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:

enter image description 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.

  • $\begingroup$ That is a really good question. Suppose that the two clocks are in a L shape are made so that they tick unison and they have the same lengthy when at rest. If, in the frame where you calculated time dilation, you assume that the horizontal clock is of some length $l$, and calculate the path length of the photon, and equate this to that of the vertical clock, you will find that $l$ must be shorter than the vertical clock's length. $\endgroup$ – Andrea Aug 26 '16 at 16:01
  • $\begingroup$ @AndreaDiBiagio That should probably be an answer $\endgroup$ – David Z Aug 26 '16 at 16:26
  • $\begingroup$ Also, this is related: physics.stackexchange.com/q/14362/124 though I'm not sure it's quite a duplicate $\endgroup$ – David Z Aug 26 '16 at 16:29
  • $\begingroup$ I had the same problem, but I drew a rough sketch using simple speeds (like 1m/s for light and 0.5m/s for the traveller) and that was enough to convince me that what I had been told was right. It's the same in any direction. I'm easily convinced. $\endgroup$ – Alan Gee Aug 26 '16 at 19:46

First: An observer traveling with both a vertical and a horizontal clock must see them tick at the same rate --- otherwise he'd know he was moving.

Second: The traveling observer and a "stationary" observer must agree about how many times each clock ticks during the time it takes the traveler to go from (say) Mars to Jupiter, because they can both simply watch the clocks and count their ticks. Therefore, since the traveling observer says they both tick an equal number of times, so must the "stationary" observer.

Putting the first and second observations together, everyone agrees that the horizontal and vertical clocks tick at the same rate.

Now if you take the vertical clock away, there's no reason for the tick-rate of the horizontal clock to change. Thus the horizontal clock must tick at the same rate as the vertical, even if the vertical clock is not there.

So: Use the vertical clock to calculate the time dilation. Recognize that the same time dilation must apply to the horizontal clock, whether or not there's actually a vertical clock on board. Now (all of this from the viewpoint of the "stationary" observer) you know the horizontal clock's tick-rate. You also know how fast the clock is moving, and you know the speed of light, so you can figure out the length of the light-beam's round-trip journey, and therefore can figure out the length of the horizontal clock.

  • $\begingroup$ How can you say which clock is "true one" ? Why you prefer vertical over horizontal or any other orientation? It looks identical to saying about 2 objects A and B that move relative to each other, that the one that is truly moving is B $\endgroup$ – Alex Burtsev Jan 31 '18 at 15:59
  • 1
    $\begingroup$ @AlexBurtsev: Neither clock is the "true one". But here is the key difference between the vertical and horizontal clocks: Motion cannot change the length of the vertical clock. Here's why: If your vertical clock and my vertical meter stick are both originally one meter long (before you start moving), then, as you fly by me, we can both check to see which is longer by holding them up against each other --- and we have to agree on the answer, because we're looking at the same sticks. If the length of the clock changes, that lets us determine who's moving --- which relativity (CONTINUED) $\endgroup$ – WillO Jan 31 '18 at 16:17
  • 1
    $\begingroup$ (CONTINUED) says is impossible. So the reasoning is this: 1) Your movement can't change my measurement of the length of your vertical clock. 2) Therefore any change (according to me) in the tick rate of your vertical clock must be entirely attributable to "time dilation". 3) Any time dilation must affect your horizontal clock the same way it affects your vertical clock, as explained in the answer. 4) Now we know how fast the horizontal clock is ticking, and can use that to figure out the length contraction of the horizontal clock. $\endgroup$ – WillO Jan 31 '18 at 16:20
  • $\begingroup$ Ok, I understood your reasoning , thank you.I agree your logic is solid if you consider length contraction a physical change, so must be time dilation. However I do not consider it physical change, but its totally different question, wchich I would llike to discuss, and would probably start or join a question about it. But in short my point is, if you consider length contraction a physical change, so must you consider optical perspective, and objects that are closer are larger becouse they undergo physical change. $\endgroup$ – Alex Burtsev Jan 31 '18 at 16:57
  • $\begingroup$ @AlexBurtsev : I'm not sure what you mean by "physical change". Length contraction is shorthand for the fact that the length of an object is frame-dependent. The length of the clock does not change in any one frame, but it is different in one frame than it is in the other. I have no idea what you mean by "optical perspective". $\endgroup$ – WillO Jan 31 '18 at 17:20

@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 $$\Delta t^\prime = \gamma\, \Delta t $$


The light clock thought experiment you are describing is a one-dimensional experiment: On the left there is an observer, on the right there is the observed object moving horizontally = in x direction. The vertical dimension has been added for measuring purposes only - with a light ray traveling up and down.

By consequence, if you "lay down" the mirror system on the right side, the experimental configuration does not change. The observed object is still traveling in horizontal x direction. The only difference is that the travel of the horizontal light ray can no longer be compared directly with the observer's vertical light ray on the left. This configuration is less clear, but it is still the same process: an object receding horizontally from the observer in x direction.

Length contraction is a corollary of time dilation, that implies that with the light clock thought experiment you can also derive length contraction.

  • $\begingroup$ But if you lay down the left clock as well then? How would you derive the formula previously derived from the Pythagorean theorem? $\endgroup$ – hampadampadoo Aug 26 '16 at 17:27
  • $\begingroup$ I tried to explain that the "Laying down" of the right (or the left) light clock is meaningless. If you laid it down you have to position it in vertical position again in order to take the measurements. Laying down causes the confusion of the experimental x dimension with the additional dimension for measuring purposes. $\endgroup$ – Moonraker Aug 26 '16 at 17:32

Does no one see the issue that both the person on the ship and the one outside the ship can't see light moving at a constant rate at the same time for the horizontal clock? If the person inside the ship sees light moving at a constant speed, then the person outside would see light traveling faster when light is moving toward the front and slower towards the back. If the person outside sees light moving at a constant speed, then the person inside would see light moving slower going towards the front and faster towards the back. They would both, however, calculate the same average speed of light.

  • 2
    $\begingroup$ It seems that no one cared to answer you shortly after you posted it, but the "can't" you issue up there is predicated on the idea that both observers share the same notion of simultaneity. Einsteinian relativity discards that notion in favor of the constancy of the speed of light (and experimental evidence agrees with Einstein). So, yeah, we noticed that problem but the resolution is not the "obvious" one. $\endgroup$ – dmckee --- ex-moderator kitten Sep 20 '18 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.