statistical mechanics - 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.

Answer

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.

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