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The chemical potential of an ideal single-occupancy lattice gas with occupied-site internal partition function $z$ and surface coverage $\theta$. The ratio $\theta /(1-\theta )$ accounts for the entropy of distributing particles among occupied and vacant adsorption sites.

If $x_{\alpha }$ counts amphiphile molecules belonging to aggregates of size $\alpha$, the corresponding dimensionless aggregate count is $n_{\alpha }=x_{\alpha }/\alpha$. Its ideal translational chemical potential contains $\ln n_{\alpha }$. Confusing these two distributions drops a factor of $\alpha$ and changes both normalization and the chemical-potential interpretation.

For a constant effective density of states and attractive coupling $\lambda$, let the linearized BCS theory gap equation integrate $\tanh (\xi /(2k_B{T}_c))/(2\xi )$ over $-\sigma <\xi <\theta$, with positive cutoffs. Its even integrand makes the result the mean of two positive-cutoff logarithms. When $k_B{T}_c\ll \sigma ,\theta$, $1/\lambda \simeq \frac{1}{2}\ln (1.13\theta /(k_B{T}_c))+\frac{1}{2}\ln (1.13\sigma /(k_B{T}_c))$, giving the displayed geometric mean of cutoffs.

If a band bottom truncates the lower side, $\sigma =\min (\mu ,\theta )$ for positive chemical potential $\mu$. The weak-coupling limit at fixed $\mu /\theta >0$ must not be interchanged with an uncontrolled $\mu \to 0$ extrapolation: the lower logarithm requires $\mu \gg k_B{T}_c$ when $\mu <\theta$.

The continuous evolution, in suitable attractive Fermi systems, from weakly paired overlapping Cooper pairs to tightly bound composite bosons undergoing Bose-Einstein condensation. Pair formation, phase coherence and the chemical potential must be determined consistently across the crossover. A weak-coupling gap-temperature formula with fixed density of states is not uniformly valid through the molecular limit.

The BCS quasiparticle spectrum is the positive excitation energy obtained by diagonalizing the paired particle-hole block. Here the weak coupling shell model has $\Delta \ll {\Omega }_D$: the minimum lies inside the pairing shell, and states outside it cost more. To derive it for a uniform isotropic real $\Delta >0$, measure the normal-state energy from the chemical potential, ${\xi }_k={ϵ}_k-\mu$, and use the same sign convention as the BCS coherence factors:Its eigenvalues are ${\epsilon }_{\pm }=\pm E_k$, where $E_k=\sqrt{{\xi }_k^2+{\Delta }^2}$. The positive $E_k$ is the cost of creating a quasiparticle above the BCS ground state. The lower branch $-E_k$ is its particle-hole partner in this doubled matrix description, not an excitation that lowers the ground state energy without bound.

The location and value of the minimum follow directly:The unique minimum as a function of $\xi$ is $E(0)=\Delta$, at the normal-state Fermi surface. On an isotropic three-dimensional Fermi surface it occurs for every direction of $k$, rather than at a unique point of momentum space. Near that minimum,Farther away, $E(\xi )\simeq |\xi |$ when $\Delta \ll |\xi |\ll {\Omega }_D$. For the illustrative $\lambda =0.25$ example, a state at $\xi =\Delta$ costs $\sqrt{2}\Delta \simeq 1.03644\mathrm{meV}$ rather than the minimum $0.732871\mathrm{meV}$.

For a nonideal chemical solution, write its thermodynamic activities on a mole fraction basis as $a_A=x{\gamma }_A$ and $a_B=(1-x){\gamma }_B$. Use composition-independent pure-component reference chemical potentials at the same temperature and pressure:The activity coefficients describe departure from the ideal solution under that reference convention.^[ref]

At fixed temperature and pressure, insert these expressions into the Gibbs-Duhem equation. The ideal solution terms cancel because $xd\ln x+(1-x)d\ln (1-x)=dx-dx=0$. What remains isThus two independent fits to measured activity coefficients can each look smooth while failing to describe one possible Gibbs free energy. This is a necessary consistency condition; satisfying it alone does not establish experimental accuracy or thermodynamic stability.

For an original algebraic example, suppose at one fixed temperature and pressure that $\ln {\gamma }_A=\alpha (1-x{)}^2$, with dimensionless constant $\alpha$. The constraint requires $d\ln {\gamma }_B/dx=2\alpha x$. With the pure-B normalization ${\gamma }_B(0)=1$, integration givesFor $\alpha =1/2$ and $x=1/4$, the derivatives of the two logarithms are $-3/4$ and $1/4$; their weighted sum is $-3/16+3/16=0$. Choosing $\ln {\gamma }_B=\alpha x$ instead would give a weighted sum $3/16$ at that composition, contradicting the Gibbs-Duhem equation. These are invented model functions, with no claim to fit a particular mixture.

The fixed-temperature, fixed-pressure restriction matters. Along a path with changing temperature or pressure, the general equation contains $SdT-Vdp$, and the reference chemical potentials and $RT$ also change. Across phase coexistence, apply the relation separately within each smooth thermodynamic phase; do not differentiate a discontinuous switch between phases as if it were a single composition curve.

The Boltzmann distribution assigns probabilities to individual microstates in a canonical ensemble, a statistical ensemble for thermal contact. The system exchanges energy with a thermal reservoir at fixed positive absolute temperature $T$, while its particle number $N$ and external parameters, such as volume, stay fixed. The thermal reservoir is large enough that these exchanges barely change its temperature; the coupling permits thermal equilibrium but contributes negligible interaction energy. All the microstates considered have the same $N$. No chemical potential term is needed in their weights; fixed $N$ does not imply that the chemical potential is zero.

For a finite discrete set of microstates with energies $E_i$, writeHere $k_B$ is the Boltzmann constant, $\beta$ is the inverse temperature, $e^{-\beta E_i}$ is a Boltzmann factor, and $Z$ is the partition function. The Boltzmann factor is a weight; dividing by the partition function makes it a probability. ^[ref]

The useful surprise is that a fixed temperature still permits thermal fluctuations in the system's energy. The Boltzmann distribution predicts their probabilities and expected values without tracking every microscopic exchange. Its exponential function also makes rare excited microstates sensitive to temperature: an energy gap much larger than $k_{B}T$ gives strong Boltzmann suppression, while warming increases their relative populations. A population ratio can therefore reveal a gap or serve as a thermometer, provided thermal equilibrium and the relevant state degeneracies are known.

Historically, Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) presents the canonical ensemble in Chapter IV. His equations 90–92 give an exponential function of energy and its normalization, in classical phase space. This classical formulation exhibits the same weighting principle. ^{[ref] [ref]}

Boltzmann suppression is the exponential reduction of an equilibrium population when creating it costs energy much greater than the temperature. For a nonrelativistic species with negligible chemical potential, the leading factor is $e^{-m/T}$, multiplied by its momentum phase-space volume. It concerns the equilibrium abundance; freeze-out can preserve a larger population than the instantaneous equilibrium value at later times.

A nonrelativistic species with $g$ states and negligible chemical potential has equilibrium density $g(mT/(2\pi ){)}^{3/2}{e}^{-m/T}$. If particle-antiparticle annihilation maintains equilibrium after the species becomes massive, its eventual freeze-out population can be much smaller than a relativistically decoupled population. The nonrelativistic annihilation rate and any asymmetry determine the final abundance.

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

In a bosonic coherent-state path integral, coherent-state resolutions between imaginary-time evolution factors turn a normal-ordered grand Hamiltonian $K=H-\mu N$ into its complex symbol. Coherent-state overlaps supply the first-order temporal term. A thermal trace makes the fields periodic in imaginary time. The chemical potential couples to number, and interactions retain their normal-ordering convention.

For equal-mass conduction and valence edges at $0,-E_g$ and fixed $\mu =0$, $n=N(T)F_{1/2}(0)$ and $p=N(T)F_{1/2}[-E_g/(k_{B}T)]$, where $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Other states pinning the chemical potential can compensate the unequal carrier charges; imposing intrinsic charge neutrality in addition would change the problem.

At chemical equilibrium, a reversible chemical reaction has equal forward and reverse reaction rates, so its macroscopic composition stays constant. Individual molecules can keep changing identity: the reaction has not stopped. This dynamic balance is the definition described by the ^[ref]. Equal reaction rates also need not mean equal concentrations. A small population converting rapidly can balance a large population converting slowly.

This distinction gives two useful questions. Which composition is favored at the specified temperature and other constraints? Thermodynamics answers through chemical potentials and an equilibrium constant. How quickly does a prepared mixture approach that composition? Chemical kinetics answers through reaction rates. Knowing the destination does not determine the journey's duration. A composition that barely changes during observation may reflect extremely slow chemical kinetics, so constancy alone does not establish chemical equilibrium.

Consider the idealized elementary chemical reactionAssume a well-mixed closed system at fixed temperature and volume, with first-order law of mass action rates in both directions and no competing chemical reactions. Write $a=[\mathrm{A}]$ and $b=[\mathrm{B}]$ for concentrations in $\mathrm{M}=\mathrm{mol}/\mathrm{L}$. The total $a+b=C=1\mathrm{M}$ is conserved because each conversion replaces one molecule by one molecule. This conservation is special to the chosen stoichiometry.

The forward and reverse reaction rates, measured as conversions per unit volume per unit time, are $J_f=k_{f}a$ and $J_r=k_{r}b$. HenceSetting this derivative to zero givesWith $a(0)=1\mathrm{M}$ and $b(0)=0$, solving the ordinary differential equation yieldsAt $t=5\mathrm{s}$, these are approximately $0.3515\mathrm{M}$ and $0.6485\mathrm{M}$. Initially $J_f=0.3M/s$ and $J_r=0$. At chemical equilibrium, both are $0.075M/s$, while the net reaction rate $J_f-J_r$ is zero. The persistent exchange is visible in the lower panel even after the upper panel has nearly flattened.

Why does chemical equilibrium also involve a free energy minimum? For a closed system at fixed temperature and pressure, without imposed non-expansion work, stable equilibrium minimizes Gibbs free energy over compositions allowed by conserved elemental inventories. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19. This is constrained Gibbs minimization for chemical equilibrium: freely minimizing over unrelated species amounts would change the inventory. At fixed temperature and volume, the appropriate potential is Helmholtz free energy. In either setting, varying an allowed extent of reaction gives the corresponding free energy slope ${\sum }_i{\nu }_i{\mu }_i$, with molar chemical potentials and signed stoichiometric vector entries. An interior equilibrium has zero slope; stability requires a minimum. For A converting to B, this says ${\mu }_{\mathrm{B}}={\mu }_{\mathrm{A}}$.

The beauty of this connection is that molecular exchange, a macroscopic composition, and a geometric minimum describe the same balance. In the ideal concentration model, let $x=b/C$ and choose the standard molar chemical potential difference consistent with $K=3$. Up to a composition-independent constant, the fixed-volume Helmholtz free energy then obeyswhere $n$ is the conserved total amount of substance and $R$ the molar gas constant. Its derivative is $\ln [x/(1-x)]-\ln 3$, vanishing at $x=3/4$. Its second derivative is $1/x+1/(1-x)>0$ inside $0<x<1$, so this balance is the unique minimum. The mixing entropy terms explain why the favored composition still contains A.

For a nondegenerate spin-one-half ideal gas with energies $\mp {\mu }_{B}B(\mathbf{r})$, the two component chemical potentials are $k_{B}T\ln (n_{\pm }/n_Q)\mp {\mu }_{B}B$. Diffusive and spin-conversion equilibrium require one spatially constant common chemical potential, provided spin conversion is allowed. Hence the displayed densities give total number density $n=2n_Q{e}^{\beta \mu }\cosh (\beta {\mu }_{B}B)$ and magnetization $M=2{\mu }_B{n}_Q{e}^{\beta \mu }\sinh (\beta {\mu }_{B}B)=n{\mu }_B\tanh (\beta {\mu }_{B}B)$. The high-temperature fixed-density limit is $M\simeq n{\mu }_B^{2}B/(k_{B}T)$, a Curie law. At fixed chemical potential the density itself varies with temperature, so the unnormalized magnetization need not have a pure $1/T$ dependence.

The chemical potential measures how a thermodynamic potential changes when matter is added under specified constraints. It plays for matter exchange the role that temperature plays for heat exchange: a difference supplies a direction, and equality characterizes thermodynamic equilibrium when the corresponding exchange is allowed. This connects diffusion, phase coexistence and chemical equilibrium through one quantity.

For a single component, the particle convention iswhere $U$ is internal energy, $N$ counts particles, $S$ is entropy and $V$ is volume. This partial derivative holds entropy and volume fixed; it measures an incremental change rather than the average energy per particle. See David Tong, Statistical Physics, section 1.4.

In chemistry the molar convention instead uses amount of substance $n$, measured in moles:Here $G$ is Gibbs free energy, $T$ is temperature, $p$ is pressure and $N_{\mathrm{A}}$ is the Avogadro constant. The units are $\mathrm{J}$ per particle versus $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use $\mu$ for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials areThey describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating $G=U-TS+pV$ and using $d{N}_i=N_{\mathrm{A}}d{n}_i$; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing $U$ by $U+{ϵ}_{0}N$ shifts ${\mu }_{\mathrm{particle}}$ by ${ϵ}_0$. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

Rigid compartments impose a different constraint: fixed volume rather than fixed pressure. With Helmholtz free energy $F=U-TS$,This follows immediately by differentiating $F=U-TS$ in the chemical potential differential above. It expresses the same chemical potential through the thermodynamic potential appropriate to fixed temperature and volume.

Consider a closed system consisting of two rigid compartments of equal volume $V$, maintained at a common fixed temperature $T$. A valve permits exchange of the same ideal gas; the partition stays in place. Initially $n_A=9\mathrm{mol}$ and $n_B=3\mathrm{mol}$, with conserved total amount of substance $n_{\mathrm{tot}}=12\mathrm{mol}$. These are macroscopic moles, modeled continuously, not twelve individual particles. By the ideal gas law, $p_B/p_A=n_B/n_A=1/3$, so initiallyAt $300\mathrm{K}$ with the stipulated $R$, this is approximately $-2740.311\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, favoring transfer from A to B.

At fixed $T$ and equal $V$, integrating the ideal-gas chemical potential gives the total Helmholtz free energy, up to a term independent of the allocation,Here $n_{\mathrm{ref}}$ is a positive reference amount of substance; any omitted common term linear in $n_A+n_B$ is constant along this exchange. Let $x=n_A/n_{\mathrm{tot}}$ be the dimensionless fraction in A. The relative Helmholtz free energy then takes the particularly simple formIts derivative along the allowed transfer and its second derivative areThus the Helmholtz free energy is a strictly convex function of the allocation for positive amounts. Its unique minimum has $x=1/2$, or $n_A=n_B=6\mathrm{mol}$, and equal chemical potentials. Initially its excess above this minimum is $n_{\mathrm{tot}}RT[\frac{3}{4}\ln (3/2)+\frac{1}{4}\ln (1/2)]\simeq 3915.474\mathrm{J}$ at $300\mathrm{K}$. Relaxation decreases the total Helmholtz free energy by this amount in the model; the final pressures are also equal because both compartments have the same volume.

The beautiful connection is that the direction of matter exchange and the slope of a free energy curve are the same calculation. The entropy gained by redistributing the ideal gas makes equal allocation favorable here. As with chemical equilibrium, a free energy minimum identifies the favored state, while transport properties determine how quickly it is reached. The fixed-temperature, fixed-pressure Gibbs free energy criterion is summarized in 16.4 Free Energy; the rigid experiment instead uses Helmholtz free energy.

Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.

For a composition free energy $F=\int [f(\varphi )+\kappa |\nabla \varphi {|}^2/2]$, the local chemical potential is the functional derivativeAt phase coexistence, exchanges between the bulk phases require equal chemical potentials. Mechanical equilibrium additionally requires equal pressure for a flat interface, or a Young–Laplace equation pressure jump for a curved interface.

The occupied-to-vacant probability ratio of the adsorption-site grand partition function is $\theta /(1-\theta )=z_s{e}^{\beta ({\mu }_s+ϵ)}$. Taking its logarithm yields the displayed chemical potential. binding energy and vibrational energies must share a consistent reference so that zero-point energy is not counted twice.

For a pure ideal gas at temperature $T$ and pressure $p$,where $R$ is the molar gas constant and $p^{\circ }$ specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace $p$ by its partial pressure. More generally write ${\mu }_{\mathrm{m},i}={\mu }_{\mathrm{m},i}^{\circ }+RT\ln a_i$ using its thermodynamic activity $a_i$ and a consistent standard state. For an ideal dilute chemical solution, $a_i=c_i/c^{\circ }$ with reference concentration $c^{\circ }$; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at $T=300\mathrm{K}$, with $p_A=2p^{\circ }$ and $p_B=p^{\circ }$. Stipulate $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. ThenHalving the pressure gives the same decrement whether starting from $2p^{\circ }$ or $20p^{\circ }$, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.