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In a "slowly" expanding Universe, suppose the number density of a particle species $A$ undergoing a equilibrium decay $A\to B+C$. The equilibrium number density depletes with the expansion and the consequent fall of temperature but in equilibrium. For example, the number density of the relativistic species depletes as $\sim T^3$ and non-relativistic species depletes as $\sim e^{-m_A/T}T^{3/2}$ with the fall of temperature.

How is this possible? Shouldn't the rate of forward and backward reaction be equal in chemical equilibrium? And if that is so, how can the number density of A change with the fall of temperature?

I think the italic statement above is true only for a given temperature. If yes, does that explain how the number density deplete in equilibrium?

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You are mixing up two eras in cosmology.

Before freeze out

At young age the universe was incredibly hot such that reactions (i.e. $A\rightarrow B+C$) were indeed in thermal equilibrium since $kT >> m_A, m_B, m_C$ there was no preferred direction for this reaction to move into !

The result is that the number densities of the particles are determined by a thermal distribution:

$$F(\vec{p})d\vec{p} = 4\pi g \frac{p^2 dp}{\exp(\frac{E-\mu}{T})\pm1} $$

With the +(-) sign for fermions (bosons). From this we get that the number densities before freeze-out were given by:

$$n(T) = \int dp F(p) \sim \begin{cases}T^3\ for\ the\ ultrarelativistic particles\ p>>m \\\exp(-m/T)\ if\ m>>p \end{cases}$$

Those are the relations that you mentioned.

After freeze out

As time passes and the universe cools there will be a point a which particles are no longer able to maintain this thermal equilibrium since $A\rightarrow B+C$ wil drop out of equilibrium !$(let's say $m(B)+m(C) < m(A)$ such that A will decay.

At this point the relations are no longer valid since there will be less A particles as expected. And more B,C type particles.

This is depicted in the figure below. At early times T is high and the number densities are given by the formulas above, this is depicted by the solid black line. As T drops we observe that the effective particle densities are higher than those expected from the equilibrium calculations (i.e. B or C type particles in our above example).

Note that the final density depends on the crossection of the decay, this is sensible since a higher crossection means that the reaction will remain in equilibrium for longer such that it will follow the full curve for longer.

I hope this helped ? :)

Freeze out in action

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  • $\begingroup$ My question is related to the era 'before' freeze-out (not 'after'). At that time, equilibrium was maintained. Agree? If at all times before freeze-out chemical equilibrium were also maintained, the rate of forward and reverse reaction should have been equal. So number density of A cannot change. But as I've mentioned, and you too, the number density falls as the temperature falls, even before freeze-out. See your 'before' freeze-out formulas. My question is, how is it possible to maintain chemical equilibrium and still reduce the number density before freeze-out? @gertian $\endgroup$ – SRS May 4 '17 at 21:33
  • $\begingroup$ (not sure but this is what I think) But I'd say that equilibrium implies that A,B and C are able to maintain the correct densities given the environment. Before freeze out he reaction is going fast enough such that the only factor determining the equilibrium densities is their momentum distribution. This does not imply that the forward/backward reaction speed will be equal, equilibrium only implies that: $n(A)\sigma_\rightarrow = n(B)n(C)\sigma_\leftarrow $ assuming equal reaction speeds to both sides we find: $n(A)/n(B)n(C) = constant$ but still, this does not imply that they are fixed... $\endgroup$ – gertian May 5 '17 at 15:14
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Yes, you are right! The statement "The rate of forward and backward reaction is equal in equilibrium" is only true at a given temperature.

So we can use equilibrium statistical mechanics to determine the statistical properties of the particles from the temperature, as long as the relevant interaction rates are much larger than than the expansion rate.

Note also that, in the expanding universe, a decrease in number density does not neccesarily imply that the number of particles changes. Relativistic species in equilibrium, for example, have a constant number of particles in a comoving volume, so the change in (physical) number density comes just from the dilution effect of the expansion.

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