Maxwell’s Relations (Part 3)

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but Emily Roach pointed out that much of this can also be found in the third chapter of Jaynes’ unpublished book: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

— 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.

10 Responses to Maxwell’s Relations (Part 3)

  1. Toby Bartels says:

    Typo (or probably edit-o): The word ‘vanishes’ should be removed from the first place where it appears. (Or else the word ‘equals’ just before it should be replaced with a minus sign.)

  2. Joseph Rizcallah says:

    Thank you for the interesting article.
    I think that the integrals on the right-hand sides of the above equalities should have no circles, which usually indicate a closed boundary.

  3. allenknutson says:

    There are two ways to quantize a Lagrangian submanifold: geometric quantization and deformation quantization.

    In geometric, we (usually) extend our symplectic structure to a Kahler structure, consider a holomorphic line bundle with a Hermitian connection whose curvature is the symplectic form, and find a section whose norm is maximized on the Lagrangian submanifold. (There are lots more options, involving splittings of the complexified tangent bundle, but let’s hope we don’t need them.)

    In deformation, we deform the coordinate ring of the symplectic manifold + Poisson bracket to a noncommutative algebra, and the Lagrangian to a left module over this algebra. In your T^*M case this would probably be the ring D of differential operators on M, and the Lagrangian to a D-module.

    Do either of these suggest themselves in this context?

    • John Baez says:

      This is actually one of the directions I’m heading with all this stuff… I’d rather announce it when it’s closer to done. But thanks, this is very helpful.

  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).


  5. Grgur says:

    I’m a bit late to the party, so let me just mention that many more relations of interest can be derived a lot more easily by using 2-forms/Jacobians. I’m talking about Bridgman’s thermodynamic equations† (also called Thermodynamic expansion formulae) that are closely related to this 2-form method.

    Following Dearden’s more modern listing of these equations††, the relation between the two is (see Table 1):

    \displaystyle{ (\partial A)_B = - (\partial B)_A = \frac{1}{V} \frac{dA \wedge dB}{dT \wedge dP} }

    From this identification all the Bridgman’s relations follow (one also needs dU = T dS - P dV , dT \wedge dS = dP \wedge dV , definitions of susceptibilities, and so on).

    So this method of deriving Maxwell’s relations isn’t a mere trick or shorthand, but the best method we have of deriving relations among first-order derivatives in general!

    It’s interesting to read how Bridgman figured that we can meaningfully isolate (\partial A)_B . It seems they weren’t aware that what they wrote can be made more rigorous and general.


    • John Baez says:

      This is great! Thanks, I didn’t know about Bridgman. He published his work in 1914, long before Jaynes or Ritchie emphasized the importance of Jacobians.

You can use Markdown or HTML in your comments. You can also use LaTeX, like this: $latex E = m c^2 $. The word 'latex' comes right after the first dollar sign, with a space after it.

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.