09 September 2016 67 4K Report

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame. Now observer A wants to know the time of clock B. For example, at 3 o'clock of clock A, observer A sends the image of clock A to observer B. When observer B receives the image, he will immediately send the image of clock B back to observe A. According to special relativity, observer B should see that the time shown on the image of clock A (3 o'clock) earlier than the time of clock B (for example 3:15) minus the time for the image to travel from A to B (for example 10 minutes calculated by the position of clock B in the reference frame of observer A divided by constant c, the speed of light) because of time dilation. In the reference frame of observer A, the distance from A to B is the same as the distance from B to A because observer B receives the image of clock A and sends the image of clock B at the same time. Thus, observer A will receive the image of clock B at 3:20. Deducted by the time for the image of clock B traveling from B to A (10 minutes), clock A should be 3:10 when observer B sends the image of clock B back to A, earlier than the time (3:15) shown on the  image of clock B. That is, clock B is faster than clock A observed by both observers, contradicting the prediction of special relativity that observer A should see that clock B slower than clock A.

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