 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'/(cv)+x'/(c+v))] = tau(x',0,0,t+x'/(cv)) (given) Doubling both sides: tau(0,0,0,t)+tau(0,0,0,t+x'/(cv)+x'/(c+v)) = 2 * tau(x',0,0,t+x'/(cv)) Taking out the t for 3:00pm on a Friday afternoon: tau(0,0,0,0)+tau(0,0,0,x'/(cv)+x'/(c+v)) = 2 * tau(x',0,0,x'/(cv)) Synchronize clocks at t = 0, we remove tau(0,0,0,0)+ tau(0,0,0,x'/(cv)+x'/(c+v)) = 2 * tau(x',0,0,x'/(cv)) 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'/(cv)+x'/(c+v)) = 2 * tau(0,0,0,x'/(cv)) Removing the superflous coordinates, all zero: tau(x'/(cv)+x'/(c+v)) = 2 * tau(x'/(cv)) Setting the time a = x'/(cv) 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. 

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