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A field of metric tensors fully characterises the curvature of a vacuum space-time. (For example, the spacetime between some single point masses which are themself not part of the manifold) The metric tensor $g_{\mu\nu}$ is a $n$ x $n$ tensor in $n$ dimensions.

With $g_{\mu\nu} = g_{\nu\mu}$, the number of obviously independent functions reduces to:

4D - 10 (spacetime)

3D - 6

2D - 3

However, the curvature of a 2D plane, which is described in this list by a field of metric tensors with three independent functions, can be fully characterised by using only one single parameter: the altitude above ground, as it is done in colormaps of geographical/topological maps.

  1. How comes that two degrees of freedom vanish by changing the form of the plot?

  2. Is this somehow generalizable, can the number of independent functions be reduced also for the metric tensor of spacetime? How and to which number?

  3. Does it help, therein, that the speed of light in vacuum is constant regardless of the curvature of spacetime?

EDIT: This question can be read as "How many scalar fields are necessary to fully characterise any possible tensorfield of metric tensors in 2, 3, 4... dimensions?"

Scibo
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  • This question (v4) is also addressed in e.g this and this Phys.SE answers. – Qmechanic Aug 29 '23 at 11:46
  • Thank you for those related links! The first one is about weak gravitational fields. My question is regarding general (merely mathematical) curvature, independent of being weak or strong. For 2D, the formula d(d-3)/2 does not make sence as it leads to a negative result. – Scibo Aug 29 '23 at 12:09
  • The second one is also very interesting, and I'm still reading. However, the formula at the end as well doesn't make sence for 2 dimensions (as it leads to a negative result). – Scibo Aug 29 '23 at 12:17
  • For $d\leq 3$ GR is a topological field theory with no propagating DOF. – Qmechanic Aug 29 '23 at 12:19
  • Okay, thank you for that comment! However, my question is ment merely regarding general curvature, not only regarding GR... GR is only one application, it's also interesting for curvature in 3D (diffeomorphisms of images), or generally in 4D, not only for spacetimes... – Scibo Aug 29 '23 at 12:39
  • Related: https://physics.stackexchange.com/q/506086/2451 – Qmechanic Aug 29 '23 at 12:43
  • My question is ment primarily to understand the metric tensor in arbitrary dimensions. – Scibo Aug 29 '23 at 12:43
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    What do you mean by "degress of freedom" if you don't want to be specific to GR? Usually, "degrees of freedom" are specific to a physical theory, not something a tensor possesses in general. For instance, a 4-vector has 4 components, but a standard massive vector field has 3 d.o.f. and a standard massless vector field has 2 d.o.f. It doesn't make sense to ask about the d.o.f. of something without specifying its dynamics. – ACuriousMind Aug 29 '23 at 15:19
  • By "degrees of freedom" I mean how many variables (at every point) are needed to characterise any possible tensorfield of metric tensors at any specific number of dimensions. I wonder why it's less than three in 2D (I truely don't understand that at the moment). And I'm interested in general diffeomorphisms: image distortions of 3D images - How many 3D scalar fields are necessary for their full description? And of course I'm interested as well in GR. This "degrees of freedom" is not about dynamics, but on how many variables at every point are necessary to fully describe the curvature. – Scibo Aug 29 '23 at 15:49
  • I added "(at every point)" to the title of the question to make that clear – Scibo Aug 29 '23 at 16:10
  • Note that the metric tensor incorporates a coordinate choice, not just the curvature, so it must carry more information (not all physical). – Sten Aug 29 '23 at 16:15
  • However, the number of degrees of freedom at every point should not be dependent on the choice of coordinates. – Scibo Aug 29 '23 at 16:18
  • To start: in 2D, how do we calculate the metric tensor out of the altitude map? Is that probably the dot product of the partial derivatives of the altitude at every point? – Scibo Aug 29 '23 at 20:00

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