# 4. General Extremum Principles and Thermodynamic Potentials

We have seen that min{$$U(S,\bar{X})$$} and min{$$U(S,\hat{X})$$} imply one another. Under certain conditions, these principles are very convenient. For example,

$dS = \dfrac{1}{T} dU - \dfrac{P}{T} dV + \sum \dfrac{\mu_i}{T} dn_i \Rightarrow max \left\{ \sum \dfrac{\mu_i}{T} dn_i \right\}$

maximizes $$S$$ at constant $$U$$ and $$V$$.

But what do we do if we are working at constant (T,V) or (T,P)? We then have open systems to deal with, where energy must flow (to keep $$T$$ constant), or the volume must change (to keep P constant), etc. This problem arises for many laboratory reactions in chemistry. As it turns out, if thermodynamic variables other than U and V are held constant, thermodynamic potentials other than entropy or energy of the open system are extremized. (Of course, $$S$$ is still maximized for a closed system containing our open system of interest.) Very conveniently, these potentials can be computed just from the properties of the open system of interest alone (e.g. one where $$T$$ is constant, and therefore energy must be allowed to flow in and out), and the added assumption that the corresponding variable (e.g. $$T$$) is always constant in the environment. The environment is thus treated as a “bath” or “reservoir” for the intensive variable of interest. That is, we assume the closed system containing our open system of interest is so vast that a change in extensive variable (e.g. $$U$$) of the small open system does not affect the conjugate intensive variable in the environment (e.g. $$T$$; the energy deposited or taken from the environments by the open system is too small to change T in the environment).

Our goal: we want potentials where our choice of intensive variable is in the dependent variables. Let’s begin by discussing an apparently obvious way of doing this, which does not work. Let’s say we have the fundamental relation for $$U$$, but we want to hold $$T$$ constant, not $$S$$:

$U = U(S,V,n_i) \Rightarrow T=\left( \dfrac{\partial U}{\partial S} \right)_{V,n_i} = T(S,V,n_i)$

solving for $$S=S(T,V,n_i)$$, and insert in $$U$$ we seem to be able to get

$U =U(T,V,n_i).$

Now we can hold $$T$$ constant. The problem with this: $$T$$ is the slope of $$U(S)$$, and expressing a function in terms of its own slope leaves the intercept indeterminate: $$U$$ is no longer completely defined, and we do not have a fundamental relation to which the laws of thermodynamics can be applied.

We do however want an equation in terms of the slope. The solution is to express the slope $$m$$ of $$y(x)$$ in terms of the intercept $$\phi$$ (or vice-versa). The intercept as a function of slope does contain all the original information about the function $$y(x)$$. The process of transforming a function $$y(x)$$ into $$\phi(m)$$ is called Legendre transform.