Take the following situation:

An astronomer is on the surface of Sun (assume he's not rotating around the Sun). He measures two stars from two locations in the universe exploding. Both stars exploded 3000 years ago. Now, the astronomer goes somewhere else but he remains stationary relative to the sun (perhaps outside the event horizon of a Schwarzschild black hole that is neither moving towards or away from the Sun). Will the astronomer still measure the stars exploding at the same time?

I read that the concept of relativity of simultaneity in general relativity is kind of meaningless, but isn't my question in the above situation valid? Does the concept of relativity of simultaneity hold in General Relativity?

There seems to be a bit of confusion on my description (as can be seen in the comment section): Essentially I am asking: if we take into account the light travel time (time the astronomer "saw" it minus the time for the light to travel to the observer), will the explosion still be simultaneous?

  • 1
    $\begingroup$ Is it universe or university on earth or planet X with aliens? $\endgroup$
    – aitfel
    Sep 21, 2019 at 19:05
  • $\begingroup$ Well spotted... It's "universe" (and definitely not planet X with aliens) :) $\endgroup$
    – user242977
    Sep 21, 2019 at 19:09
  • $\begingroup$ Well the moment you say stars exploded 3000 years ago you are by default using a data of some observer most probably in this case the astronomer stationarily located at sun. $\endgroup$
    – aitfel
    Sep 21, 2019 at 19:12
  • 1
    $\begingroup$ When you say both stars exploded 3000 years ago, are you saying they are both 3000 light years distant, but in different directions? In which case if you move to some other arbitrary position in space then of course they will not be viewed to explode simultaneously, and this has nothing to do with GR, just the finite speed of light. $\endgroup$
    – ProfRob
    Sep 21, 2019 at 22:50
  • 1
    $\begingroup$ What does simultaneous mean if you can't unambiguously translate a frame from one explosion to the other (or from yourself to either)? $\endgroup$
    – WillO
    Sep 22, 2019 at 5:12

1 Answer 1


I'll assume that you do a good job of using various clues (the time you see the light, your location when you see it, the direction of the light, and some estimate of the distance to the star) and correctly work out more or less where each explosion took place in spacetime.

In this case, no matter where you are, no matter your speed, and no matter anything else about you, you will derive the same spacetime locations for the explosions, because the locations are an objective property of the external world and we're assuming that you measured them correctly.

There are a lot of different ways you could write down the locations. Those ways are called coordinate systems. Some coordinate systems have a coordinate called "t" in them, and depending on the coordinate system, the t coordinates of the two explosions might be the same or might be different. This isn't a property of the explosions, but of the arbitrary choice of coordinates.

The choice of coordinates is really up to you. In introductions to special relativity it's common to assume that everybody picks an "egocentric" coordinate system (a term I just made up for coordinates in which they're at rest at the origin). If everyone does that, then people moving at different speeds are likely to disagree about the equality of the t coordinates of various things. But (if they're good scientists) they won't disagree about the objective locations of those things, because they'll understand that their choice of coordinates doesn't influence the actual locations. And also (if they're good scientists) they'll understand that they don't need to choose egocentric coordinates, and especially if they're collaborating with someone else they'd be better off agreeing on a common coordinate system to communicate their results.

As I said in my other recent answer, in general relativity the choice of useful coordinate systems tends to be more restricted because most spacetimes have fewer symmetries than the Minkowski spacetime of special relativity. You can still use any coordinates you want, but most will be inconvenient because the metric will have an unnecessarily complicated form. In particular, it tends to be inconvenient to use egocentric coordinates.

When you're talking about the universe on a large scale, only one time/"t" coordinate, called cosmological time, is convenient, because it's the only one that respects the large-scale spacetime symmetries of the universe we live in. When you see a "time since the big bang" in articles about cosmology or astronomy, it's cosmological time.

When you work out the coordinates of the two stars, you'll probably end up with the same t coordinates as someone else working independently on the same problem, because you both will pick the most convenient t coordinate, and that's cosmological time. It doesn't matter where you are or how fast you're moving, since it's dictated by the objective "shape of the universe" which everyone agrees on in principle, if they have accurate enough equipment to measure it well. You could pick a different coordinate system and disagree with the other experimenter, but that doesn't say anything profound about the nature of objective reality, it just says that you picked a different coordinate system.


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.