Usual textbook treatments only consider relativity of simultaneity for 2 events, i.e. that different observers don't necessarily agree which of two events happens first. In the case of two events it is possible to choose a spatial axis that contains both events, effectively reducing the problem to one spatial dimension.

However, I am interested in the case of 3 or more events: Do there exist cases of three or more events, such that every order of the events can be observed? By that I mean (for the example of 3 events A,B,C): Do there exist examples of events A, B and C such that one observer sees them in the order ABC, another one in the order BCA, another one in the order CAB, another one in the order BAC, another one in the order ACB, and the last one in the order CBA? For one spatial dimension, the answer seems to be no, but for higher dimension I do not have a clue. Does the maximal number of events whose order can be arbitrarily exchanged by choosing a reference frame depend on the spatial dimension?

In case someone knows the answer, I would really appreciate a reference, like a book or a paper.

  • $\begingroup$ Spacetime and Geometry: An Introduction to General Relativity by Sean Carroll, section 1.11, problem 3. $\endgroup$ – tparker Oct 11 '16 at 5:06
  • $\begingroup$ @tparker Really? I'll have a look so :) Sean Carroll has some nice problems in that book, hadn't really checked the SR questions $\endgroup$ – snulty Oct 11 '16 at 8:11

The key insight is that you can only freely order events if there exists a frame in which they are all simultaneous.

For any $d$ pairwise spacelike separated events in $d$ spacetime dimensions, there exists a frame in which they are all simultaneous. You can construct this frame via a Gram-Schmidt-like procedure: start in a arbitrary frame whose origin is at event $A$ and boost in the spatial direction of $B - A$ by an amount that makes them simultaneous. Then boost in the direction of the projection of $C - A$ orthogonal to $B - A$ by an amount that makes $C$ simultaneous as well, etc. You can keep doing this $d-1$ times until you run out of orthogonal spacelike directions. You'll end up in a frame in which all $d$ events are simultaneous.

From there, you can boost in a direction that will "tilt" the spacelike hypersurface of simultaneity so that the points lie in any order you want. That is, you boost toward the points (projecting out components of directions in which you've already boosted as before) in the order in which you want them to be ordered, with a larger boost every time (though each boost can be infinitesimally small). If you have more than $d$ points though, then once you've brought the first $d$ of them simultaneous, then each remaining point will be either in the future or the past of the first $d$ points. Moving any of those in between ordering of the first $d$ points would require a finite boost, removing your freedom to order the first $d$ by a sequence of infintesimal boosts of increasing magnitude.


I claim that in $N$ spacetime dimensions, there exist sets of $N+1$ points so that any possible time ordering can be realized, as long as the operation of time reversal is also allowed. That is, we must allow all Lorentz transformations, not just those continuously connected to the identity.

Claim 1. Consider any set of $N+1$ points $\vec{a}_i$ in general position in $N$-dimensional Euclidean space. Then for any desired ordering of the $N+1$ points, there exists an axis $\hat{n}$ so the quantities $c_i = \hat{n} \cdot \vec{a}_i$ are in that order.

Proof. Without loss of generality, we can take the ordering to be $1, 2, \ldots, (n+1)$, so we just want the sequence $c_i$ to be increasing. There exists a plane that passes through the first $n$ points but not through the $(n+1)^\text{th}$ point. Let $\hat{n}$ be the axis normal to this plane, directed so that $c_{n+1}$ is largest and $c_1 = c_2 = \ldots = c_n$.

We use induction. The base case is clear. Now apply the inductive hypothesis to the first $n$ points, yielding an axis $\hat{m}$. By continuity, there is some small $\epsilon > 0$ so that $\hat{n} + \epsilon \hat{m}$ gives the desired ordering, proving the result.

Claim 2. The previous result implies that there exists sets of $N+1$ points in $N$-dimensional spacetime which can be time-ordered in any possible way, provided that time reversal is allowed.

Proof. The only difference between this case and that of claim 1 is that we cannot choose axes $\hat{n}$ that are more than $45^\circ$ away from the time axis $\hat{e}_t$. Given an axis $\hat{n}$ that isn't perpendicular to $\hat{e}_t$, we may stretch all the points in the time direction to decrease the angle between $\hat{n}$ and $\hat{e}_t$ until the angle is small enough.

Remark. If time reversal is not allowed, the previous proof fails. To see this, consider the case $N = 2$ with events $A$, $B$, and $C$. Without loss of generality there exists a reference frame where $$t_A = t_B < t_C, \quad x_A < x_C < x_B.$$ It is impossible for any other observer to see $C$ before both $A$ and $B$, because the relativity-of-simultaneity time shift effect is linear in space, yet $C$ is spatially in between $A$ and $B$.


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