Date: Apr 8, 2013 8:18 PM
Author: Rock Brentwood
Subject: Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
On Apr 7, 7:14 pm, Hetware <hatt...@speakyeasy.net> wrote:

> This is the geodesic equation under discussion:

> d^2(r)/dt^2 = r(dp/dt)^2

> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).

Where the problem comes from is not important, since all you're asking

about is how this is solved.

Notice that it's independent of p and depends only on do/dt. So,

define v = dr/dt, f = dp/dt and write the system as

dv/dt = rf^2, df/dt = -2fv/r (along with dr/dt = v, dp/dt = f).

The second equation can be rewritten as

0 = 1/f df/dt + 2/r dr/dt = 1/(f r^2) d(f r^2)/dt

Therefore, f = K/r^2, for some constant K.

The first equation then becomes, dv/dt = K^2/r^3. The standard trick

is to turn this into a conservation of energy integral:

v dv/dt = K^2/r^3 v = K^2/r^3 dr/dt,

from which it follows that

d/dt (v^2/2) = d/dt (-K^2/2 1/r^2).

The solution is

v^2 = v_0^2 - K^2/r^2

for some constant v_0.

You can take it on from here; solving for r as a function of t, and

then putting this into the equation for f (i.e. dp/dt) to get p as a

function of t.

If you go back to the original problem, the following properties hold

true. The gravitational source (which I assume is the Schwarzschild

solution) has rotational symmetry and time translation symmetry. This

corresponds (by way of the Noether Theorem) to conserved quantities:

Angular Momentum for rotational symmetry and Energy for time-

translation symmetry. Therefore, the first things to look for in the

geodesic law are to extract out the integrals for angular momentum and

energy. The angular momentum part was already removed from the problem

by the time you brought the matter here in sci.math, so that left the

energy integral to take care of. That was the missing step.