# Total energy of the Universe

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?

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Assuming GR, there is a contrived sense in which that statement holds (look at ${}_{00}$ component of Einstein equations) but note that there is no simple definition of "energy of the gravitational field" and even then it's definitely not negative. – Marek Jan 14 '11 at 2:50
One of my pop-science sources is "A Brief History of Time" by Stephen Hawking. He uses this claim to suggest that the Universe could spontaneously self-generate from "absolutely nothing". – user1265 Jan 14 '11 at 3:12
But I must admit, I have never seen anything concrete supporting the claim. In fact, there seems to be a Positive Energy Theorem in GR which explicitly violates the claim, but I don't understand it well enough to comment. – user1265 Jan 14 '11 at 3:14
If you go to Youtube and search for "A universe from nothing", you'll find an awesome talk by Lawrence Krauss. It's entertaining yet deeply illuminating. Heck, I might just go and watch it again. – Lagerbaer Jan 14 '11 at 4:14
Cross-posted from mathoverflow.net/questions/38659 – Qmechanic Feb 24 '12 at 0:53

A longer popular text why energy conservation becomes trivial (or violated) in general relativity is e.g. here: http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html

To summarize four of the points:

1. In GR, spacetime is dynamical, so in general, it is not time-translational-invariant. Because it's not, one can't apply Noether's theorem to argue that there is a conserved energy.

2. One can see this in details in cosmology: the energy carried by radiation decreases as the universe expands as every photon's wavelength is getting bigger; the cosmological constant has a constant energy density while the volume is increasing, 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 OK for non-gravitating field theories, on 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 the 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.

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Thanks a lot, Lubos. If I understand you right, energy is not conserved in GR, so it doesn't matter at all whether the "total energy" is zero or not, since it can't even be defined properly. – user1265 Jan 16 '11 at 4:20
If I understand right, your point 3 is the same as Matt Reece's comment above ? – user1265 Jan 16 '11 at 4:22
I disagree with Motl's answer and have refuted it at length on my blog at blog.vixra.org/category/energy-conservation . I recommend that anyone intersted read all these discussions at length and form their own conclusions rather than relying on anyone's assumed expertise, because this issue has been controversial amongst experts for many years. – Philip Gibbs - inactive Jan 20 '11 at 10:51
@Philip Gibbs: Both your physics and your history are wrong. See MTW, p. 457, for a good explanation. Luboš Motl is correct. – Ben Crowell Aug 5 '11 at 4:48
It just means that the fields describing the geometry of the spacetime, $g_{\mu\nu}(x^\alpha)$, are not predetermined constants but rather space- and time-dependent observables analogous to other fields (or coordinates of particles) such as the electromagnetic fields. "Dynamical" is something that nontrivially evolves with time according to the laws of physics and that can be measured. – Luboš Motl Apr 4 '15 at 13:11

(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.

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Thanks, Matt. Its not that I didn't like your answer - I just wanted to see if there was any substance to the "zero energy" argument (I was rather skeptical about it), or whether it was just a hand-waving kind of claim based on the calculation you showed. Appears it was the latter. – user1265 Jan 16 '11 at 4:25

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.

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

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Thanks for the link. Am not sure if this is the same thing Matt and Lubos are talking about, so need to read the paper. – user1265 Jan 16 '11 at 4:17
The paper shows that the comments due to Matt and Lubos are wrong. See also my blog at blog.vixra.org/category/energy-conservation where I have countered all the arguments by Lubos. – Philip Gibbs - inactive Jan 16 '11 at 17:01
Tow people have voted this down without saying what they don't like. I am puzzled because I have answered the points in the question directly. As far as I know my answers to the first two points are not controversial and the third is just a link to a paper about it as requested. – Philip Gibbs - inactive Aug 5 '11 at 15:28
I have now written another paper that directly refutes all the arguments made by people who think energy is not conserved in GR. See vixra.org/abs/1305.0034 . I am happy to see that my message is slowly getting across and this answer has been voted up a little. – Philip Gibbs - inactive May 6 '13 at 12:05

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: http://www.science20.com/hammock_physicist/square_root_universe

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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.

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I think the $\Omega=1$ argument is strongest for saying zero total energy. And I think the best measurement we have on $\Omega_{tot}$ is from the location of the first acoustic peak in the CMB (coupled with the hubble constant $H_0$). – Jeremy Jan 14 '11 at 14:00
What does $\Omega = 1$ have to do with "zero energy"? I think I managed to find the relevant part of the Lawrence Krauss talk that Lagerbaer commented on above, and it involved a similar comment. But no relation between $\Omega$ and any reasonable definition of "energy" comes to mind. What am I missing? – Matt Reece Jan 15 '11 at 17:50
@Matt, I assumes that Cosmonut is relating to a pop-sci explanation of some kind, and that either he or the author misconstrued the science. Why did I assume that? Because I could think of no other way to make sense of the question as asked. The connection with "total energy" is tenuous at best, but could be a generalization of "kinetic energy of expansion" minus "potential energy of gravitation" (and yes, neither construction is rigorous, but that's pop-sci for you). – dmckee Jan 15 '11 at 18:27
Nope, my question was not regarding the flatness of the universe (at least to my knowledge). The "zero energy" claim was mentioned in Brief History of Time, Chapter 8. As far as I can see, the claim is quite clear. Alan Guth, it seems, also used it to suggest that the universe is the "ultimate free lunch". From all the answers, it looks to me like this is one of those cases where the attempt to popularize, actually leads to misleading statements. – user1265 Jan 16 '11 at 4:11

Re a recent variant of the question:

Physicist using mc² to calculate positive energy and −MmG/R to calculate negative energy to calculate total energy of the universe but I heard that one of them is relatively invariant and other one is not. Is that true?

Physicists don't use mc² to calculate positive energy, that's merely a mass-energy equivalence expression. And nor do they actually use −MmG/R to calculate negative energy. The expression relates to gravitational potential energy, which is said to be zero at infinite distance, and negative near the surface of the planet. However there is no actual negative energy, just less positive energy. A gravitational field converts gravitational potential energy (which is mass energy) into kinetic energy, which is typically dissipated into space, leaving a mass deficit. But conservation of energy applies. No energy disappears. See The Foundation of the General Theory of Relativity where Einstein said "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". Gravitational field energy is actually positive, not negative.

Than why is everywhere and everyone doing these calculations to find total energy of the universe.

I can only suggest that those who advocate a zero-energy universe have never read Einstein's original papers and do not understand gravity.

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This doesn't really seem to answer the question though. – Kyle Kanos Mar 25 '15 at 14:34
The recent variant isn't a crisp clear single question, such as is the total energy of the universe zero? If it was, my answer would have been a crisp clear no. I would doubtless have referred to the conversion of potential energy to kinetic energy, the mass deficit, and the positive energy of the gravitational field. Alternatively I might have referred to a brick. You do not create energy when you lift a brick. Nor do you destroy energy when you drop it. Conservation of energy applies, there are no perpetual motion machines, there is no free lunch. – John Duffield Mar 25 '15 at 16:51
There are three precise questions asked; you've answered half of one of them. – Kyle Kanos Mar 25 '15 at 16:55
OK how about this: 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? – John Duffield Mar 25 '15 at 17:28
That's fine (though seemingly wrong, given all the answers above having contrary information), but you should probably change your answer to actually answer the questions asked, rather than your irrelevant tangent you've opted to write. – Kyle Kanos Mar 25 '15 at 17:38

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