# 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?

• Cross-posted from mathoverflow.net/q/38659 Commented Feb 24, 2012 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. Commented Mar 9, 2020 at 19:22

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. Commented Apr 19, 2020 at 22:33
• @safesphere The definition of energy is the ability of one system to perform work on another system. If you want to keep that definition, then the Doppler shift does decrease the ability of distant systems to perform work on each other. There is no way around that. You could argue that energy is the wrong quantity to keep our eyes on in the cosmological context, but then please give us a new and better observable than energy, first. Commented Jun 17, 2023 at 1:00
• @FlatterMann Energy is frame dependent. The energy conservation law holds when the energy before and after is measured in the same frame. When you properly do this, there is no Doppler shift. In cosmology, as well as in gravity, this shift is merely the effect of measuring in different frames. Since energy is frame dependent, you get different measurements as expected. This doesn’t violate energy conservation. It is like to say that a car moving at 100 mph has no energy to hit another car moving with the same speed in the same direction. True, but this doesn’t violate energy conservation. Commented Jun 17, 2023 at 16:11
• @safesphere Frames have nothing to do with the definition of energy. If you have energy, then you can perform work, if you don't, then you can't. Like I said, energy is probably the wrong observable, but you have to come up with a better one. Complaining about frames makes no difference about the facts on the ground. Commented Jun 17, 2023 at 17:31
• @FlatterMann If you stand on the road, you can get hit by a moving car. In your terminology, the car can do work on you when it hits you. Yet when you sit in this car (different frame), you are in touch with it, but it cannot hit you or do work on you. So work directly depends on the frame. The “facts on the ground” you refer to relate to “the total energy in the universe”. On one hand, it does not depend on the frame, but on the other hand, it is not a definition of energy since energy depends on the frame. Strictly speaking, the total energy of the universe is undefined. Commented Jun 17, 2023 at 18:23

(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.

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. The total energy is also exactly zero in a closed spatially finite cosmology whether or not it is homogeneous.

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.

• Or are we? CCC takes the lemons of a lonely ending and makes the lemonade of a brand new universe out of it with a fairly trivial and appealing rescaling argument that gels perfectly with one of the other grand questions of physics: "What in nature sets scale?". If the answer is "Nothing.", then CCC sounds like a home run to me that kills a large number of flies at once. Commented Jun 17, 2023 at 1:04
• @FlatterMann How does CCC argue for a new inflationary period and deny the continuation of a universe with only massless particles? Commented Jun 17, 2023 at 4:07
• @Wookie You can interpret CCC as a phase transition. In the standard model all fields are massless. They obtain an effective mass phenomenologically by interaction. At high energies/temperatures that mass term becomes irrelevant. One can argue in reverse that we are simply at such a high temperature that the mass of the photon doesn't matter and that other effective interactions will dominate in what we would call an ultra-cold later-era universe. The inhabitants of that era will look at us as their inflationary phase. It's simple, elegant... and probably completely false. I like it anyway. Commented Jun 17, 2023 at 5:20

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.