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Topic:
Misner, Thorne and Wheeler, Exercise 8.5 (c)
Replies:
38
Last Post:
Apr 13, 2013 11:57 PM




Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted:
Apr 8, 2013 12:17 PM


"Dirk Van de moortel" wrote in message news:kjugkr$ns1$1@speranza.aioe.org...
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
====================================== Pulled sin and cos out of your arse, Dork?
For A = 0, B = 1, C =(0 to 2pi) you've drawn a circle. Is that the "probably another solution" you were looking for?
exp(i.t) = cos(t) + i.sin(t)  Euler.
> Indeed, writing LT and inverse: > x' = g ( x  v t ) [1] > t' = g ( t  v x ) [2] > x = g ( x' + v t' ) [3] > t = g ( t' + v x' ) [4]  A fucking idiot called Dork.
 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.


Date

Subject

Author

4/7/13


Guest

4/7/13


Lord Androcles, Zeroth Earl of Medway

4/8/13


Lord Androcles, Zeroth Earl of Medway

4/8/13


Hetware

4/8/13


Lord Androcles, Zeroth Earl of Medway

4/8/13


Hetware

4/9/13


Lord Androcles, Zeroth Earl of Medway

4/9/13


Hetware

4/8/13


Hetware

4/8/13


Lord Androcles, Zeroth Earl of Medway

4/9/13


Hetware

4/9/13


Lord Androcles, Zeroth Earl of Medway

4/9/13


Hetware

4/9/13


Lord Androcles, Zeroth Earl of Medway

4/9/13


Hetware

4/8/13


Dirk Van de moortel

4/8/13


Lord Androcles, Zeroth Earl of Medway

4/8/13


rotchm@gmail.com

4/9/13


Dirk Van de moortel

4/13/13


Dono

4/13/13


rotchm@gmail.com

4/13/13


Dono

4/13/13


Dono

4/13/13


Dono

4/13/13


rotchm@gmail.com

4/13/13


Dono

4/13/13


rotchm@gmail.com

4/13/13


Dono

4/13/13


Dono

4/13/13


rotchm@gmail.com

4/13/13


Dono

4/13/13


rotchm@gmail.com

4/13/13


Dono

4/9/13


Guest

4/9/13


Dirk Van de moortel

4/9/13


Lord Androcles, Zeroth Earl of Medway

4/8/13


Rock Brentwood

4/8/13


Hetware

4/8/13


Lord Androcles, Zeroth Earl of Medway


