I'm confused by the derivation of the expression for time dilation as well as the concept itself. I've read multiple answers, textbooks, watched lectures, and they confused me further due to subtleties that are not fully explained. Here's what I know tho. Let $S'$ be an inertial frame moving with velocity $v$ with respect to another frame $S$. The Lorentz transformation equation of the time coordinate is $$t'=\gamma\Bigl(t-\cfrac{v}{c^2} x\Bigr)$$ So the time interval between two events would be $$\Delta t'=\gamma\Bigl(\Delta t-\cfrac{v}{c^2}\Delta x\Bigr)$$ This is the most general expression for the time difference between two events. If the time in $S$ is taken on one and the same clock, then $\Delta x=0$ and $\Delta t'=\gamma\Delta t$ which means the time difference in the moving frame is greater than in the rest frame. But $\Delta x$ needn't be zero. Or should it be? Should the time difference in $S$ be taken on the same clock? Why not any two clocks in any two locations $x_1$ and $x_2$? If all the clocks in $S$ are synchronized, it shouldn't matter, right?

If that's right, however, then if the two events under consideration are the moving observer coinciding with point $x_1$ at time $t_1$ and $x_2$ at time $t_2$, then we can write $\Delta x=v\Delta t$ and that would give us $\Delta t'= \Delta t/\gamma$ which says that the time on the moving clock is less than on the stationary one. Is this in agreement with the case of $\Delta x=0\,$? Also, $\Delta x$ doesn't have to be either of the previous cases. Can't we talk about events far away from where the observers are located such that $\Delta x$ is neither of the two cases?

Finally, what does it mean for a clock to run slower? Does it mean that, say, the hands are moving slower, in which case it would measure less time difference between events? Or does it mean that the time difference between two events is greater on the said slower clock which gives the name "time dilation"?

  • $\begingroup$ You are not considering the same two events in the upper paragraphs and the lower paragraphs. The same two events can't have both $\Delta x = 0$ (as discussed in your first paragraphs) and $\Delta x \neq 0$ (as discussed in your latter paragraphs). See my answer for a more complete discussion. $\endgroup$
    – hft
    Oct 29, 2021 at 18:19

6 Answers 6


(a) I find talk of 'moving clocks' and 'stationary clocks' confusing. I recommend this definition of time dilation:

The time interval between two events is least in the inertial frame in which the events occur at the same place (strictly: with the same spatial co-ordinates). In any other inertial frame the events occur in different places and the time interval between them is greater by a factor of $\gamma(v)=(1-v^2/c^2)^{-1/2}$ in which $v$ is the relative velocity between the frames. [In Special Relativity we may call the shortest time 'the proper time' and the longer times 'improper times'.]

In your question you start by having the events occurring at the same place in the S frame. $\Delta t$ is therefore the proper time between the events. You have argued correctly that in the S' frame the time $\Delta t'$ between the events is longer by a factor of $\gamma(v)$.

(b) You then go on to ask: "Should the time difference in S be taken on the same clock? Why not any two clocks in any two locations 𝑥1 and 𝑥2 ? If all the clocks in S are synchronized, it shouldn't matter, right?"

The point about the S frame is that it is special for the two events, because they take place with the same spatial co-ordinates in this frame, so you need only one clock, at the place where both events occur. I suppose that you could use clocks elsewhere in S, but you'd have to allow for transit times of signals sent from the point at which the events occur to the clocks that you are using! But $\Delta x$ is still zero, because $\Delta x$ is the distance between the points at which the events themselves occur!

(c) "what does it mean for a clock to run slower?" It means that it registers a shorter time between the same two events.

I dislike this form of words in Special Relativity. It makes it seem as if the clock's mechanism has been affected, whereas time dilation is due to the interconnectedness of time and space.

  • $\begingroup$ I don't see why $\Delta x$ is still zero if we use clock(s) elsewhere. If, in S, a clock emits a light signal and after that another clock at a different location emits a light signal, how is $\Delta x=0$ in this case? Or do you mean that the measurement is carried out at the same location by one observer after information has travelled from the first and second clocks locations to the observer's location? The measurement would then be carried out by one clock at the same position. Is that what you meant? $\endgroup$
    – EM_1
    Oct 29, 2021 at 18:38
  • $\begingroup$ It's much simpler than that. I'd forget about 'observers'. They confuse the issue. We're talking about two events. The S frame is special because the events take place at a single spatial point, P. So $\Delta x = 0$. And only one clock is needed – at the point in question. Choosing to read the time interval in the S frame using clocks at points other than P doesn't alter the fact that the events occur at P so $\Delta x = 0$. But it means that you have to arrange for signals to go from P to the clocks, and you have to allow for the transit times of the signals – complicated! $\endgroup$ Oct 29, 2021 at 18:58
  • $\begingroup$ I see. So we consider the (proper) frame and compare every other frame to it instead of worrying about who's moving relative to what. Is that a fair summary of your answer? I still have one question regarding part (c) of your answer. Does it mean that the clock that measures the proper time runs the slowest? In the language you detest, the stationary/proper clock runs the slowest? $\endgroup$
    – EM_1
    Oct 29, 2021 at 19:20
  • $\begingroup$ (a) "Is that a fair summary of your answer?" To be brutally honest, I don't think I said this at all! If you are to apply the Lorentz transforms – as you did correctly at the beginning of your original post, you need to know the relative velocity between the two relevant frames. In the case of time dilation one of these frames is the one in which the events occur in the same place. What I was emphasising was that only one clock (at the place where the events occur) is needed in this special frame. (b) "the stationary/proper clock runs the slowest?" Yes, that's right (through gritted teeth). $\endgroup$ Oct 29, 2021 at 19:42
  • $\begingroup$ A personal note... I found Special Relativity much easier to understand without the terms 'moving clocks', 'stationary clocks', 'observers', 'clocks running slow', 'clocks running fast'. And dwelling too long on thought experiments using light signals tends to make one think, wrongly, that light propagation is a key part of SR. I much prefer the approach of Taylor and Wheeler in Spacetime Physics. $\endgroup$ Oct 31, 2021 at 9:08

But Δx needn't be zero. Or should it be? Should the time difference in S be taken on the same clock? Why not any two clocks in any two locations x1 and x2?

The time difference that is measured on a clock at the same location is called the proper time. This will be the smallest measured time difference of any observer.

If you choose to assume that $\Delta x = 0$ (or rather, you choose to considerer two events A and B at the same location in S) then in the S frame the clock measuring the time between the events is stationary (i.e., the events occur at the same location in S) and $\Delta t$ is the proper time.

This time $\Delta t$ will be less than that measured by a moving observer. But, don't forget that in this case the moving observer is in the S' frame and is moving at velocity $v$ as measured in the S frame.

In this case, indeed you will have $\Delta t' = \gamma \Delta t$, such that $\Delta t'$ is bigger than $\Delta t$.

However, note that the location of the moving observer in the S' frame in this case can not be the same as the location of the measured events we are discussing above, since the observer is moving in this frame and the measured event is at the same location in this frame.

however... which says that the time on the moving clock is less than on the stationary one.

If you want to consider the moving observer to be measuring the proper time interval between events, you can do that too, but in this case you will not have $\Delta x = 0$ and $\Delta t$ will not be the proper time. Now $\Delta t$ and $\Delta x$ are referring to the difference between different events than the events discussed above. The events will necessarily be different than the events discussed above. We should probably use labels like $\Delta t_{A,B}$ and $\Delta t_{C,D}$ to differentiate between the time difference between events A and B (the ones considered above) and the time difference between events C and D (the ones considered here). We usually don't do all this extra labeling though.

For example, if the S frame is fixed on earth and the S' frame is in a spaceship moving away from earth, then if we consider measurements of events C and D that occur at the same location in the spaceship, then $\Delta x'=0$ and it is $\Delta t'$ that is the proper time not $\Delta t$.

In this case, indeed, you will have $\Delta t = \Delta t' \gamma$ such that $\Delta t$ is bigger than $\Delta t'$.

This could seem quite confusing if we don't include the extra labels, but with the extra labels is it probably more clear: $$ \Delta t'_{A,B} = \gamma \Delta t_{A,B} $$ $$ \Delta t_{C,D} = \gamma \Delta t'_{C,D} $$

The events A and B are measured at the same location in frame S. The events C and D are measured at the same location in frame S'.

Update based on comments:

There is no reason why either the observer in S or the observer in S' necessarily has to measure the proper time. Indeed, the events could be spatially seperated in both frames. In this case just fall back to the general Lorenz transformation equations for the differences: $$ c\Delta t' = \gamma c\Delta t - \beta \gamma \Delta x $$ $$ \Delta x' = -\beta\gamma c\Delta t + \gamma \Delta x\;. $$

For example, suppose that the relative velocity between S and S' is v = 200m/s and the origins coincide at time zero. And suppose further that, in S, event E is at time $t_E= 0$ seconds and location $x_E = 0$ meters and event F is at time $t_F = 1$ second and location $x_F = 100$ meters. These events can also be seen to be spatially separated in S' and neither S or S' measures the proper time.

In S' Event E also occurs at $t'_E=0$ and $x'_E=0$ and we have $$ ct'_F = \gamma c t_f - \beta \gamma x_F = \gamma c 1s - \beta \gamma 100m $$ and $$ x'_F = \gamma x_f - \beta \gamma c t_F = \gamma 100m - \beta \gamma c 1s $$

Or $$ t'_F = \gamma (1 - \frac{200}{300000000} 100/300000000) = \gamma (1 - \frac{1}{4500000000000})\quad \text{seconds} $$ and $$ x'_F = -\gamma 100 \quad \text{meters}\;, $$ where I have approximated $c = 300000000m/s$ and where, in this case, $\gamma$ is approximately 1.0000000000002.

  • $\begingroup$ I'm aware that the examples don't consider the same two events. Your labeling was indeed helpful. I believe a fair summary of your question and of time dilation generally is that the proper time, the time measured between two events occurring at the same spatial location, is the shortest possible time difference between the two events. I hope I haven't misrepresented your answer. However, the last two parts of my question remain unaddressed (at least I still don't understand them). Can we not consider two events that occur in different spatial locations in both S and S'? $\endgroup$
    – EM_1
    Oct 29, 2021 at 18:27
  • $\begingroup$ In this case, neither the time measured in S nor that measured in S' will be the proper time, right? If so, what's the relationship between the time difference between the events as measured by S and S'? Is it just given by the most general formula of the lorentz transformation for the time coordinate? $\endgroup$
    – EM_1
    Oct 29, 2021 at 18:28
  • $\begingroup$ @user626542, Yes we can consider events that are in different spacial location in both S and S'. In this case the relationship between the measured time differences and space differences in the different frames will be given by the lorentz transformation equations. $\endgroup$
    – hft
    Oct 29, 2021 at 18:59
  • $\begingroup$ @user626542, I didn't see this latter part of your question. It's generally not a good a idea to include multiple question in one post, and the question might get closed for lack of focus. I can provide an additional example in my answer directed at the situation when neither observer measures the proper time. $\endgroup$
    – hft
    Oct 29, 2021 at 19:01
  • $\begingroup$ I see. Thank you for your answer and I will be careful not to include too many questions $\endgroup$
    – EM_1
    Oct 29, 2021 at 19:07

If the time in S is taken on one and the same clock, then Δx=0 and Δt′=γΔt which means the time difference in the moving frame is greater than in the rest frame.

I like the mantra: "moving clocks run slowly" to avoid becoming in doubt in SR.

The meaning of $\Delta x = 0$ is that S' compares the time interval shown by its clock with a clock on a given $x$ in S frame. In this case, we can invert the perspective and see that from the static (in his frame) clock of S', time runs slowly in a given location of S. That is the usual concept we have of clocks in SR: they run slowly in a moving frame.

But as you say, $\Delta x$ can be non zero. For example S' can compare the readings of its clock with a clock in S on $x = 0$ and later on with a syncronized (in the S frame) clock on $x = \Delta x = v\Delta t$. In this case: $$\Delta t' = \gamma(\Delta t - v^2 \Delta t) \implies \Delta t' = \frac{\Delta t}{\gamma}$$ Now, S' is not following the readings of the same moving clock. It compares the readings of 2 clocks in S. Its is reasonable now to call S' the moving frame. The conclusion is the same: the moving clock runs slower.

Of course S' can say that S is moving and one of the clocks is receding while the other is approaching. But it is like going from Earth to Alpha Centaury, and compare with a clock here and the other there. It is easier to call this a trip from here to the star, and not that the star comes to the ship, while the earth goes away.

  • $\begingroup$ Before making assumptions about the answer, I'd like you to clarify what you mean by "runs slower" first. $\endgroup$
    – EM_1
    Oct 29, 2021 at 20:20
  • $\begingroup$ @user626542 if my clock shows say $t + 1$ hour , while your clock shows $t + 0,5$ hour, your clock is runnng slowly. $\endgroup$ Oct 29, 2021 at 20:36
  • $\begingroup$ So if my clock measures less time, it's running slowly? What if my clock measures the proper time? It's, by definition, stationary. Wouldn't that mean that the stationary clock runs slower? $\endgroup$
    – EM_1
    Oct 29, 2021 at 20:53
  • $\begingroup$ If a ship travels to AlphaCentaury, and the clocks there are synchronized with the earth, the clock of the ship will show say 5 years, while AC clock 6 years. Proper time is bigger when compared to other moving clock. In this case the comparison involves 2 clocks. $\endgroup$ Oct 29, 2021 at 21:14
  • $\begingroup$ @user626542 a scientist works in a laboratory S and places two synchronized clocks in points $x_1$ and $x_2$. Then then he moves the third clock S' from point $x_1$ to $x_2$. So, the "moving" clock S' changes its spatial position in the S frame. Two spatially separated and Einstein - synchronized clocks that are located in the points $x_1$ and $x_2$ measure greater time interval that a "single" moving clock of the frame S'. That is time dilation: physics.stackexchange.com/questions/481813/… $\endgroup$
    – user139020
    Oct 29, 2021 at 21:31

The answers from Marco Ocram and Phillip Wood look good to me but I think some basic definitions might help demystify time.


  • Time is what a clock measures.

  • A clock is a system that repeats a particular process over and over and counts the iterations.

  • Two clocks are identical if they repeat the same process.

  • Any repeatable process is a clock if something is there to count it.

The universe is such that, for any given inertial reference frame, the average number of iterations of a specific repeating process (such as a clock might count) that transpire between the start and end of another specific process is identical. That is: one characteristic of any specific process is its most likely duration, expressible as a number $n$ of iterations on a sufficiently precise clock in the same inertial reference frame.

It doesn't mean much to ask how fast or slow a clock is running. Speed implies a rate of change with respect to time. A clock measures time, and the rate of change of time with respect to time is always dimensionless $1$. The clock runs at one count per iteration of its process, which has a fixed ratio (with slight statistical variation) to all iterations of all other processes.

Time Dilation

The universe is such that when a clock in some inertial reference frame has counted $n$ iterations of its repeating process, an identical clock with a fixed relative velocity $v$ will have counted $n/\gamma (v)$ iterations, as measured by the same frame that measured $v$.

That is to say, in any given inertial reference frame A,

  • Processes in all other inertial reference frames evolve, with respect to a clock in frame A, at a slower rate than an identical process in frame A would evolve with respect to a clock in frame A.

  • Processes in all other inertial reference frames evolve, with respect to a clock in their own reference frame, at the same rate an identical process in frame A would evolve with respect to a clock in frame A.

  • $\begingroup$ So if a process in $A$ takes n identical steps or iterations to be completed, then by the time it's completed in $A$, the number of steps completed in a frame moving with velocity $v$ with respect to $A$ is $\cfrac{n}{\gamma}$ which is less than n. Hence the process takes longer to be completed in the moving frame compared to how long it takes in $A$. Is that correct? Is that what it means for a clock to run slower? That any process that takes time $t$ in $A$ takes time $t'>t$ in the moving frame with respect to $A$? $\endgroup$
    – EM_1
    Oct 30, 2021 at 7:23
  • $\begingroup$ Suppose process in frame $A$ takes $n$ ticks of a clock in frame $A$. Then as measured by frame $A$, identical process in frame $B$ takes $n'=\gamma(v_{AB})n$ ticks of a clock in frame A and $n$ ticks of an identical clock in frame $B$. As measured by frame $B$, original process takes $n$ ticks of a clock in frame $A$ and $\gamma(v_{BA})n$ ticks of a clock in frame $B$. $\endgroup$
    – g s
    Oct 30, 2021 at 23:52

Time dilation does not mean that clocks slow down when they move- that is a common misconception.

Time dilation arises from the relativity of simultaneity, which is responsible for length contraction also.

The effect is simplest to explain in the case of a 2-d spacetime. In a given reference frame, lines parallel to the x-axis are lines of constant time- anywhere along such a line it is the same time t. However, when two frames move relative to each other, their time axes tilt relative to one another, so what constitutes a level line of constant time in one frame is a sloping slice through time in the other, the slope rising upwards in the direction of motion. That means a constant time t in one frame is a successively later time in the other as you move further away from the origin.

By analogy, imagine walking down a corridor at 1m/s in which there was a person holding a clock every ten metres, but those clocks are all set 1 second ahead of each other. After walking 10m you reach the first person- 10 seconds have passed on your watch, but the person's clock shows 11 seconds have passed, because it has been set 1s ahead. You walk another 10m to reach the next person- 20s are shown to have passed on your watch, but their clock shows 22s, because it was 2s ahead. As you go further and further down the corridor, time on the clocks gets increasingly ahead of the time on your watch. By the time you have walked 200m, 200 seconds have passed on your watch, but the adjacent person's clock shows 220 seconds. Compared with the clocks of the people you are passing, your watch appears to be time dilated- it is not because your watch is running slower than their clocks, but because you are moving into regions of increasingly later time in the corridor frame. Your watch and all the clocks are all running at exactly the same rate, but your watch seems time dilated because the clocks are progressively out of synch in your frame.

Now, suppose that following you down the corridor at 10m intervals are other people wearing watches, and their watches are each set a second ahead of each other. From the perspective of any of the people holding a clock, they pass successive walkers with watches, and when they compare their clock with the passing watches they see that their clock seems to be running slow, because the time it shows is increasingly behind the time on the passing watches.

So, the effect is entirely symmetrical. From the perspective of any person with a watch, their watch seems to be time dilated according to the clocks they are passing in one direction, and from the perspective of any person with a clock, their clock seems to be time dilated compared with the watches they are passing in the other direction, and yet all the clocks and watches are running at exactly the same rate.

So you should be able to see that time dilation is an effect that arises not as a result of clocks slowing down but as a consequence of the relativity of simultaneity, which means that clocks in one frame are progressively more out of synch with clocks in another the further you move away from the origin.

Another thought experiment to help you convince yourself that clocks do not slow down when they move, is to remember that speed is entirely relative. You do not have a fixed speed as you sit at your computer reading this- you simultaneously have all possible speeds relative to all possible reference frames. There are frames of reference in which your heart beats only once a year, frames in which it beats once a month, frames in which it beats once a day and so on. Clearly your heart continues to beat at the same rate in your rest frame.

As for your question about Δ𝑥 being 0, that simply refers to a case in which the elapsed time at a single point in one frame is compared to the time difference between two corresponding points in a moving frame. In that case, the formula for the Lorenz transform takes the simplest form, but there is nothing else special about it. You are free to consider other cases in which Δx is not zero, in which case you will get a different factor when you transform the elapsed time from one frame to another.


"Consider a simple light clock in its rest frame. The clock consists of two parallel mirrors with a photon bouncing between them. The photon’s period of oscillation in its rest frame is

(1) $T = 2d/c$.

As usual $c$ is the speed of light and $d$ stands for the “proper” distance between the parallel mirrors.

Now let us consider the period of this clock as measured by Observers A and B, each moving with speed $v$ toward the clock. Each of the two observers carries along a clock identical to the mirror clock for time keeping. Observer A will see the clock moving toward her with velocity $v$ and measures the period of the moving clock as the time taken by the photon to bounce up and down between the two mirrors. Let $T_{A}$ be the period of the clock measured by Observer A. Clearly $T_{A}$ is the time the photon takes for a round trip between the two mirrors, which consists of one bounce up and one down. The length of each bounce $D$ is simply the hypotenuse of a right triangle with the sides $d$ and $vT_{A}/2$. Therefore,

(2) $D = [d^2 + (vT_{A}/2)^2]^{1/2}$,

and the round trip time is given by

(3) $T_{A} = 2D/c = \{2[d^2 + (vT_{A}/2)^2]^{1/2}\}/c$


(4) $T_{A}^2 (1 – v^2/c^2) = 4d^2 /c^2$

since by Eq. (1)

(5) $4d^2 /c^2 = T^{2}$.

Observer A concludes that the period of the moving clock is

(6) $T_{A} = T / (1 – v^2/c^2)^{1/2}$

This is the usual time dilation result. So Observer A concludes that the “moving” clock runs slow compared with his own clock." (Fred Behroozi).

Where equation (6) gives is the equation used in Relative Velocity Time Dilation: change in $t'$/change in $t = 1/(1-v^{2}/c^{2})^{1/2}$ OR

$t = t_{0}/(1-v^{2}/c^{2})^{1/2}$

where: $t =$ time observed in the other reference frame

$t_{0} =$ time in observers own frame of reference (rest time)

$v =$ the speed of the moving object

$c =$ the speed of light in a vacuum

Hope it helps!

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