How Does Time Dilation Affect Clocks in Different Frames of Reference?

[JesseM]

In his clock in the spaceship mind experiment (the link is in the post) Einstein looked at the length of the path that the photons would take as compared to the length of the path for the clock at rest wrt the observer. Einstein made a simple observation, the path is longer in the moving clock so it ticks slower. In my post it looks to me as tho the path is shorter. Make the speed of the spaceship 1 meter an hour so the effects of simutenaitey are negligable the result remains the same the path taken by the photons as observed will be shorter than the clock at rest wrt the observer

Sure, the path is shorter in the observer's frame--but if it's only by two meters, then the time it takes light to cross that distance is tiny, so you can't assume that the also-tiny effects of simultaneity and time dilation are irrelevant here. I assure you that if you actually calculate what the observer predicts about the watches fixed to either end of the light clock, and you take into account that the observer sees the watches as slightly out-of-sync and slightly slowed down, you will still find that he predicts the second watch's reading when the light passes it is greater than the first watch's reading when the light passes it by precisely L'/c, where L' is the length of the clock in the ship's frame. If you don't believe me, try working the math, using http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl , where gamma is approximated as 1 + (1/2)*(v^2/c^2)). And remember to use the following:

-If length of clock is L in the observer's frame, the length L' in the ship's frame is $$L' = L/\sqrt{1 - v^2/c^2}$$

-For a time-interval of t in the observer's frame, each of the watches on the ship only advance forward by $$t*\sqrt{1 - v^2/c^2}$$

-if the watches are synchronized and a distance L' apart in the ship's frame, in the observer's frame they are out-of-sync by vL'/c^2 (if the observer sees the ship moving forward, the clock at the back of the ship will have a time that's ahead of the clock at the front by this amount)

If you think you got an answer other than L'/c for the difference between the times on the two watches, show your work!

 

[phyti]

rabb99;
Your example looks like a complicated version of the Michelson-Morley experiment.
Check it out if you have not done so already. It can be explained using only the 2nd postulate, constant c, with no length contraction. The experiment resulted in no time difference between paths.