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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.

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  • $\begingroup$ I suspect that you're supposed to take U for the combined system as constant. Then energy will pass from one subsystem to the other in such a way as to maximise the combined entropy (as this will give the most probable configuration –the one with the most microstates). The maths of maximising isn't difficult. $\endgroup$ – Philip Wood May 22 '17 at 21:50
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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.

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  • $\begingroup$ I have calculated it with the same result, thank you very much... In our exercise we also need to show later that the entropy is extensive, therefore we can not use the extensivity of the entropy. Therefore I calculated the entropy around the maximum of $U_1$ and it was the same after a few steps. Thanks again !! $\endgroup$ – Leviathan May 24 '17 at 12:46

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