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

[my_wan]

Look at it this way. You have A and B and they are going 100 m/hour wrt each other. Now A says B has a vector magnitude of 100 m/h. B says A has a vector magnitude of 100 m/h. What happens if they are both right. The vector adds up to 200 m/h! That can't be right because they are both only going 100 m/h wrt each other.

Ahh, but A thinks his vector magnitude is 0 and B vector magnitude is 100 m/h.
Also, B thinks his vector magnitude is 0 and A vector magnitude is 100 m/h.
Now suddenly the vectors add up properly.

 

[rab99]

To calculate the one-way time for the clock to go from the left end to the right end is more complicated. It is different in different directions in the frame where the rocket is moving, but it will nevertheless be the same in both directions for the observer on the rocket if he places watches at each end; and notes the times the light hits each end; to understand this part you must take into account the relativity of simultaneity, which is based on the fact that each observer synchronizes watches in different locations using the "Einstein synchronization convention" which is based on the

JesseM i forgot to mention the space ship's velocity is 1 meter per sec and the clock is 600,000,000 meters long.

the clock has moved 2 meters by the time the light strikes the detector.

As the clock is moving so slow the effects of simutenaiety are virtually zero, same for contraction and time dilation.

 

[JesseM]

The clock at rest wrt the observer is an identical clock to the one in the space ship. Why is the length of the clock even relevant they are identical. Just assume the clock in the spaceship is shorter wrt the one at rest with the observer?

But if the light clock is shorter than L in the frame of the guy on the ship, then of course it will no longer be true that he'll measure the time for the light to go from one end of the clock to the other and back is 2L/c. The point is that whatever the length L' is in the frame of the ship, the time for the light to go from one end to the other and back is 2L'/c as measured by a stopwatch the observer on the ship is holding, so the experiment doesn't help at all in determining whether the ship is "really" in motion (in relativity this question has no single right answer).

 

[JesseM]

JesseM i forgot to mention the space ship's velocity is 1 meter per sec and the clock is 600,000,000 meters long.

the clock has moved 2 meters by the time the light strikes the detector.

As the clock is moving so slow the effects of simutenaiety are virtually zero, same for contraction and time dilation.

Again, the point is that whatever the length of the light clock L' in the ship's own frame, the time for light to go from one end to the other as measured by watches at either end of the clock which have been synchronized in the ship's frame (and are at rest in this frame) will be exactly L'/c, and the time for light to go from the back end to the front end and back as measured by a watch at the back end and at rest in the ship's frame will be 2L'/c. Even if you do the analysis in the frame where the ship is moving, you'll still make this prediction about the time as measured by these watches. Do you disagree?

 

[rab99]

I knew I should have split that proposition into two posts.

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

 

[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.