Physics 31600 R. Wald Classical Mechanics Autumn, 2002

Problem Set VI Solutions

1) Consider a space M with coordinates y^{μ} and define the Poisson-bracket-like map on functions on M by

W (f, g) = ∑_{μ,ν} W^{μν} \frac{∂f}{∂y^{μ}} \frac{∂g}{∂y^{ν}}

where W^{μν} is any antisymmetric matrix (i.e., W^{νμ} = −W^{μν}) whose components are constant (i.e., independent of y^{μ}). Show that W satisfies the Jacobi identity, i.e.,

W (W (f, g), h) + W (W (h, f), g) + W (W (g, h), f) = 0

Solution: Summation will be implied over repeated indices. First:

W (W (f, g), h) = W^{μν}W^{σρ} \frac{∂}{∂y^{μ}} (\frac{∂f}{∂y^{σ}} \frac{∂g}{∂y^{ρ}})\frac{∂h}{∂y^{ν}}  (1)

So:

W (W (f, g), h) + W (W (h, f), g) + W (W (g, h), f) =

W^{μν}W^{σρ}[\frac{∂}{∂y^{μ}} (\frac{∂f}{∂y^{σ}} \frac{∂g}{∂y^{ρ}})\frac{∂h}{∂y^{ν}} +

\frac{∂}{∂y^{μ}} (\frac{∂h}{∂y^{σ}} \frac{∂f}{∂y^{ρ}})\frac{∂g}{∂y^{ν}} +

\frac{∂}{∂y^{μ}} (\frac{∂g}{∂y^{σ}} \frac{∂h}{∂y^{ρ}})\frac{∂f}{∂y^{ν}}]  (2)

Consider the terms only involving the second derivative of f:

W^{μν}W^{σρ}[\frac{∂^{2}f}{∂y^{μ}∂y^{σ}} \frac{∂g}{∂y^{ρ}} \frac{∂h}{∂y^{ν}} + \frac{∂h}{∂y^{σ}} \frac{∂^{2}f}{∂y^{μ}∂y^{ρ}} \frac{∂g}{∂y^{ν}}]  (3)

Because of the antisymmetry of the W^{μν}, we can switch σ and ρ in the first term at the cost of a minus sign (really we are replacing W^{σρ} with −W^{ρσ} and then relabeling the indices):

W^{μν}W^{σρ}[−\frac{∂^{2}f}{∂y^{μ}∂y^{ρ}} \frac{∂g}{∂y^{σ}} \frac{∂h}{∂y^{ν}} + \frac{∂h}{∂y^{σ}} \frac{∂^{2}f}{∂y^{μ}∂y^{ρ}} \frac{∂g}{∂y^{ν}}]  (4)

The term in brackets is now symmetric in μ and ρ and antisymmetric in σ and ν. Since we have

W^{μν}W^{σρ} → W^{ρσ}W^{νμ} = (−W^{σρ})(−W^{μν}) = W^{μν}W^{σρ}  (5)

all the terms involving the second derivative of f must be zero. The same applies to the second derivatives of g and h because those terms involve cyclic permutations of f, g, and h. Therefore

W (W (f, g), h) + W (W (h, f), g) + W (W (g, h), f) = 0  (6)

QED

2) Let f(q, p; t) and g(q, p; t) be (possibly time-dependent) observables on a 2ndimensional phase space with Hamiltonian H(q, p; t).

(a) Suppose that both f and g are constants of motion, i.e., df/dt = dg/dt = 0. Show that Ω(f, g) also is a constant of motion.

Solution: In general, we have:

\frac{df}{dt} = Ω(f, H) + \frac{∂f}{∂t}  (7)

where H is the Hamiltonian. So we wish to calculate:

\frac{dΩ(f, g)}{dt} = Ω(Ω(f, g), H) + \frac{∂Ω(f, g)}{∂t}  (8)

Using the Jacobi Identity, we have:

\frac{dΩ(f, g)}{dt} = −Ω(Ω(H, f), g) − Ω(Ω(g, H), f) + \frac{∂Ω(f, g)}{∂t}  (9)

But from Equation 7 and by hypothesis we have:

\frac{df}{dt} = Ω(f, H) + \frac{∂f}{∂t} = 0 ⇒ Ω(H, f) = \frac{∂f}{∂t}  (10)
\frac{dg}{dt} = Ω(g, H) + \frac{∂g}{∂t} = 0 ⇒ Ω(g, H) = −\frac{∂g}{∂t}  (11)

(Remember that Ω(f, g) is antisymmetric in f and g). So:

\frac{dΩ(f, g)}{dt} = −Ω(\frac{∂f}{∂t} , g) + Ω(\frac{∂g}{∂t} , f) + \frac{∂Ω(f, g)}{∂t}  (12)

However, because Ω is bilinear and antisymmetric in f and g, we have:

\frac{dΩ(f, g)}{dt} = −\frac{∂}{∂t}Ω(f, g) + \frac{∂Ω(f, g)}{∂t} = 0  (13)

QED

(b) Suppose that H itself is a constant of motion and f(q, p; t) is a constant of motion. Show that \frac{∂^{n}f}{∂t^{n}} also is a constant of motion.

Solution: From Equation 7, we have:

\frac{d}{dt} \frac{∂^{n}f}{∂t^{n}} = Ω(\frac{∂^{n}f}{∂t^{n}} , H) + \frac{∂}{∂t} \frac{∂^{n}f}{∂t^{n}}  (14)
\frac{dH}{dt} = Ω(H, H) + \frac{∂H}{∂t} = \frac{∂H}{∂t}  (15)

So

\frac{∂H}{∂t} = 0  (16)

since H is a constant of the motion. Equation 16 in turn implies:

\frac{∂}{∂t}Ω(f, H) = Ω(\frac{∂f}{∂t} , H) + Ω(f, \frac{∂H}{∂t}) = Ω(\frac{∂f}{∂t} , H)  (17)

Thus:

\frac{d}{dt} \frac{∂^{n}f}{∂t^{n}} = \frac{∂^{n}}{∂t^{n}} Ω(f, H) + \frac{∂^{n}}{∂t^{n}} \frac{∂f}{∂t} = \frac{∂^{n}}{∂t^{n}} \frac{df}{dt} = 0  (18)

QED

3) Consider a particle of mass m in ordinary 3-dimensional space. Let L_{x}, L_{y}, L_{z} denote the usual Cartesian components of angular momentum of the particle, viewed as functions on its 6-dimensional phase space.

(a) Show that Ω(L_{x}, L_{y}) = L_{z}. (Thus, according to problem 2, if L_{x} and L_{y} are constants of motion, then so is L_{z}.)

Solution: The angular momenta are defined by:

L_{i} = (r⃗ × p⃗)_{i} = x_{j}p_{k} − x_{k}p_{j}  (19)

for i, j, and k cyclic permutations of x, y, and z. So

Ω(L_{x}, L_{y}) = ∑_{i} \frac{∂L_{x}}{∂x_{i}} \frac{∂L_{y}}{∂p_{i}} − \frac{∂L_{x}}{∂p_{i}} \frac{∂L_{y}}{∂x_{i}}  (20)

From Equation 19, the L_{i} do not contain either x_{i} or p_{i}, therefore:

Ω(L_{x}, L_{y}) = \frac{∂L_{x}}{∂z} \frac{∂L_{y}}{∂p_{z}} − \frac{∂L_{x}}{∂p_{z}} \frac{∂L_{y}}{∂z} = (−p_{y})(−x) − (y)(p_{x}) = L_{z}  (21)

QED (b) Suppose that an observable f(x⃗, p⃗) depends on x⃗ and p⃗ only in the scalar combinations x⃗ · x⃗, x⃗ · p⃗, and p⃗ · p⃗, i.e., suppose that f can be written in the form

f(x⃗, p⃗) = h(x⃗ · x⃗, x⃗ · p⃗, p⃗ · p⃗)

for some function h. Show that Ω(f, L_{i}) = 0.

Solution: The easy way to do this problem is to note that the L_{i} are the generating functions of rotations about the x_{i} axis. Then we have:

\frac{df(x⃗, p⃗)}{dθ_{i}} = Ω(f(x⃗, p⃗), L_{i})  (22)

where θ_{i} is the angle of rotation about the x_{i} axis. From Problem 1 of Homework II we know that if we apply a rotation to the configuration variables we must apply the same rotation to the momentum variables. Therefore the combinations x⃗ · x⃗, x⃗ · p⃗, and p⃗ · p⃗ are all invariant under rotations about any axis. This implies that

Ω(f(x⃗, p⃗), L_{i}) = \frac{df(x⃗, p⃗)}{dθ_{i}} = 0  (23)

QED