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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.


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


