# Can any pair of spacelike events be simultaneous?

1. Jul 4, 2014

### johnny_bohnny

I was looking at some space-time diagrams with the simultaneity surfaces for moving inertial observers in SR and looking at the picture an interesting thought came up to my mind. It is known that the causal structure of the universe is preserved in SR, and that cause always comes before the effect. But what about spacelike events? Can any two spacelike events in Minkowskian spacetime be simultaneous?

For instance, let's imagine that we have a rod that is not very long, let's say 1m in its rest frame. Let's imagine that the rod is actually very durable in time, so let's imagine that from the perspective of its rest frame we have a situation where one event on the first end of the road (which is spacelike separated from the other) exists 1 hour after the second event on the second half of the road. Can those two events be simultaneous, despite that one trails in time by a very big margin as viewed from its rest frame. To sum up, can any two non-causally related events be simultaneous, no matter how much they are separated in time as viewed from one frame.

2. Jul 4, 2014

### WannabeNewton

For any two given events $p,p'$ that are space-like separated, there always exists an inertial frame in which they are simultaneous. This is very easy to prove. Let $O$ be any inertial frame with coordinates $(t,x)$ and choose the origin so that $x_p = t_p = 0$; we can assume $t_{p'} \neq 0$, otherwise the problem is trivial, so then $(\frac{x_{p'}}{t_{p'}})^2 > 1$. Now boost to a frame $O'$ moving past $O$ with velocity $v$ when their origins coincide so that $t'_{p'} = \gamma(t_{p'} - v x_{p'})$. We can then always make $t'_{p'} = 0$ if we choose $v = \pm \frac{t_{p'}}{x_{p'}}$ because $0 < v < 1$ given the above.

3. Jul 4, 2014

### ghwellsjr

Assuming that you mistyped "road" for "rod" and you're asking about two events separated in time by one hour and one meter, they are not spacelike separated so they cannot be simultaneous. If you can move a massive object from one event to the other event, then they are timelike separated, not spacelike separated. Two events at either end of a 1 meter rod would have to occur in less than the time it takes for light to get from one end to the other which is a very short time, about 3.3 nanoseconds.

4. Jul 4, 2014

### robphy

In 1+1 Minkowski spacetime,
draw the past and future light cones of the two spacelike related events. The worldline joining the past intersection event to the future intersection event is timelike and is in fact Minkowski-perpendicular to the line joining the spacelike related events (and is thus simultaneous according to the inertial observer along that worldline [since a radar signal sent from the past intersection arrives at the future intersection via reflections at the given spacelike events]).

5. Jul 4, 2014

### stevendaryl

Staff Emeritus
Yes, any two spacelike events are simultaneous according to some frame. The flip side is that any two timelike events are at the same spatial location, according to some frame.

6. Jul 5, 2014

### DrGreg

Just to get the terminology right, in SR there is no such thing as a "spacelike event", but you can talk about "two spacelike-separated events". "Spacelike" isn't a property of a single event, it's a property of the separation between two events.

(This is true only in SR where separations can be represented by 4-vectors. It's more complicated in GR.)

7. Jul 6, 2014

### analyst5

I have a question regarding this. From a mathematical perspective, ok.
But let's take for example me as I'm sitting on this computer on one location and me 5 years ago in some other location. How can, let's say, a spacetime point on my right hand at this moment be simultaneous (in some frame) with some spacetime point on my left hand 5 years ago, since I changed location

8. Jul 6, 2014

### stevendaryl

Staff Emeritus
Those two events have a timelike separation. So there is no frame in which they are simultaneous. But there is a frame in which they take place at the same location.

9. Jul 6, 2014

### ghwellsjr

And if they are not spacelike separated, then they cannot be simultaneous in any frame. Clearly, it is not impossible for you to move something from your right hand to your left hand in five years. Is that so hard to understand?

10. Jul 8, 2014

### johnny_bohnny

Can anybody explain this fact in terms of space-time diagram, why are events that are close in space but far in time timelike separated, how can we see that from the diagram with simultaneity surfaces?

11. Jul 8, 2014

### ghwellsjr

I'll take a scenario similar to the one you specified in your first post except that instead of the events separated by 1 meter, I make them be 1 yard or 3 feet. This will make it easier to apply meaningful units to the diagram since we'll use the speed of light to be 1 foot per nanosecond.

Here's a diagram showing four events on the two ends of the yard stick in its rest frame. The black event at one end of the yard stick is at the origin and will stay at the origin in transformed frames. The colored events are all at the other end of the yard stick but at different times as indicated by the Coordinate Time on the diagram:

Note that the red event is along the 45 degree diagonal which represents the speed of light. When we transform the red event, it will remain on the 45 degree diagonal but will move closer or farther away from the black event.

The green event happens later than the red event and this will always be the case for any transformed frame but those two events will move farther apart in time in other frames.

Likewise, the blue event happens earlier than the red event and this will always be the case for any transformed frame but those two events will move farther apart in time in other frames.

Note that the three colored events are all in a straight vertical line in this frame. In other frames, they will remain in a straight line but that line will acquire a slope approaching 45 degrees (but never reaching 45 degrees).

So let's start by transforming to a speed of 0.377c:

Note how the red event moves closer to the black event and the line going through the three colored events acquires a slope.

Let's increase the speed to 0.666c:

Now the blue event has moved down to the Coordinate Time of 0 nanoseconds making it simultaneous with the black event.

Finally, let's increase the speed to 0.75c:

Now the green event is right over the black event at a Coordinate Distance of 0 feet making them at the same location but at different times.

For the foregoing examples, we can conclude that any pair of events that happens so close together in time that light cannot get from the one event to the other event, that those two events are spacetime separated and there is a frame in which they are simultaneous.

Does that make sense to you? Any questions?

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12. Jul 8, 2014

### ghwellsjr

I don't think I made it very clear in my previous post why pairs of events that are close in space but far in time are timelike separated so let me do that now. If you look at the last diagram, I think you can see that the time interval between the black and green events is around two and a half nanoseconds and that if we had a clock that was at rest in that frame and present at both those events, it would tick off around two and a half nanoseconds. This would clearly be an inertial clock because it is not moving in that frame. But it is also inertial in any other frame we want to transform to and it will tick off the same amount of time between those events in all of them. That is why it is called a timelike separation.

13. Jul 9, 2014

### A.T.

Check this animation at 1:25. It shows that Lorentz transformation preserves the light cones at 45°. All events inside the light cone (timelike separated) stay inside the light cone, and all outside (spacelike separated) stay outside. (Note that the time axis points downwards here):

14. Jul 9, 2014

### analyst5

What about the standard diagram where the vertical axis is ct with the simultaneity lines for different observers drawn between the axes, how can we see this fact there?

15. Jul 9, 2014

### Staff: Mentor

The simultaneity line with $t=0$ is the x-axis for that observer, and the line with $x=0$ for that observer (usually the path of that observer through spacetime) is the t-axis for that observer.

The path of a light signal through spacetime is the diagonal line given by $x=t$, and it is the same for all the observers.

For any two events, you can draw a line between them. If that line is steeper than the diagonal light-signal line, the interval between them is timelike. There will be some observer whose t axis is parallel to that line and that observer will report the two events as happening at the same place at different times.

If that line is less steep than the diagonal light-signal line, then the interval is spacelike and there will be some observer whose x-axis is parallel to that line. Because the surfaces of simultaneity are parallel to the x-axis, that observer will have a surface of simultaneity passing through both events.

16. Jul 9, 2014

### A.T.

The simultaneity lines are always outside of the light cone, so the events on them are always spacelike separated.

17. Jul 11, 2014

### ghwellsjr

My diagrams are standard. Note that I defined the speed of light to be 1 foot per nanosecond. The value of c is 1 and this guarantees that the speed of light will be along a 45 degree diagonal in the diagram.

If you wanted to use some other units such as meters and seconds, the speed of light would not be on a 45 degree diagonal but at a much shallower angle. Consider how long it takes for light to travel 1 meter. It is 3.3 nanoseconds. A line depicting the speed of light on such a diagram would be indistinguishable from a horizontal line. But by scaling the time axis by c, it expands it so that now the speed of light comes out to be on a 45 degree diagonal.

But in all cases, the simultaneity lines are horizontal. It doesn't matter what the relative scaling of the units are between the vertical time axis and the horizontal space axis.