Can special relativity distort the relative order in which events occur?

Pretend you are throwing darts at a dart board. You throw dart $d_1$ at time $t_1$. After you throw your first dart, you throw your second dart $d_2$ at time $t_2$. Given that $t_2 > t_1$ in a stationary frame of reference, is it possible for a frame of reference to exist that will make an observer $O_1$ in that frame of reference see $d_2$ hit the dart board before $d_1$?

In special relativity, you think of a 4-dimensional space-time. The key point here is that two events, 1, and 2 happening at $t_{1}, x_{1}, y_{1}, z_{1}$ and $t_{2},x_{2},y_{2},z_{2}$ have a distance given by ${}^{1}$

$$(\Delta s)^{2} = -c^{2}(\Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}$$

Now, we can therefore give any two events a unique relation to each other:

1) $(\Delta s)^{2} < 0$: These events are considered timelike separated.

2) $(\Delta s)^{2} = 0$: The events are lightlike separated

3) $(\Delta s)^{2} > 0$: The events are Spacelike separated

The key point is that, in all reference frames, timelike and null events happen in the same order (this is derivable from the fact that all observers follow timelike paths).

But, you can also show that there is no unique ordering of spacelike events--events that happen in order 1 > 2 > 3 in frame A may happen in order 2 > 1 > 3 in frame B. In fact, for any two spacelike separated events, it is possible to find a reference frame where you can reverse the order in which they happen.

So, the answer to your question is "maybe", but the trick would have to be that you throw the second dart much more quickly than the first, and it hits the dartboard a distance X a way at a time T later than the first, in such a way that $T < X/c$. You do this, then there will be a reference frame where it appears that the second dart hits first.

${}^{1}$here $\Delta t = t_{2} - t_{1}$, etc

• +1: I propose that you put the "much" of much more quickly in caps, boldface and italics. – joshphysics Aug 30 '13 at 23:25

No, special relativity does not violate the order in which events take place. Causality is preserved, two observers in different frames of reference will always agree on which dart hits the board first, as the movement of a dart corresponds to a timelike curve in four-dimensional spacetime. The only quantity they will disagree on regarding the impact is the difference of the two times, this can be explained in terms of time-dilation.

• Not quite--see my answer. the darts themselves follow lightlike trajectories, and the thrower does, but it's not necessary that the two events where the darts hit the board be timelike separated. – Jerry Schirmer Aug 30 '13 at 23:08
• I implicitly assumed that the events are timelike separated (i.e. the darts hit at the same spot). But you are right, technically this might happen. – Frederic Brünner Aug 30 '13 at 23:13
• @JerrySchirmer I assume you meant "timelike" trajectories. Also, in order for the dart hits to be spacelike separated, the time separation between the throws would have to be on the order of a nanosecond (assuming the board is on the order of one meter accross): that's a pretty damn fast dart thrower. – joshphysics Aug 30 '13 at 23:22
• @joshphysics: I would argue that it's more about the precision of the thowing than the fastness of the second dart. Put the dartboard 2 m away, throw the first dart at 1 m/s, wait 1.5 s and then throw the second dart at 2 m/s. They'll hit at the same time. Tuning it so that they hit at the same time with nanosecond precision is the tricky and unlikely part. – Jerry Schirmer Sep 1 '13 at 19:16
• @JerrySchirmer Great point. I hadn't considered that scenario. – joshphysics Sep 1 '13 at 21:44