How Do Contradictory Metrics Align with GR Field Equations?

[PeterDonis]

within that metric a surface of constant R would have surface area 4\pi R^2>4\pi r^2 would it not?

Yes, but that's not why the metric is non-Euclidean. The relationship between $R$ and $r$ is irrelevant here. See below.

That would indicate R^+ \times S^2 has negative curvature and not be Euclidean

No. What indicates that a surface of constant $t$ has negative curvature is, as I said, that the coefficient of $dR^2$ in the metric is not 1. The radial coordinate in this metric is $R$, not $r$; you don't need to know anything about $r$ to determine that the spatial metric is non-Euclidean.

Put another way, what makes a surface of constant $t$ non-Euclidean in g(1916) is that a 2-sphere of constant $R$ has surface area $4 \pi R^2$, but the proper distance between a 2-sphere at $R$ and a 2-sphere at $R + dR$ is *greater* than $dR$, whereas if the space were Euclidean, that proper distance would be equal to $dR$. For small $dR$, the proper distance between the 2-spheres is $dR / \sqrt{1 - r_s / R}$. That is, the ratio of the proper distance between two nearby 2-spheres to the difference in their areas is larger than it would be if the space were Euclidean.

as in g(current).

No; g(current) is also non-Euclidean, and for the same basic reason. The only difference is that in g(current), $r$ plays the role that $R$ plays in g(1916): in g(current), the proper distance between two nearby 2-spheres at $r$ and $r + dr$ is larger than $dr$, but the surface area of a 2-sphere at $r$ is $4 \pi r^2$.

 

[StateOfTheEqn]

The metric representations are:

g=\left(\begin{array}{cccc}<br /> 1-r_s/r&amp;0&amp;0&amp;0\\<br /> 0&amp;(1-r_s/r)^{-1}&amp;0&amp;0\\<br /> 0&amp;0&amp;r^2&amp;0\\<br /> 0&amp;0&amp;0&amp;r^2sin^2\theta<br /> \end{array}\right)

and
\overline g=\left(\begin{array}{cccc}<br /> 1-r_s/R&amp;0&amp;0&amp;0\\<br /> 0&amp;(1-r_s/R)^{-1}&amp;0&amp;0\\<br /> 0&amp;0&amp;R^2&amp;0\\<br /> 0&amp;0&amp;0&amp;R^2sin^2\theta<br /> \end{array}\right)

Let \overline{\textbf{x}} be a column vector w.r.t. the basis (dt,dR,d\theta,d\phi) and \textbf{x} be the same vector w.r.t the basis (dt,dr,d\theta,d\phi). Then \overline{\textbf{x}} = D\Phi \textbf{x}. The vectors are assumed to be carrying a Minkowski signature. For example,

\textbf{x}=\left(\begin{array}{c}<br /> dt\\<br /> idr\\<br /> id\theta\\<br /> id\phi<br /> \end{array}\right)

Let &lt;.,.&gt;_g be the metric w.r.t. g and &lt;.,.&gt;_{\overline{g}} be the metric w.r.t. \overline{g}.

Then &lt;\overline{\textbf{x}},\overline{\textbf{x}}&gt;_{\overline{g}}=(\overline{\textbf{x}})^T(\overline{g}) (\overline{\textbf{x}})=(D\Phi\textbf{x})^T(\overline{g})(D\Phi\textbf{x})=\textbf{x}^T(D\Phi)^T(\overline{g})(D\Phi)\textbf{x} = &lt;\textbf{x},\textbf{x}&gt;_{(D\Phi)^T(\overline{g})(D\Phi)}

But a simple calculation shows g \neq (D\Phi)^T(\overline{g})(D\Phi) unless r=R but they are unequal by assumption. Therefore g and \overline g do not represent the same metric.

I would like to clarify the above somewhat:

The metric(s) above take an element in T_p^*M to an element in T_p^*M \otimes T_p^*M. It is somewhat an abuse of terminology to call it a metric but it is common in GR. An actual metric would be g:T_pM \times T_pM \rightarrow \mathbb{R}. Re-doing the above calculation using the actual metric would be as follows:

Let \textbf{x}=\left(\begin{array}{c} <br /> x^0\\ <br /> ix^1\\ <br /> ix^2\\ <br /> ix^3 <br /> \end{array}\right)

and \overline{\textbf{x}}=\left(\begin{array}{c} <br /> \overline{x}^0\\ <br /> i\overline{x}^1\\ <br /> i\overline{x}^2\\ <br /> i\overline{x}^3 <br /> \end{array}\right)

Then &lt;\overline{\textbf{x}},\overline{\textbf{x}}&gt;_{\overline{g}}=(\overline{\textbf{x}})^T \overline{g} (\overline{\textbf{x}})=(D\Phi\textbf{x})^T\overline{g}(D\Phi\textbf{x} )=\textbf{x}^T(D\Phi)^T\overline{g}(D\Phi)\textbf{x}=&lt;\textbf{x}, \textbf{x}&gt;_{(D\Phi)^T\overline{g}(D\Phi)}

and the same result follows.

 

[Mentz114]

The two metric representations do not represent the same metric.

Let (t,R,\theta,\phi)=\Phi (t,r,\theta,\phi) = (t,(r^3 + r_s^3)^{1/3},\theta,\phi) where r_s is a constant > 0.

Then D\Phi=\left(\begin{array}{cccc}<br /> 1&amp;0&amp;0&amp;0\\<br /> 0&amp;\partial_r R&amp;0&amp;0\\<br /> 0&amp;0&amp;1&amp;0\\<br /> 0&amp;0&amp;0&amp;1<br /> \end{array}\right)

The metric representations are:

g=\left(\begin{array}{cccc}<br /> 1-r_s/r&amp;0&amp;0&amp;0\\<br /> 0&amp;(1-r_s/r)^{-1}&amp;0&amp;0\\<br /> 0&amp;0&amp;r^2&amp;0\\<br /> 0&amp;0&amp;0&amp;r^2sin^2\theta<br /> \end{array}\right)

and
\overline g=\left(\begin{array}{cccc}<br /> 1-r_s/R&amp;0&amp;0&amp;0\\<br /> 0&amp;(1-r_s/R)^{-1}&amp;0&amp;0\\<br /> 0&amp;0&amp;R^2&amp;0\\<br /> 0&amp;0&amp;0&amp;R^2sin^2\theta<br /> \end{array}\right)

..
..

It looks like you have done a transformation of coordinates incorrectly and then proved it was incorrect.

Starting with
$ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$
and making the transformation
$r\rightarrow {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}},\ \ dr= \frac{{R}^{2}\,dR}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}}$
the result is

$ds^2={dt}^{2}\,\left( \frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}-1\right) +\frac{{dR}^{2}\,{R}^{4}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{4}{3}}\,\left( 1-\frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}\right) }+{d\phi}^{2}\,{\sin\left( \theta\right) }^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}+{d\theta}^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}$

This is the correct metric and has Ricci tensor $R_{ab}=0$.

 

[StateOfTheEqn]

It looks like you have done a transformation of coordinates incorrectly and then proved it was incorrect.

Starting with
$ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$
and making the transformation
$r\rightarrow {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}},\ \ dr= \frac{{R}^{2}\,dR}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}}$
the result is

$ds^2={dt}^{2}\,\left( \frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}-1\right) +\frac{{dR}^{2}\,{R}^{4}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{4}{3}}\,\left( 1-\frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}\right) }+{d\phi}^{2}\,{\sin\left( \theta\right) }^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}+{d\theta}^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}$

This is the correct metric and has Ricci tensor $R_{ab}=0$.

You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$. It seems that what you have shown is that the original Schwarzschild metric, what I have been calling g(1916), is not compatible with the above metric even when it is written in terms of $R$. The metric g(1916) does not have an event horizon at $r=2m$ but your metric does, even when written in terms of $R$. At $r=2m$, g(1916) has the metric
$ds^2=(1-1/\sqrt[3]{2})dt^2-(1-1/\sqrt[3]{2})^{-1}dR^2-4(2)^{2/3}m(d\theta^2+sin^2\theta d\phi^2)$.
In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.

 

[Mentz114]

You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$. It seems that what you have shown is that the original Schwarzschild metric, what I have been calling g(1916), is not compatible with the above metric even when it is written in terms of $R$. The metric g(1916) does not have an event horizon at $r=2m$ but your metric does, even when written in terms of $R$. At $r=2m$, g(1916) has the metric
$ds^2=(1-1/\sqrt[3]{2})dt^2-(1-1/\sqrt[3]{2})^{-1}dR^2-4(2)^{2/3}m(d\theta^2+sin^2\theta d\phi^2)$.
In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.

I don't know what you're trying to say but

1) if the 1916 metric is derived as a coordinate transformation from the Schwarzschild metric then it has the form I showed and makes no predictions different from Schwarzschilds metric.

2) the metric you show is not the Schwarzschild metric and is not the spherically symmetric vacuum solution.

 

[Bill_K]

You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$.

It's called a coordinate transformation. "Writing it in terms of R" is the way you do a coordinate transformation. The 1916 paper never wrote the metric explicitly as Mentz114 does, rather left it implicitly in terms of the function R(r).

In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.

The event horizon is located at r_current = 2m, while in the 1916 coordinates it is located at r₁₉₁₆ = 0. The geometry is exactly the same.

 

[StateOfTheEqn]

There is only one unique spherically symmetrical solution to the vacuum field equations (see Birkhoff's theorem), and it's the one Schwarzschild described. His description differs only in appearance from how the Schwarzschild solution is commonly presented, due to the use of an unusual radial coordinate. Changing coordinates doesn't change any of the physical properties, metrical relations, or predictions of the solution.

The proof of the Birkhoff Uniqueness Theorem depends on defining $r$ as the 'area radius'. That is, $r=area(S^2)/4\pi$. However, a negatively curved space $\mathbb{R} \times S^2$ can have $area(S^2)/4\pi$ greater than the measured scalar radius $\rho$. In Schwarzschild's 1916 paper, $area(S^2)/4\pi=R=({\rho}^3+r_s^3)^{1/3}>\rho$. At present I am not sure what this says about uniqueness. In his paper Schwarzschild states:

Die Eindeutigkeit der Lösung hat sich durch die vorstehende Rechnung von selbst ergeben.(The uniqueness of the solution resulted spontaneously through the present calculation.) English translation at http://arxiv.org/pdf/physics/9905030v1

 

[StateOfTheEqn]

It's called a coordinate transformation. "Writing it in terms of R" is the way you do a coordinate transformation. The 1916 paper never wrote the metric explicitly as Mentz114 does, rather left it implicitly in terms of the function R(r).

His coordinate transformation is to an equivalent representation of his starting metric. But that misses the point. A coordinate transformation between equivalent representations of the same metric must be a local isometry. There is no local isometry transforming the representation which I have been calling g(current) to what I have been calling g(1916). His coordinate transformation is a local isometry (it seems) but it does not produce the g(1916) representation.

The event horizon is located at r_current = 2m, while in the 1916 coordinates it is located at r₁₉₁₆ = 0. The geometry is exactly the same.

There is a difference. There is zero volume inside a sphere with radius = 0 but there is non-zero volume inside a sphere with radius = 2m if m>0.

 

[StateOfTheEqn]

I don't know what you're trying to say but

1) if the 1916 metric is derived as a coordinate transformation from the Schwarzschild metric then it has the form I showed and makes no predictions different from Schwarzschilds metric.

2) the metric you show is not the Schwarzschild metric and is not the spherically symmetric vacuum solution.

Please read the original 1916 paper by Schwarzschild.

The original is at http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie and an English translation at http://arxiv.org/pdf/physics/9905030v1

The forward to the English translation states:

This fundamental memoir contains the ORIGINAL form of the solution of Schwarzschild’s problem.
It is regular in the whole space-time, with the only exception of the origin of the spatial co-ordinates; consequently,
it leaves no room for the science fiction of the black holes.

 

[PeterDonis]

There is zero volume inside a sphere with radius = 0 but there is non-zero volume inside a sphere with radius = 2m if m>0.

But "radius" is not necessarily the same as $r$, the coordinate that appears in equations. That depends on how $r$ is defined. In what you have been calling g(1916), if you compute the surface area of a sphere at $r = 0$ (which is the same as $R = r_s$), with $r$ defined the way Schwarzschild does it in his paper, then that surface area is *not* zero, and therefore the volume enclosed by that sphere is not zero either.

In other words, you can't just arbitrarily say that $r$ is the "radius"; you have to actually look at the metric to see what it says about the geometric meaning of $r$. If you look at the metric you are calling g(1916), it is obvious that $R$ is the areal radius, *not* $r$, because the area of a 2-sphere at $R$ is $4 \pi R^2$, but the area of a 2-sphere at $r$ is *not* $4 \pi r^2$.

Modern texts usually define what they call $r$ as the areal radius, but that just means that the modern $r$ plays the same role as Schwarzschild's $R$ did, as I said before. It does *not* mean that the $r$ in g(current) is the same as the $r$ in g(1916), even though they happen to be designated by the same lower-case letter.

 

[StateOfTheEqn]

Modern texts usually define what they call $r$ as the areal radius, but that just means that the modern $r$ plays the same role as Schwarzschild's $R$ did, as I said before. It does *not* mean that the $r$ in g(current) is the same as the $r$ in g(1916), even though they happen to be designated by the same lower-case letter.

I mention that in post #57 of this thread. I grant that if $r$ in g(current) is really equal to $R$ in g(1916) then they are equivalent representations of one metric. Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$ on a line just before eqn(6) in http://arxiv.org/pdf/physics/9905030v1. Furthermore his $R=(r^3+r_s^3)^{1/3}$ where $r_s=2m$. So, if we transform $r$ in g(current) to the $R$ of g(1916) we get equivalent metric representations. Then there is no event horizon for any $r=\sqrt{x^2+y^2+z^2} >0$.

 

[Bill_K]

Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$

OMG, I'm sure he does! But that does not mean that r is in any sense a "radius". Did you think that x, y and z were Cartesian coordinates?? They have no such meaning. It does not exclude the possibility that r = 0, and even r < 0 is possible.

 

[Mentz114]

Please read the original 1916 paper by Schwarzschild.

The original is at http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie and an English translation at http://arxiv.org/pdf/physics/9905030v1

OK, I will if I have the time. But there is only one spherically symmetric static vacuum solution, and it has an event horizon (singularity) in holonomic coordinates which is not present in some local frame bases. So what would I learn ?

 

[StateOfTheEqn]

OK, I will if I have the time. But there is only one spherically symmetric static vacuum solution, and it has an event horizon (singularity) in holonomic coordinates which is not present in some local frame bases. So what would I learn ?

You might learn that his original metric representation is not the same as the one currently used. See if you can find a local isometry that transforms the one into the other. A coordinate transformation between metric representations must be a local isometry if the metrics being represented are the same metric. If you can find such an isometry I would really like to know about it.

 

[PeterDonis]

Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$

Yes, he does. So what? That just raises the question of what $x$, $y$, and $z$ actually mean, geometrically. You can't assume that they are normal Euclidean Cartesian coordinates, any more than you can assume that $r$ is the actual radius. You have to look at the metric. Since the metric makes it clear that $r$ is not the radius, then that implies that $x$, $y$, and $z$ aren't standard Euclidean Cartesian coordinates either.

his $R=(r^3+r_s^3)^{1/3}$ where $r_s=2m$. So, if we transform $r$ in g(current) to the $R$ of g(1916) we get equivalent metric representations.

Yes.

Then there is no event horizon for any $r=\sqrt{x^2+y^2+z^2} >0$.

Only for Schwarzschild's definition of $r$. But the region $r > 0$, with Schwarzschild's definition of $r$, is not the entire spacetime. Schwarzschild appears to have assumed that it was, but he never proved it; and in fact that assumption is false. One way to see that it's false is to note, as I said before, that the area of the 2-sphere at $r = 0$ ($R = r_s$) is *not* zero.

With the modern definition of $r$, there is indeed an event horizon at $r = 2M$, which corresponds to $R = r_s$ in Schwarzschild's notation.

 

[StateOfTheEqn]

Only for Schwarzschild's definition of $r$. But the region $r > 0$, with Schwarzschild's definition of $r$, is not the entire spacetime. Schwarzschild appears to have assumed that it was, but he never proved it; and in fact that assumption is false. One way to see that it's false is to note, as I said before, that the area of the 2-sphere at $r = 0$ ($R = r_s$) is *not* zero.

Clearly $4\pi R^2/4\pi r^2 \rightarrow \infty$ as $r \rightarrow 0$ but that could be due to the negative spatial curvature growing without bound near the central singularity (at $r=0$).

 

[Bill_K]

A coordinate transformation between metric representations must be a local isometry if the metrics being represented are the same metric.

I don't know where on Earth you got this idea.

 

[PeterDonis]

Clearly $4\pi R^2/4\pi r^2 \rightarrow \infty$ as $r \rightarrow 0$

So what? What does this mean, physically?

but that could be due to the negative spatial curvature growing without bound near the central singularity (at $r=0$).

Negative spatial curvature of what? What is $r$ supposed to represent? You can't just wave your hands and say it's a "radius" of something. You have to actually look at the metric and compute things from it. Where in any such computation does $4 \pi r^2$ arise?

 

[StateOfTheEqn]

I don't know where on Earth you got this idea.

If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean? Bear in mind that what is losely called a metric in GR is only a metric representation (one for each coordinate system). There is a family of metric representations for each metric and they are pairwise transformable into each other by local isometries. Does that help?

 

[Mentz114]

If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean? Bear in mind that what is losely called a metric in GR is only a metric representation (one for each coordinate system). There is a family of metric representations for each metric and they are pairwise transformable into each other by local isometries. Does that help?

Coordinate transformation can change the components of the metric but all scalars formed by tensor contractions are invariant. If any of these invariants is different between two spacetimes, they are not the same spacetime.

For the two metrics I gave earlier, the second one gives the K-invariant
$\frac{16\,{m}^{2}\,{\left( 2\,m\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}-{R}^{3}+8\,{m}^{3}\right) }^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{10}{3}}\,{\left( {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}-2\,m\right) }^{2}}+\frac{32\,{m}^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{2}}$
If we substitute $(r^3-(2m)^3)^{1/3}$ for $R$ we get $48m^2/r^6$. So for any calculation, whether we start with $R$ or $r$ will give the same answer.

 

[Bill_K]

If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean?

They mean that the equations of physics are generally covariant. Yes, the components of the metric will change. g_μν is a tensor and is naturally expected to change under coordinate transformations. By your definition it would be unacceptable to transform, say, from rectangular coordinates to polar.

Generally, solutions of Einstein's equations have no isometries.

 

[StateOfTheEqn]

It seems to me the key to understanding the comparison between g(current) and g(1916) is in comparing the spatial curvature. In both g(current) and g(1916), R^2:=area(S^2)/4\pi so R is known as the area radius. Defining R this way, we can show the spherically symmetric solution is unique (Birkhoff Uniqueness Theorem). However, that conceals a difficulty. Solutions of the required form can have different spatial curvatures but be treated as equivalent in the proof of Birkhoffs theorem.

This can happen in the spherically symmetric space as follows: If r is the Euclidean distance we could have R&lt;r (positive spatial curvature), R=r (zero spatial curvature), or R&gt;r (negative spatial curvature). In g(1916), Schwarzschild derived a solution with negative spatial curvature where R=(r^3+r_s^3)^{1/3} and r_s=2GM.

What about a Black Hole event horizon? In g(1916), R=r_s only when the Euclidean distance r=0, that is, at the central singularity itself. In g(current), when R=r_s the value of r is left undefined because the spatial curvature (and therefore the relation of R to r) is left undefined. So, in g(current), we have no way of knowing the value of r when R=r_s. It could be zero, as in g(1916), which would imply no event horizon, except at the central singularity itself.

 

[StateOfTheEqn]

By your definition it would be unacceptable to transform, say, from rectangular coordinates to polar.

Incorrect. That transformation is a local isometry and therefore preserves the metric (but not the metric representation which is obviously different).

 

[StateOfTheEqn]

Generally, solutions of Einstein's equations have no isometries.

Both g(current) and g(1916) have rotational isometries in (at least) one of the angles.

 

[stevendaryl]

Incorrect. That transformation is a local isometry and therefore preserves the metric (but not the metric representation which is obviously different).

Whether a transformation preserves the metric depends on what the new metric is. If you have one patch describing using coordinates x^i and with metric tensor with components g_{ij}, and you transform to another patch described using coordinates x^\mu and metric tensor components g_{\mu \nu}, then it is metric-preserving if the metric components are related by:

g_{\mu \nu} = \frac{\partial x^i}{\partial x^\mu} \frac{\partial x^j}{\partial x^\nu} g_{ij}

 

[stevendaryl]

What about a Black Hole event horizon? In g(1916), R=r_s only when the Euclidean distance r=0, that is, at the central singularity itself.

In the 1916 coordinates, r=0 is not the central singularity, but is the event horizon. The region r \geq 0 in the original coordinates is the region r \geq r_s in the modern coordinates.

 

[StateOfTheEqn]

Coordinate transformation can change the components of the metric but all scalars formed by tensor contractions are invariant. If any of these invariants is different between two spacetimes, they are not the same spacetime.

For the two metrics I gave earlier, the second one gives the K-invariant
$\frac{16\,{m}^{2}\,{\left( 2\,m\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}-{R}^{3}+8\,{m}^{3}\right) }^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{10}{3}}\,{\left( {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}-2\,m\right) }^{2}}+\frac{32\,{m}^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{2}}$
If we substitute $(r^3-(2m)^3)^{1/3}$ for $R$ we get $48m^2/r^6$. So for any calculation, whether we start with $R$ or $r$ will give the same answer.

Did you mean substituting $(r^3+(2m)^3)^{1/3}$ for $R$ ? Then I get $48m^2/r^6$. This could mean we should understand the $r$ in g(current) to really be $R=(r^3+(2m)^3)^{1/3}$ where $r$ is the Euclidean radius with origin at the central singularity as in g(1916).

 

[PeterDonis]

In both g(current) and g(1916), R^2:=area(S^2)/4\pi so R is known as the area radius.

No, that's not correct as you state it, because g(current) does not use $R$ (upper case) at all. It uses $r$ (lower case) to denote the area radius, where g(1916) uses $R$ to denote the area radius. This is important because $r$ (lower case) also appears in g(1916), but it does *not* denote the area radius there. You keep calling it the "Euclidean distance", but that has no physical meaning. See below.

If r is the Euclidean distance we could have R&lt;r (positive spatial curvature), R=r (zero spatial curvature), or R&gt;r (negative spatial curvature).

Sure, if you have some way to actually measure this "Euclidean distance"--or example, if the manifold whose curvature you are measuring is embedded in some higher-dimension manifold that is Euclidean, and in which you can also measure distances. But in the case of a black hole, there is no physical measurement that corresponds to this "Euclidean distance".

Furthermore, in a case where you do have a higher-dimensional manifold in which you can measure $r$ directly, when $r = 0$ the corresponding 2-sphere must have zero area, even if the manifold whose curvature you are measuring has positive or negative spatial curvature. That's not true in g(1916); see below.

In g(1916), Schwarzschild derived a solution with negative spatial curvature where R=(r^3+r_s^3)^{1/3} and r_s=2GM.

Yes, but the $r$ in this solution has no physical meaning; you can throw it away and still describe all the physics just using $R$.

In g(1916), R=r_s only when the Euclidean distance r=0, that is, at the central singularity itself.

No, $r = 0$ is *not* the central singularity in g(1916); g(1916) does not even cover the portion of the manifold that contains the central singularity. As I have said several times, the 2-sphere at $r = 0$ in g(1916) does not have zero area; it has area $4 \pi r_s{}^2$. You continue to ignore this obvious fact, and it invalidates your interpretation of what $r$ in g(1916) means: it shows that $r = 0$ in g(1916) is not the central singularity; it's the event horizon, and $r$ is therefore *not* a "Euclidean distance"; it has no physical meaning at all.

In g(current), when R=r_s the value of r is left undefined because the spatial curvature (and therefore the relation of R to r) is left undefined.

No; once again, in g(current), $r$ (lower case) means what $R$ (upper case) means in g(1916). There is no "Euclidean distance" defined in g(current) because it's physically meaningless; there's no need for it. You can describe all the physics without defining it at all.

 

[StateOfTheEqn]

In the 1916 coordinates, r=0 is not the central singularity, but is the event horizon. The region r \geq 0 in the original coordinates is the region r \geq r_s in the modern coordinates.

I do not think this is true. In his original paper at http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie (English translation at http://arxiv.org/pdf/physics/9905030v1), Schwarzschild defines little $r$ to be $r=\sqrt{x^2+y^2+z^2}$ (the Euclidean distance from the singularity at the center) and derives $R$ in the metric representation as $R=(r^3+r_s^3)^{1/3}$.

 

[PeterDonis]

I do not think this is true.

As I keep on saying, compute the area of the 2-sphere at r = 0 in g(1916). What do you get? Why do you keep ignoring this obvious fact?

In his original paper at http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie (English translation at http://arxiv.org/pdf/physics/9905030v1), Schwarzschild defines little $r$ to be $r=\sqrt{x^2+y^2+z^2}$ (the Euclidean distance

So what? As I've also said before, that just raises the question of what $x$, $y$, and $z$ actually mean, physically. The answer is: nothing. There is no physical measurement you can make that corresponds to this "Euclidean distance". Schwarzschild just didn't realize that.

from the singularity at the center)

Schwarzschild couldn't possibly have defined $r$ this way, since he didn't even know there *was* a singularity at the center. And in modern terms, $r$ is most certainly *not* the distance (Euclidean or otherwise) from the singularity at the center; that concept has no meaning, because the singularity at the center is not a "place in space"; it's a "moment of time". (A "place in space" is described by a timelike line; but the singularity is a spacelike line, which is what describes a "moment of time".)

 

[StateOfTheEqn]

Coordinate transformation can change the components of the metric but all scalars formed by tensor contractions are invariant. If any of these invariants is different between two spacetimes, they are not the same spacetime.

For the two metrics I gave earlier, the second one gives the K-invariant
$\frac{16\,{m}^{2}\,{\left( 2\,m\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}-{R}^{3}+8\,{m}^{3}\right) }^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{10}{3}}\,{\left( {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}-2\,m\right) }^{2}}+\frac{32\,{m}^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{2}}$
If we substitute $(r^3-(2m)^3)^{1/3}$ for $R$ we get $48m^2/r^6$. So for any calculation, whether we start with $R$ or $r$ will give the same answer.

I have rethought this a bit. First, I think that substituting $(r^3+(2m)^3)^{1/3}$ for $R$ gives the desired result $48m^2/r^6$. But from the point of view of g(1916), ${R}^{3}-8\,{m}^{3}$ is just $r^3$, the Euclidean distance from the origin cubed. So your little $r$ is just the Euclidean distance from the origin. Is that what you intended?

 

[StateOfTheEqn]

As I keep on saying, compute the area of the 2-sphere at r = 0 in g(1916). What do you get? Why do you keep ignoring this obvious fact?

I think I have already answered this question. $4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty$ as $r \rightarrow 0$ and this implies the negative spatial curvature grows arbitrarily large near the central singularity.

 

[Nugatory]

I would be somewhat interested in hearing what StateOfTheEqn has to say about the Schwarzschild/Droste discussion that Russell E linked to way back in #15 of this thread.

 

[PeterDonis]

I think I have already answered this question.

No, you haven't. See below.

$4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty$ as $r \rightarrow 0$

This is correct mathematically, but it's meaningless physically; $r = 0$ and $R = r_s$ label the same physical 2-sphere, and its area is $4 \pi r_s{}^2$, not zero. The ratio $R^2 / r^2$ in g(1916) has no physical meaning, because $r$ in g(1916) has no physical meaning. If you think it does, what physical measurement does $r$ in g(1916) correspond to? (And no, the answer is not $x^2 + y^2 + z^2$, because, as I've said before, that just raises the question of what physical measurements $x$, $y$, and $z$ correspond to.)

and this implies the negative spatial curvature grows arbitrarily large near the central singularity.

It implies no such thing, because $r = 0$ in g(1916) is not the central singularity; if it were, the area of the 2-sphere at $r = 0$ would be zero, not $4 \pi r_s{}^2$. I've said all this before, and you continue to ignore it.

 

[TrickyDicky]

You guys seem to be going round in circles here, most of this was already settled way back in the discussion.

@StateOfTheEqn: a local isometry doesn't preserve the metric globally, only locally. That is as far as GR can go with its general covariance. GR only demands the existence of a smooth manifold which guarantees we can always add a pseudoRiemannian structure locally on that manifold, therefore allowing any coordinate transformation that preserves the metric locally. From this point of view it is a bit arbitrary to talk about solutions of the EFE as global geometries with its global topologies as GR is not really concerned with that as its solutions are local. Therefore in GR metrics as solutions of the EFE are always local geometries(local here meaning more or less local depending on the specific case, but certainly not referring to the global geometry understood as the whole pseudoriemannian manifold with its global topology, that is outside the reach of GR by its own structure and the nature of the EFE, this is evident for instance in cosmology. So given all this it is kind of meaningless to ask whether g(1916) and g(current) represent the same global spacetime. Mathematically, by looking at the form of the line element, we can only claim they can certainly represent the same local geometry (without adding any further mathematcal conditions than the EFE, the smooth 4-dimensional manifold M and the pseudoriemannian metric tensor g defined locally(since curvature in general is a local property). I think a not sufficiently emphasized issue in GR texts is that we are far from the deceiving simplicity of the geometries we usually deal with in classical differential geometry which are often of constant curvature(like the plane the sphere, etc) so that it is irrelevant the local/global isometry distinction we are considering here, or even far from the much more complex than these but still simpler due to its constant curvature Minkowskian space of SR.
But of course one can always add arbitrarily implicit mathematical conditions, like the highly non-trivial in singular spacetimes analyticity, so that we can have a a maximal analytical extension. Bu this only means that depending on our previous (arbitrary, remember, nothing to do with anything physical yet) mathematical choices we can think of a spacetime with a singularity surrounded by an event horizon or of a spacetime with just a naked singularity. It is a mathematical choice, not physical, so by itself it doesn't "predict" anything.
There are infinite mathematical models, we pick those that more closely represent the observations and then "a posteriori" decide those models predicted the observations. Oddly enough this is quickly forgotten and most people give a deep and almost magical meaning to the mathematical model, or the specific solution.

 

[StateOfTheEqn]

I would be somewhat interested in hearing what StateOfTheEqn has to say about the Schwarzschild/Droste discussion that Russell E linked to way back in #15 of this thread.

http://www.mathpages.com/rr/s8-07/8-07.htm is a good read and brings out the issues pretty clearly. In Schwarzschild's 1916 paper he used $R=(r^3+r_s^3)^{1/3}$ in the metric. The question that arose historically was (and is) where is $r=0$? If $r$ originates at the central singularity then $R=r_s$ when $r=0$ and the sphere (which is a collapsed sphere at a single point) has surface area $A=4\pi r_s^2$ which is an absurdity as has been pointed out by a previous poster. So, historically the conclusion that was drawn is that $r$ is to be measured from the surface of the Schwarzschild sphere at $r_s$. Then the area of the Schwarzschild sphere of radius $r_s$ would be $Area(S^2)=4\pi r_s^2=4\pi R^2$ which would make perfect sense since $R$ is the area radius. Furthermore $r$ could be both positive and negative. That is, $r=\pm\sqrt{x^2+y^2+z^2}$. Positive would of course be outside the Schwarzschild sphere at $r_s$ and negative would be inside.

There is another possibility which, in my view, arises from a more 'natural' reading of the 1916 paper. $r$ could indeed be measured from the central singularity but instead of the absurdity of a single point with non-zero surface area we have the limit $4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty$ as $r \rightarrow 0$. This would be the case if negative spatial curvature increased without bound near the central singularity. The implication would be that space-time is regular on the Schwarzschild sphere at $r=r_s$ and everywhere else where $r>0$.

 

[PeterDonis]

historically the conclusion that was drawn is that $r$ is to be measured from the surface of the Schwarzschild sphere at $r_s$.

But $r$ is not a measured physical distance; that's true regardless of what you think about "where" it is. In both g(1916) and g(current), radial distances are not directly "measured" by $r$ (lower case, with its different meanings in g(1916) and g(current)).

$r$ could indeed be measured from the central singularity

But this still won't be a physical distance that anyone can measure. In fact, it won't even be along a spacelike curve at all; the central singularity at $r = 0$ (with $r$ defined as in g(current)) is to the future of the horizon, not at some spatial distance from it.

instead of the absurdity of a single point with non-zero surface area

Strictly speaking, it isn't part of the manifold; the manifold just approaches it as a limit.

we have the limit $4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty$ as $r \rightarrow 0$.

And what is the physical meaning of this limit? I've posed this question before, and you still haven't answered it.

This would be the case if negative spatial curvature increased without bound near the central singularity.

Negative spatial curvature of what? I've posed this question before as well.

Plus, on this interpretation, $r = 0$ is not the central singularity; it's the horizon at $R = r_s$. You can't arbitrarily switch between the two definitions of $r$ (lower case).

 

[stevendaryl]

There is another possibility which, in my view, arises from a more 'natural' reading of the 1916 paper. $r$ could indeed be measured from the central singularity but instead of the absurdity of a single point with non-zero surface area we have the limit $4\pi R^2/4\pi r^2=R^2/r^2=(r^3+r_s^3)^{2/3}/r^2 \rightarrow \infty$ as $r \rightarrow 0$. This would be the case if negative spatial curvature increased without bound near the central singularity. The implication would be that space-time is regular on the Schwarzschild sphere at $r=r_s$ and everywhere else where $r>0$.

I don't understand what's the significance of that limit
4\pi R^2/4\pi r^2 \rightarrow \infty

You're saying that it has to do with unbounded negative spatial curvature?

 

[StateOfTheEqn]

I don't understand what's the significance of that limit
4\pi R^2/4\pi r^2 \rightarrow \infty

You're saying that it has to do with unbounded negative spatial curvature?

Yes. I'm currently working on a more detailed reply and I hope it won't take too long.

 

[stevendaryl]

Yes. I'm currently working on a more detailed reply and I hope it won't take too long.

In any case, if a region has a nonzero area, then it seems like, by definition, it is not a point.

 

[StateOfTheEqn]

I don't understand what's the significance of that limit
4\pi R^2/4\pi r^2 \rightarrow \infty

You're saying that it has to do with unbounded negative spatial curvature?

Consider the spatial manifold $\mathbb{R}^+\times S^2$. Suppose there are two metric on $\mathbb{R}^+\times S^2$, one Euclidean and the other non-Euclidean. Define $R=Area(S^2)/4\pi$ as the area radius for the non-Euclidean and $r=Area(S^2)/4\pi$ the area radius for the Euclidean. For the Euclidean, $r=\sqrt{x^2+y^2+z^2}$ in Cartesian coordinates centered at the origin. Now, consider how radial lines diverge. The distance between where two radial lines intersect the $\{R\}\times S^2$ sphere is $Rd\theta$ in the non-Euclidean metric and where two radial lines intersect the $\{r\}\times S^2$ sphere is $rd\theta$ in the Euclidean metric. Assume $\{R\}\times S^2$ is the same sphere in $\mathbb{R}^+\times S^2$ as $\{r\}\times S^2$ and $r \neq R$ . We can make the assumption $r \neq R$ because of the different metrics. Otherwise, if the metrics were the same then $r=R$.

If $Rd\theta>rd\theta$ we can say the non-Euclidean space is curved negatively because the radial lines diverge more than their Euclidean counterparts and if $Rd\theta<rd\theta$ we can say the non-Euclidean space is curved positively because the radial lines diverge less.

In the Schwarzschild paper of 1916 he defines $R=(r^3+r_s^3)^{1/3}$ where $r$ is the Euclidean distance from the origin. So, $R>r$ and the space will be negatively curved. The limit mentioned above is the limit in the ratio of areas $4\pi R^2/4\pi r^2=R^2/r^2$ which grows arbitrarily large as $r \rightarrow 0$ and which is a result of the negative curvature growing arbitrarily large near a mass concentrated (theoretically) at a single point.

If you accept that space can be negatively curved in this way by a gravitating body then you can get rid of the irregular sphere at $r_s$ called the event horizon of the Black Hole. Then all space-time around the gravitating body is regular except at $r=0$ which is also where $R=r_s$.

 

[atyy]

A word of caution about taking the 1916 paper as entirely correct http://www.staff.science.uu.nl/~hooft101/lectures/genrel_2010.pdf :

"In his original paper, using a slightly different notation, Karl Schwarzschild replaced (r³-(2M)³)^1/3 by a new coordinate r that vanishes at the horizon, since he insisted that what he saw as a singularity should be at the origin, claiming that only this way the solution becomes "eindeutig" (unique), so that you can calculate phenomena such as the perihelion movement (see Chapter 12) unambiguously. The substitution had to be of this form as he was using the equation that only holds if g = 1 . He did not know that one may choose the coordinates freely, nor that the singularity is not a true singularity at all. This was 1916. The fact that he was the first to get the analytic form, justifies the name Schwarzschild solution."

 

[PeterDonis]

Suppose there are two metric on $\mathbb{R}^+\times S^2$

This is mathematically fine but physically meaningless; physically there can only be one metric. Having two metrics would require the same physical measurements to yield two different results, which is impossible. So your proposal is not relevant to determining the actual physical structure of Schwarzschild spacetime, since the "non-Euclidean" metric is the one we actually physically observe. (For example, for the "Euclidean" metric to be physically relevant, $r$ would have to be the actual physical distance from the origin, but our actual physical measurements say it isn't.)

 

[TrickyDicky]

This is mathematically fine but physically meaningless; physically there can only be one metric. Having two metrics would require the same physical measurements to yield two different results, which is impossible. So your proposal is not relevant to determining the actual physical structure of Schwarzschild spacetime, since the "non-Euclidean" metric is the one we actually physically observe. (For example, for the "Euclidean" metric to be physically relevant, $r$ would have to be the actual physical distance from the origin, but our actual physical measurements say it isn't.)

Peter, I think you are misinterpreting StateOfTheEqn's point. First he is only referring to the spatial part of the spacetime and by saying that certain metrics could apply to it mathematically I don't think he is saying anything about "physically having two metrics". When for instance in the FRW case we consider three possible spatial metrics nobody thinks 3 physical measurements are to be yielded, which is absurd but that only one is eventually right.
IMO the argument StateOfTheEqn clearly is referring to an actual non-euclidean case.

Another plausible interpretation related to the above: let's recall that the way the metrics were represented back then was different to the current way, once again one can think of the way the early cosmological models of Einstein, de Sitter or Friedmann were written in the 1916-1922 period, they were usually obtained by an embedding in a higher dimensional manifold and then parametrizing and constraining it to the desired geometry thru an equation that was then differentiated. In all these cases it is possible to do this because foliation of the 3-hypersurfaces slices is allowed.
To use a trivial example of the way metrics were usually constructed back then thru embeddings in higher dimensional spaces, let's imagine we want to construct a line element for a 3-sphere(S3) by using its embedding in a 4-dimensional Euclidean space: $ds^2 =dx^2+dy^2+dz^2+dw^2$ . We would consider a hypersphere equation on that space as a hypersurface constraint:$x^2+y^2+z^2+w^2 =a^2$, with the paameter a being a euclidean radius in the embeeding 4-dimensional space but actually a radius of curvature in the non-euclidean three dimensional line element obtained just by differentiating, substituting for dw and transforming to spherical coordinates. We get $ds^2 = \frac{a^2}{ a^2−r^2} dr^2+r^2d\theta^2+r^2 sin^2\theta d\phi^2$
Here it is obvious that the case r=a doesn't mean there is a physical singularity there, it is just a sign of the way the line element was constructed by a constraint from a 4-dimensional embedding space to the three dimensional hypersurface.
In this vein the original Schwarzschild line element can be interpreted for its spatial part in a similar way, as valid only for values of R bigger than $r_s$ by construction, and considering $r_s$ as a sectional curvature radius of the spatial part of the manifold. I'm not saying it must be interpreted this way but that it is mathematically possible and nothing physical that I can think of right now goes againt it.

 

[PeterDonis]

First he is only referring to the spatial part of the spacetime

Yes, I understand that, but that doesn't change the fact that he is assuming that $r$ (lower case, using his definition) has a physical meaning that it doesn't actually have. In fact, what he's trying to do with his lower case $r$ basically amounts to re-inventing the Flamm paraboloid, but he's misinterpreting what it says.

I don't think he is saying anything about "physically having two metrics".

He is assigning a physical meaning to the ratio $R^2 / r^2$, which amounts to saying that $r$ has a physical meaning. Perhaps "two metrics" is not the way he would describe what that physical meaning is, but the fact remains that that ratio does *not* correspond to anything physical; see above.

the original Schwarzschild line element can be interpreted for its spatial part in a similar way, as valid only for values of R bigger than $r_s$ by construction, and considering $r_s$ as a sectional curvature radius of the spatial part of the manifold.

But the radius of curvature of the spatial part is *not* $r_s$ (except at the horizon--see below), nor is it $R^2 / r^2$. That's the point. There isn't even a single "radius of curvature of the spatial part" at all, since the curvature is different in the radial and tangential directions (see above); but in so far as we can define a "radius of curvature" in the radial direction, I think the best expression of it is $\sqrt{R^3 / r_s}$ (where I've used upper case $R$ again to make it clear that it's the area radius), which is the square root of the corresponding value of the Riemann curvature tensor. As you can see, at $R = r_s$, this radial radius of curvature is finite--in fact it is $r_s$. So $r_s$ can be thought of as the "radial radius of curvature at the horizon"--but of course that's not at all what StateOfTheEqn is claiming.

 

[TrickyDicky]

But the radius of curvature of the spatial part is *not* $r_s$ (except at the horizon--see below).That's the point. There isn't even a single "radius of curvature of the spatial part" at all, since the curvature is different in the radial and tangential directions (see above); but in so far as we can define a "radius of curvature" in the radial direction, I think the best expression of it is $\sqrt{R^3 / r_s}$ (where I've used upper case $R$ again to make it clear that it's the area radius), which is the square root of the corresponding value of the Riemann curvature tensor. As you can see, at $R = r_s$, this radial radius of curvature is finite--in fact it is $r_s$. So $r_s$ can be thought of as the "radial radius of curvature at the horizon".

Sure, I was referring about the curvature radius at the origin(let's not use the current terminology of horizons since we are talking about the original 1916 paper), the solution is asymptotically flat so the curvature radius will grow asymptotically as R goes to infinity and curvature goes to zero.

My point (as can be seen in the example I used) was that (and remember we are always referring here to the spatial hypersurface slice of the static spacetime) $r_s$ can only be seen as a distance, that is as a radius of a true sphere in the euclidean interpretation, but if we agree that we are dealing with a non-euclidean hypersurface(i.e. that we are using a non-euclidean metric in S2XR), it can only be interpreted as a curvature radius of the hypersurface at points closest to its singular origin, that in the weak field physical interpretation is proportional to the mass introduced as boundary condition. Therefore R can only be > $r_s$ by construction.

 

[PeterDonis]

Therefore R can only be > $r_s$ by construction.

Yes, all this is true of the Flamm paraboloid construction: it's only valid for $R > r_s$. But that's *not* the same as showing that the region $R > r_s$ is the entire spacetime. Schwarzschild (I think, based on what I've read) believed it was; Einstein apparently believed it was too. But the arguments that they thought proved that, did not actually prove it.

In so far as these arguments prove anything, they prove that the region $R > r_s$ is the entire *static* portion of the spacetime; but that's not the same as proving that the entire spacetime must be static. As far as I can tell, Schwarzschild and Einstein simply did not consider the possibility that the spacetime could have an additional region that was not static. (This appears to be a similar error on Einstein's part to the error that led him to miss predicting the expansion of the universe; he wanted a static solution to describe the universe and added the cosmological constant to the EFE to get one, rather than considering the possibility that the universe as a whole was not static.)

One reason, even within the Flamm paraboloid construction, to doubt that the region $R > r_s$ is the entire spacetime, is the fact, which I've mentioned repeatedly, that the physical area of 2-spheres at $R$ does not approach zero as a limit as $R \rightarrow r_s$; it approaches $4 \pi r_s^2$. The fact that $r = 0$ when $R = r_s$ does not change that, nor does it give $r$ a physical meaning; $r$ is just an abstract coordinate in an embedding diagram. I've repeatedly asked StateOfTheEqn to say how he would physically measure $r$ and have received no response.

 

[TrickyDicky]

Yes, all this is true of the Flamm paraboloid construction: it's only valid for $R > r_s$. But that's *not* the same as showing that the region $R > r_s$ is the entire spacetime. Schwarzschild (I think, based on what I've read) believed it was; Einstein apparently believed it was too. But the arguments that they thought proved that, did not actually prove it.

In so far as these arguments prove anything, they prove that the region $R > r_s$ is the entire *static* portion of the spacetime; but that's not the same as proving that the entire spacetime must be static. As far as I can tell, Schwarzschild and Einstein simply did not consider the possibility that the spacetime could have an additional region that was not static.

The Flamm's paraboloid representation of the spatial hypersurface have problems of its own as it tries to ilustrate in two dimensión what is very difficult or impossible to visualize in 3D(S2XR) and therefore misses some subtleties, but I guess you are just using it to refer to the spatial part of the spacetime so we probably agree about the math.
My aim wasn't to show "that the region $R > r_s$ is the entire spacetime" but only that it is a possible interpretation of the line element written by Schwarzschild if one ignores the current context and for instance do not consider the condition of analyticity for all the points of the manifold preventing the analytical extensión or considers the staticity requirement strictly and not replaceable by any orthogonal killing vector field.
You simply put more emphasis on different points from the ones I stress, like you seem more concerned about what such and such proved or didn't prove( always keeping an eye on the mainstream interpretation which is fine) while I prefer to look at the math in a more agnostic way.
Schwarzschild was certainly limited when he wrote the 1916 paper by the fact he knew the EFE only in its incomplete form( previous to Nov.25th 1915).

 

[StateOfTheEqn]

The error in the notion of Black Hole 'event horizons' at r=2GM has been exposed back in 1989. The error began with Hilbert. See the paper Black Holes:The Legacy of Hilbert's Error. See also Schwarzschild's original 1916 paper in English.

We summarize the result of the preceding sections as follows. The [Kruskal-Fronsdal] black hole is the result of a mathematically invalid assumption, explains nothing that is not equally well explained by [the Schwarzschild solution], cannot be generated by any known process, and is physically unreal. Clearly, it is time to relegate it to the same museum that holds the phlogiston theory of heat, the flat earth, and other will-o’-the-wisps of physics.

 

[stevendaryl]

The error in the notion of Black Hole 'event horizons' at r=2GM has been exposed back in 1989. The error began with Hilbert. See the paper Black Holes:The Legacy of Hilbert's Error. See also Schwarzschild's original 1916 paper in English.

I think that paper is wrong, or at best, misleading. The authors write:

Since each of these space-times assigns a different number to the limiting value of a radially approaching test particle’s locally measured acceleration, it is necessary to supplement the historical postulates by one that fixes this limit.

The Schwarzschild geometry is the unique (up to equivalence under coordinate transformations) spherically symmetric solution to the vacuum Einstein field equations. The acceleration of an infalling test particle is not an input to the Schwarzschild geometry, it's an output---it's computable from the Schwarzschild geometry.

Comparing the Kruskal extension to phlogiston and flat-earth is just trolling. No serious researcher would say something like that.

The significance of the Kruskal extension is not that it's a realistic model for the collapse of realistic stars. It's just another, interesting solution to the Einstein field equations. It helps in understanding a theory to have a bag of exact solutions (which are scarce for General Relativity).