### An immortal fumble by Androcles (29-Jul-2004)

##### Sheesh... if x' is already zero, how can you have a small part of it?
 ```| > 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.``` Fumble Index Original post & context: gHXNc.10505\$EZ7.105817118@news-text.cableinet.net