Hi all, Page 94 of Special relativity by AP French has a worked out example that puzzled me. This is the example: Frame S’ has a speed v=0.6c relative to S. Clocks are adjusted so that t=t’=0 at x=x’=0. Two events occur. Event 1 occurs at x1=10m, t1=2E-7s (y1=0, z1=0). Event 2 occurs at x2=50m, t2=3E-7s (y2=0, z2=0). Question: What is the distance between the events as measured in S’? It proceeds to work out Gamma=5/4. Then using the formula given: x2’-x1’=Gamma[(x2-x1) –v(t2-t1)] Therefore we get: x2’-x1’=5/4[(50-10)-3/5(3 X 1E8)(3-2)1E-7]=27.5m At first sight it looked ok to me because the length has been reduced from 40m to 27.5m. That is until I did the same example but now with t1=0 and t2=0. In that case the formula reads: x2’-x1’=5/4[(50-10)]=50m. In this case we get length expansion. My take on this is that S’ sees S speeding away. Therefore S’ should measure a length contraction and not expansion for the case of t1=t2=0. Surely the events given are measured in S?