My professor made following statement:

The spacetime of GR is curved in the presence of strong gravitational fields. The effects of curvature
manifest themselves at large distances. Locally, one can choose a flat Minkowskian metric.

I dont get it:

I thought, gravitation is expressed by curvature. If I sit on my chair, this will be due to gravitation. But this is not an effect shown on large distances (as e.g earth and sun).

  • 2
    $\begingroup$ This is basically a duplicate of How can locally Euclidean space of zero curvature accumulate to non-zero global curvature? $\endgroup$ Jul 16, 2020 at 15:15
  • $\begingroup$ So this statement is not correct? $\endgroup$
    – nuemlouno
    Jul 16, 2020 at 15:34
  • $\begingroup$ Your professor is saying that at any point it's always possible to choose coordinates in which the Christoffel symbols are zero at that point. In these case those coordinates would be the rest frame of a freely falling observer i.e. the Fermi normal coordinates. Your rest frame when you are sitting in your chair is not a normal coordinate system and the Christoffel symbols are not zero. $\endgroup$ Jul 16, 2020 at 15:53
  • $\begingroup$ Does it make sense to talk about 'In a local neighbourhood' in this context $\endgroup$
    – nuemlouno
    Jul 16, 2020 at 16:59
  • $\begingroup$ If you choose a point in spacetime and set up your normal coordinates with the origin at that point then there will be some finite volume around the origin where the Christoffel symbols are negligibly small. I guess that would be local region with which the geometry is approximately described by the Minkowski metric. $\endgroup$ Jul 16, 2020 at 17:03

3 Answers 3


As much as the Earth looks flat it is not, however if you take a really small part of it, you can essentially say that it is flat, same is the case with space-time.

On Space-Time, you can define coordinates, which at the given point are flat and follow Minkowskian geometry, but if you start moving away from them, the difference starts to show up. This is what your professor meant when he said you could choose coordinates that are locally flat, it is that at each and every point in space time there are coordinates which are flat.


Sitting in your chair, you are not in an inertial reference frame. The equivalence principle states that a local accelerating frame is not distinguishable from the effect of gravity. This has no bearing on curvature. If you are sitting in your chair, you have chosen an accelerating frame. You have not chosen a Minkowski metric (even though your accelerating frame is flat). If you want to chose a Minkowski metric, you must chose a frame in free fall, not in a chair.

Your professor is not talking about local frames. Curvature is not seen in local frames (just as the curvature of the surface of the Earth is not seen in a town map). Curvature is seen in the gravity of the Earth because an inertial frame (one in free fall, not sitting in a chair) on one side of the Earth, does not remain in uniform motion or at rest with respect to an inertial frame on the other side of the Earth.


I think now I understand my confusion:

One can choose riemann normal coordinates to get the canonical form of the metric (which can be the minkowski metrik). So locally we can choose a flat metric.

I thought, this implies that spacetime is locally flat without curvature.

But the curvature is described by the riemann tensor, which depends on the second derivative of the metric. Which is not vanishing by choosing riemann normal coordinated!


Your Answer

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

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