9
$\begingroup$

Why is it that, at equilibrium, certain potentials are minimised?

That is, for a system at constant temperature and pressure, the Gibbs free energy is minimised, and for fixed volume and temperature, the Helmholtz free energy is minimised.

I haven't been able to find a reason for this.

$\endgroup$
2
  • 2
    $\begingroup$ Because if they are not minimized there is a drive that pushes the system towards the minimum of the potential, much like force pushes the system towards the minimum of potential energy in mechanics. $\endgroup$
    – valerio
    Feb 16, 2018 at 14:00
  • $\begingroup$ For the free energy minimization see this nice and simple answer: physics.stackexchange.com/a/369412/226902 $\endgroup$
    – Quillo
    Apr 2, 2023 at 17:18

3 Answers 3

19
$\begingroup$

You can trace energy minimization (e.g., internal energy minimization for a closed system at constant volume, enthalpy minimization for a closed system at constant pressure, Helmholtz free energy minimization at constant volume and temperature, and Gibbs free energy minimization at constant pressure and temperature) back to the Second Law, i.e., entropy maximization at constant internal energy, or $$\left(\frac{\partial S}{\partial X}\right)_U=0\,\,\,\mathrm{and}\,\,\,\left(\frac{\partial^2 S}{\partial X^2}\right)_U<0$$ for some system parameter $X$. Of course, the first term indicates that the entropy at equilibrium lies at an extremum (i.e., it doesn't change much for small changes of $X$), and the second indicates that its slope is decreasing as $X$ increases; therefore, it must be curving downwards, i.e., it is maximized.

The general approach is as follows: We can write $$\left(\frac{\partial U}{\partial X}\right)_S=-\frac{\left(\frac{\partial S}{\partial X}\right)_U}{\left(\frac{\partial S}{\partial U}\right)_X}=-T\left(\frac{\partial S}{\partial X}\right)_U=0\tag{1}$$

using the triple product rule. Furthermore (expanding $Y=\left(\frac{\partial U}{\partial X}\right)_S$, where $Y$ is a dummy variable representing a function of $U$ and $X$, as $dY=\left(\frac{\partial Y}{\partial U}\right)_XdU+\left(\frac{\partial Y}{\partial X}\right)_UdX$ and then differentiating with respect to $X$ at constant $S$), we have

$$\begin{split}\left(\frac{\partial ^2 U}{\partial X^2}\right)_S&=\left[\frac{\partial}{\partial X}\left(\frac{\partial U}{\partial X}\right)_S\right]_S=\left[\frac{\partial}{\partial U}\left(\frac{\partial U}{\partial X}\right)_S\right]_X\left(\frac{\partial U}{\partial X}\right)_S+\left[\frac{\partial}{\partial X}\left(\frac{\partial U}{\partial X}\right)_S\right]_U \end{split} $$ but the first term disappears because $\left(\frac{\partial U}{\partial X}\right)_S$ has been found to be zero. Thus, $$\begin{split}\left(\frac{\partial^2 U}{\partial X^2}\right)_S&=\frac{\partial}{\partial X}\left[-\frac{\left(\frac{\partial S}{\partial X}\right)_U}{\left(\frac{\partial S}{\partial U}\right)_X}\right]_U\\ &=-\frac{\left(\frac{\partial ^2S}{\partial X^2}\right)_U}{\left(\frac{\partial S}{\partial U}\right)_X}+\left(\frac{\partial S}{\partial X}\right)_U\frac{\left(\frac{\partial^2S}{\partial X\partial U}\right)}{\left(\frac{\partial S}{\partial U}\right)_X^2}\\ \end{split} $$ where the second term disappears because $\left(\frac{\partial S}{\partial X}\right)_U$ is initially postulated to be zero, yielding $$\left(\frac{\partial^2 U}{\partial X^2}\right)_S=-T\left(\frac{\partial^2S}{\partial X^2}\right)_U>0\tag{2}$$ Equation (1) indicates that the energy also lies at an extremum, and Equation (2) indicates that in fact it is minimized.

For more information, please see Callen's Thermodynamics and an Introduction to Thermostatics, from which this derivation was adapted.

$\endgroup$
6
  • 4
    $\begingroup$ This is the best answer. The fundamental thing is always the Second Law. $\endgroup$
    – knzhou
    Feb 16, 2018 at 17:39
  • $\begingroup$ @Chemomechanics where did you use the constant volume assumption? $\endgroup$
    – stochastic
    Feb 17, 2018 at 1:51
  • $\begingroup$ I didn't; $X$ could be volume. But you wouldn't be free to change the volume in an arbitrary way; any changes would have to be reversible to maintain constant entropy to satisfy $\left(\partial U/\partial V\right)_S$. $\endgroup$ Feb 17, 2018 at 4:22
  • $\begingroup$ Before I ask a question about it I figured I'd ask in a comment here; in your equation (2) (which, as you point out, Callen also produces) we have merely shown that the diagonal element of the $U$ Hessian are positive -- but is that sufficient to show that the Hessian is positive definite and thus that we have a genuine minimum? $\endgroup$
    – EE18
    Jun 26, 2023 at 14:06
  • $\begingroup$ That would make a good new question; if you ask it, please add a link from here for others who are curious. $\endgroup$ Jun 26, 2023 at 14:50
3
$\begingroup$

You can see this as a consequence of the very essence of a potential: it is a function from which some physical quantities derive. If an equilibrium can be put in the form $$ f(u)=0 $$ and your problem exhibits a potential $\Pi$ so that $$ f(u)=\frac{\partial \Pi}{\partial u} $$ then you can see that solving the equilibrium is equivalent to cancelling the potential derivative, which means finding an extremum. Finally, if the potential happens to be strictly convex, there is only one extremum that is the minimum.

$\endgroup$
3
$\begingroup$

Ultimately, this is a question of convention. It's trivial to add some minus signs to make it so that equivalent potentials would be maximized, instead, for example.

Where does the minimization convention come from, then? Well, it has its origins in the study of mechanical potential energy where the analogy to objects rolling in the bottom of a well and coming to rest at the bottom of them. This initial analogy for potential energy being like a physical height of objects here on Earth, and kinetic energy being positive instead of negative, is what ultimately drove the "equilibrium = minimum" convention in physics. Had early physicists found a different metaphor made things easier for them to understand, for example, one where objects are attracted to a point where some quantity is maximized instead of minimized (a money analogy, say), then we'd have a system with flipped signs.

I should add that the thermodynamic potentials are all, ultimately, actually better described as "the relevant part of the total energy." We often are so eager to get into Legendre transforms, and the thermodynamic shield, that we gloss over this basic fact. Consider the enthalpy: $$ H = E + P V. $$ This is where it comes from. Say I have some system that has internal energy $E_1$. I then put it into mechanical contact with a large system that has mechanical energy $E_2$ so that systems 1 and 2 can exchange volume. We now assume the following two statements are true:

  1. system two is so large that its pressure (and temperature) changes as system one expands and contracts are negligible, and
  2. both systems have achieved mechanical equilibrium, so $P_1 = P_2 = P$.

We now Taylor expand system two's contribution to the total energy as a function of the change in $V_1$: \begin{align} E & = E_1 + E_2 \\ & = E_1 + E_2|_{\Delta V_1 = 0} + \left[\frac{\partial E_2}{\partial \Delta V_1}\right]_{\Delta V_1 = 0} \Delta V_1 + \frac{1}{2}\left[\frac{\partial^2 E_2}{\partial (\Delta V_1)^2}\right]_{\Delta V_1 = 0} (\Delta V_1)^2 + \ldots\\ & = E_1 + E_2|_{\Delta V_1 = 0} - \left[\frac{\partial E_2}{\partial \Delta V_2}\right]_{\Delta V_1 = 0} \Delta V_1 + \frac{1}{2}\left[\frac{\partial^2 E_2}{\partial (\Delta V_2)^2}\right]_{\Delta V_1 = 0} (\Delta V_1)^2 + \ldots\\ & = E_1 + E_2|_{\Delta V_1 = 0} - \left[-P_2\right]_{\Delta V_1 = 0} \Delta V_1 + \frac{1}{2}\left[\frac{\partial^2 E_2}{\partial (\Delta V_2)^2}\right]_{\Delta V_1 = 0} (\Delta V_1)^2 + \ldots \end{align} $\Delta V_1$ is just $V_1 - V_1'$, where $V_1'$ is the equilibrium volume. By our assumptions, the second derivatives (and higher order terms) are negligible, and we can group $\left.E_2\right|_{\Delta V_1=0}$ into constant shifts to the potential energy. There is another shift that we assume is constant, the term $P_2 V_2'$. Because constant offsets to the energy have no physical consequence outside of general relativity we can get rid of these by shifting the total energy, leaving only the relevant part of the energy: $$E_{\mathrm{relevant}} = E_1 + P V_1 + \mathcal{O}\left([\Delta V_1]^2\right).$$

As you can see, the relevant part of the energy is just the enthalpy. In textbooks I've seen it explained as the energy plus the work you need to do to make space for the system, but that explanation only works for enthalpy, and not the Helmholtz or Gibbs free energies. But they are all, each and every one, just the relevant parts of the total energy.

In sum:

  1. enthalpy = the relevant part of the energy of the system plus the energy of a work bath the system is exchanging volume with,
  2. Helmholtz free energy = the relevant part of the energy of the system plus the energy of a heat bath the system is exchanging entropy with, and
  3. Gibbs free energy = the relevant part of the energy of the system plus the energy of the heat and work baths it is in contact with

I don't know the terminology for the energy exchanged in the $\mu \mathrm{d}N$ terms, but the Landau potentials are just the relevant part of the energy when you put your system in contact with another large system that can exchange entropy at constant temperature and chemical species at constant chemical potential.

$\endgroup$

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.