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Topic: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Replies: 38   Last Post: Apr 13, 2013 11:57 PM

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 Rock Brentwood Posts: 129 Registered: 6/18/10
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 8, 2013 8:18 PM

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.