Hubble's law and conservation of energy
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).
@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,
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.