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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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 Lord Androcles, Zeroth Earl of Medway Posts: 19 Registered: 4/13/13
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 8, 2013 10:03 PM

"Hetware" wrote in message
news:CeSdneS1Wf0B-_7MnZ2dnUVZ_vGdnZ2d@megapath.net...

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.
==========================================
Good. Note that what you are doing is strictly NOT anything
to do with relativity or gravitation or even physics, it is just
math.
Pity Einstein didn't work the examples or he'd have made fewer
blunders like this one:
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When the fools chicken farmer Wilson and Van de faggot present an argument I
cannot laugh at I'll retire from usenet.