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

---
 source/know/concept/canonical-ensemble/index.md    |   2 +-
 source/know/concept/euler-equations/index.md       |   2 +-
 .../index.md                                       | 326 +++++++++++++++++++++
 .../fundamental-thermodynamic-relation/index.md    |  54 ----
 .../know/concept/laws-of-thermodynamics/index.md   | 104 -------
 .../know/concept/thermodynamic-potential/index.md  |   2 +-
 source/know/concept/triple-product-rule/index.md   |  97 ++++++
 7 files changed, 426 insertions(+), 161 deletions(-)
 create mode 100644 source/know/concept/fundamental-relation-of-thermodynamics/index.md
 delete mode 100644 source/know/concept/fundamental-thermodynamic-relation/index.md
 delete mode 100644 source/know/concept/laws-of-thermodynamics/index.md
 create mode 100644 source/know/concept/triple-product-rule/index.md

diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md
index 8a96e91..da7d436 100644
--- a/source/know/concept/canonical-ensemble/index.md
+++ b/source/know/concept/canonical-ensemble/index.md
@@ -178,7 +178,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Rearranging and substituting
-the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+the [fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/)
 then gives:
 
 $$\begin{aligned}
diff --git a/source/know/concept/euler-equations/index.md b/source/know/concept/euler-equations/index.md
index 2654d2b..415e2f1 100644
--- a/source/know/concept/euler-equations/index.md
+++ b/source/know/concept/euler-equations/index.md
@@ -146,7 +146,7 @@ When the fluid gets compressed in a certain location, thermodynamics
 states that the pressure, temperature and/or entropy must increase there.
 For simplicity, let us assume an *isothermal* and *isentropic* fluid,
 such that only $$p$$ is affected by compression, and the
-[fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+[fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/)
 reduces to $$\dd{E} = - p \dd{V}$$.
 
 Then the pressure is given by a thermodynamic equation of state $$p(\rho, T)$$,
diff --git a/source/know/concept/fundamental-relation-of-thermodynamics/index.md b/source/know/concept/fundamental-relation-of-thermodynamics/index.md
new file mode 100644
index 0000000..a51c231
--- /dev/null
+++ b/source/know/concept/fundamental-relation-of-thermodynamics/index.md
@@ -0,0 +1,326 @@
+---
+title: "Fundamental relation of thermodynamics"
+sort_title: "Fundamental relation of thermodynamics"
+date: 2024-07-21 # Originally 2021-07-07, major rewrite
+categories:
+- Physics
+- Thermodynamics
+layout: "concept"
+---
+
+In most areas of physics,
+we observe and analyze the behaviour
+of physical systems that have been "disturbed" some way,
+i.e. we try to understand what is *happening*.
+In thermodynamics, however,
+we start paying attention once the disturbance has ended,
+and the system has had some time to settle down:
+when nothing seems to be happening anymore.
+
+Then a common observation is that the system "forgets" what happened earlier,
+and settles into a so-called **equilibrium state**
+that appears to be independent of its history.
+No matter in what way you stir your tea, once you finish,
+eventually the liquid stops moving, cools down,
+and just... sits there, doing nothing.
+But how does it "choose" this equilibrium state?
+
+
+
+## Thermodynamic equilibrium
+
+This history-independence suggests that equilibrium
+is determined by only a few parameters of the system.
+Prime candidates are the **mole numbers** $$N_1, N_2, ..., N_n$$
+of each of the $$n$$ different types of particles in the system,
+and its **volume** $$V$$.
+Furthermore, the microscopic dynamics
+are driven by energy differences between components,
+and obey the universal principle of energy conservation,
+so it also sounds reasonable to define a total
+**internal energy** $$U$$.
+
+Thanks to many decades of empirical confirmations,
+we now know that the above arguments can be combined into a postulate:
+the equilibrium state of a closed system with fixed $$U$$, $$V$$ and $$N_i$$
+is completely determined by those parameters.
+The system then "finds" the equilibrium
+by varying its microscopic degrees of freedom
+such that the **entropy** $$S$$ is maximized
+subject to the given values of $$U$$, $$V$$ and $$N_i$$.
+This statement serves as a definition of $$S$$,
+and explains the **second law of thermodynamics**:
+the total entropy never decreases.
+
+We do not care about those microscopic degrees of freedom,
+but we do care about how $$U$$, $$V$$ and $$N_i$$ influence the equilibrium.
+For a given system, we want a formula $$S(U, V, N_1, ..., N_n)$$,
+which contains all thermodynamic information about the system
+and is therefore known as its **fundamental relation**.
+
+The next part of our definition of $$S$$
+is that it must be invertible with respect to $$U$$,
+meaning we can rearrange the fundamental relation
+to $$U(S, V, N_1, ... N_n)$$ without losing any information.
+Specifically, this means that $$S$$ must be continuous,
+differentiable, and monotonically increasing with $$U$$,
+such that $$S(U)$$ can be inverted to $$U(S)$$ and vice versa.
+
+The idea here is that maximizing $$S$$ at fixed $$U$$
+should be equivalent to minimizing $$U$$ for a given $$S$$
+(we prove this later).
+Often it is mathematically more convenient
+to choose one over the other,
+but by definition both approaches are equally valid.
+And because $$S$$ is rather abstract,
+it may be preferable to treat it as a parameter
+for a more intuitive quantity like $$U$$.
+
+Next, we demand that $$S$$ is additive over subsystems,
+so $$S = S_1 + S_2 + ...$$, with $$S_1$$ being the entropy of subsystem 1, etc.
+Consequently, $$S$$ is an **extensive** quantity of the system,
+just like $$U$$ (and $$V$$ and $$N_i$$),
+meaning they satisfy for any constant $$\lambda$$:
+
+$$\begin{aligned}
+    S(\lambda U, \lambda V, \lambda N_1, ..., \lambda N_n)
+    &= \lambda S(U, V, N_1, ..., N_n)
+    \\
+    U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n)
+    &= \lambda U(S, V, N_1, ..., N_n)
+\end{aligned}$$
+
+For $$U$$, this makes intuitive sense:
+the total energy in two identical systems
+is double the energy of a single of those systems.
+Actually, reality is a bit hazier than this:
+dynamics are governed by energy *differences* only,
+so an offset $$U_0$$ can be added without a consequence.
+We should choose an offset and a way to split the system into subsystems
+such that the above relation holds for our convenience.
+Fortunately, this choice often makes itself.
+
+$$S$$ does not suffer from this ambiguity,
+since the **third law of thermodynamics** clearly defines
+where $$S = 0$$ should occur: at a temperature of absolute zero.
+In this article we will not explore the reason for this requirement,
+which is also known as the **Nernst postulate**.
+Furthermore, in most situations this law can simply be ignored.
+
+Since $$U$$, $$S$$, $$V$$ and $$N_i$$ are all extensive,
+the partial derivatives of the fundamental relation are **intensive** quantities,
+meaning they do not depend on the size of the system.
+Those derivatives are very important,
+since they are usually the equilibrium properties we want to find.
+
+
+
+## Energy representation
+
+When we have a fundamental relation of the form $$U(S, V, N_1, ..., N_n)$$,
+we say we are treating the system's thermodynamics
+in the **energy representation**.
+
+The following derivatives of $$U$$ are used as the thermodynamic *definitions*
+of the **temperature** $$T$$, the **pressure** $$P$$,
+and the **chemical potential** $$\mu_k$$ of the $$k$$th particle species:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            T
+            &\equiv \bigg( \pdv{U}{S} \bigg)_{V, N_i}
+            \\
+            P
+            &\equiv - \bigg( \pdv{U}{V} \bigg)_{S, N_i}
+            \\
+            \mu_k
+            &\equiv \bigg( \pdv{U}{N_k} \bigg)_{S, V, N_{i \neq k}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The resulting expressions of the form $$T(S, V, N_1, ..., N_n)$$ etc.
+are known as the **equations of state** of the system.
+Unlike the fundamental relation, a single equation of state
+is not a complete thermodynamic description of the system.
+However, if *all* equations of state are known
+(for $$T$$, $$P$$, and all $$\mu_k$$),
+then the fundamental relation can be reconstructed.
+
+As explained above, physical dynamics are driven by energy differences only,
+so we expand an infinitesimal difference $$\dd{U}$$ as:
+
+$$\begin{aligned}
+    \dd{U}
+    = \bigg( \pdv{U}{S} \bigg)_{V, N_i} \!\dd{S}
+    \:\:+\:\: \bigg( \pdv{U}{V} \bigg)_{S, N_i} \!\dd{V}
+    \:\:+\:\: \sum_{k}^{} \bigg( \pdv{U}{N_k} \bigg)_{S, V, N_{i \neq k}} \!\dd{N_k}
+\end{aligned}$$
+
+Those partial derivatives look familiar.
+Substituting $$T$$, $$P$$ and $$\mu_k$$ gives a result
+that is also called the **fundamental relation of thermodynamics**
+(as opposed to the fundamental relation of the system only,
+just to make things confusing):
+
+$$\begin{aligned}
+    \boxed{
+        \dd{U}
+        = T \dd{S} - P \dd{V} + \sum_{k}^{} \mu_k \dd{N_k}
+    }
+\end{aligned}$$
+
+Where the first term represents heating/cooling (also written as $$\dd{Q}$$),
+and the second is physical work done on the system
+by compression/expansion (also written as $$\dd{W}$$).
+The third term is the energy change due to matter transfer and is often neglected.
+Hence this relation can be treated as a form
+of the **first law of thermodynamics** $$\Delta U = \Delta Q + \Delta W$$.
+
+Because $$T$$, $$P$$ and $$\mu_k$$ generally depend on $$S$$, $$V$$ and $$N_k$$,
+integrating the fundamental relation can be tricky.
+Fortunately, the fact that $$U$$ is extensive offers a shortcut.
+Recall that:
+
+$$\begin{aligned}
+    \lambda U(S, V, N_1, ..., N_n)
+    &= U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n)
+\end{aligned}$$
+
+For any $$\lambda$$.
+Let us differentiate this equation with respect to $$\lambda$$, yielding:
+
+$$\begin{aligned}
+    U
+    &= \pdv{}{\lambda} U(\lambda S, \lambda V, \lambda N_1, ..., \lambda N_n)
+    \\
+    &= \pdv{U(\lambda S)}{(\lambda S)} \pdv{(\lambda S)}{\lambda}
+    + \pdv{U(\lambda V)}{(\lambda V)} \pdv{(\lambda V)}{\lambda}
+    + \sum_{k} \pdv{U(\lambda N_k)}{(\lambda N_k)} \pdv{(\lambda N_k)}{\lambda}
+    \\
+    &= \pdv{U(S)}{S} S
+    + \pdv{U(V)}{V} V
+    + \sum_{k} \pdv{U(N_k)}{N_k} N_k
+\end{aligned}$$
+
+Where we once again recognize the derivatives.
+The resulting equation is known as the **Euler form**
+of the fundamental relation of thermodynamics:
+
+$$\begin{aligned}
+    \boxed{
+        U
+        = T S - P V + \sum_{k} \mu_k N_k
+    }
+\end{aligned}$$
+
+Plus a constant $$U_0$$ of course,
+although $$U_0 = 0$$ is the most straightforward choice.
+
+
+
+## Entropy representation
+
+If the system's fundamental relation
+instead has the form $$S(U, V, N_1, ..., N_i)$$,
+we are treating it in the **entropy representation**.
+Isolating the above fundamental relation of thermodynamics
+for $$\dd{S}$$ yields its equivalent form in this representation:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{S}
+        = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \sum_{k}^{} \frac{\mu_k}{T} \dd{N_k}
+    }
+\end{aligned}$$
+
+From which we can then read off the standard partial derivatives of $$S(U, V, N_1, ..., N_n)$$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \frac{1}{T}
+            &= \bigg( \pdv{S}{U} \bigg)_{V, N_i}
+            \\
+            \frac{P}{T}
+            &= \bigg( \pdv{S}{V} \bigg)_{U, N_i}
+            \\
+            \frac{\mu_k}{T}
+            &= - \bigg( \pdv{S}{N_k} \bigg)_{U, V, N_{i \neq k}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Note the signs: the parameters $$U$$, $$V$$ and $$N_i$$ are implicitly related
+by our requirement that $$S$$ is stationary at a maximum,
+so the [triple product rule](/know/concept/triple-product-rule/)
+must be used, which brings some perhaps surprising sign changes.
+Reading them off in this way is easier.
+
+And of course, since $$S$$ is defined to be an extensive quantity,
+it also has an Euler form:
+
+$$\begin{aligned}
+    \boxed{
+        S
+        = \frac{1}{T} U + \frac{P}{T} V - \sum_{k} \frac{\mu_k}{T} N_k
+    }
+\end{aligned}$$
+
+Finally, it is worth proving that minimizing $$U$$
+is indeed equivalent to maximizing $$S$$.
+For simplicity, we consider a system
+where only the volume $$V$$ can change
+in order to reach an equilibrium;
+the proof is analogous for all other parameters.
+Clearly, $$S$$ is stationary at its maximum:
+
+$$\begin{aligned}
+    0
+    &= \bigg( \pdv{S}{V} \bigg)_{U, N_i}
+    = - \frac{ \bigg( \displaystyle\pdv{U}{V} \bigg)_{S, N_i} }{ \bigg( \displaystyle\pdv{U}{S} \bigg)_{V, N_i} }
+    = - \frac{1}{T} \bigg( \pdv{U}{V} \bigg)_{S, N_i}
+\end{aligned}$$
+
+Where we have used the triple product rule.
+This can only hold if $$(\ipdv{U}{S})_{S, N_i} = 0$$,
+meaning $$U$$ is also at an extremum.
+But $$S$$ is not just at any extremum: it is at a *maximum*, so:
+
+$$\begin{aligned}
+    0
+    > \bigg( \pdvn{2}{S}{V} \bigg)_{U, N_i}
+    &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{U, N_i}
+    \\
+    &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{S, N_i}
+    + \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i} \bigg( \pdv{S}{V} \bigg)_{U, N_i}
+    \\
+    &= \bigg( \pdv{}{V} \Big( \frac{P}{T} \Big) \bigg)_{S, N_i}
+    + \frac{P}{T} \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i}
+    \\
+    &= \frac{1}{T} \bigg( \pdv{P}{V} \bigg)_{S, N_i}
+    - \frac{P}{T^2} \bigg( \pdv{T}{V} \bigg)_{S, N_i}
+    + \frac{P}{T} \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i}
+    \\
+    &= - \frac{1}{T} \bigg( \pdvn{2}{U}{V} \bigg)_{S, N_i}
+    + \frac{P}{T} \bigg[ \bigg( \pdv{}{S} \Big( \frac{P}{T} \Big) \bigg)_{V, N_i}
+    - \frac{1}{T} \bigg( \pdv{T}{V} \bigg)_{S, N_i} \bigg]
+\end{aligned}$$
+
+Because $$S$$ is at a maximum, we know that $$P/T = 0$$,
+and $$T$$ is always above absolute zero
+(since we defined $$S$$ to be monotonically increasing with $$U$$),
+which leaves $$(\ipdvn{2}{U}{V})_{S, N_i} > 0$$
+as the only way to satisfy this inequality.
+In other words, $$U$$ is at a minimum, as expected.
+
+
+
+## References
+1.  H.B. Callen,
+    *Thermodynamics and an introduction to thermostatistics*, 2nd edition,
+    Wiley.
+2.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/fundamental-thermodynamic-relation/index.md b/source/know/concept/fundamental-thermodynamic-relation/index.md
deleted file mode 100644
index 0d945fa..0000000
--- a/source/know/concept/fundamental-thermodynamic-relation/index.md
+++ /dev/null
@@ -1,54 +0,0 @@
----
-title: "Fundamental thermodynamic relation"
-sort_title: "Fundamental thermodynamic relation"
-date: 2021-07-07
-categories:
-- Physics
-- Thermodynamics
-layout: "concept"
----
-
-The **fundamental thermodynamic relation** combines the first two
-[laws of thermodynamics](/know/concept/laws-of-thermodynamics/),
-and gives the change of the internal energy $$U$$,
-which is a [thermodynamic potential](/know/concept/thermodynamic-potential/),
-in terms of the change in
-entropy $$S$$, volume $$V$$, and the number of particles $$N$$.
-
-Starting from the first law of thermodynamics,
-we write an infinitesimal change in energy $$\dd{U}$$ as follows,
-where $$T$$ is the temperature and $$P$$ is the pressure:
-
-$$\begin{aligned}
-    \dd{U} &= \dd{Q} + \dd{W} = T \dd{S} - P \dd{V}
-\end{aligned}$$
-
-The term $$T \dd{S}$$ comes from the second law of thermodynamics,
-and represents the transfer of thermal energy,
-while $$P \dd{V}$$ represents physical work.
-
-However, we are missing a term, namely matter transfer.
-If particles can enter/leave the system (i.e. the population $$N$$ is variable),
-then each such particle costs an amount $$\mu$$ of energy,
-where $$\mu$$ is known as the **chemical potential**:
-
-$$\begin{aligned}
-    \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
-\end{aligned}$$
-
-To generalize even further, there may be multiple species of particle,
-which each have a chemical potential $$\mu_i$$.
-In that case, we sum over all species $$i$$:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{U} = T \dd{S} - P \dd{V} + \sum_{i}^{} \mu_i \dd{N_i}
-    }
-\end{aligned}$$
-
-
-
-## References
-1.  H. Gould, J. Tobochnik,
-    *Statistical and thermal physics*, 2nd edition,
-    Princeton.
diff --git a/source/know/concept/laws-of-thermodynamics/index.md b/source/know/concept/laws-of-thermodynamics/index.md
deleted file mode 100644
index 3605a0e..0000000
--- a/source/know/concept/laws-of-thermodynamics/index.md
+++ /dev/null
@@ -1,104 +0,0 @@
----
-title: "Laws of thermodynamics"
-sort_title: "Laws of thermodynamics"
-date: 2021-07-07
-categories:
-- Physics
-- Thermodynamics
-layout: "concept"
----
-
-The **laws of thermodynamics** are of great importance
-to physics, chemistry and engineering,
-since they restrict what a device or process can physically achieve.
-For example, the impossibility of *perpetual motion*
-is a consequence of these laws.
-
-
-## First law
-
-The **first law of thermodynamics** states that energy is conserved.
-When a system goes from one equilibrium to another,
-the change $$\Delta U$$ of its energy $$U$$ is equal to
-the work $$\Delta W$$ done by external forces,
-plus the energy transferred by heating ($$\Delta Q > 0$$) or cooling ($$\Delta Q < 0$$):
-
-$$\begin{aligned}
-    \boxed{
-        \Delta U = \Delta W + \Delta Q
-    }
-\end{aligned}$$
-
-The internal energy $$U$$ is a state variable,
-so is independent of the path taken between equilibria.
-However, the work $$\Delta W$$ and heating $$\Delta Q$$ do depend on the path,
-so the first law means that
-the act of transferring energy is path-dependent,
-but the result has no "memory" of that path.
-
-
-## Second law
-
-The **second law of thermodynamics** states that
-the total entropy never decreases.
-An important consequence is that
-no machine can convert energy into work with 100% efficiency.
-
-It is possible for the local entropy $$S_{\mathrm{loc}}$$
-of a system to decrease, but doing so requires work,
-and therefore the entropy of the surroundings $$S_{\mathrm{sur}}$$
-must increase accordingly, such that:
-
-$$\begin{aligned}
-    \boxed{
-        \Delta S_{\mathrm{tot}} = \Delta S_{\mathrm{loc}} + \Delta S_{\mathrm{sur}} \ge 0
-    }
-\end{aligned}$$
-
-Since the total entropy never decreases,
-the equilibrium state of a system must be a maximum
-of its entropy $$S$$, and therefore $$S$$ can be used as
-a [thermodynamic "potential"](/know/concept/thermodynamic-potential/).
-
-The only situation where $$\Delta S = 0$$ is a reversible process,
-since then it must be possible to return to
-the previous equilibrium state by doing the same work in the opposite direction.
-
-According to the first law,
-if a process is reversible, or if it is only heating/cooling,
-then (after one reversible cycle) the energy change
-is simply the heat transfer $$\dd{U} = \dd{Q}$$.
-An entropy change $$\dd{S}$$ is then expressed as follows
-(since $$\ipdv{S}{U} = 1 / T$$ by definition):
-
-$$\begin{aligned}
-    \boxed{
-        \dd{S}
-        = \Big( \pdv{S}{U} \Big)_{V, N} \dd{U}
-        = \frac{\dd{Q}}{T}
-    }
-\end{aligned}$$
-
-Confusingly, this equation is sometimes also called the second law of thermodynamics.
-
-
-## Third law
-
-The **third law of thermodynamics** states that
-the entropy $$S$$ of a system goes to zero when the temperature reaches absolute zero:
-
-$$\begin{aligned}
-    \boxed{
-        \lim_{T \to 0} S = 0
-    }
-\end{aligned}$$
-
-From this, the absolute quantity of $$S$$ is defined, otherwise we would
-only be able to speak of entropy differences $$\Delta S$$.
-
-
-
-## References
-1.  H. Gould, J. Tobochnik,
-    *Statistical and thermal physics*, 2nd edition,
-    Princeton.
diff --git a/source/know/concept/thermodynamic-potential/index.md b/source/know/concept/thermodynamic-potential/index.md
index b3bedda..b15c011 100644
--- a/source/know/concept/thermodynamic-potential/index.md
+++ b/source/know/concept/thermodynamic-potential/index.md
@@ -39,7 +39,7 @@ The **internal energy** $$U$$ represents
 the capacity to do both mechanical and non-mechanical work,
 and to release heat.
 It is simply the integral
-of the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/):
+of the [fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/):
 
 $$\begin{aligned}
     \boxed{
diff --git a/source/know/concept/triple-product-rule/index.md b/source/know/concept/triple-product-rule/index.md
new file mode 100644
index 0000000..16c5440
--- /dev/null
+++ b/source/know/concept/triple-product-rule/index.md
@@ -0,0 +1,97 @@
+---
+title: "Triple product rule"
+sort_title: "Triple product rule"
+date: 2024-07-21
+categories:
+- Mathematics
+- Thermodynamics
+layout: "concept"
+---
+
+Suppose we have a function $$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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