Re: Circular motion in SR
- From: "Pmb" <someone@xxxxxxxxxxxxx>
- Date: Sat, 15 Mar 2008 23:25:41 -0400
<ram.rachum@xxxxxxxxx> wrote in message
news:f22b083b-cfca-4de4-80a7-e35304163c2e@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 14, 5:23 pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:
<ram.rac...@xxxxxxxxx> wrote in message
news:2d2a9be2-d193-41c3-a979-f12d68ea7b0a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 14, 5:06 pm, Dono <sa...@xxxxxxxxxxx> wrote:
On Mar 14, 7:59 am, "Pmb" <some...@xxxxxxxxxxxxx> wrote:
I didn't say that it was. It is a matter of fact that the force is
perpendicular to the velocity.
No, it isn't. I just showed you :
F.v=qv.E
This is a statement of the problem the OP
asked.
No, the OP asked the solution for circular motion in relativity.
That's right, the solution should be an expression for v in terms of r
and F, given that the motion is circular. Express "a" as well, while
you're at it. After I'll get the final solution, I'll try to
understand what I did wrong...
I read the Cyclotron page, but it is quite confusing.
I'm rewriting it for you so that it will be easier to read. I'll try to
have
it upp as soon as possible for you.
Pete
Thanks a bundle, Pete!
Okay. I finished it earlier than I expected. Take another look at
http://www.geocities.com/physics_world/sr/cyclotron.htm
More details are included. You'll also notice that there is no difference
between the nonrelativistic derivation and the relativistic one!! :)
Regarding your particular problem, with this example you should be able to
solve it yourself now.
Best wishes
Pete
.
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