How GR Resolves the Conservation of Momentum and Energy

[Haelfix]

Laws of physics accordint to what specific theory? GR? I thought you said forget about GR, SR? Minkowski spacetimes are strictly empty, the moment you introduce Lorentz invariance breakings is no longer SR.

No not true! Again, the equations must be invariant, but a particular solution or model need not (due to either explicit or spontaneous symmetry breaking).

Another example: Consider Newton and Maxwells laws. They are invariant under time reversal t--> -t, the laws seem to be reversible. However a particular solution that corresponds to the real world need not be. So for instance an icecube that is melting in the sun is irreversible. It never unmelts! This is a consequence of explicit symmetry breaking by initial conditions (in this case, the low entropy configuration preferentially picks out a direction or arrow of time). Note that we don't say that this model violates Newtonian physics. I recommend reading the Feynman lectures volume 1, there is a chapter on symmetries in physics.

I don't think changes of scale (dilations) are usually included in the diffeomorphism group, at least as understood in GR, as the group of coordinate transformations. A dilation is not a diffeomorphism. See this thread post #4 for reference: https://www.physicsforums.com/showthread.php?t=572492

A conformal transformation (for the present purposes sometimes called a conformal isometry) is a diffeomorphism that preserves the metric up to a scale. Eg schematically: F*g = (e^2sigma) g. See Nakahara for the precise definitions.

Now there is a bit of a subtlety b/c whether a conformal transformation is viewed as a symmetry or a diffeomorphism depends on what you keep fixed and what you allow to be transformed, so it's always important to keep that in mind. However the mathematics that comes out are guarenteed to be isomorphic in the classical case of GR, provided you take the appropriate pullbacks when needed.

 

[Haelfix]

I don't have time to write much, as I'm out with my family now.

Knowing the manifold and topology isn't enough, because most manifolds admit inequivalent Lorentzian metrics. It even is possible to have non-zero Riemann curvature tensor tensor for (M,g) while (M,h) is completely (intrinsically) flat! Intrinsic curvature is not a property of the manifold and topology alone. Minkowski space with a point removed is the topological space S^3 x R, the underlying space for the manifold of closed Friedmann-Robertson-Walker universes, and Minkowski space with a straight line removed is S^2 x R^2, the underlying space for the manifold of extended Schwarzschild.

Yes! You also need to specify the differential structure as well. For instance, the space S^7 has 27 different differential structures.

 

[TrickyDicky]

No not true! Again, the equations must be invariant, but a particular solution or model need not (due to either explicit or spontaneous symmetry breaking).
Another example: Consider Newton and Maxwells laws. They are invariant under time reversal t--> -t, the laws seem to be reversible. However a particular solution that corresponds to the real world need not be. So for instance an icecube that is melting in the sun is irreversible. It never unmelts! This is a consequence of explicit symmetry breaking by initial conditions (in this case, the low entropy configuration preferentially picks out a direction or arrow of time). Note that we don't say that this model violates Newtonian physics. I recommend reading the Feynman lectures volume 1, there is a chapter on symmetries in physics.

I don't even know where to start disentangling this, you seem to be conflating classic GR with different or more general systems that include spontaneous symmetry breaking in which the laws are invariant but the system isn't because the "background" of the system is non-invariant. I'm perfectly aware of those kind of systems that are frequent in certain models in QFT and hep. But the issue we are treating is restricted to classical GR and it was in that context that I made my claims, so If they are not true it is not because of the reasons you bring up.
By the way, according to the wiki page on symmetry breaking your example is incorrect, when the symmetry breaking is explicit the laws themselves aren't invariant, which is not the case you describe, you must be referring here to spontaneous symmetry breaking.

A conformal transformation (for the present purposes sometimes called a conformal isometry) is a diffeomorphism that preserves the metric up to a scale. Eg schematically: F*g = (e^2sigma) g. See Nakahara for the precise definitions.

Now there is a bit of a subtlety b/c whether a conformal transformation is viewed as a symmetry or a diffeomorphism depends on what you keep fixed and what you allow to be transformed, so it's always important to keep that in mind. However the mathematics that comes out are guarenteed to be isomorphic in the classical case of GR, provided you take the appropriate pullbacks when needed.

Whether a certain transformation is a diffeomorphism in a determined manifold depends on certain features of the manifold. That is the only subtlety I see here, for instance a scale transformation is a diffeomorphism if the manifold has no intrinsic curvature, but in the case we are dealing with (GR) the manifold is demanded to have intrinsic curvature, in which case once again a scale transformation is not a diffeomorphism in such manifolds.

 

[Haelfix]

I don't even know where to start disentangling this, you seem to be conflating classic GR with different or more general systems that include spontaneous symmetry breaking in which the laws are invariant but the system isn't because the "background" of the system is non-invariant. By the way, according to the wiki page on symmetry breaking your example is incorrect, when the symmetry breaking is explicit the laws themselves aren't invariant, which is not the case you describe, you must be referring here to spontaneous symmetry breaking.
.

This is not quite right, but it's getting astray. My point is valid for any physical system, and GR is no exception.

Whether a certain transformation is a diffeomorphism in a determined manifold depends on certain features of the manifold. That is the only subtlety I see here, for instance a scale transformation is a diffeomorphism if the manifold has no intrinsic curvature, but in the case we are dealing with (GR) the manifold is demanded to have intrinsic curvature, in which case once again a scale transformation is not a diffeomorphism in such manifolds..

This is confusing multiple things, both from this thread and from the one you linked too! A particular map is a diffeomorphism, if it satisfies the definition of a diffeomorphism! The definition I gave above for a conformal isometry automatically implies that it is a diffeomorphism (again see Nakahara). A related idea, is the concept of a Weyl rescaling (which some authors confusingly call a scale transformation). These are different things!

 

[TrickyDicky]

This is not quite right, but it's getting astray. My point is valid for any physical system, and GR is no exception.

It is indeed a rather orthogonal discussion to the one of the OP. Out of curiosity can you give me an example of a solution of the EFE that shows spontaneous symmetry breaking?

This is confusing multiple things, both from this thread and from the one you linked too! A particular map is a diffeomorphism, if it satisfies the definition of a diffeomorphism! The definition I gave above for a conformal isometry automatically implies that it is a diffeomorphism (again see Nakahara). A related idea, is the concept of a Weyl rescaling (which some authors confusingly call a scale transformation). These are different things!where the laws are invariant but the system isn't because the background of the system, its vacuum, is non-invariant.

First of all that definition is of a conformal transformation, not exactly the same as a scale transformation, a conformal transf. is a localized scale transformation so please don't generate unwarranted confusion.
A conformal tr. is just one that preserves the angles, a scale transformation doesn't when the manifold has curvature.

Anyway please read post #16 in the previously linked thread and tell me if you disagree with it.

 

[Haelfix]

It is indeed a rather orthogonal discussion to the one of the OP. Out of curiosity can you give me an example of a solution of the EFE that shows spontaneous symmetry breaking?

Sure but it depends what symmetry you are talking about! In general the presence of any background metric, spontaneously breaks the diffeomorphism gauge 'symmetry' of GR down to a subgroup of isometries. If you are, on the other hand, talking about some specific symmetry (for instance one generated by a killing field), then its a little less obvious however it does exist (you will need to use the initial value formulation of GR). For instance, the static Einstein universe is unstable, and although its not often presented that way, you can think of it as undergoing a phase transition where the cosmological constant acts like an order parameter. Very much like balancing a pencil on its tip!

A conformal tr. is just one that preserves the angles, a scale transformation doesn't when the manifold has curvature.
Anyway please read post #16 in the previously linked thread and tell me if you disagree with it.

I don't disagree with it, but he is talking about what I just called a Weyl rescaling. In general, and particular when ever someone talks about general covariance, diffeomorphism invariance and active and passive transformations, it is vital to stick to one textbook. Names get mixed up all over the literature. I recommend Wald or Nakahara.

 

[TrickyDicky]

Sure but it depends what symmetry you are talking about! In general the presence of any background metric, spontaneously breaks the diffeomorphism gauge 'symmetry' of GR down to a subgroup of isometries. If you are, on the other hand, talking about some specific symmetry (for instance one generated by a killing field), then its a little less obvious however it does exist (you will need to use the initial value formulation of GR). For instance, the static Einstein universe is unstable, and although its not often presented that way, you can think of it as undergoing a phase transition where the cosmological constant acts like an order parameter. Very much like balancing a pencil on its tip!

Yes the Einstein universe solution is unstable, I'm not sure it works as an example of spontaneous SB (wrt what symmetry and how that particular solution is not invariant to that symmetry?) but that is perhaps a theme for a different thread.

I don't disagree with it, but he is talking about what I just called a Weyl rescaling.

You were the first to mention scale transformations as an example but let's drop it, no point arguing about it.

A question to recover the original argument: do you then think there are solutions of the EFE that are not diffeomorphism invariant?

 

[PeterDonis]

But if removing a point turns it into a FRW solution it certainly must not be a completely flat spacetime anymore, and it surely is a solution of the EFE.

Removing a point from Minkowski spacetime doesn't turn it into an FRW solution. It just creates a spacetime which happens to have the same underlying topology as an FRW solution which is spatially closed. But Minkowski spacetime with a point removed still has the Minkowski metric on it, and I don't think *that* is a solution of the EFE.

 

[TrickyDicky]

Removing a point from Minkowski spacetime doesn't turn it into an FRW solution. It just creates a spacetime which happens to have the same underlying topology as an FRW solution which is spatially closed.

Yes, my answer was too rushed, I should have been more explicit.
The spacetime that results from removing a point is in any case diffeomorphic to the FRW closed solution if the latter is a patch of the former, do you agree?

But Minkowski spacetime with a point removed still has the Minkowski metric on it, and I don't think *that* is a solution of the EFE.

Here I disagree, removing a point of a space changes its topology and that can lead to change the metric from flat to curved.
We know the closed FRW solution is a solution of the EFE, and if you agree with what I said above about it being diffeomorphic to the space that results from removing a point from Minkowski space then the latter must also be a solution, right?
If not having singularities was a condition to be an EFE solution then the FRW spacetimes wouldn't qualify, don't you think?

 

[PeterDonis]

The spacetime that results from removing a point is in any case diffeomorphic to the FRW closed solution if the latter is a patch of the former, do you agree?

The FRW solution with closed spatial slices isn't a "patch" of Minkowski spacetime with a point removed. They are different Lorentzian metrics that can be put on the same underlying topological space, S^3 x R. But not every Lorentzian metric is a solution of the EFE; the flat Minkowski metric on S^3 x R (i.e., Minkowski space with a point removed, but with metric unchanged), to the best of my knowledge, is not. Given that it's not, I'm not sure whether it would be diffeomorphic to S^3 x R with the spatially closed FRW metric on it, since the latter of course is a solution of the EFE.

Here I disagree, removing a point of a space changes its topology and that can lead to change the metric from flat to curved.

Topology and metric are separate things. You can change the topology without changing the metric, and vice versa. Whether a particular combination of topology and metric is a solution of the EFE is a different question.

If not having singularities was a condition to be an EFE solution then the FRW spacetimes wouldn't qualify, don't you think?

I didn't say a solution of the EFE couldn't have singularities, period. I said I didn't think a solution of the EFE could have a singularity at one point but have a flat metric everywhere else.

 

[TrickyDicky]

The FRW solution with closed spatial slices isn't a "patch" of Minkowski spacetime with a point removed. They are different Lorentzian metrics that can be put on the same underlying topological space, S^3 x R. But not every Lorentzian metric is a solution of the EFE; the flat Minkowski metric on S^3 x R (i.e., Minkowski space with a point removed, but with metric unchanged), to the best of my knowledge, is not. Given that it's not, I'm not sure whether it would be diffeomorphic to S^3 x R with the spatially closed FRW metric on it, since the latter of course is a solution of the EFE.

Since you were the one that asked GeorgeJones whether it is a solution, I take it you are at least not sure about it.

Topology and metric are separate things. You can change the topology without changing the metric, and vice versa. Whether a particular combination of topology and metric is a solution of the EFE is a different question.

I didn't say anything contradicting this, note I wrote "can lead to change" , not "necessarily changes". Do you not agree that a topological change (such as removing a point) can lead to a change in the metric?

I didn't say a solution of the EFE couldn't have singularities, period. I said I didn't think a solution of the EFE could have a singularity at one point but have a flat metric everywhere else.

Ah, ok. But why that bias towards flat metrics, if you accept that curved metrics like the FRW for instance do have singularities?

 

[TrickyDicky]

Oh,well , I'll throw this out on the stoop and see if the cat licks it up:

We are at all moments dealing with Riemannian (therefore differentiable) manifolds, and differentiable manifolds that share an underlying topology like the ones that we were considering in the previous post should be diffeomorphic by definition.
That means one can be obtained from the other thru an arbitrary coordinate transformation and we know that no coordinate transformation can change the intrinsic curvature of a manifold so they both must be curved since we know the closed FRW metric is curved and also we can deduce that the other manifold must be a solution of the EFE if it can be obtained by an arbitrary coordinate transformation from a solution.

Peter, I guess if you still think that the minkowski spacetime with a point removed must conserve its flat metric you can't agree with me, but I would like to know why you insist on it, can you base that reasoning on something? Maybe that would help us both.

 

[TrickyDicky]

differentiable manifolds that share an underlying topology should be diffeomorphic by definition.

Thinking it over I'm not sure this is true, could some expert confirm it or refute it?
Edit: It is possibly not true for dimension 4 or bigger, only for low dimensions.

 

[PeterDonis]

Since you were the one that asked GeorgeJones whether it is a solution, I take it you are at least not sure about it.

I wasn't sure about Minkowski space with a point removed being a solution, yes.

Do you not agree that a topological change (such as removing a point) can lead to a change in the metric?

Not really, because there's no connection between the two, so saying "can lead to" sounds misleading to me. At least, there's no connection mathematically; as I said before, you can choose the topological space and the metric independently if all you're doing is constructing a mathematical object.

If you're asking whether a change in the topology of spacetime would lead to a change in the metric *physically*, I can't really answer that, since I don't know what theoretical framework one would use to get a handle on it as a question of physics (as opposed to just mathematics).

Ah, ok. But why that bias towards flat metrics, if you accept that curved metrics like the FRW for instance do have singularities?

The only flat metric I'm aware of is the Minkowski metric. Are there others? And the only way I'm aware of that the flat Minkowski metric can be a solution of the EFE is as the most trivial vacuum solution: no matter, no cosmological constant, nothing. And that solution does not have any singularities: it's the flat Minkowski metric on R^4. So it seems to me that the flat Minkowski metric could not be a solution to the EFE when applied to any other topological space. But of course this argument is just heuristic; I haven't actually tried to prove anything mathematically.

 

[PeterDonis]

Peter, I guess if you still think that the minkowski spacetime with a point removed must conserve its flat metric you can't agree with me, but I would like to know why you insist on it, can you base that reasoning on something? Maybe that would help us both.

Just to amplify my previous post, "Minkowski spacetime with a point removed but with the same flat metric" is just a mathematical object. Call it M. It doesn't necessarily have any physical meaning. As a mathematical object, M is perfectly consistent and well-behaved, so I can consider it abstractly regardless of whether or not it is a solution of any particular equation.

"FRW spacetime with closed spatial slices" is another mathematical object. Call it F. Again, I can consider F abstractly; it doesn't have to have any physical meaning (although of course in this case it does). It just so happens that this mathematical object, F, shares some mathematical structure with M, above: both of them have the same underlying topological space, S^3 x R. But again, that's just a mathematical fact; it does not necessarily have any physical meaning.

The question I was asking is whether mathematical object M is a solution of the EFE. I don't think it is, but that's based on physical reasoning (see my previous post), not mathematical reasoning. It's also a separate question, in principle, from the question of whether M is diffeomorphic with some other mathematical object, such as F. *If* M were a solution of the EFE, then I would agree that, since it shares the same underlying topological space with another solution F, there would have to be a diffeomorphism between M and F. But if M is *not* a solution of the EFE, as I think, then whether or not M and F are diffeomorphic has nothing to do with the diffeomorphism invariance of GR; the latter is only supposed to apply to solutions of the EFE, not to arbitrary Lorentzian metrics on arbitrary topological spaces, regardless of whether they solve the EFE.

 

[Ben Niehoff]

There is a lot of confusion in this thread, and it seems like I caused some of it, so I'll jump in.


It is important to deconstruct the layers of structure we're using in GR:

1. Topology. Topology cares only about connectedness and continuous maps between spaces. Manifolds are topological objects. For example, flat space has the base manifold R^4, and the Reissner-Nordstrom geometry has the base manifold S^2 x R^2.

A sphere and a cube are the same as topological manifolds. They are both the manifold S^2.

Topologically speaking, manifolds can be given an atlas of charts; that is, a set of continuous maps of open regions of the manifold to R^n, with continuous transition functions between them, such that the whole manifold is covered by the set of maps.

2. Differential structure. This gives a definition of "smoothness" on manifolds. Now the atlas of charts is required to be smooth, with smooth transition functions.

As differentiable manifolds, the sphere and the cube are not equivalent, because the cube has 8 points where it is not smooth.

Also, the (maximally-extended) Schwarzschild and Reissner-Nordstrom geometries are not equivalent as smooth manifolds (despite the fact that they are both S^2 x R^2 topologically), because Schwarzschild has two singularities and Reissner-Nordstrom has a countably infinite number of them.

3. (Pseudo)-Riemannian structure. This is where we give the manifold a local notion of distance and angle; i.e., a metric tensor.


Now, there are three kinds of maps between manifolds worth talking about, depending on how many layers of this structure are preserved:

Homeomorphisms preserve only layer 1. The sphere and the cube are homeomorphic.

Diffeomorphisms preserve up to layer 2. The sphere and the cube are not diffeomorphic; however, the sphere is diffeomorphic to any topological sphere that is stretched and distorted in any desired way, so long as it is still everywhere smooth. Diffeomorphisms do not care about size or shape.

Metric-preserving diffeomorphisms, which are maps $\varphi : (M, g) \rightarrow (N, h)$, smooth with smooth inverse, such that $g = \varphi^* h$ (i.e. the metric on M is the pullback along $\varphi$ of the metric on N). Metric-preserving diffeomorphisms do care about size and shape, and are fully equivalent to coordinate transformations.


There is one claim I made in a previous thread which is wrong: I claimed that a sphere of radius A is not diffeomorphic to a sphere of radius B. Clearly this is wrong, because the smooth structure comes prior to the metric structure, and diffeomorphisms care only about the smooth structure. Any two round spheres, of any radius, are diffeomorphic. A sphere is also diffeomorphic to a sphere with a smooth bump on it, etc.

What is true is that a sphere of radius A and a sphere of radius B fail to be "metric-preserving diffeomorphic" (which unfortunately does not have a convenient word to describe it). This is because there is no diffeomorphism between them such that the metric on sphere A is the pullback of the one on sphere B. Hence the "metric-preserving" condition makes things quite rigid.

In general, local conformal transformations $g \mapsto e^{2 \varphi} g$ are diffeomorphisms, provided that $e^{2 \varphi}$ is everywhere finite and strictly positive. If these conditions are broken, then a local conformal transformation can change the topology, and thus fails to be even a homeomorphism (for example, all 2-dimensional orientable manifolds are locally conformal to each other).


Finally, there is the question of what kinds of transformations leave Einstein's equations invariant? First, look at vacuum solutions with cosmological constant:

$$R_{\mu\nu} - \frac12 R g_{\mu\nu} + \Lambda g_{\mu\nu} = 0.$$
Under conformal rescaling by a constant, $g \mapsto a^2 g$, we have $R_{\mu\nu} \mapsto R_{\mu\nu}$ and $R \mapsto R/a^2$, and hence the equation is invariant if we also assume $\Lambda \mapsto \Lambda/a^2$. What about more general conformal rescaling? From here:

http://en.wikipedia.org/wiki/Ricci_curvature#Behavior_under_conformal_rescaling

one has for $g \mapsto e^{2 \varphi} g$

$$\tilde{\operatorname{Ric}}=\operatorname{Ric}+(2-n)[ \nabla d\varphi-d\varphi\otimes d\varphi]+[\Delta \varphi -(n-2)\|d\varphi\|^2],$$
which indicates that Einstein's equation cannot be invariant under general such transformations. Therefore it is clear that general diffeomorphisms are not a symmetry of Einstein's equations!


One could consider moving the extra terms to the right-hand-side and treating them as matter sources, but in a sense, the equation with sources

$$R_{\mu\nu} - \frac12 R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{1}{8 \pi G} T_{\mu\nu}$$
is trivial; we are simply taking some combination of curvatures and calling it "$T_{\mu\nu}$". We are only really interested in the kinds of $T_{\mu\nu}$ that describe physically-reasonable distributions of matter, so to call it an "'invariance" of Einstein's equations where $T_{\mu\nu}$ can be arbitrarily modified is kind of silly.


Therefore I conclude that Einstein's equations are not invariant under all diffeomorphisms, but only under certain kinds. Certainly the metric-preserving diffeomorphisms are included, since they are equivalent to coordinate transformations. I suspect that global rescaling by a constant is the only other possibility.

In any case, this means the whole argument over "active" vs. "passive" diffeomorphisms is a moot point. The only maps that need considering are the maps that are equivalent to coordinate transformations. And so GR's "diffeomorphism invariance" is really a trivial fact; any theory can be written in a parametrization-invariant way.

 

[TrickyDicky]

I edited my #83, and that was a necessary link of my reasoning so right now I haven't so much confidence that "M" is a solution of the EFE, although I don't completely rule it out.

 

[TrickyDicky]

Wow, thanks Ben, that really clears up a lot of my misunderstandings.
I'll study your post and come back if I have doubts.

 

[Ben Niehoff]

Peter, I guess if you still think that the minkowski spacetime with a point removed must conserve its flat metric you can't agree with me, but I would like to know why you insist on it, can you base that reasoning on something? Maybe that would help us both.

There is nothing wrong with "Minkowski space with a point removed, and flat metric" as a mathematical object, and yes, it is a solution to Einstein's equations, because Einsteins equations are local. The base topology is "R^4 with a point removed", and the metric tensor is the obvious one.

The feature this psuedo-Riemannian manifold is missing is that it is not "complete"; i.e., there are geodesics that leave the manifold at finite affine parameter (i.e., geodesics of finite length that leave the manifold). One may consider the "maximal extension" of this manifold, which is where you follow all the geodesics to infinite length, and fill the rest of the manifold in by analytic continuation. The result is that the missing point will be put back, giving standard Minkowski space.

A similar process is used to get the maximally-extended Schwarzschild, Reissner-Nordstrom, and Kerr spacetimes from their standard solutions (whose coordinates cover only an open patch of the total spacetime). Kruskal coordinates are an example of a coordinate chart that covers the total Schwarzschild spacetime; in the RN and Kerr cases, one still needs multiple coordinate charts.

 

[PeterDonis]

Also, the (maximally-extended) Schwarzschild and Reissner-Nordstrom geometries are not equivalent as smooth manifolds (despite the fact that they are both S^2 x R^2 topologically), because Schwarzschild has two singularities and Reissner-Nordstrom has a countably infinite number of them.

In other words, Schwarzschild and R-N are *not* diffeomorphic. Presumably the same would be true of Schwarzschild and Kerr--isn't Kerr also S^2 x R^2 topologically? (George Jones said they were different topological spaces earlier in this thread, but I suspect by "topological space" he meant to include the differential structure, your level 2. Since we were talking about diffeomorphisms, that makes sense.)

any theory can be written in a parametrization-invariant way.

I seem to remember some of the literature on loop quantum gravity saying something very like this--and then proceeding to talk about "active" diffeomorphisms as though they were a big deal. Which makes me suspect that the literature is not entirely clear on terminology.

 

[PeterDonis]

There is nothing wrong with "Minkowski space with a point removed, and flat metric" as a mathematical object, and yes, it is a solution to Einstein's equations, because Einsteins equations are local.

I can certainly see that, at every point except the one that was "removed", the flat metric will solve the EFE. But...

The base topology is "R^4 with a point removed", and the metric tensor is the obvious one.

By "R^4 with a point removed", do you mean S^3 x R? That's a different topological space, even at your level 1.

The feature this psuedo-Riemannian manifold is missing is that it is not "complete"; i.e., there are geodesics that leave the manifold at finite affine parameter (i.e., geodesics of finite length that leave the manifold).

Yes, agreed. This is why I said the removed point acts like a singularity.

One may consider the "maximal extension" of this manifold, which is where you follow all the geodesics to infinite length, and fill the rest of the manifold in by analytic continuation. The result is that the missing point will be put back, giving standard Minkowski space.

I understand how this would work regarding the metric, since the manifold starts out flat, so there's zero curvature along all of the incomplete geodesics, and therefore completing them as you say doesn't involve any curvature singularities. However, it does change the underlying topological space, from S^3 x R to R^4. In other words, analytic continuation can change the underlying topological space. Is that correct?

Kruskal coordinates are an example of a coordinate chart that covers the total Schwarzschild spacetime; in the RN and Kerr cases, one still needs multiple coordinate charts.

I thought the Penrose diagrams even for R-N and Kerr could cover the entire maximal extension with a single chart. You just have to let the timelike coordinate have infinite range instead of finite. (Actually, the Penrose chart I'm used to thinking about for Kerr only covers the equatorial plane, so that one may be a bad example.)

 

[Ben Niehoff]

I seem to remember some of the literature on loop quantum gravity saying something very like this--and then proceeding to talk about "active" diffeomorphisms as though they were a big deal. Which makes me suspect that the literature is not entirely clear on terminology.

Given how many physicists claim that GR is "diffeomorphism-invariant" when that is clearly not the case, I suspect that most physicists are simply using the word "diffeomorphism" incorrectly; what they really mean is what I've been calling "metric-preserving diffeomorphism".

In this case, there really is a symmetry between "active" and "passive". It is easiest to imagine such a map is just a reparametrization. However, if you can define some invariant notion of what it means to "lock the coordinates in place", you can consider it to be a change in the shape of the manifold. I'm not really sure what this means, though.

On a tangential note, there are metric-preserving diffeomorphisms that are not continuously connected to the identity; for example, Dehn twists on a torus (which is a fancy way of saying "cut the torus, give it a 360 degree twist, and glue it back together"). Such diffeomorphisms will be neglected if you consider only the Lie algebra of vector fields on the manifold. This is very important in string theory.

 

[TrickyDicky]

Quick question on terminology: aren't metric-preserving diffeomorphisms called isometries?

 

[Ben Niehoff]

By "R^4 with a point removed", do you mean S^3 x R? That's a different topological space, even at your level 1.

Yes

Yes, agreed. This is why I said the removed point acts like a singularity.

You might say that, but it's not a singularity that will have any physical effects. It is a "removable singularity", rather like the singularity at x=1 of the function

$$f(x) = \frac{x^2 - 4x + 3}{x - 1}.$$
By "removable", of course I mean "removable by analytic continuation". In this sense one can say that GR only cares about complete metrics, because geodesics have physical meaning, and hence the physics does not distinguish between "Minkowski space with point removed" and "Minkowski space".

I understand how this would work regarding the metric, since the manifold starts out flat, so there's zero curvature along all of the incomplete geodesics, and therefore completing them as you say doesn't involve any curvature singularities. However, it does change the underlying topological space, from S^3 x R to R^4. In other words, analytic continuation can change the underlying topological space. Is that correct?

Yes, because analytic continuation attaches extra pieces of spacetime to the manifold (in this case, the missing point; in the Schwarzschild case, a white hole, an Einstein-Rosen bridge, and a whole extra universe).

I thought the Penrose diagrams even for R-N and Kerr could cover the entire maximal extension with a single chart. You just have to let the timelike coordinate have infinite range instead of finite. (Actually, the Penrose chart I'm used to thinking about for Kerr only covers the equatorial plane, so that one may be a bad example.)

You're right. But the important point to consider is that one should not always expect a maximally-extended geometry to be covered by one chart. These geometries are special.

 

[Ben Niehoff]

Quick question on terminology: aren't metric-preserving diffeomorphisms called isometries?

So they are! I don't know how that slipped past me.

GR is invariant under isometries. Now it sounds even more trivial!

 

[PeterDonis]

In this sense one can say that GR only cares about complete metrics, because geodesics have physical meaning, and hence the physics does not distinguish between "Minkowski space with point removed" and "Minkowski space".

Ok, this makes sense. Basically, there are two different mathematical objects, but GR considers them both to be descriptions of the same physical spacetime--one is just a more complete description of it than the other.

 

[Ben Niehoff]

By the way, here is how to identify what transformations preserve Einstein's equations. Perturb the metric by $g_{\mu\nu} \mapsto g_{\mu\nu} + h_{\mu\nu}$ where $h_{\mu\nu}$ is small. $h_{\mu\nu}$ has a lot of gauge freedom, so you can partially gauge-fix by imposing

$$\nabla_\mu h^\mu{}_\nu = 0.$$
This is analogous to using the Lorentz gauge in E&M. Demanding the the Einstein equations (with zero cosmological constant) remain invariant under this perturbation gives you

$$\Box h_{\mu\nu} + 2 R_\mu{}^\rho{}_\nu{}^\sigma h_{\rho\sigma} = 0.$$
Any solutions to this equation give infinitesimal perturbations that preserve Einstein's equation. However, the gauge condition on h does not fully gauge-fix, so some of the solutions will just be infinitesimal coordinate transformations; you have to throw those ones out.

Anything left will give you a non-trivial invariance of GR. These will be local diffeomorphisms which are not isometries.

 

[TrickyDicky]

I don't disagree with it, but he is talking about what I just called a Weyl rescaling. In general, and particular when ever someone talks about general covariance, diffeomorphism invariance and active and passive transformations, it is vital to stick to one textbook. Names get mixed up all over the literature. I recommend Wald or Nakahara.

After Ben clarifications you were right about diffeomorphisms, I was confused by the way the diffeomorphism term is loosely used in GR to actually mean isometry.
So all the time I was saying diffeomorphisms I was actually referring to isometries, that are a subset of the group of diffeomorphisms. All isometries are diffeomorphisms but not all diffeomorphisms are isometries.

 

[TrickyDicky]

I think an important conclusion that I take away from this thread is that there are some terminology issues in GR that only serve to confuse matters.
So when Einstein put forward the "general covariance" principle he just meant something that is considered trivial nowadays:the fact that coordinates are not physical.
The confusion came when someone called this diffeomorphism invariance, which can cause confusion because coordinate transformations in (pseudo)Riemannian manifolds are indeed diffeomorphisms but not all diffeomorphism are coordinate transformations, in GR coordinate transformations are metric-preserving diffeomorphisms, not simple diffeomorphisms due to GR being based in (pseudo)Riemannian manifolds, not simply smooth manifolds.
Wrt to the passive vs active diffeomorphism issue I can't see how it can have any physical consequence, it is just a way to allude to the fact that every diffeomorphism is bijective and therefore every active (passive)transformation that is a diffeomorphis has its inverse passive (active) transformation. Every push-forward has its pull-back if it is a diffeomorphism. So maybe the OP could come back and explain how exactly is the active-passive thing physically relevant.
Also to link with the OP concern about Killing fields and diffeomorphism invariance as understood in GR (that I guess it should be called to avoid confusion isometry invariance although maybe it is implicit in the context of Riemannian manifolds), although it was previously clarified thru a different path KV fields are precisely those that preserve the metric and generate isometries so they are obviously invariant to coordinate transformations.

 

[haushofer]

And so GR's "diffeomorphism invariance" is really a trivial fact; any theory can be written in a parametrization-invariant way.

But it is not trivial that the solutions of GR are invariant under general coordinate transformations, right? Take e.g. Newton-Cartan theory; there the "Einstein equations" are general covariant. After gauge-fixing to certain observers, these Einstein equations become the Poisson equations. The general coordinate transformations are then broken to the Galilei transformations, including time-dependent accelerations.

 

[stevendaryl]

Wrt to the passive vs active diffeomorphism issue I can't see how it can have any physical consequence, it is just a way to allude to the fact that every diffeomorphism is bijective and therefore every active (passive)transformation that is a diffeomorphism has its inverse passive (active) transformation

I think that the distinction between active and passive is perhaps clearer when you use a simpler transformation than a full diffeomorphism. So let's take invariance under Galilean transformations:

x' = x - vt
t' = t

There is an active and a passive interpretation of this transformation. The passive interpretation is about the form of the equations of motion: If you start with the equations expressed in terms of (x,t), and then you rewrite the equations in terms of (x',t'), then the equations have the same form. For example, if the equations of motion are:

m d²x/dt² = F

then since d²x/dt² = d²x'/dt'², the equations have the same form in terms of (x',t'), provided that the force F transforms as:

F' = F.

On the other hand, the active interpretation is about solutions to the equations of motion: If x = f(t) is a solution to the equations of motion, then x = f(t) - vt is also a solution to the equations of motion.

Notice the distinction: In the passive case, we are considering the same situation expressed in different coordinates. In the active case, we are consider a different situation expressed in the same coordinates. Newton's equations, with a velocity-independent force, are invariant under both types of transformations.

But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d²x/dt² = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.

 

[haushofer]

So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?

 

[stevendaryl]

So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?

I'm a little on shakier ground when it comes to diffeomorphisms. The passive case is clear enough--it's just a coordinate change, and the equations are the same for any coordinate system. The active case is the one I'm a little fuzzy on. Roughly speaking, it works like this:

Let m be a (continuous, invertible, differentiable, blah, blah, blah) function from ℝ⁴ to ℝ⁴. Let X^$\alpha$ be some coordinate system (function from spacetime points $P$ to ℝ⁴). Let S be some solution to the full equations of motion. Then if the theory is invariant under active diffeomorphisms, then we can get another solution S' in the following round-about way:

Let X'^$\alpha$ be the coordinate system obtained from X^$\alpha$ by applying the transformation m. Then pick S' so that the description of S' in terms of the original coordinate system X^$\alpha$ looks the same as the description of S in terms of the new coordinate system X'^$\alpha$.

So a very simple example is the following: we have two charged point masses A and B, each of mass M, and in the coordinate system X^$\alpha$ they are a distance L apart in the x-direction (coordinate X^$1$), with a nonzero relative velocity in the y-direction (coordinate X^$2$). Otherwise, we have an empty universe (with some specified boundary conditions at infinity, such as being asymptotically flat). So we use GR, plus electrodynamics in curved spacetime, to compute the trajectories of those two masses.

Now, consider the simple coordinate transformation x → λx, where λ is just some constant.

Under a passive transformation, this is just a rescaling of the x-coordinate, just like changing from feet to inches. Clearly, the physics is not changed by such a rescaling.

Under an active transformation, this is the same as if you physically changed the location of the two masses. This is a drastically different situation; it can make the difference between the case where the two masses are gravitational bound and the case where they are not. The claim made for General Relativity is that if you make the corresponding changes to the metric, and to the electric fields, and to trajectories, etc., then the new situation will be a solution, as well. That's a much stronger claim than invariance under passive transformations.

Where I'm a little fuzzy is exactly what does it mean to say that we change the distance between the masses and change everything else appropriately.

 

[TrickyDicky]

You raise some good points.

I think that the distinction between active and passive is perhaps clearer when you use a simpler transformation than a full diffeomorphism. So let's take invariance under Galilean transformations:

x' = x - vt
t' = t

There is an active and a passive interpretation of this transformation. The passive interpretation is about the form of the equations of motion: If you start with the equations expressed in terms of (x,t), and then you rewrite the equations in terms of (x',t'), then the equations have the same form. For example, if the equations of motion are:

m d²x/dt² = F

then since d²x/dt² = d²x'/dt'², the equations have the same form in terms of (x',t'), provided that the force F transforms as:

F' = F.

On the other hand, the active interpretation is about solutions to the equations of motion: If x = f(t) is a solution to the equations of motion, then x = f(t) - vt is also a solution to the equations of motion.

Notice the distinction: In the passive case, we are considering the same situation expressed in different coordinates. In the active case, we are consider a different situation expressed in the same coordinates. Newton's equations, with a velocity-independent force, are invariant under both types of transformations.

But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d²x/dt² = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.

Thanks, this helps me see where the physical problem with the passive/active distinction comes from. Certainly this problem shows up in GR with any solution of the EFE that is not static. Since the mathematical definition of diffeomorphism includes both types of transformation this problem would make many GR solutions simply not invariant wrt diffeomorphisms (in the GR sense that preserves the metric), even if the equations are.
It also reminds me of the situation in Hamiltonian mechanics with the difference between symplectic and contact manifolds.
Would this be linked to the spontaneous symmetry breaking concept mentioned above?

So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?

The problem is that mathematically is not very rigorous to separate passive from active transformations, it makes little sense.

I'm a little on shakier ground when it comes to diffeomorphisms. The passive case is clear enough--it's just a coordinate change, and the equations are the same for any coordinate system. The active case is the one I'm a little fuzzy on. Roughly speaking, it works like this:

Let m be a (continuous, invertible, differentiable, blah, blah, blah) function from ℝ⁴ to ℝ⁴. Let X^$\alpha$ be some coordinate system (function from spacetime points $P$ to ℝ⁴). Let S be some solution to the full equations of motion. Then if the theory is invariant under active diffeomorphisms, then we can get another solution S' in the following round-about way:

Let X'^$\alpha$ be the coordinate system obtained from X^$\alpha$ by applying the transformation m. Then pick S' so that the description of S' in terms of the original coordinate system X^$\alpha$ looks the same as the description of S in terms of the new coordinate system X'^$\alpha$.

So a very simple example is the following: we have two charged point masses A and B, each of mass M, and in the coordinate system X^$\alpha$ they are a distance L apart in the x-direction (coordinate X^$1$), with a nonzero relative velocity in the y-direction (coordinate X^$2$). Otherwise, we have an empty universe (with some specified boundary conditions at infinity, such as being asymptotically flat). So we use GR, plus electrodynamics in curved spacetime, to compute the trajectories of those two masses.

Now, consider the simple coordinate transformation x → λx, where λ is just some constant.

Under a passive transformation, this is just a rescaling of the x-coordinate, just like changing from feet to inches. Clearly, the physics is not changed by such a rescaling.

Under an active transformation, this is the same as if you physically changed the location of the two masses. This is a drastically different situation; it can make the difference between the case where the two masses are gravitational bound and the case where they are not. The claim made for General Relativity is that if you make the corresponding changes to the metric, and to the electric fields, and to trajectories, etc., then the new situation will be a solution, as well. That's a much stronger claim than invariance under passive transformations.

Where I'm a little fuzzy is exactly what does it mean to say that we change the distance between the masses and change everything else appropriately.

Do you agree that mathematically is not possible to separate active from passive transformations, because basically they are dual notions and it is the same operation only conventionally the object motion wrt the observer is the reference POV in the active case and the observer motion wrt the object is the reference in the passive one?
Once again the physical interpretation might be an instance of spontaneous symmetry breaking IMO.

 

[Ben Niehoff]

But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d²x/dt² = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.

Come now, you must be fair. The transformation must also include

u' = u - v

just as it does in the passive case. Then the transformed solution works! There is no material distinction between "active" and "passive".

You might complain that I'm not allowed to touch u because it's a constant in the equation of motion. But "active" vs. "passive" is always a matter of interpretation. I am actively changing the physical situation, not just by shifting x(t), but also by changing u.

What you are really going after is the idea of "preferred frames". With u included, there is a preferred frame. But you must make a distinction between "having no preferred frame", and "being invariant under Lorentz/Galilean/etc. symmetry", because it is always possible to specify the preferred frame in an invariant way; it is, after all, a real, geometrical object. Being invariant under some set of coordinate transformations is a property of how the equations are written down; having a preferred frame is the result of placing some real object that doesn't participate in dynamics.

The same can happen in electromagnetism if I consider a situation with a fixed, infinite current-carrying wire along the z axis. There is now a preferred frame, but I can still write everything in a Lorentz-invariant way.