Universe Hubble radius equal to Schwarzschild radius

[bcrowell]

But still, in this particular case, the statements "at $r=0$ there is a singularity" and "the domain of the metric does not include $r=0$" and "the metric is not well-defined at $r=0$" are equivalent?

No. The most fundamental reason that such a definition doesn't work is that the singularity is not a point or point-set in the spacetime manifold. Therefore it doesn't make sense to ask whether the metric is defined "there." There's "no there there" at which we could even ask whether the metric is defined. Specifying $r=0$ does not specify a set of points in the manifold; those values of the $r$ coordinate are not part of the coordinate chart.

Note that the metric could also be undefined at a certain coordinate value without there being any singularity. For example, when we express the metric of the Schwarzschild spacetime in the Schwarzschild coordinates, it's undefined at the event horizon, but this is not considered a singularity. (People describe it as a coordinate singularity, but that just means not a real singularity.)

 

[Smattering]

Yes. The technical term is "manifold with metric".

O.k., now things seem to become clearer for me.

So since this Schwarzschild case is completely symmetrical, the only parameters that the metric depends on in this particular case are that radial coordinate $r$ and some time coordinate $t$?

And inside the event horizon, there are no wordlines with constant $r$ just in the same sense as there are no wordlines with constant $t$ outside of the event horizon?

 

[PeterDonis]

the only parameters that the metric depends on in this particular case are that radial coordinate $r$ and some time coordinate ##t##?

The metric doesn't depend on $t$. The only coordinate it depends on is $r$.

inside the event horizon, there are no wordlines with constant $r$ just in the same sense as there are no wordlines with constant $t$ outside of the event horizon?

Sort of. There are no worldlines with constant $t$ inside the horizon either, so the two coordinates are not symmetric in this respect.

 

[Smattering]

No. The most fundamental reason that such a definition doesn't work is that the singularity is not a point or point-set in the spacetime manifold. Therefore it doesn't make sense to ask whether the metric is defined "there." There's "no there there" at which we could even ask whether the metric is defined. Specifying $r=0$ does not specify a set of points in the manifold; those values of the $r$ coordinate are not part of the coordinate chart.

Hm ... how does this radial coordinate $r$ translate to the spacetime manifold then?

Note that the metric could also be undefined at a certain coordinate value without there being any singularity. For example, when we express the metric of the Schwarzschild spacetime in the Schwarzschild coordinates, it's undefined at the event horizon, but this is not considered a singularity. (People describe it as a coordinate singularity, but that just means not a real singularity.)

Yes, I read that. But as far as I understood, this "coordinate singularity" can be eliminated via a coordinate transformation, whereas there exists no coordinate transformation that can eliminate the singularity at $r=0$.

 

[bcrowell]

Hm ... how does this radial coordinate $r$ translate to the spacetime manifold then?

I'm not sure what you're asking here. The coordinate r, or its specific value r=0? We can't normally cover an entire manifold with a single coordinate chart. The inability to do so is not a big crisis and occurs even in well-behaved manifolds such as a sphere.

Yes, I read that. But as far as I understood, this "coordinate singularity" can be eliminated via a coordinate transformation, whereas there exists no coordinate transformation that can eliminate the singularity at $r=0$.

Yes, that's correct, but I don't see how that affects the logic of the discussion. The point is that if you want to define a singularity in GR, it is neither necessary nor sufficient for the metric to be undefined at a particular coordinate value. This is why we define a singularity in GR in terms of geodesic incompleteness.

 

[Smattering]

I'm not sure what you're asking here. The coordinate r, or its specific value r=0? We can't normally cover an entire manifold with a single coordinate chart.

Then let me rephrase this:

1. What is the domain of the Schwarzschild metric?
2. How is this domain related to the spacetime manifold (or a specific part of it)?

Yes, that's correct, but I don't see how that affects the logic of the discussion. The point is that if you want to define a singularity in GR, it is neither necessary nor sufficient for the metric to be undefined at a particular coordinate value. This is why we define a singularity in GR in terms of geodesic incompleteness.

O.k., if that is the convention, then let's stick to it, but still I wonder whether geodesic incompleteness and the inability to eliminate a singularity by coordinate transformation is or is not equivalent.

 

[bcrowell]

Then let me rephrase this:

1. What is the domain of the Schwarzschild metric?
2. How is this domain related to the spacetime manifold (or a specific part of it)?

1. The domain is the entire spacetime manifold. The manifold is not defined in terms of a specific coordinate chart, but if you want to characterize it in terms of the Schwarzschild coordinates, $r=0$ is absent.

2. The domain is the entire spacetime manifold.

O.k., if that is the convention, then let's stick to it, but still I wonder whether geodesic incompleteness and the inability to eliminate a singularity by coordinate transformation is or is not equivalent.

You can't define a singularity as the inability to eliminate a singularity. That would be like defining a lion as a lion with four legs. Nobody would have bothered to make a definition in terms of geodesic completeness if there were some simpler definition.

The questions you're asking are all good ones, but we're spending a lot of time undoing your preconceptions (which are natural preconceptions) and reiterating material that is found in standard textbooks. There is a limit to how much we can achieve through Socratic dialog. If you could tell us a little about your background in math and physics, we could recommend a book at the right level for you that would contain a systematic treatment of these issues. A couple of good possibilities at the graduate level are Wald and Carroll. At a lower level of mathematical sophistication, my own GR book is free online: http://www.lightandmatter.com/genrel/. There is a partial version of Carroll that is free online: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html

 

[Smattering]

1. The domain is the entire spacetime manifold. The manifold is not defined in terms of a specific coordinate chart, but if you want to characterize it in terms of the Schwarzschild coordinates, $r=0$ is absent.

2. The domain is the entire spacetime manifold.

Then, I do not understand the objections you made here:
https://www.physicsforums.com/threa...hwarzschild-radius.841320/page-2#post-5282982

Because, when the domain is the entire spacetime manifold, then "not being part of the domain" should be exactly the same as "not being part of the spacetime manifold".

You can't define a singularity as the inability to eliminate a singularity. That would be like defining a lion as a lion with four legs.

But I might be able to define a coordinate singularity as a singularity that can be removed by a coordinate transformation in the same way as I can define a waterfowl as a fowl that can swim. I really do not see any principle issue here.

Nobody would have bothered to make a definition in terms of geodesic completeness if there were some simpler definition.

I am not saying that the other definition is simpler.

The questions you're asking are all good ones, but we're spending a lot of time undoing your preconceptions (which are natural preconceptions) and reiterating material that is found in standard textbooks. There is a limit to how much we can achieve through Socratic dialog. If you could tell us a little about your background in math and physics, we could recommend a book at the right level for you that would contain a systematic treatment of these issues. A couple of good possibilities at the graduate level are Wald and Carroll. At a lower level of mathematical sophistication, my own GR book is free online: http://www.lightandmatter.com/genrel/ . There is a partial version of Carroll that is free online: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html

O.k., thanks for the time you spent. I will have a look at this material.

I do not have any mentionable physics background apart from what I can remember from school (otherwise I would not be asking such questions, would I ;-) ), but I have the German equivalent to a Master's degree in computer science. So I have a bit of math background, but with an focus on discrete math and certainly no differential geometry.

 

[PeterDonis]

An online version of Carroll's lecture notes is here:

http://arxiv.org/abs/gr-qc/9712019

AFAIK this is the complete notes.

 

[PeterDonis]

when the domain is the entire spacetime manifold, then "not being part of the domain" should be exactly the same as "not being part of the spacetime manifold"

He wasn't objecting to that; he was objecting to your statement that "the metric is not well-defined" at the singularity. You can't even ask whether the metric is well-defined someplace that isn't part of the spacetime manifold.

 

[PeterDonis]

What is the domain of the Schwarzschild metric?

It's important to distinguish "the Schwarzschild metric" from "the Schwarzschild coordinate chart". The "Schwarzschild metric", properly speaking, is just another name for "the Schwarzschild geometry", a geometric object that exists independently of any choice of coordinates. (I prefer the term "Schwarzschild spacetime" for this object, to avoid confusion.) But many sources will use the term "the Schwarzschild metric", when what they really mean is "the Schwarzschild coordinate chart", the coordinates $t, r, \theta, \phi$, in which the line element looks like

$$
ds^2 = \left(1 - \frac{2M}{r} \right) dt^2 + \frac{1}{1 - \frac{2M}{r}} dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2
$$

These coordinates are often used, but they have a coordinate singularity at $r = 2M$, so strictly speaking there are actually two of these charts, one for $r > 2M$ and one for $0 < r < 2M$, and they are disconnected. To remove the coordinate singularity, we would need to choose a different chart which is nonsingular for all $r > 0$; examples are the Painleve chart and the Eddington-Finkelstein chart. But all of these charts only cover $r > 0$, because the spacetime itself--the geometric object that exists independently of any choice of coordinates--only contains points for which $r > 0$.

Another source of confusion is the fact that, as my last statement implied, $r$ has a double meaning. It is a coordinate which appears in all of the charts I mentioned; but it is also a physical property of a point in Schwarzschild spacetime. Every point in this spacetime lies on a 2-sphere with a well-defined physical area $A$, so we can label points by the area $A$ of the 2-sphere they lie on. For reasons which are too long to fit in the margin of this post, physicists prefer instead to use the label $r = \sqrt{A / 4 \pi}$, the "areal radius" (the radius that a 2-sphere in Euclidean 3-space with area $A$ would have). It is this second meaning of $r$ that I used in the last sentence of the previous paragraph, which could be rephrased as: Schwarzschild spacetime only contains 2-spheres with physical area $A$ that is positive.

 

[bcrowell]

Because, when the domain is the entire spacetime manifold, then "not being part of the domain" should be exactly the same as "not being part of the spacetime manifold".

Yes, that's correct.

But I might be able to define a coordinate singularity as a singularity that can be removed by a coordinate transformation in the same way as I can define a waterfowl as a fowl that can swim. I really do not see any principle issue here.

Using your analogy, you haven't defined "fowl," i.e., "singularity."

 

[bcrowell]

An online version of Carroll's lecture notes is here:

http://arxiv.org/abs/gr-qc/9712019

AFAIK this is the complete notes.

He wrote a GR textbook based on the notes. The notes are not the complete textbook.

O.k., thanks for the time you spent. I will have a look at this material.

Cool. Here are some of the topics you probably want to look at:

* manifold

* chart

* atlas

* singularity

* geodesic incompleteness

Wald introduces these very systematically and early on, but Wald is the hardest of the three books, and it's also not free online.

 

[Smattering]

He wasn't objecting to that; he was objecting to your statement that "the metric is not well-defined" at the singularity. You can't even ask whether the metric is well-defined someplace that isn't part of the spacetime manifold.

Fair enough since being "well-defined" implies that something is defined at all, but as $r=0$ is not part of the domain, the metric is simply undefined at $r=0$.

 

[Smattering]

One other question:

Is there a solution of the Einstein field equations for a completely homogeneous universe where all the mass/energy is uniformly distributed?

 

[bcrowell]

Is there a solution of the Einstein field equations for a completely homogeneous universe where all the mass/energy is uniformly distributed?

Yes. All the standard cosmological solutions are homogeneous and isotropic.

 

[PeterDonis]

as $r=0$ is not part of the domain, the metric is simply undefined at $r=0$.

Even that might be going further than is justified. Saying that $r = 0$ "is not part of the domain" and that the metric is "undefined" there implies that $r = 0$ is still "somewhere", it just doesn't happen to be somewhere that is "part of the domain". That's not correct. The label $r = 0$ does not refer to anywhere. At best, it can be construed as referring to limits that can be taken as $r \rightarrow 0$ along curves in the manifold that traverse all positive values of $r$. But there is no "place" to which those limits refer, not even one that "isn't part of the manifold".

 

[PeterDonis]

Is there a solution of the Einstein field equations for a completely homogeneous universe

As bcrowell says, the standard cosmological solutions are homogeneous and isotropic. However, "homogeneous" here means spatially homogeneous--more precisely, there is a family of observers in these spacetimes (these observers are called "comoving" observers) to whom, at every instant of their time, the spatial slice of the universe at that time is homogeneous (and isotropic).

If by "homogeneous" you mean "homogeneous in space and time", i.e., the density of mass/energy is the same everywhere and also at all times, then there is a solution with that property called the Einstein static universe. However, this solution is unstable against small perturbations, like a pencil balanced on its point; any tiny fluctuation in the density of mass/energy anywhere in the universe will cause it to either expand or collapse.

 

[Smattering]

Yes. All the standard cosmological solutions are homogeneous and isotropic.

I thought they were only homogeneous on cosmic scales. Because locally (e.g. our the solar system), the mass is obviously not uniformly distributed.

 

[Smattering]

As bcrowell says, the standard cosmological solutions are homogeneous and isotropic. However, "homogeneous" here means spatially homogeneous--more precisely, there is a family of observers in these spacetimes (these observers are called "comoving" observers) to whom, at every instant of their time, the spatial slice of the universe at that time is homogeneous (and isotropic).

So the cosmological solutions are just ignoring any local deviations and treat the universe as entirely homogeneous not only on cosmic scales but also locally?

Yes, then this it what I was referring to.

 

[PeterDonis]

the cosmological solutions are just ignoring any local deviations and treat the universe as entirely homogeneous not only on cosmic scales but also locally?

Mathematically, yes, the solutions treat the universe as homogeneous on all scales.

Physically, the solutions are not intended to treat the universe as homogeneous on, say, the scale of the Milky Way galaxy; they are simply not applied on that scale. The density of mass/energy in the solutions is not intended to describe an actual continuous distribution of matter; it is treated similarly to the density in fluid dynamics, where the fluid is treated as continuous mathematically but everyone understands that it's actually made up of molecules, so the fluid solution is not applicable on those scales; it's only applicable on scales much larger than the scale of molecules. In the same way, the solutions in cosmology that have the same density of mass/energy everywhere in space are only applicable on scales much larger than the scale of galaxies and galaxy clusters, which are the "molecules" of the "fluid" of matter/energy in the universe.

 

[Smattering]

If by "homogeneous" you mean "homogeneous in space and time", i.e., the density of mass/energy is the same everywhere and also at all times, then there is a solution with that property called the Einstein static universe. However, this solution is unstable against small perturbations, like a pencil balanced on its point; any tiny fluctuation in the density of mass/energy anywhere in the universe will cause it to either expand or collapse.

This is not what I actually meant, but does it imply that in an entirely homogeneous and isotropic universe, we could fine-tune the cosmological constant such that the curvature of spacetime completely disappears and it behaves like Minkowski spacetime?

 

[PeterDonis]

does it imply that in an entirely homogeneous and isotropic universe, we could fine-tune the cosmological constant such that the curvature of spacetime completely disappears and it behaves like Minkowski spacetime?

No. The only way to get flat Minkowski spacetime is to have zero density of mass/energy everywhere and zero cosmological constant everywhere. The Einstein static universe is still a curved spacetime, even though it doesn't expand or contract with time. One manifestation of this is that spatially, the Einstein static universe is a 3-sphere, not a Euclidean 3-space.

 

[Smattering]

No. The only way to get flat Minkowski spacetime is to have zero density of mass/energy everywhere and zero cosmological constant everywhere.

O.k., then when they say that *our* universe is approx. flat on large scales, this only means that our universe is too large to measure the curvature with the available methods?

The Einstein static universe is still a curved spacetime, even though it doesn't expand or contract with time. One manifestation of this is that spatially, the Einstein static universe is a 3-sphere, not a Euclidean 3-space.

So which property leads to this topology? Is it the homogenity and isotropy, or is it the cosmological constant, or is it a combination of both?

 

[bcrowell]

O.k., then when they say that *our* universe is approx. flat on large scales, this only means that our universe is too large to measure the curvature with the available methods?

When people say that our universe is approximately flat on large scales, they're referring to *spatial* flatness only. The Riemann tensor measures the curvature of spacetime, not just space. Our universe's spacetime is not even approximately flat. Spacetime curvature is how GR describes gravity. A universe with flat spacetime would be one in which there are no gravitational effects whatsoever.

So which property leads to this topology? Is it the homogenity and isotropy, or is it the cosmological constant, or is it a combination of both?

The Einstein field equations relate the curvature to the stress-energy tensor. The cosmological constant can be treated as one term in the stress-energy tensor. There are other terms as well, such as a term for dark matter and one for baryonic matter. Given the curvature of a manifold, there are theorems that in some cases uniquely determine the topology that is consistent with that curvature. This is one such case.

 

[Smattering]

Hi bcrowell,

thank you so much for your answer.

When people say that our universe is approximately flat on large scales, they're referring to *spatial* flatness only. The Riemann tensor measures the curvature of spacetime, not just space. Our universe's spacetime is not even approximately flat. Spacetime curvature is how GR describes gravity. A universe with flat spacetime would be one in which there are no gravitational effects whatsoever.

Yes, I was aware that our universe can certainly not be completely flat due to the fact that we can observe lots of gravitational effects all around us. The question was only related to the curvature on very large scales.

I was also aware that the curvature described by GR is a curvature of 3+1-dimensional spacetime. This is quite obvious, because otherwise the geodesic line of an object would be independent of its relative speed, wouldn't it?

But I have to confess that until now I would have thought that a curvature of spacetime would also imply a spatial curvature in most cases. I can even believe that there might be some special cases where the spacetime has a curvature although there is no spatial curvature. But I would not have thought that the universe can be spatially flat on cosmological scales, and still the spacetime has a curvature on the same scale. But maybe I will understand this once I have read the material you suggested.

The Einstein field equations relate the curvature to the stress-energy tensor. The cosmological constant can be treated as one term in the stress-energy tensor. There are other terms as well, such as a term for dark matter and one for baryonic matter. Given the curvature of a manifold, there are theorems that in some cases uniquely determine the topology that is consistent with that curvature. This is one such case.

Yes, that makes sense. In the case of an ordinary sphere, I can easily imagine that there might be no other topology that fits to that global curvature. However, I was rather wondering how this particular curvature results in case of Einstein's static universe. Is it important that it is static, or does a non-staic cosmological model (as you mentioned it before) also have the same topology?

 

[PeterDonis]

I was aware that our universe can certainly not be completely flat due to the fact that we can observe lots of gravitational effects all around us. The question was only related to the curvature on very large scales.

You're still confusing space curvature with spacetime curvature. The gravitational effects we observe around us don't necessarily tell us that space is curved; that depends on how we split up spacetime into space and time. But the gravitational effects we observe do tell us unequivocally that spacetime is curved.

On the scale of the entire universe, as best we can tell from current observations, the universe is spatially flat, if we use the splitting of spacetime into space and time that is the "natural" one for observers who see the universe as homogeneous and isotropic. In other words, those observers would also observe the universe to be spatially flat. However, other observers in different states of motion, who would naturally split spacetime into space and time in a different way, might not observe the universe to be spatially flat. But the spacetime of the universe is curved regardless.

otherwise the geodesic line of an object would be independent of its relative speed

I'm not sure I understand what you mean by this. If you are trying to describe geodesic deviation--the fact that geodesics that are parallel at some particular point don't stay parallel--then yes, that is the definitive sign of spacetime curvature.

I have to confess that until now I would have thought that a curvature of spacetime would also imply a spatial curvature in most cases.

Whether space is curved depends on how we split up spacetime into space and time (see above). There is no absolute sense in which space is curved or flat.

 

[Smattering]

I'm not sure I understand what you mean by this. If you are trying to describe geodesic deviation--the fact that geodesics that are parallel at some particular point don't stay parallel--then yes, that is the definitive sign of spacetime curvature.

No, I was referring to something different. Let's assume I am throwing stones in a vacuum (jsut to get rid of aerodynamic effects). Let's further assume that I throw all the stone in the eaxctly same direction and angle:

If the curvature was purely spatial, then I would expect all the stones to follow the exactly same trajectory indepently of their initial velocity, wouldn't I? But in reality, the trajectories differ with the initial velocity. Thus, I conclude that the curvature cannot pe purely spatial, but it must also have some temporal component.

Whether space is curved depends on how we split up spacetime into space and time (see above). There is no absolute sense in which space is curved or flat.

O.k., this is something I was not aware of. Now I would like to have a brain with more than 3 dimensions.

I was trying to imagine a 2+1-dimensional spacetime, but still I am lacking at least one dimension to embed the curvature. ;-)

 

[martinbn]

[QUOTE="Smattering, post: 5284699, member: 576347".
I was trying to imagine a 2+1-dimensional spacetime, but still I am lacking at least one dimension to embed the curvature. ;-)[/QUOTE]

Use 2+1 flat space-time. The usual slicing in 2d planes would be an example of flat space through time. If you slice it differently say with a small bump then it will be non flat space through time. Imagine the slices are thin metal sheets. Stack them and hit them so that all get a bump but fit together (you have to be careful not to change the space-like nature of the slices).

 

[PeterDonis]

Let's assume I am throwing stones in a vacuum (jsut to get rid of aerodynamic effects). Let's further assume that I throw all the stone in the eaxctly same direction and angle:

Same direction and angle, but different velocities, all launched at the same time, correct?

If the curvature was purely spatial, then I would expect all the stones to follow the exactly same trajectory indepently of their initial velocity, wouldn't I?

No. First, we have to assume a specific splitting of spacetime into space and time, so that we can determine that the curvature is "purely spatial". Given that, stones launched at different velocities will spend different amounts of time in a region with a given spatial curvature, so the spatial curvature will bend their trajectories by different amounts.

in reality, the trajectories differ with the initial velocity.

"In reality", there is no invariant way to split up spacetime curvature into "space curvature" and "time curvature", so there's no way to tell what part of the difference is due to "space curvature" and what part is due to "time curvature". All you can say is that it's due to spacetime curvature.