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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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 Hetware Posts: 148 Registered: 4/13/13
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 8, 2013 11:21 PM

On 4/8/2013 8:18 PM, Alfred Einstead wrote:
> 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.

I don't fully agree with that view. The context of a problem often
suggests a particular approach.

Clearly, what I posted as my effort to solve the problem suggested /an/
approach.

> 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.

I've typed all of the above into Mathematica for typesetting. I have a
hard time following math in ASCII. I appreciate your help. It did get
me further along than I was. Differential equations is not a strong
suit of mine. Unfortunately, I can't spend any more time on it tonight.

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

The exercise is titled "A sheet of paper in polar coordinates."