I have great trouble in understanding simultaneity in special relativity. Let me illustrate it with a concrete example.

Assuming there is a train, its two end points are $A$ and $B$, the length of the train is $x$. The train moves at speed $v$. Assuming the train is moving in the direction from $A$ to $B$.

For a ground observer observing the train movement, he notices that two lightnings strike simultaneously at $A$ and $B$ when the middle part of the train $O'$ passes through right in front of him. In other words, the ground observer is located at the middle part of the train($O'$) when the lightnings strike simultaneously at $A$ and $B$.

Now there is another moving observer sitting inside the train and he sits right in the middle of the train ($O$, equidistant from $A$ and $B$). Does this moving observer think that the lighting happens at the same time? If no, how much time has passed before he notices lightning at $B$, after he had observed lightning at $A$?

  • $\begingroup$ The beginning of the train is A, its end B, and yet it moves from A to B? So it moves from its own beginning to its own end? $\endgroup$ May 7, 2014 at 8:30
  • $\begingroup$ @brightmagus, I've edited the question to make it clearer-- I am talking about the direction of the train is moving from $A$ to $B$ $\endgroup$
    – Graviton
    May 7, 2014 at 9:07
  • $\begingroup$ The direction doesn't matter, but OK. $\endgroup$ May 7, 2014 at 9:15
  • $\begingroup$ Graviton: Now I see a problem - where exactly did the lightnings strike? Did they strike the ground or the train? Because if they stroke the train - both observers will see them simultaneously. If they stroke the ground, than obviously the observer on the train will not see them simultaneously. So the question is, whether the reflected light was emitted locally for the train passenger or locally for the outside observer. This is key issue here. I, somehow, understood, that the train was hit. $\endgroup$ May 7, 2014 at 17:36
  • $\begingroup$ You might want to see my answer now. I added pre-amble and introduced some changes to the rest of text. In many places I replaced the word "source of light" with "emitted light" which makes the situation more obvious $\endgroup$ May 8, 2014 at 6:48

5 Answers 5


The only safe way for beginners to answer questions in special relativity is to sit down with a large sheet of paper and work through the Lorentz tranformations:

$$\begin{align} t' &= \gamma (t - \frac{vx}{c^2}) \\ x' &= \gamma (x - vt) \end{align}$$

Let's be absolutely clear what the tranformations tell us. If we use a coordinate system $(t, x)$ to label spacetime points, and another observer moving at constant velocity $v$ relative to us uses another coordinate system $(t', x')$, the transformations convert our labels $(t, x)$ to the other observer's labels $(t' x')$.

So to answer your question we take the two spacetime points labelling the ends of the train and apply the transformations. This tells us where those two points are in the moving observer's coordinates.

In our frame at $t = 0$ the middle of train is at $(0, 0)$, so the front of the train is at $(0, d/2)$ and the rear of the train is at $(0, -d/2)$ (I've called the length of the train $d$ to avoid confusion with the $x$ coordinate):


To find the potition of the front of the train in the primed frame we just feed $t = 0$ and $x = d/2$ into the Lorentz transformations:

$$\begin{align} t' &= \gamma (- \frac{vd}{2c^2}) \\ x' &= \gamma \frac{d}{2} \end{align}$$

So in the moving frame the lightning strike at front of the train is at $(-\gamma\tfrac{vd}{2c^2}, \gamma\tfrac{d}{2})$. I won't go through the details, but same calculation puts the lightning strike at the end of the train is at $(\gamma\tfrac{vd}{2c^2}, -\gamma\tfrac{d}{2})$.

So the answer is that the observer on the train sees the lightning strike the front of the train at $t' = -\gamma\tfrac{vd}{2c^2}$ and the rear of the train at $t' = \gamma\tfrac{vd}{2c^2}$. The time between the lightning strikes is $\gamma\tfrac{vd}{c^2}$.

Quick footnote

Rereading my answer it's just occurred to me that I've called the length of the train $d$ in the rest frame of the track. The length of the train for the observers on it will be greater - you can use the Lorentz transformations to calculate this too.

Length of the train

Re Graviton's comment, the easiest way to calculate the length of the train in the train's rest frame is to work backwards. Let's call the length of the train in its rest frame $\ell$, and we'll choose our zero time so that the rear of the train is at $(0, 0)$ and the front is at $(0, \ell)$. To transform from the train frame to the track frame we just use the Lorentz transformations as before, but in this case the velocity is $-v$ because if the train is moving at $v$ wrt to the track then the track is moving at $-v$ wrt the train.

When we do the transformation $(0, 0)$ just goes to $(0, 0)$ so we just need to work out where $(0, \ell)$ is in the track frame. Plugging in $t = 0$ and $x = \ell$ we find the point in the track frame is:

$$\begin{align} t &= \gamma (t' - \frac{(-v)x'}{c^2}) = \gamma\frac{v\ell}{c^2} \\ x &= \gamma (x' - (-v)t) = \gamma\ell \end{align}$$

So in the track frame the front of the train is at $(\gamma\tfrac{v\ell}{c^2}, \gamma\ell)$. But we don't want to know where the front of the train is at time $t = \gamma\tfrac{v\ell}{c^2}$, we want to know where is was at $t = 0$. So we take our value for $x$ at time $\gamma\frac{v\ell}{c^2}$ and subtract off the distance moved in time $\gamma\frac{v\ell}{c^2}$, which is just the time multiplied by the velocity. This gives us the value for $x_0$:

$$ x_0 = \gamma\ell - \gamma\frac{v\ell}{c^2} v $$

The rest is just algebra. We write the expression out in full to get:

$$\begin{align} x_0 &= \ell \left( \frac{1 - \frac{v^2}{c^2}} {\sqrt{1 - \frac{v^2}{c^2}}} \right) \\ &= \ell \sqrt{1 - \frac{v^2}{c^2}} \\ &= \frac{\ell}{\gamma} \end{align}$$

And since the rear of the train is at $x = 0$ at time zero and the front of the train is at $x = x_0$ at time zero the length of the train is just $x_0$ so:

$$ d = \frac{\ell}{\gamma} $$

At all speeds $> 0$ the value of $\gamma > 1$, so the length of the train as observed from the track is less than the length of the train in its rest frame i.e. the train is shortened. This is the Lorentz contraction.

  • $\begingroup$ John, would you like to append the calculation for the length of the train for the observers? $\endgroup$
    – Graviton
    May 7, 2014 at 9:09
  • $\begingroup$ @Graviton: I've edited my answer to show how you calculate the length of the train $\endgroup$ May 7, 2014 at 11:34
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    $\begingroup$ Be careful about saying an observer in the middle "notices" or "sees" lightning strikes. The speed of light delay between the times a lightning strike happens and is seen matters. You need the times the strikes happen. For that, you need lightning detectors with very precise clocks at each end of the train. You need two more on the ground. $\endgroup$
    – mmesser314
    May 7, 2014 at 11:34
  • $\begingroup$ @mmesser314: yes, you're quite correct and I am regrettably careless in this respect, but the discussion can get awfully convoluted if you don't use these common verbs. $\endgroup$ May 7, 2014 at 11:36
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    $\begingroup$ @mmesser314 Einstein's way of solving that problem was to assume the lightning left marks on both the train and the embankment (it must strike right on the edge of the platform or something) and that both observers could (later) verify that they had stood halfway between them (in their own frame) and could therefore correct both observations by the same delay. Round about, but sufficient. $\endgroup$ May 7, 2014 at 21:57

$\def\cA{{\cal A}} \def\cB{{\cal B}} \def\cM{{\cal M}} \def\RA{{\rm R_\cA}} \def\RB{{\rm R_\cB}} \def\rE{{\rm E}} \def\rF{{\rm F}}$ I have retrieved two drawings I did years ago for a lecture on this topic. I wish to show them in order to illustrate the way of reasoning I find best suited to introductory relativity teaching. There are two figures, showing the same physical situation as it would appear from two reference frames: frame $\tt T$ (the train) and $\tt S$ (the station).

Of course in frame $\tt T$ the train is stationary. In figure $\tt T$ you can see cartesian axes $(x,t)$ I will never use. They are there only to remind that from one frame to the other spacetime coordinates do change. There are three vertical lines, marked $\cA$, $\cB$, $\cM$. These are what I call objects: $\cA$ is the train's head, $\cB$ its tail, $\cM$ its mid-point. Since train is still, $x$-coordinates of these objects do not change with time. Therefore their worldlines have $x=\rm const.$ and are parallel to $t$-axis.

Continuing to examine the figure we see 4 points with their marks. There is one more than would suffice for Einstein's description of relativity of simultaneity, but this will help our analysis. Einstein's idea is to consider two lightnings, simultaneously falling at the train's extremities. In figure these are two events named $\RA$, $\RB$ (why R? I'll explain presently). That these events are simultaneous in $\tt T$ is shown by line $\RA\RB$ being parallel to the $x$-axis.

Note: I chose scales on $(x,t)$ axes so that 1 second in time has the same length as 1 light-second in space. In other words, I use units where $c=1$. Then worldlines of light signals are inclined 45$^\circ$ to the right or to the left.

But to physically ensure simultaneity, I do without lightnings and introduce two flashes starting from mid-point (event $\rE$). Events $\RA$, $\RB$ are defined as the arrivals of these flashes to $\cA$, $\cB$. Simultaneity is guaranteed by $\cM$ being the mid-point and by known assumptions about propagation of light. Flashes are reflected in $\RA$, $\RB$ and come back simultaneously to $\cM$ (event $\rF$).

Another thing to be noted in the figure is the red diamond. It depicts flashes' paths, from $\cM$ (mid-train) to head and tail, and vice versa. The only reason to draw it is to compare with its appearance in frame $\tt S$.

Let's now go to the next figure. It is drawn in frame $\tt S$, where the train is moving to the right, its speed being a fraction of $c$. But not too small a fraction (a relativistic train!) so as to make relativistic effects clearly visible. Axes are named $(x',t')$ since the same event has generally different coordinates in both frames.

Worldlines of $\cA$, $\cB$, $\cM$ are still parallel straight lines (all objects, as parts of the same train have the same constant speed) and are leaning to the right. The $\cM$-line is halfway between the others, as $\cM$ remains equidistant from $\cA$ and $\cB$, even if the train is moving. (Nothing will be said about their distances: Lorentz contraction is present, but we do not need to know this, nor to know its amount.)

But something new is happening: line $\RA\RB$ is no longer parallel to $x$-axis. Let's see why. Follow the flashes emitted in $\rE$. Their worldlines are still at 45$^\circ$, since speed of light is invariant: it is the same in $\tt S$ as in $\tt T$, irrespective of whether its source is still or is moving. It's obvious that the flash aimed at the train's head will employ more time to reach its target than the other one, aimed at the tail. This is because the train's head runs away, whereas the tail goes meet the incoming light. (Note that this has nothing to do with relativity, once you accept that light has the same speed in both directions.) The same result can be arrived at by pure geometry, simply looking at the figure.

We are thus led to an inescapable conclusion: events $\RA$, $\RB$, which were simultaneous in $\tt T$, are not simultaneous in $\tt S$. More precisely, $\RB$ precedes $\RA$. Actually geometry tells something more, if we complete the flashes paths. Their meeting on $\cM$ is an event, and as such cannot change according the frame we use for our measurements. In words, the times of arrival of both flashes in $\cM$ stay always the same, i.e. the red figure is a closed polygon. Its sides are at 45$^\circ$, then the polygon is a rectangle (no longer a diamond, i.e. a square). This ensures us that diagonal $\RA\RB$ makes with $x$-axis the same angle as $\rE\rF$ makes with $t$-axis. (Going on, we could compute the delay of $\RA$ wrt $\RB$.)

Up to now I assumed light source were standing on the train, as well as mirrors reflecting the flashes. What about the opposite situation, when source and mirrors stand still on station's platform?

You have only to interchange roles between $\tt S$ and $\tt T$ i.e. exchange the figure labels, with no change in events names. To be precise, if train goes right wrt station, the station goes left wrt train. Then the slanted lines $\cA$, $\cB$,$\cM$ in the second figure should slant in the opposite direction. As a consequence, events $\RA$, $\RB$, now simultaneous in $\tt S$, aren't so in $\tt T$: $\RA$ precedes $\RB$. And that's all.


If we want to get some intuition about the notion of simultaneity behind this scenario, we can think about it like this.

To avoid any confusion in my discussion, any non-primed letter should indicate a coordinate in the ground observer's reference frame and any primed letter should indicate a coordinate in the moving observer's reference frame. enter image description here

For the purpose of easy drawing, the train is drawn much smaller than $x$, the distance between $A$ and $B$. Suppose, at time $t=t_{0}$, two lighting strikes at $A$ and $B$, and in the moving observer's frame at $A'$ and $B'$. At that time, the positions of the two observers ($O$ and $O'$) coincide in both of their reference frames. Now, as the observer is moving with speed $v$ in $A$-to-$B$ direction, the points $A'$, $B'$ and $O'$ will be moving along with the observer's frame. Now, lights from $A$ and $B$ will take some time to reach $O$ and in the meantime the moving observer would have moved some distance from $O$. That's why light from $B$ will reach him first, at $t=t_{1}$ even before reaching the ground observer. After some time, the lights from $A$ and $B$ will meet at $O$ where the ground observer is standing (not shown in the drawing). As lights from $A$ and $B$ reach the ground observer at the same time crossing the same amount of distance, the ground observer will conclude the lightnings to be simultaneous.

Now, after some more time, at $t=t_{2}$, light from $A$ would reach the moving observer. At this time, light from $B$ would have crossed the ground observer to reach some point between $O$ and $A$. The moving observer would think that lightnings stroke at points $A'$ and $B'$ in his reference frame, because during the lightnings, those points coincided with $A$ and $B$ respectively. If the ground observer were using pre-relativity physics, then s/he would have thought that the moving observer is measuring the speed of light as $c-v$(from $A$) and $c+v$(from $B$). In that case, the moving observer should have also concluded the lightnings to be simultaneous. However, in special relativity, any inertial observer will measure the speed of light to be $c$. So for the moving observer, s/he is seeing that light from point $B'$ reached him/her first and light from $A'$ reached him/her after some more time. As points $A'$ and $B'$ are moving with the moving observer, those points are still at equal distances from him/her. Therefore, s/he will conclude that lightnings stroke first at $B'$ and then at $A'$.

Now, if we want to know the time difference the moving observer would measure between the lightnings or various distances/lengths involved in this scenario, we have to use Lorentz transformation, just as exactly as Jonh Rennie did.


The Fabri's explanation is complete, but, if someone need numbers, we can use an interactive construction https://www.geogebra.org/m/duvwkba7 Image of interactive construction Now, the interactive construction is public. Sorry. The construction is the reproduction of Fabri' images (answer Oct 22 '18 at 7:08), adding numeric cohordinates. K and K' are frames T and S in Fabri's constructions, side by side instead of in sequence. I fixed, for example, a relative velocity of -0.5c of K' (or S) frame respect to K (or T). As Fabri says "the same event has generally different coordinates in both frames": not only for positions but also for different units (the arrows near O or O'). Look also to geogebra.org/m/wqf5bEum

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    $\begingroup$ You should add some explanation for how your picture answers the original question. $\endgroup$
    – Chris
    Apr 29, 2019 at 23:42
  • $\begingroup$ Actually, there is an additional issue related to the need of registration to access the linked page. $\endgroup$ Apr 30, 2019 at 4:29

(To down-voters: Could you be, please, so kind as to comment, which sentence in my reasoning you find faulty and why? It would allow everybody to learn from you)

OK, I'm not sure now where exactly did the lightnings strike. Did they strike the ground or the train? Therefore, and since Graviton asked this question to better understand the simultaneity of relativity, I will give the answer for both options.


Although at first glance it might seem that where the lightnings stroke makes no difference, it is actually of utmost importance. Because whether the observer is stationary, or moving towards (away of) the emitted light changes everything. To make it clear and easier to follow, let's consider only the passenger of the train first and replace lightnings with flashlights.

If the flashlights are located at the front and at the end of the moving train, but still inside it, they are considered stationary for our passenger. There is no movement between them and the observer, they are all equidistant for the whole time. Therefore SR allows us to disregard whatever is happening outside the window, whether we see anything moving or not. This means the passenger will see both flashes at the same time.

However, what if the flashlights are at the front and at the back of the train, but outside of it? This changes everything. Let's say that each flashlight is exactly 300,000 km away from the passenger at the moment of flashing, so we have time separation of 1 second between any of them and the passenger. He will be moving toward the front flashlight, and away from the back flashlight. Therefore he will intercept the front flashlight sooner than after 1s from sending, and the rear flash will reach him after that. Therefore, the passenger will not see the flashes simultaneously.

(If that's not clear, lets talk about our Sun. We are about 8 minutes away from it. Lets say it sends us a clearly distinguishable and short green flash. The question is, what happens if we, at the moment of sending the light, take off in a spaceship and move toward the green flash with considerable velocity - near $c$ let's say? Will we intercept it sooner, than if we stayed on Earth? Sure, we will meet the green light almost halfway from Earth to the Sun, after about 4 minutes. But then, if we took off in our racket, but moved away from the Sun, we would be caught up by the green light much later than after 8 minutes. Therefore our movement relative the source of light does matter. And yet, if we stay on Earth we will see the green flash after 8 minutes, although our Solar system is constantly moving together through space.)

So I think we can go to our 2 cases now.

1. Lightnings hit the train (tragic, but ... possible)

If both observers are exactly in the middle of the two events happening simultaneously, then the relative movement of the "source" of light (i.e. places on the moving train which the light got reflected off) makes no difference here. The passenger on the train is not moving relative to the local sources of light (the beginning and the end of the train are traveling along with him), so he must see them simultaneously. The person outside of the train was in relative movement in respect to the train when the lightnings stroke it, but this movement was perpendicular to the direction of the light he saw, so at the moment of sending the rays he can be considered "stationary" as well. After the rays were sent, he didn't move in the direction of the sources (the sources moved to his side, but it was after). The light was sent at certain moment, and the movement of the sources afterwards does not matter (if the train disappeared just after sending the light, the observer would still see the lights, right?). Therefore he will also see the lightnings simultaneously.

2. Lightnings hit the ground

The person on the ground will obviously see the lightnings simultaneously in this situation. He is exactly in the middle and stationary relative to the sources of light, so this case is straightforward. However, for the passenger on the train the situation is completely different now. He is moving toward one ray of (reflected) light and away from the other. Therefore, he will see the lightning emitted in front of the train first, he will intercept it sooner than the light from his back will reach him. No simultaneity here.

To sum up - although both situations (1 and 2) might seem similar, so at first it might seem the answers should also be the same, but there is a significant difference between them. The velocity of the observer on the ground relative to the light emitted already is zero in situation 1., but the velocity of the passenger relative to the light emitted is non-zero in situation 2. This changes everything.

One last comment - although the direction of relative movement does make the difference as we saw, yet the speed of light will always be measured as $c$.

(I'm not going to do the calculations here, as John Rennie did them in his answer already.)

  • $\begingroup$ Since the light from the strikes does not arrive when the two observers are at their closest position (by assumption), the light from the strikes can not (in any frame) arrive at both observers at the some time (because by the time it arrives they are no longer opposite each other). $\endgroup$ May 7, 2014 at 21:52
  • $\begingroup$ If the Sun sends light toward you, and then moves to the side, do you see the Sun in the position it is when the light reaches you, or where it was 8 minutes ago? Now imagine there are two Suns moving one after the other and the distance between them is constant, and if they send at the same time a short green ray of light to you - will you see those rays at the same time? And now if there is another observer located exactly in between the suns and traveling with the same speed, and if they sent those green rays to him at the same moment as they did to you - would he see them simultaneously? $\endgroup$ May 7, 2014 at 22:19
  • $\begingroup$ In other words, for the observer traveling along with the Suns with the same velocity as them (no accelerations) and always with equal distance, the Suns are not moving. Therefore any rays sent at the same time must reach him at the same time. The stationary observer, however, will see rays sent to him by the suns 8 minutes ago, when they were just "in front of" him. They are no longer in front of him, they moved to the side, but he will see rays as sent 8 minutes ago. $\endgroup$ May 7, 2014 at 22:28
  • $\begingroup$ Obviously, this is a geocentric point of view, but it doesn't matter for the purpose of this thought experiment. So going back to what you said: "because by the time it arrives they are no longer opposite each other". Sure, but they were opposite each other when the light was sent - that's what matters here. $\endgroup$ May 7, 2014 at 22:34
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    $\begingroup$ It does not matter what was hit. It only matters that the strikes were at the same time in one frame and that they were the same distance away in both frames. This is because no matter what was hit we can assign a space-time coordinate to that event in each frame. Only Einstein has rigged things to be simple in the unprimed frame. then you look at the consequences for the primed frame. $\endgroup$ May 7, 2014 at 22:42

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