Solutions of the nonlinear Field Equations

[robphy]

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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).

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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