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Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted:
Apr 9, 2013 2:39 AM


rotchm <rotchm@gmail.com> wrote: >> 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.
Yes, if you mean s(At2 + Bt + C ) = sqrt( A t^2 + B t + C ). Indeed, but unless At^2 + Bt + C is indeed a perfect square, the original equations don't produce a constant angle p(t), but probably some function involving an arctangent, which is not what MTW was looking for.
Dirk Vdm


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