Energy in GR

1. Jun 11, 2012

julian

We know that the reason enery and momentum are conserved is b/c of Noether's theorem...time translational invariance implies energy conservation and space translational invariance implies momentum conservation.

Now in a curved spacetime you can still form conserved quantities - energy and momentum if the spacetime has Killing's vector fields.

However, GR is invariant under active diffeomorphisms (Einstein's Hole argument) and general active diffeomorphisms will destroy any Killing vector field. And hence energy and momentum are no longer physically meaningfull quantities? In Rovelli's book "Qunatum Gravity" he emphasizes this. For example he talks about the problem of defing the vacuum state when energy is not meanifull.

Anyway, so GR states that gravity is determined by mass or energy...but energy doesn't have physical meaning anymore?

2. Jun 11, 2012

Staff: Mentor

Not sure what you mean by this. If a geometry has a Killing vector field, it has it regardless of which coordinate chart you describe the geometry with. Diffeomorphisms are just transformations from one coordinate chart to another; they don't change the actual geometry. A given Killing vector field might look a lot simpler in one chart than in another, but that doesn't change whether it's a Killing vector field or not.

If there are no corresponding Killing vector fields in the spacetime, yes (with some caveats, see below). But again, that's a feature of the geometry independent of coordinate charts.

I haven't read the book, but I've read a number of Rovelli's papers, as well as other literature on quantum field theory in curved spacetime. Everything I've read says what I said above: if a spacetime doesn't have a time translation Killing vector field, you can't define an associated energy, so you have a problem trying to define the quantum vacuum state. More generally, in spacetimes with more than one timelike Killing vector field, such as Minkowski spacetime, you can define different notions of energy and hence different "vacuum" states. This is how the Unruh effect arises.

GR states that curvature is determined by the stress-energy tensor, via the Einstein Field Equation. That statement has physical meaning in any spacetime; you don't need any Killing vector fields to define the curvature tensor or the stress-energy tensor. But if there is no time translation Killing vector field in a particular spacetime--i.e., a particular solution of the EFE--then there won't be a conserved total energy either. Whether or not there will be a "physically meaningful" notion of energy depends on how you want to interpret the 0-0 component of the SET, or more generally the double contraction of the SET with a given observer's 4-velocity, which defines the "energy" that is measured locally by that observer.

3. Jun 11, 2012

julian

People try to argue that active diff invafriance has no physical meaning...BUT it implies that energy and momentum are no longer viable physical quantities. But usual physics has been based on these concepts! As Rovelli points out - we have to learn how to do phyics all over again...

4. Jun 11, 2012

julian

The gauge transformations of GR are not coordinate transformations - see thread on the Hole argument. Active diffeomorphims are when you simultaneously drag the gravitational and matter fields over the blank manifold....which is most definitely NOT a coordinate transformation.

5. Jun 11, 2012

Staff: Mentor

Please re-read what I said above. Diffeomorphism invariance is a separate question from whether there is a meaningful notion of energy or momentum in a spacetime. The latter depends on the presence of appropriate Killing vector fields; if they are there, they are there regardless of which coordinate chart you use. So you can have diffeomorphism invariance with a spacetime, such as Schwarzschild spacetime, that has Killing vector fields and therefore has a viable definition of total energy (in the case of Schwarzschild spacetime it's the Komar mass).

GR isn't "based on" the concepts of energy and momentum as defined by Noether's theorem; they aren't fundamental, they are just derived quantities that appear in spacetimes (particular solutions) that have the appropriate Killing vector fields.

6. Jun 11, 2012

Staff: Mentor

Can you link to the thread? The forum search tool isn't finding anything useful.

Ah, I see; you mean something different by "diffeomorphism", something that can actually change the geometry. But if you change the geometry, you change the solution of the EFE that you are talking about, so I don't understand how the statement that "GR is invariant" under these types of transformations means anything, except that the EFE itself is still valid--you've changed the particular solution but not the general equation. But obviously if I change the solution I can change from a geometry that has Killing vector fields to one that doesn't. So what?

7. Jun 11, 2012

julian

Energy of a black hole - how do you define that? By a quasi-local notion... Energy has no local interpretation in GR.

8. Jun 11, 2012

Staff: Mentor

If by "energy" you mean "energy stored in the gravitational field", then yes, there is no "local interpretation" because there's no tensor that describes it. But there is a tensor, the SET, that describes "locally" the energy (and momentum, and pressure, and stress) due to all non-gravitational matter and fields. In the case of a black hole, the SET is zero everywhere (except inside the matter that originally collapsed in the past to form the hole).

9. Jun 11, 2012

julian

Both you and Einstien had this problem! In fact he spent years trying to get out of the 'Hole argument' argument only to return to it and resolve it. The answer is that the coincidence between the values of the gravitational field and the matter field are preserved under active diffeomorphisms and SO this has physical meaning...from this you can formulate a notion of matter being located with respect to the grav field...this is what Rovelli means when he says in GR is about fields living on top of fields.

Give me second to find the Hole argument link. Cheers.

Last edited: Jun 11, 2012
10. Jun 11, 2012

julian

But for SET to satisfy enery-momentum conservation also depends on the existense of Killing vector fields...I think we agree that sometimes these notions dont correspond to conserved quantities...part of my question was if they are not invariant quantities how are we to interpret them?

Last edited: Jun 11, 2012
11. Jun 11, 2012

Staff: Mentor

Hmm. I found this paper by Rovelli on arxiv, entitled "Loop Quantum Gravity and the Meaning of Diffeomorphism Invariance":

http://arxiv.org/pdf/gr-qc/9910079v2.pdf

Section 4 specifically talks about passive and active diffeomorphism invariance. Let me see if I understand what it's saying by giving two examples. Both examples start with the standard Schwarzschild exterior coordinate chart on the exterior vacuum region of Schwarzschild spacetime (i.e., the region outside the horizon). Call that spacetime S and that chart SC.

(1) A passive diffeomorphism is a transformation from SC to some other chart on the same manifold, for example the ingoing Painleve chart, PC. The statement that GR is invariant under passive diffeomorphisms is that the transformation SC -> PC does not change the underlying geometry of S; SC and PC may assign different coordinate 4-tuples to the same events, but all geometric invariants will be the same in both. And, of course, SC on S and PC on S will both be solutions of the EFE; on the surface they will look like different solutions, but computing the geometric invariants tells us that they both describe the same underlying geometry.

(2) An active diffeomorphism is a transformation that retains SC but changes the underlying spacetime from S to something else. For example, we could retain SC but change the underlying spacetime to Minkowski spacetime, M. After this transformation, the metric, and hence all geometric invariants, will look very different in terms of SC on M than they did in terms of SC on S. However, the statement that GR is invariant under active diffeomorphisms means that, if SC on S is a solution of the EFE, so is SC on M. They are *different* solutions, with different geometric invariants, but they're both solutions. So now we have two different geometries described using the same chart.

Am I understanding this correctly?

12. Jun 11, 2012

Staff: Mentor

If by "energy-momentum conservation" you mean *local* conservation, this is not correct. The covariant divergence of the SET is always zero, identically. That only depends on the Bianchi identities, which are automatically satisfied by the LHS of the EFE (the Einstein tensor), and therefore must be satisfied by the RHS as well (the SET).

That depends on the specific problem you're trying to solve. There isn't a single interpretation that always works. Sometimes there isn't any; sometimes there is no useful notion of "energy" beyond the locally measured one (contraction of the SET with 4-velocity).

13. Jun 11, 2012

julian

From the book 'Anvanced general relativity':

"As a second example suppose $T^{ac}$ represents the (symmetric) energy-momentum tensor of a continuous distribution of matter, satisfying $\nabla_c T^{ac} = 0$. If $K$ is a Killing covector define the associated current as $J^a = T^{ac} K_c$. Then by an almost identical caculation one may verify that $\nabla_a J^a$, i.e., the current is conserved."

About active diff invariance, could you have a look at what I said in that thread.

Last edited: Jun 11, 2012
14. Jun 11, 2012

Staff: Mentor

This says that the covariant divergence of the SET is zero, as I said. The rest holds if there is a Killing vector, which I've also agreed with.

15. Jun 11, 2012

julian

"the current is conserved" is the important part...normal physics relies on the fact that we have energy-momentum current conservation.

Not sure how to post links, it was called "What do points on the manifold correspond to in reality?".

16. Jun 11, 2012

Staff: Mentor

GR only relies on the covariant divergence of the SET being zero. If there's a conserved current, it makes the calculations easier, but that doesn't mean a conserved current is required to do physics.

If by "normal physics" you mean "physics in most common scenarios, where there is a conserved current", then your statement is true, but only because it's basically a tautology: "physics where there is a conserved current relies on current conservation".

Just copy the URL from your browser's address bar and paste it into the post; the forum software automatically surrounds it with a url tag. Like so:

17. Jun 11, 2012

julian

18. Jun 11, 2012

Staff: Mentor

Your second part doesn't follow from your first and Noether's theorem. What follows is that energy is not conserved in general, i.e. under general active diffeomorphisms. This should be obvious.

But many non-conserved quantities are nonetheless physically meaningful.

No it doesn't. It states that gravity is determined by the stress-energy tensor, which is a very different thing.

19. Jun 11, 2012

Staff: Mentor

That is my understanding also. An active diffeomorphism involves things like mass magically appearing or disappearing, so it should be no surprise that things like gravitational PE and total energy are not conserved under general active diffeomorphisms.

Last edited: Jun 11, 2012
20. Jun 11, 2012

julian

That's why there was a question mark...I'm wondering if the notions of energy and momentum have become so ambiguous that they dont have anymore physical meaning than do points of spacetime...I mean with the introduction of SR spacetime points became ambiguous to a certain extent, but in GR they lost all objective physical meaning.

Was being a bit sloppy there. Need to refine my question.