10 more pages

LL1/10-mechanical-similarity.md

@@ -0,0 +1,84 @@
+---
+title: Mechanical similarity
+---
+
+Multiplication of the Lagrangian by any constant clearly does not affect the equations of motion. This fact (already mentioned in `LL1/2`) makes possible, in a number of important cases, some useful inferences concerning the properties of the motion, without the necessity of actually integrating the equations.
+
+Such cases include those where the potential energy is a homogeneous function of the co-ordinates, i.e. satisfies the condition
+
+```load
+LL1/10.1
+```
+
+where $\alpha$ is any constant and $k$ the degree of homogeneity of the function.
+
+Let us carry out a transformation in which the co-ordinates are changed by a factor $\alpha$ and the time by a factor $\beta: \v{r}_a\rightarrow \alpha\v{r}_a, t\rightarrow \beta t$. Then all the velocities $\v{v}a = \dd{\v{r}}_a/\dd{t} are changed by a factor $\alpha/\beta$, and the kinetic energy by a factor
+$\alpha^2/\beta^2$. The potential energy is multiplied by $\alpha^k$. If $\alpha$ and $\beta$ are such that $\alpha^2/\beta^2 = \alpha^k$, i.e. $\beta = \alpha^{1-k/2}$, then the result of the transformation is to multiply the Lagrangian by the constant factor $\alpha^k$, i.e. to leave the equations of motion unaltered.
+
+A change of all the co-ordinates of the particles by the same factor corresponds to the replacement of the paths of the particles by other paths, geometrically similar but differing in size. Thus we conclude that, if the potential energy of the system is a homogeneous function of degree k in the (Cartesian) co-ordinates, the equations of motion permit a series of geometrically similar paths, and the times of the motion between corresponding points are in the ratio
+
+```load
+LL1/10.2
+```
+
+where $l'/l$ is the ratio of linear dimensions of the two paths. Not only the times but also any mechanical quantities at corresponding points at corresponding times are in a ratio which is a power of $l'/l$. For example, the velocities, energies and angular momenta are such that
+
+```load
+LL1/10.3
+```
+
+The following are some examples of the foregoing.
+
+As we shall see later, in small oscillations the potential energy is a quadratic function of the co-ordinates ($k=2$). From `LL1/10.2` we find that the period of such oscillations is independent of their amplitude.
+
+In a uniform field of force, the potential energy is a linear function of the co-ordinates (see `LL1/5.8`), i.e. $k = 1$. From `LL1/10.2` we have $t'/t = \sqrt{l'/l}$.  Hence, for example, it follows that, in fall under gravity, the time of fall is as 2the square root of the initial altitude.
+
+In the Newtonian attraction of two masses or the Coulomb interaction of two charges, the potential energy is inversely proportional to the distance apart, i.e. it is a homogeneous function of degree $k = 1$. Then $t'/t = (l'/l)^{3/2}$, and we can state, for instance, that the square of the time of revolution in the orbit is as the cube of the size of the orbit (Kepler's third law).
+
+If the potential energy is a homogeneous function of the co-ordinates and the motion takes place in a finite region of space, there is a very simple relation between the time average values of the kinetic and potential energies, known as the virial theorem.
+
+Since the kinetic energy $T$ is a quadratic function of the velocities, we have by Euler's theorem on homogeneous functions $\sum_a \v{v}a\cdot\partial T/\partial\v{v}_a = 2T$, or, putting $\partial T/\partial \v{v}_a = \v{p}_a$, the momentum,
+
+```load
+LL1/10.4
+```
+
+Let us average this equation with respect to time. The average value of any function of time $f(t)$ is defined as
+
+$$
+\bar{f}=\lim_{\tau\rightarrow\inf}\frac{1}{\tau}\int_0^\tau f(t)\dd{t}
+$$
+
+It is easy to see that, if $f(t)$ is the time derivative $\dd{F(t)}/\dd{t}$ of a bounded function $F(t)$, its mean value is zero. For
+
+$$
+\bar{f}
+=\lim_{\tau\rightarrow\inf}\frac{1}{\tau}\int_0^\tau \frac{\dd{F}}{\dd{t}}\dd{t}
+=\lim_{\tau\rightarrow\inf}\frac{F(\tau)-F(0)}{\tau}
+=0
+$$
+
+Let us assume that the system executes a motion in a finite region of space and with finite velocities. Then $\sum_a\v{p}_a\cdot\v{r}_a$ is bounded, and the mean value of the first term on the right-hand side of `LL1/10.4` is zero. In the second term we replace $\v{\dot{p}}_a$ by $-\partial U/\partial \v{r}_a$ in accordance with Newton's equations `LL1/5.3`, obtaining[^1]
+
+```load
+LL1/10.5
+```
+
+If the potential energy is a homogeneous function of degree $k$ in the radius
+vectors $\v{r}_a$, then by Euler's theorem equation `LL1/10.5` becomes the required relation:
+
+```load
+LL1/10.6
+```
+
+Since $\bar{T}+\bar{U}=\bar{E}=E$, the relation `LL1/10.6` can also be expressed as
+
+```load
+LL1/10.7
+```
+
+which express $\bar{U}$ and $\bar{T}$ in terms of the total energy of the system.
+
+In particular, for small oscillations ($k=2$) we have $\bar{T} = \bar{U}$, i.e. the mean values of the kinetic and potential energies are equal. For a Newtonian interaction $(k = - 1)$, $2\bar{T} = - \bar{U}$, and $E=-\bar{T}$, in accordance with the fact that, in such an interaction, the motion takes place in a finite region of space only if the total energy is negative (see `LL1/15`).
+
+[^1]: The expression on the right of `LL1/10.5` is sometimes called the virial of the system.

LL1/10_problems.md

@@ -0,0 +1,7 @@
+PROBLEMS
+PROBLEM 1. Find the ratio of the times in the same path for particles having different
+masses but the same potential energy.
+SOLUTION. t'/t = (m'/m).
+PROBLEM 2. Find the ratio of the times in the same path for particles having the same mass
+but potential energies differing by a constant factor.
+SOLUTION. I'/t

LL1/11-motion-in-one-dimension.md

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+---
+title: Motion in one dimension
+---
+
+The motion of a system having one degree of freedom is said to take place in one dimension. The most general form of the Lagrangian of such a system in fixed external conditions is
+L = 1a(q)i2-U(q),
+
+```load
+LL1/11.1
+```
+
+where $a(q)$ is some function of the generalised co-ordinate $q$. In particular,
+if $q$ is a Cartesian co-ordinate ($x$, say) then
+
+```load
+
+```
+
+The equations of motion corresponding to these Lagrangians can be integrated in a general form. It is not even necessary to write down the equation of motion; we can start from the first integral of this equation, which gives the law of conservation of energy. For the Lagrangian `LL1/11.2` (e.g.) we have $\mfrac{1}{2}m\dot{x}^2+U(x)=E$. This is a first-order differential equation, and can be inte- grated immediately. Since $\dd{x}/\dd{t} = \sqrt{2[E - U(x)]/m}$, it follows that
+
+```load
+LL1/11.3
+```
+
+The two arbitrary constants in the solution of the equations of motion are
+here represented by the total energy E and the constant of integration.
+Since the kinetic energy is essentially positive, the total energy always
+exceeds the potential energy, i.e. the motion can take place only in those
+regions of space where U(x) < E. For example, let the function U(x) be
+of the form shown in Fig. 6 (p. 26). If we draw in the figure a horizontal
+line corresponding to a given value of the total energy, we immediately find
+the possible regions of motion. In the example of Fig. 6, the motion can
+occur only in the range AB or in the range to the right of C.
+The points at which the potential energy equals the total energy,
+U(x) = E,
+(11.4)
+give the limits of the motion. They are turning points, since the velocity there
+is zero. If the region of the motion is bounded by two such points, then the
+motion takes place in a finite region of space, and is said to be finite. If the
+region of the motion is limited on only one side, or on neither, then the
+motion is infinite and the particle goes to infinity.
+
+A finite motion in one dimension is oscillatory, the particle moving re-
+peatedly back and forth between two points (in Fig. 6, in the potential well
+AB between the points X1 and x2). The period T of the oscillations, i.e. the
+time during which the particle passes from X1 to X2 and back, is twice the time
+from X1 to X2 (because of the reversibility property, §5) or, by (11.3),
+T(E) =
+(11.5)
+
+where X1 and X2 are roots of equation (11.4) for the given value of E. This for-
+mula gives the period of the motion as a function of the total energy of the
+particle.
+U
+A
+B
+C
+U=E
+x,
+X2
+X
+FIG. 6

LL1/11_problems.md

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

LL1/12-determination-of-the-potential-energy-from-the-period-of-oscillation.md

@@ -0,0 +1,57 @@
+---
+title: Determination of the potential energy from the period of oscillation
+---
+
+Let us consider to what extent the form of the potential energy U(x) of a
+field in which a particle is oscillating can be deduced from a knowledge of the
+period of oscillation T as a function of the energy E. Mathematically, this
+involves the solution of the integral equation (11.5), in which U(x) is regarded
+as unknown and T(E) as known.
+We shall assume that the required function U(x) has only one minimum
+in the region of space considered, leaving aside the question whether there
+exist solutions of the integral equation which do not meet this condition.
+For convenience, we take the origin at the position of minimum potential
+energy, and take this minimum energy to be zero (Fig. 7).
+U
+U=E
+121U
+X
+X1
+X2
+FIG. 7
+28
+Integration of the Equations of Motion
+§12
+In the integral (11.5) we regard the co-ordinate x as a function of U. The
+function x(U) is two-valued: each value of the potential energy corresponds
+to two different values of X. Accordingly, the integral (11.5) must be divided
+into two parts before replacing dx by (dx/dU) dU: one from x = X1 to x = 0
+and the other from x = 0 to X = X2. We shall write the function x(U) in
+these two ranges as x = x1(U) and x = x2(U) respectively.
+The limits of integration with respect to U are evidently E and 0, so that
+we have
+T(E) =
+=
+If both sides of this equation are divided by V(a - E), where a is a parameter,
+and integrated with respect to E from 0 to a, the result is
+dUdE
+or, changing the order of integration,
+dE
+The integral over E is elementary; its value is TT. The integral over U is
+thus trivial, and we have
+since x2(0) = x1(0) = 0. Writing U in place of a, we obtain the final result:
+(12.1)
+Thus the known function T(E) can be used to determine the difference
+x2(U)-x1(U). - The functions x2(U) and x1(U) themselves remain indeter-
+minate. This means that there is not one but an infinity of curves U = U(x)
+§13
+The reduced mass
+29
+which give the prescribed dependence of period on energy, and differ in such
+a way that the difference between the two values of x corresponding to each
+value of U is the same for every curve.
+The indeterminacy of the solution is removed if we impose the condition
+that the curve U = U(x) must be symmetrical about the U-axis, i.e. that
+x2(U) = 1(U) III x(U). In this case, formula (12.1) gives for x(U) the
+unique expression
+(12.2)

LL1/13-the-reduced-mass.md

@@ -0,0 +1,36 @@
+---
+title: The reduced mass
+---
+
+A complete general solution can be obtained for an extremely important
+problem, that of the motion of a system consisting of two interacting particles
+(the two-body problem).
+As a first step towards the solution of this problem, we shall show how it
+can be considerably simplified by separating the motion of the system into
+the motion of the centre of mass and that of the particles relative to the centre
+of mass.
+The potential energy of the interaction of two particles depends only on
+the distance between them, i.e. on the magnitude of the difference in their
+radius vectors. The Lagrangian of such a system is therefore
+L =
+(13.1)
+Let r III r1-r2 - be the relative position vector, and let the origin be at the
+centre of mass, i.e. M1r1+M2r2 = 0. These two equations give
+= m2I/(m1+m2),
+r2
+-M1I/(m1+m2).
+(13.2)
+Substitution in (13.1) gives
+L
+(13.3)
+where
+m=mym
+(13.4)
+is called the reduced mass. The function (13.3) is formally identical with the
+Lagrangian of a particle of mass m moving in an external field U(r) which is
+symmetrical about a fixed origin.
+Thus the problem of the motion of two interacting particles is equivalent
+to that of the motion of one particle in a given external field U(r). From the
+solution r = r(t) of this problem, the paths r1 = r1(t) and r2 = r2(t) of the
+two particles separately, relative to their common centre of mass, are obtained
+by means of formulae (13.2).

LL1/13_problem.md

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

LL1/14-motion-in-a-central-field.md

@@ -0,0 +1,33 @@
+---
+title: Motion in a central field
+---
+
+On reducing the two-body problem to one of the motion of a single body,
+we arrive at the problem of determining the motion of a single particle in an
+external field such that its potential energy depends only on the distance r
+from some fixed point. This is called a central field. The force acting on the
+particle is F = du(r)/dr = - (dU/dr)r/r; its magnitude is likewise a func-
+tion of r only, and its direction is everywhere that of the radius vector.
+As has already been shown in §9, the angular momentum of any system
+relative to the centre of such a field is conserved. The angular momentum of a
+single particle is M = rxp. Since M is perpendicular to r, the constancy of
+M shows that, throughout the motion, the radius vector of the particle lies
+in the plane perpendicular to M.
+Thus the path of a particle in a central field lies in one plane. Using polar
+co-ordinates r, in that plane, we can write the Lagrangian as
+(14.1)
+see (4.5). This function does not involve the co-ordinate explicitly. Any
+generalised co-ordinate qi which does not appear explicitly in the Lagrangian
+is said to be cyclic. For such a co-ordinate we have, by Lagrange's equation,
+(d/dt) aL/dqi = aL/dqi = 0, so that the corresponding generalised momen-
+tum Pi = aL/dqi is an integral of the motion. This leads to a considerable
+simplification of the problem of integrating the equations of motion when
+there are cyclic co-ordinates.
+In the present case, the generalised momentum is the same as
+the angular momentum M z = M (see (9.6)), and we return to the known law
+of conservation of angular momentum:
+M = mr2o = constant. =
+(14.2)
+This law has a simple geometrical interpretation in the plane motion of a single
+particle in a central field. The expression 1/2 . rdo is the area of the sector
+bounded by two neighbouring radius vectors and an element of the path

LL1/9-angular-momentum.md

@@ -104,3 +104,27 @@ LL1/9.6
 In other words, the angular momentum $\v{M}$ of a mechanical system consists of its "intrinsic angular momentum" in a frame in which it is at rest, and the angular momentum $\v{R}\times\v{P}$ due to its motion as a whole.
 
 Although the law of conservation of all three components of angular momentum (relative to an arbitrary origin) is valid only for a closed system, the law of conservation may hold in a more restricted form even for a system in an external field. It is evident from the above derivation that the component of angular momentum along an axis about which the field is symmetrical is always conserved, for the mechanical properties of the system are unaltered by any rotation about that axis. Here the angular momentum must, of course, be defined relative to an origin lying on the axis.
+
+The most important such case is that of a centrally symmetric field or central field, i.e. one in which the potential energy depends only on the distance from some particular point (the centre). It is evident that the component of angular momentum along any axis passing through the centre is conserved in motion in such a field. In other words, the angular momentum $\v{M}$ is conserved provided that it is defined with respect to the centre of the field.
+
+Another example is that of a homogeneous field in the z-direction; in such a field, the component $M_z$ of the angular momentum is conserved, whichever point is taken as the origin.
+
+The component of angular momentum along any axis (say the z-axis) can
+be found by differentiation of the Lagrangian:
+
+```load
+LL1/9.7
+```
+
+where the co-ordinate is the angle of rotation about the z-axis. This is evident from the above proof of the law of conservation of angular momentum, but can also be proved directly. In cylindrical co-ordinates $r, \phi, z$ we have (substituting $x_a = r_a\cos\phi_a, y_a = -r_a\sin\phi_a$
+
+$$
+M_z=\sum_a m_a(x_a\dot{y}_a-y_a\dot{x}_a)
+$$
+```load
+LL1/9.8
+```
+
+The Lagrangian is, in terms of these co-ordinates,
+
+and substitution of this in `LL1/9.7` gives `LL1/9.8`.

LL1/9_problems.md

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

tools/gen_index.sh

@@ -1,7 +1,7 @@
 #!/bin/bash
 
 book=LL1
-sections=$(find LL1 -name "*.md" | sort)
+sections=$(find LL1 -name "*.md" | sort -h)
 
 echo "<ul>"
 for s in $sections; do