I'm confused about this, specifically the spacetime interval.

A timelike interval is one in which 2 events can be related to each other in a given reference frame within its light cone, that is, it is an interval where the 2 events can be related by signals slower than the speed of light.

A spacelike interval is one in which an event within the frame's light cones is connected with an event outside the light cone, thus, the 2 events are entirely unrelated.

Now what I'm confused about, is that if an event happens in space (say the sun exploded) it takes the light 8 minutes to reach Earth, then it would affect Earth where there is another event inside Earth, 8 minutes later.

So that means that the sun explosion will eventually affect any event on Earth, so how are both events 'entirely unrelated'?

How I think it should be, is that the interval is spacelike before light reaches Earth then it becomes timelike once it affected Earth, is that correct? Can I have a good physical intuition about this rather than a mathematical answer, please?

  • $\begingroup$ "the sun explosion will eventually affect any event on Earth" -- how exactly will a future sun explosion affect the Battle of Hastings? $\endgroup$
    – WillO
    Jan 18, 2019 at 13:52
  • $\begingroup$ Yes, that was the problem, time being an extra coordinate was tricky, as I'm new to special relativity. $\endgroup$
    – khaled014z
    Jan 18, 2019 at 16:38

2 Answers 2


I think the ambiguity comes from our definition of the word "event." Remember that events occur at points in space AND time, not just space. For instance, the explosion of the Sun (call it event $S$) would occur at location: the Sun, and at time: when it exploded.

Now, if you want to find the interval between this and another event, you need to assign that event both a point in space and in time. For instance, if the location of the second event is the Earth, you have a choice of time coordinates: before the light reaches Earth, when the light reaches Earth, and after the light reaches Earth. Choosing a time for your event gives you a choice of different events; let's call them "before" ($E_1$), "during" ($E_2$), and "after" ($E_3$).

For each one of those events you assign it a type of interval with event $S$. Because the light from the explosion cannot have affected anything before the light hit Earth, the interval between $S$ and $E_1$ is $spacelike$. The interval between $S$ and $E_2$ is connected exactly by the speed of light, so it is a $lightlike$ interval. And since events after the light from the explosion could potentially be affected by it, the interval between $S$ and $E_3$ is $timelike$.

Note that the intervals themselves do not change over time, since they are connections not between points in space but between points in spacetime. Thus, you need to specify a time for an event to find its interval with another.

  • $\begingroup$ Exactly what I needed! Amazing explanation, thank you. $\endgroup$
    – khaled014z
    Jan 18, 2019 at 16:34

It's always useful to think about two spacetime points, and then get to their separation. We have

  1. Photons leaving the sun at some $x_1^\mu$.
  2. Photons reaching the earth at some $x_2^\mu$.

The spacetime separation between events is $s^2 = (x_1^\mu - x_2^\mu)^2$. By definition the spacetime interval corresponding to light reaching the earth from the sun is lightlike (or null), such that $s^2 = 0$. Thus is it not correct to say that the two events are entirely unrelated (without defining carefully what that means). If you picked a $y^\mu$ that share the same time coordinate as $x_2^\mu$ but was further away in space the separation would become spacelike, as even light has not yet had time to reach that point. Conversely, if we picked a $y^\mu$ that was closer to the sun spatially, the spacetime separation would be timelike.

A spacetime interval is defined using the elapsed time between two events so it will be invariant in time - you cannot have a separation starting out being spacelike and ending up timelike! For more information I recommend this similar question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.