, while its dimensions vary in inverse propor-
tion to a and to m.
INDEX
Acceleration, 1
Coriolis force, 128
Action, 2, 138ff.
Couple, 109
abbreviated, 141
Cross-section, effective, for scattering,
variable, 157
49ff.
Additivity of
C system, 41
angular momentum, 19
Cyclic co-ordinates, 30
energy, 14
integrals of the motion, 13
d'Alembert's principle, 124
Lagrangians, 4
Damped oscillations, 74ff.
mass, 17
Damping
momentum, 15
aperiodic, 76
Adiabatic invariants, 155, 161
coefficient, 75
Amplitude, 59
decrement, 75
complex, 59
Degeneracy, 39, 69, 160f.
Angle variable, 157
complete, 160
Angular momentum, 19ff.
Degrees of freedom, 1
of rigid body, 105ff.
Disintegration of particles, 41ff.
Angular velocity, 97f.
Dispersion-type absorption, 79
Area integral, 31n.
Dissipative function, 76f.
Dummy suffix, 99n.
Beats, 63
Brackets, Poisson, 135ff.
Eccentricity, 36
Eigenfrequencies, 67
Canonical equations (VII), 131ff.
Elastic collision, 44
Canonical transformation, 143ff.
Elliptic functions, 118f.
Canonical variables, 157
Elliptic integrals, 26, 118
Canonically conjugate quantities, 145
Energy, 14, 25f.
Central field, 21, 30
centrifugal, 32, 128
motion in, 30ff.
internal, 17
Centrally symmetric field, 21
kinetic, see Kinetic energy
Centre of field, 21
potential, see Potential energy
Centre of mass, 17
Equations of motion (I), 1ff.
system, 41
canonical (VII), 131ff.
Centrifugal force, 128
integration of (III), 25ff.
Centrifugal potential, 32, 128
of rigid body, 107ff.
Characteristic equation, 67
Eulerian angles, 110ff.
Characteristic frequencies, 67
Euler's equations, 115, 119
Closed system, 8
Collisions between particles (IV), 41ff.
Finite motion, 25
elastic, 44ff.
Force, 9
Combination frequencies, 85
generalised, 16
Complete integral, 148
Foucault's pendulum, 129f.
Conditionally periodic motion, 160
Frame of reference, 4
Conservation laws (II), 13ff.
inertial, 5f.
Conservative systems, 14
non-inertial, 126ff.
Conserved quantities, 13
Freedom, degrees of, 1
Constraints, 10
Frequency, 59
equations of, 123
circular, 59
holonomic, 123
combination, 85
Co-ordinates, 1
Friction, 75, 122
cyclic, 30
generalised, 1ff.
Galilean transformation, 6
normal, 68f.
Galileo's relativity principle, 6
163
164
Index
General integral, 148
Mechanical similarity, 22ff.
Generalised
Molecules, vibrations of, 70ff.
co-ordinates, 1ff.
Moment
forces, 16
of force, 108
momenta, 16
of inertia, 99ff.
velocities, 1ff.
principal, 100ff.
Generating function, 144
Momentum, 15f.
angular, see Angular momentum
Half-width, 79
generalised, 16
Hamiltonian, 131f.
moment of, see Angular momentum
Hamilton-Jacobi equation, 147ff.
Multi-dimensional motion, 158ff.
Hamilton's equations, 132
Hamilton's function, 131
Hamilton's principle, 2ff.
Newton's equations, 9
Holonomic constraint, 123
Newton's third law, 16
Nodes, line of, 110
Impact parameter, 48
Non-holonomic constraint, 123
Inertia
Normal co-ordinates, 68f.
law of, 5
Normal oscillations, 68
moments of, 99ff.
Nutation, 113
principal, 100ff.
principal axes of, 100
One-dimensional motion, 25ff., 58ff.
tensor, 99
Oscillations, see Small oscillations
Inertial frames, 5f.
Oscillator
Infinite motion, 25
one-dimensional, 58n.
Instantaneous axis, 98
space, 32, 70
Integrals of the motion, 13, 135
Jacobi's identity, 136
Particle, 1
Pendulums, 11f., 26, 33ff., 61, 70, 95,
Kepler's problem, 35ff.
102f., 129f.
Kepler's second law, 31
compound, 102f.
Kepler's third law, 23
conical, 34
Kinetic energy, 8, 15
Foucault's, 129f.
of rigid body, 98f.
spherical, 33f.
Perihelion, 36
Laboratory system, 41
movement of, 40
Lagrange's equations, 3f.
Phase, 59
Lagrangian, 2ff.
path, 146
for free motion, 5
space, 146
of free particle, 6ff.
Point transformation, 143
in non-inertial frame, 127
Poisson brackets, 135ff.
for one-dimensional motion, 25, 58
Poisson's theorem, 137
of rigid body, 99
Polhodes, 117n.
for small oscillations, 58, 61, 66, 69, 84
Potential energy, 8, 15
of system of particles, 8ff.
centrifugal, 32, 128
of two bodies, 29
effective, 32, 94
Latus rectum, 36
from period of oscillation, 27ff.
Least action, principle of, 2ff.
Potential well, 26, 54f.
Legendre's transformation, 131
Precession, regular, 107
Liouville's theorem, 147
L system, 41
Rapidly oscillating field, motion in, 93ff.
Reactions, 122
Mass, 7
Reduced mass, 29
additivity of, 17
Resonance, 62, 79
centre of, 17
in non-linear oscillations, 87ff.
reduced, 29
parametric, 80ff.
Mathieu's equation, 82n.
Rest, system at, 17
Maupertuis' principle, 141
Reversibility of motion, 9
Index
165
Rigid bodies, 96
Space
angular momentum of, 105ff.
homogeneity of, 5, 15
in contact, 122ff.
isotropy of, 5, 18
equations of motion of, 107ff.
Space oscillator, 32, 70
motion of (VI), 96ff.
Rolling, 122
Time
Rotator, 101, 106
homogeneity of, 5, 13ff.
Rough surface, 122
isotropy of, 8f.
Routhian, 134f.
Top
Rutherford's formula, 53f.
asymmetrical, 100, 116ff.
"fast", 113f.
spherical, 100, 106
Scattering, 48ff.
symmetrical, 100, 106f., 111f.
cross-section, effective, 49ff.
Torque, 108
Rutherford's formula for, 53f.
Turning points, 25, 32
small-angle, 55ff.
Two-body problem, 29
Sectorial velocity, 31
Separation of variables, 149ff.
Uniform field, 10
Similarity, mechanical, 22ff.
Sliding, 122
Variation, 2, 3
Small oscillations, 22, (V) 58ff.
first, 3
anharmonic, 84ff.
Velocity, 1
damped, 74ff.
angular, 97f.
forced, 61ff., 77ff.
sectorial, 31
free, 58ff., 65ff.
translational, 97
linear, 84
Virial, 23n.
non-linear, 84ff.
theorem, 23f.
normal, 68
Smooth surface, 122
Well, potential, 26, 54f.
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§14
Motion in a central field
31
(Fig. 8). Calling this area df, we can write the angular momentum of the par-
ticle as
M = 2mf,
(14.3)
where the derivative f is called the sectorial velocity. Hence the conservation
of angular momentum implies the constancy of the sectorial velocity: in equal
times the radius vector of the particle sweeps out equal areas (Kepler's second
law).t
rdd
dd
0
FIG. 8
The complete solution of the problem of the motion of a particle in a central
field is most simply obtained by starting from the laws of conservation of
energy and angular momentum, without writing out the equations of motion
themselves. Expressing in terms of M from (14.2) and substituting in the
expression for the energy, we obtain
E = =
(14.4)
Hence
(14.5)
or, integrating,
constant.
(14.6)
Writing (14.2) as do = M dt/mr2, substituting dt from (14.5) and integrating,
we find
constant.
(14.7)
Formulae (14.6) and (14.7) give the general solution of the problem. The
latter formula gives the relation between r and , i.e. the equation of the path.
Formula (14.6) gives the distance r from the centre as an implicit function of
time. The angle o, it should be noted, always varies monotonically with time,
since (14.2) shows that & can never change sign.
t The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.
32
Integration of the Equations of Motion
§14
The expression (14.4) shows that the radial part of the motion can be re-
garded as taking place in one dimension in a field where the "effective poten-
tial energy" is
(14.8)
The quantity M2/2mr2 is called the centrifugal energy. The values of r for which
U(r)
(14.9)
determine the limits of the motion as regards distance from the centre.
When equation (14.9) is satisfied, the radial velocity j is zero. This does not
mean that the particle comes to rest as in true one-dimensional motion, since
the angular velocity o is not zero. The value j = 0 indicates a turning point
of the path, where r(t) begins to decrease instead of increasing, or vice versa.
If the range in which r may vary is limited only by the condition r > rmin,
the motion is infinite: the particle comes from, and returns to, infinity.
If the range of r has two limits rmin and rmax, the motion is finite and the
path lies entirely within the annulus bounded by the circles r = rmax and
r = rmin- This does not mean, however, that the path must be a closed curve.
During the time in which r varies from rmax to rmin and back, the radius
vector turns through an angle Ao which, according to (14.7), is given by
Mdr/r2
(14.10)
The condition for the path to be closed is that this angle should be a rational
fraction of 2n, i.e. that Ao = 2nm/n, where m and n are integers. In that case,
after n periods, the radius vector of the particle will have made m complete
revolutions and will occupy its original position, so that the path is closed.
Such cases are exceptional, however, and when the form of U(r) is arbitrary
the angle is not a rational fraction of 2nr. In general, therefore, the path
of a particle executing a finite motion is not closed. It passes through the
minimum and maximum distances an infinity of times, and after infinite time
it covers the entire annulus between the two bounding circles. The path
shown in Fig. 9 is an example.
There are only two types of central field in which all finite motions take
place in closed paths. They are those in which the potential energy of the
particle varies as 1/r or as r2. The former case is discussed in §15; the latter
is that of the space oscillator (see §23, Problem 3).
At a turning point the square root in (14.5), and therefore the integrands
in (14.6) and (14.7), change sign. If the angle is measured from the direc-
tion of the radius vector to the turning point, the parts of the path on each
side of that point differ only in the sign of for each value of r, i.e. the path
is symmetrical about the line = 0. Starting, say, from a point where = rmax
the particle traverses a segment of the path as far as a point with r rmin,
§14
Motion in a central field
33
then follows a symmetrically placed segment to the next point where r = rmax,
and so on. Thus the entire path is obtained by repeating identical segments
forwards and backwards. This applies also to infinite paths, which consist of
two symmetrical branches extending from the turning point (r = rmin) to
infinity.
'max
min
so
FIG. 9
The presence of the centrifugal energy when M # 0, which becomes
infinite as 1/22 when r -> 0, generally renders it impossible for the particle to
reach the centre of the field, even if the field is an attractive one. A "fall" of
the particle to the centre is possible only if the potential energy tends suffi-
ciently rapidly to -00 as r 0. From the inequality
1mr2 = E- U(r) - M2/2mr2
or r2U(Y)+M2/2m < Er2, it follows that r can take values tending to zero
only if
(14.11)
i.e. U(r) must tend to - 8 either as - a/r2 with a > M2/2m, or proportionally
to - 1/rn with n > 2.
PROBLEMS
PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass
m moving on the surface of a sphere of radius l in a gravitational field).
SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the
polar axis vertically downwards, the Lagrangian of the pendulum is
1ml2(02 + 62 sin20) +mgl cos 0.
2*
34
Integration of the Equations of Motion
§14
The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the
z-component of angular momentum, is conserved:
(1)
The energy is
E = cos 0
(2)
= 0.
Hence
(3)
where the "effective potential energy" is
Ueff(0) = COS 0.
For the angle o we find, using (1),
do
(4)
The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively.
The range of 0 in which the motion takes place is that where E > Ueff, and its limits
are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots
between -1 and +1; these define two circles of latitude on the sphere, between which the
path lies.
PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a
cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational
field.
SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the
polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co-
ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is
= a.
By the same method as in Problem 1, we find
==
The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots;
these define two horizontal circles on the cone, between which the path lies.
PROBLEM 3. Integrate the equations of motion for a pendulum of mass M2, with a mass M1
at the point of support which can move on a horizontal line lying in the plane in which m2
moves (Fig. 2, §5).
SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The
generalised momentum Px, which is the horizontal component of the total momentum of the
system, is therefore conserved
Px = cos = constant.
(1)
The system may always be taken to be at rest as a whole. Then the constant in (1) is zero
and integration gives
(m1+m2)x+m2) sin = constant,
(2)
which expresses the fact that the centre of mass of the system does not move horizontally.
§15
Kepler's problem
35
Using (1), we find the energy in the form
E
(3)
Hence
Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
dulum, which moves in an arc of a circle.
$15. Kepler's problem
An important class of central fields is formed by those in which the poten-
tial energy is inversely proportional to r, and the force accordingly inversely
proportional to r2. They include the fields of Newtonian gravitational attrac-
tion and of Coulomb electrostatic interaction; the latter may be either attrac-
tive or repulsive.
Let us first consider an attractive field, where
U=-a/r
(15.1)
with a a positive constant. The "effective" potential energy
(15.2)
is
of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
r
8
it tends to zero from negative values ; for r = M2/ma it has a minimum value
Ueff, min = -mx2/2M2.
(15.3)
Ueff
FIG. 10
It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
for E > 0.
36
Integration of the Equations of Motion
§15
The shape of the path is obtained from the general formula (14.7). Substi-
tuting there U = - a/r and effecting the elementary integration, we have
o =
- constant.
Taking the origin of such that the constant is zero, and putting
P = M2/ma, e= [1 1+(2EM2/mo2)]
(15.4)
we can write the equation of the path as
p/r = 1+e coso.
(15.5)
This is the equation of a conic section with one focus at the origin; 2p is called
the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
is seen from (15.5) to be such that the point where = 0 is the point nearest
to the origin (called the perihelion).
In the equivalent problem of two particles interacting according to the law
(15.1), the orbit of each particle is a conic section, with one focus at the centre
of mass of the two particles.
It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
has been said earlier in this section. According to the formulae of analytical
geometry, the major and minor semi-axes of the ellipse are
a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
(15.6)
y
X
2b
ae
2a
FIG. 11
The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
becomes a circle. It may be noted that the major axis of the ellipse depends
only on the energy of the particle, and not on its angular momentum. The
least and greatest distances from the centre of the field (the focus of the
ellipse) are
rmin = =p/(1+e)=a(1-e),
max=p(1-e)=a(1+e). = = (15.7)
These expressions, with a and e given by (15.6) and (15.4), can, of course,
also be obtained directly as the roots of the equation Ueff(r) = E.
§15
Kepler's problem
37
The period T of revolution in an elliptical orbit is conveniently found by
using the law of conservation of angular momentum in the form of the area
integral (14.3). Integrating this equation with respect to time from zero to
T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
f = nab, and by using the formulae (15.6) we find
T = 2ma3/2-(m/a)
= ma((m2E3).
(15.8)
The proportionality between the square of the period and the cube of the
linear dimension of the orbit has already been demonstrated in §10. It may
also be noted that the period depends only on the energy of the particle.
For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
tance of the perihelion from the focus is
rmin ==pl(e+1)=a(e-1), = =
(15.9)
where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
y
p
ale-1)
FIG. 12
If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
perihelion distance rmin = 1p. This case occurs if the particle starts from rest
at infinity.
The co-ordinates of the particle as functions of time in the orbit may be
found by means of the general formula (14.6). They may be represented in a
convenient parametric form as follows.
Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
we can write the integral (14.6) for the time as
t
=
=
38
Integration of the Equations of Motion
§15
The obvious substitution r-a = - ae cos $ converts the integral to
sioant
If time is measured in such a way that the constant is zero, we have the
following parametric dependence of r on t:
r = a(1-e cos ), t =
(15.10)
the particle being at perihelion at t = 0. The Cartesian co-ordinates
x = r cos o, y = r sin (the x and y axes being respectively parallel to the
major and minor axes of the ellipse) can likewise be expressed in terms of
the parameter $. From (15.5) and (15.10) we have
ex = = =
y is equal to W(r2-x2). Thus
x = a(cos & - e),
y = =av(1-e2) $.
(15.11)
A complete passage round the ellipse corresponds to an increase of $ from 0
to 2nr.
Entirely similar calculations for the hyperbolic orbits give
r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
(15.12)
x = a(e-cosh ) y = a1/(e2-1)sinh &
where the parameter $ varies from - 00 to + 00.
Let us now consider motion in a repulsive field, where
U
(a>0).
(15.13)
Here the effective potential energy is
Utt
and decreases monotonically from + 00 to zero as r varies from zero to
infinity. The energy of the particle must be positive, and the motion is always
infinite. The calculations are exactly similar to those for the attractive field.
The path is a hyperbola (or, if E = 0, a parabola):
pr r = =1-e coso, =
(15.14)
where P and e are again given by (15.4). The path passes the centre of the
field in the manner shown in Fig. 13. The perihelion distance is
rmin =p(e-1)=a(e+1). =
(15.15)
The time dependence is given by the parametric equations
= =(ma3/a)(esinh+) =
(15.16)
x
= a(cosh & e ,
y = av((e2-1) sinh &
§15
Kepler's problem
39
To conclude this section, we shall show that there is an integral of the mo-
tion which exists only in fields U = a/r (with either sign of a). It is easy to
verify by direct calculation that the quantity
vxM+ar/r
(15.17)
is constant. For its total time derivative is v
since M = mr xv,
Putting mv = ar/r3 from the equation of motion, we find that this expression
vanishes.
y
0
(I+e)
FIG. 13
The direction of the conserved vector (15.17) is along the major axis from
the focus to the perihelion, and its magnitude is ae. This is most simply
seen by considering its value at perihelion.
It should be emphasised that the integral (15.17) of the motion, like M and
E, is a one-valued function of the state (position and velocity) of the particle.
We shall see in §50 that the existence of such a further one-valued integral
is due to the degeneracy of the motion.
PROBLEMS
PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
moving in a parabola in a field U = -a/r.
SOLUTION. In the integral
we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
the required dependence:
r=1p(1+n2),
t=
y=pn.
40
Integration of the Equations of Motion
§15
The parameter n varies from - 00 to +00.
PROBLEM 2. Integrate the equations of motion for a particle in a central field
U = - a/r2 (a > 0).
SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
(a) for E > andM 0 and
(b) for E>0 0nd and M 2/2m a,
(c) for E <0 and Ms1
In all three cases
In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
origin as
00. The fall from a given value of r takes place in a finite time, namely
PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
(14.10), which we write as
in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
the first-order term gives the required change so:
(1)
where we have changed from the integration over r to one over , along the path of the "un-
perturbed" motion.
In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.

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CHAPTER IV
COLLISIONS BETWEEN PARTICLES
§16. Disintegration of particles
IN many cases the laws of conservation of momentum and energy alone can
be used to obtain important results concerning the properties of various mech-
anical processes. It should be noted that these properties are independent of
the particular type of interaction between the particles involved.
Let us consider a "spontaneous" disintegration (that is, one not due to
external forces) of a particle into two "constituent parts", i.e. into two other
particles which move independently after the disintegration.
This process is most simply described in a frame of reference in which the
particle is at rest before the disintegration. The law of conservation of momen-
tum shows that the sum of the momenta of the two particles formed in the
disintegration is then zero; that is, the particles move apart with equal and
opposite momenta. The magnitude Po of either momentum is given by the
law of conservation of energy:
here M1 and m2 are the masses of the particles, E1t and E2i their internal
energies, and E the internal energy of the original particle. If € is the "dis-
integration energy", i.e. the difference
E= E:-E11-E2i,
(16.1)
which must obviously be positive, then
(16.2)
which determines Po; here m is the reduced mass of the two particles. The
velocities are V10 = Po/m1, V20 = Po/m2.
Let us now change to a frame of reference in which the primary particle
moves with velocity V before the break-up. This frame is usually called the
laboratory system, or L system, in contradistinction to the centre-of-mass
system, or C system, in which the total momentum is zero. Let us consider
one of the resulting particles, and let V and V0 be its velocities in the L and
the C system respectively. Evidently V = V+vo, or V -V = V0, and SO
(16.3)
where 0 is the angle at which this particle moves relative to the direction of
the velocity V. This equation gives the velocity of the particle as a function
41
42
Collisions Between Particles
§16
of its direction of motion in the L system. In Fig. 14 the velocity V is repre-
sented by a vector drawn to any point on a circle+ of radius vo from a point
A at a distance V from the centre. The cases V < vo and V>00 are shown
in Figs. 14a, b respectively. In the former case 0 can have any value, but in
the latter case the particle can move only forwards, at an angle 0 which does
not exceed Omax, given by
(16.4)
this is the direction of the tangent from the point A to the circle.
C
V
V
VO
VO
max
oo
oo
A
V
A
V
(a) V<VO
)V>Vo
FIG. 14
The relation between the angles 0 and Oo in the L and C systems is evi-
dently (Fig. 14)
tan
(16.5)
If this equation is solved for cos Oo, we obtain
V
0
(16.6)
For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus
sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however,
the relation is not one-to-one: for each value of 0 there are two values of Oo,
which correspond to vectors V0 drawn from the centre of the circle to the
points B and C (Fig. 14b), and are given by the two signs in (16.6).
In physical applications we are usually concerned with the disintegration
of not one but many similar particles, and this raises the problem of the
distribution of the resulting particles in direction, energy, etc. We shall
assume that the primary particles are randomly oriented in space, i.e. iso-
tropically on average.
t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral
section.
§16
Disintegration of particles
43
In the C system, this problem is very easily solved: every resulting particle
(of a given kind) has the same energy, and their directions of motion are
isotropically distributed. The latter fact depends on the assumption that the
primary particles are randomly oriented, and can be expressed by saying
that the fraction of particles entering a solid angle element doo is proportional
to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is
obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is
(16.7)
The corresponding distributions in the L system are obtained by an
appropriate transformation. For example, let us calculate the kinetic energy
distribution in the L system. Squaring the equation V = V0 + V, we have
2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the
kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of
particle is under consideration, and substituting in (16.7), we find the re-
quired distribution:
(1/2mvov) dT.
(16.8)
The kinetic energy can take values between Tmin = 3m(e0-V)2 and
Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed
uniformly over this range.
When a particle disintegrates into more than two parts, the laws of con-
servation of energy and momentum naturally allow considerably more free-
dom as regards the velocities and directions of motion of the resulting particles.
In particular, the energies of these particles in the C system do not have
determinate values. There is, however, an upper limit to the kinetic energy
of any one of the resulting particles. To determine the limit, we consider
the system formed by all these particles except the one concerned (whose
mass is M1, say), and denote the "internal energy" of that system by Ei'.
Then the kinetic energy of the particle M1 is, by (16.1) and (16.2),
T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the
primary particle. It is evident that T10 has its greatest possible value
when E/' is least. For this to be so, all the resulting particles except M1
must be moving with the same velocity. Then Ei is simply the sum of their
internal energies, and the difference E;-E-E; is the disintegration
energy E. Thus
T10,max = (M - M1) E M.
(16.9)
PROBLEMS
PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra-
tion into two particles.
SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling
010 simply Oo and using formula (16.5) for each of the two particles, we can put
7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two
equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then
44
Collisions Between Particles
§17
form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using
(16.2),
(m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6
PROBLEM 2. Find the angular distribution of the resulting particles in the L system.
SOLUTION. When vo > V, we substitute (16.6), with the plus sign of the radical, in (16.7),
obtaining
(0 .
When vo < V, both possible relations between Oo and 0 must be taken into account. Since,
when 0 increases, one value of Oo increases and the other decreases, the difference (not the
sum) of the expressions d cos Oo with the two signs of the radical in (16.6) must be taken.
The result is
(0 max).
PROBLEM 3. Determine the range of possible values of the angle 0 between the directions
of motion of the two resulting particles in the L system.
SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula (16.5)
(see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema
of the resulting expression gives the following ranges of 0, depending on the relative magni-
tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10
< 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by
sin =
§17. Elastic collisions
A collision between two particles is said to be elastic if it involves no change
in their internal state. Accordingly, when the law of conservation of energy
is applied to such a collision, the internal energy of the particles may be
neglected.
The collision is most simply described in a frame of reference in which the
centre of mass of the two particles is at rest (the C system). As in $16, we
distinguish by the suffix 0 the values of quantities in that system. The velo-
cities of the particles before the collision are related to their velocities V1 and
V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
where V = V1-V2; see (13.2).
Because of the law of conservation of momentum, the momenta of the two
particles remain equal and opposite after the collision, and are also unchanged
in magnitude, by the law of conservation of energy. Thus, in the C system
the collision simply rotates the velocities, which remain opposite in direction
and unchanged in magnitude. If we denote by no a unit vector in the direc-
tion of the velocity of the particle M1 after the collision, then the velocities
of the two particles after the collision (distinguished by primes) are
V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
(17.1)
§17
Elastic collisions
45
In order to return to the L system, we must add to these expressions the
velocity V of the centre of mass. The velocities in the L system after the
collision are therefore
V1' =
(17.2)
V2' =
No further information about the collision can be obtained from the laws
of conservation of momentum and energy. The direction of the vector no
depends on the law of interaction of the particles and on their relative position
during the collision.
The results obtained above may be interpreted geometrically. Here it is
more convenient to use momenta instead of velocities. Multiplying equations
(17.2) by M1 and M2 respectively, we obtain
(17.3)
P2' muno+m2(p1+p2)/(m1+m2)
where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
mv and use the construction shown in Fig. 15. If the unit vector no is along
OC, the vectors AC and CB give the momenta P1' and P2' respectively.
When p1 and P2 are given, the radius of the circle and the points A and B
are fixed, but the point C may be anywhere on the circle.
C
p'
no
P'2
B
A
FIG. 15
Let us consider in more detail the case where one of the particles (m2, say) is
at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
momentum P1 of the particle M1 before the collision. The point A lies inside
or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
46
Collisions Between Particles
§17
diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
are the angles between the directions of motion after the collision and the
direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
gives the direction of no, is the angle through which the direction of motion
of m1 is turned in the C system. It is evident from the figure that 01 and O2
can be expressed in terms of X by
(17.4)
C
p'
P2
pi
P2
0
max
10,
X
O2
O2
B
B
A
0
A
Q
0
(a) m < m2
(b) m, m m m
AB=p : AO/OB= m/m2
FIG. 16
We may give also the formulae for the magnitudes of the velocities of the
two particles after the collision, likewise expressed in terms of X:
ib
(17.5)
The sum A1 + O2 is the angle between the directions of motion of the
particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
if M1 > M2.
When the two particles are moving afterwards in the same or in opposite
directions (head-on collision), we have X=TT, i.e. the point C lies on the
diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
In this case the velocities after the collision are
(17.6)
This value of V2' has the greatest possible magnitude, and the maximum
§17
Elastic collisions
47
energy which can be acquired in the collision by a particle originally at rest
is therefore
(17.7)
where E1 = 1M1U12 is the initial energy of the incident particle.
If M1 < M2, the velocity of M1 after the collision can have any direction.
If M1 > M2, however, this particle can be deflected only through an angle
not exceeding Omax from its original direction; this maximum value of A1
corresponds to the position of C for which AC is a tangent to the circle
(Fig. 16b). Evidently
sin Omax = OC|OA = M2/M1.
(17.8)
The collision of two particles of equal mass, of which one is initially at
rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
C
p'
P2
Q2
B
A
0
FIG. 17
Then
01=1x,
A2 = 1(-x),
(17.9)
12
=
(17.10)
After the collision the particles move at right angles to each other.
PROBLEM
Express the velocity of each particle after a collision between a moving particle (m1) and
another at rest (m2) in terms of their directions of motion in the L system.
SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
Hence
for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.
48
Collisions Between Particles
§18
§18. Scattering
As already mentioned in §17, a complete calculation of the result of a
collision between two particles (i.e. the determination of the angle x) requires
the solution of the equations of motion for the particular law of interaction
involved.
We shall first consider the equivalent problem of the deflection of a single
particle of mass m moving in a field U(r) whose centre is at rest (and is at
the centre of mass of the two particles in the original problem).
As has been shown in $14, the path of a particle in a central field is sym-
metrical about a line from the centre to the nearest point in the orbit (OA
in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
say) with this line. The angle X through which the particle is deflected as it
passes the centre is seen from Fig. 18 to be
X = -200.
(18.1)
A
X
to
FIG. 18
The angle do itself is given, according to (14.7), by
(M/r2) dr
(18.2)
taken between the nearest approach to the centre and infinity. It should be
recalled that rmin is a zero of the radicand.
For an infinite motion, such as that considered here, it is convenient to
use instead of the constants E and M the velocity Voo of the particle at infinity
and the impact parameter p. The latter is the length of the perpendicular
from the centre O to the direction of Voo, i.e. the distance at which the particle
would pass the centre if there were no field of force (Fig. 18). The energy
and the angular momentum are given in terms of these quantities by
E = 1mvoo²,
M = mpVoo,
(18.3)
§18
Scattering
49
and formula (18.2) becomes
dr
(18.4)
Together with (18.1), this gives X as a function of p.
In physical applications we are usually concerned not with the deflection
of a single particle but with the scattering of a beam of identical particles
incident with uniform velocity Voo on the scattering centre. The different
particles in the beam have different impact parameters and are therefore
scattered through different angles X. Let dN be the number of particles
scattered per unit time through angles between X and X + dx. This number
itself is not suitable for describing the scattering process, since it is propor-
tional to the density of the incident beam. We therefore use the ratio
do = dN/n,
(18.5)
where n is the number of particles passing in unit time through unit area of
the beam cross-section (the beam being assumed uniform over its cross-
section). This ratio has the dimensions of area and is called the effective
scattering cross-section. It is entirely determined by the form of the scattering
field and is the most important characteristic of the scattering process.
We shall suppose that the relation between X and P is one-to-one; this is
so if the angle of scattering is a monotonically decreasing function of the
impact parameter. In that case, only those particles whose impact parameters
lie between p(x) and p(x) + dp(x) are scattered at angles between X and
+ dx. The number of such particles is equal to the product of n and the
area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
effective cross-section is therefore
do = 2mp dp.
(18.6)
In order to find the dependence of do on the angle of scattering, we need
only rewrite (18.6) as
do = 2(x)|dp(x)/dx|dx
(18.7)
Here we use the modulus of the derivative dp/dx, since the derivative may
be (and usually is) negative. t Often do is referred to the solid angle element
do instead of the plane angle element dx. The solid angle between cones
with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
(18.7)
do.
(18.8)
t If the function p(x) is many-valued, we must obviously take the sum of such expressions
as (18.7) over all the branches of this function.
50
Collisions Between Particles
§18
Returning now to the problem of the scattering of a beam of particles, not
by a fixed centre of force, but by other particles initially at rest, we can say
that (18.7) gives the effective cross-section as a function of the angle of
scattering in the centre-of-mass system. To find the corresponding expression
as a function of the scattering angle 0 in the laboratory system, we must
express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
expressions for both the scattering cross-section for the incident beam of
particles (x in terms of 01) and that for the particles initially at rest (x in terms
of O2).
PROBLEMS
PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
for r>a).
SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
the path consists of two straight lines symmetrical about the radius to the point where the
particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
a sin to = a sin 1(-x) = a cos 1x.
A
to
p
&
FIG. 19
Substituting in (18.7) or (18.8), we have
do = 1ma2 sin X do,
(1)
i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
the total cross-section o = na2, in accordance with the fact that the "impact area" which the
particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
and m2 that of the sphere) we have
do1,
where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
stituting X = 201 from (17.9) in (1).

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§18
Scattering
51
For a sphere originally at rest, X = n-202 in all cases, and substitution in (1) gives
do2 = a2|cos 02 do2.
PROBLEM 2. Express the effective cross-section (Problem 1) as a function of the energy E
lost by a scattered particle.
SOLUTION. The energy lost by a particle of mass M1 is equal to that gained by the sphere of
mass M2. From (17.5) and (17.7), € = E2' = [2m22m2/(m1+m2)2] voo2 sin21x = Emax sin21x,
whence de = 1 mex sin X dx; substituting in (1), Problem 1, we have do = na2 de/emax. The
scattered particles are uniformly distributed with respect to € in the range from zero to
Emax.
PROBLEM 3. Find the effective cross-section as a function of the velocity Uoo for particles
scattered in a field U -rrn.
SOLUTION. According to (10.3), if the potential energy is a homogeneous function of order
k = -n, then similar paths are such that p ~v-2/n, or p = Voo-2/nf(x), the angles of deflec-
tion X being equal for similar paths. Substitution in (18.6) gives do ~ Voo-4/n do.
PROBLEM 4. Determine the effective cross-section for a particle to "fall" to the centre of
a field U = -a/r2.
SOLUTION. The particles which "fall" to the centre are those for which 2a > mp20002 (see
(14.11)), i.e. for which the impact parameter does not exceed Pmax = (2a/mvoo). The
effective cross-section is therefore o = Pmax2 = 2na/mvo².
PROBLEM 5. The same as Problem 4, but for a field U = -a/rn (n 2, a > 0).
SOLUTION. The effective potential energy Ueff = depends on r in the
manner shown in Fig. 20. Its maximum value is
Ueff
U0
FIG. 20
The particles which "fall" to the centre are those for which U0 < E. The condition U0 = E
gives Pmax, whence
=
PROBLEM 6. Determine the effective cross-section for particles of mass m1 to strike a sphere
of mass m2 and radius R to which they are attracted in accordance with Newton's law.
SOLUTION. The condition for a particle to reach the sphere is that rmin < R, where r'min
is the point on the path which is nearest to the centre of the sphere. The greatest possible
value of P is given by rmin = R; this is equivalent to Ueff(R) = E or
= , where a = ymim2 (2 being the gravitational constant) and we have put m 22 M1 on
the assumption that m2 > M1. Solving for pmax2, we finally obtain o = mR2(1+2ym2/Rvv3).
52
Collisions Between Particles
§18
When
Voo
8 the effective cross-section tends, of course, to the geometrical cross-section
of the sphere.
PROBLEM 7. Deduce the form of a scattering field U(r), given the effective cross-section
as a function of the angle of scattering for a given energy E. It is assumed that U(r) decreases
monotonically with r (a repulsive field), with U(0) > E and U(00) = 0 (O. B. FIRSOV 1953).
SOLUTION. Integration of do with respect to the scattering angle gives, according to the
formula
(1)
the square of the impact parameter, so that p(x) (and therefore x(p)) is known.
We put
s=1/r,
=1/p2,
[[1-(U|E)]
(2)
Then formulae (18.1), (18.2) become
1/
(3)
where so(x) is the root of the equation xw2(so)-so2 = 0.
Equation (3) is an integral equation for the function w(s), and may be solved by a method
similar to that used in $12. Dividing both sides of (3) by (a-x) and integrating with respect
to x from zero to a, we find
so(a)
dx ds
so(a)
ds
or, integrating by parts on the left-hand side,
This relation is differentiated with respect to a, and then so(a) is replaced by s simply;
accordingly a is replaced by s2/w2, and the result is, in differential form,
=
(11/20)
or
dx
This equation can be integrated immediately if the order of integration on the right-hand
side is inverted. Since for S = 0 (i.e. r oo) we must have W = 1 (i.e. U = 0), we have,
§19
Rutherford's formula
53
on returning to the original variables r and P, the following two equivalent forms of the final
result:
=====
(dx/dp)
(4)
This formula determines implicitly the function w(r) (and therefore U(r)) for all r > rmin,
i.e. in the range of r which can be reached by a scattered particle of given energy E.
§19. Rutherford's formula
One of the most important applications of the formulae derived above is
to the scattering of charged particles in a Coulomb field. Putting in (18.4)
U = a/r and effecting the elementary integration, we obtain
whence p2 = (a2/m2voo4) tan2oo, or, putting 00 = 1(-) from (18.1),
p2 =
(19.1)
Differentiating this expression with respect to X and substituting in (18.7)
or (18.8) gives
do = (a/v2cosxdx/sin31
(19.2)
or
do =
(19.3)
This is Rutherford's formula. It may be noted that the effective cross-section
is independent of the sign of a, so that the result is equally valid for repulsive
and attractive Coulomb fields.
Formula (19.3) gives the effective cross-section in the frame of reference
in which the centre of mass of the colliding particles is at rest. The trans-
formation to the laboratory system is effected by means of formulae (17.4).
For particles initially at rest we substitute X = 77 - 202 in (19.2) and obtain
do2 = 2n(a/mvoo2)2 sin de2/cos302
=
(19.4)
The same transformation for the incident particles leads, in general, to a very
complex formula, and we shall merely note two particular cases.
If the mass M2 of the scattering particle is large compared with the mass
M1 of the scattered particle, then X 2 O1 and m 22 M1, SO that
do1 = = (a/4E1)2do1/sin4301
(19.5)
where E1 = 1M1U..02 is the energy of the incident particle.
54
Collisions Between Particles
§ 19
If the masses of the two particles are equal (m1 = M2, m = 1M1), then by
(17.9) X = 201, and substitution in (19.2) gives
do1 = 2(/E1)2 cos 01 d01/sin³01
=
(19.6)
If the particles are entirely identical, that which was initially at rest cannot
be distinguished after the collision. The total effective cross-section for all
particles is obtained by adding do1 and do2, and replacing A1 and O2 by their
common value 0:
do
=
do.
(19.7)
Let us return to the general formula (19.2) and use it to determine the
distribution of the scattered particles with respect to the energy lost in the
collision. When the masses of the scattered (m1) and scattering (m2) particles
are arbitrary, the velocity acquired by the latter is given in terms of the angle
of scattering in the C system by V2' = [2m1/(m1+m2)]%"00 sin 1x; see (17.5).
The energy acquired by M2 and lost by M1 is therefore E = 1M2U2'2
= (2m2/m2)/0002 sin21x. Expressing sin 1x in terms of E and substituting
in (19.2), we obtain
do = de/e2.
(19.8)
This is the required formula: it gives the effective cross-section as a function
of the energy loss E, which takes values from zero to Emax = 2m2voo2/m2.
PROBLEMS
PROBLEM 1. Find the effective cross-section for scattering in a field U = a/r2 (a > 0).
SOLUTION. The angle of deflection is
The effective cross-section is
do
sin
PROBLEM 2. Find the effective cross-section for scattering by a spherical "potential well"
of radius a and "depth" U0 (i.e. a field with U = 0 for r > a and U = - U0 for r < a).
SOLUTION. The particle moves in a straight line which is "refracted" on entering and leav- -
ing the well. According to §7, Problem, the angle of incidence a and the angle of refraction
B (Fig. 21) are such that sin x/sin B = n, where n = W(1+2U0/mVo2). The angle of deflection
is X = 2(a-B). Hence
=
Eliminating a from this equation and the relation a sin a p, which is evident from the
diagram, we find the relation between P and X:
cos
1x
§20
Small-angle scattering
55
Finally, differentiating, we have the effective cross-section
cos
do.
The angle X varies from zero (for p=0) to Xmax (for P = a), where cos 1xmax = 1/n.
The total effective cross-section, obtained by integrating do over all angles within the cone
Xmax, is, of course, equal to the geometrical cross-section 2
a
to
a
FIG. 21
§20. Small-angle scattering
The calculation of the effective cross-section is much simplified if only
those collisions are considered for which the impact parameter is large, so
that the field U is weak and the angles of deflection are small. The calculation
can be carried out in the laboratory system, and the centre-of-mass system
need not be used.
We take the x-axis in the direction of the initial momentum of the scattered
particle M1, and the xy-plane in the plane of scattering. Denoting by P1' the
momentum of the particle after scattering, we evidently have sin 01 = P1y'/P1'.
For small deflections, sin 01 may be approximately replaced by 01, and P1' in
the denominator by the initial momentum P1 = MIUoo:
(20.1)
Next, since Py = Fy, the total increment of momentum in the y-direction is
(20.2)
The
force Fy = - JULY (dU/dr)dr/dy = (dU/dr)y/r.
Since the integral (20.2) already contains the small quantity U, it can be
calculated, in the same approximation, by assuming that the particle is not
deflected at all from its initial path, i.e. that it moves in a straight line y = p
with uniform velocity Voo. Thus we put in (20.2) Fy = - (dU/dr)p/r,
dt = dx/voo. The result is
56
Collisions Between Particles
§20
Finally, we change the integration over x to one over r. Since, for a straight
path, r2 = x2+p2, when x varies from - 00 to + 80, r varies from 8 to P
and back. The integral over x therefore becomes twice the integral over r
from p to 80, and dx = r dr/v(r2-p2). The angle of scattering O1 is thus
given byt
(20.3)
and this is the form of the function 01(p) for small deflections. The effective
cross-section for scattering (in the L system) is obtained from (18.8) with 01
instead of X, where sin 01 may now be replaced by A1:
(20.4)
PROBLEMS
PROBLEM 1. Derive formula (20.3) from (18.4).
SOLUTION. In order to avoid spurious divergences, we write (18.4) in the form
PO
and take as the upper limit some large finite quantity R, afterwards taking the value as R
00.
Since U is small, we expand the square root in powers of U, and approximately replace
rmin by p:
dr
The first integral tends to 1/11 as R 00. The second integral is integrated by parts, giving
=
This is equivalent to (20.3).
PROBLEM 2. Determine the effective cross-section for small-angle scattering in a field
U=a/m(n) 0).
t If the above derivation is applied in the C system, the expression obtained for X is the
same with m in place of M1, in accordance with the fact that the small angles 01 and X are
related by (see (17.4)) 01 = m2x/(m1 +m2).
§20
Small-angle scattering
57
SOLUTION. From (20.3) we have
dr
The substitution p2/r2 = U converts the integral to a beta function, which can be expressed
in terms of gamma functions:
Expressing P in terms of 01 and substituting in (20.4), we obtain
do1.
3
CHAPTER V
SMALL OSCILLATIONS
$21. Free oscillations in one dimension
A VERY common form of motion of mechanical systems is what are called
small oscillations of a system about a position of stable equilibrium. We shall
consider first of all the simplest case, that of a system with only one degree
of freedom.
Stable equilibrium corresponds to a position of the system in which its
potential energy U(q) is a minimum. A movement away from this position
results in the setting up of a force - dU/dq which tends to return the system
to equilibrium. Let the equilibrium value of the generalised co-ordinate
q be 90. For small deviations from the equilibrium position, it is sufficient
to retain the first non-vanishing term in the expansion of the difference
U(q) - U(90) in powers of q-qo. In general this is the second-order term:
U(q) - U(q0) 112 1k(q-90)2, where k is a positive coefficient, the value of the
second derivative U"(q) for q = 90. We shall measure the potential energy
from its minimum value, i.e. put U(qo) = 0, and use the symbol
x = q-90
(21.1)
for the deviation of the co-ordinate from its equilibrium value. Thus
U(x) = .
(21.2)
The kinetic energy of a system with one degree of freedom is in general
of the form 1a(q)q2 = 1a(q)x2. In the same approximation, it is sufficient to
replace the function a(q) by its value at q = qo. Putting for brevity a(go) = m,
we have the following expression for the Lagrangian of a system executing
small oscillations in one dimension:
L = 1mx2-1kx2.
(21.3)
The corresponding equation of motion is
m+kx=0,
(21.4)
or
w2x=0,
(21.5)
where
w= ((k/m).
(21.6)
+ It should be noticed that m is the mass only if x is the Cartesian co-ordinate.
+ Such a system is often called a one-dimensional oscillator.
58
§21
Free oscillations in one dimension
59
Two independent solutions of the linear differential equation (21.5) are
cos wt and sin wt, and its general solution is therefore
COS wt +C2 sin wt.
(21.7)
This expression can also be written
x = a cos(wt + a).
(21.8)
Since cos(wt+a) = cos wt cos a - sin wt sin a, a comparison with (21.7)
shows that the arbitrary constants a and a are related to C1 and C2 by
tan a = - C2/C1.
(21.9)
Thus, near a position of stable equilibrium, a system executes harmonic
oscillations. The coefficient a of the periodic factor in (21.8) is called the
amplitude of the oscillations, and the argument of the cosine is their phase;
a is the initial value of the phase, and evidently depends on the choice of
the origin of time. The quantity w is called the angular frequency of the oscil-
lations; in theoretical physics, however, it is usually called simply the fre-
quency, and we shall use this name henceforward.
The frequency is a fundamental characteristic of the oscillations, and is
independent of the initial conditions of the motion. According to formula
(21.6) it is entirely determined by the properties of the mechanical system
itself. It should be emphasised, however, that this property of the frequency
depends on the assumption that the oscillations are small, and ceases to hold
in higher approximations. Mathematically, it depends on the fact that the
potential energy is a quadratic function of the co-ordinate.
The energy of a system executing small oscillations is E =
= 1m(x2+w2x2) or, substituting (21.8),
E =
(21.10)
It is proportional to the square of the amplitude.
The time dependence of the co-ordinate of an oscillating system is often
conveniently represented as the real part of a complex expression:
x = re[A exp(iwt)],
(21.11)
where A is a complex constant; putting
A = a exp(ix),
(21.12)
we return to the expression (21.8). The constant A is called the complex
amplitude; its modulus is the ordinary amplitude, and its argument is the
initial phase.
The use of exponential factors is mathematically simpler than that of
trigonometrical ones because they are unchanged in form by differentiation.
t It therefore does not hold good if the function U(x) has at x = 0 a minimum of
higher order, i.e. U ~ xn with n > 2; see §11, Problem 2(a).
60
Small Oscillations
§21
So long as all the operations concerned are linear (addition, multiplication
by constants, differentiation, integration), we may omit the sign re through-
out and take the real part of the final result.
PROBLEMS
PROBLEM 1. Express the amplitude and initial phase of the oscillations in terms of the
initial co-ordinate xo and velocity vo.
SOLUTION. a = (xx2+002/w2), tan a = -vo/wxo.
PROBLEM 2. Find the ratio of frequencies w and w' of the oscillations of two diatomic
molecules consisting of atoms of different isotopes, the masses of the atoms being M1, m2 and
'M1', m2'.
SOLUTION. Since the atoms of the isotopes interact in the same way, we have k = k'.
The coefficients m in the kinetic energies of the molecules are their reduced masses. Accord-
ing to (21.6) we therefore have
PROBLEM 3. Find the frequency of oscillations of a particle of mass m which is free to
move along a line and is attached to a spring whose other end is fixed at a point A (Fig. 22)
at a distance l from the line. A force F is required to extend the spring to length l.
A
X
FIG. 22
SOLUTION. The potential energy of the spring is (to within higher-order terms) equal to
the force F multiplied by the extension Sl of the spring. For x < l we have 81 = (12++2) -
=
x2/21, so that U = Fx2/21. Since the kinetic energy is 1mx2, we have = V(F/ml).
PROBLEM 4. The same as Problem 3, but for a particle of mass m moving on a circle of
radius r (Fig. 23).
m
&
FIG. 23

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§22
Forced oscillations
61
SOLUTION. In this case the extension of the spring is (if
= cos
The kinetic energy is T = 1mr2o2, and the frequency is therefore w = V[F(++1)/mrl].
PROBLEM 5. Find the frequency of oscillations of the pendulum shown in Fig. 2 (§5),
whose point of support carries a mass M1 and is free to move horizontally.
SOLUTION. For < 1 the formula derived in $14, Problem 3 gives
T
Hence
PROBLEM 6. Determine the form of a curve such that the frequency of oscillations of a
particle on it under the force of gravity is independent of the amplitude.
SOLUTION. The curve satisfying the given condition is one for which the potential energy
of a particle moving on it is U = 1ks2, where s is the length of the arc from the position of
equilibrium. The kinetic energy T = 1ms2, where m is the mass of the particle, and the fre-
quency is then w = (k/m) whatever the initial value of S.
In a gravitational field U = mgy, where y is the vertical co-ordinate. Hence we have
1ks2 = mgy or y = w2s2/2g. But ds2 = dx2+dy², whence
dy = SV[(g/2w2y)-1] dy.
The integration is conveniently effected by means of the substitution y = g(1-cos E)/4ww,
which yields x = g(s+sin 5)/4w2. These two equations give, in parametric form, the equation
of the required curve, which is a cycloid.
$22. Forced oscillations
Let us now consider oscillations of a system on which a variable external
force acts. These are called forced oscillations, whereas those discussed in
§21 are free oscillations. Since the oscillations are again supposed small, it
is implied that the external field is weak, because otherwise it could cause the
displacement x to take too large values.
The system now has, besides the potential energy 1kx2, the additional
potential energy Ue(x, t) resulting from the external field. Expanding this
additional term as a series of powers of the small quantity x, we have
Ue(x, t) 12 x[dUe/dx]x_0. The first term is a function of time only,
and may therefore be omitted from the Lagrangian, as being the total time
derivative of another function of time. In the second term - [dUe/dx]x_0 is
the external "force" acting on the system in the equilibrium position, and
is a given function of time, which we denote by F(t). Thus the potential
energy involves a further term -xF(t), and the Lagrangian of the system
is
L
=
(22.1)
The corresponding equation of motion is m+kx = F(t) or
(22.2)
where we have again introduced the frequency w of the free oscillations.
The general solution of this inhomogeneous linear differential equation
with constant coefficients is x = xo+x1, where xo is the general solution of
62
Small Oscillations
§22
the corresponding homogeneous equation and X1 is a particular integral of
the inhomogeneous equation. In the present case xo represents the free
oscillations discussed in $21.
Let us consider a case of especial interest, where the external force is itself
a simple periodic function of time, of some frequency y:
F(t) = f cos(yt+)).
(22.3)
We seek a particular integral of equation (22.2) in the form X1 = b cos(yt+B),
with the same periodic factor. Substitution in that equation gives
b = f/m(w2-r2); adding the solution of the homogeneous equation, we
obtain the general integral in the form
(22.4)
The arbitrary constants a and a are found from the initial conditions.
Thus a system under the action of a periodic force executes a motion which
is a combination of two oscillations, one with the intrinsic frequency w of
the system and one with the frequency y of the force.
The solution (22.4) is not valid when resonance occurs, i.e. when the fre-
quency y of the external force is equal to the intrinsic frequency w of the
system. To find the general solution of the equation of motion in this case,
we rewrite (22.4) as
x
=
a
where a now has a different value. Asy->w, the second term is indetermin-
ate, of the form 0/0. Resolving the indeterminacy by L'Hospital's rule, we
have
x = acos(wt+a)+(f/2mw)tsin(wt+B). =
(22.5)
Thus the amplitude of oscillations in resonance increases linearly with the
time (until the oscillations are no longer small and the whole theory given
above becomes invalid).
Let us also ascertain the nature of small oscillations near resonance, when
y w+E with E a small quantity. We put the general solution in the com-
plex form
= A exp(iwt) exp[i(w+t)) = [A+B exp(iet)]exp(ist)
(22.6)
Since the quantity A+B exp(iet) varies only slightly over the period 2n/w
of the factor exp(iwt), the motion near resonance may be regarded as small
oscillations of variable amplitude.t Denoting this amplitude by C, we have
= A B exp(iet)|. Writing A and B in the form a exp(ix) and b exp(iB)
respectively, we obtain
(22.7)
t The "constant" term in the phase of the oscillation also varies.
§22
Forced oscillations
63
Thus the amplitude varies periodically with frequency E between the limits
|a-b a+b. This phenomenon is called beats.
The equation of motion (22.2) can be integrated in a general form for an
arbitrary external force F(t). This is easily done by rewriting the equation
as
or
=
(22.8)
where
s=xtiwx
(22.9)
is a complex quantity. Equation (22.8) is of the first order. Its solution when
the right-hand side is replaced by zero is $ = A exp(iwt) with constant A.
As before, we seek a solution of the inhomogeneous equation in the form
$ = A(t) exp(iwt), obtaining for the function A(t) the equation À(t)
= F(t) exp(-iwt)/m. Integration gives the solution of (22.9):
& = -
(22.10)
where the constant of integration so is the value of $ at the instant t = 0.
This is the required general solution; the function x(t) is given by the imagin-
ary part of (22.10), divided by w.t
The energy of a system executing forced oscillations is naturally not con-
served, since the system gains energy from the source of the external field.
Let us determine the total energy transmitted to the system during all time,
assuming its initial energy to be zero. According to formula (22.10), with
the lower limit of integration - 00 instead of zero and with ( - 00) = 0,
we have for t
00
exp(-iwt)dt|
The energy of the system is
E = 1m(x2+w2x2)= = 1ME2.
(22.11)
Substituting we obtain the energy transferred
(22.12)
t The force F(t) must, of course, be written in real form.
64
Small Oscillations
§22
it is determined by the squared modulus of the Fourier component of the
force F(t) whose frequency is the intrinsic frequency of the system.
In particular, if the external force acts only during a time short in com-
parison with 1/w, we can put exp(-iwt) Ill 1. Then
This result is obvious: it expresses the fact that a force of short duration
gives the system a momentum I F dt without bringing about a perceptible
displacement.
PROBLEMS
PROBLEM 1. Determine the forced oscillations of a system under a force F(t) of the follow-
ing forms, if at time t = 0 the system is at rest in equilibrium (x = x = 0): (a) F = Fo,
a constant, (b) F = at, (c) F = Fo exp(-at), (d) F = Fo exp(-at) cos Bt.
SOLUTION. (a) x = (Fo/mw2)(1-cos wt). The action of the constant force results in a dis-
placement of the position of equilibrium about which the oscillations take place.
(b) x = (a/mw3)(wt-sin wt).
(c) x = - cos wt +(a/w) sin wt].
(d) x = wt + sin wt +
+exp(-at)[(wpta2-B2) cos Bt-2aB sin
This last case is conveniently treated by writing the force in the complex form
F=Foexp(-ati)t].
PROBLEM 2. Determine the final amplitude for the oscillations of a system under a force
which is zero for t<0, Fot/T for 0 <t<<, and Fo for t > T (Fig. 24), if up to time
t = 0 the system is at rest in equilibrium.
F
Fo
,
T
FIG. 24
SOLUTION. During the interval 0<+<T the oscillations are determined by the initial
condition as x = (Fo/mTw3)(wt-sin wt). For t > T we seek a solution in the form
=c1w(t-T)+c2 sin w(t - T)+Fo/mw2
The continuity of x and x at t = T gives C1 = -(Fo/mTw3) sin wT, C2 = (Fo/mTw3 X
X (1 - cos wT). The amplitude is a = = (2Fo/mTw3) sin twT. This is the smaller,
the more slowly the force Fo is applied (i.e. the greater T).
PROBLEM 3. The same as Problem 2, but for a constant force Fo which acts for a finite
time T (Fig. 25).
§23
Oscillations of systems with more than one degree of freedom
65
SOLUTION. As in Problem 2, or more simply by using formula (22.10). For t > T we have
free oscillations about x =0, and
dt
FO
f
T
FIG. 25
The squared modulus of & gives the amplitude from the relation = The result is
a = (2Fo/mw2) sin twT.
PROBLEM 4. The same as Problem 2, but for a force Fot/T which acts between t = 0 and
t = T (Fig. 26).
F
FO
,
T
FIG. 26
SOLUTION. By the same method we obtain
a = (Fo/Tmw3)/[wT2-2wT sin wT+2(1-cos - wT)].
PROBLEM 5. The same as Problem 2, but for a force Fo sin wt which acts between t = 0
and t = T = 2n/w (Fig. 27).
F
T
FIG. 27
SOLUTION. Substituting in (22.10) F(t) = Fo sin wt = Fo[exp(iwt)-exp(-iwt)]/2i and
integrating from 0 to T, we obtain a = Fon/mw2.
$23. Oscillations of systems with more than one degree of freedom
The theory of free oscillations of systems with S degrees of freedom is
analogous to that given in §21 for the case S = 1.
3*
66
Small Oscillations
§23
Let the potential energy of the system U as a function of the generalised
co-ordinates qi (i = 1, 2, ..., s) have a minimum for qi = qio. Putting
Xi=qi-qio
(23.1)
for the small displacements from equilibrium and expanding U as a function
of the xi as far as the quadratic terms, we obtain the potential energy as a
positive definite quadratic form
(23.2)
where we again take the minimum value of the potential energy as zero.
Since the coefficients kik and kki in (23.2) multiply the same quantity XiXK,
it is clear that they may always be considered equal: kik = kki.
In the kinetic energy, which has the general form () (see (5.5)),
we put qi = qio in the coefficients aik and, denoting aik(90) by Mik, obtain
the kinetic energy as a positive definite quadratic form
Emission
(23.3)
The coefficients Mik also may always be regarded as symmetrical: Mik=Mki.
Thus the Lagrangian of a system executing small free oscillations is
(23.4)
i,k
Let us now derive the equations of motion. To determine the derivatives
involved, we write the total differential of the Lagrangian:
- kikxi dxk - kikxxdxi).
i,k
Since the value of the sum is obviously independent of the naming of the
suffixes, we can interchange i and k in the first and third terms in the paren-
theses. Using the symmetry of Mik and kik, we have
dL =
Hence
k
Lagrange's equations are therefore
(i=1,2,...,s);
(23.5)
they form a set of S linear homogeneous differential equations with constant
coefficients.
As usual, we seek the S unknown functions xx(t) in the form
xx = Ak explicut),
(23.6)
where Ak are some constants to be determined. Substituting (23.6) in the
§23
Oscillations of systems with more than one degree of freedom
67
equations (23.5) and cancelling exp(iwt), we obtain a set of linear homo-
geneous algebraic equations to be satisfied by the Ak:
(23.7)
If this system has non-zero solutions, the determinant of the coefficients
must vanish:
(23.8)
This is the characteristic equation and is of degree S in w2. In general, it has
S different real positive roots W&2 (a = 1,2,...,s); in particular cases, some of
these roots may coincide. The quantities Wa thus determined are the charac-
teristic frequencies or eigenfrequencies of the system.
It is evident from physical arguments that the roots of equation (23.8) are
real and positive. For the existence of an imaginary part of w would mean
the presence, in the time dependence of the co-ordinates XK (23.6), and SO
of the velocities XK, of an exponentially decreasing or increasing factor. Such
a factor is inadmissible, since it would lead to a time variation of the total
energy E = U+: T of the system, which would therefore not be conserved.
The same result may also be derived mathematically. Multiplying equation
(23.7) by Ai* and summing over i, we have = 0,
whence w2 = . The quadratic forms in the numerator
and denominator of this expression are real, since the coefficients kik and
Mik are real and symmetrical: (kA*Ak)* = kikAAk* = k
= kikAkAi*. They are also positive, and therefore w2 is positive.t
The frequencies Wa having been found, we substitute each of them in
equations (23.7) and find the corresponding coefficients Ak. If all the roots
Wa of the characteristic equation are different, the coefficients Ak are pro-
portional to the minors of the determinant (23.8) with w = Wa. Let these
minors be . A particular solution of the differential equations (23.5) is
therefore X1c = Ca exp(iwat), where Ca is an arbitrary complex constant.
The general solution is the sum of S particular solutions. Taking the real
part, we write
III
(23.9)
where
(23.10)
Thus the time variation of each co-ordinate of the system is a super-
position of S simple periodic oscillations O1, O2, ..., Os with arbitrary ampli-
tudes and phases but definite frequencies.
t The fact that a quadratic form with the coefficients kik is positive definite is seen from
their definition (23.2) for real values of the variables. If the complex quantities Ak are written
explicitly as ak +ibk, we have, again using the symmetry of kik, kikAi* Ak = kik(ai-ibi)
X
= kikaiak kikbibk, which is the sum of two positive definite forms.
68
Small Oscillations
§23
The question naturally arises whether the generalised co-ordinates can be
chosen in such a way that each of them executes only one simple oscillation.
The form of the general integral (23.9) points to the answer. For, regarding
the S equations (23.9) as a set of equations for S unknowns Oa, as we can
express O1, O2, ..., Os in terms of the co-ordinates X1, X2, ..., Xs. The
quantities Oa may therefore be regarded as new generalised co-ordinates,
called normal co-ordinates, and they execute simple periodic oscillations,
called normal oscillations of the system.
The normal co-ordinates Oa are seen from their definition to satisfy the
equations
Oatwaia = 0.
(23.11)
This means that in normal co-ordinates the equations of motion become S
independent equations. The acceleration in each normal co-ordinate depends
only on the value of that co-ordinate, and its time dependence is entirely
determined by the initial values of the co-ordinate and of the corresponding
velocity. In other words, the normal oscillations of the system are completely
independent.
It is evident that the Lagrangian expressed in terms of normal co-ordinates
is a sum of expressions each of which corresponds to oscillation in one dimen-
sion with one of the frequencies was i.e. it is of the form
(23.12)
where the Ma are positive constants. Mathematically, this means that the
transformation (23.9) simultaneously puts both quadratic forms-the kinetic
energy (23.3) and the potential energy (23.2)-in diagonal form.
The normal co-ordinates are usually chosen so as to make the coefficients
of the squared velocities in the Lagrangian equal to one-half. This can be
achieved by simply defining new normal co-ordinates Qx by
Qa = VMaOa.
(23.13)
Then
The above discussion needs little alteration when some roots of the charac-
teristic equation coincide. The general form (23.9), (23.10) of the integral of
the equations of motion remains unchanged, with the same number S of
terms, and the only difference is that the coefficients corresponding to
multiple roots are not the minors of the determinant, which in this case
vanish.
t The impossibility of terms in the general integral which contain powers of the time as
well as the exponential factors is seen from the same argument as that which shows that the
frequencies are real: such terms would violate the law of conservation of energy
§23
Oscillations of systems with more than one degree of freedom
69
Each multiple (or, as we say, degenerate) frequency corresponds to a number
of normal co-ordinates equal to its multiplicity, but the choice of these co-
ordinates is not unique. The normal co-ordinates with equal Wa enter the
kinetic and potential energies as sums Q and Qa2 which are transformed
in the same way, and they can be linearly transformed in any manner which
does not alter these sums of squares.
The normal co-ordinates are very easily found for three-dimensional oscil-
lations of a single particle in a constant external field. Taking the origin of
Cartesian co-ordinates at the point where the potential energy U(x,y,2) is
a minimum, we obtain this energy as a quadratic form in the variables x, y, Z,
and the kinetic energy T = m(x2+yj++2) (where m is the mass of the
particle) does not depend on the orientation of the co-ordinate axes. We
therefore have only to reduce the potential energy to diagonal form by an
appropriate choice of axes. Then
L =
(23.14)
and the normal oscillations take place in the x,y and 2 directions with fre-
quencies = (k1/m), w2=1/(k2/m), w3=1/(k3/m). In the particular
case of a central field (k1 =k2=kg III three frequencies
are equal (see Problem 3).
The use of normal co-ordinates makes possible the reduction of a problem
of forced oscillations of a system with more than one degree of freedom to a
series of problems of forced oscillation in one dimension. The Lagrangian of
the system, including the variable external forces, is
(23.15)
where L is the Lagrangian for free oscillations. Replacing the co-ordinates
X1c by normal co-ordinates, we have
(23.16)
where we have put
The corresponding equations of motion
(23.17)
each involve only one unknown function Qa(t).
PROBLEMS
PROBLEM 1. Determine the oscillations of a system with two degrees of freedom whose
Lagrangian is L = (two identical one-dimensional systems of
eigenfrequency wo coupled by an interaction - axy).
70
Small Oscillations
§24
SOLUTION. The equations of motion are x+ wo2x = ay, j + wo2y = ax. The substitution
(23.6) gives
Ax(wo2-w2) = aAy,
(1)
The characteristic equation is (wo2-w2)2= a2, whence w12 = wo2-a, w22 = wo2-+x. For
w = W1, the equations (1) give Ax = Ay, and for w = w2, Ax = -Ay. Hence x =
(Q1+Q2)/V2, y = (Q1-Q2)/V2, the coefficients 1/V2 resulting from the normalisation
of the normal co-ordinates as in equation (23.13).
For a < wo2 (weak coupling) we have W1 all wo-1x, W2 ill wotla. The variation of x
and y is in this case a superposition of two oscillations with almost equal frequencies, i.e.
beats of frequency W2-W1 = a (see $22). The amplitude of y is a minimum when that of x
is a maximum, and vice versa.
PROBLEM 2. Determine the small oscillations of a coplanar double pendulum (Fig. 1, $5).
SOLUTION. For small oscillations (01 < 1, 02 < 1), the Lagrangian derived in §5, Problem
1, becomes
L =
The equations of motion are
= 0, lio +1202+802
=
0.
Substitution of (23.6) gives
41(m1+m2)(g-h1w2)-A2w2m2l2 0, = 0.
The roots of the characteristic equation are
((ma3
As m1
8 the frequencies tend to the values (g/l1) and /(g/l2), corresponding to indepen-
dent oscillations of the two pendulums.
PROBLEM 3. Find the path of a particle in a central field U = 1kr2 (called a space oscillator).
SOLUTION. As in any central field, the path lies in a plane, which we take as the xy-plane.
The variation of each co-ordinate x,y is a simple oscillation with the same frequency
= v(k/m): x = a cos(wt+a), y=b cos(wt+), or x = a cos , y = b cos(+8)
= b cos 8 cos -b sin 8 sin , where = wt +a, 8 = B-a. Solving for cos o and sin o and
equating the sum of their squares to unity, we find the equation of the path:
This is an ellipse with its centre at the origin.t When 8 = 0 or IT, the path degenerates to a
segment of a straight line.
$24. Vibrations of molecules
If we have a system of interacting particles not in an external field, not all
of its degrees of freedom relate to oscillations. A typical example is that of
molecules. Besides motions in which the atoms oscillate about their positions
of equilibrium in the molecule, the whole molecule can execute translational
and rotational motions.
Three degrees of freedom correspond to translational motion, and in general
the same number to rotation, so that, of the 3n degrees of freedom of a mole-
cule containing n atoms, 3n-6 - correspond to vibration. An exception is formed
t The fact that the path in a field with potential energy U = 1kr2 is a closed curve has
already been mentioned in $14.

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§24
Vibrations of molecules
71
by molecules in which the atoms are collinear, for which there are only two
rotational degrees of freedom (since rotation about the line of atoms is of no
significance), and therefore 3n-5 vibrational degrees of freedom.
In solving a mechanical problem of molecular oscillations, it is convenient
to eliminate immediately the translational and rotational degrees of freedom.
The former can be removed by equating to zero the total momentum of the
molecule. Since this condition implies that the centre of mass of the molecule
is at rest, it can be expressed by saying that the three co-ordinates of the
centre of mass are constant. Putting ra = rao+Ua, where ra0 is the radius
vector of the equilibrium position of the ath atom, and Ua its deviation from
this position, we have the condition = constant = or
= 0.
(24.1)
To eliminate the rotation of the molecule, its total angular momentum
must be equated to zero. Since the angular momentum is not the total time
derivative of a function of the co-ordinates, the condition that it is zero can-
not in general be expressed by saying that some such function is zero. For
small oscillations, however, this can in fact be done. Putting again
ra = rao+ua and neglecting small quantities of the second order in the
displacements Ua, we can write the angular momentum of the molecule as
= .
The condition for this to be zero is therefore, in the same approximation,
0,
(24.2)
in which the origin may be chosen arbitrarily.
The normal vibrations of the molecule may be classified according to the
corresponding motion of the atoms on the basis of a consideration of the sym-
metry of the equilibrium positions of the atoms in the molecule. There is
a general method of doing so, based on the use of group theory, which we
discuss elsewhere. Here we shall consider only some elementary examples.
If all n atoms in a molecule lie in one plane, we can distinguish normal
vibrations in which the atoms remain in that plane from those where they
do not. The number of each kind is readily determined. Since, for motion
in a plane, there are 2n degrees of freedom, of which two are translational
and one rotational, the number of normal vibrations which leave the atoms
in the plane is 2n-3. The remaining (3n-6)-(2n-3) = n-3 vibrational
degrees of freedom correspond to vibrations in which the atoms move out
of the plane.
For a linear molecule we can distinguish longitudinal vibrations, which
maintain the linear form, from vibrations which bring the atoms out of line.
Since a motion of n particles in a line corresponds to n degrees of freedom,
of which one is translational, the number of vibrations which leave the atoms
t See Quantum Mechanics, $100, Pergamon Press, Oxford 1965.
72
Small Oscillations
§24
in line is n - 1. Since the total number of vibrational degrees of freedom of a
linear molecule is 3n - 5, there are 2n-4 which bring the atoms out of line.
These 2n-4 vibrations, however, correspond to only n-2 different fre-
quencies, since each such vibration can occur in two mutually perpendicular
planes through the axis of the molecule. It is evident from symmetry that
each such pair of normal vibrations have equal frequencies.
PROBLEMS
PROBLEM 1. Determine the frequencies of vibrations of a symmetrical linear triatomic
molecule ABA (Fig. 28). It is assumed that the potential energy of the molecule depends
only on the distances AB and BA and the angle ABA.
3
B
A
(o)
(b)
(c)
FIG. 28
SOLUTION. The longitudinal displacements X1, X2, X3 of the atoms are related, according
to (24.1), by MA(X1+x3) +mBX2 = 0. Using this, we eliminate X2 from the Lagrangian of the
longitudinal motion
L =
and use new co-ordinatesQa=x1tx,Qx1-x3.The result
where u = 2mA+mB is the mass of the molecule. Hence we see that Qa and Qs are normal
co-ordinates (not yet normalised). The co-ordinate Qa corresponds to a vibration anti-
symmetrical about the centre of the molecule (x1 = x3; Fig. 28a), with frequency
wa = (k1u/mAmB). The co-ordinate Q8 corresponds to a symmetrical vibration (x1 = -x3;
Fig. 28b), with frequency
The transverse displacements y1, y2, y3 of the atoms are, according to (24.1) and (24.2),
related by mA(y1+y2) +mBy2 = 0, y1 = y3 (a symmetrical bending of the molecule; Fig. 28c).
The potential energy of this vibration can be written as 1/22/282, where 8 is the deviation of the
angle ABA from the value IT, given in terms of the displacements by 8 = [(y1-yy)+(ys-y2)]/L.
Expressing y1,y2, y3 in terms of 8, we obtain the Lagrangian of the transverse motion:
L =
whence the frequency is = V(2k2u/mAmB).
t Calculations of the vibrations of more complex molecules are given by M. V. VOL'KENSH-
TEIN, M. A. EL'YASHEVICH and B. I. STEPANOV, Molecular Vibrations (Kolebaniya molekul),
Moscow 1949; G. HERZBERG, Molecular Spectra and Molecular Structure: Infra-red and
Raman Spectra of Polyatomic Molecules, Van Nostrand, New York 1945,
§24
Vibrations of molecules
73
PROBLEM 2. The same as Problem 1, but for a triangular molecule ABA (Fig. 29).
y
A
A
3
2a
I
2
B
(a)
(b)
(c)
FIG. 29
SOLUTION. By (24.1) and (24.2) the x and y components of the displacements u of the
atoms are related by
0,
0,
(y1-y3) sin x-(x1+x3) cos a = 0.
The changes 8l1 and Sl2 in the distances AB and BA are obtained by taking the components
along these lines of the vectors U1-U2 and U3-U2:
8l1 = (x1-x2) sin cos a,
8l2 = -(x3-x2) sin at(y3-y2) cos a.
The change in the angle ABA is obtained by taking the components of those vectors per-
pendicular to AB and BA:
sin sin
a].
The Lagrangian of the molecule is
L
We use the new co-ordinates Qa = 1+x3, q81 = X1-x3, q82 = y1+y3. The components
of the vectors u are given in terms of these co-ordinates by X1 = 1(Qa+q x3 = 1(Qa-981),
X2
= -MAQa/MB, = -MAQ82/MB. The
Lagrangian becomes
L
=
qksici +
sin
a
cos
a.
74
Small Oscillations
§25
Hence we see that the co-ordinate Qa corresponds to a normal vibration antisymmetrical
about the y-axis (x1 = x3, y1 = -y3; Fig. 29a) with frequency
The co-ordinates qs1, qs2 together correspond to two vibrations symmetrical about the
y-axis (x1 = -X3, y1 y3; Fig. 29b, c), whose frequencies Ws1, W82 are given by the roots
of the quadratic (in w2) characteristic equation
1
When 2 x = 75, all three frequencies become equal to those derived in Problem 1.
PROBLEM 3. The same as Problem 1, but for an unsymmetrical linear molecule ABC
(Fig. 30).
A
FIG. 30
SOLUTION. The longitudinal (x) and transverse (y) displacements of the atoms are related
by
mAX1+mBX2+mcx3 = 0, mAy1tmBy2+mcy3= 0,
MAhy1 = mcl2y3.
The potential energy of stretching and bending can be written
where 2l = li+l2. Calculations similar to those in Problem 1 give
for the transverse vibrations and the quadratic (in w2) equation
=
for the frequencies wil, W12 of the longitudinal vibrations.
$25. Damped oscillations
So far we have implied that all motion takes place in a vacuum, or else that
the effect of the surrounding medium on the motion may be neglected. In
reality, when a body moves in a medium, the latter exerts a resistance which
tends to retard the motion. The energy of the moving body is finally dissipated
by being converted into heat.
Motion under these conditions is no longer a purely mechanical process,
and allowance must be made for the motion of the medium itself and for the
internal thermal state of both the medium and the body. In particular, we
cannot in general assert that the acceleration of a moving body is a function
only of its co-ordinates and velocity at the instant considered; that is, there
are no equations of motion in the mechanical sense. Thus the problem of the
motion of a body in a medium is not one of mechanics.
There exists, however, a class of cases where motion in a medium can be
approximately described by including certain additional terms in the
§25
Damped oscillations
75
mechanical equations of motion. Such cases include oscillations with fre-
quencies small compared with those of the dissipative processes in the
medium. When this condition is fulfilled we may regard the body as being
acted on by a force of friction which depends (for a given homogeneous
medium) only on its velocity.
If, in addition, this velocity is sufficiently small, then the frictional force
can be expanded in powers of the velocity. The zero-order term in the expan-
sion is zero, since no friction acts on a body at rest, and so the first non-
vanishing term is proportional to the velocity. Thus the generalised frictional
force fir acting on a system executing small oscillations in one dimension
(co-ordinate x) may be written fir = - ax, where a is a positive coefficient
and the minus sign indicates that the force acts in the direction opposite to
that of the velocity. Adding this force on the right-hand side of the equation
of motion, we obtain (see (21.4))
mx = -kx-ax.
(25.1)
We divide this by m and put
k/m= wo2, a/m=2x; =
(25.2)
wo is the frequency of free oscillations of the system in the absence of friction,
and A is called the damping coefficient or damping decrement.
Thus the equation is
(25.3)
We again seek a solution x = exp(rt) and obtain r for the characteristic
equation r2+2xr + wo2 = 0, whence ¥1,2 = The general
solution of equation (25.3) is
c1exp(rit)+c2 exp(r2t).
Two cases must be distinguished. If wo, we have two complex con-
jugate values of r. The general solution of the equation of motion can then
be written as
where A is an arbitrary complex constant, or as
= aexp(-Xt)cos(wta),
(25.4)
with w = V(w02-2) and a and a real constants. The motion described by
these formulae consists of damped oscillations. It may be regarded as being
harmonic oscillations of exponentially decreasing amplitude. The rate of
decrease of the amplitude is given by the exponent X, and the "frequency"
w is less than that of free oscillations in the absence of friction. For 1 wo,
the difference between w and wo is of the second order of smallness. The
decrease in frequency as a result of friction is to be expected, since friction
retards motion.
t The dimensionless product XT (where T = 2n/w is the period) is called the logarithmic
damping decrement.
76
Small Oscillations
§25
If A < wo, the amplitude of the damped oscillation is almost unchanged
during the period 2n/w. It is then meaningful to consider the mean values
(over the period) of the squared co-ordinates and velocities, neglecting the
change in exp( - At) when taking the mean. These mean squares are evidently
proportional to exp(-2xt). Hence the mean energy of the system decreases
as
(25.5)
where E0 is the initial value of the energy.
Next, let A > wo. Then the values of r are both real and negative. The
general form of the solution is
-
(25.6)
We see that in this case, which occurs when the friction is sufficiently strong,
the motion consists of a decrease in /x/, i.e. an asymptotic approach (as t ->
00)
to the equilibrium position. This type of motion is called aperiodic damping.
Finally, in the special case where A = wo, the characteristic equation has
the double root r = - 1. The general solution of the differential equation is
then
(25.7)
This is a special case of aperiodic damping.
For a system with more than one degree of freedom, the generalised
frictional forces corresponding to the co-ordinates Xi are linear functions of
the velocities, of the form
=
(25.8)
From purely mechanical arguments we can draw no conclusions concerning
the symmetry properties of the coefficients aik as regards the suffixes i and
k, but the methods of statistical physics make it possible to demonstrate
that in all cases
aki.
(25.9)
Hence the expressions (25.8) can be written as the derivatives
=
(25.10)
of the quadratic form
(25.11)
which is called the dissipative function.
The forces (25.10) must be added to the right-hand side of Lagrange's
equations:
(25.12)
t See Statistical Physics, $123, Pergamon Press, Oxford 1969.
§26
Forced oscillations under friction
77
The dissipative function itself has an important physical significance: it
gives the rate of dissipation of energy in the system. This is easily seen by
calculating the time derivative of the mechanical energy of the system. We
have
aL
=
Since F is a quadratic function of the velocities, Euler's theorem on homo-
geneous functions shows that the sum on the right-hand side is equal to 2F.
Thus
dE/dt==2-2F,
(25.13)
i.e. the rate of change of the energy of the system is twice the dissipative
function. Since dissipative processes lead to loss of energy, it follows that
F > 0, i.e. the quadratic form (25.11) is positive definite.
The equations of small oscillations under friction are obtained by adding
the forces (25.8) to the right-hand sides of equations (23.5):
=
(25.14)
Putting in these equations XK = Ak exp(rt), we obtain, on cancelling exp(rt),
a set of linear algebraic equations for the constants Ak:
(25.15)
Equating to zero their determinant, we find the characteristic equation, which
determines the possible values of r:
(25.16)
This is an equation in r of degree 2s. Since all the coefficients are real,
its roots are either real, or complex conjugate pairs. The real roots must be
negative, and the complex roots must have negative real parts, since other-
wise the co-ordinates, velocities and energy of the system would increase
exponentially with time, whereas dissipative forces must lead to a decrease
of the energy.
§26. Forced oscillations under friction
The theory of forced oscillations under friction is entirely analogous to
that given in §22 for oscillations without friction. Here we shall consider
in detail the case of a periodic external force, which is of considerable interest.
78
Small Oscillations
§26
Adding to the right-hand side of equation (25.1) an external force f cos st
and dividing by m, we obtain the equation of motion:
+2*+wox=(fm)cos = yt.
(26.1)
The solution of this equation is more conveniently found in complex form,
and so we replace cos st on the right by exp(iyt):
exp(iyt).
We seek a particular integral in the form x = B exp(iyt), obtaining for B
the value
(26.2)
Writing B = exp(i8), we have
b tan 8 = 2xy/(y2-wo2).
(26.3)
Finally, taking the real part of the expression B exp(iyt) = b exp[i(yt+8)],
we find the particular integral of equation (26.1); adding to this the general
solution of that equation with zero on the right-hand side (and taking for
definiteness the case wo > 1), we have
x = a exp( - At) cos(wtta)+bcos(yt+8)
(26.4)
The first term decreases exponentially with time, so that, after a sufficient
time, only the second term remains:
x = b cos(yt+8).
(26.5)
The expression (26.3) for the amplitude b of the forced oscillation increases
as y approaches wo, but does not become infinite as it does in resonance
without friction. For a given amplitude f of the force, the amplitude of the
oscillations is greatest when y = V(w02-2)2); for A < wo, this differs from
wo only by a quantity of the second order of smallness.
Let us consider the range near resonance, putting y = wote with E small,
and suppose also that A < wo. Then we can approximately put, in (26.2),
22 =(y+wo)(y-wo) 22 2woe, 2ixy 22 2ixwo, SO that
B = -f/2m(e-ii))wo
(26.6)
or
b f/2mw01/(22+12),
tan 8 = N/E.
(26.7)
A property of the phase difference 8 between the oscillation and the external
force is that it is always negative, i.e. the oscillation "lags behind" the force.
Far from resonance on the side < wo, 8 0; on the side y > wo, 8
-77.
The change of 8 from zero to - II takes place in a frequency range near wo
which is narrow (of the order of A in width); 8 passes through - 1/2 when
y = wo. In the absence of friction, the phase of the forced oscillation changes
discontinuously by TT at y = wo (the second term in (22.4) changes sign);
when friction is allowed for, this discontinuity is smoothed out.
§26
Forced oscillations under friction
79
In steady motion, when the system executes the forced oscillations given
by (26.5), its energy remains unchanged. Energy is continually absorbed by
the system from the source of the external force and dissipated by friction.
Let I(y) be the mean amount of energy absorbed per unit time, which depends
on the frequency of the external force. By (25.13) we have I(y) = 2F, where
F is the average value (over the period of oscillation) of the dissipative func-
tion. For motion in one dimension, the expression (25.11) for the dissipative
function becomes F = 1ax2 = Amx2. Substituting (26.5), we have
F = mb22 sin2(yt+8).
The time average of the squared sine is 1/2 so that
I(y) = Mmb2y2. =
(26.8)
Near resonance we have, on substituting the amplitude of the oscillation
from (26.7),
I(e) =
(26.9)
This is called a dispersion-type frequency dependence of the absorption.
The half-width of the resonance curve (Fig. 31) is the value of E for which
I(e) is half its maximum value (E = 0). It is evident from (26.9) that in the
present case the half-width is just the damping coefficient A. The height of
the maximum is I(0) = f2/4mx, and is inversely proportional to . Thus,
I/I(O)
/2
€
-1
a
FIG. 31
when the damping coefficient decreases, the resonance curve becomes more
peaked. The area under the curve, however, remains unchanged. This area
is given by the integral
[ ((7) dy = [ I(e) de.
Since I(e) diminishes rapidly with increasing E, the region where |el is
large is of no importance, and the lower limit may be replaced by - 80, and
I(e) taken to have the form given by (26.9). Then we have
"
(26.10)
80
Small Oscillations
§27
PROBLEM
Determine the forced oscillations due to an external force f = fo exp(at) COS st in the
presence of friction.
SOLUTION. We solve the complex equation of motion
2+wo2x = (fo/m) exp(at+iyt)
and then take the real part. The result is a forced oscillation of the form
x=bexp(at)cos(yt+8),
where
b =
tan s =
§27. Parametric resonance
There exist oscillatory systems which are not closed, but in which the
external action amounts only to a time variation of the parameters.t
The parameters of a one-dimensional system are the coefficients m and k
in the Lagrangian (21.3). If these are functions of time, the equation of
motion is
(27.1)
We introduce instead of t a new independent variable T such that
dr = dt/m(t); this reduces the equation to
d2x/d-2+mkx=0.
There is therefore no loss of generality in considering an equation of motion
of the form
(27.2)
obtained from (27.1) if m = constant.
The form of the function w(t) is given by the conditions of the problem.
Let us assume that this function is periodic with some frequency y and period
T = 2n/y. This means that w(t+T) = w(t), and so the equation (27.2) is
invariant under the transformation t t+ T. Hence, if x(t) is a solution of
the equation, so is x(t+T). That is, if x1(t) and x2(t) are two independent
integrals of equation (27.2), they must be transformed into linear combina-
tions of themselves when t is replaced by t + T. It is possible to choose X1
and X2 in such a way that, when t t+T, they are simply multiplied by
t A simple example is that of a pendulum whose point of support executes a given periodic
motion in a vertical direction (see Problem 3).
+ This choice is equivalent to reducing to diagonal form the matrix of the linear trans-
formation of x1(t) and x2(t), which involves the solution of the corresponding quadratic
secular equation. We shall suppose here that the roots of this equation do not coincide.

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§27
Parametric resonance
81
constants: x1(t+T) = 1x1(t), x2(t+T) = u2x2(t). The most general functions
having this property are
(t) = 111t/TII1(t), x2(t) = M2t/T112(t),
(27.3)
where II1(t), II2(t) are purely periodic functions of time with period T.
The constants 1 and 2 in these functions must be related in a certain way.
Multiplying the equations +2(t)x1 = 0, 2+w2(t)x2 = 0 by X2 and X1
respectively and subtracting, we = = 0, or
X1X2-XIX2 = constant.
(27.4)
For any functions x1(t), x2(t) of the form (27.3), the expression on the left-
hand side of (27.4) is multiplied by H1U2 when t is replaced by t + T. Hence
it is clear that, if equation (27.4) is to hold, we must have
M1M2=1.
(27.5)
Further information about the constants M1, 2 can be obtained from the
fact that the coefficients in equation (27.2) are real. If x(t) is any integral of
such an equation, then the complex conjugate function x* (t) must also be
an integral. Hence it follows that U1, 2 must be the same as M1*, M2*, i.e.
either 1 = M2* or 1 and 2 are both real. In the former case, (27.5) gives
M1 = 1/1*, i.e. /1112 = 1/22/2 = 1: the constants M1 and 2 are of modulus
unity.
In the other case, two independent integrals of equation (27.2) are
x2(t) = -/I2(t),
(27.6)
with a positive or negative real value of u (Iu/ # 1). One of these functions
(x1 or X2 according as /x/ > 1 or /u/ <1) increases exponentially with time.
This means that the system at rest in equilibrium (x = 0) is unstable: any
deviation from this state, however small, is sufficient to lead to a rapidly
increasing displacement X. This is called parametric resonance.
It should be noticed that, when the initial values of x and x are exactly
zero, they remain zero, unlike what happens in ordinary resonance (§22),
in which the displacement increases with time (proportionally to t) even from
initial values of zero.
Let us determine the conditions for parametric resonance to occur in the
important case where the function w(t) differs only slightly from a constant
value wo and is a simple periodic function:
w2(1) = con2(1+h cosyt)
(27.7)
where
the
constant h 1; we shall suppose h positive, as may always be
done by suitably choosing the origin of time. As we shall see below, para-
metric resonance is strongest if the frequency of the function w(t) is nearly
twice wo. Hence we put y = 2wo+e, where E < wo.
82
Small Oscillations
§27
The solution of equation of motion+
+wo2[1+hcos(2wot)t]x
(27.8)
may be sought in the form
(27.9)
where a(t) and b(t) are functions of time which vary slowly in comparison
with the trigonometrical factors. This form of solution is, of course, not
exact. In reality, the function x(t) also involves terms with frequencies which
differ from wother by integral multiples of 2wo+e; these terms are, how-
ever, of a higher order of smallness with respect to h, and may be neglected
in a first approximation (see Problem 1).
We substitute (27.9) in (27.8) and retain only terms of the first order in
€, assuming that à ea, b ~ eb; the correctness of this assumption under
resonance conditions is confirmed by the result. The products of trigono-
metrical functions may be replaced by sums:
cos(wot1e)t.cos(2wote)t =
etc., and in accordance with what was said above we omit terms with fre-
quency 3(wo+1e). The result is
= 0.
If this equation is to be justified, the coefficients of the sine and cosine must
both be zero. This gives two linear differential equations for the functions
a(t) and b(t). As usual, we seek solutions proportional to exp(st). Then
= 0, 1(e-thwo)a- - sb = 0, and the compatibility condition
for these two algebraic equations gives
(27.10)
The condition for parametric resonance is that S is real, i.e. s2 > 0.1 Thus
parametric resonance occurs in the range
(27.11)
on either side of the frequency 2wo.ll The width of this range is proportional
to h, and the values of the amplification coefficient S of the oscillations in the
range are of the order of h also.
Parametric resonance also occurs when the frequency y with which the
parameter varies is close to any value 2wo/n with n integral. The width of the
t An equation of this form (with arbitrary y and h) is called in mathematical physics
Mathieu's equation.
+ The constant u in (27.6) is related to s by u = - exp(sn/wo); when t is replaced by
t+2n/2wo, the sine and cosine in (27.9) change sign.
II If we are interested only in the range of resonance, and not in the values of S in that
range, the calculations may be simplified by noting that S = 0 at the ends of the range, i.e.
the coefficients a and b in (27.9) are constants. This gives immediately € = thwo as in
(27.11).
§27
Parametric resonance
83
resonance range (region of instability) decreases rapidly with increasing N,
however, namely as hn (see Problem 2, footnote). The amplification co-
efficient of the oscillations also decreases.
The phenomenon of parametric resonance is maintained in the presence
of slight friction, but the region of instability becomes somewhat narrower.
As we have seen in §25, friction results in a damping of the amplitude of
oscillations as exp(- - At). Hence the amplification of the oscillations in para-
metric resonance is as exp[(s-1)t] with the positive S given by the solution
for the frictionless case, and the limit of the region of instability is given by
the equation - X = 0. Thus, with S given by (27.10), we have for the resonance
range, instead of (27.11),
(27.12)
It should be noticed that resonance is now possible not for arbitrarily
small amplitudes h, but only when h exceeds a "threshold" value hk. When
(27.12) holds, hk = 4X/wo. It can be shown that, for resonance near the fre-
quency 2wo/n, the threshold hk is proportional to X1/n, i.e. it increases with n.
PROBLEMS
PROBLEM 1. Obtain an expression correct as far as the term in h2 for the limits of the region
of instability for resonance near 2 = 2wo.
SOLUTION. We seek the solution of equation (27.8) in the form
x = ao cos(wo+1e)t +bo (wo+le)t +a1 cos 3( (wo+le)t +b1 sin 3(wo+le)t,
which includes terms of one higher order in h than (27.9). Since only the limits of the region
of instability are required, we treat the coefficients ao, bo, a1, b1 as constants in accordance
with the last footnote. Substituting in (27.8), we convert the products of trigonometrical
functions into sums and omit the terms of frequency 5(wo+1) in this approximation. The
result is
[
- cos(wo+l)
cos 3(wo+1e)tt
sin 3(wo+1e)t = 0.
In the terms of frequency wothe we retain terms of the second order of smallness, but in
those of frequency 3( (wo+1) only the first-order terms. Each of the expressions in brackets
must separately vanish. The last two give a1 = hao/16, b1 = hbo/16, and then the first two
give woe +thwo2+1e2-h2wo2/32 = 0.
Solving this as far as terms of order h2, we obtain the required limits of E:
= theo-h20003.
PROBLEM 2. Determine the limits of the region of instability in resonance near y = wo.
SOLUTION. Putting y = wote, we obtain the equation of motion
0.
Since the required limiting values of ~~h2, we seek a solution in the form
ao cos(wote)t sin(wote)t cos 2(wo+e)t +b1 sin 2(wo+e)t- +C1,
84
Small Oscillations
§28
which includes terms of the first two orders. To determine the limits of instability, we again
treat the coefficients as constants, obtaining
cos(wote)t-
+[-2woebo+thwo861] sin(wo+e)t.
+[-30002a1+thanoPao] cos 2(wote)t+
sin 2(wote)t+[c1wo+thwo2ao] 0.
Hence a1 = hao/6, b1 = hbo/6, C1 = -thao, and the limits aret € = -5h2wo/24, € = h2wo/24.
PROBLEM 3. Find the conditions for parametric resonance in small oscillations of a simple
pendulum whose point of support oscillates vertically.
SOLUTION. The Lagrangian derived in §5, Problem 3(c), gives for small oscillations
( < 1) the equation of motion + wo2[1+(4a/1) cos(2wo+t)) = 0, where wo2 = g/l.
Hence we see that the parameter h is here represented by 4all. The condition (27.11), for
example, becomes |
§28. Anharmonic oscillations
The whole of the theory of small oscillations discussed above is based on
the expansion of the potential and kinetic energies of the system in terms of
the co-ordinates and velocities, retaining only the second-order terms. The
equations of motion are then linear, and in this approximation we speak of
linear oscillations. Although such an expansion is entirely legitimate when
the amplitude of the oscillations is sufficiently small, in higher approxima-
tions (called anharmonic or non-linear oscillations) some minor but qualitatively
different properties of the motion appear.
Let us consider the expansion of the Lagrangian as far as the third-order
terms. In the potential energy there appear terms of degree three in the co-
ordinates Xi, and in the kinetic energy terms containing products of velocities
and co-ordinates, of the form XEXKXI. This difference from the previous
expression (23.3) is due to the retention of terms linear in x in the expansion
of the functions aik(q). Thus the Lagrangian is of the form
(28.1)
where Nikl, liki are further constant coefficients.
If we change from arbitrary co-ordinates Xi to the normal co-ordinates Qx
of the linear approximation, then, because this transformation is linear, the
third and fourth sums in (28.1) become similar sums with Qx and Qa in place
t
Generally, the width AE of the region of instability in resonance near the frequency
2wo/n is given by
AE =
a result due to M. BELL (Proceedings of the Glasgow Mathematical Association 3, 132, 1957).
§28
Anharmonic oscillations
85
of the co-ordinates Xi and the velocities Xr. Denoting the coefficients in these
new sums by dapy and Hapy's we have the Lagrangian in the form
(28.2)
a
a,B,Y
We shall not pause to write out in their entirety the equations of motion
derived from this Lagrangian. The important feature of these equations is
that they are of the form
(28.3)
where fa are homogeneous functions, of degree two, of the co-ordinates Q
and their time derivatives.
Using the method of successive approximations, we seek a solution of
these equations in the form
(28.4)
where Qa2, and the Qx(1) satisfy the "unperturbed" equations
i.e. they are ordinary harmonic oscillations:
(28.5)
Retaining only the second-order terms on the right-hand side of (28.3) in
the next approximation, we have for the Qx(2) the equations
(28.6)
where (28.5) is to be substituted on the right. This gives a set of inhomo-
geneous linear differential equations, in which the right-hand sides can be
represented as sums of simple periodic functions. For example,
cos(wpt + ag)
Thus the right-hand sides of equations (28.6) contain terms corresponding
to oscillations whose frequencies are the sums and differences of the eigen-
frequencies of the system. The solution of these equations must be sought
in a form involving similar periodic factors, and so we conclude that, in the
second approximation, additional oscillations with frequencies
wa+w
(28.7)
including the double frequencies 2wa and the frequency zero (corresponding
to a constant displacement), are superposed on the normal oscillations of the
system. These are called combination frequencies. The corresponding ampli-
tudes are proportional to the products Axap (or the squares aa2) of the cor-
responding normal amplitudes.
In higher approximations, when further terms are included in the expan-
sion of the Lagrangian, combination frequencies occur which are the sums
and differences of more than two Wa; and a further phenomenon also appears.
86
Small Oscillations
§28
In the third approximation, the combination frequencies include some which
coincide with the original frequencies W Wa+wp-wp). When the method
described above is used, the right-hand sides of the equations of motion there-
fore include resonance terms, which lead to terms in the solution whose
amplitude increases with time. It is physically evident, however, that the
magnitude of the oscillations cannot increase of itself in a closed system
with no external source of energy.
In reality, the fundamental frequencies Wa in higher approximations are
not equal to their "unperturbed" values wa(0) which appear in the quadratic
expression for the potential energy. The increasing terms in the solution
arise from an expansion of the type
which is obviously not legitimate when t is sufficiently large.
In going to higher approximations, therefore, the method of successive
approximations must be modified so that the periodic factors in the solution
shall contain the exact and not approximate values of the frequencies. The
necessary changes in the frequencies are found by solving the equations and
requiring that resonance terms should not in fact appear.
We may illustrate this method by taking the example of anharmonic oscil-
lations in one dimension, and writing the Lagrangian in the form
L =
(28.8)
The corresponding equation of motion is
(28.9)
We shall seek the solution as a series of successive approximations:
where
x(1) = a cos wt,
(28.10)
with the exact value of w, which in turn we express as w=wotw1)+w(2)+....
(The initial phase in x(1) can always be made zero by a suitable choice of the
origin of time.) The form (28.9) of the equation of motion is not the most
convenient, since, when (28.10) is substituted in (28.9), the left-hand side is
not exactly zero. We therefore rewrite it as
(28.11)
Putting x(1)+x(2), w wotwi and omitting terms of above the
second order of smallness, we obtain for x(2) the equation
= aa2 cos2wt+2wowlda cos wt
= 1xa2-1xa2 cos 2wt + 2wow1)a cos wt.
The condition for the resonance term to be absent from the right-hand side
is simply w(1) = 0, in agreement with the second approximation discussed
§29
Resonance in non-linear oscillations
87
at the beginning of this section. Solving the inhomogeneous linear equation
in the usual way, we have
(28.12)
Putting in (28.11) X wo+w(2), we obtain the equa-
tion for x(3)
= -
or, substituting on the right-hand side (28.10) and (28.12) and effecting
simple transformation,
wt.
Equating to zero the coefficient of the resonance term cos wt, we find the
correction to the fundamental frequency, which is proportional to the squared
amplitude of the oscillations:
(28.13)
The combination oscillation of the third order is
(28.14)
$29. Resonance in non-linear oscillations
When the anharmonic terms in forced oscillations of a system are taken
into account, the phenomena of resonance acquire new properties.
Adding to the right-hand side of equation (28.9) an external periodic force
of frequency y, we have
+2x+wo2x=(fm)cos = yt - ax2-Bx3;
(29.1)
here the frictional force, with damping coefficient A (assumed small) has also
been included. Strictly speaking, when non-linear terms are included in the
equation of free oscillations, the terms of higher order in the amplitude of
the external force (such as occur if it depends on the displacement x) should
also be included. We shall omit these terms merely to simplify the formulae;
they do not affect the qualitative results.
Let y = wote with E small, i.e. y be near the resonance value. To ascertain
the resulting type of motion, it is not necessary to consider equation (29.1)
if we argue as follows. In the linear approximation, the amplitude b is given
88
Small Oscillations
§29
near resonance, as a function of the amplitude f and frequency r of the
external force, by formula (26.7), which we write as
(29.2)
The non-linearity of the oscillations results in the appearance of an ampli-
tude dependence of the eigenfrequency, which we write as
wo+kb2,
(29.3)
the constant K being a definite function of the anharmonic coefficients (see
(28.13)). Accordingly, we replace wo by wo + kb2 in formula (29.2) (or, more
precisely, in the small difference y-wo). With y-wo=e, the resulting
equation is
=
(29.4)
or
Equation (29.4) is a cubic equation in b2, and its real roots give the ampli-
tude of the forced oscillations. Let us consider how this amplitude depends
on the frequency of the external force for a given amplitude f of that force.
When f is sufficiently small, the amplitude b is also small, so that powers
of b above the second may be neglected in (29.4), and we return to the form
of b(e) given by (29.2), represented by a symmetrical curve with a maximum
at the point E = 0 (Fig. 32a). As f increases, the curve changes its shape,
though at first it retains its single maximum, which moves to positive E if
K > 0 (Fig. 32b). At this stage only one of the three roots of equation (29.4)
is real.
When f reaches a certain value f k (to be determined below), however, the
nature of the curve changes. For all f > fk there is a range of frequencies in
which equation (29.4) has three real roots, corresponding to the portion
BCDE in Fig. 32c.
The limits of this range are determined by the condition db/de = 8 which
holds at the points D and C. Differentiating equation (29.4) with respect to
€, we have
db/de =
Hence the points D and C are determined by the simultaneous solution of
the equations
2-4kb2e+3k264+2 0
(29.5)
and (29.4). The corresponding values of E are both positive. The greatest
amplitude is reached where db/de = 0. This gives E = kb2, and from (29.4)
we have
bmax = f/2mwod;
(29.6)
this is the same as the maximum value given by (29.2).
§29
Resonance in non-linear oscillations
89
It may be shown (though we shall not pause to do so heret) that, of the
three real roots of equation (29.4), the middle one (represented by the dotted
part CD of the curve in Fig. 32c) corresponds to unstable oscillations of the
system: any action, no matter how slight, on a system in such a state causes
it to oscillate in a manner corresponding to the largest or smallest root (BC
or DE). Thus only the branches ABC and DEF correspond to actual oscil-
lations of the system. A remarkable feature here is the existence of a range of
frequencies in which two different amplitudes of oscillation are possible. For
example, as the frequency of the external force gradually increases, the ampli-
tude of the forced oscillations increases along ABC. At C there is a dis-
continuity of the amplitude, which falls abruptly to the value corresponding
to E, afterwards decreasing along the curve EF as the frequency increases
further. If the frequency is now diminished, the amplitude of the forced
oscillations varies along FD, afterwards increasing discontinuously from D
to B and then decreasing along BA.
b
(a)
to
b
(b)
f<f
b
(c)
f>tp
B
C
Di
A
E
F
€
FIG. 32
To calculate the value of fk, we notice that it is the value of f for which
the two roots of the quadratic equation in b2 (29.5) coincide; for f = f16, the
section CD reduces to a point of inflection. Equating to zero the discriminant
t The proof is given by, for example, N.N. BOGOLIUBOV and Y.A. MITROPOLSKY, Asymp-
totic Methods in the Theory of Non-Linear Oscillations, Hindustan Publishing Corporation,
Delhi 1961.
4
90
Small Oscillations
§29
of (29.5), we find E2 = 3X², and the corresponding double root is kb2 = 2e/3.
Substitution of these values of b and E in (29.4) gives
32m2wo2x3/31/3k.
(29.7)
Besides the change in the nature of the phenomena of resonance at fre-
quencies y 22 wo, the non-linearity of the oscillations leads also to new
resonances in which oscillations of frequency close to wo are excited by an
external force of frequency considerably different from wo.
Let the frequency of the external force y 22 two, i.e. y = two+e. In the
first (linear) approximation, it causes oscillations of the system with the same
frequency and with amplitude proportional to that of the force:
x(1)= (4f/3mwo2) cos(two+e)t
(see (22.4)). When the non-linear terms are included (second approximation),
these oscillations give rise to terms of frequency 2y 22 wo on the right-hand
side of the equation of motion (29.1). Substituting x(1) in the equation
= -
using the cosine of the double angle and retaining only the resonance term
on the right-hand side, we have
= - (8xf2/9m2w04) cos(wo+2e)t.
(29.8)
This equation differs from (29.1) only in that the amplitude f of the force is
replaced by an expression proportional to f2. This means that the resulting
resonance is of the same type as that considered above for frequencies
y 22 wo, but is less strong. The function b(e) is obtained by replacing f by
- 8xf2/9mwo4, and E by 2e, in (29.4):
62[(2e-kb2)2+12] = 16x2f4/81m4w010.
(29.9)
Next, let the frequency of the external force be 2= 2wote In the first
approximation, we have x(1) = - (f/3mwo2) cos(2wo+e)t. On substituting
in equation (29.1), we do not obtain terms representing an
external force in resonance such as occurred in the previous case. There is,
however, a parametric resonance resulting from the third-order term pro-
portional to the product x(1)x(2). If only this is retained out of the non-linear
terms, the equation for x(2) is
=
or
(29.10)
i.e. an equation of the type (27.8) (including friction), which leads, as we
have seen, to an instability of the oscillations in a certain range of frequencies.

6094
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#!/bin/bash
# 1. ./aws-textract.sh 1_create_batch pdfs/1-mechanics.pdf
# 1a. upload batch to aws textract
# 1b. download batch results
# 2. ./aws-textract.sh 2_extract_rawtext pdfs/1-mechanics.pdf
# will generate textract/1-mechanics.txt
# 3. ./aws-textract.sh 3_rawtext_to_chapters textract/1-mechanics.txt
FILE=${2}
# exit if file does not exist
if [ ! -f "$FILE" ]; then
echo "File does not exist: $FILE"
exit 1
fi
batch_seq() {
LAST_PAGE=$(pdfinfo "$FILE" | grep Pages | awk '{print $2}')
for i in $(seq 0 10 $LAST_PAGE); do
if [ $i -eq 0 ]; then
i=1
fi
j=$((i+9))
if [ $((i+9)) -gt $LAST_PAGE ]; then
j=$LAST_PAGE
fi
echo "${i}-${j}"
done
}
1_create_batch() {
pdfseparate "$FILE" "${FILE%.pdf}-page-%d.pdf"
for BATCH in batch_seq; do
pdfunite $(seq -w -f "${FILE%.pdf}-page-%g.pdf" $i $j) "${FILE%.pdf}-batch-$BATCH.pdf"
done
rm pdfs/*page*.pdf
}
2_extract_rawtext() {
BOOK=$(basename "$FILE")
TXT_FILE=textract/${BOOK%.pdf}.txt
truncate -s 0 ${TXT_FILE}
echo '======= GENERATED FROM ./tools/aws-textract.sh 2_extract_rawtext =======' >> ${TXT_FILE}
for BATCH in $(batch_seq); do
# zip files come from the aws textract console
unzip textract/"${BOOK%.pdf}-batch-$BATCH.zip" rawText.txt
cat rawText.txt >> ${TXT_FILE}
mv rawText.txt textract/${BOOK}-rawText-$BATCH.txt
done
}
3_rawtext_to_chapters() {
# modify for each book
BOOK=1
for chapter in $(seq 15 50); do
CHAPTER_NAME=$(grep -o "$chapter"'-[^.]*' $BOOK/index.md)
CHAPTER_FILE=${FILE%.txt}-${CHAPTER_NAME}.txt
echo " $CHAPTER_FILE"
set -x
sed "1,/^§$chapter/d;/^§$((chapter+1))/,\$d" $FILE > $CHAPTER_FILE
done
}
$@