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I understand that energy conservation is not a rule in general relativity, but I'd like to know under what circumstances it can still be possible. In other words, when is it possible to associate a potential energy to the gravitational field, so that the energy is constant in the evolution of the system?

Here are some examples, is there a convenient way to define energy in these scenarios?

  • Just a system of gravitational waves.
  • A point mass moving in a static (but otherwise arbitrary) space-time. Equivalent (if I'm not mistaken) to a test mass moving in the field of a second much larger mass, the larger mass wouldn't move.
  • Two rotating bodies of similar mass.

Overall, I'm trying to understand what keeps us from associating a potential energy to the metric. When we break the time translation symmetry of a system by introducing an electromagnetic field, we can still conserve energy by defining an electromagnetic potential energy. Why can't we do the same when we break TT symmetry by making space-time curved?

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Good question, one I've been pondering many times. Certain people hold that the conservation of energy is vacuous in GR while others think that concepts like ADM energy make the conservation non-trivial. See this article and discussions therein. –  Marek Jan 7 '11 at 13:43
    
I am not sure, but I think it has something to do with the global properties of the metric, especially how it behaves asymptotically. –  Sklivvz Jan 7 '11 at 16:34
    
@Skli: I think so too. It must be some property of the metric that keeps us from defining a potential energy. And it probably comes from Einstein's field equations. But I have no idea what it could be. –  Malabarba Jan 7 '11 at 16:54
    
@Sklivvz : it actually comes directly from the metric. The problem is that GR has the metric serving a dual role of being a measuring stick for objects AND also being the thing that defines what the gravitational field is. A dynamical field means that your measuring stick is changing, and then what do you use to measure THAT. –  Jerry Schirmer Jan 7 '11 at 19:56

3 Answers 3

up vote 13 down vote accepted

There are a few different ways of answering this one. For brevity, I'm going to be a bit hand-wavey. There is actually still some research going on with this.

Certain spacetimes will always have a conserved energy. These are the spacetimes that have what is called a global timelike (or, if you're wanting to be super careful and pedantic, perhaps null) Killing vector. Math-types will define this as a vector whose lowered form satisfies the Killing equation: $\nabla_{a}\xi_{b} + \nabla_{b} \xi_{a} = 0$. Physicists will just say that $\xi^{a}$ is a vector that generates time (or null) translations of the spacetime, and that Killing's equation just tells us that these translations are symmetries of the spacetime's geometry. If this is true, it is pretty easy to show that all geodesics will have a conserved quantity associated with the time component of their translation, which we can interpret as the gravitational potential energy of the observer (though there are some new relativistic effects--for instance, in the case of objects orbiting a star, you see a coupling between the mass of the star and the orbiting objects angular momentum that does not show up classically). The fact that you can define a conserved energy here is strongly associated with the fact that you can assign a conserved energy in any Hamiltonian system in which the time does not explicitly appear in the Hamiltonian--> time translation being a symmetry of the Hamiltonian means that there is a conserved energy associated with that symmetry. If time translation is a symmetry of the spacetime, you get a conserved energy in exactly the same way.

Secondly, you can have a surface in the spacetime (but not necessarily the whole spacetime) that has a conserved killing tangent vector. Then, the argument from above still follows, but that energy is a charge living on that surface. Since integrals over a surface can be converted to integrals over a bulk by Gauss's theorem, we can, in analogy with Gauss's Law, interpret these energies as the energy of the mass and energy inside the surface. If the surface is conformal spacelike infinity of an asymptotically flat spacetime, this is the ADM Energy. If it is conformal null infinity of an asymptotically flat spacetime, it is the Bondi energy. You can associate similar charges with Isolated Horizons, as well, as they have null Killing vectors associated with them, and this is the basis of the quasi-local energies worked out by York and Brown amongst others.

What you can't have is a tensor quantity that is globally defined that one can easily associate with 'energy density' of the gravitational field, or define one of these energies for a general spacetime. The reason for this is that one needs a time with which to associate a conserved quantity conjugate to time. But if there is no unique way of specifying time, and especially no way to specify time in such a way that it generates some sort of symmetry, then there is no way to move forward with this procedure. For this reason, a great many general spacetimes have quite pathological features. Only a very small proprotion of known exact solutions to Einstein's Equation are believed to have much to do with physics.

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I should point out that the essential difference with E&M here is that, although the electromagnetic field is dyanmic, the hamiltonian of the E&M field still does not explicitly include time--it only includes time in the dependence of $\vec E$ and $\vec B$ on time. This makes time translation a symmetry of its hamiltonian. For a spacetime with no timelike or null killing vector, there is no vector that generates time translations in this way. And with no such symmetry, there is no energy defined. –  Jerry Schirmer Jan 7 '11 at 19:59
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Sorry I am nit-picky, but math-types would definitely define the Killing vector as ${\mathcal L}_{\xi}\, g = 0$ :-P –  Marek Jan 7 '11 at 20:30
    
Heh. Fair enough. But they're equivalent if you have a metric-compatible connexion, and I didn't want to get into what the Lie Derivative was. –  Jerry Schirmer Jan 7 '11 at 20:37
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@jerry, added "small" in "only a very small proportion" ;-) Hopefully that's what you meant (maybe it was "tiny") –  Sklivvz Jan 7 '11 at 21:06
    
@Jerry: Thanks. I knew it had to be about breaking time translation, but I didn't know what broke it. –  Malabarba Jan 8 '11 at 23:46

Energy conservation does work perfectly in general relativity. The overall Lagrangian is invariant under time translations and Noether's Theorem can be used to derive a non-trivial and exact conserved current for energy. The only thing that makes general relativity a little different from electromagnetism is that the time translation symmetry is part of a larger gauge symmetry so time is not absolute and can be chosen in many ways. However there is no problem with the derivation of conserved energy with respect to any given choice of time translation.

There is a long and interesting history to this problem. Einstein gave a valid formula for the energy in the gravitational field shortly after publishing general relativity. The mathematicians Hilbert and Klein did not like the coordinate dependence in Einstein's formulation and claimed it reduced to a trivial identity. They enlisted Noether to work out a general formalism for conservation laws and claimed that her work supported their view.

The debate continued for many years especially in the context of gravitational waves which some people claimed did not exist. They thought that the linearised solutions for gravitational waves were equivalent to flat space via co-ordinate transformations and that they carried no energy. At one point even Einstein doubted his own formalism, but later he returned to his original view that energy conservation holds up. The issue was finally resolved when exact non-linear gravitational wave solutions were found and it was shown that they do carry energy. Since then this has even been verified empirically to very high precision with the observation of the slowing down of binary pulsars in exact agreement with the predicted radiation of gravitational energy from the system.

The formula for energy in general relativity is usually given in terms of pseudo tensors such as those proposed by Laundau & Lifshitz, Dirac, Weinberg or Einstein himself. Wikipedia has a good article on these and how they confirm energy conservation. Although pseudotensors are mathematically rigorous objects which can be understood as sections of jet bundles, some people don't like their apparent co-ordinate dependence. There are other covariant approaches such as the Komar Superpotential or a more general formula of mine which gives the energy current in terms of the time translation vector $k^{\mu}$ as

$ J^{\mu}_G = \frac{1}{16\pi G} (k^{\mu}R - 2k^{\mu}\Lambda - 2{{k^{\alpha}}_{;\alpha}}^{\mu} + {{k^{\alpha}}_{;}}^{\mu}_{\alpha}+ {{k^{\mu}}_{;}}^{\alpha}_{\alpha})$

Despite these general formulations of energy conservation in general relativity there are some cosmologists who still take the view that energy conservation is only approximate or that it only works in special cases or that it reduces to a trivial identity. In each case these claims can be refuted either by studying the formulations I have referenced or by comparing the arguments given by these cosmologists with analogous situations in other gauge theories where conservation laws are accepted and follow analogous rules.

One area of particular contention is energy conservation in a homogeneous cosmology with cosmic radiation and a cosmological constant. Despite all the contrary claims, a valid formula for energy conservation in this case can be derived from the general methods and is given by this equation.

$ E = Mc^2 + \frac{\Gamma}{a} + \frac{\Lambda c^2}{\kappa}a^3 - \frac{3}{\kappa}\dot{a}^2a - Ka = 0$

$a(t)$ is the universal expansion factor as a funcrtion of time normalised to 1 at the current epoch.

$E$ is the total energy in an expanding region of volume $a(t)^3$. This always comes to zero in a perfectly homogeneous cosmology.

$M$ is the total mass of matter in the region

$c$ is the speed of light

$\Gamma$ is the density of cosmic radiation normalised to the current epoch

$\Lambda$ is the cosmological constant, thought to be positive.

$\kappa$ is the gravitational coupling constant

$K$ is a constant that is positive for spherical closed space, negative for hyperbolic space and zero for flat space.

The first two terms describe the energy in matter and radiation with the matter energy not changing and the radiation decreasing as the universe expands. Both are positive. The third term is "dark energy" which is currently though to be positive and contributing about 75% of the non-gravitational energy, but this increases with time. The final two terms represent the gravitational energy which is negative to balance the other terms.

This equation holds as a consequence of the well-known Friedmann cosmological equations, that come from the Einstein field equations, so it is in no sense trivial as some people have claimed it must be.

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Seems like a very thorough work to. –  Robert Filter Jan 20 '11 at 10:56
    
Ok, so you can always define a gravitational energy so that the overall energy of the system conserves. But, is this definition of gravitational energy always a same predefined function of the metric, or do you need to a new way to define it whenever you work with a new gravitational field? For instance, there's a single definition I can use for electromagnetic potential energy (in terms of the electric and magnetic field density), and this definition will guarantee conservation of energy for any system with any electromagnetic field. Is there similar definition for gravitaty energy? –  Malabarba Jan 20 '11 at 13:19
    
Even for the electromagnetic field the energy depends on the reference frame. In general relativity there is a greater choice of valid reference frames which define different quantities for the energy. This is not a problem either in practice or in principle. For example in the case of the cosmological solutions the choice of reference frame is usually taken to be comoving co-ordinates which leads to the formula above. Other choices would also be valid. –  Philip Gibbs Jan 20 '11 at 13:33
    
I gave a general expression above for the energy current 4-vector that works for any gravitational field. Perhaps I should have clarified that getting the energy in a region of space from this current just requires integrating the time component of the current over the region. –  Philip Gibbs Jan 20 '11 at 13:36
    
Nah, it's clear enough. I just had to read your answer a couple more times. :-) –  Malabarba Jan 20 '11 at 13:40

o carry on with Jerry Schirmer, the Killing vector defines isometries on a manifold. If there exists a Killing vector $K_t~=~\partial/\partial t$ this means the momentum $K_t\cdot P~=~$ constant. This then is a statement which can be interpreted as the constancy of an observable labeled energy. As a rule of thumb, if a metric component involves time in an explicit way, and for example $K_t~\propto~\sqrt{g_{tt}(t)}$ is not proper or the action of this vector is not an isometry.

This happens with the FLRW equation of cosmology. In a de Sitter form we have $$ ds^2~=~dt^2~-~e^{\sqrt{\Lambda/3}t}(dr^2~+~r^2d\Omega^2), $$ which has a time dependency. So we are not able to derive a conservation of energy from first principles. The Ricci curvature is $R_{\mu\nu}~=~\Lambda g_{\mu\nu}$, and for $k~=~0$ the spatial curvature is zero. The cosmological constant is dependent on the vacuum energy density, plus pressure terms. With the equation of state $p~=~-\rho$, which approximates observational data pretty well, one can do some detailed balance work to show the universe is a net nothing and remains so.

Does this connects with something deeper than just a "detailed balance? It might, and I suspect that Phillip's analysis connects with this. The deSitter metric is a time dependent conformal theory of a flat metric. For $g'~=~\Omega^2g$ the line element for $g'$ $$ ds^2~=~\Omega^2(du^2~-~d\sigma_{space}^2). $$ However for the time variable $du^2~=~\Omega^{-2}dt^2$ $\Omega$ is time dependent and $$ ds^2~=~dt^2~-~\Omega^2(t)d\sigma_{space}^2. $$ This recovers the de Sitter metric for $\Omega^2(t)~=~exp(\sqrt{\Lambda/3}t)$. The de Sitter spacetime is then conformally equivalent to a flat spacetime which trivially has a $K_t~=~\partial/\partial t$. So the spacetime we observe with the equation of state $p~=~-\rho$ is a class of spacetimes which are conformal to flat spacetime and which also preserve $E~=~constant$. I think that Phillip’s work on this matter projects out this special case of conformal spacetimes.

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