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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 7, 2013 11:34 PM

"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.

-- 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.