Maxwell Relations [Thermal & Statistical Mechanics I Studied #28]

In the previous post, we threw out new concepts and definitions, and even figured out what they mean. Alright, then let’s scroll back up and look at this picture again.

Sum to the mary: U(S,V) with dU = TdS − pdV, H(S,p) with dH = TdS + Vdp, A(T,V) with dA = −SdT − pdV, G(T,p) with dG = −SdT + Vdp, and the partial derivatives that give T, p, S and V

Sum to the mary again, now with the two partial-derivative relations of each function circled: U(S,V) in green, H(S,p) in orange, A(T,V) in red, G(T,p) in blue

By bundling them up a few at a time like this, we can make certain relations out of the connections between them. These relations are called the Maxwell Relations, and well then, let’s get started~~~

…is what I’d love to say, but wait, before that there’s one thing we need to prove first. I’ll prove just that one thing, and then derive the Maxwell Relations.

First, we need to dig one more layer into state functions. One of the characteristics of a state function was that it “does not depend on the path.”

Alright, so let’s say we have a ‘state function’ $f$ defined by variables $x$ and $y$. For that $f$, the change $df$ is

(I’ve said this many times already…? For those who don’t buy it, go look at the ‘partial derivatives’ part of your math textbook — it’s derived in detail there.)

Since this $f$ is a state function, it couldn’t care less about paths and such, so the integral of $df$ over a closed loop is zero.

Around a closed loop ∮df = 0, that is ∮((∂f/∂x)_y dx + (∂f/∂y)_x dy) = 0. Let’s write this with vectors: ∮((∂f/∂x)_y, (∂f/∂y)_x)·(dx, dy) = 0. I’ll write it like this: ∮F·dr = 0

By Stokes’ theorem, this result and ∇ × F = 0 mean the same thing

Curl of F as a determinant with rows i, j, k / ∂/∂x, ∂/∂y, ∂/∂z / (∂f/∂x)_y, (∂f/∂y)_x, 0, which gives ∇ × F = i(−∂/∂z (∂f/∂y)_x) + j(∂/∂z (∂f/∂x)_y) + k(∂/∂x (∂f/∂y)_x − ∂/∂y (∂f/∂x)_y) = 0

We could say the condition for path-independence is that each of these must all~~ be zero. But for $i$ and $j$, since $f$ is a function of $x$ and $y$ to begin with, the derivative with respect to $z$ is 0. Setting that aside, a state function $f$ that is path-independent means

The k component, k(∂/∂x (∂f/∂y)_x − ∂/∂y (∂f/∂x)_y) = 0, gives ∂/∂x (∂f/∂y)_x = ∂/∂y (∂f/∂x)_y; f depends only on x and y, so let’s drop the subscripts: ∂/∂x ∂f/∂y = ∂/∂y ∂f/∂x. For a path-independent (two-variable) state function, which partial you take first doesn’t affect the value. That’s what I wanted to say!

Kyaa!!!! $T, V, U, S, p, H, A, G$ — they’re all~~ state functions, right?

When you take partial derivatives with respect to two variables, the order of the partial derivatives doesn’t matter~~~!!!

Why?!?! Because they’re state functions!!!

Then let’s head into the derivation of the Maxwell Relations.

Relation i), for U(S,V): T = (∂U/∂S)_V and −p = (∂U/∂V)_S. Blue one: hold S on both sides, take the partial w.r.t. V; red one: hold V on both sides, take the partial w.r.t. S. U is a state function of S and V, so ∂/∂V ∂U/∂S = ∂/∂S ∂U/∂V, and so (∂T/∂V)_S = −(∂p/∂S)_V

Relation ii), for H(S,p): T = (∂H/∂S)_p and V = (∂H/∂p)_S. Blue one: hold S on both sides, take the partial w.r.t. p; red one: hold p on both sides, take the partial w.r.t. S. Likewise, the mixed partials of H are equal, ∂/∂p ∂H/∂S = ∂/∂S ∂H/∂p, so (∂T/∂p)_S = (∂V/∂S)_p

Relation iii), for A(T,V): S = −(∂A/∂T)_V and p = −(∂A/∂V)_T. Blue one: hold T on both sides, take the partial w.r.t. V; red one: hold V on both sides, take the partial w.r.t. T. Likewise, the mixed partials of A are equal, ∂/∂V ∂A/∂T = ∂/∂T ∂A/∂V, so (∂S/∂V)_T = (∂p/∂T)_V

Relation iv), for G(T,p): S = −(∂G/∂T)_p and V = (∂G/∂p)_T. Blue one: hold T on both sides, take the partial w.r.t. p; red one: hold p on both sides, take the partial w.r.t. T. Likewise, the mixed partials of G are equal, ∂/∂p ∂G/∂T = ∂/∂T ∂G/∂p, so −(∂S/∂p)_T = (∂V/∂T)_p

The four Maxwell relations: i. (∂T/∂V)_S = −(∂p/∂S)_V, ii. (∂T/∂p)_S = (∂V/∂S)_p, iii. (∂S/∂V)_T = (∂p/∂T)_V, iv. −(∂S/∂p)_T = (∂V/∂T)_p

From here on out, I’m gonna milk these relations for all they’re worth~~ ^^*

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

On the two summary cards, the pressure from $U(S,V)$ is $p = -\left(\frac{\partial U}{\partial V}\right)_S$, since $dU = TdS - pdV$.
In part ii), the outer subscript of the red line is $p$: $\left(\frac{\partial V}{\partial S}\right)_p = \left(\frac{\partial}{\partial S}\left(\frac{\partial H}{\partial p}\right)_S\right)_p$.
In parts ii) to iv), the line "U is a state function of S and V" was carried over from part i): it is $H$ of $S$ and $p$, $A$ of $T$ and $V$, and $G$ of $T$ and $p$.

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