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I have a basical question on the mathematical condition characterizing a system beeing in thermodynamic equilibrium:

Let $P(A, B, C)$ is a thermodynamic potential with natural variables $A, B, C$ describing a given system

(e.g. : internal energy $U(S, V, N)$, Entropy $S(U, V, N)$, Helmholtz free energy $F(T, V, N)$ or Gibbs energy $G(T, p, N)$

then for a system described by $P(A, B, C)$ is that is in thermodynamic equilibrium the necessary condition is

$$d P= \frac{\partial P}{\partial A} dA +\frac{\partial P}{\partial B} dB +\frac{\partial P}{\partial C} dC=0$$

i.e. in TD equilibrium the potential is extremal (but recall that it depends on concrete potential if the desired extremum is a maximum or minimum).

From basic analysis course we know that a neccessary condition on a function $P: \mathbb{R}^3 \to \mathbb{R}$ to have at point a $(A_0,B_0,C_0)$ an extremum tells that it's gradient at $(A_0,B_0,C_0)$ vanish, ie

$$(\nabla P)_{(A, B, C)=(A_0,B_0,C_0)}= \left(\frac{\partial P}{\partial A}, \frac{\partial P}{\partial B}, \frac{\partial P}{\partial C}\right)_{(A, B, C)=(A_0,B_0,C_0)}= (0,0,0)$$

This imply obviously that at extremal points neccessarily all partial derivatives in natural variables vanish.

On the other hand if we focus on the example of internal energy $P(A,B,C)=U(S, V, N)$, then using

$$\mathrm{d} U = \frac{\partial U}{\partial S} \mathrm{d} S + \frac{\partial U}{\partial V} \mathrm{d} V + \sum_i\ \frac{\partial U}{\partial N_i} \mathrm{d} N_i\ = T \,\mathrm{d} S - p \,\mathrm{d} V + \sum_i\mu_i \mathrm{d} N_i\,$$

we have

$$T = \frac{\partial U}{\partial S},\quad p = \frac{\partial U}{\partial V},\quad \mu = \frac{\partial U}{\partial N}.$$

On the other hand there obviuosly exist systems in thermodinal equilibrium with $T, p, \mu \neq 0$.

And this exactly my Question: if I have a TD system described by natural variables $A, B, C$ and TD potential $P(A,B,C)$ and I want to determine it's locus where it has it's TD equilibrium, how one should read & interpret the condition

$$dP= \frac{\partial P}{\partial A} dA +\frac{\partial P}{\partial B} dB +\frac{\partial P}{\partial C} dC=0$$

correctly?

The mathematical criterium that demands that all partial derivatives have to vanish at extremal points lead me to an absurd statement as I tried to described above. That is I think that the condition $d P=0$ has in this case to be interpreted in another way as I did above. Can somebody explain what is the right interpretation of $d P=0$ plausibly compatible with physics and analysis. Where is the error in my reasonings above?

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    $\begingroup$ Is the condition $\nabla P_{(A,B,C)} = 0$ equivalent to $\text{d}P=0$? It looks to me that mathematically $\nabla P = 0\Longrightarrow \text{d}P = 0$ but the converse is not true. $\endgroup$ Jan 12, 2020 at 8:01

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At equilibrium, a thermodynamic potential is maximized or minimized with respect not to its natural variables but to whatever variables that are not externally constrained. An unconstrained variable could be: the extent to which chemical reaction has proceeded; the energy of subsystem 1 when the system is divided into two subsystems that can exchange heat with each other; etc.

Consider a thermodynamic relation concerning the Helmholtz free energy: $$ dA = -SdT - PdV + \bigg(\frac{\partial A}{\partial X}\bigg)_{T, V} dX. $$ Here, temperature ($T$) and volume ($V$) are natural variables of $A$, and X is some unconstrained variable. If we fix $T$ and $V$ (i.e., $dT = dV = 0$) and let the system reach equilibrium, the system will adjust itself such that $A$ is minimized with respect to $X$, making $(\partial A / \partial X)_{T, V} = 0$, i.e., $dA = 0$ for an arbitrary $dX$.

Note that if we lift the constraint $dV = 0$ as well when minimizing $A$, we will have $P = 0$, and OP found this strange. But this result isn't strange at all! Intuitively, a system with no constraint on its volume will expand forever to become infinitely large, in which case $P = 0$.

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    $\begingroup$ I think I understand... so the partial derivatives $\partial A /\partial X$ vanishes if there is no imposed restriction condition on natural variable $X$. On the other hand if a natural variable $Y$ is constrained by assumption, ie for example $Y=T$ and the system is connected with thermal reservoir with fixed temperature $T_0$ or $Y=V$ and the system sits in fixed volume $V_0$, then there is no reason why $\partial A /\partial Y=0$. Is that the issue? So the neccessary conditions for a system in TD equilibrium with given constrains is that only partials of non constrained variables vanish? $\endgroup$
    – user267839
    Jan 12, 2020 at 22:38
  • $\begingroup$ @KarlPeter That's right! I would also like to emphasize that an unconstrained variable doesn't need to be any of the natural variables. It could be the number of particles in the left half of the box, or even more drastically, the number of particles in a given energy level, as in the following: en.m.wikipedia.org/wiki/… $\endgroup$
    – higgsss
    Jan 13, 2020 at 1:57
  • $\begingroup$ One aspect I not fully understand. I would like to present an example where occurs exactly that kind of pathology which still confuses me. Assume we have a potential $P(A,B,C)$ of a system beeing in thermodynamical equilibrium. Assume that $A,B$ are constrained by assumption, e.g. a system in fixed volume and connected with thermal reservoir. that means expectially $dA=dB=0$ but not neccessary $\partial P/\partial A, \partial P/\partial B =0$ by the "logic" above. The variable $C$ is not constrained $\endgroup$
    – user267839
    Jan 13, 2020 at 14:51
  • $\begingroup$ and the potential behaves with respect to $C$ as Mexacan hat potential en.wikipedia.org/wiki/… That is since all states of $C$ that are allowed coincide with the circle orbit along the lowest possible potential value, i.e. the bottom of the hat. Then definitely $dC \neq 0$ and thus in order to preserve the condition $dP = \frac{\partial P}{\partial A} dA +\frac{\partial P}{\partial B} dB +\frac{\partial P}{\partial C} dC=0$ we need $\frac{\partial P}{\partial C}=0$, right? But this imply that the thermodynamic $\endgroup$
    – user267839
    Jan 13, 2020 at 14:51
  • $\begingroup$ potential $P$ is independent of $C$. Isn't this physically absurd or is this exactly what can going on in a thermodynamical equilibrium? $\endgroup$
    – user267839
    Jan 13, 2020 at 14:52

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