Schwarzschild's Outer Metric Is Incompatible with Kepler's III Law

Brown, in his Reflections on Relativity attempts "to give a very plausible (if not entirely rigorous) derivation of Schwarzschild's metric purely from [a] knowledge of the inverse square characteristic of gravity, Kepler's third law for circular orbit, and the null intervals of light paths."

Brown bases his derivation on the space-time metric

$$d\tau^2=g_{tt}dt^2-r^2d\phi^2,$$

for purely circular motion in the equatorial plane ($\theta=\pi/2$), and at constant radius ($dr=0$). However, for null intervals of light paths $d\tau=0$ so his demonstration reduces to $g_{tt}=r^2\omega^2$, where we introduced the angular speed as $\omega=d\phi/dt$, with $t$ lab time. There is no constant of integration, and, even if we were to introduce Kepler's law

$$r^3\omega^2=GM,$$

we would get the desired result, $g_{tt}=1-2GM/c^2r$, which is the time component of the Schwarzschild metric. 

Poor Kepler didn't know the difference between local time, $\tau$, and coordinate time $t$, and, even if he had known, it wouldn't have changed any of his law, except maybe for small relativistic corrections. But, we want to use Kepler's III as it stands. Consequently, we work with the space part of the metric,

$$dt^2=g_{rr}dr^2+r^2d\phi^2.$$

One could object at this point and say we put $g_{tt}=1$ to derive the above equations. However, it will turn out that

$$g_{tt}g_{rr}=1,$$

so that multiplying each of the metric coefficient by the same factor does not change the curvature. 

For the metric

$$dt^2=E(r)dr^2+G(r)d\phi^2,$$

where the metric coefficient depend only on the radial coordinate $r$, the Gaussian curvature is

$$K_G=-\frac{1}{2\sqrt{EG}}\frac{d}{dr}\left[\frac{dG/dr}{\sqrt{EG}}\right].$$

Putting $E=(1-2GM/r)^{-1}$ and $G=r^2$ into the expression for $K_G$, we find

$$K_G=-\frac{GM}{r^3c^2}=-\frac{\omega^2}{c^2}<0,$$

where we used Kepler's III to get the second equality.

We want to compare this with Kepler's laws. Newton found the radius of curvature is given by

$$\rho=r/\sin\alpha=c/\omega>0,$$

where $\sin\alpha=r\omega/c$. The radius of curvature appear's in Newton's expression for the centripetal acceleration, $v^2/\rho$. (Please refer to a previous blog for the derivation.) 

The expression for Gaussian curvature says that Schwarzschild's outer metric is dealing with a hyperbolic point,  whereas Newton's radius of curvature says that it is an elliptic point. This is apparent from Kepler's I. The two are incompatible with one another, and we had no right to use Kepler's III to reduce the Gaussian curvature to a (negative) constant. Schwarzschild's outer solution is a hyperbolic metric of non-constant curvature.

Instead, if we set $E=(1-r^2\omega^2/c^2)^{-1}$ and $G=r^2$, we come out with a positive Gaussian curvature,

$$K_G=\frac{\omega^2}{c^2}>0.$$

Thus, we can set 

$$K_G=1/\rho^2,$$

showing that the metric of the inner solution is compatible with Kepler's laws. In fact, we were wrong to apply Kepler's III to the outer solution, which has a non-constant curvature---and there is absolutely nothing that can make it constant. 

We have remarked that if we divide the metric coefficients by the same factor, it will not change the curvature. Conformal transformations of the metric keep angles constant but change the magnification so that should not change the curvature. So if we take $E=(1-r^2\omega^2)^{-2}$ and $G=r^2/(1-r^2\omega/c^2)$, we come out with the well-known Beltrami metric, which I have shown to describe a uniformly accelerating disc. There is no need to apply Kepler's III since the metric is one of constant (positive) curvature.