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Definition of chemical equilibrium According to the definition of chemical equilibrium in Wikipedia, it is a situation where the rate of the forward reaction is same as the rate of backward reaction and the concentrations of the reactants and the products do not change with time. I hope that the same definition applies to particle physics processes such as a 2-2 scattering of the form $A+B\leftrightarrow C+D$ i.e., in chemical equilibrium concentrations or number densities of each particle species remains unchanged with time.

Consistency of the definition with Boltzmann equation Boltzmann equation is used to study baryogenesis, relic abundance of dark matter and so on. According to Boltzmann equation, the number density $n$ of the species $A$ in an expanding Universe changes as: $$\frac{dn}{dt}+3Hn=\int d\pi_A d\pi_B d\pi_C d\pi_D[f_Af_B|\mathcal{M}|^2_{AB\to CD}+f_Cf_D|\mathcal{M}|^2_{CD\to AB}]$$ where $\mathcal{M}_{AB\to CD}$ and $\mathcal{M}_{CD\to AB}$ are the amplitudes of the forward and backward processes respectively, $H$ is the Hubble parameter defined as $H=\frac{\dot{a}}{a}$ where $a$ is the exapsnion factor. $d\pi_i$ and $f_i$ ($i=A,B,C,D$) are respectively the integrals over phase space and phase space distributions. Due to the $+$ sign on the RHS, the RHS is nonzero in chemical equilibrium even when $$\mathcal{M}_{AB\to CD}=\mathcal{M}_{CD\to AB},~\text{(assuming CP or T invariance)}$$ Therefore, $$\frac{dn}{dt}+3Hn\neq 0.$$ Therefore, the number density of species $A$ can change with time even when the dilution due to expansion is zero i.e., $H=0$. Therefore, the number density or the concentration of $A$ changes in even in chemical equilibrium.

Question Shouldn't the RHS of the Boltzmann equation be zero in chemical equilibrium in order to keep the concentration of each species unchanged with time?

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  • $\begingroup$ I don't really understand the question - by definition, in a chemical equilibrium you would have $\mathrm{d}n/\mathrm{d}t = 0$. You exhibit an equation where $\mathrm{d}n/\mathrm{d}t \neq 0$, therefore it describes a system that is not in chemical equilibrium. What exactly do you want to know? $\endgroup$
    – ACuriousMind ♦
    Commented Nov 17, 2018 at 13:42
  • $\begingroup$ @ACuriousMind Let me say it differently. If $dn/dt=0$ with $H=0$ (assuming no expansion), the RHS must be zero. But the RHS cannot be zero. Chemical equilibrium only requires $\mathcal{M}_{AB\to CD}=\mathcal{M}_{CD\to AB}$. The RHS is zero only in absence of scattering in which case the number density dilutes as $n\propto a^{-3}$ (if $H\neq 0$) due to expansion or stays constant in absence of expansion (when $H=0$). $\endgroup$
    – SRS
    Commented Nov 17, 2018 at 13:51
  • $\begingroup$ @ACuriousMind Do you mean Boltzmann equation cannot describe chemical equilibrium? $\endgroup$
    – SRS
    Commented Nov 17, 2018 at 13:58
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    $\begingroup$ The signs in your Boltzmann equation look wrong. Surely, the “in” and “out” rates have opposite sign $\endgroup$
    – Thomas
    Commented Nov 17, 2018 at 21:23
  • $\begingroup$ Thanks. What about $(f^{\rm eq}_Af^{\rm eq}_B-f^{\rm eq}_Cf^{\rm eq}_D)?$ Will it vanish? @Thomas $\endgroup$
    – SRS
    Commented Nov 17, 2018 at 21:38

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The first term on the right hand side should have a minus sign because the process $A+B \rightarrow C +D$ depletes the species $A$. So if all the $f$'s have their equilibrium form, $dn/dt$ should be zero.

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  • $\begingroup$ Thanks for pointing out the sign mistake. What about $(f^{\rm eq}_Af^{\rm eq}_B-f^{\rm eq}_Cf^{\rm eq}_D)?$ How will it vanish? @EricDavidKramer $\endgroup$
    – SRS
    Commented Nov 17, 2018 at 21:40
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    $\begingroup$ For the simplest example, suppose the chemical potential is zero and take the statistics to be Maxwell-Boltzmann (you are already assuming that in your Boltzmann equation by not including $1-f$ and $1+f$ factors for FD or BE). Then in equilibrium, $f_A = e^{-E_A/T}$ and similarly for the other species. The products are equal because $E_A + E_B = E_C + E_D$ by energy conservation. $\endgroup$
    – Eric David Kramer
    Commented Nov 18, 2018 at 12:07

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