Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
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).
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!
