Meaning of r in Schwarzchild coordinates

[PeterDonis]

It seems we have explored this topic before, in a thread I forgot all about

So had I. Thanks for digging it up!

if you have a congruence orthogonal to a foliation, why can't you then diagonalize the 3-manifolds of the foliation, and then build the 4th basis as the surface orthogonal congruence tangent, and end up with coordinates everywhere orthgonal?

This is an interesting question, because that's pretty much what you're doing in FRW spacetime to construct the standard FRW chart, and it works. The key difference in Kerr spacetime is that the 3-manifolds in question have only one spacelike KVF (the axial one), whereas the ones in FRW spacetime have six. In that other thread we had proposed that maybe there is some minimum number of spacelike KVFs that the 3-manifold has to have for the method you describe to produce a diagonal line element. (The number we proposed was three, since that's how many Schwarzschild spacetime has, and your method works for that spacetime.)

To put what I just said another way, consider how the "direction" in spacetime of the 4th basis vector changes as we move through the 3-manifold. In FRW spacetime, the change is isotropic; it's the same if we move in any direction in the 3-manifold. In Kerr spacetime, the change is not isotropic. This anisotropy (which I'm sure there's a slicker way of describing, perhaps in terms of the extrinsic curvature of the 3-manifold as embedded in spacetime) might be the reason there has to be at least one cross term in the line element.

 

[PWiz]

The number we proposed was three, since that's how many Schwarzschild spacetime has

I don't get this. I thought there were only two Killing vector fields in the Schwarzschild metric (using standard Schwarzschild coordinates) : $∂_t$ (timelike) and $∂_{\phi}$ (spacelike).

Which is the the third KVF here? Both $r$ and $\theta$ appear in the metric coefficients!

 

[PAllen]

I have a couple thoughts on the issue of why a hypersurface orthogonal congruence is not sufficient to ensure diagonalizability in 4-d (though it is necessary). [call this condition A ]

1) There is a good reason this is hard to visualize. Precisely because every 3-manifold is diagonalizable (over a finite region; we are not talking strictly global), it is impossible to construct a 2x1 analog of the insufficiency of this condition. Since all of our intuition is 3 dimensional, we can't readily visualize the breakdown.

2) I am not convinced that killing vectors have anything to do with the issue. Looking at 3-manifolds, they are diagonalizable even with no killing vector fields (yes, it is possible for a 3-manifold to have no killing vector fields). From Ben Niehoff in the other thread: "However, keep in mind: In order to write a diagonal metric, the twist-free congruences do not have to be Killing (and they do not have to be geodesic). Also, a Killing congruence is not necessarily twist-free (consider that generated by ∂t+∂ϕ in Minkowski space)."

3) My best current attempt at an understanding of the 4-d insufficiency of condition A, is as follows, by analogy to the 2X1 case. Diagonalizing a 2 surface uses only one of two available gauge degrees of freedom. That means any 2-surface in the 2X1 orthogonal split can have a set of orthogonal coordinates conformally stretched as needed. Imagine coordinates on successive 2-surfaces. You need to align them so same coordinates on successive surfaces line up with the a member of the orthogonal congruence. Besides just scaling, you have this conformal degree of freedom to accomplish this. In a 3X1 split of 4-d, all gauge degrees of freedom [within the 3-surface] are needed to diagonalize a 3-surface (in general). This leaves you with nothing to play with to get coordinates on successive 3-surfaces to line up with the orthogonal congruence. Sure, you can still scale, but you can't stretch in any way.

 

[PAllen]

I don't get this. I thought there were only two Killing vector fields in the Schwarzschild metric (using standard Schwarzschild coordinates) : $∂_t$ (timelike) and $∂_{\phi}$ (spacelike).

Which is the the third KVF here? Both $r$ and $\theta$ appear in the metric coefficients!

See:

http://d.umn.edu/~vvanchur/2013PHYS5551/Chapter5.pdf

Killing vectors don't have to be (and usually are not) manifest in coordinates. In this case, conventional spherical coordinates directly manifest only one of the spatial killing vector fields.

 

[PeterDonis]

I thought there were only two Killing vector fields in the Schwarzschild metric (using standard Schwarzschild coordinates) : ∂_t (timelike) and ∂ϕ (spacelike).

No, there are three spacelike ones (plus the timelike $\partial_t$ -- actually it's only timelike outside the horizon, but that's a whole other discussion...). You can't always "read off" all the KVFs from the line element, even in "adapted" coordinates. In any spherically symmetric spacetime, there are three spacelike KVFs corresponding to the spherical symmetry; that's what "spherical symmetry" means. Think of them as rotating a 2-sphere about three mutually perpendicular axes; all three of those transformations leave the 2-sphere invariant. Only one of them happens to have a simple expression in Schwarzschild coordinates.

(Actually, the more precise way to say this is that there is a 3-parameter family of spacelike KVFs, all leaving every 2-sphere invariant; the rotations about the three mutually perpendicular axes are just the "basis vectors" of the family--all the other rotations can be expressed as linear combinations of them.)

 

[PWiz]

am not convinced that killing vectors have anything to do with the issue. Looking at 3-manifolds, they are diagonalizable even with no killing vector fields (yes, it is possible for a 3-manifold to have no killing vector fields).

An amateurish question here: by 3-manifolds, do you mean a 2 spatial+1 temporal dimension manifold?
If you're talking about Lorentzian manifolds, I must say that after thinking about it, I agree with you. You've already said before that staticity and diagonalizability are two different things; the former is not essential for the latter to be true. If we consider the region within the EH of a non-rotating, uncharged black hole, we would find that no timelike KVFs exist there, yet the metric is diagonal, proving that diagonalization of a metric has nothing to do with the existence of timelike KVFs (I'm not sure what to conclude about spacelike KVFs and diagonalizability over here). Now this could be some really faulty reasoning, so I'd like to have some opinion on this.

 

[PWiz]

See:

http://d.umn.edu/~vvanchur/2013PHYS5551/Chapter5.pdf

Killing vectors don't have to be (and usually are not) manifest in coordinates. In this case, conventional spherical coordinates directly manifest only one of the spatial killing vector fields.

OK, I'll go through the link.

Actually, the more precise way to say this is that there is a 3-parameter family of spacelike KVFs, all leaving every 2-sphere invariant; the rotations about the three mutually perpendicular axes are just the "basis vectors" of the family--all the other rotations can be expressed as linear combinations of them.

The more precise way actually appears to be simpler to understand! I think I got it - I can easily visualize spacelike KVFs now. This is exactly why the Kerr metric ( in Boyer-Lindquist coordinates) only 1 spacelike KVF right? Only a rotation around the axis of rotation would leave the 2-sphere invariant.

 

[PAllen]

An amateurish question here: by 3-manifolds, do you mean a 2 spatial+1 temporal dimension manifold?
If you're talking about Lorentzian manifolds, I must say that after thinking about it, I agree with you. You've already said before that staticity and diagonalizability are two different things; the former is not essential for the latter to be true. If one considers regions within the EH of a non-rotating, uncharged black hole, we would find that no timelike KVFs exist there, yet the metric is diagonal. Now this could be some really faulty reasoning, so I'd like to have some opinion on this.

No, the 3-manifolds I refer to are pure Reimannian manifolds. Every possible spacelike 3-surface in a 4-d Lorentzian manifold can be considered a 3-d pure Riemannian manifold with 3-metric induced from the 4-metric. Because it is spacelike, the induced metric will be positive definite.

 

[PWiz]

No, the 3-manifolds I refer to are pure Reimannian manifolds. Every possible spacelike 3-surface in a 4-d Lorentzian manifold can be considered a 3-d pure Riemannian manifold with 3-metric induced from the 4-metric. Because it is spacelike, the induced metric will be positive definite.

Alright, so you were talking about a hypersurface in spacetime with constant t right?
And what do you think about the 2nd paragraph in my post?

 

[PeterDonis]

This is exactly why the Kerr metric ( in Boyer-Lindquist coordinates) only 1 spacelike KVF right? Only a rotation around the axis of rotation would leave the 2-sphere invariant.

For Kerr spacetime, it's better to visualize the rotation you describe as leaving a cylinder invariant, since Kerr spacetime is not spherically symmetric.

 

[PWiz]

For Kerr spacetime, it's better to visualize the rotation you describe as leaving a cylinder invariant, since Kerr spacetime is not spherically symmetric.

OK, I'll do that. What do you think about my reasoning in paragraph 2 of #41 though?

 

[PeterDonis]

What do you think about my reasoning in paragraph 2 of #41 though?

There's only one paragraph in post #41 (at least, only one that you wrote). Do you mean this?

If we consider the region within the EH of a non-rotating, uncharged black hole, we would find that no timelike KVFs exist there, yet the metric is diagonal, proving that diagonalization of a metric has nothing to do with the existence of timelike KVFs (I'm not sure what to conclude about spacelike KVFs and diagonalizability over here).

You are correct that being able to find coordinates in which the metric is diagonal does not require a timelike KVF. FRW spacetime is another counterexample; there is no timelike KVF in that spacetime, but the metric is diagonal in standard FRW coordinates.

 

[PWiz]

There's only one paragraph in post #41 (at least, only one that you wrote). Do you mean this?

Yes, that is what I was referring to. I thought I'd put a line break while typing, but it looks like it didn't pan out in the post.

So spacelike KVFs have no role in this issue as well? (I can't think of a spacetime metric without any spacelike KVFs for comparitive purposes, unfortunately.)

 

[PAllen]

Yes, that is what I was referring to. I thought I'd put a line break while typing, but it looks like it didn't pan out in the post.

So spacelike KVFs have no role in this issue as well? (I can't think of a spacetime metric without any spacelike KVFs for comparitive purposes, unfortunately.)

Exact solutions need lots of symmetry, or they couldn't be exact. Any N body metric (which can only be numerically approximated) would have no killing vector fields. In other words, the real world has no exact killing vectors. Of course, cosmology closely approximates a solution with killing vectors, and for some purposes you can consider a star, as isolated and approximately Kerr far away (but not so much nearby, even if isolated).

 

[PWiz]

Exact solutions need lots of symmetry, or they couldn't be exact. Any N body metric (which can only be numerically approximated) would have no killing vector fields. In other words, the real world has no exact killing vectors. Of course, cosmology closely approximates a solution with killing vectors, and for some purposes you can consider a star, as isolated and approximately Kerr far away (but not so much nearby, even if isolated).

Alright, I understood. Thanks.

 

[pervect]

I was doing some more reading about the general case, and I suspect that Frobenius' theroem may be involved in the more general solution.

But it's all terribly abstract, I'm not getting much of a "physical" feel.

"Hypersurface orthogonality" is mentioned as a special case of Frobenius' theorem (see pg 434, for instance).

 

[PWiz]

I was doing some more reading about the general case, and I suspect that Frobenius' theroem may be involved in the more general solution.

But it's all terribly abstract, I'm not getting much of a "physical" feel.

"Hypersurface orthogonality" is mentioned as a special case of Frobenius' theorem (see pg 434, for instance).

I don't know much about this, but after some reading on the net, it would appear that the theorem only applies to homogenous, linear 1st order PDEs. The EFEs are coupled, non-linear 2nd order PDEs. Can the theorem be applied to GR?

 

[PeterDonis]

Can the theorem be applied to GR?

Yes. See, for example, Wald, Appendix B.3. Google Books has it here:

https://books.google.com/books?id=9...=frobenius theorem general relativity&f=false

 

[PeterDonis]

it would appear that the theorem only applies to homogenous, linear 1st order PDEs.

Can you give a reference? The discussions I've seen (such as the one in Wald) don't talk about differential equations at all; they talk about manifolds and vector fields or differential forms. (Also, in GR the theorem isn't applied to the EFE; it is applied after you've already solved the EFE, to the manifold with metric that you obtain as the solution.)

 

[PWiz]

...the theorem addresses the problem of finding a maximal set of independent solutions of a regular system of first-order linear homogeneouspartial differential equations.

I read this. I'll emphasize again - I know nothing about this theorem. I've yet to go through a completely formal treatment in GR and topology.

 

[PeterDonis]

I read this.

Yes; the differential equations referred to are not Einstein's Field Equations, so the fact that the EFE is second-order and nonlinear is irrelevant. The equations referred to are equations describing integral curves of vector fields in a spacetime manifold that is a solution of Einstein's Field Equations. AFAIK those equations are first order and linear. (The Wiki page also describes a geometric interpretation of what is going on, which is much closer to how the theorem is applied in GR.)

 

[aleazk]

Can you give a reference? The discussions I've seen (such as the one in Wald) don't talk about differential equations at all; they talk about manifolds and vector fields or differential forms. (Also, in GR the theorem isn't applied to the EFE; it is applied after you've already solved the EFE, to the manifold with metric that you obtain as the solution.)

Check "Introduction to Smooth Manifolds" by J.Lee. From page 361: "...explicitly finding integral manifolds boils down to solving a system of PDEs, we can interpret the Frobenius theorem as an existence and uniqueness result for such equations." And then it goes to show this connection.