There is the standard definitive statement of the conservation of energy momentum:


Which one expects to be upheld everywhere on some spacetime region M i.e four-volume (covariant divergence if not flat).

$$\intop_{M}\nabla_{\mu}T^{\mu\nu}dM=0$$ This has been shown, for the covariant (non-flat) case, (The theory of Relativity, Moller, section 126) to be equivalent to the statement:


Where $\tau$ is the gravitational stress-energy-pseudotensor. Which allows one to apply Gauss's theorem, bringing our integral to some bounding three-surface $\partial M_{\nu}$:

$$\intop_{M}\partial_{\mu}\left(T^{\mu\nu}+\tau^{\mu\nu}\right)=\intop_{\partial M}\left(T^{\mu\nu}+\tau^{\mu\nu}\right)d(\partial M)_{\mu}=0$$

For a non compact spacetime and spacelike boundary (choosing $\nu=0$) one will in general, have two spacelike surfaces of infinite extent ( $\partial M_{1}$,$\partial M_{2}$) at different times to bound some 4-region, leading to either the conclusion that energy momentum is the same over each surface, cancelling to zero over both, due to opposite surface normals (energy momentum conserved), or that it vanishes on each surface individually (energy still conserved).

For the compact case, one can enclose a region M with one hypersurface (think a comoving frame at one moment in cosmological time in the closed FRW universe for instance).

leading to the conclusion that the total energy momentum of a closed universe is always zero. I've seen arguments both ways for the closed case, is this a valid argument?


1 Answer 1


No. All the equation tells you is that the total energy momentum (as expressed by the integral of matter tensor and the gravity pseudotensor) is conserved, for the two hypersurfaces case. The energy momentum is the same when integrated at one time as at another time.

The compact case is treated below.

[deleted text on flux. The argument is more complex for the compact case, and more material was added in]

Before we do that it is clear that the pseudotensor is not covariant. However, it's been proven that the integral is Lorentz invariant, IF the integrals are done at infinity and spacetime is asymptotically flat (AF) at infinity. The AF condition is needed to impose a Lorentz structure at infinity, which then can be used for the integral. If you had some preferred coordinate in the whole spacetime such as what you get with a timelike Killing vector then it's also ok. But the integral only needs the AF condition.

Still, it's fair to note that the flat asymptotic spacetime does NOT hold for cosmology, where the FLRW metric is never flat spacetime (yes, it looks like flat space, NOT flat spacetime). It is known that energy is not conserved in cosmology. It's not only not zero, it keep growing ad infinitum

Anyway, whether you do the integral with Lorentz covariance or not, that energy is not zero, in general.


Note that a compact manifold may mean with boundaries, or without boundaries. In cosmology the closed universe has compact spatial surfaces (a 3 sphere) but without boundaries. Boundaries means edges, and we haven't seen any.

FROM WIKIPEDIA at https://en.wikipedia.org/wiki/Shape_of_the_universe#Bounded_or_unbounded:

When cosmologists speak of the universe as being "open" or "closed", they most commonly are referring to whether the curvature is negative or positive. These meanings of open and closed are different from the mathematical meaning of open and closed used for sets in topological spaces and for the mathematical meaning of open and closed manifolds, which gives rise to ambiguity and confusion. In mathematics, there are definitions for a closed manifold (i.e., compact without boundary) and open manifold (i.e., one that is not compact and without boundary). A "closed universe" is necessarily a closed manifold. An "open universe" can be either a closed or open manifold. For example, in the Friedmann–Lemaître–Robertson–Walker (FLRW) model the universe is considered to be without boundaries, in which case "compact universe" could describe a universe that is a closed manifold.


First, doing your integral: you can do the integral at two different times, and use the two 3 spheres (of a closed universe) as the boundaries at those times, but you also need to deal with what happens (ie, the integral) at spatial infinity. Unfortunately the spacetime at infinity (spatial or otherwise) is not Minkowski, and there is no Lorentz covariant structure you can impose there, nor use for your integral. So it would not be a Lorentz covariant entity. The energy density is finite at spatial infinity, so is momentum density. If you integrate the T tensor you would not get zero. If you integrate any version of a $\tau$ pseudotensor, you'll get arbitrary numbers depending on which you choose and the coordinate system.

The bottom line here is that there is nothing definitive that you can say about the integral when it is not Lorentz covariant.


It has its history, with Newtonian arguments about matter and gravitational energy cancelling out, and other proposals:

In https://en.wikipedia.org/wiki/Zero-energy_universe the history of it is that:


Pascual Jordan first suggested that since the positive energy of a star’s mass and the negative energy of its gravitational field together may have zero total energy, conservation of energy would not prevent a star being created by a quantum transition of the vacuum. George Gamow recounted putting this idea to Albert Einstein: “Einstein stopped in his tracks and, since we were crossing a street, several cars had to stop to avoid running us down”.[3]

The zero-energy universe theory originated in 1973, when Edward Tryon proposed in the Nature journal that the universe emerged from a large-scale quantum fluctuation of vacuum energy, resulting in its positive mass-energy being exactly balanced by its negative gravitational potential energy.[4]

Free-lunch interpretation

A generic property of inflation is the balancing of the negative gravitational energy, within the inflating region, with the positive energy of the inflaton field to yield a post-inflationary universe with negligible or zero energy density.[5][6] It is this balancing of the total universal energy budget that enables the open-ended growth possible with inflation; during inflation, energy flows from the gravitational field (or geometry) to the inflation field—the total gravitational energy decreases (i.e. becomes more negative) and the total inflation energy increases (becomes more positive). But the respective energy densities remain constant and opposite since the region is inflating. Consequently, inflation explains the otherwise curious cancellation of matter and gravitational energy on cosmological scales, which is consistent with astronomical observations.[7]


In General Relativistic cosmology there is no conserved pseudotensor since there is no asymptotically flat spacetime from which to extract some Lorentz covariance. It is known that the energy in the cosmological spacetimes keeps increasing. There is also the cosmological constant which keep increasing the energy proportionally to the volume of the universe. When the universe approaches empty spacetime, it'll still have the cosmological constant and then there is no conserved energy either.


You might know that if one talks about lightlike infinity being asymptotically flat, the energy momentum is also conserved, but it is different than if you use spacelike hypersurfaces. In the spacelike case it is the ADM mass (which for gravitational radiation emitted by a black hole is the mass of both the black hole plus the radiation), whereas for lightlike infinity it is only the black hole mass, as the flux of energy through those lightlike hypersurfaces takes the gravitational radiation energy out. That's the so called Bondi mass, and the black hole looses it as it radiates. That is what is used to figure out the mass or energy lost by a Black Hole (a compact spacetime inside, with boundaries, but also singularities), and carried away by gravitational radiation. The sum of the two is not zero, and it can radiate all its mass away, eventually.

I don't have the Weinberg book to know or interpret exactly what he said, but the treatments for the kinds of mass (or energy in the sense that they are the same) that are conserved is also treated very well, both physically and mathematically, in Wald.

So, it gets tricky, but unless you define the total energy momentum as T-G (which you can trivially get from a Lagrangian variational treatment) you don't get zero in general, and you only have pseudotensors and generally not Lorentz invariance.

  • $\begingroup$ @Bob_Bee Thanks Bob, of course you're right in that it's just expressing energy conservation. I'll pull out my Wald and check out what you mentioned (: I've another angle in working on I'll post later (: $\endgroup$
    – R. Rankin
    Commented May 27, 2017 at 5:24
  • $\begingroup$ @Bob_Bee Apologies Bob, I'm unsatisifed here in that the lattermost expression is an integral over a three space, not a surface, so flux is not what it is measuring. (for example for the 0-0 componentit's integrating energy density over a volume). The pseudotensor as a density has little physical sense, but it's integral should be at least Lorentz covariant according to most all of my texts. $\endgroup$
    – R. Rankin
    Commented May 28, 2017 at 23:15
  • $\begingroup$ @R.Rankin. I will check and comment back either way. $\endgroup$
    – Bob Bee
    Commented May 29, 2017 at 2:13
  • $\begingroup$ @R.Rankin You followed up with a fair point, and I had to be more clear and more complete, so the answer was edited. $\endgroup$
    – Bob Bee
    Commented May 29, 2017 at 6:59
  • $\begingroup$ @Bob_Bee That answer is much appreciated! (: $\endgroup$
    – R. Rankin
    Commented May 29, 2017 at 8:08

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