# Two events "connected by propagation of a ray of light" have $(\Delta s)^2=0$?

This says that $(\Delta s)^2 = -c^2(t_2-t_1)^2 + (\vec{x_2}-\vec{x_1})^2=0$.

What I'm trying to imagine here (as per my understanding of the statement) is that two events are in the same light cone. Which means that two events may be separated by time and space. How does $(\Delta s)^2$ become zero then?

Can someone please explain what this means conceptually and show how $(\Delta s)^2=0$ mathematically? Would really appreciate that.

• I see in a previous question that you are still struggling to understand the place that interval holds in special relativity and its importance. Lunging forward to the general theory with that uncertainty intact is going to be difficult. Commented Jan 8, 2017 at 19:58
• I think I cleared the previous question for myself. Commented Jan 8, 2017 at 19:59

"Connected by propagation of a ray of light" means that the two events have a separation in spacetime such that if a ray of light were emitted from the location of event 1 at the same time as the occurrence of event 1, then that ray of light would reach the location of event 2 at the same time that event 2 occurs. If event 1 forms the vertex of the light cone, then in this scenario, event 2 lies on the surface of the light cone, not inside of it. The math you've already written shows how it equals zero - since light travels at speed $c$, $\left| \Delta \vec{x}\right| = \left|\vec{x}_2 - \vec{x}_1\right| = ct_2 - ct_1 = c\Delta t$. This exactly cancels the first term.
If event 2 were inside the light cone, then the interval between the two events would be $\left(\Delta s\right)^2 < 0$, since the new $\left|\Delta \vec{x}\right|$ would be smaller than the distance that light would travel, $c\Delta t$. Such events are called time-like and can be linked causally, since messages between the events need not exceed the speed of light, which forms the surface of the light cone.