## Maxwell’s Relations (Part 2)

The Maxwell relations are some very general identities about the partial derivatives of functions of several variables. You don’t need to understand anything about thermodyamics to understand them, but they’re used a lot in that subject, so discussions of them tend to use notation from that subject.

Last time I went through a standard way to derive these relations for a function of two variables. Now I want to give a better derivation, which I found here:

• David J. Ritchie, A simple method for deriving Maxwell’s relations, American Journal of Physics 36 (1958), 760–760.

This paper is just one page long, and I can’t really improve on it, but I’ll work through the ideas in more detail. It again covers only the case of a function of two variables, and I won’t try to go beyond that case now—maybe later.

So, remember the setup. We have a smooth function on the plane $U \colon \mathbb{R}^2 \to \mathbb{R}$

We call the coordinates on the plane $S$ and $V,$ and we give the partial derivatives of $U$ funny names: $\displaystyle{ T = \left.\frac{\partial U}{\partial S} \right|_V \qquad P = - \left. \frac{\partial U}{\partial V} \right|_S }$

None of these funny names or the minus sign has any effect on the actual math involved; they’re just traditional in thermodynamics. So, mathematicians, please forgive me! If I ever generalize to the n-variable case, I’ll switch to more reasonable notation.

We instantly get this: $dU = T dS - P dV$

and since $d^2 = 0$ we get $0 = d^2 U = dT \wedge dS - dP \wedge dV$

so $dT \wedge dS = dP \wedge dV$

Believe it or not, this simple relation contains all four of Maxwell’s relations within it!

To see this, note that both sides are smooth 2-forms on the plane. Now, the space of 2-forms at any one point of the plane is a 1-dimensional vector space. So, we can divide any 2-form at a point by any nonzero 2-form at that point and get a real number.

In particular, suppose $X$ and $Y$ are functions on the plane such that $dX \wedge dY \ne 0$ at some point. Then we can divide both sides of the above equation by $dX \wedge dY$ and get $\displaystyle{ \frac{dT \wedge dS}{dX \wedge dY} \; = \; \frac{dP \wedge dV}{dX \wedge dY} }$

at this point. We can now get the four Maxwell relations simply by making different choices of $X$ and $Y.$ We’ll choose them to be either $T,S,V$ or $P.$ The argument will only work if $dX \wedge dY \ne 0,$ so I’ll always assume that. The argument works the same way each time so I’ll go faster after the first time.

### The first relation

Take $X = V$ and $Y = S$ and substitute them into the above equation. We get $\displaystyle{ \frac{dT \wedge dS}{dV \wedge dS} \; = \; \frac{dP \wedge dV}{dV \wedge dS} }$

or $\displaystyle{ \frac{dT \wedge dS}{dV \wedge dS} \; = \; - \frac{dP \wedge dV}{dS \wedge dV} }$

Next we use a little fact about differential forms and partial derivatives to simplify both sides: $\displaystyle{ \frac{dT \wedge dS}{dV \wedge dS} \; = \; \left.\frac{\partial T}{\partial V} \right|_S }$

and similarly $\displaystyle{ \frac{dP \wedge dV}{dS \wedge dV} \; = \; \left. \frac{\partial P}{\partial S}\right|_V }$

If you were scarred as a youth when plausible-looking manipulations with partial derivatives turned out to be unjustified, you might be worried about this—and rightly so! Later I’ll show how to justify the kind of ‘cancellation’ we’re doing here. But anyway, it instantly gives us the first Maxwell relation: $\boxed{ \displaystyle{ \left. \frac{\partial T}{\partial V} \right|_S \; = \; - \left. \frac{\partial P}{\partial S} \right|_V } }$

### The second relation

This time we take $X = T, Y = V.$ Substituting this into our general formula $\displaystyle{ \frac{dT \wedge dS}{dX \wedge dY} \; = \; \frac{dP \wedge dV}{dX \wedge dY} }$

we get $\displaystyle{ \frac{dT \wedge dS}{dT \wedge dV} \; = \; \frac{dP \wedge dV}{dT \wedge dV} }$

and doing the same sort of ‘cancellation’ as last time, this gives the second Maxwell relation: $\boxed{ \displaystyle{ \left. \frac{\partial S}{\partial V} \right|_T \; = \; \left. \frac{\partial P}{\partial T} \right|_V } }$

### The third relation

This time we take $X = P, Y = S.$ Substituting this into our general formula $\displaystyle{ \frac{dT \wedge dS}{dX \wedge dY} \; = \; \frac{dP \wedge dV}{dX \wedge dY} }$

we get $\displaystyle{ \frac{dT \wedge dS}{dP \wedge dS} \; = \; \frac{dP \wedge dV}{dP \wedge dS} }$

which gives the third Maxwell relation: $\boxed{ \displaystyle{ \left. \frac{\partial T}{\partial P} \right|_S \; = \; \left. \frac{\partial V}{\partial S} \right|_P }}$

### The fourth relation

This time we take $X = P, Y = T.$ Substituting this into our general formula $\displaystyle{ \frac{dT \wedge dS}{dX \wedge dY} \; = \; \frac{dP \wedge dV}{dX \wedge dY} }$

we get $\displaystyle{ \frac{dT \wedge dS}{dP \wedge dT} \; = \; \frac{dP \wedge dV}{dP \wedge dT} }$

or $\displaystyle{ -\frac{dS \wedge dT}{dP \wedge dT} \; = \; \frac{dP \wedge dV}{dP \wedge dT} }$

giving the fourth Maxwell relation: $\boxed{ \displaystyle{ \left. \frac{\partial V}{\partial T} \right|_P \; = \; - \left. \frac{\partial S}{\partial P} \right|_T }}$

You can check that other choices of $X$ and $Y$ don’t give additional relations of the same form.

### Determinants

So, we’ve see that all four Maxwell relations follow quickly from the equation $dT \wedge dS = dP \wedge dV$

if we can do ‘cancellations’ in expressions like this: $\displaystyle{ \frac{dA \wedge dB}{dX \wedge dY} }$

when one of the functions $A,B \colon \mathbb{R}^2 \to \mathbb{R}$ equals one of the functions $X,Y \colon \mathbb{R}^2 \to \mathbb{R}.$ This works whenever $dX \wedge dY \ne 0.$ Let’s see why!

First of all, by the inverse function theorem, if $dX \wedge dY \ne 0$ at some point in the plane, the functions $X$ and $Y$ serve as coordinates in some neighborhood of that point. In this case we have $\displaystyle{ \frac{dA \wedge dB}{dX \wedge dY} = \det \left( \begin{array}{cc} \displaystyle{ \frac{\partial A}{\partial X} } & \displaystyle{ \frac{\partial A}{\partial Y} } \\ \\ \displaystyle{ \frac{\partial B}{\partial X} } & \displaystyle{\frac{\partial B}{\partial Y} } \end{array} \right) }$

Yes: the ratio of 2-forms is just the Jacobian of the map sending $(X,Y)$ to $(A,B).$ This is clear if you know that 2-forms are ‘area elements’ and the Jacobian is a ratio of area elements. But you can also prove it by a quick calculation: $\displaystyle{ dA = \frac{\partial A}{\partial X} dX + \frac{\partial A}{\partial Y} dY }$ $\displaystyle{ dB = \frac{\partial B}{\partial X} dX + \frac{\partial B}{\partial Y} dY }$

and thus $\displaystyle{ dA \wedge dB = \left(\frac{\partial A}{\partial X} \frac{\partial B}{\partial Y} - \frac{\partial A}{\partial Y} \frac{\partial B}{\partial X} \right) dX \wedge dY}$

so the ratio $dA \wedge dB/dX \wedge dY$ is the desired determinant.

How does this help? Well, take $\displaystyle{ \frac{dA \wedge dB}{dX \wedge dY} }$

and now suppose that either $A$ or $B$ equals either $X$ or $Y.$ For example, suppose $A = X.$ Then we can do a ‘cancellation’ like this: $\begin{array}{ccl} \displaystyle{ \frac{dX \wedge dB}{dX \wedge dY} } &=& \det \left( \begin{array}{cc} \displaystyle{ \frac{\partial X}{\partial X} } & \displaystyle{ \frac{\partial X}{\partial Y} } \\ \\ \displaystyle{ \frac{\partial B}{\partial X} } & \displaystyle{\frac{\partial B}{\partial Y} } \end{array} \right) \\ \\ &=& \det \left( \begin{array}{cc} \displaystyle{ 1 } & \displaystyle{ 0 } \\ \\ \displaystyle{ \frac{\partial B}{\partial X} } & \displaystyle{\frac{\partial B}{\partial Y} } \end{array} \right) \\ \\ &=& \displaystyle{\frac{\partial B}{\partial Y} } \end{array}$

or to make it clear that the partial derivatives are being done in the $X,Y$ coordinate system: $\displaystyle{ \frac{dX \wedge dB}{dX \wedge dY} = \left.\frac{\partial B}{\partial Y} \right|_X }$

This justifies all our calculations earlier.

### Conclusions

So, we’ve seen that all four Maxwell relations are unified in a much simpler equation: $dT \wedge dS = dP \wedge dV$

which follows instantly from $dU = T dS - P dV$

This is a big step forward compared to the proof I gave last time, which, at least as I presented it, required cleverly guessing a bunch of auxiliary functions—even though these auxiliary functions turn out to be incredibly important in their own right.

So, we should not stop here: we should think hard about the physical and mathematical meaning of the equation $dT \wedge dS = dP \wedge dV$

And Ritchie does this in his paper! But I will talk about that next time.

Part 1: a proof of Maxwell’s relations using commuting partial derivatives.

Part 2: a proof of Maxwell’s relations using 2-forms.

Part 3: the physical meaning of Maxwell’s relations, and their formulation in terms of symplectic geometry.

For how Maxwell’s relations are connected to Hamilton’s equations, see this post:

### 6 Responses to Maxwell’s Relations (Part 2)

1. Toby Bartels says:

If you were scarred as a youth when plausible-looking manipulations with partial derivatives turned out to be unjustified, you might be worried about this

Ironically, ‘this’ ―writing partial derivatives as ratios of top-rank exterior forms― is the safe way to do only justified manipulations. The curly ∂ is a warning that something funny is going on and you should not trust the notation. (Unfortunately, you can’t always trust a straight d either.)

2. scentoni says:

Of course, not all situations involve simple hydrostatic pressure; some require a full stress tensor and instead of $-p dV$ one has something like $\sigma_{i j} \eta_{i j}$ (where hydrostatic pressure is the special case $\sigma_{i j} =-p \delta_{i j}$). Notation tends to vary between linear elasticity theory, fluid dynamics, and other fields where the stress and deformation tensors appear.

3. Joseph Rizcallah says:

We can take as independent any pair of variables: $(S,T); (P,V); (S,P); (T,P); (S,V); (T, V)$. The first two choices do not mix the variables on the two sides of the equation $dT \wedge dS = dP \wedge dV$ and simply yield $\left| \begin{array}{cc} \frac{\partial P}{\partial S} & \frac{\partial P}{\partial T} \\ \frac{\partial V}{\partial S} & \frac{\partial V}{\partial T} \end{array} \right|=1$. Geometrically this means the area elements in the PV and TS planes are equal. Suppose, for example, we choose $(S,P)$ as the independent pair of variables. Then $dV = \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial P}dP$ and $dT = \frac{\partial T}{\partial S}dS + \frac{\partial T}{\partial P}dP$. Substituting in $dT \wedge dS = dP \wedge dV$ we get $\frac{\partial T}{\partial P}dP \wedge dS = \frac{\partial V}{\partial S}dP \wedge dS$, whence $\frac{\partial T}{\partial P}=\frac{\partial V}{\partial S}$, which is the fourth relation. The other three relations can be derived in an analogous way using the other three pairs.

• Toby Bartels says:

Geometrically this means the area elements in the PV and TS planes are equal.

The paper by Ritchie that John cited, stresses the importance of this, giving it this physical interpretation: that the work done by the system is equal to the heat absorbed by the system, when undergoing any cyclic process.

• John Baez says:

Yes! I was going to talk about the physical and mathematical interpretation next time. This is a very nice physical interpretation.

4. Maxwell Relations are commonly known as a set of four partial differential equations between four thermodynamic quantities or potentials: pressure (P), volume (V), temperature (T), and entropy (S).

[….]

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