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In other words, if events a and b are time-like separated, and events b and c are time-like separated, does it follow that events a and c are time-like separated?

Is there a straightforward proof of transitivity here?

(Thanks a lot for the awesome explanations. I really appreciate them. Both the diagram-based explanation and the interval-based explanation were very helpful. I swear this is not a homework question! I took an STR class in college 15 years ago but went into a different field of study. I just forgot how to test for this and was curious. :) )

user356083
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2 Answers2

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A lot of questions in special relativity become surprisingly simple if you draw the correct diagram. Consider this:

enter image description here

All the events timelike connected to the point B lie in B's light cone i.e. in the pink triangle. So we can see at a glance that A and C are timelike connected to B.

However I have also drawn the light cones of A and C - I have shaded the forward light cones and drawn the reverse light cones with a dashed line. Again we can see at a glance that neither A nor C lies in the forward or reverse light cones of the other. Hence A and C are not timelike connected.

With a diagram like this you can play around moving the positions of A and C, and it is immediately obvious whether they are timelike connected or not.

John Rennie
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No. Consider the following three events.

$(c t_a, x_a) = (1, -0.9)$

$(c t_b, x_b) = (0, 0)$

$(c t_c, x_c) = (1, 0.9)$

Then using $$ \Delta_{ij} = -(ct_i - ct_j)^2 + (x_i-x_j)^2 $$

We have

$\Delta_{ab}^2 = -0.19 < 0$

$\Delta_{bc}^2 = -0.19 < 0$

$\Delta_{ac}^2 = 3.24 > 0$

In other words, $b$ is timelike separated from $a$ and $c$, but $a$ and $c$ are spacelike separated.

On the other hand, if you restrict $t_a < t_b < t_c$, and if $ab$ and $bc$ are timelike separated, then it does follow that $ac$ are also timelike separated. The reason is that you know you can follow a timelike trajectory from $a$ to $b$ and another one from $b$ to $c$, and so the combined trajectory is everywhere timelike. It might be possible to find an even more direct timelike path, but it suffices to find one timelike path connecting $a$ and $c$ to conclude they are timelike separated. My example above avoids this argument by choosing the ordering of $t_a, t_b, t_c$ such that you couldn't follow a timelike path forward in time from $a$ to $b$ to $c$.

Andrew
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