# Is the total energy of the universe zero?

In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, "because the positive energy of matter is cancelled out by the negative energy of the gravitational field".

But I can't find anything concrete to substantiate this claim. As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius $R$ using Newton's Laws and threw in $E=mc^2$ for energy of the sphere, and it was by no means obvious that the answer is zero!

So, my questions:

1. What is the basis for the claim – does one require General Relativity, or can one get it from Newtonian gravity?

2. What conditions do you require in the model, in order for this to work?

3. Could someone please refer me to a good paper about this?

• Cross-posted from mathoverflow.net/q/38659 – Qmechanic Feb 24 '12 at 0:53
• Although Guth and Motl have both used GR to arrive at opposite conclusions about the question stated in this question's title (with Guth's viewpoint supported in Gibbs' answer), I think the issue may be seen more clearly thru Einstein-Cartan Theory, which Nikodem J. Poplawski applied to inflationary cosmology in numerous papers, available free on Arxiv, between 2009 & 2019. Relying heavily on causal separations between regions on declining spatial & temporal scales, his cosmology uses an infinite divisibility of space & time to sustain G.'s viewpoint, albeit without use of a scalar field. – Edouard Mar 9 '20 at 19:22

## 6 Answers

On my blog, I published a popular text why energy conservation becomes trivial (or is violated) in general relativity (GR).

To summarize four of the points:

1. In GR, spacetime is dynamical, so in general, it is not time-translation invariant. One therefore can't apply Noether's theorem to argue that energy is conserved.

2. One can see this in detail in cosmology: the energy carried by radiation decreases as the universe expands since every photon's wavelength increases. The cosmological constant has a constant energy density while the volume increases, so the total energy carried by the cosmological constant (dark energy), on the contrary, grows. The latter increase is the reason why the mass of the universe is large - during inflation, the total energy grew exponentially for 60+ $e$-foldings, before it was converted to matter that gave rise to early galaxies.

3. If one defines the stress-energy tensor as the variation of the Lagrangian with respect to the metric tensor, which is okay for non-gravitating field theories, one gets zero in GR because the metric tensor is dynamical and the variation — like all variations — has to vanish because this is what defines the equations of motion.

4. In translationally invariant spaces such as Minkowski space, the total energy is conserved again because Noether's theorem may be revived; however, one can't "canonically" write this energy as the integral of energy density over the space; more precisely, any choice to distribute the total energy "locally" will depend on the chosen coordinate system.

• (1) "every photon's wavelength increases" - It does not. In the frame of the emitter, the photon doesn't redshift. In the frame of the receiver, the photon is emitted already redshifted and doesn't redshift in flight. (2) "The cosmological constant has a constant energy density" - There is no cosmological constant. It is not required by the equivalence principle. The sole purpose of this constant is to save the Friedmann model, but this model has failed anyway. (3) "during inflation, the total energy grew exponentially" - There is no evidence for inflation. It is only a speculation. – safesphere Apr 19 '20 at 22:33

(Now I notice you're the same person who asked this at MathOverflow, where I've previously answered something similar -- if you didn't like the answer then, you won't like it now.)

This is really just expanding on Marek's comment:

How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.

What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the whole action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.

This is what "canceled out by the negative energy of the gravitational field" means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.

The claim that the total energy in the universe is zero can be rigorously justified.

To answer your specific questions:

1. General Relativity is required. It does not apply for Newtonian gravity.

2. It has to be assumed that classical general relativity, with or without cosmological constant, is correct and that the universe is spatially homogeneous on sufficiently large scales. If the universe is infinite the total energy is not really defined, but it is still true that the total energy in an expanding volume of space is asymptotically zero when the region is large enough for the homogeneity of the universe to be a good enough approximation.

3. Here is a link to a paper as requested.

I'm a little out of my depths here, but I suspect you're asking about the density parameter, and will proceed on that assumption.

In the accepted big-bang-and-inflation scenario, and before for we had evidence for the existence of dark energy, it was possible to talk about the possible fate of the universe (open or closed) in terms of the initial expansion as balanced by the total mass only.

Now, in that model, for the universe to be as big, as dense, and as old as we see it, that balance must have been very nearly at the critical value between open and close (a geometry called "flat").

This claim was allowed by measurement, and preferred on a philosophical basis by some theorists.

Try the wikipedia article on Friedmann equations for some more discussion. You're looking for $\Omega = \rho/\rho_c \approx 1$. Or there may be better links.

Note, however that the issues are changed rather a lot by the presence of dark energy in the universe. There is no chance of a closed geometry, and we are doomed to a cold and lonely ending in the far distance future as accelerating expansion rips the regions of low entropy ever further apart.

General relativity has difficulty in defining what is energy. In loose terms the difficulty boils down to the fact that gravitational energy can not be localized.

For a speculative blog about these matters, see: https://www.science20.com/hammock_physicist/square_root_universe

I will try to answer in the view of General Relativity.

I quote directly from Einstein and Rosen paper :

The four-dimensional space is described mathematically by two congruent parts or "sheets", corresponding to $u > 0$ and $u < 0$, which are joined by a hyperplane $r = 2m$ or $u = 0$ in which $g$ vanishes. We call such a connection between the two sheets a "bridge". We see now in the given solution, free from singularities, the mathematical representation of an elementary particle (neutron or neutrino). Characteristic of the theory we are presenting is the description of space by means of two sheets. A bridge, spatially finite, which connects these sheets characterizes the presence of an electrically neutral elementary particle. With this conception one not only obtains the representation of an elementary particle by using only the field equations, that is, without introducing new field quantities to describe the density of matter; one is also able to understand the atomistic character of matter as well as the fact that there can be no particles of negative mass. The latter is made clear by the following considerations. If we had started from a Schwarzschild solution with negative $m$, we should not have been able to make the solution regular by introducing a new variable $u$ instead of $r$; that is to say, no "bridge" is possible that corresponds to a particle of negative mass.

A.Einstein, N.Rosen - "The Particle Problem in the General Theory of Relativity"

This was the paper that investigated ER bridges (wormholes). It would be useful to read the paper to understand further more why there can be no negative mass particles.

So, the total energy of the universe can not be zero, because there can be no particles of negative mass. So, in the light of mass/energy equivalence, the energy of the universe cannot be zero.