Is there always a frame in which spatially separated events are simultaneous?

In relativity, if two events are simultaneous in a specified frame, they cannot be simultaneous in any other frame.

My question is this: given any two events, is there always a frame in which these two events are simultaneous? For example, if I drop a blue ball on one side of a tennis court, and my friend drops a red ball on the opposite side of the court one day later -- from my frame one day later -- is there a frame in which the blue and red balls hit the ground simultaneously?

• Your title and question body are inconsistent. The title says "spatially separated" but the body doesn't - which is it? – David Z May 28 '14 at 19:44
• In my question body, I specify that the two people dropping the balls are on opposite sides of the tennis court, so they are spatially separated. – user7348 May 28 '14 at 19:45
• Oh... wait, I misread that. I thought you meant spacelike separated. – David Z May 28 '14 at 19:46
• Are you assuming a normal-sized tennis court? :) See my answer below for an explanation why that is important. – NeutronStar May 28 '14 at 19:50
• "In relativity, if two events are simultaneous in a specified frame, they cannot be simultaneous in any other frame. " Are you talking about the simplified case with one space dimension? – JiK Oct 26 '16 at 7:56

Is there always a frame in which spatially separated events are simultaneous?

The answer is no.

Two events that are spatially separated in one frame of reference

(1) will be co-located in another frame of reference and not simultaneous in any frame if the interval is time-like

(2) will be simultaneous in another frame of reference and not co-located in any frame if the interval is space-like .

(3) will be neither co-located nor simultaneous in any other frame if the interval is light-like.

Time-like interval

If the interval is time-like, the separation in time, $|c\Delta t|$, is larger than the separation in space $|\Delta x|$:

$$|c\Delta t| \gt |\Delta x|$$

Thus, there is a frame of reference in which $\Delta x' = 0$; the two events are co-located in this frame.

Space-like interval

If the interval is space-like, the separation in time is less than the separation in space:

$$|c\Delta t| \lt |\Delta x|$$

Thus, there is a frame of reference in which $c\Delta t' = 0$; the two events are simultaneous in this frame.

Light-like interval

If the interval is light-like the separation in time equals the separation in space:

$$|c\Delta t| = |\Delta x|$$

Thus, in all frames of reference, the events are neither co-located nor simultaneous, i.e.,

$$|c\Delta t'| = |\Delta x'|$$

All of this follows directly from the Lorentz transformation. Let's take your example of two events with spatial separation of a tennis court so

$$|\Delta x| = 78\mathrm m$$

Light travels this distance in $\Delta t_c = \frac{78}{300 \cdot 10^6} = 260\mathrm{ns}$

Thus, if the two events occur within 260ns in this frame of reference, the events have space-like interval and are thus simultaneous in another, relatively moving reference frame of reference.

Since, in your example, the events occur 1 day apart, the events have time-like interval and cannot be simultaneous in any reference frame.

• Just one question, the answer you gave would change if we were to allow information to travel faster than light right? – user7348 May 29 '14 at 3:07
• @user7348 if you allow FTL, you destroy Relativity. – Davidmh May 29 '14 at 16:14

We must look at the spacetime interval between the two events. Two events separated by a distance magnitude $\Delta r$ and a time interval $\Delta t$ in a frame are said to be space-like if $c^2\Delta t^2 < \Delta r^2$ (i.e., the distance between the events is too great for light emitted by one event to be seen by the other) and are said to be time-like if $c^2\Delta t^2 > \Delta r^2$ (i.e., the distance between the events is small enough that light emitted by one event can be seen by the other). Events with $c^2\Delta t^2 = \Delta r^2$ are called light-like. These events are separated by just the right amount of spatial distance that light from one event will reach the other right as it is happening.

Space-like events cannot have a causal relationship (i.e., one event is influence by the other) because the spatial separation between the two is too great for information from one of the events to reach the other before it happens. For these events, there is a reference frame where the two events happen simultaneously but no reference frame where the two events happen at the same location.

Time-like events can have a causal relationship since the spatial separation is small enough that light/information can travel from one event to the other. For such events there exists a reference frame where the two events happen at the same location but no reference frame where they happen at the same time.

So, in short, if two events are separated by a space-like interval there does exist a frame where they happen simultaneously. If two events are separated by a time-like interval there does not exist a frame where they happen simultaneously. In your tennis ball example, since the spatial separation is much smaller than the time separation (i.e., enough time elapses between the events that light from one event can be viewed by the other) we have a time-like interval and the events cannot be viewed simultaneously in any reference frame. At least, that is assuming you're talking about a standard tennis court. If you had a tennis court that was longer than one light-day then you could have a space-like interval between the two events.

• Good answer. Somehow this violates my intuition though. I feel like I see simultaneous events all the time with only a small amount of distance between the two events. – user7348 May 28 '14 at 20:21
• @user7348, That's because the light/time separation is even smaller. If, in your tennis ball example, you and your friend dropped the ball at the exact same time then the time separation is zero and thus the spatial separation (although small) is greater than the time separation and we have a space-like situation. This is a bit trivial because the frame at which the two events are simultaneous is already your frame. – NeutronStar May 28 '14 at 20:30
• @user7348 If, for instance, your friend waited half the time it takes for light to cross the court to release their ball after you have already released yours then you still have a space-like separation between events but you'll still need to find a frame in which the two events are simultaneous using the math in John Rennie's answer (but such a frame does exist). – NeutronStar May 28 '14 at 20:31
• OK, but my original question is whether given two events with arbitrary time delay in the rest frame between the two events, and arbitrary separation between the positions of the events, can we then always find a frame in which they are simultaneous? I think the answer you have given is no. This is good, because for example, it's absurd that me typing this message is simultaneous with say George Washington's birth from some frame of reference. – user7348 May 28 '14 at 20:48
• If you were typing your post a couple of light-centuries away from Washington's birthplace... – Kevin May 29 '14 at 1:14

Joshua's answer is correct, and I see you have accepted it, but let me try and show how this emerges from the Lorentz transformations.

Let's choose our coordinates so that the earlier point is at the origin, then the second point is at $(t, x)$. That is the two events are separated by a time $t$ and a distance $x$. Now let's try and find a frame where the two points are simultaneous. The Lorentz transformations tell us:

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

The first point $(0, 0)$ just transforms to $(0, 0)$ so the condition for the events to be simultaneous is that $t' = 0$, so we have:

$$t' = \gamma (t - \frac{vx}{c^2}) = 0$$

and this is true only if:

$$t = \frac{vx}{c^2}$$

or:

$$v = c^2\frac{t}{x}$$

But the maximum velocity $v$ can have is $c$, so for the frame to exist we have the inequality:

$$c^2\frac{t}{x} < c$$

or with a quick rearrangement:

$$\frac{x}{t} > c$$

And here is our result. The ratio $x/t$ has the dimensions of a velocity, and this velocity must be greater than the speed of light. In other words a frame where the two events are simultaneous can only exist if the two events are spacelike separated.

• So from the moon, which is 1.38 light-second from the earth, events that occur within 1 second are simultaneous? – ja72 May 28 '14 at 21:03
• @ja72, there is a frame in which the events that occur 1 second apart are simultaneous since they have space-like interval. Since the interval is $1 - (1.38)^2 = 0.9044$, there is a relatively moving frame where the events are simultaneous and the Earth-Moon distance is measured to be $\sqrt{0.9044} = 0.951$ light-seconds and this frame has a relative speed of $\frac{1}{1.38}c = 0.725c$ – Alfred Centauri May 28 '14 at 22:25
• @John Rennie is it possible to have some object for which $t=t'$ – Alexander Cska Dec 23 '18 at 9:30

No. Causally related events can never be viewed as simultaneous in any frame.

• People are probably downvoting this because it doesn't explain much, and that does mean it could be a much better answer, but technically it is correct. – David Z May 28 '14 at 19:47
• Well, there's no causality in my question at all, which is why I down voted. – user7348 May 28 '14 at 19:49
• @user7348 but causally related events can be taken as examples of spatially separated events for which there is no frame in which they are simultaneous. So this answer is a counterexample that shows the answer to your question is "no." Of course you're free to vote as you like, but I think the issue here is more the failure to explain why causality is relevant; the fact that the answer mentions causality is not itself a problem. – David Z May 28 '14 at 22:04

This is only possible if both events are independent of each other and satisfy the condition...

$∆t/∆x=v/c^2$
$∆t$ implies time interval between events,
$∆x$ implies difference in coordinates of events.

• Since you aready have an answer bellow, please consider merging the two into onw answer. – Mikael Fremling Oct 26 '16 at 10:00

protected by rob♦Oct 26 '16 at 9:57

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