# Hubble's law and conservation of energy

If all distances are constantly increasing, as Hubble's law say, then lots of potential energies of form ~$\frac{1}{r}$ changes, so how is the total energy of the Universe conserved with Hubble's expansion?

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@kalle43,

There is such a thing as a law of conservation of energy. And IMHO it is a fundamental law of nature - though its statement admittedly becomes murky in a general relativistic or quantum setting. This remains true no matter what background metric describes your spacetime or whether you are talking about equilibrium or non-equilibrium processes.

In physics we describe systems, and processes which affect those systems, by understanding the conserved quantities of the system in question. So to answer your question about energy conservation in the context of Hubble expansion, we first have to identify the system we are dealing with and its dynamics.

In the presence of homogenous and isotropic matter with stress-energy tensor:

$$T_{\mu\nu} = diag(-\rho+\kappa\Lambda,p-\kappa\Lambda,p-\kappa\Lambda,p-\kappa\Lambda)$$

where $\rho$ and $p$ are the density and pressure of our matter distribution; $\Lambda$ is the cosmological constant and $\kappa = 1/8\pi G$

With the ansatz of of a homogenous and isotropic metric $g = a(t)^2 diag(-1,1,1,1)$, Einstein's equations $G_{\mu\nu} = \kappa T_{\mu\nu}$ yield the two Friedmann equations. They are:

$H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2}$ (First Friedmann Eq.)

$\frac{\ddot a}{a} = -\frac{4\pi G}{3}(\rho + 3p) + \frac{\Lambda}{3}$ (Second Friedmann Eq.)

and $k \in {-1,0,1}$ determines whether the spacetime is open ($k = -1$), flat ($k=0$) or closed ($k=+1$). For our purposes we can set $k$ to zero. Also for our purposes $p=0$ in the second equation. $H = \dot a/a$ is the Hubble parameter.

Now taking the time-derivative of the first equation gives:

$$2 H \dot H = 2 \left( \frac{\dot a}{a} \right) \left( \frac{\ddot a}{a} - \frac{\dot a^2}{a^2} \right) = \frac{d}{dt}\left( \frac{\kappa}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2} \right)$$

Substituting the r.h.s of the 2nd Friedmann Equation in the lhs of the above expression, we get:

$$\frac{d}{dt}\left( \kappa \rho + \Lambda - \frac{3k}{a^2} \right) = - 3 H \left( \rho + p - \frac{k}{a^2} \right)$$

This is the statement of energy conservation for our system. For $H > 0$ ($H < 0$) the energy density (the r.h.s.) in a given volume is a decreasing (increasing) quantity. This in line with our intuitive expectation regarding an expanding (contracting) cosmology.

Apologies for the long-winded answer. You could also have found this material on the relevant wikipedia page. I wanted to make my answer self-contained.

Of course, the Universe as we observe it consists of bound systems - solar systems, galaxies, galaxy clusters - which are seemingly not affected by cosmic expansion. For these cases one needs to do a little more work.

Cheers,

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That's not the correct form for a homogenous and isotropic metric. you get the Friedmann equations from $ds^{2}=-dt^{2}+(a(t))^{2}\left(\frac{1}{1-kr^{2}}dr^{2}+r^{2}d\theta^{2}+r^{2}\‌​sin^{2}\theta d\phi^{2}\right)$ –  Jerry Schirmer Nov 27 '10 at 18:44
You're right. My bad. I left out the $1/(1-kr^2)$ part. –  user346 Nov 27 '10 at 22:49
I'm confused by the basic logic here: your final equation tells you that the time derivative of the quantity you prefer to call energy is non-zero. Where is the conservation part? –  user566 Jan 16 '11 at 17:04
Good point @Moshe. I'll have to think about it and get back to you. –  user346 Jan 16 '11 at 18:22
Hi @Moshe. I was traveling the past few days. In the meantime it seems others have made my job much easier. In particular you should take a look at the answer given by @[Philip Gibbs] to this question. As for your first comment - in general, gravitational systems are locally non-equilibrium systems so conservation of energy must take these external forces into account. The hubble rate $H$ plays the role of an external driving force. If $H=0$ then energy is constant and conserved trivially. –  user346 Jan 20 '11 at 12:14

Short answer: conservation of energy is not a fundamental law.

Noether's theorem tells us that whenever there is some symmetry in physical laws, one obtains some conserved quantity. For translations in space one obtains conserved momentum. For rotations one gets angular momentum. And for translations in time, one gets the law of conservation of energy.

That means that conservation of energy only holds for systems that are governed by laws which are invariant in time, i.e. static. Most systems one encounters in daily life are of this sort (even friction, when looked upon closely, conserves energy; the missing energy is just transformed into kinetic energy of atoms as heat). But this is only the consequence of living in an idealized static Minkowski space-time.

The moment one leaves this nice static place and considers dynamical universe, one has to dispose of the simple concept of conservation of energy. One can consider various notions of energy in General Relativity but these concepts are pretty messy and one is pretty much forced to talk only about energy locally in some small volume. But you can't really say anything about universe as a whole.

So, first one has to note that conservation of some quantity means that the quantity is constant through the evolution of the system in time. But what is time? Already Special Relativity tells us that every observer carries his own local notion of time. General Relativity complicates this notion a great deal further. So it doesn't even make sense to talk about constancy of some quantity in time unless we specify what time we mean! To make sense of this one usually restricts to small volume (so that time has almost same meaning for each point of the volume) or else one has a good notion of what time is globally. This second point is fortunately true in our universe (and also in flat Minkowski space-time) because it can be described quite well by some FRLW solutions of Einstein equations.

So the general notion of time complicates things but can be dealt with. The worse problem is that in General Relativity it is really hard to say what gravitational energy (i.e. energy stored as curvature of space-time) is. It turns out that different observers won't really agree on this (so the concept is not covariant) and except for some special situations one can't say anything useful.

Note: To elaborate on a small volume in the above, one can often limit oneself to a volume where the system is approximately isolated from the rest of the universe. E.g. the solar system is to a good degree isolated and the energy would be conserved if it wasn't for incoming (and outgoing) particles (like light) and bodies (like asteroids).

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What about conservation of energy from a specific reference frame of, say, a galaxy? Is energy in that reference frame still not conserved? –  Sklivvz Nov 26 '10 at 18:52
@Marek: I sincerely don't understand, because as far as I know energy conservation is pretty much a balance law, so in your example energy would be still conserved if, e.g. the energy of a galaxy decreases by an amount equal to the energy irradiated and increases by an amount equal to the energy received by gravitational waves. –  Sklivvz Nov 27 '10 at 10:38
@Marek, it's not what I meant. I meant that energy is conserved if all variations are accounted for. As far as I know, energy conservation is not the same as constant energy!? –  Sklivvz Nov 27 '10 at 11:16
@Marek: conservation of energy means that energy is constant during the evolution in time in a closed system. If galaxies are not a closed system, as you said above, the conservation of energy doesn't mean constant energy (for the galaxy). That said - thanks a lot for the explanation re: what is hard to account for, it makes more sense to me now. –  Sklivvz Nov 27 '10 at 11:26
@Marek: can we agree that conservation of energy means that no energy is created or destroyed out of nothing? :-) This accounts for both open and closed systems (e.g. an open system can lose energy to the environment, but it may not lose or gain energy without interacting with the environment) –  Sklivvz Nov 27 '10 at 11:41

There are a few answers to this one:

1) Conservation of energy in relativity is only a local law. Energy is conserved in a local reference frame, but it is not conserved globally.

2) The cosmological constant has a negative pressure associated with it, so as the universe expands, the matter in the universe has work done on it by the cosmological constant, which then causes more expansion, which causes more work done, etc. So conservation of energy is still there, it's just enforced by the novel thermodynamics of a cosmological fluid.

3) Marek's answer--a completely coordinate-free definition of energy doesn't exist for Big Bang Cosmology, and so this isn't a well-defined question.

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Expansion is found between galaxies and thought as a uniform law in the universe. But it doesn't mean that everything is expanding, especially not for gravitational bounded systems (Peacock "Cosmological physics" 3.3 ).

THE NATURE OF THE EXPANSION An inability to see that the expansion is locally just kinematical also lies at the root of perhaps the worst misconception about the big bang. Many semi-popular accounts of cosmology contain statements to the effect that space itself is swelling up'' in causing the galaxies to separate. This seems to imply that all objects are being stretched by some mysterious force: are we to infer that humans who survived for a Hubble time would find themselves to be roughly four metres tall? Certainly not. Apart from anything else, this would be a profoundly anti-relativistic notion, since relativity teaches us that properties of objects in local inertial frames are independent of the global properties of spacetime. If we understand that objects separate now only because they have done so in the past, there need be no confusion. A pair of massless objects set up at rest with respect to each other in a uniform model will show no tendency to separate (in fact, the gravitational force of the mass lying between them will cause an inward relative acceleration). In the common elementary demonstration of the expansion by means of inflating a balloon, galaxies should be represented by glued-on coins, not ink drawings (which will spuriously expand with the universe).

So conservation of energy can still be used to study the evolution of solar system, Galactic system, galaxy clusters etc. But the WHOLE universe is something different. What we observed is only a part of it, we can not simply take it as the whole vision. Since the age of the universe is limited (13.7Gyr), there may be much more objects outside our sight. Then all the conservation laws of a closed system can not be used.

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Contrary to some of the answers above, energy conservation does hold exactly in general relativity. It can be derived using Noether's Theorem and the time symmetry of the gravitational field equations when all fields including the gravitational metric are treated as dynamical.

Of course the Newtonian 1/r energy law is not applicable on a cosmological scale but it is true that the gravitational contribution to the energy equation is negative in general relativity as it is in Newtonian physics. Matter, radiation and dark energy contribute positive terms. Overall the energy in a given region of space which might expand over time changes only by an amount calculated as the flux of energy over the boundary of the region. This statement embodies the law of conservation of energy.

There are many fallacies that are often repeated that make people think that energy conservation in general relativity is not exactly conserved. The most common is to think that Noether's theorem can only work in a static gravitational field. This is not true so long as the variable energy of the gravitational field itself is included in the equation.

For a fuller explanation the entries I posted on my blog can be consulted. Alternativly the Wikipedia entry about the "Stress-energy-momentum pseudotensor" is another valid analysis, although I prefer the covariant formalisms myself.

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Dear Phil, just if you wanted to know whom you received the first negative vote for, it was me - because your comment is just incorrect. There is no covariant expression for energy density whose integral would be conserved in GR. In backgrounds with generic asymptotic conditions, there's even no globally conserved energy. In particular, any realization of a valid "conservation law" in cosmology for a compact universe is equivalent to 0=0, a completely vacuous tautology. That's different from non-gravitational setting where the conservation law is always nontrivial. –  Luboš Motl Jan 15 '11 at 22:30
Concerning the LL pseudotensor, just look at en.wikipedia.org/wiki/… - The full stress-energy pseudotensor, including the matter, also contains the $T_{\mu\nu}$ term that totally cancels the $G_{\mu\nu}$ term. So using the equations of motion, the total LL-tensor may be written as the simple combination of the partial derivatives of the metric whose conservation is an identity that requires no other eqns of motion. That's different from proper conservation laws that always require one to use eqns of motion again to be verified. –  Luboš Motl Jan 15 '11 at 22:38
Because in the normal theories with conserved energy, one can define the energy - or its tensor - in terms of all/most fields, and to prove that it is conserved, one must still use the equations of motion for all the fields, it follows that in normal theories, the conserved energy may actually tell us something about the solutions of the theory - about the final state from the known initial state. That's not the case of GR which is why we say that there's no non-vacuous energy conservation law in GR in generic (e.g. compact) backgrounds. –  Luboš Motl Jan 15 '11 at 22:40
While Lubos and I agree on many things, conservation of energy in general relativity is not one of them. In case anyone wants to look into it, we have argued about it at length on vixra log at blog.vixra.org/category/energy-conservation. I have contradicted the points he makes above over there so there is no need to go through it all again here. Sorry Lubos but since you have given me a negative mark here I will have to do the same for you because you are really wrong about this! –  Philip Gibbs - inactive Jan 16 '11 at 16:53
Well if people are just going to vote down my good answers I wont stick around, bye. –  Philip Gibbs - inactive Jan 16 '11 at 19:03