Schwarzschild's Metric 1916

[TrickyDicky]

You are mistaken. It does not.

Sorry, wrote too fast.
Deleted

 

[TrickyDicky]

Before I got muddled up with the KVF switches at the event horizon(sorry about that), I was trying to achieve a better understanding of the Birkhoff's theorem and spherical symmetry in the Schwarzschild spacetime (this is not exactly the topic of the thread although highly related so I''ll try and keep it brief).
There is an element of confusion(at least for me) that I briefly mentioned in a previous post, namely the distinction between local isometries(local KVFs) versus global isometries(global KVFs) in spacetimes with somewhat complex 4D topologies like the Schwarzschild spacetime.
Certainly the foliation of spheres allows to have spherical symmetry(the 3 rotational KVFs) in this spacetime locally seemingly at any point. And since GR is only concerned with the local geometry I guess that is considered sufficient for practical matters. I admit I'm not at all certain that the spacetime has spherical symmetry as a global feature due to global geometrical-topological reasons but that is probably outside the scope of GR.

In any case, let's say the premise is fulfilled in what concerns to the Birkhoff's theorem, if the theorem states that spherical symmetry implies staticity (and asymptotic flatness), I don't know how to interpret the theorem in the regions where there is spherical symmetry but not staticity (inside the event horizon).

I'm still left with the unchallenged to this date claim(but I would like to hear the possible mathematical obstructions to this) that I can fabricate mathematically a curved Lorentzian manifold(unrelated to the current Schwarzschild spacetime) with a naked singularity at the origin to which I can assign as boundary condition a constant mass, that is a solution of the EFE in vacuum and has of course spherical symmetry around the origin and with a line element apparently similar to the one Schwarzschild found in his original paper.

 

[PeterDonis]

if the theorem states that spherical symmetry implies staticity

It doesn't. It only states that spherical symmetry + vacuum implies a 4th KVF whose integral curves are orthogonal to the integral curves of 3 KVFs arising from spherical symmetry. It does *not* say the 4th KVF has to be timelike. (That was part of the point of the blog post I mentioned--I think--in an earlier post in this thread: to do the proof of the theorem in a way that makes it absolutely obvious that the 4th KVF does not have to be timelike.)

 

[PAllen]

.. I was trying to achieve a better understanding of the Birkhoff's theorem and spherical symmetry in the Schwarzschild spacetime (this is not exactly the topic of the thread although highly related so I''ll try and keep it brief).
There is an element of confusion(at least for me) that I briefly mentioned in a previous post, namely the distinction between local isometries(local KVFs) versus global isometries(global KVFs) in spacetimes with somewhat complex 4D topologies like the Schwarzschild spacetime.
Certainly the foliation of spheres allows to have spherical symmetry(the 3 rotational KVFs) in this spacetime locally seemingly at any point. And since GR is only concerned with the local geometry I guess that is considered sufficient for practical matters. I admit I'm not at all certain that the spacetime has spherical symmetry as a global feature due to global geometrical-topological reasons but that is probably outside the scope of GR.

In any case, let's say the premise is fulfilled in what concerns to the Birkhoff's theorem, if the theorem states that spherical symmetry implies staticity (and asymptotic flatness), I don't know how to interpret the theorem in the regions where there is spherical symmetry but not staticity (inside the event horizon).

This has been explained before. The correct statement of the theorem is more like (from a paper I link later):

"Birkhoff's theorem shows that any spherically symmetric solution of the vacuum Einstein equations is locally
isometric to a neighborhood in Schwarzschild spacetime. Hence it is a local uniqueness theorem whose corollary is
that locally spherically symmetric solutions exhibit an additional local Killing vector field; however this fi eld is not
necessarily timelike."

It is only when the extra local killing field is timelike that you have staticity.

I'm still left with the unchallenged to this date claim(but I would like to hear the possible mathematical obstructions to this) that I can fabricate mathematically a curved Lorentzian manifold(unrelated to the current Schwarzschild spacetime) with a naked singularity at the origin to which I can assign as boundary condition a constant mass, that is a solution of the EFE in vacuum and has of course spherical symmetry around the origin and with a line element apparently similar to the one Schwarzschild found in his original paper.

If one admits unusual topologies, as described in the linked paper, you can sort of do this. You can think of these as sections of Kruskal, glued in various ways, or with various identifications (analogous to how you get an every where metrically flat torus by cutting and gluing the flat plane; or the non orientable Mobius strip or Klein bottle by similar operations). This must be done very carefully to ensure that the resulting manifold is everywhere spherically symmetric. However, you can get non-orientable manifolds consistent with Birkhoff by these techniques. At least one of their examples (see page 6) has singularities with no horizons (but is a closed universe). It is nowhere static, but is still consistent with Birkhoff - it has an extra translational killing vector field.

http://arxiv.org/abs/0910.5194

 

[TrickyDicky]

It doesn't. It only states that spherical symmetry + vacuum implies a 4th KVF whose integral curves are orthogonal to the integral curves of 3 KVFs arising from spherical symmetry. It does *not* say the 4th KVF has to be timelike. (That was part of the point of the blog post I mentioned--I think--in an earlier post in this thread: to do the proof of the theorem in a way that makes it absolutely obvious that the 4th KVF does not have to be timelike.)

Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.

 

[PAllen]

Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.

Actually, by mathematical criteria (not physical), the Kruskal extension is the most natural extension, while those in the paper I linked are contrived. In exactly the sense that if you have a flat open disc, the natural extension is the Euclidean plane, while extension to a torus or Klein bottle is more contrived. Also true is that just as with the open flat disc, you know that there opportunity for extension, otherwise you have removable geodesic incompleteness.

I don't think it is interesting to worry about Birkhoff's paper. Many generalizations and new proof methods have been done over the years. It is accepted terminology to lump them all together and call the Birkhoff (who was preceded by two years anyway, in deriving 'his' theorem).

Here is another reference:

http://arxiv.org/abs/0908.4110

Note the statement:

"Although Birkhoff's theorem is a classic result, many current textbooks and review articles on general relativity
no longer provide a proof or even a careful statement of the theorem. Frequently it is cited as proving that the
spherically symmetric vacuum solution is static. This is clearly not the case..."

 

[PeterDonis]

I would appreciate a reference to a published paper that shows the wording you use and that includes a proof.

As I noted in the blog post, MTW's discussion makes it clear that the KVF does not have to be timelike, and they clearly state the wording of the theorem without using the word "static" or implying staticity. So far, they're the only source I'm aware of that never slips up about this (except for the minor wart in the details of their proof, which is why I wrote the post).

Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.

I believe MTW's statement of the theorem is the "local" version, which basically says (my wording, not theirs) that any open neighborhood of a spacetime which is spherically symmetric and vacuum is isometric to an open neighborhood of Schwarzschild spacetime. That's not the same as the stronger claim that the entire spacetime must be the maximally extended Schwarzschild spacetime; I think you're correct that the latter claim requires specifying the global topology as well.

I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment).

I agree this is likely, but I'm not sure it's impossible: the Painleve chart was discovered in 1921 (by Painleve, and independently in 1922 by Gullstrand, according to Wikipedia), and AFAIK Painleve did include the extension of the chart to the future interior region (which was simple in his chart since it's nonsingular at the horizon).

Unfortunately I haven't been able to find the original paper by Birkhoff online.

Neither have I; I also haven't been able to find either Painleve's or Gullstrand's original papers online. It would be really interesting to see just what they were thinking at the time.

 

[PAllen]

Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.

From http://arxiv.org/abs/0908.4110, the following is a rigorous statement of the theorem with cosmological constant aspects removed:

Theorem 1. The only locally spherically symmetric solutions to the vacuum EFE are locally isometric to the Schwarzschild family of solutions. Furthermore, these solutions are
real analytic in each local coordinate chart.

Corollaries established are that the M parameter of each local coordinate chart must match. Also, it is clear from prior context that in the theorem statement, they mean Kruskal geometry, in that a local chart may be isometric to any part of Kruskal: interior, exterior, some of both, white hole, black hole. This statement also clarifies confusion over analyticity. Birkhoff does imply something about analyticity, but only for each chart, not globally. Thus (as shown especially in the other paper I linked) there are global solutions that cannot be geodesically completed further, that are analytic in each chart, but not globally analytic, and that have completely different topologies than Kruskal, including ones with naked singularities and also non-orientable topologies.

 

[stevendaryl]

I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment).

By the "timelike to spacelike" transition, I assume you mean the character of the r and t coordinates of the Schwarzschild solution at the event horizon? In that case, that transition is blatantly obvious from the Schwarzschild metric.

What's interesting about the Kruskal coordinates is that they DON'T have such a transition. You have one coordinate that is timelike everywhere, and another coordinate that is spacelike everywhere.

 

[TrickyDicky]

By the "timelike to spacelike" transition, I assume you mean the character of the r and t coordinates of the Schwarzschild solution at the event horizon? In that case, that transition is blatantly obvious from the Schwarzschild metric.

What's interesting about the Kruskal coordinates is that they DON'T have such a transition. You have one coordinate that is timelike everywhere, and another coordinate that is spacelike everywhere.

I was actually thinking about the local Killing vector fields that as you know are coordinate independent.

 

[TrickyDicky]

@PAllen: Thanks for the interesting references.

 

[StateOfTheEqn]

Can you give some specific references? In the "current" metric you give in your OP, spacelike slices of constant time are certainly not Euclidean; that's obvious just from looking at the line element you wrote down.

The specific reference is http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie .

The line element I wrote down is $d\sigma^2=R^2d\Omega^2=R^2(d\theta^2+sin^2\theta d\phi^2)$ where $R=(r^3+r_s^3)^{1/3}$. Indeed that is not Euclidean but neither is it a slice of constant time. Would that not have the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$

 

[PeterDonis]

The specific reference is http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie .

That's Schwarzschild's 1916 paper. I was asking about the modern derivations of g(current) that you referred to, which you said assumed space was Euclidean. I've never seen a modern derivation that makes that assumption, and indeed, as I said, it's obvious from the g(current) you wrote down in the OP of this thread that slices of constant time in that metric are not Euclidean.

The line element I wrote down is $d\sigma^2=R^2d\Omega^2=R^2(d\theta^2+sin^2\theta d\phi^2)$ where $R=(r^3+r_s^3)^{1/3}$.

That's not what you have been calling g(current), unless I'm completely misunderstanding your terminology; it's part of what you have been calling g(1916).

Indeed that is not Euclidean

As a metric on the 2-sphere, no, of course not, because the 2-sphere is not flat. But that is not what makes the metric of a 3-dimensional slice of constant time non-Euclidean. See below.

but neither is it a slice of constant time.

It's a 2-sphere which is a portion of a slice of constant time, in the g(1916) metric.

Would that not have the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$

That would be a complete slice of constant time in what you are calling g(1916), yes. And as a metric on a 3-dimensional space, it is also not Euclidean. But the reason it is not Euclidean is not that $R^2(d\theta^2+sin^2\theta d\phi^2)$ is not Euclidean; the reason is that the $dR^2$ term has a coefficient that depends on $R$, instead of just being 1. If the 3-metric were

$$
d\sigma^2 = dR^2 + R^2(d\theta^2+sin^2\theta d\phi^2)
$$

then that *would* be a Euclidean 3-metric; it would just be the standard Euclidean metric on 3-space expressed in spherical polar coordinates. But the factor of $\left( 1 - r_s / R \right)^{-1}$ in front of the $dR^2$ makes the metric, as a metric on 3-space, non-Euclidean.

 

[StateOfTheEqn]

Just to be clear, my interest is in the event horizon. I accept that there is a real physical singularity at r=0. In g(current) there is at the event horizon what I and perhaps some others call a coordinate singularity given that time slows down and, from the point of view of an external observer, seems to stop. With respect to the metric g(1916) that does not happen. I do not know whether g(current) and g(1916) reopresent the same or different metrics. I strongly suspect the latter is true.

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 $$D\Phi=\left(\begin{array}{cccc} 1&0&0&0\\ 0&\partial_r R&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)$$

The metric representations are:

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

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

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,

$$\textbf{x}=\left(\begin{array}{c} dt\\ idr\\ id\theta\\ id\phi \end{array}\right)$$

Let $<.,.>_g$ be the metric w.r.t. $g$ and $<.,.>_{\overline{g}}$ be the metric w.r.t. $\overline{g}$.

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T(\overline{g}) (\overline{\textbf{x}})=(D\Phi\textbf{x})^T(\overline{g})(D\Phi\textbf{x})=\textbf{x}^T(D\Phi)^T(\overline{g})(D\Phi)\textbf{x} = <\textbf{x},\textbf{x}>_{(D\Phi)^T(\overline{g})(D\Phi)}$$

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.

Calculating $r_s$:

In his 1916 paper Schwarzschild derives the formula ${\dot \theta}^2=r_s/2(r^3+r_s^3)=r_s/2R^3$. The equation $2{\dot \theta}^2R^3=r_s$ imposes a constraint on $\dot \theta$ given $R$ (and implicitly $r$) similar to Kepler's 3rd law: $T^2=Kr^3$. For a circular orbit, $K=4\pi^2/GM$

We can rewrite Kepler's 3rd law as $(2\pi/\dot\theta)^2=Kr^3$ and then $8\pi^2/K=2\dot\theta^2r^3$

So, we have
$$\begin{eqnarray} 2{\dot \theta}^2r^3&=&8\pi^2/K\\ 2{\dot \theta}^2R^3&=&r_s \end{eqnarray}$$
Since $r_s$ is a constant we have $R/r \rightarrow 1$ as $r \rightarrow \infty$. Then $r_sK/8\pi^2=1$
and so $r_s=8\pi^2/K=2GM$

No event horizon at $r_s$:

Setting $r=r_s$ we have:

$R=(r^3+r_s^3)^{1/3}=(2r_s^3)^{1/3}=\sqrt[3]{2}r_s$ and
$ds^2=(1-1/\sqrt[3]{2})dt^2-(1-1/\sqrt[3]{2})^{-1}dR^2-R^2d\Omega^2$ with $(1-1/\sqrt[3]{2}) >0$

In fact, $(1-r_s/R)>0$ for all $r>0$

 

[StateOfTheEqn]

But the reason it is not Euclidean is not that $R^2(d\theta^2+sin^2\theta d\phi^2)$ is not Euclidean; the reason is that the $dR^2$ term has a coefficient that depends on $R$, instead of just being 1. If the 3-metric were

$$
d\sigma^2 = dR^2 + R^2(d\theta^2+sin^2\theta d\phi^2)
$$

then that *would* be a Euclidean 3-metric; it would just be the standard Euclidean metric on 3-space expressed in spherical polar coordinates. But the factor of $\left( 1 - r_s / R \right)^{-1}$ in front of the $dR^2$ makes the metric, as a metric on 3-space, non-Euclidean.

Yes, you are right. It is the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$ that makes $\mathbb{R}^+ \times S^2$ non-Euclidean, not the previous one. But within that metric a surface of constant $R$ would have surface area $4\pi R^2>4\pi r^2$ would it not? That would indicate $\mathbb{R}^+ \times S^2$ has negative curvature and not be Euclidean as in g(current).

 

[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:

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

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

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,

$$\textbf{x}=\left(\begin{array}{c} dt\\ idr\\ id\theta\\ id\phi \end{array}\right)$$

Let $<.,.>_g$ be the metric w.r.t. $g$ and $<.,.>_{\overline{g}}$ be the metric w.r.t. $\overline{g}$.

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T(\overline{g}) (\overline{\textbf{x}})=(D\Phi\textbf{x})^T(\overline{g})(D\Phi\textbf{x})=\textbf{x}^T(D\Phi)^T(\overline{g})(D\Phi)\textbf{x} = <\textbf{x},\textbf{x}>_{(D\Phi)^T(\overline{g})(D\Phi)}$$

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 $\textbf{x}=\left(\begin{array}{c} x^0\\ ix^1\\ ix^2\\ ix^3 \end{array}\right)$

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

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T \overline{g} (\overline{\textbf{x}})=(D\Phi\textbf{x})^T\overline{g}(D\Phi\textbf{x} )=\textbf{x}^T(D\Phi)^T\overline{g}(D\Phi)\textbf{x}=<\textbf{x}, \textbf{x}>_{(D\Phi)^T\overline{g}(D\Phi)}$$

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 $$D\Phi=\left(\begin{array}{cccc} 1&0&0&0\\ 0&\partial_r R&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)$$

The metric representations are:

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

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

..
..

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.