```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 askingabout 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 asdv/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)/dtTherefore, f = K/r^2, for some constant K.The first equation then becomes, dv/dt = K^2/r^3. The standard trickis 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^2for some constant v_0.You can take it on from here; solving for r as a function of t, andthen putting this into the equation for f (i.e. dp/dt) to get p as afunction of t.If you go back to the original problem, the following properties holdtrue. The gravitational source (which I assume is the Schwarzschildsolution) has rotational symmetry and time translation symmetry. Thiscorresponds (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 thegeodesic law are to extract out the integrals for angular momentum andenergy. The angular momentum part was already removed from the problemby the time you brought the matter here in sci.math, so that left theenergy integral to take care of. That was the missing step.
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