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In B's frame, the signal is sent when the front of A's rod arrives. Since the rod rest length was 6 ly, its length in B's frame is 4.8 ly and at the same point in time in B's frame, A is therefore 4.8 ly away. The separation speed of the signal and A in B's frame is -1.6c and so it takes the signal 4.8/1.6 = 3 years to arrive in B's frame. The proper time elapsed for A between these events is time dilated and therefore equal to 3/1.25 = 2.4 years.david316 said:What is the time difference in the reference frame of A between the event on the world line of A that is simultaneous with the sending of the signal in B's rest frame and the event of A receiving the signal?
The time elapsed between the sending of the signal and the arrival of the signal based on A's simultaneity convention (ie, not B's) is 6 years.