diff --git a/1/15-keplers-problem.md b/1/15-keplers-problem.md
index 9adda1a..0e60f9b 100644
--- a/1/15-keplers-problem.md
+++ b/1/15-keplers-problem.md
@@ -1,3 +1,379 @@
+---
+title: 15-keplers-problem
+---
+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
@@ -214,3 +590,5 @@ 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
diff --git a/1/16-disintegration-of-particles.md b/1/16-disintegration-of-particles.md
index 83fe807..ccec2f2 100644
--- a/1/16-disintegration-of-particles.md
+++ b/1/16-disintegration-of-particles.md
@@ -1,4 +1,6 @@
-§16. Disintegration of particles
+---
+title: 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
@@ -126,283 +128,3 @@ SOLUTION. In the C system, the corresponding angles are related by 010 = n-020.
 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).
diff --git a/1/17-elastic-collisions.md b/1/17-elastic-collisions.md
index cae51e6..567745d 100644
--- a/1/17-elastic-collisions.md
+++ b/1/17-elastic-collisions.md
@@ -1,3 +1,26 @@
+---
+title: 17-elastic-collisions
+---
+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
@@ -138,3 +161,5 @@ SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The mo
 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
diff --git a/1/18-scattering.md b/1/18-scattering.md
index 76df87a..5a4554d 100644
--- a/1/18-scattering.md
+++ b/1/18-scattering.md
@@ -1,4 +1,6 @@
-§18
+---
+title: 18-scattering
+---
 §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
@@ -115,3 +117,80 @@ 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,
diff --git a/1/19-rutherfords-formula.md b/1/19-rutherfords-formula.md
index e69de29..09b1b16 100644
--- a/1/19-rutherfords-formula.md
+++ b/1/19-rutherfords-formula.md
@@ -0,0 +1,88 @@
+---
+title: 19-rutherfords-formula
+---
+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
diff --git a/1/20-small-angle-scattering.md b/1/20-small-angle-scattering.md
new file mode 100644
index 0000000..cbe21e6
--- /dev/null
+++ b/1/20-small-angle-scattering.md
@@ -0,0 +1,117 @@
+---
+title: 20-small-angle-scattering
+---
+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
diff --git a/1/21-free-oscillations-in-one-dimension.md b/1/21-free-oscillations-in-one-dimension.md
new file mode 100644
index 0000000..4f6b54a
--- /dev/null
+++ b/1/21-free-oscillations-in-one-dimension.md
@@ -0,0 +1,80 @@
+---
+title: 21-free-oscillations-in-one-dimension
+---
+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
diff --git a/1/22-forced-oscillations.md b/1/22-forced-oscillations.md
new file mode 100644
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--- /dev/null
+++ b/1/22-forced-oscillations.md
@@ -0,0 +1,171 @@
+---
+title: 22-forced-oscillations
+---
+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).
diff --git a/1/23-oscillations-of-systems-with-more-than-one-degree-of-freedom.md b/1/23-oscillations-of-systems-with-more-than-one-degree-of-freedom.md
new file mode 100644
index 0000000..542ce5e
--- /dev/null
+++ b/1/23-oscillations-of-systems-with-more-than-one-degree-of-freedom.md
@@ -0,0 +1,205 @@
+---
+title: 23-oscillations-of-systems-with-more-than-one-degree-of-freedom
+---
+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
diff --git a/1/24-vibrations-of-molecules.md b/1/24-vibrations-of-molecules.md
new file mode 100644
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--- /dev/null
+++ b/1/24-vibrations-of-molecules.md
@@ -0,0 +1,181 @@
+---
+title: 24-vibrations-of-molecules
+---
+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.
+74
+Small Oscillations
diff --git a/1/25-damped-oscillations.md b/1/25-damped-oscillations.md
new file mode 100644
index 0000000..4725080
--- /dev/null
+++ b/1/25-damped-oscillations.md
@@ -0,0 +1,129 @@
+---
+title: 25-damped-oscillations
+---
+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.
diff --git a/1/26-forced-oscillations-under-friction.md b/1/26-forced-oscillations-under-friction.md
new file mode 100644
index 0000000..0d745be
--- /dev/null
+++ b/1/26-forced-oscillations-under-friction.md
@@ -0,0 +1,129 @@
+---
+title: 26-forced-oscillations-under-friction
+---
+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
diff --git a/1/27-parametric-resonance.md b/1/27-parametric-resonance.md
new file mode 100644
index 0000000..6b4a240
--- /dev/null
+++ b/1/27-parametric-resonance.md
@@ -0,0 +1,178 @@
+---
+title: 27-parametric-resonance
+---
+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
diff --git a/1/28-anharmonic-oscillations.md b/1/28-anharmonic-oscillations.md
new file mode 100644
index 0000000..a3db212
--- /dev/null
+++ b/1/28-anharmonic-oscillations.md
@@ -0,0 +1,125 @@
+---
+title: 28-anharmonic-oscillations
+---
+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
diff --git a/1/29-resonance-in-non-linear-oscillations.md b/1/29-resonance-in-non-linear-oscillations.md
new file mode 100644
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@@ -0,0 +1,250 @@
+---
+title: 29-resonance-in-non-linear-oscillations
+---
+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.
diff --git a/1/30-motion-in-a-rapidly-oscillating-field.md b/1/30-motion-in-a-rapidly-oscillating-field.md
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@@ -0,0 +1,149 @@
+---
+title: 30-motion-in-a-rapidly-oscillating-field
+---
+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
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--- /dev/null
+++ b/1/31-angular-velocity.md
@@ -0,0 +1,54 @@
+---
+title: 31-angular-velocity
+---
+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
diff --git a/1/32-the-inertia-tensor.md b/1/32-the-inertia-tensor.md
new file mode 100644
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--- /dev/null
+++ b/1/32-the-inertia-tensor.md
@@ -0,0 +1,298 @@
+---
+title: 32-the-inertia-tensor
+---
+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,
diff --git a/1/33-angular-momentum-of-a-rigid-body.md b/1/33-angular-momentum-of-a-rigid-body.md
new file mode 100644
index 0000000..e387276
--- /dev/null
+++ b/1/33-angular-momentum-of-a-rigid-body.md
@@ -0,0 +1,93 @@
+---
+title: 33-angular-momentum-of-a-rigid-body
+---
+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
diff --git a/1/34-the-equations-of-motion-of-a-rigid-body.md b/1/34-the-equations-of-motion-of-a-rigid-body.md
new file mode 100644
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+++ b/1/34-the-equations-of-motion-of-a-rigid-body.md
@@ -0,0 +1,130 @@
+---
+title: 34-the-equations-of-motion-of-a-rigid-body
+---
+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
diff --git a/1/35-eulerian-angles.md b/1/35-eulerian-angles.md
new file mode 100644
index 0000000..31d8d96
--- /dev/null
+++ b/1/35-eulerian-angles.md
@@ -0,0 +1,167 @@
+---
+title: 35-eulerian-angles
+---
+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
+114
+Motion of a Rigid Body
diff --git a/1/36-eulers-equations.md b/1/36-eulers-equations.md
new file mode 100644
index 0000000..dda0066
--- /dev/null
+++ b/1/36-eulers-equations.md
@@ -0,0 +1,76 @@
+---
+title: 36-eulers-equations
+---
+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
diff --git a/1/37-the-asymmetrical-top.md b/1/37-the-asymmetrical-top.md
new file mode 100644
index 0000000..1f64b8f
--- /dev/null
+++ b/1/37-the-asymmetrical-top.md
@@ -0,0 +1,251 @@
+---
+title: 37-the-asymmetrical-top
+---
+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
+122
+Motion of a Rigid Body
diff --git a/1/38-rigid-bodies-in-contact.md b/1/38-rigid-bodies-in-contact.md
new file mode 100644
index 0000000..2c024f2
--- /dev/null
+++ b/1/38-rigid-bodies-in-contact.md
@@ -0,0 +1,171 @@
+---
+title: 38-rigid-bodies-in-contact
+---
+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
diff --git a/1/39-motion-in-a-non-inertial-frame-of-reference.md b/1/39-motion-in-a-non-inertial-frame-of-reference.md
new file mode 100644
index 0000000..ada07ad
--- /dev/null
+++ b/1/39-motion-in-a-non-inertial-frame-of-reference.md
@@ -0,0 +1,189 @@
+---
+title: 39-motion-in-a-non-inertial-frame-of-reference
+---
+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
diff --git a/1/40-hamiltons-equations.md b/1/40-hamiltons-equations.md
new file mode 100644
index 0000000..0bfbf19
--- /dev/null
+++ b/1/40-hamiltons-equations.md
@@ -0,0 +1,83 @@
+---
+title: 40-hamiltons-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
diff --git a/1/41-the-routhian.md b/1/41-the-routhian.md
new file mode 100644
index 0000000..5139cb4
--- /dev/null
+++ b/1/41-the-routhian.md
@@ -0,0 +1,39 @@
+---
+title: 41-the-routhian
+---
+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
diff --git a/1/42-poisson-brackets.md b/1/42-poisson-brackets.md
new file mode 100644
index 0000000..5b77cbb
--- /dev/null
+++ b/1/42-poisson-brackets.md
@@ -0,0 +1,169 @@
+---
+title: 42-poisson-brackets
+---
+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
diff --git a/1/43-the-actions-as-a-function-of-the-co-ordinates.md b/1/43-the-actions-as-a-function-of-the-co-ordinates.md
new file mode 100644
index 0000000..9539c09
--- /dev/null
+++ b/1/43-the-actions-as-a-function-of-the-co-ordinates.md
@@ -0,0 +1,83 @@
+---
+title: 43-the-actions-as-a-function-of-the-co-ordinates
+---
+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
diff --git a/1/44-maupertuis-principle.md b/1/44-maupertuis-principle.md
new file mode 100644
index 0000000..29de623
--- /dev/null
+++ b/1/44-maupertuis-principle.md
@@ -0,0 +1,106 @@
+---
+title: 44-maupertuis-principle
+---
+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:
diff --git a/1/45-canonical-transformations.md b/1/45-canonical-transformations.md
new file mode 100644
index 0000000..0fb3b65
--- /dev/null
+++ b/1/45-canonical-transformations.md
@@ -0,0 +1,141 @@
+---
+title: 45-canonical-transformations
+---
+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
diff --git a/1/46-louivilles-theorem.md b/1/46-louivilles-theorem.md
new file mode 100644
index 0000000..6a20d7b
--- /dev/null
+++ b/1/46-louivilles-theorem.md
@@ -0,0 +1,40 @@
+---
+title: 46-louivilles-theorem
+---
+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,
diff --git a/1/47-the-hamilton-jacobi-equations.md b/1/47-the-hamilton-jacobi-equations.md
new file mode 100644
index 0000000..ce61484
--- /dev/null
+++ b/1/47-the-hamilton-jacobi-equations.md
@@ -0,0 +1,91 @@
+---
+title: 47-the-hamilton-jacobi-equations
+---
+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.
diff --git a/1/48-separation-of-the-variables.md b/1/48-separation-of-the-variables.md
new file mode 100644
index 0000000..5dda7f5
--- /dev/null
+++ b/1/48-separation-of-the-variables.md
@@ -0,0 +1,193 @@
+---
+title: 48-separation-of-the-variables
+---
+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
+The Canonical Equations
diff --git a/1/49-adiabatic-invariants.md b/1/49-adiabatic-invariants.md
new file mode 100644
index 0000000..31238ff
--- /dev/null
+++ b/1/49-adiabatic-invariants.md
@@ -0,0 +1,174 @@
+---
+title: 49-adiabatic-invariants
+---
+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
diff --git a/1/50-general-properties-of-motion-in-s-dimensions.md b/1/50-general-properties-of-motion-in-s-dimensions.md
new file mode 100644
index 0000000..45f0a36
--- /dev/null
+++ b/1/50-general-properties-of-motion-in-s-dimensions.md
@@ -0,0 +1,571 @@
+---
+title: 50-general-properties-of-motion-in-s-dimensions
+---
+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.
+PHYSICS
+The enormous increase in the number
+and size of scientific journals has led to a
+qualitative change in the problem of
+scientific communication. The policies
+of most journals are based on the old
+need to ensure that no valid science
+was lost to the scientific public by being
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+PHYSICS will help its readers to find
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+Pergamon Press
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+Printed in Great Britain/Bradley
+COURSE OF THEORETICAL PHYSICS
+by L.D. LANDAU and E.M. LIFSHITZ
+Institute of Physical Problems, USSR Academy of Sciences
+The complete Course of Theoretical Physics by Landau and Lifshitz, recognized as two
+of the world's outstanding physicists, is being published in full by Pergamon Press. It
+comprises nine volumes, covering all branches of the subject translations from the Russian
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+"The titles of the volumes in this series cover a vast range of topics, and there seems to be
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+Proceedings of the Physical Society
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+MECHANICS
+"The entire book is a masterpiece of scientific writing. There is not a superfluous sentence
+and the authors know exactly where they are going
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+QUANTUM MECHANICS (Non-relativistic Theory)
+throughout the five hundred large pages, the authors' discussion proceeds with the
+clarity and succinctness typical of the very best works on theoretical physics."
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+FLUID MECHANICS
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+under the headings of shock waves, one-dimensional gas flow and flow past finite bodies.
+There is a chapter on the fluid dynamics of combustion while unusual topics discussed are
+relativistic fluid dynamics, dynamics of superfluids and fluctuations of fluid dynamics
+a
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+THE CLASSICAL THEORY OF FIELDS (Second Edition)
+"This is an excellent and readable volume. It is a valuable and unique addition to the
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+STATISTICAL PHYSICS
+stimulating reading, partly because of the clarity and compactness of some of the
+treatments put forward, and partly by reason of contrasts with texts on statistical mechanics
+and statistical thermodynamics better known to English sciences
+Other features
+attract attention since they do not always receive comparable mention in other textbooks."
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+THEORY OF ELASTICITY
+"I shall be surprised if this book does not come to be regarded as a masterpiece."
+Journal of the Royal Institute of Physics
+ELECTRODYNAMICS OF CONTINUOUS MEDIA
+"Within the volume one finds everything expected of a textbook on classical electricity
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+08 006466 3
\ No newline at end of file
diff --git a/tools/aws-textract.sh b/tools/aws-textract.sh
index f7d3d91..e97334d 100755
--- a/tools/aws-textract.sh
+++ b/tools/aws-textract.sh
@@ -59,6 +59,12 @@ batch_seq() {
         echo "  $CHAPTER_FILE"
         set -x
         sed "1,/^§$chapter/d;/^§$((chapter+1))/,\$d" $FILE > $CHAPTER_FILE
+        cat <<EOF > $BOOK/$CHAPTER_NAME.md
+---
+title: $CHAPTER_NAME
+---
+EOF
+        cat $CHAPTER_FILE >> $BOOK/$CHAPTER_NAME.md
     done
 }