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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 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