| > Yes, he's right and you are wrong.
| > Reference :
| >     http://www.fourmilab.ch/etexts/einstein/specrel/www/
| > "But the ray moves relatively to the initial point of k, when measured
| > in the stationary system, with the velocity c-v, so that x'/(c-v) = t."
| > Since x' is defined as x'=x-vt, and v = x/t, it follows that x' = x-x = 0,
| > and hence t = 0.
| > Einstein has a divide by zero in
| >
| >     1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt
| >
| > at dtau/dx' and the Lorentz transforms cannot be derived.
|
|
| And just where is there a division by zero here?

At the term dtau/dx', because x' = 0.
 Sheesh... if x' is already zero, how can you have a small part of it?
f'(x) = [f(x+h) - f(x)]/h as h tends to zero. You can't do that if h is
already zero!

Androcles.