## Wednesday, November 18, 2009

### Irodov Problem 1.354

Since all clocks (A-G) in frame K have been synchronized to each other, they will show the same time as seen by any observer in Frame K. While all clocks (A'-G') in K' have also been synchronized to each other, as seen from Frame K, these clocks run slower as K' is moving with respect to K. Clocks D and D' are perfectly synchronized as seen from K. Thus, all clocks in Frame K' to the right of D' i.e. (E'-G') will progressively fall behind. Similarly, all clocks to the left of D' i.e. (A'-C') must be progressively ahead so as to catch up with D by the time they reach the center despite their slower rate.

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