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Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted:
Apr 8, 2013 9:08 PM


> 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]
Yes, thats a "trivial" solution and as you hinted, there are others. The general solution to this is r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
I do like Alfred's elegant approach!


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