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

[TrickyDicky]

I was wondering, if a metric known to be a solution of Eintein field equation can be a static or expanding spacetime depending on a change of coordinates, what does this tells us about that metric? Can it be physical?
I mean in the sense that it is usually said that to check if something is phyisical or the result of a coordinate peculiarity the best way is to see whether that property changes with a change of coordinates, for instance like it is often done to see if a singularity is physical or can be made to go away by a change of coordinates.
To be more specific I was thinking of de Sitter's and Milne's models. They share the property that both are empty universes, which maybe is giving us the clue that empty universes are not physical?

 

[Bill_K]

If peculiar coordinate systems meant the solution itself was unphysical that would rule out the Schwarzschild solution.

 

[TrickyDicky]

If peculiar coordinate systems meant the solution itself was unphysical that would rule out the Schwarzschild solution.

I don't understand this. The Schwarzschild solution is a vacuum solution, in which Riemannian curvature doesn't vanish (only Ricci curvature), not exactly the same as an "empty" universe solution in which Riemannian curvature vanishes and therefore is a flat spacetime.
So I don't know in what sense you think it has a peculiar coordinate system. Schwarzschild solution is static under any coordinate transformation, in the OP I referred specifically to solutions that can switch from static to expanding with a coordinate transformation.
AFAIK the vacuum is something physical.

 

[JesseM]

I was wondering, if a metric known to be a solution of Eintein field equation can be a static or expanding spacetime depending on a change of coordinates, what does this tells us about that metric? Can it be physical?

Not a "static or expanding spacetime" but a static or expanding space. The way that the shape of space changes dynamically over time depends entirely on your choice of simultaneity convention (slicing the spacetime into a series of spacelike surfaces in different ways). Once you have a simultaneity convention, for each spacelike surface you can calculate the frame-invariant proper distance along spacelike curves confined entirely to that surface, and use that to construct an http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag.html , and for each surface (represented by a dotted curve on the Kruskal-Szekeres diagram) there's a corresponding embedding diagram on the right:

It's exactly the same with the expanding Milne universe--because you are choosing a different simultaneity convention than the one in any inertial frame, the embedding diagram of a spacelike surface of simultaneity looks different (the spatial curvature in each surface is different), but the spacetime curvature is flat either way. From Ned Wright's cosmology tutorial, here's a diagram showing surfaces of simultaneity in the Milne universe (gray curves) as well as lines of constant cosmoving position (black line), when plotted in an ordinary Minkowski diagram where the horizontal axis is a surface of constant time in an inertial frame, and the vertical axis is a line of constant position in the inertial frame:

 

[Bill_K]

Schwarzschild solution is static under any coordinate transformation, in the OP I referred specifically to solutions that can switch from static to expanding with a coordinate transformation.

Take a look at Novikov coordinates, in which the Schwarzschild solution is time-dependent. Discussed in MTW on p.826.

 

[TrickyDicky]

Not a "static or expanding spacetime" but a static or expanding space.

Thanks for your nice explanation.
But actually I'm referring to spacetimes not to the spatial spacelike surface.
So I'm actually asking about "static spacetimes" (basically a space-time is static if it admits a hypersurface orthogonal timelike Killing vector field) like Schwarzschild or Minkowski, versus "non-static spacetimes" like those called "expanding":e.g. our universe (even if in these what expands is only the space).
The way you explain it makes it seem like expansion is a coordinate property of the spacelike surfaces that can go away with a coordinate transformation.
In fact, at least for Minkowski and de Sitter models , looks like even though they are static spacetimes they can acquire "expansion" (FRW versions of Minkowski (Milne) and de Sitter) thru a coordinate transformation.

 

[JesseM]

Thanks for your nice explanation.
But actually I'm referring to spacetimes not to the spatial spacelike surface.
So I'm actually asking about "static spacetimes" (basically a space-time is static if it admits a hypersurface orthogonal timelike Killing vector field) like Schwarzschild or Minkowski, versus "non-static spacetimes" like those called "expanding":e.g. our universe (even if in these what expands is only the space).

But under that definition an empty Milne universe is a static spacetime, no? It "admits a hypersurface orthogonal to the timelike Killing vector field" even if that hypersurface doesn't happen to be one of the hypersurfaces of simultaneity in the cosmological coordinate system used to describe the Milne universe (instead it could just be a hypersurface of simultaneity in an inertial frame in the same spacetime).

In fact, at least for Minkowski and de Sitter models , looks like even though they are static spacetimes they can acquire "expansion" (FRW versions of Minkowski (Milne) and de Sitter) thru a coordinate transformation.

But the "expansion" in this case wouldn't be a matter of making them non-static as I understand it, it would just be a property of one particular set of spacelike hypersurfaces.

 

[TrickyDicky]

But under that definition an empty Milne universe is a static spacetime, no? It "admits a hypersurface orthogonal to the timelike Killing vector field" even if that hypersurface doesn't happen to be one of the hypersurfaces of simultaneity in the cosmological coordinate system used to describe the Milne universe (instead it could just be a hypersurface of simultaneity in an inertial frame in the same spacetime).

Absolutely, that is my point.

But the "expansion" in this case wouldn't be a matter of making them non-static as I understand it, it would just be a property of one particular set of spacelike hypersurfaces.

Exactly, That is what I'm saying, they are still static spacetimes and yet they can have that "expanding" property with the right coordinates, isn't that odd?
I mean compare with the way a coordinate property (like a coordinate singularity) is usually treated.

 

[JesseM]

Exactly, That is what I'm saying, they are still static spacetimes and yet they can have that "expanding" property with the right coordinates, isn't that odd?

Well, it doesn't really seem odder to me than the relativity of simultaneity itself.

I mean compare with the way a coordinate property (like a coordinate singularity) is usually treated.

I suppose coordinate singularities are treated in a "dismissive" way precisely because in the past some authors thought they were real and some work had to be done to show this was a mistake, while there isn't really an analogous type of mistake that needs correcting in the case of expanding space. And you can also say that while simultaneity is coordinate-dependent and therefore arbitrary, once you have fixed a simultaneity convention, the notion of "distance" in any given hypersurface of simultaneity does have a coordinate-independent meaning in terms of proper distance along spacelike curves confined to that hypersurface.

 

[TrickyDicky]

Well, it doesn't really seem odder to me than the relativity of simultaneity itself.

If one intuitevely understands Lorentz transformations, I can see nothing odd in the relativity of simultaneity. In any case I don't see the relation between this SR concept and what I'm trying to clarify.

I suppose coordinate singularities are treated in a "dismissive" way precisely because in the past some authors thought they were real and some work had to be done to show this was a mistake, while there isn't really an analogous type of mistake that needs correcting in the case of expanding space.

But what "authors think should be real" is not a very rigorous way to solve things in mathematical physics, is it?
I mean that if a mathematical argument can be used to dismiss a singularity due to its coordinate dependence regardless what authors in general may prefer, why is this argument not valid to dismiss another coordinate property?
Consider Milne universe, according to its scale factor, radiation in this universe should undergo cosmological redshift. And yet we know this universe is a patch of static Minkowski spacetime, and static spacetimes by definition can't have cosmological redshift.
So is cosmological redshift in this universe a spurious coordinate property and there is really no redshift? Or alternatively we must admit that static spacetimes can have cosmological redshift?

 

[grey_earl]

In the Milne universe, radiation undergoes redshift because an "inertial" observer (what I really mean is an observer with constant spatial coordinates) in this universe is accelerating, which is (after the Einstein equivalence principle) equivalent to a gravitational field in what is seen and felt by observers. So this observer sees redshift because he himself is accelerating through spacetime, not because spacetime is expanding per se.
BTW, de Sitter spacetime isn't globally static. There is the so-called static coordinate patch, but this doesn't cover the whole de Sitter spacetime. Just like the Milne universe doesn't cover the whole Minkowski spacetime.
When you want to discuss things like "Is something physical?" _and_ have to use coordinate systems, it's best to use global ones. In the right coordinates, the Schwarzschild spacetime has no interior, so there is no black hole. But that's just because those coordinates aren't global.

 

[Samshorn]

... if a mathematical argument can be used to dismiss a singularity due to its coordinate dependance... why is this argument not valid to dismiss another coordinate property?

Coordinate singularities aren't "dismissed", they are simply recognized for what they are. What we "dismiss" is the erroneous belief that a coordinate singularity must represent a singularity of the manifold, such as singular intrinsic curvature. (By the way, this isn't a "mathematical argument", it is a physical argument; the application of coordinates to physical entities involves physical reasoning.)

Consider Milne universe, according to its scale factor, radiation in this universe should undergo cosmological redshift. And yet we know this universe is a patch of static Minkowski spacetime, and static spacetimes by definition can't have cosmological redshift. So is cosmological redshift in this universe a spurious coordinate property and there is really no redshift? Or alternatively we must admit that static spacetimes can have cosmological redshift?

You're confused. Redshift is a coordinate independent prediction of any given model, and nothing you've said conflicts with this fact. The "Milne universe" supposes that all substance originated at a single event in Minkowski spacetime, and is expanding outward from that event, with speeds distributed in a Lorentz-invariant way. On this basis we would expect each straight worldline emanating from the origin event to be receiving redshifted light from the surrounding (finite and spatially expanding) cloud of particles. On the other hand, we can describe this "universe" in terms of a curved spatial foliations, so that it is of infinite spatial extent with constant negative spatial curvature, we again find exactly the same predicted red shift.

It doesn't matter what coordinate system we use to evaluate a situation, the invariant predictions (like observed redshift) come out the same, because the choice of coordinates doesn't affect physically invariant things.

 

[TrickyDicky]

Coordinate singularities aren't "dismissed", they are simply recognized for what they are. What we "dismiss" is the erroneous belief that a coordinate singularity must represent a singularity of the manifold, such as singular intrinsic curvature. (By the way, this isn't a "mathematical argument", it is a physical argument; the application of coordinates to physical entities involves physical reasoning.)

lol

You're confused. Redshift is a coordinate independent prediction of any given model, and nothing you've said conflicts with this fact. The "Milne universe" supposes that all substance originated at a single event in Minkowski spacetime, and is expanding outward from that event, with speeds distributed in a Lorentz-invariant way. On this basis we would expect each straight worldline emanating from the origin event to be receiving redshifted light from the surrounding (finite and spatially expanding) cloud of particles. On the other hand, we can describe this "universe" in terms of a curved spatial foliations, so that it is of infinite spatial extent with constant negative spatial curvature, we again find exactly the same predicted red shift.

It doesn't matter what coordinate system we use to evaluate a situation, the invariant predictions (like observed redshift) come out the same, because the choice of coordinates doesn't affect physically invariant things.

Oh, this is ok with me but doesn't have anything to do with my post and my questions. Do you consider Milne universe (a patch of Minkowski spacetime) as a static spacetime or not?

 

[grey_earl]

Since the Milne universe is a patch of the static Minkowski spacetime, it must be static also, as I said before. However, it's not geodesically complete, that's one thing.
The second thing is that an observer "at rest" in the Milne universe (i.e. constant spatial coordinates) is accelerated, and so sees an apparent redshift. That's the second thing. (See p.ex. http://iopscience.iop.org/0264-9381/16/10/323/pdf/0264-9381_16_10_323.pdf or http://prd.aps.org/abstract/PRD/v55/i10/p6061_1)
This redshift is real in the sense that such a observer sees it, but its origin is a simple Doppler effect and not the expansion of spacetime per se.

 

[TrickyDicky]

In the Milne universe, radiation undergoes redshift because an "inertial" observer (what I really mean is an observer with constant spatial coordinates) in this universe is accelerating, which is (after the Einstein equivalence principle) equivalent to a gravitational field in what is seen and felt by observers. So this observer sees redshift because he himself is accelerating through spacetime, not because spacetime is expanding per se.
BTW, de Sitter spacetime isn't globally static. There is the so-called static coordinate patch, but this doesn't cover the whole de Sitter spacetime. Just like the Milne universe doesn't cover the whole Minkowski spacetime.
When you want to discuss things like "Is something physical?" _and_ have to use coordinate systems, it's best to use global ones. In the right coordinates, the Schwarzschild spacetime has no interior, so there is no black hole. But that's just because those coordinates aren't global.

Since the Milne universe is a patch of the static Minkowski spacetime, it must be static also, as I said before. However, it's not geodesically complete, that's one thing.
The second thing is that an observer "at rest" in the Milne universe (i.e. constant spatial coordinates) is accelerated, and so sees an apparent redshift. That's the second thing. (See p.ex. http://iopscience.iop.org/0264-9381/16/10/323/pdf/0264-9381_16_10_323.pdf or http://prd.aps.org/abstract/PRD/v55/i10/p6061_1)
This redshift is real in the sense that such a observer sees it, but its origin is a simple Doppler effect and not the expansion of spacetime per se.

Thanks, earl_gray, I found your posts very informative and to the point.
Some details that I think are worth clarifying further:
I agree on the mechanism of redshift (Doppler) in the Milne model, we actually only know two basic mechanisms of redshift anyway: gravitational and Doppler. In FRW cosmology, the cosmological redshift mechanism depends on the coordinates chosen and here at PF there have been endless discussions about whether it was Doppler, Gravitational or a mixture of the two depending on what coordinate system is used and the state of motion of the observers.
That said my point was that in the Milne model that you have stated in your post is a static spacetime, an observer sees a "cosmological redshift" (whatever the mechanism), the distinction of apparent versus real doesn't apply to the redshift since it is a simple measure.
So we have a static model that has both cosmological redshift, and "aparent" expansion, in this case the term apparent does apply since the expansion in this model is a coordinate effect in a static spacetime and expansion is not directly measurable as redshift is.
It is usually stated that static spacetimes can't have cosmological redshift (regardless of the mechanism of production of the redshift), is the Milne universe some kind of exception to this rule?

I also completely agree with you that when discussing the "physicality of something" one must consider it from the global coordinate system to get the maximally extended manifold.

 

[grey_earl]

One could argue that Doppler redshift is no "real" redshift, but I don't think that discussion is necessary. And gravitational redshift can have two causes: the expansion of spacetime or the "climbing out" from a static mass source in a static background. Both should be distinguished in a detailed discussion.

It is usually stated that static spacetimes can't have cosmological redshift (regardless of the mechanism of production of the redshift), is the Milne universe some kind of exception to this rule?

I haven't seen this claim before, but I also didn't do much cosmology. Maybe it was meant that in a static universe, if the observer's four-velocity agrees (to within a factor) with the Killing vector of stationarity, then there is no redshift, which would exclude the Milne universe. Maybe one has to substitute "geodesically complete static spacetime", which also would exclude the Milne universe. Do you have a source for this statement?

 

[TrickyDicky]

One could argue that Doppler redshift is no "real" redshift, but I don't think that discussion is necessary. And gravitational redshift can have two causes: the expansion of spacetime or the "climbing out" from a static mass source in a static background. Both should be distinguished in a detailed discussion.

spcetime expansion is not usually considered a cause of gravitational redshift in standard cosmology, it is considered the cause of "cosmological redshift". The specific mechanism (Doppler, Gravitational or both, depends on coordinates choices as I said.

I haven't seen this claim before, but I also didn't do much cosmology. Maybe it was meant that in a static universe, if the observer's four-velocity agrees (to within a factor) with the Killing vector of stationarity, then there is no redshift, which would exclude the Milne universe. Maybe one has to substitute "geodesically complete static spacetime", which also would exclude the Milne universe. Do you have a source for this statement?

Well, it's a consequence of attributing cosmological redshift to expansion.
Maybe those reasons exclude Milne universe from the general claim.

 

[TrickyDicky]

So it would be interesting to know if the general claim "static spacetimes can't have cosmological redshift" is an accurate statement according to current cosmology or it needs to be constrained to specific conditions and is not true in general. Anyone?

 

[Mentz114]

So it would be interesting to know if the general claim "static spacetimes can't have cosmological redshift" is an accurate statement according to current cosmology or it needs to be constrained to specific conditions and is not true in general. Anyone?

I don't know if this helps, but the general formula for the frequency ratio between emitter and receiver is (A,B label two observers)
$$\frac{\nu_A}{\nu_B}=\frac{(g_{ab}k^au^b)_A}{(g_{ab}k^au^b)_B}$$
where $k^\mu$ is the null geodesic connecting A and B and $u^\mu$ the four-velocity of the observer. The conditions for $k^\mu$ to be a null geodesic are that $\nabla_\nu k_\mu k^\nu = 0$ and $g^{\mu\nu} k_\mu k_\nu = 0$ .

It is clear that the contributions from the velocity and the metric cannot be separated in a unique way. But if A and B have the same velocity then only contributions from the metric will be present.

That formula gives the special relativistic Doppler effect if $k^\mu=(1,1,0,0)$, $u^\mu_{(A)}=(1,0,0,0)$ and $u^\mu_{(B)}=(\gamma,\gamma\beta,0,0)$ ( with $g^{\mu\nu}=\eta^{\mu\nu}$).

 

[TrickyDicky]

It is clear that the contributions from the velocity and the metric cannot be separated in a unique way. But if A and B have the same velocity then only contributions from the metric will be present.

Sure, but this only rules out a purely Doppler cosmological redshift in a static spacetime. Most people will not agree that cosmological redshift is exactly the same as the Doppler effect.
Contributions from the metric seem to take some part in cosmological redshift too.

 

[TrickyDicky]

When you want to discuss things like "Is something physical?" _and_ have to use coordinate systems, it's best to use global ones. In the right coordinates, the Schwarzschild spacetime has no interior, so there is no black hole. But that's just because those coordinates aren't global.

How do you make this distinction between local and global coordinates? GR is a local geometry theory and the metrics that are solutions to the GR equations are always expressed in "local" coordinates (3,1) because they are describing the intrinsic spacetime curvature. Global cordinates would have to belong to a embedding in higher dimension coordinates, at least that is what they do with de Sitter and anti de Sitter spaces, they are often described in terms of a 5 coordinate embedding to cover the whole spacetimes.

 

[Mentz114]

How do you make this distinction between local and global coordinates? GR is a local geometry theory and the metrics that are solutions to the GR equations are always expressed in "local" coordinates (3,1) because they are describing the intrinsic spacetime curvature. Global cordinates would have to belong to a embedding in higher dimension coordinates, at least that is what they do with de Sitter and anti de Sitter spaces, they are often described in terms of a 5 coordinate embedding to cover the whole spacetimes.

My understanding is that the metric is expressed in global coordinates that cover some or all of the manifold. Local coordinates are related to worldlines, with a local Minkowski frame at every point on a timelike worldline. The transformation between global and local frame coordinates is done with a change of basis using the frame field ( a tetrad ).

 

[TrickyDicky]

My understanding is that the metric is expressed in global coordinates that cover some or all of the manifold.

There seems to be some terminology confusion here. If the coordinates cover a patch of the manifold they are not global, in this sense global coordinates refer to those that cover the whole manifold, which in the case of a curved manifold unlike in flat manifolds would need to have more coordinates than the original manifold patch coordinates.

 

[TrickyDicky]

Local coordinates are related to worldlines, with a local Minkowski frame at every point on a timelike worldline. The transformation between global and local frame coordinates is done with a change of basis using the frame field ( a tetrad ).

These local coordinates refer to local inertial coordinates, or Minlowski local frames at infinitesimal points of a GR manifold by the Equivalence principle. When I said local coordinates I meant what you call global coordinates, or the usual coordinates of the line element that cover a patch of the manifold. And by global coordinates I meant those that determine the topology and cover the whole manifold.

 

[grey_earl]

How do you make this distinction between local and global coordinates? GR is a local geometry theory and the metrics that are solutions to the GR equations are always expressed in "local" coordinates (3,1) because they are describing the intrinsic spacetime curvature. Global cordinates would have to belong to a embedding in higher dimension coordinates, at least that is what they do with de Sitter and anti de Sitter spaces, they are often described in terms of a 5 coordinate embedding to cover the whole spacetimes.

(To avoid confusion: space in the following is short for spacetime.)

Global coordinates for me are coordinates that cover the whole manifold, p.ex. cartesian coordinates for Minkowski space. Every point of Minkowski space is described by one coordinate tuple (x,y,z,t), and to every coordinate tupel there belongs a point of Minkowski space. Coordinates which are not global for Minkowski space are for example Rindler coordinates: every tuple of Rindler coordinates corresponds to a point in Minkowski space, but there are points in Minkowski space which do not have a corresponding tuple. "Local" coordinates don't have meaning in that concept; coordinates are either global or not. This does not imply that we cannot describe space locally, in the neighbourhood of one point, with a coordinate system and in the neighbourhood of another point with another.

An example of global coordinates for de Sitter space is given by ds² = - dt² + H^{-2} cosh²(H t) dΩ², where dΩ² is the metric of the 3-sphere. We don't need an embedding in a higher-dimensional space for global coordinates; but there are spaces which don't admit global coordinates (I think the Kerr black hole falls in this class).

 

[TrickyDicky]

(To avoid confusion: space in the following is short for spacetime.)

Global coordinates for me are coordinates that cover the whole manifold, p.ex. cartesian coordinates for Minkowski space. Every point of Minkowski space is described by one coordinate tuple (x,y,z,t), and to every coordinate tupel there belongs a point of Minkowski space. Coordinates which are not global for Minkowski space are for example Rindler coordinates: every tuple of Rindler coordinates corresponds to a point in Minkowski space, but there are points in Minkowski space which do not have a corresponding tuple.

I think it is useful to make the distinction I made earlier between flat and curved manifolds. The example of Minkowski space is relevant, being a flat manifold you can cover the whole manifold with a 4 coordinate metric, the same happens with Euclidean space that can be covered with three coordinates or the plane with two. However in the case of a two sphere for instance the line elements expressed in 2 coordinates cover patches of the total manifold and there are always points like at the equator or at the poles dpending on the specific coordinates where one finds coordinate singularities, where you have to transform coordinates to a different overlapping patch of the manifold.
So in curved manifolds, (unless some hidden symmetry is exploited) I would say global coordinates need one more coordinate to cover the whole manifold.

An example of global coordinates for de Sitter space is given by ds² = - dt² + H^{-2} cosh²(H t) dΩ², where dΩ² is the metric of the 3-sphere. We don't need an embedding in a higher-dimensional space for global coordinates; but there are spaces which don't admit global coordinates (I think the Kerr black hole falls in this class).

Yes, those are de sitter global coordinates, and are a Minkowkian 5-dimensional ambient space restricted to dS^4 (see http://www.bourbaphy.fr/moschella.pdf equations 3-8)
Note de sitter geometry is very special, being maximally symmetric.

 

[grey_earl]

However in the case of a two sphere for instance the line elements expressed in 2 coordinates cover patches of the total manifold and there are always points like at the equator or at the poles dpending on the specific coordinates where one finds coordinate singularities, where you have to transform coordinates to a different overlapping patch of the manifold.

True; technically this is a coordinate singularity is you use polar coordinates for the sphere. However, you can for example obtain its area by integrating dA = sin θ dθ dφ over the whole sphere, and so it's an integrable singularity. Those typically don't cause any problems, so let's include them in the definition of global coordinates: coordinates which at most have integrable singularities.

Yes, those are de sitter global coordinates, and are a Minkowkian 5-dimensional ambient space restricted to dS^4 (see http://www.bourbaphy.fr/moschella.pdf equations 3-8)
Note de sitter geometry is very special, being maximally symmetric.

You can define de Sitter space by the embedding into a 5-dimensional manifold and can of course get global coordinates by restricting 5-dimensional global coordinates to the de Sitter manifold in a reasonable way, but the point is that you don't need to do it. You can as well define de Sitter space by saying that topologically it is R x S³ and by giving its symmetry group. Using coordinates adapted to the topology you then get the coordinates I gave you, unambigously and without the need to embed de Sitter in any higher-dimensional space.

 

[TrickyDicky]

True; technically this is a coordinate singularity is you use polar coordinates for the sphere. However, you can for example obtain its area by integrating dA = sin θ dθ dφ over the whole sphere, and so it's an integrable singularity. Those typically don't cause any problems, so let's include them in the definition of global coordinates: coordinates which at most have integrable singularities.

We basically agree, as always there is a nagging problem of terminology.
I was using the word "gobal" in a more topological than geometrical way. And in this sense the coordinate systems on n=manifold dimensionality can never be global, you need the embedding. But I admit this is kind of trivial and in the definition of manifold.
The conclusion is that within the dimensionality n of the manifold at hand, in the case of the manifold being intrinsically curved, one can never cover all the points of the manifold with coordinates of the same dimensionality n of the manifold, this follows from the fact that the line element is a infinitesimal Euclidean (or Minkowskian, but flat in any case) local representation of the manifold, and one can't make something curved fit a flat space without tearing it as cartographers know well. However, within the choice of coordinates for a given manifold the choice goes from coordinates that cover a small part of the manifols to coordinates that cover almost all the points of the manifold depending on the manifold and the ability to find coordinates.

You can define de Sitter space by the embedding into a 5-dimensional manifold and can of course get global coordinates by restricting 5-dimensional global coordinates to the de Sitter manifold in a reasonable way, but the point is that you don't need to do it. You can as well define de Sitter space by saying that topologically it is R x S³ and by giving its symmetry group. Using coordinates adapted to the topology you then get the coordinates I gave you, unambigously and without the need to embed de Sitter in any higher-dimensional space.

You are right. But note that even if you don't need to embed it, actually the coordinates in the line element of the manifold you gave was defined from the Minkowskian 5-dimensional ambient pace, the parameter H^-2 or 1/R^2 belongs to 5-dimension space.
Like in the 2-sphere case you find the trivial singularities for the r=0, r=pi, Theta=0, Theta=pi... if you express the 3-sphere spatial part in spherical coordinates or the corresponding ones if you express it in other coordinates.
BTW,you mentioned in a previous post that in the case of the Schwazschild manifold:"In the right coordinates, the Schwarzschild spacetime has no interior, so there is no black hole. But that's just because those coordinates aren't global." 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"

 

[grey_earl]

The parameter H is intrinsically defined in de Sitter space: it's R/12 in the four-dimensional case, where R is the curvature scalar. I don't need any embedding for it, period.
If you want to embed de Sitter space, then H is the radius as seen from the embedding space, but H is not a five-dimensional parameter, as you put it. That's just not true.

To counter your statement that no curved manifolds exist without coordinate singularities: 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.

The case θ=0 is very different from the case r=2M. Calculate, for example, the geodesic equation for a particle traveling outwards, from r = r_0 to infinity, at any θ=const, including θ=0. You don't have any problem as long as r_0 > 2M, it doesn't matter if θ=0 or θ=π/2, or any other value. For r_0 = 2M, however, you have a problem.
That's why the coordinate singularity at θ=0 is ignorable, but the one at r=2M is not.
(calculation of Christoffel symbols usually is a mess, so here they are, for your convenience: http://theory.gsi.de/~vanhees/faq/gravitation/node65.html)

 

[TrickyDicky]

The parameter H is intrinsically defined in de Sitter space: it's R/12 in the four-dimensional case, where R is the curvature scalar. I don't need any embedding for it, period.
If you want to embed de Sitter space, then H is the radius as seen from the embedding space, but H is not a five-dimensional parameter, as you put it. That's just not true.

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 didn't deny the 4-dimensionality of the topology of the manifold, I was just alluding to the way that line element is usually defined in order to be easily visualized. Just see the WP page for de Sitter space or "The large scale structure of space-time" by Hawking and Ellis pg. 124.

To counter your statement that no curved manifolds exist without coordinate singularities: 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.

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.

The case θ=0 is very different from the case r=2M. Calculate, for example, the geodesic equation for a particle traveling outwards, from r = r_0 to infinity, at any θ=const, including θ=0. You don't have any problem as long as r_0 > 2M, it doesn't matter if θ=0 or θ=π/2, or any other value. For r_0 = 2M, however, you have a problem.
That's why the coordinate singularity at θ=0 is ignorable, but the one at r=2M is not.

I'm not saying those two cases are the same , 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.

 

[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?