In thermodynamics the potentials are typically only a function of 2 variables, say $$U=U(S,V)$$ with entropy $S$ and volume $V$. I see that conjugate pairs $S,T$ or $p,V$ always have the unit of energy when multiplied. But what is the reason that for example $S$ and $T$ can not be independent leading to the potentials only depending on 2 of the 5 variables.


1 Answer 1


Indeed it is possible to express the internal energy of a thermodynamic system as a function of two conjugate variables. However the resulting function is not as useful as the thermodynamic potential expressed as a function of the independent variables $S,V,N$(in general don't forget N for fluid systems).

First of all, let me remark that it is not by chance that the product of two conjugate variables has the physical dimension of energy. This is a consequence of the energy being a homogeneous function of its extensive variables $S,V,N$. Indeed, Euler's theorem ensures that $$ U= TS - PV + \mu N $$

Let's use now the ideal gas as a simple example to show that it is possible to use as two independent variables the two conjugate quantities $P$ and $V$.

$$ U = \frac{3}{2}N k_BT = \frac{3}{2} PV $$

where use has been done of the equation of state.

Although this expression of the internal energy as a function of $P,V$ provides the correct value of the internal energy for any thermodynamic state, just using the corresponding values of $P$ and $V$, it cannot be defined a thermodynamic potential. A reason for this name is the possibility of obtaining explicitly all the thermodynamic quantities (by partial derivatives). This is true only if a specific set of variables is used (the so called natural variables for that potential). From the function $U=\frac{3}{2}PV$ is not possible to achieve such a result. The reason is evident in the present case because in order to obtain, say the dependence of $U$ on entropy, one should be able to get it from $$ P=-\left( \frac{\partial U}{\partial{V}}\right)_{S,N}= \frac{2}{3}\frac{U}{V}. $$ However, the integration with respect to $V$ of the previous equation provides $$ U=V^{-\frac{2}{3}}\phi(S,N) $$ where $\phi(S,N)$ is an arbitrary function of $S$ and $N$. Spoiling this result from any practical use.

  • $\begingroup$ So does this mean that $p$ and $V$ as a set of variables are not the only combinations that lead to an ill-posed potential? What about $U(T,V)$? In that case I would obtain $$U=\phi(V,N) \, {\rm e}^{\frac{2S}{3Nk_B}}$$ which doesn't yield the volume dependence? $\endgroup$
    – Diger
    Apr 25, 2019 at 18:41
  • $\begingroup$ But combining the last result with yours yields $$U=\phi(N) V^{-\frac{2}{3}} \, {\rm e}^{\frac{2S}{3Nk_B}}$$ or? $\endgroup$
    – Diger
    Apr 25, 2019 at 19:52
  • $\begingroup$ @Diger You have to keep in mind that in this example we know everything. In order to understand the general difficulty you have to assume that all we know would be $P=\frac{2}{3}\frac{U}{V}$. It is a consequence of the Euler's theorem that if one knows other relations, like the relation between $T$ and $U$ it would be possible to recover the thermodynamic potential. $\endgroup$ Apr 25, 2019 at 20:03
  • $\begingroup$ And why can a potential not depend on the variables like $U(S,V,N,T)$? Put differently, if it depends on the variables in such a form you can always reduce the number of variables to 3? $\endgroup$
    – Diger
    Apr 25, 2019 at 20:14
  • $\begingroup$ @Diger Four independent variables are not allowed by the physics of a one component fluid system. If it would be allowed to have four independent variables, instead of a unique triple point, people would find a three phases coexistence line. This is not what people measure in labs. $\endgroup$ Apr 25, 2019 at 20:20

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