People have been saying that a global frame of reference does not exist in General Relativity.

However, from Wikipedia:

In Schwarzschild coordinates ${\displaystyle (t,r,\theta ,\phi )}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form $${\displaystyle g=-c^{2}\,{d\tau }^{2}=-\left(1-{\frac {r_{\mathrm {s} }}{r}}\right)c^{2}\,dt^{2}+\left(1-{\frac {r_{\mathrm {s} }}{r}}\right)^{-1}\,dr^{2}+r^{2}g_{\Omega },} $$ where ${\displaystyle g_{\Omega }}{\displaystyle g_{\Omega }}$ is the metric on the two sphere, i.e. ${\displaystyle g_{\Omega }=\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right)}.$

"Schwarzschild metric", Wikipedia

Isn't $\left(t,r,\theta,\phi\right)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?

  • $\begingroup$ related: physics.stackexchange.com/questions/458854/… $\endgroup$
    – user4552
    Commented Sep 22, 2019 at 2:21
  • 1
    $\begingroup$ Isn't $\left(t,r,\theta,\phi\right)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does? $\endgroup$
    – user4552
    Commented Sep 22, 2019 at 2:22
  • $\begingroup$ @BenCrowell Assuming the observer is at infinity and is co-moving with the black hole, the observer will be stationary in those coordinates and the time coordinate "t" will flow at the same rate as the a clock of the observer. Am I right or wrong? $\endgroup$
    – user242977
    Commented Sep 22, 2019 at 3:59
  • $\begingroup$ Indeed @The Last StyleBender, however this is also the case for Gullstrand-Painleve and Eddington-Finkelstein coordinates! But the way these "time" coordinates extend through the rest of spacetime is different. I completely concur with Crowell. $\endgroup$ Commented Sep 25, 2019 at 8:38

3 Answers 3


People have been saying that a global frame of reference does not exist in General Relativity.

Can you give an example? I don't really know what this means.

There are no global inertial reference frames in general relativity, in general.

[...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?

The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.

In special relativity, "observer" usually means "inertial reference frame."

In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).

In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.

Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.

  • $\begingroup$ I've posted a related question here and I'm wondering if you can answer this physics.stackexchange.com/questions/504024/…. $\endgroup$
    – user242977
    Commented Sep 22, 2019 at 3:44
  • $\begingroup$ Regarding your answer to this question: wouldn't the Schwarzschild metric be a coordinate system in which the observer (co-moving with the black hole at infinity) will be at rest and the "t" coordinate representing the time flow of the observer? $\endgroup$
    – user242977
    Commented Sep 22, 2019 at 3:50
  • $\begingroup$ @the-last-stylebender If you're at rest in Schwarzschild coordinates (dr/dt = dθ/dt = dϕ/dt = 0) then you're at rest relative to the black hole. If you're far from the hole then the t coordinate will more or less agree with your proper time (the speed at which your watch ticks). If you're closer, you need to look at the metric to determine the ratio between them. $\endgroup$
    – benrg
    Commented Sep 22, 2019 at 5:31
  • $\begingroup$ See also the almost identical discussion under Nat's answer $\endgroup$ Commented Sep 25, 2019 at 8:40
  • $\begingroup$ Although I know the Schwarzschild black hole's felt to be much simpler, mathematically, than the Kerr or Kerr-Newman BH's, its non-rotating characteristic seems astronomically absurd, so I appreciate your giving its coordinate system some epistemological utility. $\endgroup$
    – Edouard
    Commented Jan 19, 2021 at 5:03

when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime

A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.

To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures

  • $\begingroup$ Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not? $\endgroup$
    – user242977
    Commented Sep 21, 2019 at 18:14
  • $\begingroup$ @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate. $\endgroup$
    – aitfel
    Commented Sep 21, 2019 at 19:23
  • $\begingroup$ A "frame of reference" has to do with observers, which are timelike worldlines (or similar), not charts and an atlas. Certainly this is a physical interpretation, at least $\endgroup$ Commented Sep 25, 2019 at 8:27
  • $\begingroup$ @aitfel, actually we require 8 Schwarzschild coordinate charts to cover the maximally extended spacetime. (I mean, cover everywhere except $r=2M$.) Each quadrant of spacetime needs 2 charts, because of the $\theta$ and $\phi$ coordinates $\endgroup$ Commented Sep 25, 2019 at 8:32

Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.

  • 2
    $\begingroup$ Why is the Euclidean metric relevant to a question abour general relativity? $\endgroup$
    – PM 2Ring
    Commented Sep 21, 2019 at 20:12
  • $\begingroup$ It's close to a minkowski metric it's also non curved $\endgroup$
    – Kugutsu-o
    Commented Sep 22, 2019 at 18:53

Your Answer

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