I have taken a relativity course, and am not quite clear as to whether the notion of a frame of referene is independent of the coordinate system. At the moment, I think that any frame of reference (of which a special subset are intertial- i.e. those related by lorentz boosts or free falling in GR) can choose any coordinate system, i.e. any way of labelling spacetime events, that they like.

Specific examples of where this difference (?) might beimportant:

  1. I found myself arging that $\nabla_a u^a$ must always vanish for dust because if one goes into the local rest frame of dust and chooses local minkowki coordinates so that the components of the metric connection vanish, then this clearly vanishes. However I am not sure if I have applied too many restrictions here.

  2. Secondly, In the Schwarzschild geometry, whose coordinates are those that are used in the scwarzschild line element? I'm not sure if such a question even makes sense? I think: they are the coordinate system of an observer 'at rest' at infinity who decides to have local Minkowski coords i.e. the line elemnt at infinity reduces to local cartesian coords.

  • 3
    $\begingroup$ related or duplicate: physics.stackexchange.com/q/458854 $\endgroup$
    – user4552
    May 23, 2019 at 15:43
  • $\begingroup$ @BenCrowell I added an answer to your post quoted above. $\endgroup$
    – Elio Fabri
    Jun 2, 2019 at 14:40

1 Answer 1


I found myself arging that $\nabla_a\,u^a$ must always vanish for dust

Right. "Dust" means non-interacting particles. So each one follows a geodesic.

In the Schwarzschild geometry, whose coordinates are those that are used ...

It's true that Schwarzschild spacetime is asymptotically flat and usual Schwarzschild coordinates aptly reflect this fact. But there are an infinity of coordinates with the same property.

Just a special point. If you start from infinity with $t$ defined as Minkowsky time of an inertial frame, how could you extrapolate it at every distance from the symmetry centre?

Actually the most important porperty of Schwarzschild's $t$ is that the metric doesn't depend on that coordinate, so that an isometry of spacetime is immediately apparent - Schwarzschild spacetime is static. Think of how easily gravitational redshift can be derived using that isometry.

See also my comment to Ben Crowell.


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