"Greg Neill" escreveu na mensagem
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"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.
The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.
I've been doing those kind of derivatives a few time
ago and I know that *arcsin(u)" and *sqrt(1-u^2)* cancel
each other out easy in the process.
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)
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.
I was just testing the limits of the solution/example.
In practice v can be only a very, very, small fraction of c.
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).
Even so, my experience tell me that one shouldn't
ignore the full solution of the problem.
One must not assume simplifications to avoid
surprises.
There's an *arcsin(v/c)* function, a *v/c* function
and the whole solution is times *1/2*.
In general yes. The solution is of the form of
the usual Lorentz equation.
Lorentz equation is trivial for circular motion,
gyroscopes, and whenever Pitagoras Theorem
is a good working tool - electricity, and so on.
Gamma and Lorentz equation are trivial for
everything that cycles, circles, and so on.
\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
The derivative of *arcsin(u)* is *1/sqrt(1-u^2)*du/dt
and the integral is perfect.