cosmology - 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?

Answer

@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,