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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
MTW, Exercise 8.5 (c), Crotchm reasserts his imbecility
Posted: Apr 13, 2013 11:05 AM

On Apr 13, 7:55 am, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 1:54 am, "Dono." <sa...@comcast.net> wrote:
>

> > 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).

>
> 1) ?? Didnt you see DvDM's reply to me? He agreed that what I put down
> is a solution
> to r r''' + 3 r'' r' = 0.
>
> Try it. You can do it by direct substitution!!  TRY IT !!
>

No imbecile, he didn't agree, he simply did not notice your utter
cretinism.

> 2) You seem to impose that r'' must equal zero. Why? It need not be.
>