special relativity - Why is time-order invariant in timelike interval?

Why do two observers measure the same order of events if we are inside the light cone?

(e.g. if $ds^2 > 0$ time-order is preserved according to the classical mechanics book I am reading, but it doesn't give any proof of this) I assume there is some simple geometrical argument that I am missing. And why do two observers measure possible different order of events if we are outside the light cone?

Answer

For a geometrical argument, you're looking for basically what Ron posted. But you can also argue this one mathematically: as you may know, the difference between two spacetime events is represented by a time difference $\Delta t$ and a spatial difference $\Delta x$. Under a Lorentz boost, these quantities transform like this:

$$\begin{align}c\Delta t' &= \gamma(c\Delta t - \beta\Delta x) \\ \Delta x' &= \gamma(\Delta x - \beta c\Delta t)\end{align}$$

Now, the spacetime interval is $\Delta s^2 = c^2\Delta t^2 - \Delta x^2$. For a timelike interval, $\Delta s^2 > 0$, this means $c\Delta t > \Delta x$, assuming that both differences are positive (and you can always arrange for that to be the case). Using the Lorentz boost equations, you can see that in this case, $c\Delta t'$ has to be positive. So for two events separated by a timelike interval, if one observer (in the unprimed reference frame) sees event 2 later than event 1, any other observer (in the primed reference frame) will also see event 2 later than event 1.

On the other hand, suppose you have a spacelike interval, $\Delta s^2 < 0$. In this case, $\Delta x > c\Delta t$, so it is possible to get $c\Delta t' < 0$ for a specific velocity (namely $\beta > \frac{c\Delta t}{\Delta x}$). So if one observer (in the unprimed reference frame) sees event 2 later than event 1, it's still possible for another observer (in the primed reference frame) to see them in the reverse order.