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Did the law of conservation of energy apply to the earliest moments of the Big Bang? If so, what theoretical physics supports this?

I hear that Einstein's theory of relativity disputes the law of conservation of energy- so does that mean the law is false, or only some aspect of it?

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The second part of the question (v1) seems related to and –  Qmechanic May 24 '11 at 8:44

5 Answers 5

Yes, the energy conservation law fails not only right after the Big Bang but in any cosmological evolution. See e.g.

The time-translational invariance is broken, so via Noether's theorem, one doesn't expect a conserved quantity. Also, if one defines the "total" stress energy tensor as a variation of the action with respect to the metric tensor, it vanishes in GR because the metric tensor is dynamical and the variation has to vanish because it's an equation of motion (Einstein's equations).

If the space is asymptotically flat or AdS or similarly simple, a conservation law - for the ADM energy - may be revived.

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I think this is a language game. What is the definition of "energy" if not "that which is conserved"? –  Dan Brumleve May 24 '11 at 8:28
@Lubos, about this non-conservation of energy in expanding universes, i have just one question: where it is calculated the energetic (initial and ongoing) cost of setting a universe in expansion at a given rate $\kappa$? in your article you mention the energy of matter and cosmological constant but you don't include this term in your balance –  lurscher May 24 '11 at 11:54
Dear @Dan, it is not a language game and you can only stand by your position if you're either blinded or unable to follow elementary arguments. On the other hand, you're right that the most sensible definition of energy is that it is a quantity that is conserved as a result of the time-translational invariance. However, what I am explaining to you is that there is no (conserved) energy in GR - obviously, the symmetry isn't there either. It's like the only sensible definition of a cold fusion reactor is a fusion reactor that is cold - but this doesn't imply that it actually exists. ;-) –  Luboš Motl May 25 '11 at 9:07
It is not silly. There is no nontrivial concept of energy in GR. Standard definitions of "really total" or "conserved" energy density in GR are identically equal to zero (or, more generally, they're independent of the degrees of freedom, or dependent on the gauge). $H=0$ is really a constraint or an equation of motion in GR because $H$ generates translations in time and all diffeomorphisms, including translations in time, are gauge symmetries that must keep physical states invariant (using quantum mechanics language, and assuming that there is no asymptotic region). –  Luboš Motl May 25 '11 at 9:10
Well, when I say that I think that someone is an idiot, it is also just a word, and that's what I will choose to say right now. Why are you writing on this server if you admit that your texts are just sequences of words that don't mean anything? The question asked about a well-defined thing using pretty much well-defined words, my answer is right, your answer is wrong - right and wrong are also words, by the way - so what's the purpose of hiding and diluting the wrongness of your answer by trying to pretend that words don't mean anything? They do, otherwise we wouldn't use them. –  Luboš Motl May 25 '11 at 9:20

Energy is indeed conserved perfectly well in General Relativity, even in the extreme conditions of the early big bang. It is not merely true by definition, it is true as a result of the dynamics of the field equations.

This question has been answered before e.g. at Energy conservation in General Relativity and should probably be closed as repetition, so I'll just add a link to a long discussion on the subject at

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I will also add a pointer to another paper at where I refute all the known arguments against energy conservation in GR –  Philip Gibbs - inactive May 6 '13 at 12:15

Conservation laws are established in general relativity if there is a Killing vector $K_a$, where for some values of the index $a$ there may be zero entries, so that for a momentum vector $p^a$ $=~(p_t,~{\vec p})$ the inner product $p^aK_a~=~constant$. The Killing vector is then an isometry such that a vector along a parallel translation defines a conserved quantity relative to the Killing vector. That conserved quantities are variables conjugate to the components of the Killing vector. How to find Killing vectors is somewhat involved, but as a rule, if a metric coefficient $g_{ab}~=~g_{ab}(t)$ then there is no Killing vector with a component along that coordinate direction. The general line element for a cosmology involves a scale factor $a~=~a(t)$, $$ ds^2~=~-dt^2~+~a^2(t)(dr^2~+~r^2d\Omega^2 $$ which is a pretty good clue that this spacetime has not fundamental conservation of energy. There is no timelike directed part of a Killing vector, therefore conservation of the energy conjugate variable can’t be established fundamentally.

So is energy absolutely not conserved in our universe? The answer to this depends upon upon some other conditions; for it does turn out that our universe may have a unique condition which recovers energy conservation.. The FLRW equation for the scale parameter $a~=~a(t)$ is $$ \big(\frac{\dot a}{a}\Big)^2~=~8\pi G\rho/3~–~\frac{k}{a^2} $$ where ${\dot a}/a~=~H$, the Hubble parameter, and flatness means $k~=~0$, spherical geometry is $k~=~1$ and hyperbolic geometry is $k~=~-1$. There is an equation of state for the mass-energy and pressure in the spacetime $$ \frac{d(\rho a^3)}{dt}~+~p\frac{da^3}{dt}~=~0 $$

I will consider an approximate de Sitter spacetime, which has $\rho~=~constant$, and is identified in ways not entirely understood with the quantum field vacuum. The FLRW equation for $k~=~0$ is then $$ \frac{da}{dt}~=~\sqrt{8\pi G\rho/3}~a $$ which has the solution $a~=~\sqrt{3/8\pi G\rho}~exp(\sqrt{8\pi G\rho /3}t)$. Using the Einstein field equation we then have that the stress energy is $T^{00}~=~8\pi G\rho~=~\Lambda$, which is the cosmological constant. Returning the first equation, the FLRW equation, we then see that $H^2~=\Lambda/3$. The dynamical equation for the dS spacetime with $\rho~=~const$ gives

ρd(a^3)/dt + pda^3/dt = 0

or $p~=~-\rho$. This is the equation of state for $p~=~w\rho$, and $w~=~-1$. This corresponds to a case where the total energy is zero and the first law of thermodynamics is $dF~=~dE~–~pdV~=~0$ means the energy that is increased in a unit volume of the universe under expansion is compensated for by a negative pressure which removes work from the system. Further $pdV~=~d(NkT)$, and for a constant thermal energy for the vacuum and $Nk~=~S$ the entropy of the universe.

For this particular special case we do have an equation of state which gives a conservation of energy. Another way of seeing this is this spacetime has a time dependent conformal factor $a(t)$, and this metric is conformally equivalent to a flat spacetime, where one can define an ADM mass that is conserved.

Of course the question might be raised whether this pertains to our physical universe. The inflationary period had a huge exponential acceleration, or equivalently a scale factor which grew extremely rapidly. This period should then have had conservation of energy. As for the time period before then, who knows? After the reheating period the universe became radiation dominated and energy conservation is not immediately apparent. Energy conservation may only then be established in our universe in the very distant future as it approaches an empty de Sitter vacuum state.

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great overview. My question is; how much energy contributes/costs inflation itself? Somehow there has to be a mechanism that switches on and off inflationary expansion, so, instead of writing ill-defined things like $\dot \Lambda$ lets assume some quintaessence powering down dynamically the values for $\dotdot a$, this is not derived from GR or thermodynamics alone since it relies on details of the quintaessence physics. How do we compute how much potential energy is stored when the inflation is turned off (present era)? –  lurscher May 26 '11 at 1:14
if we are saying something about energy conservation in a expanding universe, i'm sure we should have something to say too about what energy costs are associated to setting up the dynamical fields that create a universal expansion in the first place –  lurscher May 26 '11 at 1:17
btw @Lawrence, i always appreciate that you are one of the few posters that consistently belong to the "shut up and calculate" crowd; avoiding to place complex discursive arguments in their answers morally over the equations that are more capable of explain it with less ambiguity. thanks! –  lurscher May 26 '11 at 1:32
There is this issue of reheating, where after about 63 efolds the inflaton potential went from a constant minus some slow linear dependent term of the field to a quadratic term. This change in the potential is rather odd, and it may be some sort of phase transition. The reheating is similar to a latent heat. There is also the very early phase of this which is a bit mysterious as well. –  Lawrence B. Crowell May 26 '11 at 1:50
I tend to work somewhat in string and branes, but frankly I think the fundamental things to consider are degrees of freedom and quantum information. It is my sense that string theory and M-theory represents structures which are most appropriate to physics, but there are things maybe missing. The whole business has turned into the vast unwieldy arena that is become impossible to track. Trying to tie down cosmology with this seems maybe premature. This gets into my sense about quintessence and brane-brane interaction ideas, in that they are premature. –  Lawrence B. Crowell May 26 '11 at 1:51

It is my, perhaps inadequate, understanding that there is indeed something, and it could be called energy, which is conserved by GR. Whether I am right or not, prof. Motl is certainly wrong to say that the Big Bang breaks time-invariance: this is a misunderstanding of time-invariance symmetry.

GR is indeed time-invariant.

The kind of time-invariance required to apply Noether's theorem is simply the physical fact that if we perform our experiment at time $t=t_o$ and wait $\Delta t$ seconds (or you could use units of age of the Universe) and measure the results, the answer will be the same as if we performed the experiment at time $t=t_1$ and waited $\Delta t$ seconds, thus measuring the results at time $t=t_1+\Delta t$. Physically, this means you have to arrange that the Big Bang also happened $t_1-t_o$ seconds later than it did (unless you have physical reasons for knowing that the time elapsed from the Big Bang is not relevant to the quantities you are going to measure). Mathematically, this is guaranteed as long as the Lagrangian does not depend explicitly on time.

(Effective Lagrangians often do vary with time, this means the system is not a closed system but open, and of course energy is not necessarily conserved in an open system.)

Therefore there is a corresponding quantity which is conserved. Whether it should be called energy might be worth discussion, but I always assumed it was...

For a useful, balanced, discussion of some of the many issues relevant to the OP, although it is from a mathematics department, see even though I am not sure I agree with them.

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Certainly yes, and Einstein's theories support this. Conservation of energy is a fundamental principle and it is true essentially by definition.

Wikipedia's article on the history of energy includes this quote from Feynman:

"There is a fact, or if you wish, a law, governing natural phenomena that are known to date. There is no known exception to this law—it is exact so far we know. The law is called conservation of energy; it states that there is a certain quantity, which we call energy that does not change in manifold changes which nature undergoes. That is a most abstract idea, because it is a mathematical principle; it says that there is a numerical quantity, which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number, and when we finish watching nature go through her tricks and calculate the number again, it is the same."

The Wikipedia article on physical cosmology has this to say:

There is no unambiguous way to define the total energy of the universe in the current best theory of gravity, general relativity. As a result it remains controversial whether one can meaningfully say that total energy is conserved in an expanding universe. For instance, each photon that travels through intergalactic space loses energy due to the redshift effect. This energy is not obviously transferred to any other system, so seems to be permanently lost. Nevertheless some cosmologists insist that energy is conserved in some sense.

I interpret this to mean that while in some cases it is not meaningful to say that energy is conserved, in these same cases it is no more meaningful to say that it isn't. The issue is the well-definedness of energy, not whether or not it is conserved.

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Not sure why the downvote. Did Einstein make any argument that energy is not conserved? Luboš seems to be using tools that would not have been available to Einstein. –  Dan Brumleve May 24 '11 at 8:37
Can we push philosophy.SE past the beta threshold? There we can all talk about Popper's theories which would be inappropriate here. –  Dan Brumleve May 24 '11 at 9:04
You got the downvote because what you say is just plain wrong. Wether Einstein had the tools or not is irrelevant by the way. We are talking about GR and it's known that for large classes of solutions, among which Friedman-Walker metrics, that there is no energy conservation. –  Raskolnikov May 24 '11 at 11:51
@Dan Relax, not conserving energy at the Big Bang is not the end of the world. –  dbrane May 24 '11 at 14:25
Dear @Dan, you're not right about physics but you're also wrong about the history (independently of the fact that, as I agree, the history is totally irrelevant for the validity of the conservation law). Emmy Noether actually found her Noether's theorem in the very same year when Einstein completed his general relativity, 1915, see - so it was totally available to Einstein. –  Luboš Motl May 24 '11 at 15:03

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