Re: Time dilatation in circular motion

"El Enrrabadore-mor" (enrrabadore@xxxxxxxxxxxxxx) writes:
"Greg Neill" <gneillREM@xxxxxxxxxxxxxxx> escreveu na mensagem
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"El Enrrabadore-mor" <enrrabadore@xxxxxxxxxxxxxx> wrote in message
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"Tom Roberts" <tjroberts137@xxxxxxxxxxxxx> escreveu na mensagem
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\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):
\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?

Definition of proper time? Velocity of a particle in its rest frame?

[...]

--John Park
.

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