From 96447d884e02012a4ed9146dc6c00d186a201038 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Jul 2024 17:52:57 +0200
Subject: Improve knowledge base

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+---
+title: "Triple product rule"
+sort_title: "Triple product rule"
+date: 2024-07-21
+categories:
+- Mathematics
+- Thermodynamics
+layout: "concept"
+---
+
+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.
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