Relativity says that two events simultaneous in one reference frame might not be simultaneous in another reference frame. Can we extend this idea a little and say that the order of two events might be be reversed in going from one reference frame to another? And if so, what if they have a cause-effect relation ship in the first frame. Isn't reversing the two in the second frame a contradiction?

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    $\begingroup$ The important thing to keep in mind is that time ordering can only be reversed for space-like separated events. From that you can draw the correct conclusion about the possibility of causality reversal in special relativity. $\endgroup$ Feb 13 '14 at 3:49
  • $\begingroup$ I'm sorry @dmckee, I don't know what space-like separated events are. Can you explain more if you have the time? $\endgroup$ Feb 13 '14 at 3:52

No, ordering of timelike-separated events is always preserved in special relativity.

First, let's agree on some definitions. Say event $O$ occurs at the origin of our 1D coordinate system. Consider another event $A$ in spacetime, at time $t$ and position $x$ in some pre-fixed coordinates. There possibilities for how $x$ and $t$ relate are as follows:

  • $\lvert t \rvert < \lvert x \rvert$: $A$ is spacelike separated from $O$;

  • $\lvert t \rvert = \lvert x \rvert$: $A$ is lightlike (or null) separated from $O$:

    • $t = \lvert x \rvert$: $A$ is in $O$'s future;
    • $t = -\lvert x \rvert$: $A$ is in $O$'s past;
  • $\lvert t \rvert > \lvert x \rvert$: $A$ is timelike separated from $O$:

    • $t > \lvert x \rvert$: $A$ is in $O$'s future;
    • $t < -\lvert x \rvert$: $A$ is in $O$'s past.

Now apply a boost of velocity $v$, where $\lvert v \rvert < c \equiv 1$. In this new frame, we have \begin{align} t' & = \frac{t-vx}{\sqrt{1-v^2}}, \\ x' & = \frac{x-vt}{\sqrt{1-v^2}}. \end{align} As you can check, $(x')^2 - (t')^2 = x^2 - t^2$ independent of $v$. This quantity is the invariant interval, and the fact that it doesn't change between reference frames means the relation between $O$ and $A$ cannot shift amongst the different classifications given above.

The important lesson here is that the future is the future, the past is the past, and everything else is everything else (note I'm not calling this the "present," which doesn't mean anything in relativity), and these definitions are frame-independent.

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    $\begingroup$ Probably useful to note that this work is done in $c =1$ units. Using SI units every occurrence of $t$ in the above should be read as $ct$ and every occurrence of $v$ as $v/c$. $\endgroup$ Feb 13 '14 at 14:11

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