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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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 Dono Posts: 368 Registered: 11/6/06
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 13, 2013 1:54 AM

On Apr 8, 6:08 pm, rotchm <rot...@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.
>

No, pretentious imbecile, what you put down is not a solution since r"
isn't null for all t (while r"'=0). You are not only a pretender, you
are an outright idiot. And to think that the other french idiot, YBM,
rushes to suck your cock and lick your ass every time you post one of