### Sixth Assignment Solutions

2. (problem 13.12*) Let r be the bead's position along the rod. The PE is zero, so L = 1/2m(r·2 + r2ω2). Note that ω is not a dynamical variable, it is fixed by the constraint. Therefore
H = pr· - L = /2m – ½mr²ω²
This is not the total KE, which is /2m + ½mr²ω².

3. (Problem 13.23 ***)

(a) The gravitational PE is Ugr = Mgy - mgy - mg(x+y) + const = -mgx if we drop the uninteresting constant. The spring PE is harder. If we let lo denote the spring's natural, unloaded length, then k(le - lo) = mg and if x¢ denotes the spring's true extension (from its unloaded length), then lo+ x¢ = le + x so

 x¢ = x + (le - lo) = x + mg k
Thus the spring PE is

 Uspr = 1/2k x¢2 = 1/2k æè x + mg k öø 2 = 1/2k x2 + mgx + const.
If we add this to Ugr = -mgx, the terms in mgx cancel and (dropping another uninteresting constant) we get U = Ugr + Uspr = 1/2k x2 as claimed.
(b) The KE is T = 1/2M y·2 + 1/2m y·2 + 1/2m (x· + y·)2 = 1/2m [3y·2 + (x· + y·)2], from which we find the momenta,

px = T

 ¶ × x
= m(
×
x

+
×
y

)      and     py = T

 ¶ × y
= m(
×
x

+ 4
×
y

)
whence

 × x + × y = px m and × y = 1 3m (py - px).
From these we can calculate the Hamiltonian,

 H= T + U = 1 2m éë (px-py)2 3 + px  2 ùû + 1 2 kx2.
Because this doesn't depend on y, the coordinate y is ignorable. This is traceable to the fact that the total mass on each side is the same.
(c) The Hamilton equations for x are

 × x = ¶H ¶px = 1 3m (4px - py)      and × p x = - ¶H ¶x = -kx
(1)
and those for y

 × y = ¶H ¶py = 1 3m (py - px)      and × p y = - ¶H ¶y = 0.
(2)
The initial conditions are that x(0) = xo, y(0) = yo, and x·(0) = y·(0) = 0. These imply that px(0) = py(0) = 0, and, because py is constant, py = 0 for all time. Combining the two equations (1) and setting py = 0, we find that x·· = 4p·x/3m = -4kx/3m. Therefore x = xocoswt, where w = Ö{4k/3m}. Next, from the first of Eqs.(1) (with py = 0) we find that px = [3/4]m x· = - [3/4]mwxosinwt and finally, from the first of Eqs.(2), y· = -px/3m = [1/4]wxosinwt, so y = - [1/4] xocoswt + const = yo+ [1/4] xo(1-coswt).

4. (problem 13.25***)
(a) For example,
 ¶H ¶Q = ¶H ¶q ¶q ¶Q + ¶H ¶p ¶p ¶Q = - × p ( Ö 2P cos Q) - × q ( Ö 2P sin Q)

 = -( × p p + × q q) = -1/2 d/dt(p2 + q2) = - × P
The other Hamilton equation is proved in a similar way.

(b) obvious

(c) H = 1/2(p2 + q2) = 1/2(2P cos2Q + 2P sin2Q) = P. So Q is ignorable.

(d) Q· = H/P = 1 so Q = t - t0. Put into the problem's first equation with P = E to get q = (2E)½ sin(t-t0) which is correct for this SHO.

5.
(a) This is something you grind out. It simplifies a little if you think of the Poisson bracket as the skew product of two 4D gradients, namely (in the order x, y, px, py)

ÑL = (py, -px, -y, x)     and     ÑH = (x, y, px, py)

The skew product says to multiply the first two components of the first gradient with the last two component of the second gradient, and vice versa with opposite sign. So you get
[L,H] = pypx - pxpy + yx - xy = 0.

(b)
dx/dt = -y     dy/dt = x     dpx/dt = -py     dpy/dt = px.
This is solved by
x = r cos t     y = r sin t     px = p cos t     py = p sin t
which is a rotation by an angle t.

(c) By integrating each equation, for example
py = G/px ® G = pxpy + f(x, y, py)
you find a consistent solution of all four equations G = pxpy + xy.
To verify by substitution: x = A cos t, y = 0 gives trivially G = 0. More interesting is, say, x = A cos t, y = B sin t. From x· = H/px = px etc we find that then px = -A sin t,  py = B cos t and G = AB sin t cos t - AB cos t sin t = 0.

(d) We have

H + G = 1/2[(x+y)2 + (px+py)2] = 1/2[S2 + S·2] = E+A
which is solved by S = [2(E+A)]½ sin(t-t1). Similarly we find D = [2(E-A)]½ sin(t-t2). Solving for x and y gives the general solution for the two oscillators, with four constants of integration, E±A, t1,2.