I am confused about whether the temperature of an isolated system in thermal equilibrium is finite or infinite.

Consider an isolated system made up of two components: $A$ and $B$. The total energy of the system $U_{tot}+U_A+U_B$ is constant. When this system is in thermal equilibrium,

$$\frac{\partial S}{\partial U_A}=0$$

$$\frac{\partial S_A}{\partial U_A}=\frac{\partial S_B}{\partial U_B}$$

where $S=S_A+S_B$ is the total entropy of the system. I know that temperatures of $A$ and $B$ are equal, $T_A=T_B$. But from the first equation, the temperature of the system as a whole is practically infinite because $\beta=1/k_BT=0$?

Consider a related isolated system: two-state paramagnet described in section 3.3 in An Introduction to Thermal Physics by Schroeder, its entropy vs. energy graph is shown below


I am uncertain whether the paramagnet is an isolated system or not. It seems to reach thermal equilibrium when its total energy is zero. This is when the temperature is infinite.

Are there any examples when system's temperature is finite in thermal equilibrium?


1 Answer 1


The thermodynamic beta equation you are referring to is

$$ \beta = \frac{1}{T} = \frac{\partial S}{\partial E} $$

and arguably in your situation, all the $E$ is $U$. However, note that your first equation is

$$ \frac{\partial S}{\partial U_{A}} = 0 $$

and not

$$ \frac{\partial S}{\partial U} = 0 $$

If it were the second, then I guess the $\beta$ equation applies, but $U_{A}$ is only a portion of $U$, even if $U = E$. In fact, $U_{A}$ and $U_{B}$ are not independent.

What these equations say is different. Since, as you pointed out, $U = U_{A} + U_{B}$, we have $dU = dU_{A} + dU_{B}$. By nature of the system's isolation, $dU = 0$ for all allowed processes, so

$$ 0 = dU_{A} + dU_{B} $$

If we now consider the total entropy, $S = S_{A} + S_{B}$,

$$ \begin{align} dS &= \frac{\partial S_{A}}{\partial U_{A}}dU_{A} + \frac{\partial S_{A}}{\partial U_{B}}dU_{B} + \frac{\partial S_{B}}{\partial U_{A}}dU_{A} + \frac{\partial S_{B}}{\partial U_{B}}dU_{B} \\ dS &= \frac{\partial S_{*}}{\partial U_{*}}\left( dU_{A} + dU_{B} \right) + \frac{\partial S_{A}}{\partial U_{B}}dU_{B} + \frac{\partial S_{B}}{\partial U_{A}}dU_{A} \\ dS &= \frac{\partial S_{B}}{\partial U_{A}}dU_{A} + \frac{\partial S_{A}}{\partial U_{B}}dU_{B} \end{align} $$

Where for clarity I gave a single name to the common value of $dS_{i}/dU_{i}$ from the second assumption and the $dU_{i}$ sum to zero as we said. Continuing, we substitute to a common differential of $dU_{A}$, and for the first (positive) term we substitute $S_{B} = S - S_{A}$ and for the second term we use the fact that differentiating with $\partial U_{B}$ is the same as differentiating with $-\partial U_{A}$:

$$ \begin{align} dS &= \left( \frac{\partial S_{B}}{\partial U_{A}} - \frac{\partial S_{A}}{\partial U_{B}}\right)dU_{A} \\ dS &= \left( \frac{\partial }{\partial U_{A}} \left( S - S_{A}\right) - - \frac{\partial }{\partial U_{A}}S_{A}\right)dU_{A} \\ dS &= \left(\frac{\partial S}{\partial U_{A}} - \frac{\partial S_{A}}{\partial U_{A}} + \frac{\partial S_{A}}{\partial U_{A}} \right) dU_{A} \\ dS &= 0 \end{align} $$

The last step used the first hypothesis of your question. Thus, changes in $U_{A}$ cause no change in system entropy conditioned upon adhering to processes within this isolated system of elements A and B. $dU_{A}$ and $dU_{B}$ are related to one another (just negatives for a two component system), and along that curve of possible processes there is a directional zero derivative. Obviously, by adding energy to the system, we can increase entropy as well, but this will necessarily break the $dU_{A} = -dU_{B}$ relation while energy is being added.

Note that since adding energy to the system can still change entropy, you do not have a system at infinite temperature. It is also worth noting that:

$$ \frac{\partial S}{\partial S_{B}} = - \frac{\partial S}{\partial S_{A}} = 0 $$

as one would expect.

  • $\begingroup$ I don't understand what you mean by $\beta$ only applies to $\frac{\partial S}{\partial U}$. $U=U_A+U_B$ is constant here. It does not make any sense to me to differentiate with respect to a constant quantity. $\endgroup$
    – Jimmy Yang
    Feb 28, 2023 at 6:23
  • $\begingroup$ I was suspicious that I did not use the first assumption, investigated, and sure enough, for a total derivative, it is needed. We are not differentiating with respect to $dU$, we are differentiating along the $U_{A}$, $U_{B}$ curve. What I am saying is that is the definition of $\beta$. $\endgroup$ Feb 28, 2023 at 6:25
  • $\begingroup$ Basically, we are taking the gradient of $S$ for variables $U_{A}$ and $U_{B}$, but we are restricted to the direction $dU_{A} = -dU_{B}$. At the two conditions hypothesized in your question, you can show that $dS = 0$. $\endgroup$ Feb 28, 2023 at 6:29
  • $\begingroup$ Thank you for the detailed answer! But there is one thing that is bothering me... If we take a look at the total entropy S vs. total energy U graph for this system, how it will evolve as the system approaches thermal equilibrium? So the system will try to increase its entropy to the maximum while its total energy stays constant? The S vs. U curve will become taller? $\endgroup$
    – Jimmy Yang
    Feb 28, 2023 at 6:51
  • 1
    $\begingroup$ The "total U" curve is multivalued (not a well defined function). You should think of this as is $S$ vs $U_{1}$ or $U_{2}$. These graphs will have a maximum like the image you included. You can also think of the 3D graph of $S$ vs $U_{1}$ and $U_{2}$. This is the behavior across all energy levels. Any particular energy level, U, would be a "minus 45 degree" line $dU_{1} = - dU_{2}$. The total U line the system is currently on defines the total energy, and it will move along that line at -45 degree slope to a maximum S. Then, increasing S will be perpendicular (at a +45 degree slope). $\endgroup$ Feb 28, 2023 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.