The concept of relative simultaneity is a major part of understanding the special theory of relativity. If you don't know anything about special relativity yet you might want to check out either the node of the same name or theory of relativity. There isn't really much math here beyond algebra, and a lot of that can be skipped if you don't like it. I hope that just reading the descriptions you can get an idea of what's going on.

In special relativity, the same event will be observed to have happened at different times by different observers if they are moving relative to one another. The basic idea of relative simultaneity is that two events that are simultaneous to observer Alice will not be simultaneous according to observer Bob if Bob is moving relative to Alice. This comes from the fact that the speed of light is the same for both Alice and Bob (in fact, all inertial observers). This doesn't destroy the concept of simultaneity altogether. All observers in the same reference frame (ie. stationary with respect to one another, in special relativity) will still agree on whether two events are simultaneous or not. Furthermore, there is still a definite relationship between the timing of events in one frame of reference and in another frame of reference.

The relationship is given by the Lorentz transformations and tells us that if an event happens at position x and time t in reference frame S, then in reference frame S' (which has t' = 0 and x' = 0 when t = 0 and x = 0) moving at velocity u in the positive x direction the event occurs at t' = γ(t - ux/c2). That means that if S' is moving toward the event according to S, then it will occur at an earlier time in S' than it did in S. Likewise, if S' is moving away from the event, then it will occur later in S' than in S. For two events that are a distance Δx apart in space and Δt in time as observed in frame S, we can use the Lorentz transformation we just listed to find that in S' the time difference between the events is

Δt' = γ(Δt - uΔx/c2)

### Understanding Relative Simultaneity

Suppose that Alice is sitting in a space station that is exactly between two stars, each 5 light years away. Alice is approximately at rest relative to the stars. Let us suppose that Alice sees each star has a flare simultaneously. Since the light from each flare reached Alice simultaneously and she knows it would have taken light from each star 5 years to reach her, Alice can then determine that the two stars had flares simultaneously 5 years ago. Let's call that time when the flares occurred according to Alice t = 0. So remember, that means Alice observed the flares simultaneously at t = 5 years.

Now, suppose at t = 0 Bob passed Alice's space station flying in a space ship at 2/3 c (meaning 2/3 the speed of light in vacuum). If Bob is moving toward star 1, then from Alice's point of view he will hit the light from star 1 first, since it is also moving toward him. Specifically, at t = 3 years, the light from star 1 will have traveled 3 light years toward Alice, and Bob will have traveled 2 light years toward star 1. Since the total distance is 5 light years, that means that Bob will be met at that point by the light from star 1. Meanwhile, he is moving away from star 2, so according to Alice it will take the light from star 2 longer to catch up with Bob. Specifically, the light from star 2 won't catch Bob until t = 15 years, when Bob has traveled 10 light years away from Alice and the light from star 2 has traveled the 5 light years to Alice and then the extra 10 to get to Bob. That may be confusing. I've made some crude ASCII diagrams below to help explain. All the action is happening along one line, but I've put things offset one line vertically at times, in order that you can make out the separate light beams.

```Alice    A
Bob      B
Stars    *
Light    ~~

t = 0        *          A          *
B->

t = 3 yrs    *~~~~~~
A
~~~~~~*
B

t = 5 yrs    *~~~~~~~~~~
A
~~~~~~~~~~*
B

t = 15 yrs   *~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A
B
```

In the last diagram I've removed star 1, since it's not of any interest in that diagram.

Ok, so all we've determined so far is that according to Alice, light from star 1 must reach Bob before light from star 2. So we know that Bob will see star 1 flare before star 2. If we wanted to know exactly how much before, though, we'd have to account for time dilation, which makes time pass more slowly for Bob, according to Alice. Still, we can't dispute that Bob will see star 1 flare first.

Now we must use our rules of relativity to figure out what this means for Bob. When Bob passes Alice in her space station, he sets his clock to t'= 0, so his watch is synchronized initially with Alice's. According to him, the two stars also have the same distance from him at t'= 0. For Bob, star 1 appears to be moving toward him at 2/3 c, and star 2 appears to be moving away from him at 2/3 c. Now, if we were to tell Bob that star 1 and star 2 each had a flare at t' = 0, then he would conclude the following: Each star had the same distance at t' = 0, and the light each star lets off will approach Bob at the speed of light in his frame of reference, regardless of how the star is moving. So, just as Alice did, Bob concludes that light traveling from two equidistant sources at the same speed must reach him at the same time. Since Bob sees the light from star 1 first, he must therefore conclude that star 1 actually flared first.

At first it might seem like this is just screwy reasoning, but in fact it is a direct consequence of our statement that light always travels the same speed. The result is unintuitive because light does not obey the normal addition of velocities. By that I mean that if a stationary star emits light, it goes at c. If a star moving at speed v emits light, it still moves at c not at v+c! That may well seem strange and counterintuitive, but it's the way nature works. See special relativity and relativistic addition of velocities for more details.

### A possible misconception

I know that when I first learned about this idea, I thought that the idea was the following: The time of an event A for an observer S is the time when the light from A reaches S. In that case Alice would say the flares happened at t = 5 years in the example above. I want to emphasize that that is not what we have been saying. When an observer receives light from an event, she takes the time the light arrived and how far away the event was (we assume she can tell this from the light she receives), and then she uses the fact that the speed of light is always the same to figure out when the light must have originated from that event. That is why Alice knows the flares happened at t = 0 years, not at t = 5 years. Again, the weird effects with time here are from the unintuitive fact that the speed of a light beam will be the same for any two observers, even if one is moving relative to the other.

### What can we say about temporal relationships?

Δt' = γ(Δt - uΔx/c2)

That is generally what we know about the relationship in time for two events according to different observers. We can get a bit more information from this formula, though. First, let's re-write it as

Δt' = γΔt(1 - (u/c)*(Δx/Δt)/c)

So, if you can't get between two events going at less than c (meaning Δx/Δt > c), then there is an observer with a velocity u such that the two events are simultaneous in his frame. Mathematically, if (Δx/Δt)/c > 1, then there's a u < c where (u/c)*(Δx/Δt)/c = 1, making Δt' = 0. More than that, you could choose u large enough that Δt' has the opposite sign from Δt. That means there's a frame where the two events have the other order in time. This isn't as bad as it might seem at first, because at least if no one can get between the events going less than the speed of light, then the two events can't influence one another.

On the other hand, in the case that you can get between two events going at less than c (in which case Δx/Δt < c), then there's no u value that makes Δt' go to zero or switch signs, meaning that the two events have the same order in time for all observers (though the actual length of time between them might be different). When this is the case and the time order is unambiguous, the events are said to have a time-like separation or be connected by a time-like trajectory. Two events like this, where you can get from one to the other going less than the speed of light, can be causally connected (meaning event A could have caused event B), so it's good that all observers agree on their order in time, otherwise we could end up with all sorts of time paradoxes.

### A geometric view

One can also take a geometric view of all this. I won't attempt to draw the spacetime diagram here (due to the limitations of ASCII art), but, if you were to make a spacetime diagram in terms of the coordinates of a frame of reference S with time on the vertical axis and space on the horizontal, then a pair of simultaneous events would both fall on the same horizontal line, representing the same value of t. Now, using the Lorentz transformations, we could draw on another set of axes, representing the coordinates of S'. Lines of constant time t' would have the form γ(t - ux/c2) = constant, which would be a line with a positive slope. In that case, the two events that fall on the same horizontal line (and are, thus, simultaneous in S) will not fall on the same line of constant t', so they will not be simultaneous in S'.

### Relative simultaneity in everyday life

You might ask, "If this is all true, how come I've never noticed this?" As with many things in relativity, part of the answer is that the effects don't appear until you start dealing with velocities comparable to the speed of light, which we almost never do (except in particle accelerators, etc). If we go back to the formula for the relationship of the time difference in different frames of reference, written in the form

Δt' = γΔt(1 - (u/c)*(Δx/Δt)/c)

then we see that for u small enough Δt' and Δt are approximately the same. Specifically, it looks like we need u*(Δx/Δt) << c2 for the order in time not to switch. The change in the length of the time interval also depends on γ, which becomes 1 for u << c. Generally, unless Δx/Δt is large compared to c, u << c will make this effect of relativity disappear, and we know we can always choose a range of u small enough so that the effect is negligible.

Let's try some real world numbers to get an idea for the size of this effect. Let's say that according to an observer stationary relative to the earth's surface two events happen simultaneously at opposite sides of the earth, that's about 12,000 km. That will be our reference frame S, which is approximately inertial if we only consider a short period of time and neglect gravitational effects. Let us also suppose there's an SR71 jet carrying a second observer that is flying directly above the first observer at a speed of 1 km/s. This is probably the one of the situations in which the effects of relative simultaneity should be most apparent to people on earth. In that case, it turns out that Δt' ≅ 0.13 μs, so it would only be detectable with a fairly sensitive time measurement. Needless to say that in our day to day lives, these effects are completely imperceptible.

Want to test out whether you've understood all this? Then check out The Barn And The Pole: A Relativity Paradox, which is a writeup of a classic relativity paradox. You may want to check out the writeups on the Lorentz transformations and on length contraction.

Sources: My own knowledge of special relativity.

Though I didn't really consult any sources to write this, if you want to look at some, here are some suggestions:

The book I originally learned from (not necessarily recommended)

A. P. French, Special Relativity

A well respected introduction to special relativity from the view of geometry (spacetime diagrams) at an introductory/intermediate college level

Taylor and Wheeler, Spacetime Physics

Finally, most general physics text books will have a chapter somewhere near the back about relativity, which generally will contain a section on this subject. For example

Serway and Jewett, Physics for Scientists and Engineers 6th ed. chapter 39, section 4.

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