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Mar2-12, 04:01 PM
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P: 4,781
Quote Quote by Adel Makram View Post
Thank u for your care to clarify things for me. But my scenario is different from just assigning events to different coordinates in different frames. I posted here a clear detailed diagram and calculation based on my idea. Even from the diagram it is clear that A and B are still spacelike for the train observer because the source is put near to A than B which means that Signal must reaches B in no later than a signal would have taken to reach B coming from A. But how about the math! I see no contradictory to insert ∆t``= (AB`)/c and equating it with ∆t`= (AB`v/c2)/(1-(v2/c2)) can you? In other words, if A and B are spacelike for the train observer, why the math of LT allows the time difference between receiving signal at both ends to equal the time difference a light signal would take to reaches B from A?
Could you explain what the single and double back quotes mean?