Are the original Schwarzschild solution and Hilbert's solution exactly the same?

Question

Schwarzschild original metric solution can be found here in equation (14): http://old.phys.huji.ac.il/~barak_kol/Courses/Black-holes/reading-papers/SchwarzschildTranslated.pdf

Hilbert's metric solution is the one we are all familiar with, and the one that wikipedia shows under the name "Schwarzschild solution": https://en.wikipedia.org/wiki/Schwarzschild_metric

It is clear that in the metrics (Im not talking about their derivation), they only differ in what they refer as R and r respectively. In Schwarzschild's original metric, $R$ is just an "auxiliary quantity" which follows $R^3=r^3 + α^3$, with $r$ been the distance marker and $\alpha$ being the well-known $α=2GM$. One can easily see that in the Hilbert's metric, he substitues $R$ for $r$. For both, $r$ takes values from zero to infinity. But $R$ and $r$ do not follow a linear relationship!

This has been noted in https://arxiv.org/abs/physics/0310104, and it not only affects the metric but there are also differences between both derivations of the metrics, I quote in page 5: "What is not legitimate, although first done by Hilbert and subsequently handed down to the posterity, is to assume without justification that the range of the “new” $r$ is still $0 < r < ∞$, as it was for the “old” $r$, because this is tantamount to setting sqrt(G(0)) = 0, an arbitrary choice [5], equivalent to setting $ρ = 0$ in Schwarzschild’s result, reported in equation (5)."

Even Schwarzschild himself stated that "Actually Mr. Einstein’s approximation for the orbit goes into the exact solution when one substitutes for $r$ the quantity $R$", suggesting that both $r$-metric and $R$-metric do not result in the exact same orbits for the Mercury problem which Einstein was addresing. You may want to see that 1915 Einstein's aproximation uses $g_{tt}= 1-(α/r)$.

Moreover, I have found other papers claiming that both solutions are different and they do not resemble the same geodesics: https://www.researchgate.net/publication/331936281_Schwarzschild's_family

Thanks for the edit.

Answer 1

$R$ and $r$ are just coordinates. They're completely equivalent because there's a bijective mapping between them. The $r>0$ and $R>α$ regions are the same, and so are the $-α<r<0$ and $0<R<α$ regions. Being negative doesn't make $r$ unphysical. Coordinates labeled by the letter $x$ can be negative, and so can coordinates labeled by the letter $r$. From now on I'll just use $R$.

Schwarzschild coordinates are singular at $R=0$ and $R=α$. (In particular, $R=α$ does not cover the event horizon, contrary to semi-popular belief.) But they are well behaved for all other $R$. For $R>α$, they of course cover the exterior. For $0<R<α$ they cover a "vacuum big crunch" manifold. For $R<0$ they cover a manifold describing a negative point mass (with no event horizon) in a Minkowski background.

As it turns out, the $0<R<α$ manifold can be interpreted as the black hole interior, but given that there's no way to reach it from the exterior in Schwarzschild coordinates, Schwarzschild can be forgiven for not realizing that. The $R<0$ manifold really is disconnected from the others and is physically irrelevant—although a similar region shows up in the maximally extended Kerr geometry, where it's accessible in principle by going through the ring singularity.

The arXiv paper claims that Hilbert and others wrongly assumed that $R$ is valid in $(0,\infty)$. That would indeed be a wrong assumption since the coordinates are singular at $α$, and some people do make that mistake. But it's not wrong to say that $R$ is valid in $(0,α)$. It doesn't "have to be" valid, but in point of fact it is.

[...] suggesting that both $r$-metric and $R$-metric do not result in the exact same orbits for the Mercury problem which Einstein was addresing.

Orbits are the same whether expressed in terms of $r$ or $R$, just like temperatures are the same whether expressed in Fahrenheit or Celsius. If mathematics is internally consistent then calculations related by a substitution of variables can't produce inconsistent results.

Things can get tricky when relating a calculated to a measured orbit. Observations of Mercury's orbit predated general relativity, and they were interpreted in a Newtonian flat-space model. To really work out the correct experimental orbit for comparison, you'd have to reanalyze the raw observational data in a more accurate general-relativistic framework. You might find that the radial distances calculated in the Newtonian model match the general-relativistic $r$, or that they match $R$, or (more likely) that they're different from either one.

If you analyze everything consistently, then every physically meaningful result that you calculate will be independent of the coordinates that you choose. But if you equate a Newtonian $r$ and a Schwarzschild $r$ (or $R$) just because they have similar names, then you may get into trouble. (Although in this case you probably won't because the difference between them is probably minuscule.)

Answer 2

The solutions are the same: they describe the same spacetime, but using slightly different coordinates. But the early workers, especially Schwarzschild, were not sure what to make of the region inside the horizon, or whether there was such a region. Droste's work showed more insight on this point, I think, and as I understand it his contribution was somewhat more pioneering than Hilbert's (it was a little earlier and quite thorough).