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Assume this question:

Three events A, B, C are seen by observer O to occur in the order ABC. Another observer O$^\prime$ sees the events to occur in the order CBA. Is it possible that a third observer sees the events in the order ACB?

I could draw spacetime diagram for three arbitrary-ordered events for O and O$^{\prime}$. But I couldn't draw spacetime diagram for third observer. Could someone help me figure this out?

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2 Answers

up vote 2 down vote accepted

Provided the intervals between all events are spacelike they can appear in any order. See this article for a popular science level description, or this paper for the full details.

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Might make a difference that I wouldn't interpret "seen by observer" in terms of spacelike planes in spacetime, but via light cones with the sharp end up. –  NiftyKitty95 Oct 25 '12 at 8:55
    
Agreeed, and I should probably add that this is an exceptional circumstance and in the majority of cases different observers can't see all six possible orderings. –  John Rennie Oct 25 '12 at 9:23
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One can design a situation that obeys your conditions – both ABC and CBA ordering may be realized in some inertial systems – but the ordering ACB is impossible.

Simply imagine that A, B, C are three events in spacetime that lie on a spacelike line in the spacetime. Then B is inevitably in the middle between A, C, chronologically and spatially, regardless of the frame, so only ABC and CBA orderings are possible.

However, when A,B,C do not belong to the same line, and we add this proposition as an extra assumption, then one may show that your additional assumptions are enough for any ordering to be realized in some inertial systems.

When both ABC and CBA ordering is possible, it proves that A-B have to be spatially separated (because the ordering of A,B appears in both ways, depending on the frame), B-C are spatially separated (the same reason), A-C are spatially separated (the same).

With the extra assumption that A, B, C don't lie at the same line in the spacetime, we see that ABC is a spacelike triangle in the spacetime. You may therefore go to the inertial system in which all events A, B, C occur simultaneously. Now, when you boost the system in directions along the vertices of the triangle, or the opposite direction, you may get all $3!=6$ orderings.

You probably failed to draw a spacetime diagram because one needs at least 2 spatial dimensions plus 1 time to realize all the situations: it's not a problem that may be fully solved and visualized in the 1+1-dimensional spacetime.

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If A, B, C are three events in spacetime that lie on a spacelike line, BCA and BAC are authorized, but not ACB and CAB. –  Trimok Oct 25 '12 at 10:45
    
Sorry, the "authorized" assumed orderings are written by the OP, you can't change them, and they are ABC and CBA, so your comment makes no sense. B is in the middle by assumption when they're on a line. –  Luboš Motl Oct 25 '12 at 10:48
    
I was not enough precise. It is OK with the assumptions ABC and CBA of the OP, and I do agree that B is the middle when A, B, C are on a line. But, in this case, the orders BCA and BAC are authorized (you can find an observer who sees this order). So you have 2 forbidden orders (ACB, CAB), and 2 possible orders (BCA, BAC) with the hypothesis (ABC, CBA). –  Trimok Oct 25 '12 at 13:08
    
How can you exactly realize BCA in a reference frame if B is in between A and C?? –  Luboš Motl Oct 25 '12 at 21:06
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Fatima, it was a pleasure. I tried to explain why AB, AC, and BC had to be spacelike separated. It's because you said that in two reference frames, the ordering was ABC and CBA. So in one of them, A occurred before B, and in the other, B was before A. But the "both orderings are possible" is only possible if A,B are spacelike separated: if two events E,F are timelike or null separated, all observers agree which of them occurred first. Analogously for AC and BC pairs, they also have to be spacelike separated because the two observers see different orderings of the two members of each pair. –  Luboš Motl Oct 27 '12 at 7:30
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