How Do Contradictory Metrics Align with GR Field Equations?

[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.