# Problem Set #5 Solutions

## Problem 1. Symmetries of the 3 Pseudosphere.

### (i)

First we show that the symmetries of the 4 dimensional spacetime in which
the pseudosphere is embedded include things like

R = x d/dy - y d/dx

X = x d/dt + t d/dx
To be a symmetry, a vectorfield must satisfy Killing's equation:

g_mn,a k^a + g_an k^a_,m + g_ma k^a_,n = 0,

where the underline indicates subscrits following, the caret superscripts
following, and the comma indicates partial derivatives. We verify this equation
for the vectorfield X, with spacetime metric

g_tt = -1; g_xx = +1, etc.

Since all components of the metric are constant, the first term vanishes.
Furthermore, only x and t dependence in the Killing vector, so we need consider
only three equations:

mn = tt:

g_at k^a,t + g_ta k^a,t,

since g is diagonal, this forces the dummy variable in the sum to also be
t, and this is solved identically. Ditto for the xx equation.

mn = xt:

g_at k^a,_x + g_xa k^a,_t = g_tt k^t,_x + g_xx k^x_,t = 0.

### (ii)

Now the problem asks for you to "pull back" the vectors from the 4-D spacetime.
This is not possible in general, hence the quotes. One can only do this if
one can find a unique solution v to the pushforward equation:

\psi_* v = X,

and so on. Since the pushforward of any curve will lie in the subspace, the
pushforward of the vector will be tangent to the subspace. Thus we have to
verify that a vectorfield like X is tangent to the subspace:

t^2 - x^2 - y^2 - z^2 - 1 = 0.
If you think of the vectorfield as a directional derivative, then we just have
to show that the derive of the above function in the X direction is zero.
Easy.

The actual computation of the vectorfield X in the \psi, \theta, \phi
coordinates is quite a mess.

### (iii)

This is straightforward. The metric on the 3-Manifold is pulled back from
that of spacetime. This is equivalent to saying that the distance between
points, and the lengths of vectors, can be computed in either space. Hence
a symmetry of one space is a symmetry of the other.
Because the symmetries are so simple in 4-space and so messy in 3-space, one
might find it advantageous in computations to represent points in the 3-space
with their four dimensional coordinates.

## Problem 2. The double pendulum.

### (i)

The manifold that describes the space where the pendula can both swing in full
circles is a torus.
### (ii)

Both angles will run 0 to 2 Pi.
### (iii)

Potential energy due to gravity will be proportional to the vertical position
of each bob. Lifting the top bob also lifts the bottom bob, hence the factor
of two:

V = mg ( 1 - 2 cos u - cos v )

### (iv)

F sin u

### (v)

As the coordinates are defined, moving u also moves the bottom bob. Thus

F (sin u + sin v)
### (vi)

To find the equilibrium, we also have to include the constraint forces of the
pendulum rods. To be continued.