Exercises

22.9 EXERCISES
22.9 Exercises
22.1
22.2
22.3
22.4
A surface of revolution, whose equation in cylindrical polar coordinates is ρ =
ρ(z), is bounded by the circles ρ = a,
z = ±c (a > c). Show that the function
that makes the surface integral I = ρ−1/2 dS stationary with respect to small
2
variations is given by ρ(z) = k + z /(4k), where k = [a ± (a2 − c2 )1/2 ]/2.
Show that the lowest value of the integral
B
(1 + y 2 )1/2
dx,
y
A
√
where A is (−1, 1) and B is (1, 1), is 2 ln(1 + 2). Assume that the Euler–Lagrange
equation gives a minimising curve.
The refractive index n of a medium is a function only of the distance r from a
fixed point O. Prove that the equation of a light ray, assumed to lie in a plane
through O, travelling in the medium satisfies (in plane polar coordinates)
2
1 dr
r2 n2 (r)
= 2 2
− 1,
2
r
dφ
a n (a)
where a is the distance of the ray from O at the point at which dr/dφ = 0.
If n = [1 + (α2 /r2 )]1/2 and the ray starts and ends far from O, find its deviation
(the angle through which the ray is turned), if its minimum distance from O is a.
The Lagrangian for a π-meson is given by
L(x, t) = 12 (φ̇2 − |∇φ|2 − µ2 φ2 ),
22.5
where µ is the meson mass and φ(x, t) is its wavefunction. Assuming Hamilton’s
principle, find the wave equation satisfied by φ.
Prove the following results about general systems.
(a) For a system described in terms of coordinates qi and t, show that if t does
not appear explicitly in the expressions for x, y and z (x = x(qi , t), etc.) then
the kinetic energy T is a homogeneous
quadratic function of the q̇i (it may
also involve the qi ). Deduce that i q̇i (∂T /∂q̇i ) = 2T .
(b) Assuming that the forces acting on the system are derivable from a potential
V , show, by expressing dT /dt in terms of qi and q̇i , that d(T + V )/dt = 0.
22.6
For a system specified by the coordinates q and t, show that the equation of
motion is unchanged if the Lagrangian L(q, q̇, t) is replaced by
dφ(q, t)
,
dt
where φ is an arbitrary function. Deduce that the equation of motion of a particle
that moves in one dimension subject to a force −dV (x)/dx (x being measured
from a point O) is unchanged if O is forced to move with a constant velocity v
(x still being measured from O).
In cylindrical polar coordinates, the curve (ρ(θ), θ, αρ(θ)) lies on the surface of
the cone z = αρ. Show that geodesics (curves of minimum length joining two
points) on the cone satisfy
L1 = L +
22.7
ρ4 = c2 [β 2 ρ + ρ2 ],
2
where c is an arbitrary constant, but β has to have a particular value. Determine
the form of ρ(θ) and hence find the equation of the shortest path on the cone
between the points (R, −θ0 , αR) and (R, θ0 , αR).
[ You will find it useful to determine the form of the derivative of cos−1 (u−1 ). ]
797
CALCULUS OF VARIATIONS
22.8
22.9
22.10
Derive the differential equations for the plane-polar coordinates, r and φ, of a
particle of unit mass moving in a field of potential V (r). Find the form of V if
the path of the particle is given by r = a sin φ.
You are provided with a line of length πa/2 and negligible mass and some lead
shot of total mass M. Use a variational method to determine how the lead shot
must be distributed along the line if the loaded line is to hang in a circular arc
of radius a when its ends are attached to two points at the same height. Measure
the distance s along the line from its centre.
Extend the result of subsection 22.2.2 to the case of several dependent variables
yi (x), showing that, if x does not appear explicitly in the integrand, then a first
integral of the Euler–Lagrange equations is
F−
n
yi
i=1
22.11
∂F
= constant.
∂yi
A general result is that light travels through a variable medium by a path which
minimises the travel time (this is an alternative formulation of Fermat’s principle).
With respect to a particular cylindrical polar coordinate system (ρ, φ, z), the speed
of light v(ρ, φ) is independent of z. If the path of the light is parameterised as
ρ = ρ(z), φ = φ(z), use the result of the previous exercise to show that
v 2 (ρ + ρ2 φ + 1)
2
22.12
22.13
2
is constant along the path.
For the particular case when v = v(ρ) = b(a2 + ρ2 )1/2 , show that the two Euler–
Lagrange equations have a common solution in which the light travels along a
helical path given by φ = Az + B, ρ = C, provided that A has a particular value.
Light travels in the vertical xz-plane through a slab of material which lies between
the planes z = z0 and z = 2z0 , and in which the speed of light v(z) = c0 z/z0 .
Using the alternative formulation of Fermat’s principle, given in the previous
question, show that the ray paths are arcs of circles.
Deduce that, if a ray enters the material at (0, z0 ) at an angle to the vertical,
π/2 − θ, of more than 30◦ , then it does not reach the far side of the slab.
A dam of capacity V (less than πb2 h/2) is to be constructed on level ground next
to a long straight wall which runs from (−b, 0) to (b, 0). This is to be achieved by
joining the ends of a new wall, of height h, to those of the existing wall. Show
that, in order to minimise the length L of new wall to be built, it should form
part of a circle, and that L is then given by
b
dx
,
2 2 1/2
−b (1 − λ x )
where λ is found from
sin−1 µ (1 − µ2 )1/2
V
=
−
hb2
µ2
µ
22.14
22.15
and µ = λb.
In the brachistochrone problem of subsection 22.3.4 show that if the upper endpoint can lie anywhere on the curve h(x, y) = 0, then the curve of quickest descent
y(x) meets h(x, y) = 0 at right angles.
The Schwarzchild metric for the static field of a non-rotating spherically symmetric black hole of mass M is given by
2GM
(dr)2
(dt)2 −
(ds)2 = c2 1 − 2
− r2 (dθ)2 − r 2 sin2 θ (dφ)2 .
cr
1 − 2GM/(c2 r)
Considering only motion confined to the plane θ = π/2, and assuming that the
798
22.9 EXERCISES
22.16
path of a small test particle is such as to make ds stationary, find two first
integrals of the equations of motion. From their Newtonian limits, in which
GM/r, ṙ 2 and r2 φ̇2 are all c2 , identify the constants of integration.
Use result (22.27) to evaluate
1
(1 − x2 )Pm (x)Pn (x) dx,
J=
−1
22.17
where Pm (x) is a Legendre polynomial of order m.
Determine the minimum value that the integral
1
[x4 (y )2 + 4x2 (y )2 ] dx
J=
0
22.18
22.19
22.20
22.21
can have, given that y is not singular at x = 0 and that y(1) = y (1) = 1. Assume
that the Euler–Lagrange equation gives the lower limit, and verify retrospectively
that your solution makes the first term on the LHS of equation (22.15) vanish.
Show that y − xy + λx2 y = 0 has a solution for which y(0) = y(1) = 0 and
λ ≤ 147/4.
Find an appropriate, but simple, trial function and use it to estimate the lowest
eigenvalue λ0 of Stokes’ equation,
d2 y
+ λxy = 0,
with y(0) = y(π) = 0.
dx2
Explain why your estimate must be strictly greater than λ0 .
Estimate the lowest eigenvalue, λ0 , of the equation
d2 y
− x2 y + λy = 0,
y(−1) = y(1) = 0,
dx2
using a quadratic trial function.
A drumskin is stretched across a fixed circular rim of radius a. Small transverse
vibrations of the skin have an amplitude z(ρ, φ, t) that satisfies
1 ∂2 z
c2 ∂t2
in plane polar coordinates. For a normal mode independent of azimuth, z =
Z(ρ) cos ωt, find the differential equation satisfied by Z(ρ). By using a trial
function of the form aν − ρν , with adjustable parameter ν, obtain an estimate for
the lowest normal mode frequency.
[ The exact answer is (5.78)1/2 c/a. ]
Consider the problem of finding the lowest eigenvalue, λ0 , of the equation
∇2 z =
22.22
(1 + x2 )
d2 y
dy
+ 2x
+ λy = 0,
dx2
dx
y(±1) = 0.
(a) Recast the problem in variational form, and derive an approximation λ1 to
λ0 by using the trial function y1 (x) = 1 − x2 .
(b) Show that an improved estimate λ2 is obtained by using y2 (x) = cos(πx/2).
(c) Prove that the estimate λ(γ) obtained by taking y1 (x) + γy2 (x) as the trial
function is
λ(γ) =
64/15 + 64γ/π − 384γ/π 3 + (π 2 /3 + 1/2)γ 2
.
16/15 + 64γ/π 3 + γ 2
Investigate λ(γ) numerically as γ is varied, or, more simply, show that
λ(−1.80) = 3.668, an improvement on both λ1 and λ2 .
799
CALCULUS OF VARIATIONS
22.23
For the boundary conditions given below, obtain a functional Λ(y) whose stationary values give the eigenvalues of the equation
(1 + x)
dy
d2 y
+ (2 + x)
+ λy = 0,
dx2
dx
y(0) = 0, y (2) = 0.
Derive an approximation to the lowest eigenvalue λ0 using the trial function
y(x) = xe−x/2 . For what value(s) of γ would
y(x) = xe−x/2 + β sin γx
22.24
be a suitable trial function for attempting to obtain an improved estimate of λ0 ?
This is an alternative approach to the example in section 22.8. Using the notation
of
section, the expectation value of the energy of the state ψ is given by
that
of H by ψi , so that Hψi = Ei ψi , and, since
ψ ∗ Hψ dv. Denote the eigenfunctions
H is self-adjoint (Hermitian), ψj∗ ψi dv = δij .
(a) By writing any function ψ as
cj ψj and following an argument similar to
that in section 22.7, show that
∗
ψ Hψ dv
E= ∗
≥ E0 ,
ψ ψ dv
the energy of the lowest state. This is the Rayleigh–Ritz principle.
(b) Using the same trial function as in section 22.8, ψ = exp(−αx2 ), show that
the same result is obtained.
22.25
22.26
This is an extension to section 22.8 and the previous question. With the groundstate (i.e. the lowest-energy) wavefunction as exp(−αx2 ), take as a trial function
the orthogonal wave function x2n+1 exp(−αx2 ), using the integer n as a variable
parameter. Use either Sturm–Liouville theory or the Rayleigh–Ritz principle to
show that the energy of the second lowest state of a quantum harmonic oscillator
is ≤ 3ω/2.
The Hamiltonian H for the hydrogen atom is
−
2 2
q2
∇ −
.
2m
4π0 r
For a spherically symmetric state, as may be assumed for the ground state, the
only relevant part of ∇2 is that involving differentiation with respect to r.
(a) Define the integrals Jn by
∞
Jn =
rn e−2βr dr
0
and∗ show that, for ∗a trial wavefunction of the form exp(−βr) with β > 0,
ψ Hψ dv and ψ ψ dv (see exercise 22.24(a)) can be expressed as aJ1 − bJ2
and cJ2 respectively, where a, b and c are factors which you should determine.
(b) Show that the estimate of E is minimised when β = mq 2 /(4π0 2 ).
(c) Hence find an upper limit for the ground-state energy of the hydrogen atom.
In fact, exp(−βr) is the correct form for the wavefunction and the limit gives
the actual value.
22.27
The upper and lower surfaces of a film of liquid, which has surface energy per
unit area (surface tension) γ and density ρ, have equations z = p(x) and z = q(x),
respectively. The film has a given volume V (per unit depth in the y-direction)
and lies in the region −L < x < L, with p(0) = q(0) = p(L) = q(L) = 0. The
800