### An immortal fumble by Androcles (17-Aug-2005)

##### Androcles takes a limit
 ```| It's a possibility but I still can prove that Einstein's 1905 paper has | no mistakes in it. No you cannot. I can prove it does, and will now do so. Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ 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)) (given) 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 Einstein 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) | One might object to it on philosophical grounds but | not on mathematical ones (its mathematics is very easy, BTW, this is | not where the difficulty with this paper lies, and this is not where | you'll ever find anything wrong). | | -- | Jan Bielawski And I did. Androcles. ``` Fumble Index Original post & context: VFxMe.8150\$Wq4.5620@fe1.news.blueyonder.co.uk