A Statistical View of Entropy — Question 6

Question 6

Equal temperatures emerge by maximizing the combined entropy

Two model systems exchange energy in isolation, with fixed $U_A+U_B=10\varepsilon$, where $\varepsilon>0$ is an energy scale. Their dimensionless state-count models are $\Omega_A=C_A(U_A/\varepsilon)^2$ and $\Omega_B=C_B(U_B/\varepsilon)^3$, with positive constants $C_A,C_B$. Treat energy as continuous and use $S=k_B\ln\Omega$ and $1/T=\partial S/\partial U$.

1. Maximize the combined entropy to find the equilibrium energy division, and verify that it is a maximum.

2. Find both temperatures at that division. Explain why equilibrium does not require equal energies.

Question 6 - Solution

Equal temperatures emerge by maximizing the combined entropy

(a) Method and calculation

$$\frac{dS_{AB}}{dU_A}=k_B\left(\frac2{U_A}-\frac3{10\varepsilon-U_A}\right)=0.$$$$U_A=4\varepsilon,\quad U_B=6\varepsilon,\quad\frac{d^2S_{AB}}{dU_A^2}=-k_B\left(\frac2{U_A^2}+\frac3{U_B^2}\right)<0.$$

(b) Interpretation and further calculation

$$T_A=U_A/(2k_B),\quad T_B=U_B/(3k_B).$$$$T_A=T_B=2\varepsilon/k_B.$$The different energy dependences of the state counts give different energies at the same temperature. Constants $C_A,C_B$ shift entropy but do not change the maximizing division.

Physical check

The plotted entropy has an arbitrary vertical offset; its maximum and curvature are the relevant features.