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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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 Dirk Van de moortel Posts: 157 Registered: 12/6/11
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 8, 2013 9:27 AM
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Hetware <hattons@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).
>
> 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).

Normally we'd write dotted variables, but with quotes it's easier.
So write
r' = dr/dt
r'' = d^2(r)/dt^2
p = dp/dt
p'' = d^2(p)/dt^2
then you have
r'' = r (p')^2 [1]
p'' = -2/r p' r' [2]

Deriving [1] gives
r''' = r' (p')^2 + 2 r p' p''
which with [2] gives
r''' = -3 r' (p')^2
which again with [1] gives
r r''' + 3 r'' r' = 0

So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
you're done. Easy. Take
r(t) = A t + B [3]
then
r'(t) = A
r''(t) = 0
r''"(t) = 0
so indeed
r r''' + 3 r'' r' = 0

Now, from [1] and [3] you get
0 = ( A t + B ) (p')^2
so
p' = 0
so
p(t) = C [4]

So you get
r(t) = A t + B
p(t) = C
Check it out with [1] and [2]. Trivial.

So
x(t) = (A t + B ) cos(C)
y(t) = (A t + B ) sin(C)

There's probably another solution, but seems to be the one they're
after.

Dirk Vdm

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