Followup from this question,

It seems like the total temperature of the composite system $T=T_A+T_B$ can be found using this relation:

$$\frac{1}{T}=\frac{\partial S}{\partial U}$$

where $S=S_A+S_B$ is total entropy and $U=U_A+U_B$ is total energy. In particular, I want to show that


I started out by applying chain rule to $\frac{\partial S}{\partial U}$:

$$\frac{1}{T}=\frac{\partial S}{\partial U_A}\frac{\partial U_A}{\partial U} + \frac{\partial S}{\partial U_B} \frac{\partial U_B}{\partial U}$$

Then I expand $S$ and $U$:

$$\frac{1}{T}=\frac{\partial}{\partial U_A}(S_A+S_B)\frac{\partial}{\partial U} (U-U_B) + \frac{\partial}{\partial U_B} (S_A+S_B) \frac{\partial U_B}{\partial U}$$

$$\frac{1}{T}=\frac{\partial}{\partial U_A}(S_A+S_B)\frac{\partial}{\partial U} (U-U_B) + \frac{\partial}{\partial U_B} (S_A+S_B) \frac{\partial U_B}{\partial U}$$

Since $\frac{1}{T_A}=\frac{\partial S_A}{\partial U_A}$ and $\frac{1}{T_B}=\frac{\partial S_B}{\partial U_B}$,

$$\frac{1}{T}=(\frac{1}{T_A}+\frac{\partial S_B}{\partial U_A}) (1-\frac{\partial U_B}{\partial U}) + (\frac{\partial S_A}{\partial U_B}+\frac{1}{T_B}) \frac{\partial U_B}{\partial U}$$

I am not sure what to do next.

  • $\begingroup$ You don't add temperatures. The composite system doesn't have a temperature until the two subsystems are in thermal equilibrium with each other; then they both have the same temperature. It doesn't make sense to add temperatures of two systems. If you brought an ice cube ($T=273$ K) into contact with boiling water ($T=373$ K), is the temperature of the composite system the sum of the temperatures, i.e. 646 K? Nope! $\endgroup$
    – march
    Commented Feb 28, 2023 at 9:19
  • $\begingroup$ @march I see, how about when the objects are in thermal equilibrium? Is it possible to show that $1/T=1/(2T_A)=1/(2T_B)$? At equilibrium, $\frac{\partial S}{\partial S_{B}} = - \frac{\partial S}{\partial S_{A}} = 0$, so I got $1/T=0$ instead. $\endgroup$
    – Jimmy Yang
    Commented Feb 28, 2023 at 9:33

1 Answer 1


Assuming your system is isolated you want to maximize $S$ with fixed $U$ according to the second law. More concretely, you want to maximize $S=S_A+S_B$ by only varying $U_A,U_B$ under the constraint $U_A+U_B=U$.

Adding some mathematical assumptions, you just apply the method of Lagrange multipliers. You can thus get rid of the constraint by minimizing instead: $$ S-\beta (U_1+U_2) $$ with $\beta$ the Lagrange multiplier. Looking for stationary points gives you: $$ \frac{\partial S_A}{\partial U_A}=\frac{\partial S_B}{\partial U_B}=\beta $$ And you identify the first two members as $$ \frac{\partial S_A}{\partial U_A}=\frac{1}{T_A} \\ \frac{\partial S_B}{\partial U_B}=\frac{1}{T_B} $$ respectively and the Lagrange multiplier as $1/T$, the inverse temperature of the whole system.

Actually, the Lagrange function is related to free energy. To summarize you have a system of equations: $$ T_A=T_B \\ U_A+U_B=U $$ for two unknowns $U_A,U_B$, so (up to mathematical subtleties) you have all the information needed to calculate the equilibrium state. They simply translate thermal equilibrium as the homogeneity of temperature. Unless you specify the $S_A,S_B$ dependences on energy, you cannot go further in general. Note that you do not add the temperatures, you need to equate them.

Hope this helps.


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