"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
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.[...]
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

This cannot possibly be correct. Differentiate your expression by t
and you must get a factor of v'(t) (i.e. dv(t)/dt), but there is no
such factor in the original integral.

One thing is for sure. The derivative of the solution must
give the original function (inside the integral).

So let's do the derivative of your solution (and mine)
and see who got the correct solution.

1 - Your solution:
\tau = sqrt(1-|v|^2/c^2)

No, it's \tau = sqrt(1-|v|^2/c^2)*t

where t is the time in the observer frame.
Note that here the sqrt() term is a constant since
you chose a fixed velocity, v.


2 - My solution:
\tau = 1/2*(v/c)*sqrt(1-v^2/c^2) + 1/2*arcsin(v/c)

The original integral:
\tau = \integral sqrt(1-v(t)^2/c^2) dt

If you are to be correct, then:
d/dt [sqrt(1-|v|^2/c^2)] = sqrt(1-|v|^2/c^2)
which obviously looks wrong.

Because it should be:

d/dt [sqrt(1-|v|^2/c^2)]*t = sqrt(1-|v|^2/c^2)