Meaning of r in Schwarzchild coordinates

[PeterDonis]

If the coefficients are independent of theta, phi and t, then things should look unchanged if we flip the sign of any of these coordinates. So such a metric should exhibits PT symmetry.

And T symmetry, and P symmetry. PT symmetry is a weaker condition than P or T symmetry; PT only means invariance if P and T are both applied, but not necessarily if either one is applied individually. P symmetry means symmetry if just P is applied; T symmetry means symmetry if just T is applied.

If a diagonal metric employs the same coordinates but the coefficients are only independent of t, then the metric should show T symmetry.

Yes, but not P symmetry. It might still show PT symmetry.

Could one explore this chain of reasoning onto non-diagonal metrics?

Yes, by looking at which non-diagonal terms are present. For example, in Kerr spacetime (in appropriate coordinates), there is only one non-diagonal term, and it is a term in $dt d \phi$. So reversing the sign of both $t$ and $\phi$ (which a PT transformation does) leaves the metric invariant. But reversing $t$ or $\phi$ individually (i.e., P alone or T alone) does not. (You also have to look at the metric coefficients, of course; in Kerr spacetime those are all functions of $r$ and $\theta$ only, not $t$ or $\phi$.)

 

[PWiz]

And T symmetry, and P symmetry. PT symmetry is a weaker condition than P or T symmetry

Hmmm, interesting, I did not previously know that.

Yes, but not P symmetry. It might still show PT symmetry

OK. I think I finally wrapped my head around this.

Thanks @fzero @PAllen @stevebd1 @pervect @PeterDonis for helping me out! I appreciate it

 

[pervect]

FRW spacetime is neither T symmetric nor PT symmetric, but it has a diagonal metric in appropriate coordinates.

I thought we were talking Schwarzaschild vs Kerr? At least that's what I was talking about.

As PAllen said, the condition for being able to find coordinates in which the metric is diagonal is that there is a family of worldlines filling the spacetime and a family of hypersurfaces that are everywhere orthogonal to that family of worldlines. Schwarzschild spacetime and FRW spacetime both meet that requirement, but Kerr spacetime does not (it has a family of worldlines filling it, but they aren't hypersurface orthogonal).

I'm not quite following - which family of worldlines are we talking about? If we're talking about a static or stationary metric, I'd assume we were talking about the worldlines whose tangents were timelike Killing vectors, but if I'm following, you're trying to extend the analysis to include more general cases than I was initially considering. But if we are considering a more general case, then it's not clear to me what worldlines we are talking about here.

The extra condition that brings in T or PT symmetry is the presence of a timelike Killing vector field. (Note that we are now restricting attention to the region outside the horizon; at and inside the horizon, in both Schwarzschild and Kerr spacetime, the KVF in question is not timelike.) '

Right, I agree with this.

If you have a timelike KVF that is hypersurface orthogonal (i.e., a static spacetime), then you have T symmetry, as in the Schwarzschild case. If you have a timelike KVF but it isn't hypersurface orthogonal (i.e., a stationary spacetime that is not static), you have PT symmetry, as in the Kerr case.

That's pretty much the condition I was assuming, analyzing only the case where a timelke KVF was present. Rather than use the language of Killing vectors, though, I tried to make the same point used different language, The Killing vectors are basically about the symmetries of the space-time, I thought that the argument in terms of the symmetries would be more accessible to a wider audience than an argument based on the Killing vector fields. To my mind, though, the arguments are basically the same, only the manner in which it is presented changes, as long as one agrees that Killing vector fields represent underlying symmetries.

A few more little nits. I think that I've really only shown that for our example, we can eliminate the time-space terms like dt*dr, dt*phi, dt*dtheta by my time reversal argument. We need a different argument than I presented (probably involving the other KVF's and/or their associated radial symmetries) to eliminate terms like dphi*dtheta., space-space terms that are not diagonal.

Additionally I glossed over the whole issue of equivalence classes , i.e. metrics related by diffeomorphisms. As an example, one could use the non-diagonal Painleve metric to represent the same space-time as the Schwarzschild metric. So we don't really have to use a diagonal metric to represent the Schwarzschild space-time, it's just that it's a coordinate choice that respects the underlying coordinate-independent symmetries.

 

[PeterDonis]

thought we were talking Schwarzaschild vs Kerr? At least that's what I was talking about.

If we're restricting discussion to just those spacetimes, then yes, what you say about symmetries is correct. In the thread as a whole we have been considering more general conditions for a diagonal metric (see the recent posts between me and PWiz).

which family of worldlines are we talking about?

The family of worldlines that have some particular symmetry property. In the case of spacetimes with a timelike KVF, the worldlines are the orbits of the KVF, as you say. In the case of spacetimes which are homogeneous and isotropic, such as FRW spacetime, the worldlines are those of observers for whom those properties are manifest, i.e., they actually see the spacetime as homogeneous and isotropic. We could also express this in terms of spacelike KVFs; see below.

The Killing vectors are basically about the symmetries of the space-time

Yes, and these can be spacelike as well as timelike. For example, spherical symmetry means that there is a 3-parameter family of spacelike KVFs whose orbits generate 2-spheres. Homogeneity and isotropy means that there is a 6-parameter family of spacelike KVFs (3 parameters for spherical symmetry, plus 3 for spatial translation symmetry) whose orbits generate spacelike hypersurfaces; the "comoving" worldlines in FRW spacetime are the ones that are everywhere orthogonal to those hypersurfaces.

We need a different argument than I presented (probably involving the other KVF's and/or their associated radial symmetries) to eliminate terms like dphi*dtheta., space-space terms that are not diagonal.

Yes, to show that those terms can always be eliminated, you would look at the 3-parameter family of spacelike KVFs corresponding to spherical symmetry.

one could use the non-diagonal Painleve metric to represent the same space-time as the Schwarzschild metric. So we don't really have to use a diagonal metric to represent the Schwarzschild space-time, it's just that it's a coordinate choice that respects the underlying coordinate-independent symmetries.

More precisely, that respects all of them (including the spacelike as well as the timelike ones) and their relationships. Painleve coordinates still respect the timelike KVF: the metric is independent of $t$ and the integral curves of the timelike KVF are curves with constant spatial coordinates. And Painleve coordinates still respect spherical symmetry; the angular part of the metric is the same as for Schwarzschild. The only difference is that the surfaces of constant time are not orthogonal to the integral curves of the timelike KVF; in other words, the coordinates do not "respect" the hypersurface orthogonality of the timelike KVF.

 

[PAllen]

It seems we have explored this topic before, in a thread I forgot all about - finding more definitive answers:

https://www.physicsforums.com/threads/when-can-a-metric-be-put-in-diagonal-form.798380/

In particular, see Ben Niehoff's comment #6, and Bcrowell's #12, and the unusual case Peter mentioned in #5. This means the condition I gave in this thread is necessary but not sufficient for diagonalizability. This ZAMO exception seem really mysterious to me since (as covered in http://arxiv.org/abs/0809.3327, linked to by bcrowell), any 3-manifold is diagonalizable. Thus, 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? Of further relevance is that ZAMO congruence is twist free, so that is not a problem. All quite mysterious to me, but it is well established that there is no diagonalization of Kerr in a finite region.

 

[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.