Finding Gamma by Experiment In this animation, I show two views of of light propagating along the +x direction from a source for a nominal time t. One view is seen by a viewer, A, where he/she is moving in the +x direction and the other view of the same propagation is seen by a viewer, B, that is moving at -v with respect to the source. Since light travels at speed c in any frame the equations for the two viewers are but the times in the two viewers' frames may be different we have the following transformation equations for viewers A and B: c*tA=c*gamma*(tB-v*tB/c) (A) c*tB=c*gamma*(tA+v*tA/c) (B) where the expansion factor gamma, which is the only difference between the Galilean and Lorentz transformations, must be computed. Multiplying equation A by equation B we easily find that gamma=1/sqrt(1-v^2/c^2) so that equations A and B become c*tA=c*tB(1-v/c)/sqrt(1-v^2/c^2) (A1) c*tB=c*tA(1+v/c)/sqrt(1+v^2/c^2) (B1) Note that when we substitute the result for tA from equation A1 into equation B1, we obtain c*tB=c*tB as we should expect. Also note that gamma=1 will not work unless v=0. What is the difference tB-tA? tB-tA=gamma*(tA-tB)+gamma*(tA+tB)v/c (1+gamma)(tB-tA)=gamma(tA+tB)*v/c dt=tB-tA=gamma(tA+tB)*v/c/(1+gamma) For symmetry let tA+tB=2*t0 where t0 is a measure of time in the frame of the light source. Since we can choose any increasing value for tB, we choose tB=t0+dt/2=t0+gamma*t0*v/c/(1+gamma)=t0*[1+(v/c)gamma/(1+gamma)] Then using equation A1 we obtain: tA=tB(1-v/c)*gamma or equivalently tA=2*t0-tB This results in a certain symmetry with repect to t0 beween tA and tB. So how will this experiment be carried out?. We will need a pulsed light source that is stationary as well as two space ships to observed the pulse from the source. The space ships must fly along routes parallel to the x axis and displaced from it by a small distance so as not to block the light. The space along the x axis must contain some very fine space dust that will scatter a small fraction of the light from the light pulse so that it can be seen by the space chips. You might comment that this would be an expensive experiment but you must realize that it is just a way to visualize the effect of gamma on the light pulses. After pulse B has gone 800 units, the results of gamma for both pulse B and pulse A are stated and they agree with the definition of gamma.
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