Suppose a clock, located at point $x'$ in the inertial frame $S'$, registers two events $t_1'$ and $t_2'$. Let $\Delta t'=t_2'-t_1'$. The same two events will be registered by two different (synchronized in their rest frame) clocks, located in two different points of the inertial frame $S$ (which moves with a certain speed relative to $S'$). Let $\Delta t=t_2-t_1$ the elapsed time between these two events as seen from $S$. From the theory of relativity we know that $\Delta t' = \sqrt{1-\beta^2}\Delta t$.

At first glance, the time dilation formula looks asymmetrical. We know that all of the clocks in $S$ are synchronized. Thus, it is tempting to assume that $\Delta t$, which is calculated by two different clocks in $S$, can be calculated by one clock in $S$. But then we have a situation where two perfect and equivalent clocks - one from $S$ and one from $S'$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

My textbook says that in order to resolve this "paradox", one must realize that clocks which are synchronized in one inertial frame need not be synchronized in another inertial frame. That is - the clocks "belonging" to the system $S$ aren't showing the same time when seen by an observer in $S'$.

However I don't quite understand this argument. Why do we care what the clocks of $S$ show to an observer in $S'$? As far as I understand, the time interval $\Delta t$ in the formula above is the time elapsed according to the clocks in $S$ as seen by an observer in $S$. In other words - the formula connects what an observer in $S'$ sees to what an observer in $S$ sees = two different observers, not one.

  • $\begingroup$ The situation is symmetrical. For the unprimed observer to measure the elapsed time of the primed observer's moving clock, he has to use two different but synchronized clocks. Vice versa for the primed observer if he wants to measure the unprimed observer's clocks. Each will measure the other's clock ticking more slowly. The key to the apparent paradox is realizing that "same time" measurements made by one observer are not "same time" measurements according to the other observer. I suggest Mermin's "It's about Time: Understanding Einstein's Relativity". $\endgroup$
    – user93237
    Commented Jan 3, 2016 at 21:03
  • 2
    $\begingroup$ There's a subtle difference. To the observer in S', the two events occur at the same location. To the observer in S, the two events occur at two different locations. So the same formula used in going from S' to S cannot be applied in going from S to S'. $\endgroup$ Commented Jan 3, 2016 at 21:12
  • $\begingroup$ @ChesterMiller - this does not answer OP's question. $\endgroup$
    – cth
    Commented Jan 4, 2016 at 9:13
  • 1
    $\begingroup$ I do not understand your paradox clearly. $\endgroup$
    – Alice
    Commented Jun 2, 2020 at 7:42

6 Answers 6


Perhaps you might find it helpful to consider an analogy involving distances instead of time. Suppose in your coordinate system your x axis is horizontal, but mine slopes down to the right somewhat, and that your frame of reference is moving to the right while mine is still.

Suppose you measure the height at which which two events occur at a certain point along your x axis, and you find a height difference dh between them. In my reference frame the the second event will occur further along my x-axis than the first (as my axis is moving relative to yours), and since my x-axis is sloping down compared with yours I will find that the second event is higher than I would had my x-axis been parallel to yours, so I find a different value for dh. The effect is nothing to do with how I measure the height, or whether I use one ruler for both measurements or two- it arises from the fact that the events are spread along my axis, and my axis is at a different slope to yours.

Exactly the same situation arises with simultaneity, except with height (ie a measurement in the y axis) replaced by time (a measurement in the t axis). You measure the times of two events at one point on your spatial axis. I measure them at two points along mine, and mine is sloping relative to yours so I get different values for dt (whether I use one clock/ruler or two).


Your question is not about relativity in particular so much as it's about perceived asymmetry in general. So here's an example that captures exactly the same kind of "asymmetry".

Stand on the equator, facing South. The South Pole is roughly 6000 miles in front of you and the North Pole is roughly 18000 miles in front of you. (That is, you'd have to walk 18,000 miles in the forward direction to make it 3/4 of the way around the globe to the North Pole.)

I'm standing next to you, facing North. In my coordinate system, the North Pole is 6000 miles in front and the South Pole is 18,000 miles ahead.

Our measurement systems are equivalent in the sense that they make exactly the same predictions for the consequences of any particular journey. For example, if we both go due South, we agree that we'll get to the South Pole in 6000 miles, though you'll describe that journey as walking forward and I'll describe it as walking backward.

So according to you, the South Pole is closer (in the forward direction) than the North. According to me, the opposite is true.

Now to paraphrase your argument:

Here we have a situation where two perfect and equivalent measurements are no longer equivalent (they give different values for the distance from where we're standing to the South Pole). But this contradicts the equivalence of our measurement systems!

Or does it?


At first glance, the time dilation formula looks asymmetrical.

You need to use the complete Lorentz transformation for time to see if the formula is symmetric. In your question, let $\Delta x^\prime=0$, Lorentz transformation implies:

$$\Delta t=\frac{\Delta t^\prime+(v/c^2)\Delta x^\prime}{\sqrt{1-\beta^2}}=\frac{\Delta t^\prime+(v/c^2)×0}{\sqrt{1-\beta^2}}=\frac{\Delta t^\prime}{\sqrt{1-\beta^2}} \space ,$$

which is your result. However, Lorentz transformation for time in $S^\prime$, yields:

$$\Delta t^\prime=\frac{\Delta t-(v/c^2)\Delta x}{\sqrt{1-\beta^2}} \space ,$$

Remember that in, $S$, we have $\Delta x=v\Delta t$, which implies:

$$\Delta t^\prime=\frac{\Delta t-(v/c^2)\Delta x}{\sqrt{1-\beta^2}}=\frac{\Delta t-(v^2/c^2)\Delta t}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}\Delta t \space ,$$

This equation is similar to the first one, which shows the symmetry in relativity.

But then we have a situation where two perfect and equivalent clocks - one from $S$ and one from $S′$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

I think you have mistaken time rates (fast or slow) for the elapsed time (readings of the clocks). If you consider the first one, according to the symmetry, each observer measures, say, the angular velocity of the other clock slower (one says $\omega^\prime/\omega=\gamma$, and the other one deduces $\omega/\omega^\prime=\gamma$ instead.), however, both observers agree that the readings of the clocks ($\Delta t$s) are similar in both frames as I showed through the equations above. This similarity is due to clocks synchronization. That is, if two clocks starts ticking simultaneously in one frame; from the viewpoint of other frame, one of the clocks starts ticking earlier than the other one does despite each observer sees the other clock slower.

I hope that I've understood your question correctly.


But then we have a situation where two perfect and equivalent clocks - one from S and one from S′ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

Whenever you say something, like ...(times) are no longer equivalent/one of them is slower than another.... You automatically put yourself into one of the inertial frames, here it is frame S.

On the flip side, the statement is still true if you are within the context of frame S'. You would still say the other clock seems slower than yours, within the context this S' frame. Therefore, the symmetry holds, in that sense.

To generalize, whenever you describe something in the language of the theory of special relativity, unless it is something invariance like the spacetime interval, be aware that most statement comes automatically with a certain inertial frame.


But then we have a situation where two perfect and equivalent clocks - one from $S$ and one from $S′$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

The two clocks are not equivalent. For $S$, one of them is at rest and the other is moving at a speed $v=\beta c$.

What $S$ does is that he uses two different clocks stationary in his frame to measure the time elapsed on a single moving clock of $S′$.

$S′$ does the same, but with two different clocks stationary in his own frame.

So, the situation is not symmetrical in the absolute sense. When $S$ or $S'$ talk about measuring how a moving clock runs slow, they are doing different experiments. Therefore, even though each feels the clocks of the other run slower, there is no contradiction.

Indeed, what the principle of relativity tells you is that the situation is symmetrical, but only in the "relative sense" - i.e., $S$ will come to exactly the same conclusions about the clocks of $S'$, as $S'$ will come to for clocks of $S$.

Physical Resolution

There is no paradox if you do the analysis correctly. So this is technically not a resolution, but rather a way to see that there is no conflict with the conclusions of special relativity. So here goes a more physical explanation using measurements:

We first recall two important principles when dealing with time measurements (these can be derived from simple thought experiments):

  1. A moving clock run slows by a factor of $\gamma = \frac{1}{\sqrt{1-\beta^2}}$.
  2. Clocks of S are unsynchronized in the frame of $S′$ (and vice versa), with the one at the rear operationally leading the one at the front by an amount $vl/c^2$ ($l$ being the rest separation between the clocks, and $v$ the relative speed).

Now let us try to understand why $S′$ can reconcile with the fact that $S$ thinks that the clocks of $S′$ run slower (while it is indeed "true" that the clocks of $S$ run slower for him).

Here is the claim of $S$: "While $\Delta t$ time elapsed for me, a moving clock of $S'$ only advanced by $\Delta t/\gamma$."

Let us imagine that $S′$ is moving to the right of S with speed $v$.

$S′$ notes that the measurement process through which $S$ claimed that the clocks of $S'$ are slower involve measurement of the time shown by one single clock of $S'$ by two different clocks of $S$. $S'$ will then process this information as follows: "The time elapsed between the two measurements for $S$ was $\Delta t$. Silly $S$ $-$ he did the measurement with two unsynchronized clocks with rest separation $l=v\Delta t$. That means the two clocks used in his measurement were already unsynchronized by an amount $vl/c^2 = \beta^2 \Delta t$. Moreover, because the clock at the rear leads the one at the front, he measured an extra time difference $\beta^2 \Delta t$ purely because of clock asynchronization. The correct time measurement would have been $\Delta t'= (1-\beta^2) \Delta t$. But, since his clocks are always slow by a factor of $\gamma$, the 'actual time' elapsed for me is $\gamma \times \Delta t' = \Delta t/\gamma$ $-$ which is indeed what I measure!"

  • $\begingroup$ Thank you for your response. I'm assuming you meant to say: "$S'$ notes that the measurement process by which $S$ claimed his clocks are faster involve measurement of the time shown by one single clock of his by two different clocks of $S$", since $\Delta t > \Delta t / \gamma$, i.e. more time elapsed for $S$ than for $S'$ and thus $S$ can claim his clocks are faster. $\endgroup$
    – grjj3
    Commented Jun 4, 2020 at 12:46
  • $\begingroup$ It's a pronoun issue. I used 'he/him' for both. I'll edit it shortly for clarity. But for now, indeed, every observer basically claims that the clocks of the other guy run slow. $\endgroup$
    – Vivek
    Commented Jun 4, 2020 at 12:55
  • $\begingroup$ @grjj3 Done the edits. Let me know if anything is unclear. $\endgroup$
    – Vivek
    Commented Jun 4, 2020 at 14:05

Relativity is a theory of frames of reference. The question says that $\Delta t$ is the time measured between two events by two synchronized clocks in frame $S$. But now, here's the catch: $\Delta t$ is actually the time between the two events as observed by $S$, as observed by $S'$.

Usually, in SR, we try and compare the flow of time and the measure of lengths, with respect to another frame. So, the $\Delta t$ in the equation above is what the person standing in $S'$ sees in his frame. Let me try and explain.

Suppose you look at it from the perspective of $S'$, with $S$ travelling relative to $S'$ with velocity $v$. Let's give them both stopwatches. Now, let two events $A$ and $B$ happen in $S$. When $A$ happens, $s$ (the observers are denoted by small letters here) starts the stopwatch. When $B$ happens, he stops the stopwatch. The time between them is $\Delta t$, as we know from the question. Now, will the time between them, as seen by $s'$ be the same? No. The stopwatch will start and stop at a different time, so you get $\Delta t'$.

But now, say $s'$ is trying to see how much time elapsed according to $s$. He will see that $$\Delta t = \frac{\Delta t'}{\sqrt{1 - \beta^2}}$$

That's where we have to compare two observers instead of one, and think about how the clocks in two separate frames are synchronized.

Coming back to the question the OP asked, it does matter how clocks are 'ticking' in other frames, when you are comparing some quantity in your frame to the others. That's what the word 'relativity' means. How $S'$ behaves relative to $S$ or vice versa.

When $S'$ tries to measure what $\Delta t$ is going to be, of course he cares about how the clocks are synchronized in that frame. Because, he is trying to find out how much time has elapsed in the other frame.

$\Delta t$ in this context isn't the time that $s$ measures. Instead, it is the time that $s'$ sees has passed for $s$. And how does he 'see' how much time has passed for $s$? By finding out what time the clocks (or stopwatches) in $S$ are showing in between the two events (which includes measuring whether they are synchronized or not to get a better idea), and comparing those values with that of his own clocks.

So, that is why it matter to know what is going on in other frames. You may see a lot of bold and italics in this answer, and that is because this is a delicate concept, and you need to know what's going on exactly and not get caught up on the long sentences.

  • $\begingroup$ So $\Delta t$ is the elapsed time in frame $S$ as seen by the observer $s'$ (located in frame $S'$), and not the elapsed time in frame $S$ as seen by the observer $s$ (located in $S$)? Then what's the connection between these two observations? I.e., what is the relationship between $\Delta t$ and the elapsed time in frame $S$ as seen by the observer $s$? And if both $\Delta t$ and $\Delta t'$ are elapsed times in different frames as seen by one observer ($s'$) in a specific frame ($S'$), why does measuring $\Delta t$ makes sense if from his viewpoint clocks in $S$ aren't synchronized? $\endgroup$
    – grjj3
    Commented Jun 4, 2020 at 12:20
  • $\begingroup$ @grjj3 Yes, sort of. See, $\Delta t$ is the time as observed by $s$, but the catch is that $s'$ is also observing $\Delta t$ (he is trying to find out how much time has elapsed for $s$). So, it's a double observation of the same phenomenon. Time measured by $s$, which is 'seen' (measured) by $s'$. One thing you have to keep in mind is that $\Delta t$ is going to be different when you ask $s$ (i.e. if you ask him how much time has elapsed according to him, his answer is going to differ). Same for $\Delta t'$. So, long story short: both the time intervals are not measured in one frame; $\endgroup$
    – PNS
    Commented Jun 4, 2020 at 12:38
  • $\begingroup$ Continued. Both the intervals can be measured by both the frames. That is the point. So, even $s$ can try and measure $\Delta t'$, and get a different value. It is all about observers trying to observe what is happening in other people's frames. $\endgroup$
    – PNS
    Commented Jun 4, 2020 at 12:40
  • $\begingroup$ Thank you. Can you recommend a good literature that touches on this delicate subject? (With emphasis on the viewpoint of different observers on clocks in different reference frames). $\endgroup$
    – grjj3
    Commented Jun 5, 2020 at 14:15
  • $\begingroup$ I can't pinpoint any good ones off the top of my mind. I think the most rigorous explanation for this comes from some of the works of Einstein and Lorentz themselves. Otherwise, I can't say that reading a few books can help you with this delicacy. You just have to read as much as you can and absorb it, to get the essence of it. Einstein especially was very fond of this phenomena, and although I haven't read every of his papers on relativity, I think you can get a very nice grasp of this idea from his writings (though sometimes he can be hard to understand). $\endgroup$
    – PNS
    Commented Jun 6, 2020 at 6:01

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