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Classical Mechanics versus Thermodynamics (Part 3)

23 September, 2021

There’s a fascinating analogy between classical mechanics and thermodynamics, which I last talked about in 2012:

• Classical mechanics versus thermodynamics (part 1).
• Classical mechanics versus thermodynamics (part 2).

I’ve figured out more about it, and today I’m giving a talk about it in the physics colloquium at the University of British Columbia. It’s a colloquium talk that’s supposed to be accessible for upper-level undergraduates, so I’ll spend a lot of time reviewing the basics… which is good, I think.

I don’t know if the talk will be recorded, but you can see my slides here, and I’ll base this blog article on them.

Hamilton’s equations versus the Maxwell relations

Why do Hamilton’s equations in classical mechanics:

$\begin{array}{ccr} \displaystyle{ \frac{d p}{d t} } &=& \displaystyle{- \frac{\partial H}{\partial q} } \\ \\ \displaystyle{ \frac{d q}{d t} } &=& \displaystyle{ \frac{\partial H}{\partial p} } \end{array}$

look so much like the Maxwell relations in thermodynamics?

$\begin{array}{ccr} \displaystyle{ \left. \frac{\partial T}{\partial V}\right|_S } &=& \displaystyle{ - \left. \frac{\partial P}{\partial S}\right|_V } \\ \\ \displaystyle{ \left. \frac{\partial S}{\partial V}\right|_T } &=& \displaystyle{ \left. \frac{\partial P}{\partial T} \right|_V } \end{array}$

William Rowan Hamilton discovered his equations describing classical mechanics in terms of energy around 1827. By 1834 he had also introduced Hamilton’s principal function, which I’ll explain later.

James Clerk Maxwell is most famous for his equations describing electromagnetism, perfected in 1865. But he also worked on thermodynamics, and discovered the ‘Maxwell relations’ in 1871.

Hamilton’s equations describe how the position $q$ and momentum $p$ of a particle on a line change with time $t$ if we know the energy or Hamiltonian $H(q,p)$ :

Two of the Maxwell relations connect the volume $V$ , entropy $S$ , pressure $P$ and temperature $T$ of a system in thermodynamic equilibrium:

$q \to S \qquad p \to T$
$t \to V \qquad H \to P$

Hamilton’s equations:

become these relations:

$\begin{array}{ccr} \displaystyle{ \frac{d T}{d V} } &=& \displaystyle{- \frac{\partial P}{\partial S} } \\ \\ \displaystyle{ \frac{d S}{d V} } &=& \displaystyle{ \frac{\partial P}{\partial T} } \end{array}$

These are almost like two of the Maxwell relations! But in thermodynamics we always use partial derivatives:

$\begin{array}{ccr} \displaystyle{ \frac{\partial T}{\partial V} } &=& \displaystyle{ - \frac{\partial P}{\partial S} } \\ \\ \displaystyle{ \frac{\partial S}{\partial V} } &=& \displaystyle{ \frac{\partial P}{\partial T} } \end{array}$

and we say which variables are held constant:

If we write Hamilton’s equations in the same style as the Maxwell relations, they look funny:

$\begin{array}{ccr} \displaystyle{ \left. \frac{\partial p}{\partial t}\right|_q } &=& \displaystyle{ - \left. \frac{\partial H}{\partial q}\right|_t } \\ \\ \displaystyle{ \left. \frac{\partial q}{\partial t}\right|_p } &=& \displaystyle{ \left. \frac{\partial H}{\partial p} \right|_t } \end{array}$

Can this possibly be right?

Yes! When we work out the analogy between classical mechanics and thermodynamics we’ll see why.

We can get Maxwell’s relations starting from this: the internal energy $U$ of a system in equilibrium depends on its entropy $S$ and volume $V.$

Temperature and pressure are derivatives of $U:$

$\displaystyle{ T = \left.\frac{\partial U}{\partial S} \right|_V \qquad P = - \left. \frac{\partial U}{\partial V} \right|_S }$

Maxwell’s relations follow from the fact that mixed partial derivatives commute! For example:

$\displaystyle{ \left. \frac{\partial T}{\partial V} \right|_S \; = \; \left. \frac{\partial}{\partial V} \right|_S \left. \frac{\partial }{\partial S} \right|_V U \; = \; \left. \frac{\partial }{\partial S} \right|_V \left. \frac{\partial}{\partial V} \right|_S U \; = \; - \left. \frac{\partial P}{\partial S} \right|_V }$

To get Hamilton’s equations the same way, we need a function $W$ of the particle’s position $q$ and time $t$ such that

$\displaystyle{ p = \left. \frac{\partial W}{\partial q} \right|_t \qquad H = -\left. \frac{\partial W}{\partial t} \right|_q }$

Then we’ll get Hamilton’s equations from the fact that mixed partial derivatives commute!

The trick is to let $W$ be ‘Hamilton’s principal function’. So let’s define that. First, the action of a particle’s path is

$\displaystyle{ \int_{t_0}^{t_1} L(q(t),\dot{q}(t)) \, dt }$

where $L$ is the Lagrangian:

$L(q,\dot{q}) = p \dot q - H(q,p)$

The particle always takes a path from $(q_0, t_0)$ to $(q_1, t_1)$ that’s a critical point of the action. We can derive Hamilton’s equations from this fact.

Let’s assume this critical point is a minimum. Then the least action for any path from $(q_0,t_0)$ to $(q_1,t_1)$ is called Hamilton’s principal function

$W(q_0,t_0,q_1,t_1) = \min_{q \; \mathrm{with} \; q(t_0) = q_0, q(t_1) = q_1} \int_{t_0}^{t_1} L(q(t),\dot{q}(t)) \, dt$

A beautiful fact: if we differentiate Hamilton’s principal function, we get back the energy $H$ and momentum $p$ :

$\begin{array}{ccc} \displaystyle{ \frac{\partial}{\partial q_0} W(q_0,t_0,q_1,t_1) = -p(t_0) } && \displaystyle{ \frac{\partial}{\partial t_0} W(q_0,t_0,q_1,t_1) = H(t_0) } \\ \\ \displaystyle{ \frac{\partial}{\partial q_1} W(q_0,t_0,q_1,t_1) = p(t_1) } && \displaystyle{ \frac{\partial}{\partial t_1}W(q_0,t_0,q_1,t_1) = -H(t_1) } \end{array}$

You can prove these equations using

$L = p\dot{q} - H$

which implies that

$\displaystyle{ W(q_0,t_0,q_1,t_1) = \int_{q_0}^{q_1} p \, dq \; - \; \int_{t_0}^{t_1} H \, dt }$

where we integrate along the minimizing path. (It’s not as trivial as it may look, but you can do it.)

Now let’s fix a starting-point $(q_0,t_0)$ for our particle, and say its path ends at any old point $(q,t)$ . Think of Hamilton’s principal function as a function of just $(q,t)$ :

$W(q,t) = W(q_0,t_0,q,t)$

Then the particle’s momentum and energy when it reaches $(q,t)$ are:

$\displaystyle{ p = \left. \frac{\partial W}{\partial q} \right|_t \qquad H = -\left. \frac{\partial W}{\partial t} \right|_q }$

This is just what we wanted. Hamilton’s equations now follow from the fact that mixed partial derivatives commute!

So, we have this analogy between classical mechanics and thermodynamics:

Classical mechanics	Thermodynamics
action: $W(q,t)$	internal energy: $U(V,S)$
position: $q$	entropy: $S$
momentum: $p = \frac{\partial W}{\partial q}$	temperature: $T = \frac{\partial U}{\partial S}$
time: $t$	volume: $V$
energy: $H= -\frac{\partial W}{\partial t}$	pressure: $P = - \frac{\partial U}{\partial V}$
$dW = pdq - Hdt$	$dU = TdS - PdV$

What’s really going on in this analogy? It’s not really the matchup of variables that matters most—it’s something a bit more abstract. Let’s dig deeper.

I said we could get Maxwell’s relations from the fact that mixed partials commute, and gave one example:

But to get the other Maxwell relations we need to differentiate other functions—and there are four of them!

• $U$ : internal energy
• $U - TS$ : Helmholtz free energy
• $U + PV$ : enthalpy
• $U + PV - TS$ : Gibbs free energy

They’re important, but memorizing all the facts about them has annoyed students of thermodynamics for over a century. Is there some other way to get the Maxwell relations? Yes!

In 1958 David Ritchie explained how we can get all four Maxwell relations from one equation! Jaynes also explained how in some unpublished notes for a book. Here’s how it works.

Start here:

$dU = T d S - P d V$

Integrate around a loop $\gamma$ :

$\displaystyle{ \oint_\gamma T d S - P d V = \oint_\gamma d U = 0 }$

$\displaystyle{ \oint_\gamma T d S = \oint_\gamma P dV }$

This says the heat added to a system equals the work it does in this cycle

Green’s theorem implies that if a loop $\gamma$ encloses a region $R,$

$\displaystyle{ \oint_\gamma T d S = \int_R dT \, dS }$

Similarly

$\displaystyle{ \oint_\gamma P d V = \int_R dP \, dV }$

But we know these are equal!

So, we get

$\displaystyle{ \int_R dT \, dS = \int_R dP \, dV }$

for any region $R$ enclosed by a loop. And this in turn implies

$d T\, dS = dP \, dV$

In fact, all of Maxwell’s relations are hidden in this one equation!

Mathematicians call something like $dT \, dS$ a 2-form and write it as $dT \wedge dS$ . It’s an ‘oriented area element’, so

$dT \, dS = -dS \, dT$

Now, starting from

$d T\, dS = dP \, dV$

We can choose any coordinates $X,Y$ and get

$\displaystyle{ \frac{dT \, dS}{dX \, dY} = \frac{dP \, dV}{dX \, dY} }$

(Yes, this is mathematically allowed!)

If we take $X = V, Y = S$ we get

$\displaystyle{ \frac{dT \, dS}{dV \, dS} = \frac{dP \, dV}{dV \, dS} }$

and thus

$\displaystyle{ \frac{dT \, dS}{dV \, dS} = - \frac{dV \, dP}{dV \, dS} }$

We can actually cancel some factors and get one of the Maxwell relations:

$\displaystyle{ \left. \frac{\partial T}{\partial V}\right|_S = - \left. \frac{\partial P}{\partial S}\right|_V }$

(Yes, this is mathematically justified!)

Let’s try another one. If we take $X = T, Y = V$ we get

$\displaystyle{ \frac{dT \, dS}{dT \, dV} = \frac{dP \, dV}{dT \, dV} }$

Cancelling some factors here we get another of the Maxwell relations:

$\displaystyle{ \left. \frac{\partial S}{\partial V} \right|_T = \left. \frac{\partial P}{\partial T} \right|_V }$

Other choices of $X,Y$ give the other two Maxwell relations.

In short, Maxwell’s relations all follow from one simple equation:

$d T\, dS = dP \, dV$

Similarly, Hamilton’s equations follow from this equation:

$d p\, dq = dH \, dt$

All calculations work in exactly the same way!

By the way, we can get these equations efficiently using the identity $d^2 = 0$ and the product rule for $d$ :

$\begin{array}{ccl} dU = TdS - PdV & \implies & d^2 U = d(TdS - P dV) \\ \\ &\implies& 0 = dT\, dS - dP \, dV \\ \\ &\implies & dT\, dS = dP \, dV \end{array}$

Now let’s change viewpoint slightly and temporarily treat $P$ and $V$ as independent from $S$ and $T.$ So, let’s start with $\mathbb{R}^4$ with coordinates $(S,T,V,P)$ . Then this 2-form on $\mathbb{R}^4$ :

$\omega = dT \, dS - dP \, dV$

is called a symplectic structure.

Choosing the internal energy function $U(S,V)$ , we get this 2-dimensional surface of equilibrium states:

$\displaystyle{ \Lambda = \left\{ (S,T,V,P): \; \textstyle{ T = \left.\phantom{\Big|} \frac{\partial U}{\partial S}\right|_V , \; P = -\left. \phantom{\Big|}\frac{\partial U}{\partial V} \right|_S} \right\} \; \subset \; \mathbb{R}^4 }$

Since

$\omega = dT \, dS - dP \, dV$

we know

$\displaystyle{ \int_R \omega = 0 }$

for any region in the surface $\Lambda$ , since on this surface $dU = TdS - PdV$ and our old argument applies.

This fact encodes the Maxwell relations! Physically it says: for any cycle on the surface of equilibrium states, the heat flow in equals the work done.

Similarly, in classical mechanics we can start with $\mathbb{R}^4$ with coordinates $(q,p,t,H)$ , treating $p$ and $H$ as independent from $q$ and $t .$ This 2-form on $\mathbb{R}^4$ :

$\omega = dH \, dt - dp \, dq$

is a symplectic structure. Hamilton’s principal function $W(q,t)$ defines a 2d surface

$\displaystyle{ \Lambda = \left\{ (q,p,t,H): \; \textstyle{ p = \left.\phantom{\Big|} \frac{\partial W}{\partial q}\right|_t , \; H = -\left.\phantom{\Big|} \frac{\partial W}{\partial t} \right|_q} \right\} \subset \mathbb{R}^4 }$

We have $\int_R \omega = 0$ for any region $R$ in this surface $\Lambda.$ And this fact encodes Hamilton’s equations!

Summary

In thermodynamics, any 2d region $R$ in the surface $\Lambda$ of equilibrium states has

$\displaystyle{ \int_R \omega = 0 }$

This is equivalent to the Maxwell relations.

In classical mechanics, any 2d region $R$ in the surface $\Lambda$ of allowed $(q,p,t,H)$ 4-tuples for particle trajectories through a single point $(q_0,t_0)$ has

$\displaystyle{ \int_R \omega = 0 }$

This is equivalent to Hamilton’s equations.

These facts generalize when we add extra degrees of freedom, e.g. the particle number $N$ in thermodynamics:

$\omega = dT \, dS - dP \, dV + d\mu \, dN$

or more dimensions of space in classical mechanics:

$\omega = dp_1 \, dq_1 + \cdots + dp_{n-1} dq_{n-1} - dH \, dt$

We get a vector space $\mathbb{R}^{2n}$ with a 2-form $\omega$ on it, and a Lagrangian submanifold $\Lambda \subset \mathbb{R}^{2n}$ : that is, a n-dimensional submanifold such that

$\int_R \omega = 0$

for any 2d region $R \subset \Lambda.$

This is more evidence for Alan Weinstein’s “symplectic creed”:

EVERYTHING IS A LAGRANGIAN SUBMANIFOLD

As a spinoff, we get two extra Hamilton’s equations for a point particle on a line! They look weird, but I’m sure they’re correct for trajectories that go through a specific arbitrary spacetime point $(q_0,t_0).$

$\begin{array}{ccrcccr} \displaystyle{ \left. \frac{\partial T}{\partial V}\right|_S } &=& \displaystyle{ - \left. \frac{\partial P}{\partial S}\right|_V } & \qquad & \displaystyle{ \left. \frac{\partial p}{\partial t}\right|_q } &=& \displaystyle{ - \left. \frac{\partial H}{\partial q}\right|_t } \\ \\ \displaystyle{ \left. \frac{\partial S}{\partial V}\right|_T } &=& \displaystyle{ \left. \frac{\partial P}{\partial T} \right|_V } & & \displaystyle{ \left. \frac{\partial q}{\partial t}\right|_p } &=& \displaystyle{ \left. \frac{\partial H}{\partial p} \right|_t } \\ \\ \displaystyle{ \left. \frac{\partial V}{\partial T} \right|_P } &=& \displaystyle{ - \left. \frac{\partial S}{\partial P} \right|_T } & & \displaystyle{ \left. \frac{\partial t}{\partial p} \right|_H } &=& \displaystyle{ - \left. \frac{\partial q}{\partial H} \right|_p } \\ \\ \displaystyle{ \left. \frac{\partial V}{\partial S} \right|_P } &=& \displaystyle{ \left. \frac{\partial T}{\partial P} \right|_S } & & \displaystyle{ \left. \frac{\partial p}{\partial H} \right|_q } &=& \displaystyle{ \left. \frac{\partial t}{\partial q} \right|_H } \end{array}$

6 Comments | mathematics, physics | Permalink
Posted by John Baez

Maxwell’s Relations (Part 3)

22 September, 2021

In Part 2 we saw a very efficient formulation of Maxwell’s relations, from which we can easily derive their usual form. Now let’s talk more about the meaning of the Maxwell relations—both their physical meaning and their mathematical meaning. For the physical meaning, I’ll draw again from Ritchie’s paper:

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

but Emily Roach pointed out that much of this can also be found in the third chapter of Jaynes’ unpublished book:

• E. T. Jaynes, Thermodynamics, Chapter 3: Plausible reasoning.

First I’ll do the case of 2 dimensions, and then the case of n dimensions. In the 2d case I’ll talk like a physicist and use notation from thermodynamics. We’ll learn the Maxwell relations have these meanings, apart from their obvious ones:

• In any thermodynamic cycle, the heat absorbed by a system equals the work it does.

• In any thermodynamic cycle, energy is conserved.

• Any region in the surface of equilibrium states has the same area in pressure-volume coordinates as it does in temperature-entropy coordinates.

In the n-dimensional case I’ll use notation that mathematicians will like better, and also introduce the language of symplectic geometry. This will give a general, abstract statement of the Maxwell relations:

• The manifold of equilibrium states is a Lagrangian submanifold of a symplectic manifold.

Don’t worry—I’ll explain it!

Maxwell’s relations in 2 dimensions

Suppose we have a physical system whose internal energy $U$ is a smooth function of its entropy $S$ and volume $V.$ So, we have a smooth function on the plane:

$U \colon \mathbb{R}^2 \to \mathbb{R}$

and we call the coordinates on this plane $(S,V).$

Next we introduce two more variables called temperature $T$ and pressure $P.$ So, we’ll give $\mathbb{R}^4$ coordinates $(S,T,V,P).$ But, there’s a 2-dimensional surface where these extra variables are given by the usual formulas in thermodynamics:

$\displaystyle{ \Lambda = \left\{(S,T,V,P) : \; T = \left.\frac{\partial U}{\partial S} \right|_V, \; P = - \left. \frac{\partial U}{\partial V} \right|_S \right\} \; \subset \mathbb{R}^4 }$

We call this the surface of equilibrium states.

By the equations that $T$ and $P$ obey on the surface of equilibrium states, the following equation holds on this surface:

$dU = T d S - P d V$

So, if $\gamma$ is any loop on this surface, we have

$\displaystyle{ \oint_\gamma ( T d S - P d V) \; = \; \oint_\gamma \; dU \; = 0 }$

where the total change in $U$ is zero because the loop ends where it starts! We thus have

$\displaystyle{ \oint_\gamma T d S = \oint_\gamma P d V }$

In thermodynamics this has a nice meaning: the left side is the heat absorbed by a system as its state moves around the loop $\gamma,$ while the right side is the work done by this system. So, the equation says the heat absorbed by a system as it carries out a cycle is equal to the work it does.

We’ll soon see that this equation contains, hidden within it, all four Maxwell relations. And it’s a statement of conservation of energy! By the way, it’s perfectly obvious that energy is conserved in a cycle: our loop takes us from a point where $U$ has some value back to the same point, where it has the same value.

So how do we get Maxwell’s relations out of this blithering triviality? There’s a slow way and a fast way. Since the slow way provides extra insight I’ll do that first.

Suppose the loop $\gamma$ bounds some 2-dimensional region $R$ in the surface $\Lambda.$ Then by Green’s theorem we have

$\displaystyle{ \oint_\gamma T d S = \int_R dT \wedge dS }$

That is, the heat absorbed in the cycle is just the area of the region $R$ in $(T,S)$ coordinates. But Green’s theorem also says

$\displaystyle{ \oint_\gamma P dV = \int_R dP \wedge dV }$

That is, the work done in the cycle, which is minus the left hand side, is minus the area of the region $R$ in $(P,V)$ coordinates!

That’s nice to know. But since we’ve seen these are equal,

$\displaystyle{ \int_R dT \wedge dS = \int_R dP \wedge dV }$

for every region $R$ in the surface $\Lambda.$

Well, at least I’ve shown it for every region bounded by a loop! But every region can be chopped up into small regions that are bounded by loops, so the equation is really true for any region $R$ in the surface $\Lambda.$

We express this fact by saying that $dT \wedge dS$ equals $dP \wedge dV$ when these 2-forms are restricted to the surface $\Lambda.$ We write it like this:

$\displaystyle{ \left. dT \wedge dS \right|_{\Lambda} = \left. dP \wedge dV \right|_{\Lambda} }$

Now, last time we saw how to quickly get from this equation to all four Maxwell relations!

(Back then I didn’t write the $|_{\Lambda}$ symbol because I was implicitly working on this surface $\Lambda$ without telling you. More precisely, I was working on the plane $\mathbb{R}^2,$ but we can identify the surface $\Lambda$ with that plane using the $(S,V)$ coordinates.)

So, given what we did last time, we are done! The equation

$\displaystyle{ \left. dT \wedge dS \right|_{\Lambda} = \left. dP \wedge dV \right|_{\Lambda}}$

expresses both conservation of energy and all four Maxwell equations—in a very compressed, beautiful form!

Maxwell’s relations in n dimensions

Now we can generalize everything to n dimensions. Suppose we have a smooth function

$U \colon \mathbb{R}^n \to \mathbb{R}$

Write the coordinates on $\mathbb{R}^n$ as

$(q^1, \dots, q^n)$

and write the corresponding coordinates on the cotangent bundle $T^\ast \mathbb{R}^n$ as

$(q^1, \dots, q^n, p_1, \dots, p_n)$

There is a submanifold of $T^\ast \mathbb{R}^n$ where $p_i$ equals the partial derivative of $U$ with respect to $q^i:$

$\displaystyle{ p_i = \frac{\partial U}{\partial q^i} }$

Let’s call this submanifold $\Lambda$ since it’s the same one we’ve seen before (except for that annoying little minus sign in the definition of pressure):

$\displaystyle{ \Lambda = \left\{(q,p) \in T^\ast \mathbb{R}^n : \; p_i = \frac{\partial U}{\partial q^i} \right\} }$

In applications to thermodynamics, this is the manifold of equilibrium states. But we’re doing math here, and this math has many applications. It’s this generality that makes the subject especially interesting to me.

Now, $U$ started out life as a function on $\mathbb{R}^n,$ but it lifts to a function on $T^\ast \mathbb{R}^n,$ and we have

$\displaystyle{ dU = \sum_i \frac{\partial U}{\partial q^i} dq^i }$

By the definition of $\Lambda$ we have

$\displaystyle{ \left. \frac{\partial U}{\partial q^i} \right|_\Lambda = \left. p_i \right|_\Lambda }$

$\displaystyle{ \left. dU \right|_\Lambda = \left. \sum_i p_i dq^i \right|_\Lambda }$

The 1-form on the right-hand side here:

$\displaystyle{ \theta = \sum_i p_i \, dq^i }$

is called the tautological 1-form on $T^\ast \mathbb{R}^n.$ Its exterior derivative

$\displaystyle{ \omega = d\theta = \sum_i dp_i \wedge dq^i }$

is called the symplectic structure on this cotangent bundle. Both of these are a big deal in classical mechanics, but here we are seeing them in thermodynamics! And the point of all this stuff is that we’ve seen

$\displaystyle{ \left. dU \right|_\Lambda = \left. \theta \right|_\Lambda }$

Taking $d$ of both sides and using $d^2 = 0$ , we get

$\displaystyle{ \left. d\theta \right|_\Lambda = 0 }$

or in other words

$\displaystyle{ \left. \omega \right|_\Lambda = 0 }$

And this is a very distilled statement of Maxwell’s relations!

Why? Well, in the 2d case we discussed earlier, the tautological 1-form is

$\theta = T dS - P dV$

thanks to the annoying minus sign in the definition of pressure. Thus, the symplectic structure is

$\omega = dT \wedge dS - dP \wedge dV$

and the fact that the symplectic structure $\omega$ vanishes when restricted to $\Lambda$ is just our old friend

$\displaystyle{ \left. dT \wedge dS \right|_{\Lambda} = \left. dP \wedge dV \right|_{\Lambda} }$

As we’ve seen, this equation contains all of Maxwell’s relations.

In the n-dimensional case, $\Lambda$ is an n-dimensional submanifold of the 2n-dimensional symplectic manifold $T^\ast M.$ In general, if an n-dimensional submanifold $S$ of a 2n-dimensional symplectic manifold has the property that the symplectic structure vanishes when restricted to $S,$ we call $S$ a Lagrangian submanifold.

So, one efficient abstract statement of Maxwell’s relations is:

The manifold of equilibrium states is a Lagrangian submanifold.

It takes a while to see the value of this statement, and I won’t try to explain it here. Instead, read Weinstein’s introduction to symplectic geometry:

• Alan Weinstein, Symplectic geometry, Bulletin of the American Mathematical Society 5 (1981), 1–13.

You’ll see here an introduction to Lagrangian submanifolds and an explanation of the “symplectic creed”:

EVERYTHING IS A LAGRANGIAN SUBMANIFOLD.
— Alan Weinstein

The restatement of Maxwell’s relations in terms of Lagrangian submanifolds is just another piece of evidence for this!

• 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:

• Classical mechanics versus thermodynamics.

7 Comments | physics, mathematics | Permalink
Posted by John Baez

Maxwell’s Relations (Part 2)

18 September, 2021

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

$0 = d^2 U = dT \wedge dS - dP \wedge dV$

$dT \wedge dS = dP \wedge dV$

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

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

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

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

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

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

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 in 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:

• Classical mechanics versus thermodynamics.

5 Comments | mathematics, physics | Permalink
Posted by John Baez

Maxwell’s Relations (Part 1)

17 September, 2021

The Maxwell relations are equations that show up whenever we have a smooth convex function

$U \colon \mathbb{R}^n \to \mathbb{R}$

They say that the mixed partial derivatives of $U$ commute, but also that the mixed partial derivatives of various functions constructed from $U$ commute. So in a sense they are completely trivial except for the way we construct these various functions! Nonetheless they are very important in physics, because they give generally valid relations between the derivatives of thermodynamically interesting quantities.

For example, here is one of the Maxwell relations that people often use when studying a thermodynamic system like a cylinder of gas or a beaker of liquid:

$\displaystyle{ \left. \frac{\partial T}{\partial V} \right|_S \; = \; - \left. \frac{\partial P}{\partial S} \right|_V }$

where:

• $S$ is the entropy of the system
• $T$ is its temperature
• $V$ is its volume
• $P$ is its pressure

There are also three more Maxwell relations involving these quantities. They all have the same general look, though half contain minus signs and half don’t. So, they’re quite tiresome to memorize. It’s more interesting to figure out what’s really going on here!

I already gave one story about what’s going on: we start with a function $U$ , which happens in this case to be a function of $S$ and $V$ called the internal energy of our system. If we say its mixed partial derivatives commute:

$\displaystyle{ \frac{\partial^2 U}{\partial V \partial S} = \frac{\partial^2 U}{\partial S \partial V} }$

we get the Maxwell relation I wrote above, simply by using the standard definitions of $T$ and $P$ as partial derivatives of $U.$ We can then build a bunch of other functions from $U, S, T, V$ and $P,$ and write down other equations saying that the mixed partial derivatives of these functions commute. This gives us all the Maxwell relations.

Let me show you in detail how this works, but without explaining why I’m choosing the particular ‘bunch of other functions’ that I’ll use. There is a way to explain that using the concept of Legendre transform. But next time I’ll give a very different approach to the Maxwell relations, based on this paper:

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

This sidesteps the whole business of choosing a ‘bunch of other functions’, and I think it’s very nice.

What follows is a bit mind-numbing, so let me tell you what to pay attention to. I’ll do the same general sort of calculation four times, first starting with any convex smooth function $U \colon \mathbb{R}^2 \to \mathbb{R}$ and then with three other functions built from that one. The only clever part is how we choose those other functions.