Maxwell's Relations

I recently found out that thermodynamics can be recast in terms of exterior calculus. One of the many things made easier is Maxwell's relations. Anyone who's taken thermodynamics has at some point run into Maxwell's relations. They're usually derived using the equality of mixed partial derivatives and Legendre transforms. Some people instead come up with all sorts of funny ways to remember them like this one.

Try explaining how those diagrams work. Go ahead I'll wait.

I'm assuming you know at least a little bit of exterior calculus. The only rules we really need are some of the rules for the exterior derivative and the wedge product:

\begin{equation}
d^2 f = 0
\end{equation}

\begin{equation}
d( f dg ) = df \wedge dg
\end{equation}

\begin{equation}
a \wedge b = - b \wedge a
\end{equation}

\begin{equation}
a \wedge a = 0
\end{equation}

Firstly we'll look at a basic one-form

\begin{equation}
df = \left(\frac{\partial f}{\partial x}\right)_y dx + \left(\frac{\partial f}{\partial y}\right)_x dy
\end{equation}

Now take the wedge product of both sides with dx (with the first term on the right cancelling)

\begin{equation}
df \wedge dx = \left(\frac{\partial f}{\partial y}\right)_x dy \wedge dx
\end{equation}

rearranging to get the ratio of wedge products

\begin{equation}
\frac{df \wedge dx}{dy \wedge dx} = \left(\frac{\partial f}{\partial y}\right)_x
\label{surform}
\end{equation}

This rule needs a name. I tried searching for it but couldn't find anything. Surface form? Surface derivative? Whatever time for the real show to start. Let's begin with the first law of thermodynamics. Then we'll take the exterior derivative of both sides.

\begin{equation}
dU = T dS - p dV
\end{equation}

\begin{equation}
0 = d^2 U = dT \wedge dS - dp \wedge dV
\end{equation}

\begin{equation}
dT \wedge dS = dp \wedge dV
\end{equation}

Divide both sides by the quantity $ dS \wedge dV $ and substitute for the correct partial derivatives using \eqref{surform} to get

\begin{equation}
-\left(\frac{\partial T}{\partial V}\right)_S = \left(\frac{\partial p}{\partial S}\right)_V
\end{equation}

where the minus sign came from flipping the wedge product. To get the other three relations we need only divide by one differential from each side $ dS \wedge dp $, $ dT \wedge dV $, and $ dT \wedge dp $. We easily get the four Maxwell relations for $ (p, V, T, S) $. To get any others just start with the first law with any new variables in it (like $ (\mu,N) $ or $ (M,B) $).

The reciprocity theorem is also easy to see

\begin{equation}
\frac{dy \wedge dx}{dz \wedge dx} = \frac{1}{\frac{dz \wedge dx}{dy \wedge dx}} \implies
\left(\frac{\partial y}{\partial z}\right)_x = \frac{1}{\left(\frac{\partial z}{\partial y}\right)_x}
\end{equation}

as well as the cyclic rule

\begin{equation}
\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1
\end{equation}

since we have to switch three of the wedge products to get the numerators and denominators to cancel - picking up three negative signs along the way. There are more applications of exterior calculus to thermodynamics, but I have to learn them first before I can write about them :P