I take two mirrors parallel and facing each other.

A light pulse is being infinitely reflected between the two mirrors $M_1,M_2$. $E_1,E_2$ are the events when the pulse is reflected by $M_1,M_2$, and $\Delta t$ is the time between $E_1$ and $E_2$ for an observer that is not moving w.r.t. the mirrors, while $\Delta t'$ is the time between $E_1$ and $E_2$ for an observer that is moving with velocity $v$ w.r.t. the mirrors. We have

$$\Delta t'=\gamma \Delta t$$

and, as $v \to c$, $\gamma \to \infty$. Does that imply that $E_2$ does not take place?

  • 1
    $\begingroup$ No because as you know the speed of light is independent of the speed of the reference frame so the guy who is moving with the clock wouldn't notice a thing. $\endgroup$
    – Gonenc
    Apr 22, 2015 at 15:11
  • $\begingroup$ Why does your thought experiment need all the stuff with the mirrors? Why not just say, how do I interpret $t^\prime = \gamma(v) t$ as $v\to c$? $\endgroup$
    – innisfree
    Apr 22, 2015 at 17:22
  • $\begingroup$ as @gonenc suggests, are you interpreting $v$ in $\gamma(v)$ as the speed of the light bouncing between the mirrors? If so, you are mistaken... $\endgroup$
    – innisfree
    Apr 22, 2015 at 18:00
  • $\begingroup$ @innisfree The experiment was the basis on which the book I was reading introduced me to special relativity in like 20 pages. And v is the observer's velocity wrt the mirrors $\endgroup$
    – Anubhav
    Apr 23, 2015 at 14:41

2 Answers 2


It is worthwhile to note that $\Delta t' = \gamma \Delta t$ only for events with the same $x$ coordinate. The more general relation (i.e. Lorentz transformation) is: $$\Delta t' = \gamma \left(\Delta t - \frac{v}{c^2}\Delta x \right)$$

where we assume that $v$ is the $x$-component of the velocity of the observer moving with respect to the mirrors, so that to that observer, the mirror setup is moving at $-v$. By considering the time between successive $E_1$s (say), we can avoid having to consider $\Delta x = 0$, and in that case, we get, for the overall $E_1$ to $E_1$ cycle, $$(2\Delta t') = \gamma (2\Delta t)$$

Otherwise, the time between $E_1$ and $E_2$ will be different for the 'onward' and 'return' journey. For light going from $M_1$ to $M_2$, assuming that $M_2$ is in the positve $x$-direction from $M_1$, $\Delta x = c\Delta t$, and since $\gamma = (1-\frac{v^2}{c^2})^{-\frac{1}{2}}$, $$\Delta t'_{12} = \sqrt{\frac{c-v}{c+v}}\Delta t$$ For light going from $M_2$ to $M_1$, $\Delta x = -c\Delta t$, and $$\Delta t'_{21} = \sqrt{\frac{c+v}{c-v}}\Delta t$$ The former goes to zero, and the later goes to infinity as $|v| \rightarrow c$, with $v \gt 0$. For $v < 0$, this behavior is inverted.

Also note that in the moving observer's frame, the separation between $M_1$ and $M_2$ goes to zero. The situation looks like a pair of flat mirrors in contact with the light pulse at the interface, all travelling at the same speed. It takes no time for light to go from one mirror to another in one direction, but infinite time in the other. We can therefore get the same results by considering length contraction instead of time dilation (this may also be somewhat more intuitive here).

Therefore, if $M_2$ is in the positive $x$ direction from $M_1$, and $v \gt 0$, then it turns out that $E_2$ happens instantaneously after $E_1$, but it takes an infinite time for $E_1$ to occur again. This obviously changes (i.e. $E_1$ and $E_2$ may be exchanged) with the 'signs' of the mirror separation and observer velocity. This is one of the reasons that, for simple thought experiments with 'light clocks', the mirror separation is chosen in a direction perpendicular to the relative motion: so that arguments about 'time elapsed' are uniform.

As an additional remark, note that there is no problem with the overall idea that some things may be infinite when you approach the speed of light; that is built into special relativity almost by design.

  • $\begingroup$ Thanks for the explanation. I understood the second part about the distance contracting to 0.im still very doubtful about how the time for the 'onward' and 'return' journey would be different. Could you also suggest a good book for special relativity for beginners? $\endgroup$
    – Anubhav
    Apr 23, 2015 at 14:55
  • $\begingroup$ Let's put ourselves in a rest frame in which the mirror setup is moving with velocity $v$. When the light goes in the direction of positive $v$, the mirrors also move in this direction, and light has to 'catch up' to the second mirror. When the light goes in the other direction, the mirror comes towards it, and therefore, it takes less time. As for some experimental verification of this, a rather similar effect results in the Doppler shift, where the time period of a light wave is less when it moves in the direction of the motion of the source, as opposed to the opposite direction. $\endgroup$
    – AV23
    Apr 23, 2015 at 15:51
  • $\begingroup$ You're best off choosing a book that you feel suits your needs, but maybe you can try Vol 1 of the Feynman Lectures on Physics, or its relativity-specific sub-book, Six Not So Easy Pieces. $\endgroup$
    – AV23
    Apr 23, 2015 at 15:53

I take two mirrors parallel and facing each other.

For simplicity, I'll first discuss the case that these two mirrors were and remained at rest to each other.

A light pulse is being infinitely reflected between the two mirrors $M_1$, $M_2$. $E_1$, $E_2$ are the events when the pulse is reflected by $M_1$, $M_2$ [...]

Accordingly (and as far as I understand your notation), the (constant) distance between these two mirrors can be expressed as

$$\frac{c}{2}~\tau_{\text{ping}} = \frac{c}{2}~\sqrt{s^2[~E_{(k)},E_{(k + 2)}~]}$$

(with the suitable sign convention of the interval between two time-like separated events being positive).

[snip elaborations involving additional participants ...] Does that imply that $E_2$ does not take place?

Reflection event $E_2$ did take place due to your own setup description; therefore as a matter of fact (for any "valid trial" matching to your description) for any and all conceivable observers, and of course also independent and regardless of any possible subsequent assignments of coordinates (such as $t$, or $t'$) to events.

Further, if it is indeed (meant as) part of your setup prescription that the two mirrors were and remained at rest to each other (throughout any "valid trial"), then subsequent reflection events $E_{(k)}$ are of course prescribed outright to have taken place, too (perhaps only up to some interger $k_{\text{max}}$, if you so specify, for a trial to be valid).

Otherwise, if the mirrors may accelerate away from each other (in the course of a "valid trial", by your setup prescription), then it is of course not guaranteed that a reflection event "$E_3$" (or indeed any subsequent reflection) would have occured (see for instance "hyperbolic motion").


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