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Old May 8th 08 posted to sci.physics.relativity,sci.physics
Greg Neill
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Posts: 1,680
Default Time dilatation in circular motion

"El Enrrabadore-mor" wrote in message


Wait, now that I'm thinking on the derivatives, I notice
that I've screwed the integral solution.
I have some square terms which shouldn't be squared.
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)


[snip]


The derivative of *arcsin(u)* is *1/sqrt(1-u^2)*du/dt
and the integral is perfect.


So where is your dv in the original integral? You've
only got dt.

Integral _T1^T2 sqrt(1 - (v(t)/c)^2) dt

If you use a form such as Integral _a^b u*du then you
better make sure that you have that du in place.
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