I summarised the question in the final para, if you prefer to skip all my rationale and trig.

btw, I have looked at

I have also done some off site research, (not to mention the luxury of being taught some of this - a long long time ago).

None of which inputs seem to cover my precise point. I hope you will also find this is not a duplicate. There is necessarily some scene setting and context.


Consider two light clocks, A and B, in the same moving frame of reference and an observer, O, at rest.

These are initially one way light clocks in which the light travels from an emitter to a detector. They are not two way clocks in which light travels back to its origin, although that is discussed as a check and to show the reasoning can derrives the Lorentz transform correctly.

Clock A: Tangential to Velocity

The treatment of light clock A is the familiar clock from every special relativity text book, I recap it here.

I used this diagram even though it is a two way clock.

Two way clock picture

Light clock A is oriented so its light moves tangentially to the observed direction of motion. Light clock B is oriented so its light moves parallel to the observed direction of motion.
In special relativity the length contraction effect can be derived by applying basic trigonometry to the light path. This is well established in text books, the distance $D$ that O observes the light in clock A to travel is $$D^2=\frac{L^2}{\bigg(1-\frac{v^2}{c^2}\bigg)}$$
$L$ is the distance from light clock detector to mirror (in its own frame of reference)
$D$ is the distance the light travels in the observer's rest frame $v$ is the frame's velocity
$c$ is the assumed speed of light

O's necessary assumption the speed of light is constant means that he observes clock A running slower in proportion to the greater path length; and so we get length contraction.

Clock B: Parallel to Velocity

I now try to apply this logic to clock B. The distance that O observes the light in clock B to travel.
In the forward direction the distance that the detector moves is $$v\delta t$$ The distance the light moves must be this plus L (but length contracted) $$D=L/\gamma+v\delta t$$ where for the observer again $$\delta t=D/c$$ Unfortunatly that returns $$D=\frac{L}{\gamma\bigg(1-\frac{v}{c}\bigg)}$$

D is bigger than L, in O's frame light takes longer to travel D, and so time in the clock frame appears dilated to O.

I go on and do the two way clock.

If Clock B was Two Way Clock Parallel to Velocity

OK let's determine the length of the return journey of the light and add it to length of the outbound journey of the light, $D$ in the previous equation. I can say $$c\delta t+v\delta t=L/\gamma$$ (two "objects" moving together from a separation of L, one at v, one at c) This gives us $\delta t$ and so the length of the light's return journey is $$D_{return}=\frac{cL}{(c+v)}=\frac{L}{\gamma(1+v/c)}$$ Adding this up, then $$D_{bothways}=\frac{L}{(1+v/c)}+\frac{L}{\gamma(1-v/c)}$$ so this works out to $$D_{bothways}=\frac{2L}{\gamma(1-v^2/c^2)}$$ Now I didn't derrive it here but we can do that eleswhere $$\gamma=1/\sqrt(1-v^2/c^2)$$ So we get, which is twice as big for Clock A and correct. $$D_{bothways}=\frac{2L}{\sqrt(1-v^2/c^2)}$$


The clock A and clock B equations don't agree... so that under certain (contrived?) circumstance, if

  • the effect being viewed is mono-directional and
  • runs parallel to the the motion of the frame, (depending on whether it is in the direction or counter the direction of motion)

You will get different magnitudes of time dilation effects. OK, I better ask about this...

The Question

As a community we've already accepted length contraction as directional. Given that time dilation is a manifestation of the same effect, it seems only logical to accept that (under some conditions) relativistic temporal effects can also be directional e.g. one way light clocks (if such a thing can exist).

What is wrong with my application of maths?


Just to try and make this sem relevant, we now start to get weird behavior if we take two differently oriented directional light clocks on a hypothetical relativistic journey. They will start to tell different times. I understand that GPS satellite clocks are relativisitcally compensated. If the (I believe atomic) clocks have a "directional" component, is it possible this effect could be significant?

  • $\begingroup$ Can you explain what a "one way light clock" is? From your brief description of it, I'm not sure it functions as a clock in the usual sense of the word. $\endgroup$
    – knzhou
    Jan 11 '17 at 5:44
  • $\begingroup$ A "one way light clock" most certainly does not function as expected in the context of special relativity. The relativity of simultaneity kills the notion from the get-go. $\endgroup$ Jan 11 '17 at 7:11
  • $\begingroup$ like this en.wikipedia.org/wiki/One-way_speed_of_light $\endgroup$
    – JMLCarter
    Jan 11 '17 at 21:05
  • $\begingroup$ @dmckee That may be true, but it's not a very scientific answer. Just makes me feel a bit smaller. $\endgroup$
    – JMLCarter
    Jan 11 '17 at 21:10
  • $\begingroup$ It tells you what to spend some more time reading and thinking about to understand the problem. Coming to terms with the relativity of simultaneity is a hurdle for everyone when they start learning relativity, not the least because it often seems unmotivated at first. Now you have a reason to put the effort in: to see why this is a problem. $\endgroup$ Jan 11 '17 at 21:12

Well , I see some deficiencies in your calculations and explanations:

1- I think in your first equation you must replace $L$ with $D$ and vice versa or you have to use $(1-\frac{v^2}{c^2})^{-1}$ instead of $(1-\frac{v^2}{c^2})$, because from O's viewpoint $D$ should be greater than $L$ according to your denotations for $D$ and $L$.

2- A one way light clock is meaningless. A clock must have its own "Tick"s and "Tack"s that represent a complete period. Moreover, tick and tack must both occur in one point where the point observer is located, regardless of the path which light travels, to illustrate a point clock for a point observer. Remember that the path of light can be any arbitrarily chosen path not essentially a straight line even it can be a function of a closed curvilinear using a set of infinitesimal mirrors and I proved this long ago that such a clock complies with Einstein's time dilation factor (the traditional gamma factor)!

  • $\begingroup$ 1 - Let's be clear why it's called "Length contraction". check this out en.wikipedia.org/wiki/Ladder_paradox, very clear - whether in the ladder or the garage frame distances in the other frame appear contracted. I add the diagram I used for the trig. $\endgroup$
    – JMLCarter
    Jan 11 '17 at 21:16
  • $\begingroup$ 2 - Interested to know more here, this could be the heart of the matter (I think mostly in the UK we would say "tick" and "tock", btw). So what "goes wrong" if the clock is not a "point clock", if the light path is not closed. Can you explain why this is necessary? $\endgroup$
    – JMLCarter
    Jan 11 '17 at 21:30
  • $\begingroup$ (I've no doubt this works for a closed path, that's not the question) $\endgroup$
    – JMLCarter
    Jan 11 '17 at 21:37
  • $\begingroup$ OK I also needed keyword Relativity of Simultaineity. I can find my answer thanks. $\endgroup$
    – JMLCarter
    Jan 11 '17 at 23:00
  • 1
    $\begingroup$ I correct that now I seem to have "length dilation", which must be wrong. $\endgroup$
    – JMLCarter
    Jan 11 '17 at 23:37

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.