"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
t...
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
1 - Your solution:
\tau = sqrt(1-|v|^2/c^2)

You're not paying attention, and you are MISQUOTING. I said "For
circular motion, v(t) is constant," (TYPO: I of course meant |v(t)|
is constant.) \tau = sqrt(1-|v|^2/c2) (T2 - T1)

Just MISquoting the formula is not sufficient, you must also read the
words that give the conditions under which the formula is valid.

I didn't misquote your formula by propose.
I just didn't realise that you've simplified that much the problem.
Such simplification makes your solution ridiculous, that's all.
So, I never realise that one could make such ridiculous
simplification: What you say is that:
1 - The velocity v is constant.
2 - Hence, the sqrt(1-(v/c)^2) is a constant term K.
3 - Therefore, \integral K = K t

Without such ridiculous simplification, to include "t" or not doesn't
make any difference, since it is obvious that dt/dt = 1.
Because dt/dt = 1 is so obvious, I didn't care about it.
Hence, I didn't misquote your formula by propose.

If you want to consider the general case where v(t) does
not have a constant magnitude, then you must integrate
over the actual function. This is obviously more
difficult that assuming constant speed (either linear or
circular motion). In fact, for most functions v(t) there
will not be an easily obtained closed form solution, or
perhaps no closed form solution. In such cases one
usually resorts to numerical integration or an approximation
method.

When you did the integral you made the error of assuming
that dv/dt = dt, which it won't be in general.

If you have an integral such as \Integral _a^b U(t)dt,
you cannot assume that dU(t) = t. You need to have
U(t)du.





[...]
Can't see what's wrong.

What is wrong is that it is not possible to perform that integral
without knowing the functional form of v(t). When |v(t)| is
constant, the integral is trivial, and has the value given above.
When |v(t)| is not constant, you must know its functional form in
order to do the integral.

That's the ridiculous part of relativity approach to the
rigid disk.

Absurd. You're miffed because the math for the general case
might be hard.

When |v(t)| is constant, it doesn't matter if it's
circular motion or linear motion, you always got Lorentz equation.

Indeed. What a pleasing result! Wouldn't it be more
strange if, despite the obvious analogs between the
mathematical physics of linear and circular motion
(v ~ w, T ~ F, L ~ momentum, etc., and the functional forms
of rotational and linear energy), that relativity would
predict different forms for the time dilation effect?


To compare, or to say, that circular motion is the same
thing as linear motion, sounds ridiculous to me.

They are clearly not the same thing, as rotational and
linear momenta are separately conserved. Yet the
mathematical form for dealing with them run along parallel
lines.


I'm about to say that, relative to the XXI Century, relativity
theory is in its Stone Age. ...and people stoned with it.

Uh oh.