I recently started studying Special Relativity an my book discusses the following:

Say I have synchronized two separated clocks in a reference frame S, if then an observer in another reference frame S' for whom the clocks are moving sees the clocks he would say those clocks are out of synchronization for him, as for him the light that was used to sync the clocks reaches one of them before the other.

Up to this point, I think I understand what is happening, but if then the observer in S' knowing about the relativity of simultaneity applies a Lorentz transformation to try and "see" the clocks as an observer in S would, it will, nevertheless, still find a non-zero difference between the times "shown" by the clocks, proportional to the distance separating them.

What does this mean? Isn't the Lorentz transformation an attempt to find the space and time coordinates as they would be seen from other reference frames, and if so, why is it unsuccessful here?

What's the physical interpretation of the difference in time the observer in S' just found, considering is neither the difference he sees from his reference frame or the one seen by the original observer in S?

Thank you, and excuse me for not showing the math I don't how to write it here in a readable way.

edit: I'll try to write the equations for further clarification

Suppose that the clocks are placed in $$x_1$$ and $$x_2$$ in S they are in sync so, $$ \Delta t=0 $$ However in S', which is moving with speed u relative to S, we have for each clock $$t_1'=\gamma(t_1-\frac{u}{c^2}x_1) $$ $$t_2'=\gamma(t_2-\frac{u}{c^2}x_2) $$ then for this observer the time interval between them is $$\Delta t' = t_2'-t_1'=\gamma(\Delta t - \frac{u}{c^2}\Delta x) $$ with $$ \Delta t=0 $$ because the clocks are in sync in S, then $$\Delta t' =- \gamma \frac{u}{c^2}\Delta x $$ Therefore, the clocks are out of sync for the observer in S'. But if I now tried to apply $$\Delta t'=\gamma \Delta t$$ I would get $$\Delta t=-\frac{u}{c^2}\Delta x $$ Which is different to 0. This is the result I find troubling.

  • $\begingroup$ If by "out of synchronization" you mean that the two clocks would have a time offset, then you are correct. Technically they are still synchronized as they are running at the same rate, even when seen from another moving observer. The time offset is physically irrelevant, of course, just as in Galilean relativity. What you are tripping over is that Lorentz transformations are local. If you want to find absolute differences between two events (and "what it would look like" physics), you still have to factor in the distance between events, which is not contained in the Lorentz transformations. $\endgroup$
    – CuriousOne
    Mar 26, 2016 at 22:25
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    $\begingroup$ It sounds like you are running into the distinction between what David Griffiths in his Introduction To Electrodynamics calls the "difference between what we see and what we observe." It's the idea that if you are watching an event through a powerful telescope, say, then you have to deal with the delay caused by the time of flight of light from the event location to you, but if you are deducing what time coordinate the other observer will assign to that event, you are thinking of a different thing. $\endgroup$
    – user55515
    Mar 26, 2016 at 23:21
  • $\begingroup$ Thanks for your replies, but I think I didn't really make myself clear, I'll try and write the equations. $\endgroup$ Mar 27, 2016 at 0:37
  • $\begingroup$ To put it another way: Lorentz transformations and Doppler effect are not the same thing. I think you may be hung up on the phenomenology of the Doppler effect, where what you "see" depends on which way you are going/looking (towards/away) from a source. The Lorentz transformation does not have that artifact. There is nothing wrong thinking this trough really well, by the way. It is an important distinction. $\endgroup$
    – CuriousOne
    Mar 27, 2016 at 0:55
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    $\begingroup$ Your equation $\Delta t'=\gamma\Delta t$ is not correct because neither observer considers both events to be at the same location. $\endgroup$
    – WillO
    Mar 27, 2016 at 1:00

2 Answers 2


To see what's happening, pretend for a minute that the earth is stationary with respect to the sun. You and your friend on the sun have synchronized your clocks.

When your clock reads 12:08, you look through your telescope and see your friend's clock reading 12:00 -- because light from the sun takes 8 minutes to get here. But you are perfectly well aware that light takes 8 minutes to get here, so you say "Yes, sure enough, my friend's clock reads 12:08 right now, though the light I am seeing is from eight minutes ago."

The Lorentz transformation in this case is the identity --- your velocity with respect to your friend is zero, so the Lorentz transformation is $x'=x$, $t'=t$. That is, the Lorentz transformation tells you that when you say it's 12:08, so does your friend. It does not (directly) tell you anything about what you'll see through your telescope. Nor is it supposed to. It tells you what's happening at your friend's house right now according to your frame, not what you're seeing through your telescope right now.


The interpretation that the time difference $\Delta t'$ is related to "light signal propagation delay" is not quite right.

The reason is that two events that are spacelike separated in S, in particular simultaneous in S, necessarily remain spacelike in S'. Hence they cannot be causally related, and the time difference between them must necessarily be shorter then the propagation time required by a light signal.

Let us verify this the hard way. The event coordinates observed in S' read $$ x'_1 = \gamma(x_1 - ut_1)\\ x'_2 = \gamma(x_2 - ut_2) $$ and so for $\Delta t = 0$ we find $$ \Delta x' = \gamma \Delta x $$ But this means $$ \Big | \frac{\Delta x'}{\Delta t'} \Big | = \Big | \frac{\gamma \Delta x}{\gamma \frac{u}{c^2} \Delta x} \Big | = \frac{c}{u}c > c, \;\;\; \forall \; u<c $$ In other words, the propagation time for a signal between events 1 and 2 would have to be superluminal, which cannot be.

And just to wrap everything up in one place: As WillO already mentioned, the time delay $\Delta t'$ is not due to time dilation either unless the two events take place at the same location in S. To see this, simply consider a 2nd pair of events occurring simultaneously in S at the same locations $x_3 = x_1$, $x_4 = x_2$, but at a slightly later time $\delta t > 0 $. In S' events 3 and 4 are observed at (using $t_1 = t_2 = t$) $$ t'_3 = \gamma \left( t + \delta t - \frac{u}{c^2}x_1\right)\\ t'_4 = \gamma \left( t + \delta t - \frac{u}{c^2}x_2\right) $$ If we now check the time differences in S' between events 1 and 3 (at $x_1$ in S), and 2 and 4 (at $x_2$), we find that they are indeed time dilated compared to $\delta t$: $$ t'_3 - t'_2 = \gamma \delta t \\ t'_4 - t'_1 = \gamma \delta t $$ although the time difference between 3 and 4 is again $\Delta t' = - \gamma \frac{u}{c^2}\Delta x$.

To sum up, the time delay introduced by relativity of simultaneity is due neither to "light signal propagation delay", nor to time dilation. It is actually a distinct phenomenon.


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