The Big Bang — Question 3

Question 3

An interaction freezes out when its rate falls below expansion

Let thermal energy T be expressed in MeV and rates be $\Gamma=2(T/{\mathrm{MeV}})^5$ s$^{-1}$ and $H=.50(T/{\mathrm{MeV}})^2$ s$^{-1}$.

1. Find the freeze-out thermal energy from Gamma=H.

2. Compute Gamma/H at T=2 MeV and .20 MeV and identify which regime can maintain equilibrium.

Question 3 - Solution

An interaction freezes out when its rate falls below expansion

(a) Method and calculation

$$2T^5=.50T^2\Rightarrow T^3=.25\quad(T\text{ numerically in MeV}).$$$$\boxed{T_f=.25^{1/3}=.629961\ {\mathrm{MeV}}}.$$

(b) Interpretation and further calculation

$$\boxed{\Gamma/H=4(T/{\mathrm{MeV}})^3:\quad32\ \text{at }2,\quad.032\ \text{at }.20}.$$$$\boxed{\Gamma/H\gg1:\ \text{rapid reactions};\quad\Gamma/H\ll1:\ \text{freeze-out}}.$$

Physical check

Freeze-out is gradual in a full kinetic treatment. The equality is a characteristic scale, not an instantaneous switch.