# Conservation of energy in general relativity

• I
Qwet
Hello.
I have a question about the law of energy conservation in GR.
As time is inhmogeneous, we don't have energy-momentum 4-vector which would be preserved during system's dynamical change. It is only possible to define 4-vector locally. And next, the problem regarding how to sum this vectors arises.

Do I understand correctly that this problem can be solved only if there is Killing vector field? As the space has to be isometrical according to Noether's theorem (conservation of energy-momentum is given in relation to space-time movement which means we have to consider isometry group). Or it is sufficient to have any kind of symmetry (not just isometry)?
I considered the case where there is no Killing field. And as I understand, it is still possible to construct values satisfying a differential continuity equation. And using Gauss's theorem we can get an integral equation to correspond to the law of conservation. But, again, it is only possible if we have Killing field because the derivatives are covariant in Gauss's theorem (for energy-momentum tensor: DiTik = 0, where Di is covariant). And the equation for covariant divergence doesn't correspond to any conservation law.

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Homework Helper
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The conservation of energy is a rather subtle thing in general relativity. The covariant form of the conservation laws in general relativity can be expressed in the form

(1/√g) ∂j (√gTij) + ΓimnTmn = 0

The square root of the determinant of the metric √g commonly occurs in the form

1/2 ln det g

because

exp ( 1/2 ln det g ) = √g.

This expression 1/2 ln det g is sometimes called the 'dilaton' and it has some role in the fundamental formulas of general relativity. For example

(1/√g) ∂i√g = 1/2 ∂i ln det g = Γjij

Qwet
The conservation of energy is a rather subtle thing in general relativity
Right. And that is the reason for my question.

2022 Award
Local conservation of energy in GR is expressed by the zero divergence of the stress-energy tensor, ##\nabla_\mu T^{\mu\nu}=0##. Roughly speaking, that is the statement that in any small region of spacetime, the stress-energy that enters is the same as the stress-energy that leaves. This is always true as a trivial consequence of the Einstein field equations - ##G^{\mu\nu}=8\pi GT^{\mu\nu}##, and ##\nabla_\mu G^{\mu\nu}=0## is the Bianchi identity.

The complexity is when you try to integrate that and talk about global conservation of energy. That isn't always possible. It is possible in a stationary spacetime, which is one that has a timelike Killing vector field. It's also possible in asymptotically flat spacetimes, which need not have a timelike Killing vector field. But there are interesting spacetimes (e.g. the FLRW spacetime used in cosmology) which are neither asymptotically flat nor stationary. In these, there's no generally agreed method to define a global conservation of energy. The local law still holds.

• Abhishek11235, Demystifier, anuttarasammyak and 5 others
Homework Helper
Gold Member
Right. And that is the reason for my question.

Maybe you can think about this using this analogy. Take the Faraday tensor F. If we have a spatial region Ω, then the magnetic flux through ∂Ω is given by

∂Ω F = magnetic flux

The electric flux through ∂Ω is given by

∂Ω *F = electric flux

The electric charge contained in Ω is given by -∫Ω *j. The idea of pullback is central to understanding these types of formulas, and also the whole subject of integrating differential forms. Since the conservation law in electrodynamics is d*j = 0 is a consequence of d*F = *j, I suspect we have to understand what * and pullback is in the context of T. This tensor can be considered to be a vector valued 1-form (energy 4-current, momentum 4-current), and then we can apply ideas like * and d to it. The pull-back of the energy 4-current to space is the energy, and the flux of this energy is the space momentum. The fact that T is symmetric implies that the space component of the momentum 4-current is the flux of energy in space. This kind of argument about pullbacks etc should give some insight into the meaning of objects like T and *T. Then we can get insight into how to think about the equation d*T = 0. (This equation is of course correct, it is given in MTW, but I'm talking about how to interpret it) I am saying all this in the context of special relativity, but I think it will be useful in the more general case.

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• vanhees71
2022 Award
@love_42 - your maths would be a lot easier to read if you used LaTeX. There's a guide here, also linked immediately below the reply box

• vanhees71
Qwet
It's also possible in asymptotically flat spacetimes
Yes, thank you. I did't mention this fact because in this case the problem is trivial. As I understand, it just simplifies the calculations.
But I am interested if the law still holds its physical sense in the case of no Killing vector field?
As you say,
In these, there's no generally agreed method to define a global conservation of energy
doesn't that mean that there is no physical sense globally (there is no preserved global 4-vector)?

Qwet
Then the conservation law can be written d*T = 0
Yes, but in general relativity we can only write the equality of energy-momentum tensor covariant divergence to zero which is the equation of matter motion. And this gives the first law of thermodynamics. But this is not the energy-momentum conservation law itself, because there is no appropriate 4-vector that would be preserved.

2022 Award
Yes, thank you. I did't mention this fact because in this case the problem is trivial. As I understand, it just simplifies the calculations.
I don't think it's that simple. The asymptotic flatness criterion gives you a way to define global energy that isn't available in a general spacetime.
But I am interested if the law still holds its physical sense in the case of no Killing vector field?
Yes - look up the ADM mass. It can be defined in, for example, the case of a radiating star which is not a stationary spacetime since the central mass is radiating. The ADM mass is not constant, but as far as I understand the change is how you calculate the energy radiating from the star as measured at infinity.
Yes, but in general relativity we can only write the equality of energy-momentum tensor covariant divergence to zero which is the equation of matter motion.
No - it's the equations of motion of stress-energy, not just matter.

Qwet
I don't think it's that simple
I didn't consider it simple. But it seems to be simpler than in case of Killing vectors.
Yes - look up the ADM mass. It can be defined in, for example, the case of a radiating star which is not a stationary spacetime since the central mass is radiating.
Thank you.

Another question - is it possible to use Gauss's theorem to get integral equation for the conservation law in case of no Killing field?

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2022 Award
Another question - is it possible to use Gauss's theorem to get integral equation for the conservation law in case of no Killing field?
You need to define "space" to be able to do integrals over regions of space. I think you can do that arbitrarily in any spacetime, but it's not physically significant since the choice of space is arbitrary.

A stationary spacetime comes with a natural notion of "space", and hence you can find a meaningful measure (the Komar mass) by this approach. However, my understanding of ADM mass is that you can't define it in finite regions (only over the whole spacetime) because it relies on the asymptotically flat region for an unambiguous notion of time.

I could be wrong - @PeterDonis, among others, will know.

Mentor
A stationary spacetime comes with a natural notion of "space", and hence you can find a meaningful measure (the Komar mass) by this approach. However, my understanding of ADM mass is that you can't define it in finite regions (only over the whole spacetime) because it relies on the asymptotically flat region for an unambiguous notion of time.

What you say of the ADM mass is correct. The same thing is also true of the Komar mass; it can only be defined over the whole spacetime, because the integral involved needs to be taken over the entire stationary space, not just a portion of it.

There is also a third concept called the Bondi mass, which, like the ADM mass, is only valid in asymptotically flat spacetimes, but in which the integral is taken to future null infinity, not spacelike infinity.

is it possible to use Gauss's theorem to get integral equation for the conservation law in case of no Killing field?

Yes, if the spacetime is asymptotically flat; the ADM mass and the Bondi mass both provide ways of doing that for asymptotically flat spacetimes.

Note that an asymptotically flat spacetime will have, at the very least, an asymptotic timelike Killing vector field--namely, the timelike KVF of the flat region at infinity. So in a sense, the reason you can define a global "total energy" (mass) for an asymptotically flat spacetime is that there is a timelike Killing vector field, or at least enough of one for that purpose.

• vanhees71 and Ibix
Qwet
Note that an asymptotically flat spacetime will have, at the very least, an asymptotic timelike Killing vector field--namely, the timelike KVF of the flat region at infinity
Is that because derivatives are covariant in Gauss's theorem?
As I understand, if they would be partial, curvature would not prevent us from defining integral equation corresponding to the law?

So in a sense, the reason you can define a global "total energy" (mass) for an asymptotically flat spacetime is that there is a timelike Killing vector field
That means that the spacetime has to be isometric, and there is no another symmetry, energy conservation law would be correct for?
And if so, if we have isometric group in this spacetime (preserving isotropic quadratic form in case of non-euclidean geometry) and dynamical field corresponding to spacetime geometry, we can get global conservation law?
In case of asymptotically flat spacetime this group is Poincaré group, its 10 parameters match maximum number of linearly independent Killing vectors.

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and I actually don't like this. He defines nonconservation of a quantity that does not match total energy for the cases that are well defined. That is, what he says 'changes with change of spacetime' is not the quantity that is conserved in the well accepted special cases. To me, this makes no sense at all. What makes more sense is to say that without a temporal isometry (timelike killing field) or asymptotic flatness, there is no invariant way to compute a total energy. Without being able to compute such a quantity, the question of conservation cannot even be asked.

[added: In recently reviewing Wald, I encountered his argument that the asymptotic flat case can be connected to Noether's theorem in a generalization of the stationary case. That is that it can be shown that there is an asymptotic kvf at infinity, justifying the ADM and Bondi total energies via Noether similar to they way the stationary metric required for Komar energy more directly ties to Noether's theorem.]

• vanhees71
Yes - look up the ADM mass. It can be defined in, for example, the case of a radiating star which is not a stationary spacetime since the central mass is radiating. The ADM mass is not constant, but as far as I understand the change is how you calculate the energy radiating from the star as measured at infinity.
The ADM mass is constant. It is the Bondi mass that decreases, and allow a definition of energy lost to gravitational radiation, among other things.

• PeterDonis and Ibix
Qwet
Only if we define it in static spacetime. If defined in non-static one, it can change with time.

Only if we define it in static spacetime. If defined in non-static one, it can change with time.
This is just false. The ADM mass (or energy) is constant for no matter what dynamics occur as long as you have asymptotic flatness.

• vanhees71
Qwet
It is only valid in static spacetime. It actually makes no physical sense otherwise. It would be like applying a formula valid only in thermal equilibrium to a situation far from equlibrium i.e. nonsense.
While I was editing that I wasn't right, you posted Thank you!

• vanhees71
While I was editing that I wasn't right, you posted Thank you!
Let's just be clear we're on the sampe page:

Komar mass/energy: only defined for stationary spacetimes. Computing otherwise is akin to trying to compute an equilibrium temperature for a non-equilibrium system. Constant by construction. Does not require asymptotic flatness. Computing for a cosmological solution is simply nonsense.

ADM mass/energy: defined only for spacetimes with asymptotic flatness. There is no requirement for stationary isometry, yet this energy is always constant. Establishing that it is constant, totally generally, given this condition is not a trivial result, and is discussed in any of books properly defining and motivating it. However, it is also wholly irrelevant to cosmology because FLRW solutions are not asymptotically flat.

• Qwet
Mentor
Is that because derivatives are covariant in Gauss's theorem?

No, it's because of what "asymptotically flat" means.

if they would be partial, curvature would not prevent us from defining integral equation corresponding to the law?

Using partial derivatives instead of covariant derivatives in a curved spacetime gives you wrong answers.

That means that the spacetime has to be isometric, and there is no another symmetry, energy conservation law would be correct for?
And if so, if we have isometric group in this spacetime (preserving isotropic quadratic form in case of non-euclidean geometry) and dynamical field corresponding to spacetime geometry, we can get global conservation law?
In case of asymptotically flat spacetime this group is Poincaré group, its 10 parameters match maximum number of linearly independent Killing vectors.

I'm not sure what all this means, but if you're asking if there is some other concept analogous to "asymptotically flat" that enables a global energy to be defined and involves other asymptotic Killing vector fields besides the particular timelike one used in asymptotically flat spactimes, the answer is no.

• vanhees71
Mentor
In case of asymptotically flat spacetime this group is Poincaré group

No, it isn't. The asymptotic timelike Killing vector field used in asymptotically flat spacetimes is just one of the ten KVFs in the Poincare group, the "time translation" one. It's not even the only timelike one (the boost KVFs are also timelike).

• vanhees71
Gold Member
2022 Award
Yes, and I would define energy as the conserved "Noether charge" for time-translation invariance, i.e., energy is conserved if and only if the considered spacetime allows for translation invariance along a time-like vector.

What of course always holds is the local conservation law from SR, because in GR at each spacetime point there are always local inertial reference frames, where the laws of SR hold locally. This is what's described by the covariant derivative ##\nabla_{\mu} T^{\mu \nu}=0##, which must be fulfilled due to the Einstein equation alone. It's a Bianchi identity of the Einstein tensor and is related to the fact that GR is a gauge theory.

It's analogous to the well known fact from classical electrodynamics that the local charge conservation law follows from the Maxwell equations alone as a necessary integrability condition and is also related to the fact that classical Maxwell electromagnetism is a gauge theory.

Qwet
Using partial derivatives instead of covariant derivatives in a curved spacetime gives you wrong answers
What do you mean by "wrong"? Partial derivatives can be used in curved spacetime.
Gauss's theorem can be used even in space with undefined metric.

Mentor
What do you mean by "wrong"?

That you obtain wrong equations that make wrong predictions for experimental results.

Partial derivatives can be used in curved spacetime.

Not if you want to make correct predictions.

Gauss's theorem can be used even in space with undefined metric.

Gold Member
2022 Award
Stokes's theorem is valid independent from any metric, because it can be formulated in terms of alternating differential forms, for which you don't need a metric nor a connection. a "plain vanilla differentiable manifold" is sufficient.

In the Ricci calculus that's, because the generalized Kronecker symbols ##\delta_{\mu_1,\ldots,\mu_k}^{\nu_1,\ldots,\nu_k}##, which are defined to be totally antisymmetric when permuting ##\mu_j## or ##\nu_j## indices, respectively and the symbol is 1 if the tuples of indices are the same, i.e., if ##(\mu_1,\mu_2,\ldots,\mu_k)=(\nu_1,\ldots,\nu_k)##, behave like tensor components under arbitrary diffeomorphisms. Here, of course, ##k \leq d##, when ##d## is the dimension of the manifold.

To formulate Gauss's theorem you need, however, Hodge dualization and the Levi-Civita tensor, for which you need a metric. The components of the Levi-Civita tensor are
$$\epsilon^{\mu_1,\ldots,\mu_d}=\sqrt{|g|} \bar{\epsilon}^{\mu_1,\ldots,\mu_d},$$
where ##\bar{\epsilon}^{\mu_1,\ldots,\mu_d}## is the usual Levi-Civita symbol of flat space described in (pseudo-)orthogonal coordinates, i.e., ##\bar{\epsilon}^{12\ldots d}=1## and otherwise totally antisymmetric under permutations of indices. To be a (pseudo-)tensor you need the factor ##\sqrt{|g|}##, where ##g=\mathrm{det} g_{\mu \nu}##, where ##g_{\mu \nu}## are the (pseudo-)metric components.

For details for the case of Minkoswki space, see App. A.6 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

• • PAllen, atyy and etotheipi
Qwet
I do not understand how partial derivatives (from calculus of function of multiple variables) relate to "space", and Gauss's theorem to metric. In any book in a section about Gauss's theorem, there should not be any word about metric.

To formulate Gauss's theorem you need, however, Hodge dualization and the Levi-Civita tensor, for which you need a metric
Why? We only need the "usual" Levi-Civita symbol which is "pseudo"-tensor (tensor density) by itself.
Dualization is given exactly by it and by "pseudo"-tensor reciprocal to it. Metric is an extra structure.

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I don't know, how you want to even formulate Gauss's theorem without Hodge duality and how to define the Levi-Civita tensor without a metric. It's clear that you don't need a metric for Stokes's theorem which works within the calculus of alternating differential forms, where you don't need a metric nor a connection.

The usual Levi-Civita symbol do not provide tensor components under general diffeomorphisms. That's the case only in flat spacetime and for pseudo-orthonormal bases.

Homework Helper
Gold Member
Why? We only need the "usual" Levi-Civita symbol which is "pseudo"-tensor (tensor density) by itself.
Dualization is given exactly by it and by "pseudo"-tensor reciprocal to it. Metric is an extra structure.

The hodge dual of a p-form β can be defined as

*β(vp+1, ..., vn) = β(v1, ..., vp)

where v1, ..., vn are oriented orthonormal vectors in TxM.

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How can be vectors orthonormal without a (pseudo-)metric?

Homework Helper
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I was not suggesting that they can be. Just sharing a useful way of defining it.