Entropy of two subsystems exchanging energy

Question

I consider two subsystems with the number of microstates $\Gamma_1(U_1)$ and $\Gamma_2(U_2)$ and their energies $U_1$ and $U_2$ respectively. The subsystems can exchange energy so that the total energy of the system is $U=U_1+U_2$. It is also given that $\ln \Gamma (U) = S(U)/k_\mathrm{B}$

Now we have the entropies given by $S_1(U_1) = -a_1 (U_1 - U_1^{0})^{2}$ and $S_2(U_2) = -a_2(U_2-U_2^{0})^{2}$ with $a_1 > 0$, $a_2 >0$. And I want to calculate the entropy of the system $S(U)$.

My idea:

\begin{align*} S(U) = k_\mathrm{B} \left( S_1 (U_1) + S_2 (U_2) \right) = k_\mathrm{B} \left(-a_1 (U_1 - U_1^{0})^{2}+ -a_2(U_2-U_2^{0})^{2}\right) \end{align*}

Is it that easy?

Please help me, because I think it's too easy my solution.

Answer 1

At equilibrium the two subsystems will adjust their energies to maximize the total entropy of the system \begin{align} S = -a_1 (U_1 - U_1^{0})^{2}-a_2(U_2-U_2^{0})^{2} \end{align} We have that \begin{align*} dS = -2a_1 (U_1 - U_1^{0})dU_1-2a_2(U_2-U_2^{0})dU_2 \end{align*} Taking into account that the total energy of the system remains constant ($U=U_1+U_2$) we have $dU_1=-dU_2$. At thermodynamic equilibrium we have $dS=0$ and we get
\begin{align*} a_1 (U_1 - U_1^{0})=a_2(U_2-U_2^{0}) \end{align*} Inserting $U_2=U-U_1$ in the above equation we get \begin{align*} U_1-U_1^{0}=\frac{a_2}{a_1+a_2}\left( U-(U_2^{0}+U_1^{0}) \right) \end{align*} and \begin{align*} U_2-U_2^{0}=\frac{a_1}{a_1+a_2}\left( U-(U_2^{0}+U_1^{0}) \right) \end{align*} Plugging these eqs into the initial eq we have the entropy at equilibrium \begin{align} S = -\frac{a_1a_2}{a_1+a_2}\left( U-(U_2^{0}+U_1^{0}) \right)^2 \end{align} This is the correct (equilibium) entropy of the system.