Calculating an arbitrary metric tensor (field) in vacuum: Make use of the constant speed of light?

The metric (=field of metric tensors) is the solution of Einstein's field equations when a special distribution of matter is given. It is among the unsolved problems of physics to calculate the metric (and, thus, solve Einstein's field equations) for an arbitrary distribution of matter. Special solutions are known (Schwarzschild, Kerr,... ), however, to the best of my knowledge, even the metric for the two-body problem isn't exactly known yet and cannot be simply constructed out of the single-body solutions.

Within the known derivation of a spherical symmetric vacuum metric (Schwarzschild), there is only one coefficient for the space (A) and only one for the time (B). It is derived therein, that A = 1/B. To me, that (AB=1) looks a bit like Einstein's postulate that the speed of light is constant.

I'm wondering:

(I) is it true that in the case of the Schwarzschild metric, "AB = 1" expresses the constancy of the speed of light in vacuum?

(II) if the answer to (I) is "yes", why isn't that used from the very beginning, why is it derived?

(III) And, finally, independent from (I) and (II), is it possible to use somehow the constancy of the speed of light in vacuum to help to derive a more arbitrary metric in vacuum? Or has that possibly already been done somewhere?

Addendum: Does it have something to do with the infinitesimal volume in 4D? The product of the sidelength in all 4 coordinate directions, isn't that invariant?

• Some issues: (1) you mean AB=1 and not A/B=1, right? (2) Kerr isn't spherically symmetric.
– Sten
Commented Sep 17, 2023 at 23:25
• I corrected that, thank you! Commented Sep 18, 2023 at 4:21

1 Answer

The Schwarzschild metric is of the form

$$ds^2=-f(r)dt^2+\frac{1}{f(r)}dr^2+r^2d\Omega^2,$$

where $$f(r)$$ is some function of radius $$r$$. You are asking if the appearance of $$f(r)$$ and its reciprocal in the temporal and radial terms, respectively, is associated with the speed of light being constant.

This is not the case, as can be shown by direct evaluation. For light, $$ds=0$$, and let's assume it is radially directed, so that $$d\Omega=0$$. Then

$$f(r)dt^2=\frac{1}{f(r)}dr^2,$$

i.e.

$$\frac{dr}{dt}=\pm f(r).$$

Evidently, the "speed of light" is not constant!

However, $$dr/dt$$ is only the coordinate speed. In general, there is no restriction on the coordinate speed of light or anything else.

• I don't mean coordinate speed. That's arbitrary. I made an addendum to my question: Probably my question has something to do with the 4D volume (instead of your ds^2) being constant? Commented Sep 18, 2023 at 4:55