Maxwell relations — derivation from each potential: Visual walkthrough

Step 1 — A function of two things is a landscape

WHAT. Suppose one number $z$ depends on two knobs, $x$ and $y$ . Write $z = z (x, y)$ . Picture $x$ running east, $y$ running north, and $z$ being the height of the ground above each spot on that east–north floor.

WHY. Everything in thermodynamics is a quantity that depends on two others (energy depends on how much disorder and how much room there is). Before we touch physics we build the only picture we need: a hill sitting over a flat map.

PICTURE. Look at the figure. The flat grid on the bottom is the $(x, y)$ floor. Above it, the curved sheet is the height $z$ . One spot on the floor → one height on the sheet.

Figure — Maxwell relations — derivation from each potential

Step 2 — "Steepness" in one chosen direction: the partial derivative

WHAT. Stand on the hill. Face east (increase $x$ , keep $y$ frozen). How fast does your height climb per step east? That number is the partial derivative: $steepness facing east (\frac{\partial z}{\partial x})_{y}$

WHY. We need a tool that measures steepness in one direction only, because a hill can be steep east yet flat north at the same spot. The subscript $y$ is the tool's whole point.

$\partial z / \partial x$ — "change in height per change in $x$ "
the subscript $_{y}$ — "while $y$ is held frozen" (you walk a straight east line, never drifting north)

PICTURE. The blue arrow is the east-facing tangent slope; the yellow arrow is the north-facing tangent slope. Same point, two different steepnesses.

Step 3 — The differential: total height change from a small step

WHAT. Take a tiny step: a bit east ( $d x$ ) and a bit north ( $d y$ ). Your total height change is east-steepness × east-step plus north-steepness × north-step: $d z = call it M (\frac{\partial z}{\partial x})_{y} d x + call it N (\frac{\partial z}{\partial y})_{x} d y = M d x + N d y$

WHY. Any small move is "some east + some north." We add the height gained on each leg. Naming the two slopes $M$ and $N$ lets us forget they came from a height and just treat them as two functions on the floor.

$M$ — the east slope, itself a function of position $(x, y)$
$N$ — the north slope, also a function of $(x, y)$
$d z$ — the total rise from the little diagonal step

PICTURE. The little step is split into a red east leg and a green north leg; the height climbed on each leg is stacked to give the total $d z$ .

Step 4 — The magic question: does climbing order matter?

WHAT. Two ways to reach the same diagonally-opposite corner of a tiny square:

Path A: go east first, then north.
Path B: go north first, then east.

On a genuine hill both land you at the same corner, so both climb the same total height.

WHY. This is the geometric seed of every Maxwell relation. If the height is real (a state function), the two paths must agree — and forcing them to agree will pin down a relationship between the slopes $M$ and $N$ .

PICTURE. The square on the floor with the two colored paths (east-then-north vs north-then-east) meeting at the same top corner.

Step 5 — Turning "order doesn't matter" into an equation

WHAT. Ask: how does the east slope $M$ change as I nudge north? That's $(\frac{\partial M}{\partial y})_{x}$ . And: how does the north slope $N$ change as I nudge east? That's $(\frac{\partial N}{\partial x})_{y}$ . The "order doesn't matter" fact forces these two to be equal: $(\frac{\partial M}{\partial y})_{x} = (\frac{\partial N}{\partial x})_{y}$

WHY. Both sides are literally the same second derivative $\frac{\partial ^{2} z}{\partial x \partial y}$ — the hill's cross-curvature — measured in the two possible orders. A smooth hill has one cross-curvature, so the orders match. This is the Schwarz / equality-of-mixed-partials theorem (see Equality of mixed partial derivatives (Schwarz theorem)).

left side — "take east slope $M$ , see how it tilts going north"
right side — "take north slope $N$ , see how it tilts going east"
the two are equal because both = $\partial^{2} z / \partial x \partial y$

PICTURE. The tiny square again, now showing that "extra rise on the far corner" is the same number whether you attribute it to $M$ steepening northward or $N$ steepening eastward.

Recall Why the two second derivatives must match

Extra height at the far corner ::: it equals (change in east-slope as you go north) and also (change in north-slope as you go east) — both describe the same corner, so they're one number.

Step 6 — Load the real hill: internal energy $U$

WHAT. Now name the abstract pieces with physics. The height is the internal energy $U$ ; the floor coordinates are entropy $S$ (disorder) running east and volume $V$ (room) running north. The First+Second Laws give (see First and Second Laws of Thermodynamics): $d U = M T d S N - P d V$

WHY. This is our $d z = M d x + N d y$ with a specific hill. Matching term-by-term:

$z \to U$ (the height is energy)
$x \to S$ , $y \to V$ (floor axes are disorder and room)
$M \to T$ (east slope of energy vs. disorder is temperature)
$N \to - P$ (north slope of energy vs. room is minus the pressure)

The minus on $P$ is not a choice — energy drops when the gas expands and pushes out, so the north slope is negative.

PICTURE. The same landscape as Step 1, now re-labelled: axes $S$ and $V$ , height $U$ , with the east slope tagged $T$ and the north slope tagged $- P$ .

Step 7 — Cross-differentiate: the relation drops out

WHAT. Plug $M = T$ and $N = - P$ into the Step-5 box: $(\frac{\partial M}{\partial y})_{x} = (\frac{\partial N}{\partial x})_{y} ⟹ (\frac{\partial T}{\partial V})_{S} = (\frac{\partial ( - P )}{\partial S})_{V}$ Pull the minus out front: $(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V}$

WHY. We did nothing physical here — we only renamed the pure-calculus identity of Step 5. The physics was already loaded in Step 6. The minus sign is inherited directly from the $- P$ ; we never guessed it.

$(\frac{\partial T}{\partial V})_{S}$ — how temperature changes as you make room, at fixed disorder
$- (\frac{\partial P}{\partial S})_{V}$ — minus how pressure changes as you add disorder, at fixed room
these two hard-to-picture quantities are forced equal by the hill's smoothness

PICTURE. The Step-5 square with the physics labels: east-slope $T$ tilting as we move north ( $V$ ) equals north-slope $- P$ tilting as we move east ( $S$ ).

Step 8 — Edge & degenerate cases: when does the picture break?

WHAT. The equal-climb argument needs a smooth hill. Where might that fail?

A kink / phase transition. During boiling, adding heat at fixed pressure grows volume with the temperature stuck flat — the surface develops a sharp crease. At the crease the slope isn't single-valued, so the mixed partials aren't guaranteed equal at that exact edge. Away from the transition (pure phase) the hill is smooth again and the relation holds.
A flat direction (degenerate slope). If a slope is exactly zero — e.g. the ideal gas where $(\frac{\partial U}{\partial V})_{T} = 0$ — the relation still holds; a zero is a perfectly good, smooth value. Zero is not a break.
Boundaries ( $V \to 0$ , $T \to 0$ ). As $T \to 0$ the surface flattens (Third Law: slopes involving $S$ tend to $0$ ); the relation survives as a limit, both sides going to zero together.

WHY. The reader must never meet a scenario we skipped. The rule is simple: smooth ⇒ relation holds; kink ⇒ re-examine only at the kink.

PICTURE. Two mini-surfaces side by side: a smooth dome (relation valid) and a creased ridge at a phase line (relation valid on each smooth face, ambiguous exactly on the crease).

The one-picture summary

Everything above collapses into one diagram: a smooth energy hill $U (S, V)$ , a tiny square on its floor, and the single statement that both climbing orders reach the same corner — which, after labelling slopes $T$ and $- P$ , is the Maxwell relation.

Recall Feynman: the whole walkthrough in plain words

Picture a smooth grassy hill. Your altitude only depends on where you stand, not how you walked there. So if you go a tiny bit east then a tiny bit north, you end up at the same height as going north then east — the hill doesn't play favourites with routes. Now, "how steep is the east slope, and does that steepness itself change as I drift north?" It turns out that number has to match "how steep is the north slope, and does it change as I drift east?" — because both are secretly measuring the same little bit of extra rise at the far corner. That's the whole trick. Then we say: let the hill's height be energy, let east be disorder and north be room. The east slope of energy is temperature; the north slope is minus pressure. Feed those names into the "orders must match" rule, and out pops a fact connecting how temperature reacts to more room with how pressure reacts to more disorder — two things almost impossible to picture directly, tied together for free by the shape of a smooth hill.

Connections

Maxwell relations — derivation from each potential — the parent, all four relations
Equality of mixed partial derivatives (Schwarz theorem) — the Step-5 engine
First and Second Laws of Thermodynamics — source of $d U = T d S - P d V$
Thermodynamic potentials & Legendre transforms — the other three hills
Joule expansion and internal energy — where $(\partial U / \partial V)_{T} = 0$ pays off