"El Enrrabadore-mor" wrote in message

"Greg Neill" escreveu na mensagem
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"El Enrrabadore-mor" wrote in message

"Tom Roberts" 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

Right, where v(t) is some function of time.


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

I don't think you can get away with this unless v(t) is
a linear function of t or constant.


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

The original problem claimed that v = 0.999c = c
so that (for v = c):

v cannot equal c, at least not in our universe.

If you're going to assume a constant v(t) then sqrt(1 - (v/c)^2)
is also a constant. So let k = sqrt(1 - (v/c)^2) and you've
got:

\tau = \integral_T1^T2 k dt

\tau = _T1^T2 k*t

\tau = (T2 - T1)*k = DT*k where k = sqrt(1 - (v/c)^2)

which is the usual Lorentz equation for time dilation
(as expected for a body moving with constant velocity
with respect to an inertial frame).


\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

1. Dubious evaluation of an integral
2. Assuming a velocity equal to the speed of light