2
$\begingroup$

I've recently been trying to understand the mathematical postules of classical thermodynamics and I've been following Callen's Introduction to Thermodynamics and Thermostatics. However, his derivation of the homogeneity of entropy doesn't seem to line up for me. Postulate III states:

Postulate III. The entropy of a compositive system is additive over the constituent subsystems. The entropy is continuous and differentiable and is a monotonically increasing function of the energy.

Then he goes on to state

The additivity property applied to spatially separate subsytems requires the following property: The entropy of a simple system is a homogeneous first-order function of the extensive parameters.

which seems like a non-sequitor to me. Why would additivity over separate systems require homogeneity of a fundamental relation for entropy? This seems to make a similar conceptual mistake I made in this question, it seems like Callen is interpreting additivity over constituent systems as implying linearity for the entropy function i.e. $S(U_1+U_2,\vec{X}_1+\vec{X}_2) = S(U_1,\vec{X}_1) + S(U_2,\vec{X}_2)$ which implies homogeneity. But this doesn't hold for all $(U_1,\vec{X}_1)$ and $(U_2,\vec{X}_2)$, it only holds for equilibrium systems. It doesn't even hold for ideal gases. Am I missing something?

$\endgroup$
1

1 Answer 1

5
$\begingroup$

Callen's condition of homogeneity of degree one for the entropy as a function of its extensive variables is a direct consequence of his third postulate.

One has to apply the postulate to a compound system made by $n$ equal systems. The fact that they are equal means that they are characterized by the same values of temperature, pressure, and chemical potential. Therefore, if we would remove the walls between them, we will remain with a system still at equilibrium characterized by an energy that is $n$ times the energy of a subsystem, and similarly for volume and number of particles (or moles). In the formula, in such a case, due to the equilibrium condition, it is true that $$ S(nU,nV,nN)= n S(U,V,N) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [1] $$ for all positive integer values of $n$ and for all ($U,V,N$) in the domain of $S$. Thus, let's introduce $\tilde U= nU,\tilde V = nV,\tilde N = nN$. Equation [$1]$ becomes $$ S(\tilde U,\tilde V,\tilde N)= n S\left(\frac{\tilde U}{n},\frac{\tilde V}{n},\frac{\tilde N}{n} \right) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [2] $$ i.e., taking into account the arbitrariness of the arguments $$ S\left(\frac{U}{m},\frac{ V}{m},\frac{ N}{m} \right) = \frac{1}{m}S( U, V, N),~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [2^{\prime}] $$ again for all positive integer values of $m$ and for all ($U,V,N$) In the domain of $S$.

Combining [$1$] and [$2^{\prime}$] we get $$ S\left(\frac{nU}{m},\frac{n V}{m},\frac{n N}{m} \right) = \frac{n}{m}S( U, V, N),~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [2''] $$ for all integer positive values of $n$ and $m$ and for all ($U,V,N$) in the domain of $S$. At this point, it is enough to work with sequences of ratios $n_i/m_i$ and using the continuity of entropy (again in the third postulate) to prove homogeneity of degree one $$ S\left(\lambda U, \lambda V,\lambda N \right) = \lambda S( U, V, N) $$ for all real positive values of $\lambda$ and for all ($U,V,N$) in the domain of $S$.

As you can see, there is no mistake since the total system and subsystems are all at mutual equilibrium.

A word of caution is required for the domain of $S$. Callen does not discuss it, but it is clear that homogeneity is meaningful only if $U, V,$ and $N$ are positive. That is not a problem for $V$ and $N$. However, also $U$ is not an issue if we require that internal energy is limited below. Since it is always possible to add an arbitrary constant to energy, we can take the lower limit as the new zero of energy. Similarly for the entropy, which is limited below as a consequence of the fourth principle of thermodynamics.

As a final remark, I would add that I have played for a while with alternative formulations of Callen's postulates. My partial conclusion is that Callen's choice of postulates looks wise and carefully planned.

$\endgroup$
5
  • $\begingroup$ Suppose we denote composition of systems $s_1,s_2$ by $\oplus$. Then the third postulate could be states as $S(s_1\oplus s_2) = S(s_1)+S(s_2)$. But entropy is a function of the extensive variables of a system, so if the extensive variables of $s_1,s_2$ are $\vec{X}_1,\vec{X}_2$ then this postulate becomes $S(\vec{X}_1+\vec{X}_2)=S(\vec{X}_1)+S(\vec{X}_2)$ which is linearity. Am I missing something? $\endgroup$ Commented Dec 25, 2019 at 3:13
  • $\begingroup$ You are missing mathematics and physics of the composition. Math, because you are translating your "composition" of entropies into sum of their argument, which is not the mathematical translation of Postulate III. Physics, because in order to justify the equality between entropy of the sum of the variables as sum of the separate entropies you need thermodynamic equilibrium between the separate parts. Which is not true for all possible choices of the arguments. $\endgroup$ Commented Dec 25, 2019 at 8:59
  • $\begingroup$ @KPomykala an example of correct mathematical translation of the "composition" is the form I wrote Postulate III here: physics.stackexchange.com/questions/481879/… $\endgroup$ Commented Dec 25, 2019 at 9:02
  • $\begingroup$ So does this mean the homogeneity holds only for "combining" perfectly identical systems? This is what Callen probably means by simple systems when he concludes that homogeneity follows right? $\endgroup$ Commented Dec 13, 2020 at 10:22
  • $\begingroup$ @pindakaas, yes, to show that fundamental equations are homogeneous functions of their extensive variables, one must combine perfectly identical systems. Missing that, you still have a thermodynamic system, but it is useless to arrive at the homogeneity condition. $\endgroup$ Commented Dec 13, 2020 at 11:26

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.