0
$\begingroup$

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?"

$\endgroup$
13
  • $\begingroup$ This question (v4) is also addressed in e.g this and this Phys.SE answers. $\endgroup$
    – Qmechanic
    Aug 29, 2023 at 11:46
  • $\begingroup$ 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. $\endgroup$
    – Scibo
    Aug 29, 2023 at 12:09
  • $\begingroup$ 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). $\endgroup$
    – Scibo
    Aug 29, 2023 at 12:17
  • $\begingroup$ For $d\leq 3$ GR is a topological field theory with no propagating DOF. $\endgroup$
    – Qmechanic
    Aug 29, 2023 at 12:19
  • 1
    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Aug 29, 2023 at 15:19

0

Your Answer

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