A typical Lorentz Contraction proof relies on the axiom that "the speed of light is constant" and goes as follows. Given:

  • Frame $F_1$ moves at speed $v$ relative to frame $F_0$. In frame $F_1$ sit 2 parallel mirrors.
  • Distance between mirrors is measured as $l_0$ in $F_1$ (at rest relative to mirrors).
  • Distance between mirrors is measured as $l$ in $F_0$ (while mirrors move past in $F_1$ at speed $v$).
  • Time for light to make "roundtrip" between mirrors measured as $t_0$ in $F_1$ (at rest relative to mirrors).
  • Time for light to make "roundtrip" between mirrors measured as $t$ in $F_0$ (while mirrors move past in $F_1$ at speed $v$).
  • Already proved $t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t_0$ (time dilation).

A "roundtrip" of light passing between mirrors takes two trips; measured from $F_0$, those trips take times $t_1$ and $t_2$. During those trips, the ship travels $vt_1$ and $vt_2$, meaning light travels $l+vt_1$ and $l-vt_2$ when light moves in the same and opposite directions as $F_1$, respectively, all measured in $F_0$. The constancy of the speed of light gives:

  • Trip 1 (light moves same direction as $F_1$ relative to $F_0$): $c = \frac{l + vt_1}{t_1}$ $\Rightarrow$ $t_1 = \frac{l}{c-v}$
  • Trip 2 (light moves opposite direction as $F_1$ relative to $F_0$): $c = \frac{l - vt_2}{t_2}$ $\Rightarrow$ $t_2 = \frac{l}{c+v}$
  • So, $\color{red}{t = t_1 + t_2 = \frac{l}{c-v} + \frac{l}{c+v} =\frac{2lc}{c^2-v^2}= \frac{2l/c}{1-\frac{v^2}{c^2}} = \frac{2\gamma^2}{c} l}$.

Measured in $F_1$, the "roundtrip" distance is simply $2l_0$, and so $c=\frac{2l_0}{t_0} \Rightarrow t_0 = \frac{2l_0}{c}$.

Combining this with time dilation yields $t=\gamma t_0 = \gamma\frac{2l_0}{c} = \frac{2\gamma}{c}l_0$.

Putting it all together yields $$\frac{2\gamma^2}{c}l =t = \frac{2\gamma}{c}l_0 \Rightarrow l = \frac{l_0}{\gamma} \tag*{$\Box$}$$


Can I shorten this proof to just use "one trip" between the mirrors instead of a "round trip"? I have tried, and cannot! I $\color{red}{\text{have highlighted in red}}$ the part of the proof where the round trip yields some nice cancellation.

What am I missing?

  • There are proofs that rely on axioms other than "the speed of light is constant", but I'm looking for a proof that just relies on that.

  • The typical proof for time dilation $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$ involves light bouncing between two mirrors that are PERPENDICULAR to the motion of the reference frames. I went through this proof, and it absolutely does NOT break down when only one trip between the mirrors is considered. The proof in this question involves mirrors separated by a distance PARALLEL to the motion of the reference frames.

  • If "roundtrip" is unclear, here are two animations, each depicting two "roundtrips":

Roundtrips with tracers

2 roundtrips

First image made by me. Second image from Help Me Gain an Intuitive Understanding of Lorentz Contraction , which goes through this same proof based on the speed of light being constant.

A proof I’ve seen for time dilation is as follows, and only seems to require a single trip of a light beam:

Suppose a pair of mirrors separated by distance $L$ is moving past at speed $v$, such that the displacement between the mirrors is perpendicular to the motion of the mirrors. In the reference frame of the mirrors, light bouncing between the mirrors travels distance $L$ in $t_0$ seconds at speed $c=\frac{L}{t_0}$. In the reference frame relative to which the mirrors are moving at speed $v$, however, light bouncing between the mirrors takes time $t$ to do so and travels $\sqrt{(vt)^2 + L^2}$. So, $c = \frac{\sqrt{(vt)^2 + L^2}}{t}$ as well because the speed of light is constant to all observers. Solving for $t$ and substituting $t_0=\frac{L}{c}$ yields $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$.

light bouncing between mirrors separated by L

Image used is from what about doing the laser beam in a moving reference frame but with a ball

  • $\begingroup$ Hey Zach, in order to comply with our referencing guidelines you should really quote that image and indicate its source. If it's not clear how to do that, feel free to ask for help. Welcome to the site! $\endgroup$
    – David Z
    Jun 6, 2018 at 21:58
  • 4
    $\begingroup$ I wonder if this is related to the inability of measuring the one-way speed of light. $\endgroup$
    – Mark H
    Jun 6, 2018 at 23:21
  • 1
    $\begingroup$ A derivation of the Lorentz transformation doesn't have to talk about light at all. See, e.g., arxiv.org/abs/physics/0302045 $\endgroup$
    – user4552
    Jun 7, 2018 at 0:02
  • $\begingroup$ Zach, I note that your edit doesn't change the fact that you have an unattributed image. That still needs to be fixed. $\endgroup$
    – David Z
    Jun 7, 2018 at 0:17
  • $\begingroup$ @DavidZ spent some time making a similar gif in Mathematica (the red/blue one made by ME), but I actually like having them both here. In the Markdown I posted the link to the phsyics.stackexchange post from which I copied the other image. $\endgroup$ Jun 7, 2018 at 0:21

2 Answers 2


It's possible to do this with one trip, but it's a bit trickier to identify what's happening. Ultimately it amounts to effectively using Lorentz transformations, albeit in a more physical way in terms of the functioning of clocks in SR.

Consider one way trip from left to right mirror.

Let's note:

  1. Frame $F_0$ will claim it takes time $l/(c-v)$ between the two events.

  2. Frame $F_1$ will claim it takes time $l_0/c$.


Frame $F_0$ will insist that the clocks of $F_1$ are unsynchronized, with the one on the left leading the one on the right by an amount $\frac{v l_0}{c^2}$. Plus, moving clocks of $F_1$ are all running slow by a factor of $\gamma$ (you can find this in any basic book on SR $-$ eg. Resnick-Halliday-Krane or Resnick's book on SR).

$F_0$ will mumble to himself: Hmm! Silly guy, $F_1$ $-$ used two different clocks to measure the time elapsed between two events, but forgot to sync his clocks! That's why the time elapsed for $F_1$ wasn't $l_0/c$, but rather $l_0/c + vl_0/c^2$. Plus, these moving clocks of $F_1$ were all running slow by a factor of $\gamma$. So, the "actual time elapsed" between the two events was $\gamma\big(\frac{l_0}{c}+\frac{v l_0}{c^2} \big)$. And, of course, being the meticulous observer that I'm (who has all his clocks synced), this is what I measure. So $F_0$ will conclude,

$$ \frac{l}{c-v} = \gamma\bigg(\frac{l_0}{c}+\frac{v l_0}{c^2} \bigg). $$

Which leads to,

$$ l = \frac{l_0}{\gamma}.$$

So, what's up with a round trip? We can now understand that when you let the light do a round trip, $F_0$ cannot argue that two different (unsynchronized) clocks were used by $F_1$ in his measurement (because only one clock, sitting at, say, the left mirror, records both the events). The only grudge $F_0$ can now have is that the single clock used by $F_1$ was running slow by a factor of $\gamma$.

The great thing about relativity is that all observers have these grudges for one another $-$ and they can "fix" space & time measurements of another observer by noting when & where the events took place, & hence perpetually be under the delusion that they all measured space & time the right way! :-)

A very nice book which has amazing thought experiments in SR is the book by N. D. Mermin. In this book, he basically explores SR through light signals & doppler shifts, and I recommend it for the sheer joy it brings to the mind through numerous insights & the way it molds the intuition of a person used to non-relativistic mechanics.

  • $\begingroup$ How did you come with the $vl_0/c^2$? If these considerations hold when the ligth does just one trip, then why one can derive the right expressions when the mirrors are perpendicular to the motion of the frames without checking the synchronization of the clocks? $\endgroup$
    – Saoirse
    May 28, 2023 at 20:45

If you use the time dilation result you can use the fact that both static and moving observers agree on the relative speed of motion (by symmetry) to get the reciprocal factor for length contraction.


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