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For time-like separated events all observers will see the same order of events from any reference frame. So I will always be seen opening the refrigerator before I take something out of it. For space-like, I can be opening my refrigerator and my neighbor could be turning on their TV at the same time and these events are space-like separated and one event can never be the cause of the other. So a third observer can see these events in either order.

Let's say you have event A which is JFK assassination. Event B is a Secret Service Agent back in Washington watching his kids baseball game. Here's the question. Could a 3rd observer see the Secret Service Agent watching the game, warn the Secret Service Agent that J.F.K. will be shot, the SSA calls and warns other SSA and in the 3rd observers worldline Kennedy wasn't assassinated. So can space-like separated events ever become time-like?

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    $\begingroup$ When you use the word "become", what precisely do you mean? $\endgroup$ – Alfred Centauri Jul 5 '18 at 1:32
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    $\begingroup$ If Event A is the Kennedy assassination, then the Kennedy assassination exists, so Kennedy is assassinated, so he can't be not-assassinated. So.....what are you asking? $\endgroup$ – WillO Jun 8 '19 at 4:33
  • $\begingroup$ Time cones are open convex sets. Even in the case of piecewise smooth causal curve, the curve doesn't switch causal cones at the breaks. $\endgroup$ – Cinaed Simson Jun 9 '19 at 2:28
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Time-like and space-like are invariants. That is, they are the same for different observers even if those observers are moving relative to each other, or the events.

It's a lot easier to see when you look at it in special relativity than in generl relativity.

So start with natural units with c=1. Consider an event at $(x,t)$ and ask if that point is space-like, time-like, or null separated from the origin. These are based on the value of $x^2-t^2$. If this is greater than $0$ then it is space-like. Less than $0$ it is time-like. Equal to $0$ is null. A null path is a path light can follow.

So look at this from the point of view of a second observer who happens to be moving in the x-direction, and happens to be coincident with the first observer at t=0. That is, they have the same origin for x and t. So you get the Lorentz transformed value of the interval as so.

$$x^\prime = \gamma (x-vt) $$ $$t^\prime = \gamma (t-vx) $$ where $$\gamma =\frac{1}{1-v^2}.$$ So the interval will transform as follows.

$$ (x^\prime)^2 - (t^\prime)^2 =\gamma^2 \{ (x-vt)^2-(t-vt)^2 \} = \gamma^2 \{ x^2(1-v^2) - t^2(1-v^2) \} = x^2-t^2 $$

That is, the Lorentz transform leaves the interval unchanged. All observers see it the same. So space-like is space-like, and time-like is time-like, and changing observes does not change it.

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