# Thermodynamics partial derivative brackets

Today I half-remembered a wonderful trick to doing thermodynamical partial derivatives, but I neither recall the full set of rules, nor where I got them from! I was hoping someone else knows where they come from or where I can find them written out.

The rules, as far as I remember them, is to define a bracket $[A,B]$ defined on thermodynamical potentials and variables such that:

$[X,X] = 1$,

$[X,Y] = -[Y,X]$,

If $dX = AdB + CdD$, then $[X,Y] = [B,Y][D,Y]$,

and most importantly,

$\left(\frac{\partial X}{\partial Y}\right)_Z = \frac{[X,Z]}{[Y,Z]}$.

As an example of this working (I'm dropping the commas), we have

$$\left(\frac{\partial P}{\partial V}\right)_S = \frac{[PS]}{[VS]}\\ \\ = \frac{[PS]}{[VS]}\cdot \frac{[TP]}{[TP]}\cdot \frac{[TV]}{[TV]} \cdot \frac{[PP]}{[VV]} = \frac{[TP]}{[TV]}\cdot \frac{[HP]}{[TP]}\cdot \frac{[TV]}{[UV]}\\ \\ = \frac{[TP]}{[TV]}\cdot \frac{[HP]}{[TP]} / \frac{[UV]}{[TV]}\\ \\ = \left(\frac{\partial P}{\partial V}\right)_T \cdot \left(\frac{\partial H}{\partial T}\right)_P / \left(\frac{\partial U}{\partial T}\right)_V\\ \\ = \left(\frac{\partial P}{\partial V}\right)_T \cdot \left(\frac{C_p}{C_V}\right)$$

The second equality involves multiplying by 1 several times. The second is based on the third rule.

Edit:

As a user pointed out, the first identity is trivially wrong, and should be

$[X,X] = 0$.

The role of Jacobians in thermodynamics is presented nicely by Jaynes. It also contains the rules of manipulation. The physical meaning of Jacobian $\frac{[X,Y]}{[U,V]}$ in which $X,Y,U,V,$ are thermodynamic variables is as follows: Take an infinitesimal closed oriented thermodynamic path. The same path can be drawn on either $X$-$Y$ thermodynamic diagram or on $U$-$V$ thermodynamic diagram. The Jacobian $\frac{[X,Y]}{[U,V]}$ gives the ratio of oriented areas enclosed by the same thermodynamic path on the $X$-$Y$ and $U$-$V$ thermodynamic diagrams. In other words: \begin{align} \int dX~dY=\int dU~dV\frac{[X,Y]}{[U,V]} \end{align}