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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.

Qmechanic
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Diger
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1 Answers1

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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.

GiorgioP-DoomsdayClockIsAt-90
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  • 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? – Diger Apr 25 '19 at 18:41
  • But combining the last result with yours yields $$U=\phi(N) V^{-\frac{2}{3}} , {\rm e}^{\frac{2S}{3Nk_B}}$$ or? – Diger Apr 25 '19 at 19:52
  • @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. – GiorgioP-DoomsdayClockIsAt-90 Apr 25 '19 at 20:03
  • 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? – Diger Apr 25 '19 at 20:14
  • @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. – GiorgioP-DoomsdayClockIsAt-90 Apr 25 '19 at 20:20
  • So it can not be proofed by theoretical means, but follows from experiment? – Diger Apr 25 '19 at 20:29
  • @Diger It depends what is deemd to be a proof by theoretical means. Within pure thermodynamics, the choice of the correct number of independent variable for a specific system is only an experimental fact. In a Statistical Mechanics approach it can be considered a trivial consequence of the independent global parameters one chooses to characterize a macrostate. – GiorgioP-DoomsdayClockIsAt-90 Apr 25 '19 at 20:45
  • But also in statistical mechanics for e.g. the canonical ensemble you make use of $${\rm d}F=-S{\rm d}T - p{\rm d}V + \mu{\rm d}N$$ to calculate the observables from $$F=-k_BT \log Z , $$ which already implies a specific number of independent variables for the free energy. – Diger Apr 25 '19 at 21:31
  • @Diger That's what I wrote (the choice of the independent parameters characterizing a amcrostate). However in that case it is a purely theoretical recollection of which parameter characterize the macrostate. in the case of the canonical ensemble there is the thermostat temperature, the volume in which particle are confined (which can be seen as a part of the Hamiltonian) and the number of particles (again a basic specification for the Hamiltonian). Anyway, this is becoming a little too long series of comments and would be better a discussion in the chat room. Unfortunately here is late. – GiorgioP-DoomsdayClockIsAt-90 Apr 25 '19 at 21:38
  • Yes ok, but as far as I remember the temperature in the canonical ensemble arises by comparison of the equation $$S+\frac{F}{T}-\frac{U}{T}=0$$ with the equation which arises from maximizing the Gibbs-functional for entropy $S_{\text Gibbs}$ under the constraint of a normalized density function $\varrho$ and mean internal energy $U$. The normalization fixes the first lagrangian parameter $\lambda_1$ which typically is combined in a new constant (not depending on phase space parameters) $Z$ (the density of states). – Diger Apr 25 '19 at 22:28
  • The second lagrangian parameter $\lambda_2$ remains in the equation $$S_{\text Gibbs} - k_B \log Z + \lambda_2 U=0$$ which means $\lambda_2=-\frac{1}{T}$ by comparison. So again pure thermodynamics seems to be used. You can also answer in chat. – Diger Apr 25 '19 at 22:29