---

Categories: Mathematics, Thermodynamics.

Triple product rule

Suppose we have a function $f(x, y, z)$, whose stationary points we want to find. This is simple: we take the differential $\dd{f}$ and set it to zero:

$$\begin{aligned} 0 = \dd{f} &= \bigg( \pdv{f}{x} \bigg)_{y, z} \dd{x} + \bigg( \pdv{f}{y} \bigg)_{x, z} \dd{y} + \bigg( \pdv{f}{z} \bigg)_{x, y} \dd{z} \end{aligned}$$

But what if we have a constraint of the form $f(x, y, z) = C$, for some constant $C$? In that case, $f$ must be stationary everywhere, so the above still holds, but the coordinates $(x, y, z)$ are no longer independent: there exists an implicit relation $z(x, y)$ to satisfy the constraint.

Then $z$ can be regarded as a height function, in which case we can vary $(x, y)$ such that $z$ stays constant, i.e. it is possible to choose $\dd{x}$ and $\dd{y}$ such that $\dd{z} = 0$, leaving:

$$\begin{aligned} 0 &= \bigg( \pdv{f}{x} \bigg)_{y, z} \dd{x} + \bigg( \pdv{f}{y} \bigg)_{x, z} \dd{y} \end{aligned}$$

We divide this by $\dd{y}$. Note the subscript $(f, z)$, which says those variables are kept constant for that derivatives, to indicate that $x$ and $y$ are not independent:

$$\begin{aligned} 0 &= \bigg( \pdv{f}{x} \bigg)_{y, z} \bigg( \pdv{x}{y} \bigg)_{f, z} + \bigg( \pdv{f}{y} \bigg)_{x, z} \end{aligned}$$

Rearranging this gives a form of the triple product rule heavily used in thermodynamics:

$$\begin{aligned} \boxed{ \bigg( \pdv{x}{y} \bigg)_{f, z} = - \frac{ \bigg( \displaystyle\pdv{f}{y} \bigg)_{x, z} }{ \bigg( \displaystyle\pdv{f}{x} \bigg)_{y, z} } } \end{aligned}$$

If we had divided by $\dd{x}$ instead of $\dd{y}$, we would have arrived at an equivalent result:

$$\begin{aligned} \bigg( \pdv{y}{x} \bigg)_{f, z} = - \frac{ \bigg( \displaystyle\pdv{f}{x} \bigg)_{y, z} }{ \bigg( \displaystyle\pdv{f}{y} \bigg)_{x, z} } \end{aligned}$$

Comparing the two previous relations, we see that $\ipdv{y}{x}$ is simply one over $\ipdv{x}{y}$, just like in an unconstrained problem:

$$\begin{aligned} \bigg( \pdv{y}{x} \bigg)_{f, z} = \bigg( \displaystyle\pdv{x}{y} \bigg)_{f, z}^{-1} \end{aligned}$$

You may think this is obvious, but it was worth checking that it holds here too. Applying this to either of our earlier relations yields the standard form of the triple product rule:

$$\begin{aligned} \boxed{ -1 = \bigg( \pdv{x}{y} \bigg)_{f, z} \bigg( \pdv{f}{x} \bigg)_{y, z} \bigg( \displaystyle\pdv{y}{f} \bigg)_{x, z} } \end{aligned}$$

Many authors write this relation with $f(x, y, z) = z(x, y)$, in which case it becomes:

$$\begin{aligned} -1 = \bigg( \pdv{x}{y} \bigg)_{z} \bigg( \pdv{z}{x} \bigg)_{y} \bigg( \displaystyle\pdv{y}{z} \bigg)_{x} \end{aligned}$$

References

1. H.B. Callen, Thermodynamics and an introduction to thermostatistics, 2nd edition, Wiley.

---