Chemical potential μ = (∂G - ∂N)_{T,P}: Visual walkthrough

Step 0 — The three things we are even allowed to change

Figure — Chemical potential μ = (∂G - ∂N)_{T,P}

In the figure the box has three controls: a thermometer (you set $T$ ; heat flow then moves entropy $S$ ), a piston (controls volume $V$ ), and a hatch (controls particle number $N$ ). Note the honest wiring: you turn $T$ directly, and $S$ responds — you cannot grab entropy with your hands. Everything on this page is about what happens when we nudge the hatch — the $N$ dial that the old first law ignored.

Step 1 — Energy stored in the box, and its "response slopes"

WHAT. We say $U$ depends on the three state variables: $U = U (S, V, N)$ .

WHY. Because those are the only things that can change, $U$ can only change if one of them changes. So $U$ is a surface sitting over the three variables.

PICTURE. Look at the figure: $U$ is drawn as a curved sheet. If you walk along it, its steepness in each direction is a slope. Each slope gets a name:

$d U = slope in S T d S - slope in V P d V + slope in N μ d N$

Term by term, right where each sits:

$d U$ — a tiny change in the box's total energy.
$T d S$ — the reversible heat that flows in when entropy changes by $d S$ ; the energy responds with slope $T$ (the temperature). That slope IS what temperature means.
$- P d V$ — nudge the volume by $d V$ ; energy responds with slope $- P$ . The minus sign says: let the box expand ( $d V > 0$ ) and it loses energy by pushing on the world.
$+ μ d N$ — nudge the hatch by $d N$ particles; energy responds with slope $μ$ .

Why this is the honest definition: we did not assume anything extra: we just noticed $U$ has a slope in the $N$ direction and gave it a symbol. Everything else is bookkeeping.

Step 2 — Why $U$ is the wrong potential for a lab

WHAT. We want a new energy bookkeeper whose natural variables are $T, P, N$ .

WHY. So that "freeze $T$ and $P$ , wiggle $N$ " is a thing you can actually do at the bench.

PICTURE. In the figure we trade the "hard" variables $S, V$ for the "easy" variables $T, P$ . This trade has a name and a recipe.

Step 3 — Building Gibbs free energy $G$

WHAT. We defined a new energy $G$ as $U$ with two correction terms.

WHY. Each correction cancels a "hard" variable and hands us an "easy" one. This is the standard Legendre swap — trade a variable for its slope. See Euler Relation and Gibbs-Duhem for the deeper machinery.

PICTURE. In the figure the tall bar $U$ has a chunk $T S$ shaved off the top and a chunk $P V$ stacked on; the leftover height is $G$ .

Step 4 — Differentiate $G$ (just the product rule)

WHAT. Take a tiny change of $G = U - T S + P V$ .

WHY. We want to see which variables $G$ naturally responds to. To find its slopes we differentiate.

Using the product rule on $- T S$ (it becomes $- T d S - S d T$ ) and on $+ P V$ (it becomes $+ P d V + V d P$ ):

$d G = from Step 1 d U product rule on - T S - T d S - S d T product rule on + P V + P d V + V d P$

PICTURE. The figure shows the five little rectangles the product rule creates — two from each product term, plus the $d U$ block waiting to be replaced.

Nothing physical happened yet — this is pure calculus. The magic is the next substitution.

Step 5 — Substitute the first law: the cancellation

WHAT. Replace $d U$ using Step 1, $d U = T d S - P d V + μ d N$ .

WHY. Because $d U$ is the only term still carrying the hard variables $S, V$ . Once we plug it in, watch what dies.

$d G = (A T d S B - P d V + μ d N) - A - T d S - S d T - B + P d V + V d P$

The $T d S$ from the first law cancels the $- T d S$ from the product rule ( $A$ with $- A$ ). The $- P d V$ cancels $+ P d V$ ( $B$ with $- B$ ). This cancellation is the whole point — it is why $G$ was built with exactly those two correction terms.

PICTURE. In the figure, matching amber blocks annihilate in pairs; only three survive.

What survives:

Step 6 — Read off μ by freezing two dials

WHAT. Hold $T$ fixed ( $d T = 0$ ) and $P$ fixed ( $d P = 0$ ).

WHY. That is what the water bath and open atmosphere do for you automatically. Freezing those two variables kills the first two terms of the master differential.

$d G = - S d T^{0} + V d P^{0} + μ d N = μ d N$

Divide by $d N$ :

PICTURE. The figure shows the $G$ surface over the $(N)$ axis with the $T, P$ floor held level; the tangent line's steepness is $μ$ .

Recall

Why does the subscript $T, P$ matter? ::: Because a different slope of the same surface (e.g. holding $S, V$ ) is a different physical quantity; only the $T, P$ -frozen slope of $G$ equals $μ$ .

Step 7 — Edge case: the pure-substance shortcut μ = G/N

WHAT. For a single pure substance, scaling the amount of stuff by any factor $λ$ (e.g. $λ = 2$ means "twice as much", $λ = \frac{1}{2}$ means "half as much") scales $G$ by the same $λ$ , at fixed $T, P$ . Here $λ$ is just a dimensionless scale factor, a plain positive number you multiply the amount by. In symbols:

$G (T, P, λ N) = λ G (T, P, N) .$

WHY it collapses to a per-particle number. If $G$ is strictly proportional to $N$ at fixed $T, P$ , then the slope $\partial G / \partial N$ equals the average $G / N$ — a straight line through the origin has slope = height ÷ base. Euler's theorem (Euler Relation and Gibbs-Duhem) states this cleanly:

$G = N (\frac{\partial G}{\partial N})_{T, P} = N μ ⟹ μ = \frac{G}{N}$

PICTURE. In the figure, $G$ vs $N$ is a straight ray from the origin. The slope $μ$ and the ratio $G / N$ are the same dashed line — that equality is only true for the straight ray of a pure substance. For a mixture the curve bends and you must use partial molar quantities $μ_{i} = (\partial G / \partial N_{i})_{T, P, N_{j \neq = i}}$ .

Step 8 — Degenerate limits: does μ still make sense?

Worked example Three boundary cases, checked

$N \to 0$ (nearly empty box). For an ideal gas $μ = μ^{\circ} (T) + k_{B} T ln (P / P^{\circ})$ , and pressure $\to 0$ as the box empties, so $ln (P / P^{\circ}) \to - \infty$ : $μ \to - \infty$ . Picture: an empty room is infinitely "cheap" to enter — particles rush in. This is why a vacuum sucks.
$d N = 0$ (sealed hatch). The $μ d N$ term vanishes and $G$ changes only through $T, P$ . $μ$ still exists as a slope; it is just not exercised. The old first law is recovered.
Two boxes, $μ_{A} = μ_{B}$ (equilibrium). No slope difference means no drive: $d G = (μ_{A} - μ_{B}) d N_{A} = 0$ for any $d N_{A}$ . Nothing flows. This is diffusive (and chemical) equilibrium: whenever particles can move between connected systems, they stop only when chemical potentials are equal. (The related but distinct question of how a phase boundary line tilts in the $P$ – $T$ plane is the subject of Phase Equilibrium and Clausius-Clapeyron.)

The figure shows all three: the diverging $μ$ as $N \to 0$ (left), a sealed hatch (middle), and a level "μ-landscape" between two connected boxes at equilibrium (right, flat = no flow).

The one-picture summary

This single figure compresses the whole page: start at $d U = T d S - P d V + μ d N$ (top), swap hard variables for easy ones by adding $- T S + P V$ to get $G$ (middle), cancel the matching terms, and read off the surviving $N$ -slope as $μ = (\partial G / \partial N)_{T, P}$ (bottom). The dashed amber arrow traces particle flow always downhill in $μ$ .

(In the diagram, "chem-pot" is the plain-text stand-in for the Greek letter $μ$ used everywhere else on this page — Mermaid labels must stay plain text.)

Recall Feynman: the walkthrough in plain words

We started with a box that has three controls — a temperature knob, a piston, and a hatch for adding particles. Turning the temperature knob makes heat flow, and that heat quietly moves a bookkeeping quantity called entropy (roughly, the logarithm of how many ways the particles can be arranged). The old energy rulebook only tracked heat and squeeze, so we gave the hatch its own slope and named it $μ$ : the energy that rides in with one new particle. But that slope was written in quantities nobody can hold fixed in a lab (entropy, volume). So we built a new energy scorecard, $G$ , by shaving off the heat-energy $T S$ and adding the push-energy $P V$ — a bookkeeping trick that magically swaps the awkward variables for the easy ones, temperature and pressure. When we recomputed the slopes of $G$ , the heat and squeeze terms cancelled perfectly (that cancellation is the whole reason we chose $- T S + P V$ ), leaving three clean terms. Freeze temperature and pressure — like sitting in a lab — and only the particle term survives: $μ$ is the slope of $G$ as you add particles. For one pure ingredient, that slope is just $G$ divided by $N$ , because doubling the stuff doubles $G$ in a straight line. And when two connected boxes reach the same $μ$ , the slopes match, nothing flows — that's equilibrium.