Solutions of the nonlinear Field Equations

[notknowing]

Einsteins field equations are nonlinear but I guess that nobody has already found solutions to the full nonlinear equations (because of course it is very hard to do so). Nevertheless, such solutions could (I think) hold a number of surprises. Instead of linear gravitational waves, one could think of soliton like waves with strange properties. Solutions near spinning black holes could also be siginificantly modified by taking into account nonlinear terms. Finally, even cosmological models could have to be modified because of these nonlinearities. Maybe that there is no need then to invoke dark energy and other strange stuff.
What do you think ?

 

[George Jones]

The Friedmann-Robertson-Walker and de Sitter cosmological solutions, as well as the Schwarzschild, Reissner-Nordstron, Kerr and Kerr-Newman black hole solution are all solutions of the full non-linear Einstein equations.

 

[notknowing]

The Friedmann-Robertson-Walker and de Sitter cosmological solutions, as well as the Schwarzschild, Reissner-Nordstron, Kerr and Kerr-Newman black hole solution are all solutions of the full non-linear Einstein equations.

This shows indeed my ignorance on this subject.
But what about nonlinear gravitational waves ?
Or could there maybe exist stable structures in spacetime in which the curvature is itself the source of the curvature ?

 

[George Jones]

But what about nonlinear gravitational waves?

A project underway, LIGO, hopes to detect gravitational radiation in the next decade or so. In order to be detectable, the gravitational radiation will have to be produced by processes in which there are strong gravitational interactions, like collisions that involve black holes and neutron stars. In analyzing such strong interactions, non-linearity must be included. This is done numerically on computers.

Off the top of my head, don't know if any exact soliton-like solutions erxist, and I only have one GR book (Hartle) with me right now.

Or could there maybe exist stable structures in spacetime in which the curvature is itself the source of the curvature ?

I'm not sure what you mean here. What is a stable structure?

 

[notknowing]

I'm not sure what you mean here. What is a stable structure?

Take the field equations in vacuum, with no mass whatsoever. Then, try to find a stationary solution of the nonlinear field equations. One could image some kind of peculiar curvature of spacetime (for instance with spherical symmetry) in which this curvature is in fact the source of its own curvature. This could for instance look a bit like the space-curvature around a mass, but then without the presence of the mass. Maybe a far-fetched fantasy but can it really be excluded that such a thing exists ?

 

[Chris Hillman]

Exact solutions to the fully nonlinear EFE

Hi again, notknowing,

Einsteins field equations are nonlinear but I guess that nobody has already found solutions to the full nonlinear equations (because of course it is very hard to do so).

Waay back in 1916, Einstein did say something gloomy to the effect that he feared exact solutions would never be found, but within a month or so Schwarzschild produced a very simple exact solution of fundamental importance (the static spherically symmetric vacuum solution). Levi-Civita soon found dozens of new exact one or two parameter solutions, and Kottler and Reissner and Nordstrom provided important generalizations of the Schwarzschild vacuum.

Next, in 1918 Weyl found ALL the axisymmetric static vacuum solutions of the EFE (they turn out to correspond to axisymmetric harmonic functions), and over the next few decades many more solutions were found, including many important cosmological models such as the FRW dusts and Kasner dusts, as well as the pp waves discovered by Brinkmann in 1925 and an important special case, the gravitational plane waves found by Baldwin and Jeffery in 1926.

All these were found using some symmetry Ansatz. Then in 1963, Kerr made a sensational discovery, the Kerr vacuum, and within a decade after that zillions of new solutions were found, including the Ernst family of all stationary axisymmetric electrovacuum solutions. Even better, this work, and new developments in the field nonlinear PDEs generally, inspired the introduction of fundamental new techniques for finding exact solutions in gtr, as well as techniques for drawing reliable inferences are the solution space itself.

A comparatively recent and rather spectacular achievement, which may turn out to be almost as important as the Kerr vacuum, is the discovery of an exact rigidly rotating disk of dust found by Neugebauer and Meinel in 1993. (The exterior region of) this solution belongs to the Ernst vacuum family, but their achievement was to single out this particular solution from that rather large family. This has since been generalized in various directions.

(Things called "counter-rotating disks of dust" range from the profoundly unrealistic to the unphysical, and are not very interesting compared to the true rotating disk solutions.)

Nevertheless, such solutions could (I think) hold a number of surprises. Instead of linear gravitational waves, one could think of soliton like waves with strange properties.

Aha! One of the developments in PDEs I was thinking of was the introduction of the inverse scattering transform for solving the famous KdV equation, and which was soon adapted to attack the EFE. See Belinsky and Verdaguer, Gravitational Solitons. (I should caution you that most "gravitational solitons" are not true solitons and do not exhibit the properties you would expect from n-soliton solutions of the KdV equation. However, current research may yet uncover true solitons lurking in solutions related to the well-known Robinson-Trautman null dusts.)

But what about nonlinear gravitational waves ?

To elaborate on what George Jones said, the Brinkmann pp-waves form a large class of exact null dust or vacuum solutions of the EFE, which model exact gravitational waves, possibly accompanied by other massless radiation such as EM radiation. The Baldwin-Jeffery plane waves are a large subclass of pp-waves, which include both the curved space generalization of purely electromagnetic plane waves (with not accompanying gravitational radiation) and their gravitational analogs.

In particular, exact null electrovacuum solutions modeling "sinusoidal" EM plane waves with linear or circular polarization are known and can be written in closed form. (See the version of "Monochromatic electromagnetic plane wave" listed at http://en.wikipedia.org/wiki/User:Hillman/Archive; I can't vouch for more recent versions, however!) Similarly, exact vacuum solutions (Petrov type N, to use the jargon) modeling "sinusoidal" gravitational plane waves with linear or circular polarizations are known and can be written in closed form (using standard special functions, to be sure, the "Mathieu cosine functions"). Comparing these solutions is very instructive, particularly for anyone who insists that gtr "unifies" electromagnetism and gravitation! (Gtr is a fine theory which has attained impressive accomplishments, but unification is not one of them.)

Or could there maybe exist stable structures in spacetime in which the curvature is itself the source of the curvature ?

You might be groping toward something like Wheeler's notion of a "geon".

Off the top of my head, don't know if any exact soliton-like solutions exist, and I only have one GR book (Hartle) with me right now.

Some interesting colliding plane wave (CPW) solutions (and some other types of solutions) have been found using the inverse scattering method. The monograph by Belinsky and Verdaguer provides an exposition of this method and gives many detailed examples.

Take the field equations in vacuum, with no mass whatsoever. Then, try to find a stationary solution of the nonlinear field equations. One could image some kind of peculiar curvature of spacetime (for instance with spherical symmetry) in which this curvature is in fact the source of its own curvature. This could for instance look a bit like the space-curvature around a mass, but then without the presence of the mass. Maybe a far-fetched fantasy but can it really be excluded that such a thing exists ?

Definitely sounds like you should read about geons. Unfortunately, I am not sure that I can recommend readable sources. The Wikipedia article "geons" in the version listed at http://en.wikipedia.org/wiki/User:Hillman/Archive is probably the best I can offer right now, but I am familiar with its limitations because I wrote it, in the spirit of "something being better than nothing". More recent versions might be much better or much, much worse--- that's just the nature of Wikipedia, so always be very wary of what you read there.

Chris Hillman

 

[Jheriko]

I would imagine the most general solutions of the EFE would be numerical, although, numerical solvers have limited accuracy and require computer programming skill as well as the maths.

Still it is probably possible to write a program which would tend towards an exact solution if allowed infinite running time and infinite precision arithmetic.

 

[Chris Hillman]

Hi, jheriko,

Surprisingly enough, the number of papers presenting new exact solutions easily outnumbers the number of papers presenting numerical studies, although the latter group is by no means small.

Chris Hillman

 

[cesiumfrog]

Take the field equations in vacuum, with no mass whatsoever. Then, try to find a stationary solution of the nonlinear field equations. [...] a bit like the space-curvature around a mass, but then without the presence of the mass.

This makes me think of the Schwarzschild solution, where there is in no place anything but vacuum.

 

[notknowing]

This makes me think of the Schwarzschild solution, where there is in no place anything but vacuum.

This is not really true. To obtain the Schwarzschild solution you really need a mass (as source of curvature). The Schwarzschild solution is than the solution of the Einstein Field Equations in the region where there is no mass present (in contrast to the region inside the mass).

 

[cesiumfrog]

In exactly what sense do you need a mass?

 

[notknowing]

In exactly what sense do you need a mass?

The Schwarzschild solution is the most symmetric solution (spherically symmetric) outside a spherically symmetric source of curvature(thus a mass). Some input of the experts would be nice here. When you see the derivation of the Schwarzschild solution, one usually does not start with the full Einstein field equations with an energy-stress tensor on the right side. Instead, one uses arguments based on symmetry to arrive at a solution which leads in the weak field limit to geodesics corresponding to Newtonian motion. So, in this last step the requirement for a mass clearly comes in.

 

[George Jones]

An eternal Schwarzschild black hole is a vacuum solution to Einstein's equation, i.e., one for which the stress-energy tensor is zero everywhere. I think this is what cesiumfrog means.

This motivated Chandrasekhar to write the following in the prologue to his book The Mathematical Theory of Black Holes.

The black holes of nature are the most perfect mathematical objects there are in the universe: the only elements in their construction are our concepts of space and time.

(These are not a spacetimes for stellar collapse.)

However, these black hole soultions all have singularities - colloquially, they all have edges, rips, or tears in their spacetimes. These black hole singularites are, again colloquially, regions that have excised from spacetime in which one would like to place infinitely dense matter.

However, when considered less colloquially and more rigorously, the actual spacetimes don't contain any matter.

These black hole solutions are not soliton-like. So, notknowing, I think you're looking for solutions that don't have (strong) curvature singularities, and this rules out black hole solutions.

 

[notknowing]

An eternal Schwarzschild black hole is a vacuum solution to Einstein's equation, i.e., one for which the stress-energy tensor is zero everywhere. I think this is what cesiumfrog means.

This motivated Chandrasekhar to write the following in the prologue to his book The Mathematical Theory of Black Holes.



(These are not a spacetimes for stellar collapse.)

However, these black hole soultions all have singularities - colloquially, they all have edges, rips, or tears in their spacetimes. These black hole singularites are, again colloquially, regions that have excised from spacetime in which one would like to place infinitely dense matter.

However, when considered less colloquially and more rigorously, the actual spacetimes don't contain any matter.

These black hole solutions are not soliton-like. So, notknowing, I think you're looking for solutions that don't have (strong) curvature singularities, and this rules out black hole solutions.

Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature ?? I would tend to think that the stress-energy tensor becomes singular or so, but not zero. And what does "M" then mean in the Schwarzschild solution ? Is this the mass of an object with zero energy-stress tensor ?

 

[George Jones]

Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature ?? I would tend to think that the stress-energy tensor becomes singular or so, but not zero.

For an eternal Schwarzschild black hole, components of the curvatiure tensor diverge as the the singularity is approached, but the stress-energy tensor is congruently zero everywhere.

And what does "M" then mean in the Schwarzschild solution ? Is this the mass of an object with zero energy-stress tensor ?

Yes.

Carroll has a good discusion of this, and maybe pervect and/or Chris will say more.

cesiumfrog was correct when he said

This makes me think of the Schwarzschild solution, where there is in no place anything but vacuum.

You are also right when you say that M is the mass of Schwarzschild black hole.

 

[robphy]

Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature ??

The Weyl tensor.

 

[notknowing]

For an eternal Schwarzschild black hole, components of the curvatiure tensor diverge as the the singularity is approached, but the stress-energy tensor is congruently zero everywhere.



Yes.

Carroll has a good discusion of this, and maybe pervect and/or Chris will say more.

cesiumfrog was correct when he said



You are also right when you say that M is the mass of Schwarzschild black hole.

Thanks for this clarification. So it seems that I have learned something and that cesiumfrog was right. I was mislead by the popular view that in a black hole a point of infinite density is obtained. So, I recapitulate things to see if I understand it correctly. First you need a sufficient amount of mass to create a black hole in the first place ("feed" it). Then, through some magic, the matter is crunched together in such a way that it is transformed into pure gravitational energy such that "the curvature itself becomes the source of curvature" (the idea which I had originally to create a kind of "geon"). Of course, in a quantum theory of gravitation, things would probably turn out to be very different.

But what then about a charged black hole ? If the stress-energy tensor is also congruently zero everywhere in this case, such that one can not speek of mass in the usual sense, where is then the charge located ? It can probably not be considered to be "smeared out" in the region inside the horizon ?

 

[George Jones]

First you need a sufficient amount of mass to create a black hole in the first place ("feed" it). Then, through some magic, the matter is crunched together in such a way that it is transformed into pure gravitational energy such that "the curvature itself becomes the source of curvature" (the idea which I had originally to create a kind of "geon").

I wouldn't put it this way.

For a black hole that forms from stellar collapse, there is a region of spacetime in which the stress-energy tensor is non-zero. See the diagrams on pages 28 and 36 of Chapter 7 of http://preposterousuniverse.com/grnotes/".

In posts #13 and #15, I was careful, and I said that eternal Scwharzschild black holes are vacuum solutions, and that eternal black holes are not the black holes which result from stellar collapse.

Also, for eternal black holes, I don't think I'd say that curvature is the source of curvature. I prefer to talk in the imprecise (and possibly wrong) language that I used in post #13.

But what then about a charged black hole ? If the stress-energy tensor is also congruently zero everywhere in this case, such that one can not speek of mass in the usual sense, where is then the charge located ? It can probably not be considered to be "smeared out" in the region inside the horizon ?

In this case of a charged, eternal black hole, the stress-energy tensor is not zero - it has the usual form for an electromagetic field. In fact, E_r = Q/r^2, so, again speaking very loosely, it look like the field for charge "residing" at the singularity r = 0. Again, r = 0 is not actually part of the spacetime manifold. Although the black hole has a charge Q, no actual "physical charges" exist anywhere in the spacetime manifold, either inside or outside the horizons.

 

[swimmingtoday]

<<Take the field equations in vacuum, with no mass whatsoever. Then, try to find a stationary solution of the nonlinear field equations. One could image some kind of peculiar curvature of spacetime (for instance with spherical symmetry) in which this curvature is in fact the source of its own curvature. >>

You seem to be using "curvature" interchangably with "non-linear terms". That is not correct.

In GR, a curvature quantity is set equal to a quantity which involves the mass-density. The "curvature quantity" is not exactly the curvature--we will get to that later--, and the mass-density quantity is a tensor with 16 components, the most prominent one under normal conditions being the mass density.

The curavature quantity contains a non-linear part, which is small under normal conditions (e.g. conditions found on Earth and in the Solar System) So the equations of GR are of a form

mass density tensor = linear portion of curvature quantity plus non-linear portion of curvature quantity.

This is numerically equivalent to

Linear portion of curvature quantity= mass density tensor + non-limear terms from curvature quantity.

In this ladder form, numerically, the behavior is as if both the mass density and a non-linear thing are drivers of the gravitational field. In a sense the non-linear terms are like the energy (actually "stress-energy" would be more accurate, but don't worry about that) of the gravitational field; and the source of gravitation is thus the mass of the particle plus the mass equaivalent (via E = m (c squared)) of the field.

But the curvature quantity is not acting as a source of curvature, rather it is the non-linear terms in the curvarture quantity that are acting like a source for the linear portion of the curvature quantity.

As for the other part of what yoiu were saying, you can indeed have curvature without mass. To see this you need to understand some of the geometry of GR. There is a quantity known as the curvarture tensor. It is a 4th order tensor--it is of the form Rabcd, where a,b,c,and d are subscripts. From it you can construct another tensor known as the contraction of the curvature tensor. It is formed from the curvature tensor as follows. Rac= Ra1c1 + Ra2c2 + Ra3c3 + Ra4c4. (Some of the plus signs are minus signs, but I'm trying to keep this simple). In a region where the mass density tensor is zero, the contracted Rieman tensor is zero. This comes very directly from the equations of GR. But it is quite possible to chave the contraction of a tensor be zero without the original tensor being zero.

 

[swimmingtoday]

<<Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature >>

The stress-energy is not proportional to the curvation. It is proportional to a quantity involving the contraction of the curvature (and the contraction of the contraction). The lack of stress-energy at a point implies the CONTRACTED curvature tensor is zero. It does not imply that the curvature tensor is zero.

Indeed, in the vacuum region of the Schwarzschild solution the curvature tensor is non-zero. In general, it turns out that a region with tidal forces (e.g, the derivative in the x direction of the x component of the gravitational force) will have a non-vanishing curvature tensor. It's doubtful you can have curvature in a spot without a tidal force, but quite possible to have curvature at a point with no matter.

 

[notknowing]

<<Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature >>

The stress-energy is not proportional to the curvation. It is proportional to a quantity involving the contraction of the curvature (and the contraction of the contraction). The lack of stress-energy at a point implies the CONTRACTED curvature tensor is zero. It does not imply that the curvature tensor is zero.

Indeed, in the vacuum region of the Schwarzschild solution the curvature tensor is non-zero. In general, it turns out that a region with tidal forces (e.g, the derivative in the x direction of the x component of the gravitational force) will have a non-vanishing curvature tensor. It's doubtful you can have curvature in a spot without a tidal force, but quite possible to have curvature at a point with no matter.

Thank you swimmingtoday for this explanation. I'll need some time to digest it though ...

 

[pervect]

Einstein's equation says:

G_uv = 8 pi T_uv, which can also be written as

R_uv = 8 pi T_uv - 4 pi T g_ab

which gives the Ricci tensor R_uv directly as an equation from the stress-energy tensor T_uv.

The important thing to note is that R_uv = 0 does not imply no curvature. The general curvature tensor is the Riemann curvature tensor, R_abcd, not the Ricci tensor, which is a contraction of the Riemann.

The significance of this is that the Riemann curvature can be non-zero even when the Ricci tensor is zero. To expand a bit on robphy's rermarks, the Riemann curvature tensor can be decomposed into the Ricci tensor, and the Weyl tensor. If you have a zero stress-energy tensor, the Ricci will be zero, but the Weyl does not have to be zero, which implies that the Riemann tensor also may be nonzero.

 

[swimmingtoday]

<<If the stress-energy tensor is zero everywhere, what is then causing the curvature ??...The Weyl tensor>>

I doubt this is true. Suppose that at some point, R1010= -(R2020 + R3030), with all three of these curvature tensor quantities non-zero. And suppose that the Weyl tensor is zero. Hopefully these conditions are self-consistant. If they are, then in this case, R00 will be zero, yet we have 3 components of the Riemann tensor which are non-zero, and so the Riemann Tensor does not vanish, depite both the Ricci Tensor and the Weyl tensor vanishing. (BTW, I suspect that the hypothetical situation I constructed is the actual situation for the vacuum region of the Schwarzschild Solution.)

 

[robphy]

<<If the stress-energy tensor is zero everywhere, what is then causing the curvature ??...The Weyl tensor>>

I doubt this is true. Suppose that at some point, R1010= -(R2020 + R3030), with all three of these curvature tensor quantities non-zero. And suppose that the Weyl tensor is zero. Hopefully these conditions are self-consistant. If they are, then in this case, R00 will be zero, yet we have 3 components of the Riemann tensor which are non-zero, and so the Riemann Tensor does not vanish, depite both the Ricci Tensor and the Weyl tensor vanishing. (BTW, I suspect that the hypothetical situation I constructed is the actual situation for the vacuum region of the Schwarzschild Solution.)

For concreteness, here is the Weyl tensor.
<br /> C_{abc}{}^d=<br /> R_{abc}{}^d <br /> - \frac{2!}{n-2}g^{de}\left(\left(g_{a[c}R_{e]b}- <br /> g_{b[c}R_{e]a}\right) <br /> - \frac{1}{n-1}Rg_{a[c}g_{e]b}\right) <br />
Weyl is the traceless part of the Riemann tensor.

Does your proposed Riemann tensor (with its claimed vanishing Weyl and Ricci [and therefore scalar-curvature] satisfy this?

(Does your proposed Riemann tensor satisfy the Bianchi identities?)

 

[swimmingtoday]

For concreteness, here is the Weyl tensor.
<br /> C_{abc}{}^d=<br /> R_{abc}{}^d <br /> - \frac{2!}{n-2}g^{de}\left(\left(g_{a[c}R_{e]b}- <br /> g_{b[c}R_{e]a}\right) <br /> - \frac{1}{n-1}Rg_{a[c}g_{e]b}\right) <br />
Weyl is the traceless part of the Riemann tensor.

Does your proposed Riemann tensor (with its claimed vanishing Weyl and Ricci [and therefore scalar-curvature] satisfy this?

(Does your proposed Riemann tensor satisfy the Bianchi identities?)

<<Does your proposed Riemann tensor (with its claimed vanishing Weyl and Ricci [and therefore scalar-curvature] satisfy this?>>

Obviously it does, because I chose only 3 components of the Riemann tensor to be non-zero (actually by symetry conditions some other ones will be non-zero too) and those quantities (and the other quantities implied by the symetry conditions) are not part of the traceless part.


<<(Does your proposed Riemann tensor satisfy the Bianchi identities?)>>

Obviously, because I just specified the Riemann tensor at a single point. The Bianchi requirements deal with *derivatives* of the Riemann components. Had I tried to specify a Riemann field over a region larger than a point or had I tried to specify derivatives of the Riemann tensor at that point, then I would have had to be concerned about Bianchi requirements. For example, if I specify an electric field vector and a magnetic field vector at a single point, I do not need to worry about Maxwell's Equations being satisfied, but if I specified them over a finite region, or if I said something about the derivatives of the electric and magnetic field at that point, I would need to be concerned that my arbitrary specification was not one allowed by Maxwell's Equations.

 

[robphy]

<<(Does your proposed Riemann tensor satisfy the Bianchi identities?)>>

Obviously, because I just specified the Riemann tensor at a single point. The Bianchi requirements deal with *derivatives* of the Riemann components. Had I tried to specify a Riemann field over a region larger than a point or had I tried to specify derivatives of the Riemann tensor at that point, then I would have had to be concerned about Bianchi requirements. For example, if I specify an electric field vector and a magnetic field vector at a single point, I do not need to worry about Maxwell's Equations being satisfied, but if I specified them over a finite region, or if I said something about the derivatives of the electric and magnetic field at that point, I would need to be concerned that my arbitrary specification was not one allowed by Maxwell's Equations.

There are two Bianchi identities:
a differential one... \nabla_{[a}R_{bc]d}{}^e=0 the one associated with the Einstein tensor being divergence-free;
and
an algebraic one... R_{[abc]}{}^d=0
(of course, Riemann has other symmetries due to metric compatibility and the torsion-free condition).

 

[swimmingtoday]

There are two Bianchi identities:
a differential one... \nabla_{[a}R_{bc]d}{}^e=0 the one associated with the Einstein tensor being divergence-free;
and
an algebraic one... R_{[abc]}{}^d=0
(of course, Riemann has other symmetries due to metric compatibility and the torsion-free condition).

The one you refer to as "the algebraic one" is not known as a Bianchi Identity. It is a true symmetry property of the Riemann Tensor, but iy is not a bianchi identity.

In my previous post I explained why the thing you refer to as the "divergence-free" one is satisfied. I'll show here that the other symmetry property (what you refer to as the "algebraic one" is satisfied)

We need to show that R0i0i + R00ii +Ri00i =0 for the Riemann tensor I choose. By invoking the fact that the Riemann tebsor is anti-symmetric in the first two indices (Rabcd = -Rbacd) we see that R00ii = 0, (We of course could have alternately used the fact that it is anti-symetric in the last two indices) So we are now needing to show that R0i0i +Ri00i =0. Again, we use the anti-symmetry of the first two indices, and so the claim is proven.

 

[robphy]

The one you refer to as "the algebraic one" is not known as a Bianchi Identity. It is a true symmetry property of the Riemann Tensor, but iy is not a bianchi identity.

http://www.google.com/search?&q="algebraic+bianchi
http://www.google.com/scholar?&q="algebraic+bianchi

In my previous post I explained why the thing you refer to as the "divergence-free" one is satisfied. I'll show here that the other symmetry property (what you refer to as the "algebraic one" is satisfied)

We need to show that R0i0i + R00ii +Ri00i =0 for the Riemann tensor I choose. By invoking the fact that the Riemann tebsor is anti-symmetric in the first two indices (Rabcd = -Rbacd) we see that R00ii = 0, (We of course could have alternately used the fact that it is anti-symetric in the last two indices) So we are now needing to show that R0i0i +Ri00i =0. Again, we use the anti-symmetry of the first two indices, and so the claim is proven.

That is merely one (1) of the set of equations implied by the "algebraic Bianchi" identity.


Your claim

I doubt this is true. Suppose that at some point, R1010= -(R2020 + R3030), with all three of these curvature tensor quantities non-zero. And suppose that the Weyl tensor is zero. Hopefully these conditions are self-consistant. If they are, then in this case, R00 will be zero, yet we have 3 components of the Riemann tensor which are non-zero, and so the Riemann Tensor does not vanish, depite both the Ricci Tensor and the Weyl tensor vanishing. (BTW, I suspect that the hypothetical situation I constructed is the actual situation for the vacuum region of the Schwarzschild Solution.)

seems to be in conflict with a counting argument: in 3+1 dimensions, Riemann (from a torsion-free metric-compatible derivative operator) has 20 algebraically-independent components, Weyl has 10, and Ricci has 10. If Weyl and Ricci are both zero, then Riemann must be zero (in accord with the expression for Weyl [a.k.a. decomposition of the Riemann tensor] given above).

Given the three nonzero components you specified, what are ALL of the other nonzero components of Riemann (implied by the algebraic symmetries of Riemann)?

Once written, we can then see explicitly what the resulting Ricci, scalar-curvature, and Weyl tensors are. I suspect it will not satisfy the expression for the Weyl tensor given earlier.

Let me ask this: Are you claiming that this expression for Weyl [a.k.a. decomposition of the Riemann tensor] is incorrect?

 

[swimmingtoday]

<<That is merely one (1) of the set of equations implied by the "algebraic Bianchi" identity.>>

THERE IS NO "ALGEBRAIC" BIANCHI IDENTITIES. There is only a Bianchi Identities Ra [bcd;e]=0 and it's rsulting (Ruv- 1/2 guvR);v= 0 I explained this to you before, and you just keep repeating the same incorrect claim. The identity R[abc]d =0 is not any sort of"agebraic" Bianchi Identities. It is an identity, but it was not done by Bianchi. How many times am I going to have to tell you this?

<<Weyl and Ricci are both zero, then Riemann must be zero>>

Another thing you are incapable of learning. If the Riemann tensor is such that R0101 = -(R0202 + R0303) then the Weyl tensor vanishes because these components are not poart of the traceless part, and the Ricci tensor also vanishes by simple direct calculation. So it it easy to construct a situation that contradicts your "counting argument. This is yet another example of something I explained to you in great detail before , but you just are not able to get.

<<Given the three nonzero components you specified, what are ALL of the other nonzero components of Riemann (implied by the algebraic symmetries of Riemann)?>>

They are all ZERO. I explained that to you ALSO before.

 

[swimmingtoday]

#########################

There are two Bianchi identities:
a differential one... \nabla_{[a}R_{bc]d}{}^e=0 the one associated with the Einstein tensor being divergence-free;
and
an algebraic one... R_{[abc]}{}^d=0
(of course, Riemann has other symmetries due to metric compatibility and the torsion-free condition).

####################

It does NOT have any other identities due to the torsion-free condition. You just never know what you are talking about. Be sure to keep repesting this incorrect claim too, even after you are corrected.

 

[robphy]

#########################

There are two Bianchi identities:
a differential one... \nabla_{[a}R_{bc]d}{}^e=0 the one associated with the Einstein tensor being divergence-free;
and
an algebraic one... R_{[abc]}{}^d=0
(of course, Riemann has other symmetries due to metric compatibility and the torsion-free condition).

####################

It does NOT have any other identities due to the torsion-free condition. You just never know what you are talking about. Be sure to keep repesting this incorrect claim too, even after you are corrected.

I think that you are correct that "It does NOT have any other identities due to the torsion-free condition." Zero torsion already implies R_{[abc]}{}^d=0.

THERE IS NO "ALGEBRAIC" BIANCHI IDENTITIES. There is only a Bianchi Identities Ra [bcd;e]=0 and it's rsulting (Ruv- 1/2 guvR);v= 0 I explained this to you before, and you just keep repeating the same incorrect claim. The identity R[abc]d =0 is not any sort of"agebraic" Bianchi Identities. It is an identity, but it was not done by Bianchi. How many times am I going to have to tell you this?

It may very well be that "algebraic Bianchi" is a misnomer for
R_{[abc]}{}^d=0.
If so, it is, of course, unfortunate that the term has been used by some
http://www.google.com/search?&q="algebraic+bianchi
http://www.google.com/scholar?&q="algebraic+bianchi
I am not a historian of mathematics. Maybe it is an identity of Ricci.
In any case, we understand that the issue on this point is the set of equations R_{[abc]}{}^d=0, whatever we call them.But now to focus back on the main issue you first raised, where you have quoted my reply (and my quote of another poster's question):

<<If the stress-energy tensor is zero everywhere, what is then causing the curvature ??...The Weyl tensor>>

I doubt this is true. Suppose that at some point, R1010= -(R2020 + R3030), with all three of these curvature tensor quantities non-zero. And suppose that the Weyl tensor is zero. Hopefully these conditions are self-consistant. If they are, then in this case, R00 will be zero, yet we have 3 components of the Riemann tensor which are non-zero, and so the Riemann Tensor does not vanish, depite both the Ricci Tensor and the Weyl tensor vanishing. (BTW, I suspect that the hypothetical situation I constructed is the actual situation for the vacuum region of the Schwarzschild Solution.)

where you say "Hopefully these conditions are self-consistant."

To which I replied [to help check consistency],

For concreteness, here is the Weyl tensor.
<br /> C_{abc}{}^d=<br /> R_{abc}{}^d <br /> - \frac{2!}{n-2}g^{de}\left(\left(g_{a[c}R_{e]b}- <br /> g_{b[c}R_{e]a}\right) <br /> - \frac{1}{n-1}Rg_{a[c}g_{e]b}\right) <br />
Weyl is the traceless part of the Riemann tensor.

Does your proposed Riemann tensor (with its claimed vanishing Weyl and Ricci [and therefore scalar-curvature] satisfy this?

which led up to your reply (again quoting me):

<<Weyl and Ricci are both zero, then Riemann must be zero>>

Another thing you are incapable of learning. If the Riemann tensor is such that R0101 = -(R0202 + R0303) then the Weyl tensor vanishes because these components are not poart of the traceless part, and the Ricci tensor also vanishes by simple direct calculation. So it it easy to construct a situation that contradicts your "counting argument. This is yet another example of something I explained to you in great detail before , but you just are not able to get.

It may be that I don't understand your claim that "the Weyl tensor vanishes because these components are not poart of the traceless part, and the Ricci tensor also vanishes by simple direct calculation" and your explanation "in great detail".

To help me, could you answer this unanswered question I posed earlier

Are you claiming that this expression for Weyl [a.k.a. decomposition of the Riemann tensor] is incorrect?

 

[swimmingtoday]

I changed my mind and now agree with you that the Weyl tensor does not vanish. But while explaining the ability of the Riemann tensor to be non-zero in a situation with a vanishing Ricci tensor as being due to the Weyl tensor non-vanishing is technically true, it appears to be trivial (in the mathematical sense) rather than really explaining in a good way what is going on.

If the Ricci tensor vanishes despite a non-vanishing Riemann tensor, then of course a quantity (the Weyl tensor) which is the Riemann tensor minus terms involving the Ricci tensor will be non-zero, and of course the Riemann tensor will be equal to this Weyl tensor. I apologize for saying you were incorrect about the Weyl tensor being non-vanishing.

But I think the way to actually understand in a non-trivial way what is going on is to note that the Ricci tensor components are sums of Riemann tensor components (i.e Rab = R0a0b + R1a1b + R2a2b + R3a3b, where the first tensor index is contravariant and the others are covariant indices), and that the sum of a set of quantities can be zero without each of the individual quantities in the sum being zero. For example, if R1010= negative (R2020 + R3030), then the sum R1010 + R2020 + R3030 will be zero, even if the three individual pieces composing the sum are not zeros. (The other term in the sum for R00 is R0000, which is zero by the anti-symmetry conditions )

 

[swimmingtoday]

<<It may very well be that "algebraic Bianchi" is a misnomer for
R_{[abc]}{}^d=0.>>

I had not seen your earlier post with the google links when I went on my tirade. While it is a misnomer, I should not have claimed you were unequivocally wrong, being that the misnomer is apparently somewhat standard; or that you were just ignoring what I had said to you previously.

 

[Chris Hillman]

How does the gravity get out of a black hole?

Hi, notknowing,

Now, I'm really strongly confused . If the stress-energy tensor is zero everywhere, what is then causing the curvature ?? I would tend to think that the stress-energy tensor becomes singular or so, but not zero. And what does "M" then mean in the Schwarzschild solution ? Is this the mass of an object with zero energy-stress tensor ?

That's a good question, and one which is frequently asked. The short answer is that in the case of a black hole formed by the gravitational collapse of some massive object, spacetime was already curved up when that object was originally formed (by concentrating mass-energy by some process, probably involving gravitation) in the very distant past. While this concentration was happening, the field (curvature) farther away was being "updated" (with a suitable time delay) to take account of the changing distribution of mass-energy. After the collapse, "the fact that the matter which was the original source of the field can no longer send signals to the exterior prevents the rest of the universe from having to worry about what might have happened to it after it fell past the event horizon", so to speak.

If http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html or http://www.math.ucr.edu/home/baez/PUB/Escape don't clear up any remaining confusion, please ask again.

Chris Hillman

 

[Chris Hillman]

Two Useful Decompositions of Riemann tensor

Hi,

<<If the stress-energy tensor is zero everywhere, what is then causing the curvature ??...The Weyl tensor>>

I doubt this is true. Suppose that at some point, R1010= -(R2020 + R3030), with all three of these curvature tensor quantities non-zero. And suppose that the Weyl tensor is zero. Hopefully these conditions are self-consistant. If they are, then in this case, R00 will be zero, yet we have 3 components of the Riemann tensor which are non-zero, and so the Riemann Tensor does not vanish, depite both the Ricci Tensor and the Weyl tensor vanishing. (BTW, I suspect that the hypothetical situation I constructed is the actual situation for the vacuum region of the Schwarzschild Solution.)

For what it is worth, there are two decompositions of the Riemann tensor (defined irrespective of any physical interpretation of our Lorentzian manifold) which are particularly useful in gtr:

1. The Ricci decomposition expresses R_{abcd} as a sum of three terms, a scalar part (a four index tensor with the algebraic symmetries of the Riemann tensor, built out of the Ricci scalar R), a trace part (another four index tensor with the algebraic symmetries of the Riemann tensor, built out of the traceless Ricci tensor S_{ab}), and the Weyl tensor C_{abcd}, or conformal curvature tensor. Therefore, the Einstein tensor vanishes (the condition in the vacuum EFE) iff the Riemann and Weyl tensors agree.

2. The Bel decomposition, taken with respect to some timelike congruence (a family of timelike curves which fills up spacetime without self-intersection; think of a bunch of parallel lines and distort this picture without creating any intersections between any pair of distorted lines)--- this congruence need be neither geodesic nor irrototational and is defined by giving some timelike vector field \vec{X}--- expresses R_{abcd} as a sum of three tensors (for D=1+3) which are essentially "three-dimensional tensors living on the spatial hyperplane element at each event which is orthogonal to the timelike curve there). These tensors are respectively:

a. the tidal or electrogravitic tensor E[X]_{ab} = R_{ambn} \, X^m \, X^n, which is symmetric (6 algebraically independent components),

b. the topogravitic tensor L[X]_{ab} = {{}^\star R_{ambn}^\star} \, X^m \, X^n, which is also symmetric (6 components)

c. the magnetogravitic tensor B[X]_{ab} = {}^\star {R_{ambn}} \, X^m \, X^n, which is traceless (8 algebraically independent components)

Here, the star is the usual dual (see any good textbook; Misner, Thorne & Wheeler offer a particularly nice exposition).

These account for the 20 algebraically independent components of the Riemann tensor. In a vacuum, E[X]_{ab} = -L[X]_{ab} and B[X]_{ab} are now traceless symmetric (5 algebraically independent components each) and account for the 10 algebraically independent components of the Weyl=Riemann tensor.

This implies for example that for vacuum solutions, the Petrov classification of the possible symmetries of the Weyl tensor (types I,II,III, D, N, O; see for example the textbook by D'Inverno) can be reformulated in terms of saying that in the case of a type D vacuum (for example), one can find an \vec{X} such that the tidal and magnetogravitic tensors assume a special form (indeed, the tidal tensor can be put in the "Coulomb form" familiar from Newtonian gravitation in the classical field theory formulation wherein the potential satisfies the Laplace equation), and similarly for the other Petrov types. This is completely analogous to characterizations of the electromagnetic field in Maxwell's theory of electromagnetism, in which we can project the field tensor F_{ab}, with respect to a given timelike congruence, onto two three-dimensional vectors (3 components each, accounting for the 6 algebraically independent components of the field tensor, which is antisymmetric).

If you are wondering why I said "algebraic symmetries", that is because the differential Bianchi identities imply that a differential equation identity relates the Ricci tensor (or equivalently, the Einstein tensor) and the Weyl tensor. This identity is again irrespective of any physical interpretation of our Lorentzian manifold, so it does not require postulating an additional equation in gtr (or any competing classical field theory of gravitation).

This is of course how the presence of matter or other nongravitational mass-energy HERE (in a nonvacuum region!) can produce curvature THERE (Weyl curvature, in a vacuum region possibly distance from the mass-energy); the matter directly produces Ricci curvature via the EFE, and this Ricci curvature in general will produce some Weyl curvature via the differential identity, which then produces more Weyl curvature further away, and so on.

(Caveat: in the case of a highly symmetric matter distribution, e.g. in the FRW cosmological models, which are dust or perfect fluids, possibly with a Lambda contribution to the stress-energy tensor, no Weyl curvature is produced, so these models exhibit "purely Ricci curvature"; we say that they are "conformally flat" because the Weyl tensor vanishes identically.)

In this way, whenever (in a model universe, in gtr) we concentrate mass-energy in some region, say while forming a star, we gradually curve up spacetime as mentioned in "How does gravity get out of a black hole?"

Chris Hillman

 

[Chris Hillman]

Electrovacuum =/= Vacuum

Hi, notknowing,

I was mislead by the popular view that in a black hole a point of infinite density is obtained.

Well, if you know the math such statements are trying to describe, they have some value, but yes, this is just one illustration of why popular descriptions are almost always terribly misleading in some ways.

I recapitulate things to see if I understand it correctly. First you need a sufficient amount of mass to create a black hole in the first place ("feed" it). Then, through some magic, the matter is crunched together in such a way that it is transformed into pure gravitational energy such that "the curvature itself becomes the source of curvature" (the idea which I had originally to create a kind of "geon"). Of course, in a quantum theory of gravitation, things would probably turn out to be very different.

Well, I don't know why you say "magic" unless gravitation is "magical". No-one was claiming (or if they did, they probably should not have) that matter is "transformed to pure gravitational energy", unless that is how you want to summarize the discussion in the FAQ "How does gravity get out of a black hole?" and in my long extinct but archived post to sci.physics (both cited in a post to this forum made earlier today, 25 November--- happy birthday, gtr!).

I am not sure what you mean by your remark about quantum gravity. If you meant that quantum gravity should prevent the formation of an event horizon, that is probably not true. If you mean that quantum gravity should drastically alter physics as sectional curvatures approach the reciprocal of the Planck length squared, that is almost certainly true, although we won't know the details until a successful theory appears (and is understood at least in its most basic respects).

But what then about a charged black hole ? If the stress-energy tensor is also congruently zero everywhere in this case, such that one can not speek of mass in the usual sense, where is then the charge located ? It can probably not be considered to be "smeared out" in the region inside the horizon ?

The stress-energy tensor is NOT identically zero outside a charged black hole! That is because the electromagnetic field contributes a nonzero term to the stress-energy. This is why we speak of "electrovacuum" rather than "vacuum" in a region where an electromagnetic field but no matter or other non-gravitational fields are present.

In the case of a charged object such as a hollow spherical metallic shell with uniformly distributed charge (for example), the charge is located on the surface just as you would expect from Maxwell. This would be modeled by an electrovacuum exterior (a portion of the Reissner-Nordstrom electrovacuum) matched to a charged spherical shell (which we can model as "infinitely thin" for convenience) and to a vacuum interior (a portion of the Minkowksi vacuum, since there is no electric or gravitational field inside the shell, as in pre-relativistic physics).

Chris Hillman

 

[Chris Hillman]

Bianchi identities

Hi, swimmingtoday,

THERE IS NO "ALGEBRAIC" BIANCHI IDENTITIES. There is only a Bianchi Identities Ra [bcd;e]=0 and it's rsulting (Ruv- 1/2 guvR);v= 0

Careful, while this terminology is incompletely standardized, most authors DO distinguish between algebraic and differential Bianchi identities. These are of course completely different, but all happen to have been discussed by Luigi Bianchi, hence the names. See for example the fine discussion in MTW.

Someone, I think Robphy, said <<Weyl and Ricci are both zero, then Riemann must be zero>>, and of course that is true. The easiest way to remember this is to recall the Ricci decomposition I described in another post I made today in this thread.

If the Riemann tensor is such that R0101 = -(R0202 + R0303) then the Weyl tensor vanishes

That's not true as stated, of course (but maybe I have simply lost track of the context of the original remark?), since you only gave one algebraic condition, reducing the number of algebraically independent components from 20 to 19, which leaves plenty of choice for a nonzero Weyl tensor since the Ricci tensor can only account for 10 of these 19.

So it it easy to construct a situation that contradicts your "counting argument. This is yet another example of something I explained to you in great detail before , but you just are not able to get.

I think you'd need to clarify exactly what it is that you are claiming. Rob's counting argument is perfectly OK, but maybe we are both misunderstanding what you are trying to say.

<<Given the three nonzero components you specified, what are ALL of the other nonzero components of Riemann (implied by the algebraic symmetries of Riemann)?>>

They are all ZERO. I explained that to you ALSO before.

OH! Well, that makes a big difference! So your condition was actually suplemented by "all components which are algebraically independent of
R_{0101}, \, R_{0202}, \, R_{0303) also vanish? Under some reasonable assumptions about your indexing convention and your choice of frame field, these would be associated with the electrogravitic tensor in the Bel decomposition taken with respect to \vec{X}=\vec{e}_0, the timelike unit vector in the frame, so you appear to be discussing a "nonrotating vacuum field", i.e. one in which the magnetogravitic tensor, taken with respect to a "spacetime symmetry adapted congruence" vanishes. Since the tidal tensor is traceless in a vacuum, the identity you wrote down holds automatically. The Schwarzschild vacuum solution, but not the Kerr vacuum solution, would answer to this description.

Well, I am still confused about what you were trying to get at, but I hope this post helps to clear the air!

Chris Hillman

 

[swimmingtoday]

<<OH! Well, that makes a big difference! So your condition was actually suplemented by "all components which are algebraically independent of
R_{0101}, \, R_{0202}, \, R_{0303) also vanish? >>

Yes. I think I stated that.

<< The Schwarzschild vacuum solution, ... would answer to this description.>>

I stated that I thought my hypothetical situation was what was going in in the Schwarzschild vacuum solution

<<Well, I am still confused about what you were trying to get at>>

Someone was puzzled at how you could have a situation where the Ricci tensor vanished in a region but the Riemann tensor did not. I gave a simple hypothetical situation as a concrete example.

 

[Chris Hillman]

OK, I think we are all on the same page now--- whew!

Chris Hillman