since t'' is determined then t'= t'' + d(y,z)/v Note that t'' ---> 0 as v--> 0 and this is the reason why guys incorrectly interpreted the result as rigid body will diliver information instaneously! However when v --> 0, the second term blows up. If somebody want to receive morse signal at the other end of the rod, he MUST detect displacement which is d(y,z). Hence sending v --> 0 means nothing! To measure the signal sent, the end of the rod y must move AT least d(y,z) >= minimum theoritical mesurement error perhaps Plank length? Therefore the second term is very important! Now you can show t' >= d(x,y)/c + d(y,z)/c for any 0 < v < c well the inequality is not that hard. you can bound the inequality below. I hope this posting will settle the problem. Therefore, you can't disprove the nonexistance of the rigidbody problem by simply using the special relativity. The nonexistance of rigidbody is just the experimental fact. CAN YOU PROVE nonexistance of a circle x^2+ y^2 = r^2. In reality, you won't find such circle! But TRYING to prove such thing exist or does not exist is just pure nonsense. I believe what A. Einstein said in his book about "rigid body does not exist which is proved by EXPERIMENT..bla bla..." is samething as me saying "a perfect mathematical circle does not exist which can be easily demonstrated by experiment. bla bla... " iLUSiON http://www.math.mcgill.ca/~chung