How GR Resolves the Conservation of Momentum and Energy

[julian]

I don't see this.

Commutators of vector fields are preserved by push-forwards of diffeomporphisms, so push-forwards preserve Lie derivatives and thus Killing's equation.

It occurred to me yesterday that the transformation rule under active diffeomorphims is very similar in mathematical form to coordinate transformations and that I could be wrong on this point...thanks.

 

[julian]

No time evolution means no dynamics. I hope you know that GR and all other gauge theories DO describe time evolution.

Sam

No GR is a fully constrained system...it predicts the relationship between measurable quantities...see Rovelii 'partial and complete observables' or whatever it's called. This relationship is determined by the constraints imposed on the system.

 

[pervect]

No GR is a fully constrained system...there are no dymanics in the usual sense. Like Thiemann stated about the Friedmann equations for cosmology ... they are gauge transformation equations and not evolution equations as most people think...

You can choose philosophically to view any deterministic system that evolves with time as fully constrained, I think. I don't like to discuss philosophy a lot because the arguments tend to be endless.

It's probalby best to say that ADM methods certainly treat GR as a system that evolves with time, where you specify it as an initial value problem along some particular space-like hypersurface, and see how the geometry evolves with time.

 

[julian]

You can choose philosophically to view any deterministic system that evolves with time as fully constrained, I think. I don't like to discuss philosophy a lot because the arguments tend to be endless.

It's probalby best to say that ADM methods certainly treat GR as a system that evolves with time, where you specify it as an initial value problem along some particular space-like hypersurface, and see how the geometry evolves with time.

It's not philosophy, any dynamical system can be written as a constrained system. This is called parameterisation. The opposite process is called deparameterisation...not all constrained systems are deparameterisable - they have to be of a specific mathematical form.

 

[julian]

It's probalby best to say that ADM methods certainly treat GR as a system that evolves with time, where you specify it as an initial value problem along some particular space-like hypersurface, and see how the geometry evolves with time.

But as Thiemann comments on the Friedmann equations - they are gauge transformation equations and not real evolution equations. Let me find the reference.

 

[julian]

From the paper "Solving the Problem of Time in General Relativity and Cosmology with Phantoms and k -- Essence" - http://arxiv.org/pdf/astro-ph/0607380.pdf

Couple of quotes:

"By “the problem of time” in General Relativity (GR) one means that GR is a completely parametrised system. That is, there is no natural notion of time due to the diffeomorphism invariance of the theory
and therefore the canonical Hamiltonian which generates time reparametrisations vanishes."

and

"...the time evolution described by the FRW equations is obtained from the Hamiltonian equations of motion generated by the Hamiltonian constraint and not by an actual Hamiltonian. This is due to the fact that the “Hamiltonian” used to derive the FRW equations is actually constrained to vanish by one of the Einstein equations. The “evolution equations” generated by a constraint must therefore be interpreted as gauge transformations and those,..."

 

[julian]

The Hole arguemnt isn't just philosophy...to have a natural notion of time the constraint equations have to be of a specific mathematical form...but the GR equations, in general, aren't of this mathematical form...I think Rovelli claims this is the whole point of his book...maybe...dont quote me.

 

[julian]

Plus if you are to identify some dynamical variable that would work locally as a clock, this would necessarily be a real material object and thus be subject to quantum fluctuations itself - there is no absolute time that ticks away and which we could use to forumulate QM in the usual way - again probably see Rovelli.

 

[Dale]

I don't see this.

Commutators of vector fields are preserved by push-forwards of diffeomporphisms, so push-forwards preserve Lie derivatives and thus Killing's equation.

I think I am either misunderstanding you or misunderstanding what an active diffeomorphism is.

My understanding is that an active diffeomorphism is a change in the manifold without a change in the coordinates. E.g. Changing from a Schwarzschild spacetime to a Kerr spacetime, but both in Schwarzschild coordinates. Clearly the Killing vectors are different.

 

[George Jones]

I think I am either misunderstanding you or misunderstanding what an active diffeomorphism is.

My understanding is that an active diffeomorphism is a change in the manifold without a change in the coordinates. E.g. Changing from a Schwarzschild spacetime to a Kerr spacetime, but both in Schwarzschild coordinates. Clearly the Killing vectors are different.

Let there be a diffeomorphism between differentiable manifolds M and N, and let g be a "metric" tensor field for M. w is a Killing vector field on M iff L_w g = 0 (Lie derivative of the metric in the direction of w is zero).

The active diffeomorphism pushes forward fields w and g to fields w* and g* on N. Becasuse w is a Killing vector field for (M , g), w* is a Killing vector field for (N , g*), i.e., L_w* g* = 0. It is true that when N = M, w* will not, in general, be a Killing vector for g, but all fields get shuffled around (including g to g*) by the diffeomorphism, not just w to w*.

 

[PeterDonis]

The active diffeomorphism pushes forward fields w and g to fields w* and g* on N. Becasuse w is a Killing vector field for (M , g), w* is a Killing vector field for (N , g*), i.e., L_w* g* = 0. It is true that when N = M, w* will not, in general, be a Killing vector for g, but all fields get shuffled around (including g to g*) by the diffeomorphism, not just w to w*.

To pick up on DaleSpam's example, this would mean that, if M is Schwarzschild spacetime and N is Kerr spacetime, the same active diffeomorphism that pushes forward the Schwarzschild metric on M (g) to the Kerr metric on N (g*) would also push forward every Killing vector field on M to a Killing vector field on N?

This seems plausible for the time translation Killing vector field, $\partial_{t}$; but does it also work for the angular ones? That seems odd, because in Schwarzschild spacetime there is a full 2-sphere's worth of angular Killing vector fields, so to speak, whereas in Kerr spacetime there is only one, $\partial_{\phi}$, since the spacetime is only axisymmetric, not spherically symmetric, correct?

 

[George Jones]

Changing from a Schwarzschild spacetime to a Kerr spacetime, but both in Schwarzschild coordinates. Clearly the Killing vectors are different.

To pick up on DaleSpam's example, this would mean that, if M is Schwarzschild spacetime and N is Kerr spacetime, the same active diffeomorphism that pushes forward the Schwarzschild metric on M (g) to the Kerr metric on N (g*) would also push forward every Killing vector field on M to a Killing vector field on N?

Kerr and Schwarzschild are not the same as topological spaces, and therefore they can't be diffeomorphic.

 

[PeterDonis]

Kerr and Schwarzschild are not the same as topological spaces, and therefore they can't be diffeomorphic.

Ah--so that makes me wonder which spacetimes *could* be transformed into one another via an active diffeomorphism. Could you do Schwarzschild to Minkowski or vice versa? I'm guessing not because of the singularity in the former. If so, the conditions for being able to do an active diffeomorphism between different spacetimes would seem to be pretty restrictive.

Also, could you still do an active diffeomorphism taking just a portion of one manifold to a portion of another? For example, the portions of Schwarzschild and Kerr exterior to their respective horizons? Or does it have to be entire manifolds?

 

[TrickyDicky]

Also, where does GR's "diffeomorphism invariance" among the EFE solutions go, if you don't even obtain topologically equivalent spaces?

 

[TrickyDicky]

Also, could you still do an active diffeomorphism taking just a portion of one manifold to a portion of another? For example, the portions of Schwarzschild and Kerr exterior to their respective horizons? Or does it have to be entire manifolds?

I don't think so, that would work if you just wanted to do a local diffeomorphism, wouldn't it? I believe "active diffeomorphisms" are not local.

 

[Dale]

Kerr and Schwarzschild are not the same as topological spaces

Oh. Is that because the Schwarzschild spacetime has a point singularity and the Kerr spacetime has a ring singularity?

EDIT: if so, then can't we restrict the manifolds e.g. to the region outside the EH. Then Kerr and Schwarzschild should have the same topology (that of R4 minus an open ball) while preserving the killing vector fields of each. Then we could do an active diffeomorphism between these two restricted manifolds.

 

[PeterDonis]

Also, where does GR's "diffeomorphism invariance" among the EFE solutions go, if you don't even obtain topologically equivalent spaces?

Well, if you can't even do a diffeomorphism in the first place between topologically inequivalent spacetimes, then the question of invariance doesn't even arise, so it can't exactly be violated.

But it does make me wonder what all the fuss over "active diffeomorphisms" is about, if the conditions for even being able to do them in the first place are so strict.

 

[TrickyDicky]

Well, if you can't even do a diffeomorphism in the first place between topologically inequivalent spacetimes, then the question of invariance doesn't even arise, so it can't exactly be violated.

But it does make me wonder what all the fuss over "active diffeomorphisms" is about, if the conditions for even being able to do them in the first place are so strict.

I wonder too, but it can't be overlooked that diffeomorphism invariance (a.k.a. general covariance) is in all texts about GR as a defining property of the theory, what's the point if diffeomorphisms aren't even allowed? It's like saying that in Newton theory Lorentz invariance can't be violated because in a Euclidean space you can't even make a Lorentz transformation and then go on to say Newton's theory main feature is Lorentz invariance, pretty absurd ain't it?

It's an intriguing situation (perhaps George Jones can clarify) or at least I surely don't get it, how is it possible that physicists in general and relativists in particular don't even have a mnimum consensus on whether the distinction between active and passive diffeomorphism has any physical consequences, I have read about the same number of mainstream experts saying it is an important distinction physically (most LQG researchers like Rovelli) as those saying it is obvious it is exactly the same thing.

 

[PeterDonis]

I wonder too, but it can't be overlooked that diffeomorphism invariance (a.k.a. general covariance) is in all texts about GR as a defining property of the theory, what's the point if diffeomorphisms aren't even allowed?

The general statements about diffeomorphism invariance apply to passive diffeomorphisms as well as active ones. Passive diffeomorphisms are just changes in coordinate chart; AFAICT everybody agrees that GR solutions are invariant under changes of coordinate chart. It's only active diffeomorphisms that appear to cause issues.

 

[Dale]

I think that it is clear that GR is invariant over active diffeomorphisms also, but I have previously simply been of the opinion that active diffeomorphisms are non-physical. Now I am not so sure, but I don't think that I understand what they are so well.

 

[TrickyDicky]

The general statements about diffeomorphism invariance apply to passive diffeomorphisms as well as active ones. Passive diffeomorphisms are just changes in coordinate chart; AFAICT everybody agrees that GR solutions are invariant under changes of coordinate chart. It's only active diffeomorphisms that appear to cause issues.

Yes, that is the situation that has me perplexed with physicists not agreeing about. My own opinion is that this issue is mostly artificial, I don't see the distinction active/passive when applied to diffeomorphisms, by definition they are bijective and every diffeomorphism can have associated a coordinate transformation and its inverse (call'em active and passive if you like), at least this is how mathematicians seem to view it and if you look to math texts the active/passive distinction wrt diffeomorphism almost never shows up.
Leaving this aside and going back to my previous question I still can't see how certain GR solutions can have diffeomorphism invariance if as George Jones said they are not diffeomorphic manifolds. If two manifolds are not diffeomorphic I'd say one can't be obtained from the other thru a coordinate transformation, is this correct?

 

[Haelfix]

Please distinguish between a statement that the laws of physics are invariant under a certain symmetry group (actually it should be called a dynamic symmetry to be precise), and a particular solution being invariant.

Obviously you cannot deform a sheet of paper into a donut, no matter how many transformations you make, so of course coordinate charts in one will not map to coordinate charts in the other.

There is no big mystery between active and passive diffeomorphisms, they are essentially related by pullbacks and pushforwards, and only in the case where you have some complicated mathematical object that is not a manifold (without a metric structure like what might show up in quantum gravity) where you have to be a little careful b/c there might be a difference. This is analogous mathematically to how tangent and cotangent bundles arise in differential geometry.

 

[TrickyDicky]

Please distinguish between a statement that the laws of physics are invariant under a certain symmetry group (actually it should be called a dynamic symmetry to be precise), and a particular solution being invariant.

Well, I'm not sure what you are asking for here. General relativity is a statement about the laws of physics (a true statement if you don't reject the theory), and that is made concrete thru the EFE in tensorial form. Solutions of those equation should share that feature of the theory, shouldn't they? Please clarify if you are not referring to this.

Obviously you cannot deform a sheet of paper into a donut, no matter how many transformations you make, so of course coordinate charts in one will not map to coordinate charts in the other.

Exactly, but we are dealing with spaces that are solutions of the EFE here.

There is no big mystery between active and passive diffeomorphisms, they are essentially related by pullbacks and pushforwards, and only in the case where you have some complicated mathematical object that is not a manifold (without a metric structure like what might show up in quantum gravity) where you have to be a little careful b/c there might be a difference. This is analogous mathematically to how tangent and cotangent bundles arise in differential geometry.

Completely agree with you about this, and yet there are still endless discussions about this among experts especially those in the Loop quantum gravity field.

 

[PeterDonis]

If two manifolds are not diffeomorphic I'd say one can't be obtained from the other thru a coordinate transformation, is this correct?

Correct; there is no diffeomorphism that converts Schwarzschild into Kerr spacetime, for example. But I can certainly define diffeomorphisms that convert Schwarzschild under one coordinate chart (say the Schwarzschild chart) into Schwarzschild under another coordinate chart (say the Painleve chart). And both are solutions of the EFE. I realize that that seems like a tautology, since both charts describe the same geometry; but the point is that formally, the metric written in Schwarzschild coordinates solves the EFE written in those coordinates, and the metric written in Painleve coordinates solves the EFE written in those coordinates. The metric and the coordinate chart transform together to keep the underlying geometry invariant. So GR is diffeomorphism invariant in this sense.

 

[Haelfix]

Solutions of those equation should share that feature of the theory, shouldn't they? .

Absolutely not! Forget about GR for a second. It is not even true about classical mechanics.

For instance, laws of physics (eg special relativity) are invariant under translations! However a particular solution typically is NOT invariant. For instance, if you are in an everywhere empty universe except for one room with a wall, that particular solution explicitly breaks Lorentz invariance in one direction.

Another example more pertinent the real world. In general, the particular solution of the EFE that we live in is decidedly not invariant under all diffeomorphisms. It is not even invariant under changes of scale (which is a subgroup of the diffeomorphism group) due to the presence of massive particles (which explicitly break conformal symmetry).

Another example the Minkowski metric breaks an infinite amount of diffeomorphisms, and only leaves a finite amount of isometries unbroken (four rotations, three translations and three boosts)

In fact the only tensor that is invariant under all diffeomorphisms is the trivial metric with all coefficients zero, but this fails to be a metric b/c it is not invertible.

See what I mean?

 

[TrickyDicky]

Correct; there is no diffeomorphism that converts Schwarzschild into Kerr spacetime, for example. But I can certainly define diffeomorphisms that convert Schwarzschild under one coordinate chart (say the Schwarzschild chart) into Schwarzschild under another coordinate chart (say the Painleve chart). And both are solutions of the EFE. I realize that that seems like a tautology, since both charts describe the same geometry; but the point is that formally, the metric written in Schwarzschild coordinates solves the EFE written in those coordinates, and the metric written in Painleve coordinates solves the EFE written in those coordinates. The metric and the coordinate chart transform together to keep the underlying geometry invariant. So GR is diffeomorphism invariant in this sense.

This was my line of reasoning too, but then what happens with the Kerr metric? isn't it also a solution of the EFE?

 

[TrickyDicky]

See what I mean?

Hmm, I'm afraid we re either talking about different things or else I detect several confusing issues in your statements.

Absolutely not! Forget about GR for a second. It is not even true about classical mechanics.

For instance, laws of physics (eg special relativity) are invariant under translations! However a particular solution typically is NOT invariant. For instance, if you are in an everywhere empty universe except for one room with a wall, that particular solution explicitly breaks Lorentz invariance in one direction.

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. That is precisely what happens in GR for translations, due to the intrinsic curvature of the spacetime. Lorentz invariance is only local there.

Another example more pertinent the real world. In general, the particular solution of the EFE that we live in is decidedly not invariant under all diffeomorphisms. It is not even invariant under changes of scale (which is a subgroup of the diffeomorphism group) due to the presence of massive particles (which explicitly break conformal symmetry).

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

Another example the Minkowski metric breaks an infinite amount of diffeomorphisms, and only leaves a finite amount of isometries unbroken (four rotations, three translations and three boosts)

I don't even know what you mean by "breaks an infinite amount of diffeomorphisms", it seems a wrong statement.

In fact the only tensor that is invariant under all diffeomorphisms is the trivial metric with all coefficients zero, but this fails to be a metric b/c it is not invertible.

Please, go thru your notion of diffeomorphism again, we might not be taking about the same thing. Not every transformation you might think of is a diffeomorphism, it is mostly restricted to coordinate charts changes AFAIK.

 

[PeterDonis]

This was my line of reasoning too, but then what happens with the Kerr metric? isn't it also a solution of the EFE?

Yes, and you can express it in different charts as well: Boyer-Lindquist, Doran, and Kerr-Schild all come to mind. All of those can be inter-converted via diffeomorphisms. But no diffeomorphism can change Kerr into Schwarzschild or vice versa.

I see what you are getting at: one could interpret the term "diffeomorphism invariance" as requiring that diffeomorphisms can convert between *any* pair of solutions. But that interpretation is obviously too strong, because different solutions can have different topologies. But the other obvious interpretation, limiting "diffeomorphism" to just passive diffeomorphisms, changes of coordinate chart on the same spacetime, seems too weak, since it leaves no scope for "active diffeomorphisms" at all. I'm struggling with that too; I don't see any obvious middle ground between the "too strong" version and the "too weak" version, because I can't come up with any examples of spacetimes which are non-trivially different but still have the same topology.

 

[George Jones]

Also, could you still do an active diffeomorphism taking just a portion of one manifold to a portion of another?

I can't come up with any examples of spacetimes which are non-trivially different but still have the same topology.

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.

 

[TrickyDicky]

Yes, and you can express it in different charts as well: Boyer-Lindquist, Doran, and Kerr-Schild all come to mind. All of those can be inter-converted via diffeomorphisms. But no diffeomorphism can change Kerr into Schwarzschild or vice versa.

I see what you are getting at: one could interpret the term "diffeomorphism invariance" as requiring that diffeomorphisms can convert between *any* pair of solutions. But that interpretation is obviously too strong, because different solutions can have different topologies. But the other obvious interpretation, limiting "diffeomorphism" to just passive diffeomorphisms, changes of coordinate chart on the same spacetime, seems too weak, since it leaves no scope for "active diffeomorphisms" at all. I'm struggling with that too; I don't see any obvious middle ground between the "too strong" version and the "too weak" version, because I can't come up with any examples of spacetimes which are non-trivially different but still have the same topology.

Glad you see what I mean.
As I said diffeomorphisms in their mathematical use include both active and passive transformations (bijectivity), so they only leave room for the "strong version interpretation" that is not compatible with different topologies.

After some reading it is obvious to me this is still controversial after 98 years of GR.

 

[PeterDonis]

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.

George, thanks, this helps to clarify things and to narrow the focus of the issue that I'm struggling with. Consider your examples:

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

The latter are solutions of the EFE, but is Minkowski space with a point removed a solution of the EFE? I know Minkowski space itself is, but wouldn't removing a point from it mess up the solution? The removed point would act like a singularity, but a solution of the EFE can't have a singularity in an otherwise completely flat spacetime, can it?

, and Minkowski space with a straight line removed is S^2 x R^2, the underlying space for the manifold of extended Schwarzschild.

Same question as above, or more generally: are there any other solutions of the EFE, besides extended Schwarzschild, which have S^2 x R^2 as the underlying space?

 

[PeterDonis]

Same question as above, or more generally: are there any other solutions of the EFE, besides extended Schwarzschild, which have S^2 x R^2 as the underlying space?

After thinking about this some more, it seems to me that Reissner-Nordstrom spacetime should have this topology. If so, a diffeomorphism between Schwarzschild and R-N spacetime would be an "active" diffeomorphism.

It also seems like FRW spacetime with k = 0 (i.e., flat spatial slices) should have topology R^4, as Minkowski spacetime does, so there should be an "active" diffeomorphism between those two as well.

 

[TrickyDicky]

is Minkowski space with a point removed a solution of the EFE? I know Minkowski space itself is, but wouldn't removing a point from it mess up the solution? The removed point would act like a singularity, but a solution of the EFE can't have a singularity in an otherwise completely flat spacetime, can it?

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.


One thing we didn't mention before is that different boundary conditions or geometric constraints like cilindrical or spherical symmetry for the EFE obviously give rise to geometrically inequivalent solutions, that helps explain why the Kerr solution is not diffeomorphic or even homeomorphic to solutions that are demanded to be spherically symmetric.

 

[Dale]

One thing we didn't mention before is that different boundary conditions or geometric constraints like cilindrical or spherical symmetry for the EFE obviously give rise to geometrically inequivalent solutions, that helps explain why the Kerr solution is not diffeomorphic or even homeomorphic to solutions that are demanded to be spherically symmetric.

But as far as I know coffee mugs and donuts have the same topological group and are therefore diffeomorphic, even though they have different symmetries. So I am not sure that the symmetry constraints are that strong.

 

[TrickyDicky]

But as far as I know coffee mugs and donuts have the same topological group and are therefore diffeomorphic,

They share the same topology, yes, so restricting to their topology we say that they are homeomorphic, being diffeomorphic alludes to their putative additional differential (smooth) structure.

even though they have different symmetries.

Not sure exactly what specific symmetries you are referring to, and also the 2-dimensional case is very particular among manifolds and very different from the 4-dimensional case. A donut and a coffee mug both share axis symmetry. Which is the one that matters topologically in this case.

So I am not sure that the symmetry constraints are that strong.

Certain global symmetries are that strong, and in the case we were discussing for instance the difference between a spherical symmetry and a cylindrical symmetry is strong in a topological level. The usual example is that of a cylinder versus a sphere when they are spread out in a plane, the sphere cannot be smooth out without tearing while the cylinder can.

My conclusion is that whenever one compares solutions of the EFE one must consider also the specific geometric conditions that are used, because if they don't share certain different global symmetries, they will hardly be solutions topologically equivalent, which is a condition to be difeomorphic.
This renders the demand for diffeomorphism invariance a bit empty of content IMO,(it would have only the weak interpretion mentioned by PeterDonis) being only an almost trivial constraint that is not peculiar to GR but that can be applied to any physical theory in as much as it is agreed that coordinates are not physical.