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.


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

$[X,X] = 0$.

For more details, see the link in the accepted answer.

  • $\begingroup$ see the letters and comments in the February 2002 and April 2001 issues of the AJP by Ritchie, Mermin, Ambegaokar, Leff, Landsberg, etc., all about this version of Maxwell's relation using the Jacobi identity. $\endgroup$
    – hyportnex
    Sep 7, 2017 at 21:13
  • $\begingroup$ Quoting Leff: Ritchie concluded: ‘‘The (Maxwell) relations are simply the mathematical statement that in any cyclic process the work done (the area in the closed cycle in the pV plane) by the system must equal the heat absorbed (the area in the closed cycle in the TS plane).’’ $\endgroup$
    – hyportnex
    Sep 7, 2017 at 21:15
  • 3
    $\begingroup$ This is terribly interesting and you are correct to observe that you cannot possibly be remembering all of these rules correctly, as [X, X] = 1 and [X, Y] = -[Y, X] seem to be at odds with each other (clearly [X, X] must be equal to its own negative, which 1 does not seem to be). $\endgroup$
    – CR Drost
    Sep 7, 2017 at 21:46

1 Answer 1


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}

  • $\begingroup$ That's exactly the PDF I had in mind! And I had no idea that Jaynes was the author until now. Thanks! $\endgroup$
    – F. Bardamu
    Sep 8, 2017 at 19:22

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