Exploring the Physicality of Metric Solutions in Einstein's Field Equation

[grey_earl]

I don't want to offend you in any way, if you felt I have done this. But it seems to me that you don't understand some (subtle) points, and I want to explain them to you.

I'm not sure if we are talking at cross purposes or you are not getting what I'm saying. I didn't "put it" like that at all.

I may have misinterpreted your statement

the parameter H^-2 or 1/R^2 belongs to 5-dimension space

you gave in your post before the last, but to me you were stating that H is a five-dimensional parameter which doesn't really belong into four-dimensional de Sitter space. And that's false. It has a five-dimensional interpretation, but that's all.

Well, I specified "intrinsically" curved manifolds, if you give me the example of the topology of a cylinder which only has extrinsic curvature, you are not really countering any statement.

What I gave you is two-dimensional de Sitter space, not a cylinder. Topologically both are R x S¹, but de Sitter space has intrinsic curvature, namely in the case I gave you R = 2. Take the metric and calculate to see it yourself, please.

... it is stated in most GR books that the r=2m singularity is similar to the θ=0 singularity in that they are both coordinate singularities. If you disagree with it that is fine with me.

I have explicitely said that θ=0 is a coordinate singularity, but an ignorable one, and I gave you my reasons. I did it in detail because you were stating before

I find this statement slightly contradicting your inclusion of coordinates with trivial singularities (like r=2m in this case) in the definition of "global coordinates"

, but that's not what I said neither meant, and I wanted to clarify it. r=2M is not a trivial singularity, but θ=0 is, and I explained you why I think this is so. And all GR specialists I have met until today agree on this, although of course I haven't met all ;)

 

[TrickyDicky]

Oh, by all means, enlighten me, I'm pleased that you are willing to explain what I don't understand

What I gave you is two-dimensional de Sitter space, not a cylinder. Topologically both are R x S¹, but de Sitter space has intrinsic curvature, namely in the case I gave you R = 2. Take the metric and calculate to see it yourself, please.

So it has the topology of a cylinder, but it has intrinsic curvature, this is a bit over my head, I thought you needed more than one chart to cover a whole intrinsically curved manifold, maybe this belongs to the topology subforum rather than the relativity one anyway.

I have explicitely said that θ=0 is a coordinate singularity, but an ignorable one, and I gave you my reasons. I did it in detail because you were stating before
, but that's not what I said neither meant, and I wanted to clarify it. r=2M is not a trivial singularity, but θ=0 is, and I explained you why I think this is so. And all GR specialists I have met until today agree on this, although of course I haven't met all ;)

This is interesting, I was interpreting "trivial" in the sense of physically trivial, and everybody seems to agree there is nothing physically special at the BH event horizon, do you opine otherwise? It seems like the whole BH notion rests upon that singularity being trivial in this sense.
From Ryder GR book:"We may note here, however, that the ‘singularity’ r=2m is more like a coordinate singularity than an actual singularity in the geometry. By analogy, at the
north and south poles on a sphere ... The (contravariant components of the) metric tensor may become singular, but nothing unusual appears at the poles; a sphere is, after all, a homogeneous space, and no one point is different from any other point."

 

[TrickyDicky]

take the manifold R x S¹, with coordinate system ds² = dx² + cosh² x dφ², with -∞ < x < ∞ and 0 ≤ φ < 2π. This is regular everywhere, it has not even integrable singularities.

I'm not saying that is not regular, I'm just saying that it can't be completely covered with one chart of polar coordinates.

 

[grey_earl]

So now you know that there are intrinsically curved manifolds where you only need one chart. The topology doesn't fully specify the metric; there are infinitely many possibilities to bestow a given topological manifold with a metric.

We didn't understand each other saying "trivial" :)
There are two types of singularities: coordinate singularities and physical singularities.
A physical singularity is a point (or curve, or any submanifold) where a curvature invariant (a scalar, like R or R_{abcd} R^{abcd}, or R_{ac} R_{bd} R^{abcd}, or whatever you can think of - as long as it is a scalar) diverges. Since scalars don't change under a change of coordinates, you cannot get rid of such singularities. An example for a physical singularity is r=0 in the Schwarzschild metric.
A coordinate singularity is a submanifold where a component of a tensorial quantity diverges, like g_{tt} at r=2M in the Schwarzschild metric, although scalar curvature invariants are perfectly finite. Since the components of tensorial objects change under a change of coordinates, we simply have chosen bad coordinates, and in another coordinate system g_{tt} at this point of the manifold is perfectly fine. Now for coordinate singularities there are two subtypes: ignorable ones and ones which significate something. An ignorable coordinate singularity is, as I said, θ=0; r=2M is of the special type. Ryder may be a bit overstating, but since in the first half of the last century many physicists, including Einstein at the beginning, confused physical singularities with coordinate singularities, I think for him it is important to insist on the fact that r=2M is not a physical singularity. And it's not only "more like a coordinate singularity", it is a coordinate singularity :)

For example, take the following Gedankenexperiment: You hover with a spacecraft far away from a black hole. You then lower your outward acceleration, sink towards the black hole and up the acceleration a bit more to hover a bit nearer to the black hole. The point where you need infinite acceleration, you have reached r=2M. However, if you hover at an r>2M and spin around the black hole, nothing, absolutely nothing will change if you reach θ=0.
So r=2M is not a physical singularity since you won't be crushed when you reach this point (this will happen at the physical singularity r=0), but it by no means is a trivial singularity like θ=0.

"Trivial" is a word you need to use with caution; you have physical singularities (r=0) and coordinate singularities, and from the coordinate singularities you have ignorable ones (θ=0) and non-ignorable ones (r=2M).

 

[grey_earl]

I'm not saying that is not regular, I'm just saying that it can't be completely covered with one chart of polar coordinates.

Which point is not covered by the coordinate system I gave you?

 

[TrickyDicky]

So now you know that there are intrinsically curved manifolds where you only need one chart. The topology doesn't fully specify the metric; there are infinitely many possibilities to bestow a given topological manifold with a metric.

I might have misinterpreted this from some notes on diff.geometry:
"Note finally that any manifold M admits a finite atlas consisting of dimM +1 (not
connected) charts. This is a consequence of topological dimension theory [Nagata,
1965], a proof for manifolds may be found in [Greub-Halperin-Vanstone, Vol. I]."
I took this to meanthat any manifold with n-dimensions needs at least n+1 charts to cover all its points.

We didn't understand each other saying "trivial" :)
...
"Trivial" is a word you need to use with caution; you have physical singularities (r=0) and coordinate singularities, and from the coordinate singularities you have ignorable ones (θ=0) and non-ignorable ones (r=2M).

Ok, just one thing, are the terms "trivial" and "ignorable" referred to singularities as you use them standard mathematical notions? I had never heard about ignorable and non-ignorable singularities, but then I'm just a layperson.

 

[George Jones]

I might have misinterpreted this from some notes on diff.geometry:
"Note finally that any manifold M admits a finite atlas consisting of dimM +1 (not
connected) charts. This is a consequence of topological dimension theory [Nagata,
1965], a proof for manifolds may be found in [Greub-Halperin-Vanstone, Vol. I]."
I took this to meanthat any manifold with n-dimensions needs at least n+1 charts to cover all its points..

I am not familiar with this result, but here is my guess. dimM + 1 refers to the "dimensionality" of the charts, not to the actual number of charts. Suppose M has dimension n. Standard charts on M are maps from subsets of M into R^n. A dimM + 1 chart is something like a homeomorphism between an open subset U of M and its image in R^(n+1).

 

[grey_earl]

With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are examples where you can take less. Minkowski space (of dimension n) is such an example, it is a manifold of dimension n where you only need one chart to cover it completely (but of course you can find n+1 charts, or more). It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

"Ignorable" is not a standard mathematical term, it's physics parlor. Trivial is a standard mathematical term, but as in "that's trivial to show", not for coordinate singularities. Non-ignorable coordinate singularities define a horizon (which may be one of: event horizon, black hole horizon, null horizon, Killing horizon, ...), and that's a standard mathematical term you can look up.

 

[TrickyDicky]

With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are examples where you can take less. Minkowski space (of dimension n) is such an example, it is a manifold of dimension n where you only need one chart to cover it completely (but of course you can find n+1 charts, or more). It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

"Ignorable" is not a standard mathematical term, it's physics parlor. Trivial is a standard mathematical term, but as in "that's trivial to show", not for coordinate singularities. Non-ignorable coordinate singularities define a horizon (which may be one of: event horizon, black hole horizon, null horizon, Killing horizon, ...), and that's a standard mathematical term you can look up.

Thanks for the info. You are right that the theorem doesn't mean what I thought.
I know now that there are curved manifolds that can be specified with jst one chart, you may want to look up the thread on manifold charts I started in the Topology subforum.

Which point is not covered by the coordinate system I gave you?

Hmm... How about point phi=pi?, I believe the maps have to be bijective, and a single chart can't have both phi=0 and phi=pi.

 

[grey_earl]

Hmm... How about point phi=pi?, I believe the maps have to be bijective, and a single chart can't have both phi=0 and phi=pi.

True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there.
(if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})

 

[George Jones]

What I wrote was silly, but I guessed (wrongly) that the number was too small, e.g., that there probably exist 3-dimensional manifolds not coverable by four charts.

[EDIT]

It is an interesting theorem, though; I can see why for some manifolds you need at least n+1 charts, but not why this is sufficient. Let's look up this Greub-Halperin-Vanstone!

[/EDIT]

Not available to me.

 

[TrickyDicky]

True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there.
(if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})

No, I actually meant two opposite points in the S^1 part of the your manifold (if you picture it as a one-sheet hyperboloid, think of a point at the front and the diametrically opposite point in the back side of the hyperboloid)
I know for sure that S^1 needs two charts to specify all its points (proof in Tensor calculus by D. Kay, page 244) due to the bijective (one-one) character of the manifold maps and so I figured that if the slices of constant x have circular shape in the hyperboloid, it seems logical that this manifold also needs more than one chart. Only one chart would for some applications map the 2 opposite points in the circumference of a slice of constant x to the same real number.