On May 8, 12:34 pm, "El Enrrabadore-mor"
wrote:
"Greg Neill" escreveu na ting.com..."El Enrrabadore-mor" wrote in message





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

"Tom Roberts" escreveu na mensagem
. ..

\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.

For Greg's "v(t)" to be a constant he is using
a rotating spatial CS mixed with a Lorentz
Transform, something that is less than rigorous.
(BTW I use that myself, carefully).

It's a bit more complicated, (as El ref'd),
http://hermes.physics.adelaide.edu.a...vity/SR/rigid_...

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.

GR is the working tool,
SR provides an approximation.
I did a GR analysis in thread
"Speed of Gravity revisited" that shows
how to apply GR in rotating CS's.
(I can copy them to this thread if you want).

\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.

So what? A "perfect" solution needs a
painstaking evaluation of GR's geodesic
equations.
Regards
Ken S. Tucker