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
    Commented Aug 26, 2016 at 16:01
  • $\begingroup$ @AndreaDiBiagio That should probably be an answer $\endgroup$
    – David Z
    Commented Aug 26, 2016 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
    Commented Aug 26, 2016 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
    Commented Aug 26, 2016 at 19:46

7 Answers 7


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

  • $\begingroup$ In many introductory treatments of SR, one uses the time dilation formula obtained from a "perpendicular light clock" to get the length-contraction formula. Is it possible to derive the time-dilation equation for a "parallel light clock" without assuming the length-contraction formula? Or does one just have to take it on faith in this account? $\endgroup$ Commented Sep 12, 2021 at 13:02
  • $\begingroup$ Lots of skipped algebra (and associated reasoning) here. It all looks technically correct but it's hard to verify certain things and see why they're being done. The last step took some effort to figure out and I still don't know how you get from ℓ′+vΔto′ to ℓ′/(1+v/c). I had to plug in numbers to verify that those two things are equal. Would you (or someone) mind adding some detail to those parts? $\endgroup$ Commented Jan 16, 2022 at 8:34
  • 1
    $\begingroup$ @MichaelSeifert The "longitudinal light clock" is developed in my paper "Relativity on Rotated Graph Paper" AmJPhy 84, 344 (2016); doi.org/10.1119/1.4943251 using radar measurements (Bondi) and the relativity principle. This implies the invariance of the area of a causal diamond (Mermin). This area is equal to the square interval. Written in light-cone coordinates, this is the essentially the product of radar-times formula (Geroch, Synge). Time dilation and length contraction follow as consequences. (See physicsforums.com/insights/relativity-rotated-graph-paper ) $\endgroup$
    – robphy
    Commented Jan 17, 2022 at 4:21
  • $\begingroup$ @MichaelSeifert I just contributed an answer using a related method that doesn't directly use the Bondi k-calculus. $\endgroup$
    – robphy
    Commented Jan 17, 2022 at 19:02

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 $\endgroup$ Commented Jan 31, 2018 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
    Commented Jan 31, 2018 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
    Commented Jan 31, 2018 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. $\endgroup$ Commented Jan 31, 2018 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
    Commented Jan 31, 2018 at 17:20

The reason why a transverse clock is typically used in teaching SR is that the associated maths takes a simpler form in such a case.

As other answers have shown, you can derive the same result from considering a clock oriented along the direction of motion. In some respects this provides more insight into the nature of time dilation as it more explicitly involves a consideration of the relativity of simultaneity.

Specifically, the clock in the moving frame ticks unevenly, as the path length of the light on the outbound tick is longer than the path for the return tick. If you consider that for a moment you will see that while the overall time for both ticks is time dilated by the familiar formula, the outbound tick is time dilated by another amount entirely and the return tick is actually time contracted.

This example is a reminder that the time dilation formula is applicable only to the interval of time between two events that occur in one place.

A more interesting result is the case of two moving back-to-back longitudinal light clocks which send off light in each direction from a common centre. Here, the outbound tick of the clock which sends out light in the direction of motion of the clocks is longer than the outbound tick of the other clock, while the reverse is true of their return clicks. Neither time dilation or length contraction alone can explain this- what it illustrates is the relativity of simultaneity. Where you have two moving reference frames, a plane of constant time in one frame is a sloping slice through time in the frame through which it is moving, the slope being upwards in the direction of motion. Both time dilation and length contraction follow from that rotation of the planes of constant time between the two reference frames.


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? $\endgroup$ Commented Aug 26, 2016 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
    Commented Aug 26, 2016 at 17:32

Any two events in the "moving" frame (the train) that occur in the same place and are separated by some time $ \Delta t $, as measured from that frame, will be separated by a longer time $ \Delta t^\prime = \gamma \Delta t $ when seen from the "stationary" frame (the train station). It's important that the events happen in the same place (e.g., light being emitted by a source and then arriving back at that source (after being reflected)) or at least in the same plane that's orthogonal to the direction of motion; otherwise, the calculation of time dilation will be confounded by differences in simultaneity between the two frames.

Clearly, the "longitudinal light clock" (rather than the more traditional "transverse light clock"), as some call it, in your example will work just fine. And the math can be very simple. I'll use similar conventions as Paul T.'s answer, so $ \Delta t_o $ is the outbound time and $ \Delta t_r $ is the return time. And I'll assume that the outbound light beam travels in the direction of the train's motion toward the mirror in the front part of the train.

Remember that $ \ell^\prime = \frac \ell\gamma $ due to length contraction and that $ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $.

So from the moving frame, the time it takes the light to travel to the mirror and back is

\begin{align} \Delta t & = \Delta t_o + \Delta t_r = \frac{\ell}{c} + \frac{\ell}{c} \\ & = \frac{2\ell}{c}. \end{align}

Now since the stationary observer sees the train moving at $v$ and the outbound light moving in the same direction at $c$, he sees the light moving relative to the train at $c - v$ and thus traversing the distance $\ell^\prime$ at that speed. He then sees the returning light traverse that same distance at $ c + v $ (again, relative to the train). Therefore, the time it takes the light to travel to the mirror and back from his frame is

\begin{align} \Delta t^\prime & = \Delta{t_o}^\prime + \Delta{t_r}^\prime = \frac{\ell^\prime}{c - v} + \frac{\ell^\prime}{c + v} \\ & = \frac{\ell^\prime\left(c + v\right)}{\left(c - v\right)\left(c + v\right)} + \frac{\ell^\prime\left(c - v\right)}{\left(c - v\right)\left(c + v\right)} \\ & = \frac{2\ell^\prime c}{c^2 - v^2} = \frac{2\ell^\prime}{c\left(1 - \frac{v^2}{c^2}\right)} \\ & = \frac{2\ell^\prime \gamma^2}{c} = \frac{2\gamma^2}{c} \frac \ell\gamma = \frac{2\ell\gamma}{c} \\ & = \gamma\Delta t. \end{align}

  • The Bondi k-calculus (an algebra-based method) develops the basic ideas of special-relativity in the $tx$-plane (without using the transverse direction). This approach is presented in his book “Relativity and Common Sense” (1962, 1964). In addition, Bondi presented “E=mc2: Thinking Relativity Through”, a series of ten lectures on BBC TV running from Oct 5 to Dec 7, 1963. (Read more at my contributed article to a blog at https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/ )

I have used it to develop the "longitudinal light clock" without appealing to the standard textbook "transverse light clock". In particular, I draw the light signals in the longitudinal light clock to form the "light-clock diamonds" (the "causal diamond" between consecutive tick events). All light-clock diamonds for all inertial observers have equal area since the Lorentz boost transformation has determinant one. This is developed using the Bondi k-calculus in my article:

Relativity on rotated graph paper,
AmJPhy 84, 344 (2016); https://doi.org/10.1119/1.4943251 .
An early draft is at https://arxiv.org/abs/1111.7254 .

Rather than use "time-dilation" from the transverse clock,
I use the Principle of Relativity in the $tx$-plane.

The key figure (using $v=(3/5)c$) is


Once the light-clock diamond size is established (using the Bondi k-calculus in the above), the spacelike-diagonal of the light-clock diamond determines the reflection events on the mirror worldlines of the longitudinal light-clock. So, draw through those events parallels to the timelike diagonal (the observer's worldline). This shows length-contraction as viewed in the lab frame. (The construction can be shown to be symmetric between the observers.)

  • I have a different approach (that doesn't use the Bondi $k$-calculus directly) which appears in my recent contributed chapter

    Introducing relativity on rotated graph paper
    Ch 7 in Teaching Einsteinian Physics in Schools
    Kersting and Blair, Routledge 2021, https://doi.org/10.4324/9781003161721

I will describe it below.
You can play with the ideas in this visualization: https://www.geogebra.org/m/HYD7hB9v#material/UBXdQaz4 (make sure BOB's diamonds are shown)

The key idea is that the diamond size is determined by the Principle of Relativity. (The diamond shape is determined by the Speed of Light Principle and the velocity of the observer.) The two observers perform the same experiment and should expect the same results:
2 seconds after they meet, send a light signal to the other.

Assuming absolute time and absolute space fails to satisfy the principle, but the third configuration works.

Using $v=(3/5)c$... and assuming the Speed of Light Principle.. we have the shape of Bob's diamonds... but what is the correct size?

Assuming absolute time (so that the heights of the diamonds are equal),
Alice (red) receives Bob's signal at 3.2, whereas Bob (blue) receives Alice's at 5.
Bob's ticks need to be scaled up from this size. This hints that there is time-dilation... but by how much?

Assuming absolute space (so the lengths of the cross-sections of the light-clocks are equal),
Alice (red) receives Bob's signal at 5, whereas Bob (blue) receives Alice's at 3.2.
Bob's ticks need to be scaled down from this size. This hints that there is length-contraction... but by how much?

By playing around (taking a hint from the geometric mean?),
we get agreement with the Principle of Relativity by each receiving the other signal at 4 ticks.

The ratio $(4\mbox{ ticks})/(2\mbox{ ticks})$ is the Doppler factor $k=2$ for $v=(3/5)c$, where we have used the Principle of Relativity and the Speed of Light Principle...
...and, as a consequence, we now know the factor for time-dilation and length-contraction.

Try it for $v=(4/5)c$.
https://www.geogebra.org/m/kvfsq664 (updated)... make sure BOB's diamonds are shown

By the way, we find that the areas of Alice's clock diamonds are equal to Bob's clock diamonds. It turns out that the area of a causal diamond (in units of clock-diamonds) is equal to the square-interval between the corners of its diagonal.

For more information, consult my article and chapter above.
See also


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.

  • 4
    $\begingroup$ 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. $\endgroup$ Commented Sep 20, 2018 at 23:54

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