I have seen and understood the classical thought experiment where you imagine a "light clock" sending a light ray between two mirrors while moving in a perpendicular direction to the lights direction in the reference frame of the clock, as shown here:
What I don't understand is that the formula for an observers perceived time, $\Delta t'$, of the clock is derived from the Pythagorean theorem which only works because the light is being reflected in a direction perpendicular to the direction of the velocity of the clock (from the clock's point of view). If the clock reflected the light in the same direction as it was itself moving, that is in the animation above the clock would be flipped 90 degrees "laying down", then it would still be a clock because it would still have a fixed period but I don't see how one would derive the same result for how a bystander perceives the clock:
$$\Delta t' = \dfrac{\Delta t}{\sqrt{1-(v/c)^2}}$$
I am asking this because in the example I've seen of length contraction, the clock was moving in the same direction as the light was being reflected, but in the derivation of the equation of the contraction effect they still used the formula for time dilation, which was derived when the clock was "standing" as in the animation above.
Best Answer
@WillO gives a good conceptual explanation. For completeness it's possible to show that the same time dilation results in either case.
A horizontal clock would be moving in the direction of its length, so we need to worry about length contraction as well. According to the stationary observer, the horizontal clock is $\ell^\prime = \frac{1}{\gamma}\ell$ long, and $$ \gamma = \frac{1}{\sqrt{1-\left( \frac{v}{c}\right)^2}} $$ is the Lorentz factor.
Stationary Clock
It takes the light $\Delta t = 2\ell / c$ to make a round trip for the stationary clock. Another way to put it is that the total round trip distance is $$ c\, \Delta t = 2 \ell .$$
Moving Clock
For the moving clock break the motion of the light up into two parts: the outgoing part (before reflection) and the returning part (after reflection).
outgoing time
For the outgoing part the distance traveled by the light in time $\Delta {t_\mathrm{o}}^\prime$ is $$c \, \Delta {t_\mathrm{o}}^\prime = \ell^\prime + v\,\Delta {t_\mathrm{o}}^\prime .$$
The light traveled speed $c$ for time $\Delta {t_\mathrm{o}}^\prime$. The light needed to move the length of the clock plus the amount the far end moved while the light was in transit. Anticipating the end result, rewrite this as
$$ c \, \Delta {t_\mathrm{o}}^\prime = \frac{\ell^\prime}{1-\frac{v}{c}} .$$
returning time
For the returning part the distance traveled by the light in time $\Delta {t_\mathrm{r}}^\prime$ is
$$c \, \Delta {t_\mathrm{r}}^\prime = \ell^\prime - v\,\Delta {t_\mathrm{r}}^\prime .$$
The light traveled speed $c$ for time $\Delta {t_\mathrm{r}}^\prime$. This time the light needed to move less than the length of the clock, because the front of the clock moved towards the light while it was in transit. Or
$$ c \, \Delta {t_\mathrm{r}}^\prime = \frac{\ell^\prime}{1+\frac{v}{c}} .$$
total time
The total distance for the light to travel out and back is $$ c\,\Delta t^\prime = c\,\Delta {t_\mathrm{o}}^\prime + c\,\Delta {t_\mathrm{r}}^\prime = \frac{\ell^\prime}{1-\frac{v}{c}} + \frac{\ell^\prime}{1+\frac{v}{c}} $$ $$ = \ell^\prime \left( \frac{1+\frac{v}{c}}{\left(1-\frac{v}{c}\right)\left(1+\frac{v}{c}\right)} + \frac{1-\frac{v}{c}}{\left(1-\frac{v}{c}\right)\left(1+\frac{v}{c}\right)} \right)$$ $$ = \frac{2\, \ell^\prime}{1-\left(\frac{v}{c}\right)^2} $$
or $$ c\,\Delta t^\prime = 2\, \gamma^2\, \ell^\prime.$$
Putting together the length contraction and the two time results gives the expected $$\Delta t^\prime = \gamma\, \Delta t $$