Does light still slow down in the direction of motion?

Question

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 :

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:

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

Answer 1

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.