Re: Circular motion in SR

On Mar 15, 7:25 pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:

More details are included. You'll also notice that there is no difference
between the nonrelativistic derivation and the relativistic one!! :)


Of course there isn't, this is what I told you all along: if you
insist on having E=0 you factor out \gamma so you get the Newtonian
solution. This is why I told you that the solution that you keep
pushing is irrelevant.

Now, try E<>0. If you are able to solve the problem, you will find out
that the relativistic solution is indeed different from the Newtonian
solution. For that, you will need to understand that d(\gamma)/dt is
NOT 0.
.