"Uncle Al" wrote in message
...


Hey cretin!

Let's see if you know any high school algebra.

Given:
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))

Need a reference, phuckwit? No need, you've quoted it yourself before,
stoooopid.

Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(c-v))

Taking out the t for 3:00pm on a Friday afternoon:

tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Synchronize clocks at t = 0, we remove tau(0,0,0,0)+

tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Taking coordinate x' as infinitessimally small, as he says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.

tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))

Removing the superflous coordinates, all zero:

tau(x'/(c-v)+x'/(c+v)) = 2 * tau(x'/(c-v))

Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity

tau(a+b) = 2*tau(a)

Renaming tau as f,

f(a+b) = 2f(a) or

½f(a+b) = f(a)

Now tell me that's a linear function, a b.

"In the first place it is clear that the equations must be linear
on account of the properties of homogeneity which we attribute to
space and time." -- Albert Phuckwit/Huckster Einstein.

In the second place tau is not a linear function. -- Androcles.

In the third place there are no coordinates to transform.

In the fourth place you've been had! (and not by me either)

Hahahahahahaha!!

Stoopid Schwartz is a phuckwit and she's been had!

Androcles.