SOrry for writing in English but I am in a library and can't wait this happy news. As usuall my instinct was right. You can't send any information faster than the speed of light using the rigid rod. I was almost mistaken yesterday. proof. |-------------------|-----------| x y z Let xy be the rigid rod and somebody will push the rod xy toward z. Suppose an observer is observing this event in the coordinate which is at rest relative to the moving coordinate. ( you may take x as the origin of the moving frame. ) We will measure the time in the rest coordinate. when t=0, the someone will push x, the end of the rod xy. when t=t', the other end of the rod y will cross the point z. We assume that the rod xy will be moving at the constant speed v. I WILL PROVE THAT t' >= d(x,z)/c where d(x,z) is the distance between x and z and c is the speed of light. Hence casuality principle will be conserved. Lorentz contraction of length and TIME DILATION will be observed to the observer. xy will be contracted to d(x,y)*(1-(v/c)^2)^0.5 Let t'' be the time when the end of the rod y start to move. y WILL NOT move at the same time when the someone pushes the other end of the rod x. If x,y moves instaneously, the observer NEVER observe Lorentz contraction! Hence there must be some time delay and t'' excatly corresponds to the time dilation of special relativity. It can be computed easily. Let alpha(v) = (1- (v/c)^2)^0.5 then t''= d(x,y)(1-alpha(v))/(v*alpha(v)) woops.. I will save this posting and post the second part soon! to be continued~ iLUSiON http://www.math.mcgill.ca/~chung