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fhisicsstudnt
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I'm learning about relativity and, by extension, the classic Michelson-Morley setup. I cannot see why a "fringe effect" was expected and could use some discussion. Here is my reasoning.
Firstly we imagine the light signal to propagate from the SM (silvered mirror, in the "center") to M1 and M2 and back at c (distance for each of 2*x), arriving at the detector simultaneously. This is taking the whole thing to be "stationary". The data is recorded in the interferometer's own reference frame, the scientist is in that frame, and so the whole thing considers itself to be at rest consistent with both "new" and "old" mechanics. There is no "fringe effect".
Next we imagine the whole thing to be in uniform translation at velocity v. The only way to accomplish this is to set up a 2nd reference frame which we consider stationary, and which we consider the interferometer to be in uniform translation with respect to. For simplicity we imagine the interferometer to be translating along the axis of one of the arms, call it the a axis.
With respect to the stationary frame the light signal propagates at v1a = c+v one way and v2a = c-v the other. The distance for the signal to travel relative to the stationary frame is x-v*t1a one way and x+v*t2a the other. t1a = (x+v*t1a)/(c+v) = x/c and t2a = (x-v*t2a)/(c-v) = x/c. Total transit time is Ta = 2*x/c
The other signal propagates at vb= (c2+v2)1/2 both ways relative to the stationary frame. The distance for the signal to travel, relative to the stationary frame, is xb = (x2+v2*tb2)1/2 both ways. The transit times are both tb = xb/vb. Tota transit time is Tb = 2*tb = 2*x/c.
So I don't get it. Even without doing the math out, all we are doing is taking the same situation and analyzing it with everything in uniform translation. Whether we are thinking "classically" or in the context of SR, uniformly moving frames and stationary frames are indiscernible from each other. If we consider the interferometer as stationary and the two light beams return to the silvered mirror simultaneously, then any uniformly translating frame will consider the two light beams to return simultaneously also, this is true either classically or relativistically. The only difference between classical and relativistic is that the two beams will classically arrive at M1 and M2 simultaneously in any moving frame whereas relativistically they will not necessarily appear to arrive at M1 and M2 simultaneously, depending on the frame.
Firstly we imagine the light signal to propagate from the SM (silvered mirror, in the "center") to M1 and M2 and back at c (distance for each of 2*x), arriving at the detector simultaneously. This is taking the whole thing to be "stationary". The data is recorded in the interferometer's own reference frame, the scientist is in that frame, and so the whole thing considers itself to be at rest consistent with both "new" and "old" mechanics. There is no "fringe effect".
Next we imagine the whole thing to be in uniform translation at velocity v. The only way to accomplish this is to set up a 2nd reference frame which we consider stationary, and which we consider the interferometer to be in uniform translation with respect to. For simplicity we imagine the interferometer to be translating along the axis of one of the arms, call it the a axis.
With respect to the stationary frame the light signal propagates at v1a = c+v one way and v2a = c-v the other. The distance for the signal to travel relative to the stationary frame is x-v*t1a one way and x+v*t2a the other. t1a = (x+v*t1a)/(c+v) = x/c and t2a = (x-v*t2a)/(c-v) = x/c. Total transit time is Ta = 2*x/c
The other signal propagates at vb= (c2+v2)1/2 both ways relative to the stationary frame. The distance for the signal to travel, relative to the stationary frame, is xb = (x2+v2*tb2)1/2 both ways. The transit times are both tb = xb/vb. Tota transit time is Tb = 2*tb = 2*x/c.
So I don't get it. Even without doing the math out, all we are doing is taking the same situation and analyzing it with everything in uniform translation. Whether we are thinking "classically" or in the context of SR, uniformly moving frames and stationary frames are indiscernible from each other. If we consider the interferometer as stationary and the two light beams return to the silvered mirror simultaneously, then any uniformly translating frame will consider the two light beams to return simultaneously also, this is true either classically or relativistically. The only difference between classical and relativistic is that the two beams will classically arrive at M1 and M2 simultaneously in any moving frame whereas relativistically they will not necessarily appear to arrive at M1 and M2 simultaneously, depending on the frame.