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This is quite a naive question however I hope to learn from this - I had always learnt a light clock in a space ship is placed like so :

enter image description here

That said as the light moves the light will seem to slow down due to the velocity of light compensating for the horizontal movement of the space ship.

Now if I were to keep the light like so would it still slow down:

enter image description here

I feel that the light would not slow down because of the fact that light is simply bouncing about and the ship also contracts in the direction leading to the interval of bounce not changing, but somehow I feel I am incorrect so could someone explain why even a horizontal light clocks light slows down.

Thanks

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    $\begingroup$ Strange but true: the speed of light is always the same, no matter in what direction, or how fast you (source of light) move. $\endgroup$ – Gyro Gearloose Feb 26 '16 at 22:20
  • $\begingroup$ @GyroGearloose Yeah I am quite confident with that however for the observer the 1st image would actually show a "slow" light clock as the lights vertical component will be lower due to its horizontal component. On the 2nd picture there would be no change (in my view however) that said how can we distinguish between stationary time and movement time if the clocks adjusted to the direction of travel. $\endgroup$ – LogicProgrammer Feb 26 '16 at 22:23
  • $\begingroup$ en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment $\endgroup$ – Gyro Gearloose Feb 26 '16 at 22:28
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Let's calculate it then to gain some intuition. For a stationary horizontal light clock, the time for one tick will be given by $$t=\dfrac{2l}{c}$$ where $l$ is the distance between its two ends.

What about the moving one? First, as you pointed out $l$ will be length contracted by a factor of $\gamma=\dfrac{1}{\sqrt{1-v^2/c^2}}$

So the time taken for one tick(as measured in the frame at rest) will be given by the sum of two times $t_1$ and $t_2$, $t_1$ being the time the light took from the start to the end, $t_2$ is from the end to the start

$$t'=t_1 +t_2 =\dfrac{l/ \gamma}{c-v} + \dfrac{l/ \gamma}{c+v}=\dfrac{2lc}{\gamma (c^2-v^2)}$$

Now $$\dfrac{t}{t'}=\dfrac{\gamma (c^2-v^2)}{c^2}=\dfrac{1-v^2/c^2}{\sqrt{1-v^2/c^2}}=\sqrt{1-v^2/c^2}$$ so $\dfrac{t}{\sqrt{1-v^2/c^2}}= \gamma t =t'$ and since $\gamma$ is always greater than one,then the one tick for $t'$ will always be greater than that of $t$.

To give you a feeling for this, if $\gamma=2$ and the clock at rest reads $t=10$ secs, if we were to look at the moving clock, its reading would be $t'=5$ secs, since its 'tick' is half as slow as $t$, that is $t'$ always needs double the time $t$ needs to register one tick(or 1 sec), so $t'$ will always lag behind $t$ by a factor of 2.

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  • $\begingroup$ Maybe I miss something: I see you compute the relativistic change of length , but I don't see where you include the relativistic change of time. $\endgroup$ – Gyro Gearloose Feb 26 '16 at 23:47
  • $\begingroup$ You wanna calculate the clock reading on the frame at rest that is watching the horizontal clock passing by, so you only need to account for change of length. $\endgroup$ – Omar Nagib Feb 26 '16 at 23:52
  • $\begingroup$ You've got a math error. The last step should result in $t_1 + t_2 = \gamma 2 l / c$, which by the way is greater than $2l/c$, not less. $\endgroup$ – Red Act Feb 27 '16 at 4:40

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