# Understanding Sakharov's third condition for baryogenesis (departure from thermal equilibrium)

I'm struggling to understand Sakharov's third condition for baryogenesis, i.e., a departure from thermal equilibrium. I don't understand what thermal equilibrium means in this context or proofs that it is neccessary for baryogenesis.

Trodden and Carroll's TASI lecture notes outline a common argument. They begin by writing the expected baryon number in thermal equilibrium as $$\langle B \rangle = \text{Tr}\left(e^{\beta H} B\right) = \cdots = 0$$ The result that it equals zero follows quickly for a CPT invariant Hamiltonian. However, why does thermal equilibrium imply Gibbs' canonical ensemble, as in the above formula?

Gibbs' ensemble is useful in situations in which only the expected energy (or temperature) adequately describe, for our purposes, the probability of states. I cannot see why Gibbs' ensemble would be appropriate for understanding systems with more macroscopic parameters of interest, in this case temperature and baryon number, nor why it would be the only distribution for thermal equilibrium.

## 1 Answer

Local thermal equilibrium implies that expectation values are given by the grand canonical ensemble characterized by Lagrange multipliers for all conserved charges. In the standard model these are energy, fixed by T, and baryon number, lepton number, and electric charge, governed by suitable chemical potential. If you make the baryon chemical potential non-zero then you are just putting in baryon number by hand, so in order to discuss baryogenesis we consider $\mu_B=0$. The universe is electrically neutral (if it was not, the Coulomb interaction would make it locally neutral). There is a subtlety here, which is that the SM (at high T) only conserves B-L, so you can move L to B (this is called leptogenesis).

Postscript: The most convenient way to look at this is the grandcanonical ensemble. The baryon chemical potential $\mu_B$ is zero, and the average baryon number is zero $\langle B\rangle=0$. Fluctuations of the baryon number are not zero $\langle B^2\rangle\neq 0$. Today, these fluctuations are extremely small because baryons are heavy, and the temperature is very small. In the canonical ensemble we consider some large volume in which $B=0$. Local fluctuations arise because baryon number can be exchanged between volume elements. Fluctuations $\langle B^2/V\rangle$ are exactly what the $\mu_B=0$ grandcanonical ensemble predicts.

As an example, consider relativistic non-interacting baryons or quarks. We have $$\langle B\rangle = \frac{\partial P}{\partial \mu}\;\;\;\; \langle B^2\rangle = \frac{\partial^2 P}{\partial \mu^2}$$ with $$P =\cosh(\mu_B b/T) \frac{gT^4}{2\pi^2}\frac{m^2}{T^2}K_2(m/T)$$ where $b=1$ for baryons and $b=1/3$ for quarks, $m$ is the mass, and $g$ is the degeneracy. For $\mu_B=0$ we get $\langle B\rangle =0$ as expected and $$\langle B^2\rangle = gb^2m^2 K_2(m/T)$$ which is not zero and of order $\exp(-m/T)$ in the non-relativistic limit. Note that today $m=1$ GeV and $T=3$ K ($T=2\cdot 10^{-4}$ eV), so this is irrelevant.

• Is the first sentence a definition? Why does it imply it? Is Sakharov's condition actually departure from Gibbs' canonical ensemble? If so, it just seems somewhat trivial, since for a CPT Hamiltonian Gibbs' ensemble obviously doesn't describe a system with B. – innisfree Oct 9 '16 at 22:52
• 1) The distribution that maximizes the entropy for given values of the conserved charges is the Gibbs distribution, so yes, thermal equilibrium implies Gibbs. 2) Arguably, all three conditions are trivial'', but the need for non-equilibrium is maybe the most non-intuitive one. – Thomas Oct 10 '16 at 3:25
• What do you mean by thermal equilibrium in this context? – innisfree Oct 10 '16 at 4:34
• I find it confusing as Gibbs' is an epistemological description of our state of knowledge of a system (which is often perfectly adequate for making predictions). So when it says departure from thermal equilibrium, it isn't clear whether this is a property about the system or our knowledge about it. If the latter, it's trivial since it means: in order to create baryon number, we cannot believe the system is described by Gibbs' (but why would we, since systems described by Gibb' are adequately described by only temperature). – innisfree Oct 10 '16 at 4:47
• Short answer: It has nothing to do with our ignorance. The part of the universe that we have access to is certainly described by a mixed state, an unless there was something very unusual about the initial state of the universe, time evolution will drive this state towards the Gibbs ensemble. We need an actual deviation from thermal equilibrium at the EW phase transition (or some earlier point in time) to generate B. – Thomas Oct 10 '16 at 16:26