# Baryon number conservation in early universe when T<m

1. May 7, 2015

### karlzr

When temperature of the universe falls below nucleon mass $T<<1$ GeV, the number densities of nucleons (proton and neutron) which are in kinetic equilibrium can be obtained as
$n_i=g_i (\frac{m_i T}{2\pi})^{3/2} e^{\frac{\mu_i-m_i}{T}}$. Since baryon number should be conserved, then I expect $n_p+n_n \propto a^{-3} \propto T^{3}$, which is not obvious from the above formula. I have taken for granted that there are no anti-particles for nucleons and all baryon number is in proton and neutron. So what 's going wrong? Does chemical potential have anything to do with it?

What do we know about antiproton/antineutron? Do they annihilate with proton/neutron around $T\approx m_i$ or what? I am trying to relate this process with that in Tevatron.

2. May 8, 2015

### George Jones

Staff Emeritus
For a species that was non-relativistic at decoupling, $T \propto a^{-2}$. Consequently, if the chemical potential is small compared the mass, and the mass is small compared to $T$, the exponential is unity, and

$$n_i \propto T^{3/2} \propto \left(a^{-2}\right)^{3/2} \propto a^{-3},$$

as expected.

3. May 8, 2015

### karlzr

But when temperature is lower than mass the exponential dies out rather than being unity.
Do you mean nucleons have different temperature as photons during nucleosynthesis?
Thanks.

4. May 8, 2015

### Chalnoth

I don't think it's possible to obtain the number density of baryons purely from thermodynamic arguments. You also need to know the level of asymmetry between matter and anti-matter generated by CP violation in the very early universe.

5. May 8, 2015

### George Jones

Staff Emeritus

6. May 8, 2015

### karlzr

My question is not about baryogenesis. I am considering the era from some temperature when QCD phase transition has already finished, saying 100 MeV to nucleosynthesis at $T\approx1$ MeV. The correct density of baryons has already been generated by some unknown mechanism before this period.

I know during this period the total baryon number must be conserved and I am asking why based on the discussion from thermodynamics. Now I think the answer lies in chemical potential which depends on temperature to ensure baryon number conservation.

7. May 8, 2015

### Chalnoth

I don't think it has anything to do with that. The number of remaining baryons is simply not thermodynamically-determined at all.

One simple way to look at it is this:
In a co-moving volume, you have $N$ anti-baryons and $N + \epsilon$ baryons early-on.
$N$ is determined by thermodynamics (the formulas you're using above).
$\epsilon$ is exclusively determined by the properties of baryogenesis. After baryogenesis has finished, we can simply take it as a fixed parameter.

Conservation of baryon number is a function of how the strong and weak nuclear forces behave. There's no reason to worry about a chemical potential, as it's not a thermodynamic process. Just take some leftover density as a fixed parameter and be done with it.

That is to say, you can simply write:

$$n_i = g_i\left({m_i T \over 2\pi}\right)^{3 \over 2} e^{\mu_i - m_i \over T} + {\epsilon_i \over a^3}$$

This equation will not be accurate while baryogenesis is occurring, and nucleosynthesis will mix things up a little bit, but it provides a rough idea of how this works.

The chemical potential really has nothing to do with this as that refers to the potential energy associated with reactions.

8. May 9, 2015

### karlzr

Do you mean thermodynamics or statistical distribution function doesn't work very well in this case? As to chemical potential, it is not constant. So how does it change with temperature?

I didn't say the net baryon number is determined from thermodynamics. I am trying to understand how the universe keeps this number unchanged after baryogenesis while nucleons are still in equilibrium with cosmic plasma. And my conclusion was: assuming the validity of distribution function, the only explanation lies in chemical potential.

Last edited: May 9, 2015
9. May 10, 2015

### Chalnoth

The distribution function.

That distribution function was constructed with the assumption that as T goes to zero, so does the number density, that is, that there are an equal number of matter and anti-matter particles. If you *really* want to do it right, you would have to go back and look at how the equation was derived in the first place, and add in the fact that there is some imbalance in matter/anti-matter. I'm sure this has been done, of course, but I'm really not sure how I'd go searching for it.