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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 8:53 PM

On 4/7/2013 11:34 PM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> news:AuGdnVu7TbVslv_MnZ2dnUVZ_g6dnZ2d@megapath.net...
>
> 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).
>
> r is radius in polar coordinates, p is the angle, and t is a path
> parameter.
>
> The authors ask me to "[S]olve the geodesic equation for r(t) and p(t),
> and show that the solution is a uniformly parametrized straight
> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for some
> j and k).
>
> I tried the following:
>
> (d^2(p)/dt^2)/(dp/dt) = -(2/r)(dr/dt)
>
> f=dp/dt
>
> (df/dt)/f = -(2/r)(dr/dt)
>
> -1/2 ln(f) + k = ln(r)
>
> a(f^(1/2)) = r
>
> a(dp/dt)^(1/2) = r
>
> And substitute for r in:
>
> d^2(r)/dt^2 = r(dp/dt)^2
>
> to get
>
> d^2(r)/dt^2 = a(dp/dt)^(3/2)
>
> But there I'm stuck.
>
> How should the problem be handled?
> =============================================
> What you have is a second order differential equation.
> Unlike the solution to the general polynomial equation,
> ax +bx^2 + cx^3 + ... + kx^n = 0, where you seek a value
> for x given values for a,b,c etc., the solution for a
> differential equation is a FUNCTION.
> In other words you cannot obtain a numerical or algebraic
> value (you don't have enough information and that is not
> the idea anyway) but you can find functions r(t) and p(t) .
> The authors have already told you the solution is a straight
> line, which is of course a function.
> http://search.snap.do/?q=solving+differential+equations&category=Web
> HTH, because we don't do homework for you.

That was the kick in the head that I needed.

I was getting close. I remembered what Wheeler told me: "Anytime you
are struggling to understand or explain something, draw a picture."

http://www.speakeasy.org/~hattons/chapter08.nb

I worked the problem in pencil today while away from my computer. Just
use p=arctan(t/r_0) where r_0 is a radial vector perpendicular to the
"curve". r=(r_0^2+t^2)=r_0/cos(p). The rest is just a matter of
plugging these into the geodesic equation.