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Old May 8th 08 posted to sci.physics.relativity,sci.physics
Tom Roberts
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Default Time dilatation in circular motion

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
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El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
Do you have an equation, dimensionally consistent, where one
could see what you are talking about?

Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame, and
t is the time coordinate of the frame); v(t) can be an arbitrary function
of time. Its elapsed proper time between t=T1 and t=T2 is:

\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"?


T1 is the lower limit and T2 is the upper limit of the \integral.

Around here, "_" introduces a subscript and "^" introduces
a superscript (which can be a power).


Meanwhile:
integral_sqrt(1-v(t)^2/c^2) dt =
= 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)


That's hopeless and wrong -- you cannot begin to do the integral until
v(t) is specified. For circular motion, v(t) is constant, the integral
is trivial:

\tau = sqrt(1-|v|^2/c^2) (T2 - T1)

Note that there is no term related to acceleration, and no term related
to "centrifugal force"; all that matters is SPEED (|v|) relative to the
inertial frame.


Tom Roberts
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