Einstein's Famous Thought Experiment Contradiction Putting Special Relativity into the General Relativity category as is current practices submerges important aspects of Einstein's 1905 paper, which I recently read in a 1952 Dover paperback (The Principle of Relativity). That paper is totally unlike the modern presentations in texts. I noticed a marked discrepancy of the analysis in that first paper with what was actually described as happening. This has to do with the famous thought experiment. I modified that experiment and still found a discrepancy that I would like explained.
For the thought experiment, Einstein calculates a round trip time for a pulse of light from inception to reflection back to the inception point as measured in a moving frame. The round trip time is given as $T = (2D/c)/\sqrt {1 - {{(v/c)}^2}} ,$  where D is the distance from the inception point to a mirror moving in the same frame. Einstein’s hypothesis is that the on-board time is dilated to the same extent as the length is contracted, such that the on-board RT time as measured is independent of whether the frame is moving or not, hence, giving rise to the notion that a moving frame cannot know it is moving without an outside reference.
To the contrary, any symmetrical effect, such as the dilation and contraction, no matter what their forms, cancel in the above equation. This leaves an RT time discrepancy that is clearly measurable by the on board system, which knows what the speed of light is and what the distance D is. This then contradicts the hypothesis that the moving observer cannot know or measure that they are moving. Furthermore, they can tell how fast they are moving.
Let us modify the thought experiment. I put a clock at both the inception and the reflection points, both synchronized, and perform the same experiment. I measure when the pulse is reflected. All dilation and contraction effects are the same for all on-board systems and paths. I launch the pulse, measure the outbound transit time and the reflected transit time. I then compare the results. The two times are asymmetrical. Their combined time is the total time calculated above. Hence, the on-board system knows how fast it is moving.
If one further modifies the thought experiment into a radar experiment with external retro-reflectors on an aircraft, one finds the same asymmetries and the same time differences comparing an approaching aircraft with a receding aircraft. In this case, only the Lorentz contraction would be occurring, but the asymmetry is still present.
How is it possible to reconcile the above conclusions with the modern notions that the moving platform cannot know it is moving? I'd prefer an answer using the same math and logic Einstein used in his first paper.
 A: You are confused because you are not taking the relativity of simultaneity into account in your analysis. When you have two mirrors moving to the right with velocity v, you are thinking that it takes light a long time to go to the right, but a very short time to go to the left, because in one direction, the mirror is going along with the light, and in the other direction, the mirror is going toward the light.
You conclude that the fellow who is riding along with the two mirrors can't see the same time for the forward leg of the light circuit between the two mirrors as for the backward leg. This conclusion is wrong, because the fellow riding along with the two mirrors has a failure of simultaneity--- the notion of "right now" is altered, so that the "right now" moment at a distance x along the direction of motion is further into the future by vx.
This asymmetrical simultaneity failure means that as the light is going to the right, it is going into the future of the moving observer slower than when it is going to the left, so that from the point of view o the moving observer, both the forward motion and the backward motion of the light go an equal number of time steps into the future, relative to the notion of simultaneity appropriate for the moving observer.
This is explained with a diagram in this answer: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman
If you understand why the line of simultaneous events for the moving observer slopes up as viewed by the stationary observer, you will immediately understand why the forward and backward motions are symmetric for the moving observer. This is explained in the first and second diagram of the linked answer, which explicitly use right-moving and left-moving light to establish the slope of the simultaneity line.
