Interacting Einstein Solids

Two Einstein solids are weakly coupled and isolated from the rest of the world. Solid $A$ has $N_A$ oscillators and $q_A$ energy units. Solid B has $N_B$ oscillators and $q_B$ energy units. Since the two solid can exchange energy with each other but not surrounding, the total energy units $$q_{\text{tot}} = q_A + q_B$$

Let $\Omega_A$ and $\Omega_B$ be the the number of microstates of solid A and B respectively. For each microstates of solid $A$, there are $\Omega_B$ microstates available to solid B, so the mulitiplicity of the whole system is $$\Omega_{\text{total}}=\Omega_A\Omega_B.$$ For large solids, $$\Omega(N,q)\approx \frac{(q+N)!}{q!N!}.$$ Taking logarithm, $$\begin{align} \ln\Omega &= (q+N)\ln(q+N)-q\ln q-N\ln N \\ \frac{\partial}{\partial q}\ln\Omega &= \ln(q+N)+\frac{q+N}{q+N}-\ln q - \frac{q}{q}\\ &=\ln\left(\frac{q+N}{q}\right). \end{align}$$ Since $\Omega_{\text{tot}}=\Omega_A\Omega_B$, $$\begin{align} \ln\Omega_{\text{tot}} &= \ln\Omega_A + \ln\Omega_B\\ \frac{d}{d q_A}\ln\Omega_{\text{tot}} &= \ln\left(\frac{q_A+N_A}{q_A}\right)-\ln\left(\frac{q_{\text{tot}}-q_A+N_B}{q_{\text{tot}}-q_A}\right)\\ &= \ln\frac{(q_A+N_A)(q_{\text{tot}}-q_A+N_B)}{q_A(q_{\text{tot}}-q_A)}. \end{align}$$ Since $$\frac{d\Omega_{\text{tot}}}{dq_A}=\frac{d\Omega_{\text{tot}}}{d \ln\Omega_{\text{tot}}}\frac{d\ln\Omega_{\text{tot}}}{dq_A}=\Omega_{\text{tot}}\frac{d}{d q_A}\ln\Omega_{\text{tot}}$$ and $\Omega_{\text{tot}}$ is a positive integer by definition, $\frac{d\Omega_{\text{tot}}}{dq_A}=0$ when $$\frac{(q_A+N_A)(q_{\text{tot}}-q_A+N_B)}{q_A(q_{\text{tot}}-q_A)}=1.$$ So, extremum of multiplicity occus at $$\bbox[5px,border:2px solid #666] { q_A=\frac{N_Aq_{\text{tot}}}{N_A+N_B}. }$$ Since this is the macrostate with highest probability, at equilibrium, it is most likely to be found at this macrostate.