The thought experiment starts with the basic experiment shown here: http://en.wikipedia.org/wiki/Relativity_of_simultaneity "A flash of light is given off at the center of the traincar just as the two observers pass each other. The observer on board the train sees the front and back of the traincar at fixed distances from the source of light and as such, according to this observer, the light will reach the front and back of the traincar at the same time. The observer standing on the platform, on the other hand, sees the rear of the traincar moving (catching up) toward the point at which the flash was given off and the front of the traincar moving away from it. As the speed of light is finite and the same in all directions for all observers, the light headed for the back of the train will have less distance to cover than the light headed for the front. Thus, the flashes of light will strike the ends of the traincar at different times." Now my modification: Problem is we have different simultaneities. So let's put photon detectors on each end A and B. Same photon that hits the back of the wagon in the ground frame it does before the first photon hits the front. In the wagon frame at the same time. I said to move the light source slightly to the front so that in the wagon frame the sequence from the detectors will get "01" and from the ground "10" ... I thought that if the information from both ends (from the detectors) is transferred to a sequence analyzer let's say in the middle of the wagon, we can argue that it may compensate the sequence as if this information is transmitted at "c". But then I thought of eliminating that by doing the following. When each detector triggers, a very slow mechanical device (let's say a plastic pin with Mickey face from the front and another one with Minnie face from the back) goes from each end at the same speed. So if the detector in the front triggers first, Mickey will reach the center of the wagon before Minnie. If Mickey wins "boom", if Minnie wins no boom. Some of my own math: In the frame of the wagon, light travels at "c". The time to reach each end from the middle is the b/2/c. Where b is the length of the wagon. Time is distance travelled/velocity. In the ground frame and noticing that b' will have some shortening due to the length compression if the speed of the wagon is high you have: Let's call A the left edge of the wagon, B the right edge of the wagon and C the center. Time from C to A (left) is = b'/2/ (c+v) (speed of light on this frame is also c). v is the velocity of the wagon. tCA=b'/2/(c+v) Time from C to B is tCB=b'/2/(c-v) You notice that tCB can be a very large number if v approaches c and tCA approaches to b'/4/c Now I said both photon detectors launch a slow device (not anything travelling at c), so after the triggering there is no relativity of simultaneity due to c being constant at both frames. In other words you won't have the (c+v) or (c-v) terms to compensate the time to reach the edges. The devices that run from the edges to the center go at a speed vD that is much slower than c and we can say than vW as well. The time for each device from the edge to the center will always be the same since there is no light effect. It is always b/2/vD. I still see this stands to force SR to be wrong and to have a winner frame of reference. I'd like opinions, see if I made mistakes, if it is a valid good experiment to force a preferred simultaneity. Thank you.