# Why are thermodynamic potentials minimised?

Why is it that, in equilibrium, certain potentials get minimised?

ie, for a system at constant temperature and pressure the Gibbs Free energy is minimised and for fixed volume and temperature the internal energy is minimised.

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

• 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. – valerio Feb 16 '18 at 14:00

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, 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.

• 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 $\left(\partial U/\partial V\right)_S$. – Chemomechanics Feb 17 '18 at 4:22
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.