Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Replies: 38   Last Post: Apr 13, 2013 11:57 PM

 Messages: [ Previous | Next ]
 rotchm@gmail.com Posts: 58 Registered: 5/27/07
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 13, 2013 10:55 AM

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 !!

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

3) Now, you seem to have also misinterpreted (your incapacity to
deduce) the notation of s(...). DvDM did understand it. You didnt read
those who replied to me? That is quite sloppy on your part.

4) Try again.