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An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

For a single quantum state, the equilibrium occupations areBoth reduce to $e^{(\mu -ϵ)/(k_{B}T)}$ when $ϵ-\mu \gg k_{B}T$.

For one three-dimensional quadratic branch, the density of states per unit volume isWith ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, changing integration variable to $x=ϵ/(k_{B}T)$ givesA degeneracy factor multiplies both densities if additional equivalent branches are present. The first expression is the noncondensed density: a macroscopically occupied ground state must be counted separately.

Bose-Einstein condensation occurs when the excited states can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total density,For field-induced triplon condensation, write $\Delta =\nu n-\eta (H-H_g)$, so $ϵ(k)={\hslash }^2{k}^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic chemical potential is zero, the Bose distribution requires $\Delta \ge 0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.

For the triplon condensation field scaling, the critical density at onset is $n_T=\zeta (3/2)/{\lambda }_T^3$. ThereforeThe low-temperature phase boundary rises from $(T,H)=(0,H_g)$; the condensed phase lies on the higher-field side.

For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the condensate density in a shifted-parabola triplon model, added condensate density keeps the minimum at zero: $\nu (n_0+n_T)=\eta (H-H_g)$. At fixed temperature this giveswith $n_0=0$ below onset. At zero temperature it becomes $n_0=\eta (H-H_g)/\nu$ within that model.

This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed phase. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{{ϵ}_k^0({ϵ}_k^0+2g{n}_0)}$, with ${ϵ}_k^0={\hslash }^2{k}^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. The gap closes and the condensate density grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model. A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.^[ref]

Differentiate the Euler relation for a single-component simple compressible system:The thermodynamic fundamental relation for the same system is $dU=TdS-pdV+\mu dN$. Subtraction givesDividing by $N$, with $s=S/N$ and $v=V/N$, proves the Gibbs-Duhem equationThe underlying Euler relation assumes extensive bulk thermodynamics and no additional work coordinates. The result says that the intensive quantities are constrained; the chemical potential cannot be varied independently of temperature and pressure within one homogeneous phase.

Bose-Einstein condensation is macroscopic occupation of the lowest one-particle state. It is a change in occupation statistics, not ordinary liquid formation and not a requirement that all atoms occupy one spatial point. For ideal bosons,Set the lowest one-particle energy to zero. Then $\mu \le 0$, and the population of excited states in a uniform three-dimensional box isHere $g_s(z)={\sum }_{j\ge 1}{z}^j/j^s$. As $\mu$ approaches zero, the excited-state population saturates at $V\zeta (3/2)/{\lambda }^3$. Any excess particles occupy the ground state. In the thermodynamic limit,Thus the onset occurs when the thermal de Broglie wavelength is comparable to the mean interparticle spacing. At finite $N$, the chemical potential remains slightly below the ground-state energy; setting it equal to zero is a thermodynamic-limit statement.

The 1995 rubidium experiment used magnetic confinement and evaporative cooling of atoms. Absorption images after trap manipulation revealed a narrow, anisotropically expanding component superposed on the broader thermal distribution; the new component grew as cooling proceeded. The thermal component supplied an internal comparison through the same imaging optics. Trap decompression and release geometry mattered, and finite imaging resolution and interactions limited direct comparison with an ideal-gas ground state.^[ref]

A narrow cloud alone is not sufficient evidence. One should check its emergence across a controlled cooling sequence, atom-number and temperature calibration, the thermal background and the predicted expansion anisotropy. These are tests of an occupation transition rather than a simple focusing artifact. The box formula above is not the exact critical-temperature or condensate-fraction formula for the trapped experiment: the trap changes the density of states, and interactions also matter. Condensation and superfluid response are related but distinct properties; the former definition is ground-state occupation.

The Fermi-Dirac distribution gives the mean occupancy of a single fermion state:Its maximum occupancy of one expresses the Pauli exclusion principle. A spatial orbital with two spin states can hold two electrons because they are different single-particle states. At zero temperature, the occupation is a step at the Fermi energy $E_F$; at low nonzero temperature, occupation changes appreciably only in an energy shell of width of order $k_{B}T$ about the chemical potential.

Consequently, most conduction electrons cannot absorb small thermal energies into already occupied states. Only a fraction of order $T/T_F$ participates, where $T_F=E_F/k_B$. For a normal metal with a smooth density of states $g(E_F)$, including both spin states and using the total-system normalization, the Sommerfeld expansion at fixed particle number givesand thereforeFor a three-dimensional free electron gas, $g(E_F)=3N/(2E_F)$, so $C_{\mathrm{el}}=({\pi }^2/2)N{k}_B(T/T_F)$. This is much smaller than a classical equipartition estimate at $T\ll T_F$. The chemical potential must adjust with temperature to maintain $N$; treating it as fixed while differentiating an energy integral gives the wrong fixed-$N$ calculation in general.

The lattice adds a Debye contribution $C_{\mathrm{ph}}={\beta }_D{T}^3$ at temperatures well below the Debye temperature. HenceA plot of measured $C/T$ against $T^2$ can separate the electronic intercept from the phonon slope. calorimetry must account for the heat capacity of the sample holder, heat leaks, thermal equilibration and thermometer calibration. A magnetic impurity contribution or a superconducting transition can invalidate this two-term fit; the prediction is for the normal metallic state in an appropriate low-temperature interval. Differences from a free-electron coefficient can reveal quasiparticle effective-mass enhancement or nontrivial electronic structure rather than a failure of Fermi statistics.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

For a complete single-particle state, the Fermi-Dirac distribution isHere ${\mu }_{\mathrm{ch}}$ denotes the chemical potential, distinct from the magnetic-moment symbol in (b). At zero temperature it approaches the Fermi energy $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an energy interval of order $k_{B}T$ about that edge.

The Fermi-edge heat-capacity counting argument uses the fact that only order $g(E_F)k_{B}T$ electrons can be rearranged because states well below the edge are blocked by the Pauli exclusion principle. Each active excitation carries energy of order $k_{B}T$, so $U(T)-U(0)\sim g(E_F)(k_{B}T{)}^2$ and $C_{V,e}\propto T$. For a smooth density of states, the Sommerfeld expansion givesHere $g$ is the total single-particle density of states, including spin. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=({\pi }^2/2)N{k}_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice heat capacity or a description across a superconducting transition.^[ref]

Measure energies from the single-particle ground state, and assume the three-dimensional ideal Bose gas in the large-particle-number limit. The harmonic-trap chemical-potential pinning discussed here uses a chemical potential satisfying ${\mu }_{\mathrm{ch}}\le 0$. On cooling, it approaches zero; below the condensation temperature it remains at zero to leading thermodynamic order. Above that temperature it is negative. The detailed negative-chemical-potential branch shape is determined by the density of states; the figure uses the harmonic trap analysed below. The finite-size condensate chemical potential stays slightly below zero rather than being exactly pinned there.

A Bose-Einstein condensate has macroscopic occupation of a single quantum state. A dilute atomic gas needs low temperature and high phase-space density while remaining metastable against ordinary liquid or solid formation. Laser cooling followed by evaporative cooling of atoms in a weak trap can achieve the required regime. For the trapped ideal gas the quantitative threshold is derived below; a uniform ideal gas has $n{\lambda }_T^3=\zeta (3/2)$ at onset. An experimental signature is a narrow low-momentum component appearing within the broader thermal distribution after time-of-flight expansion, as observed in magnetically trapped rubidium-87.^[ref]

Spinless cubic-box wave-vector counting. Impose periodic boundary conditions on a cube of side $L$. The allowed vectors are $\mathbf{k}=(2\pi /L)(n_x,n_y,n_z)$ with integer components. Each state occupies $(2\pi /L{)}^3$ in wave-vector space. For a spinless particle, counting the spherical shell givesThere is no spin multiplicity. This is continuum shell counting; a finite box has discrete levels. Using $ϵ={\hslash }^2{k}^2/(2m)$ gives the free-particle density of states per unit energy and volume

Semiclassical density of states in a central potential. At position $\mathbf{r}$, the available kinetic energy is $ϵ-U(r)$. The local-density approximation integrates the preceding free-particle result over the classically allowed region:This form assumes the allowed radial region is the connected interval $0\le r\le r_{ϵ}$. A nonmonotonic potential would require integrating all allowed shells.

For the isotropic harmonic trap, $r_{ϵ}=\sqrt{2ϵ/(m{\omega }^2)}$. Substituting $r=r_{ϵ}\sin \vartheta$ givesConsequentlyThe exact oscillator ground energy is $3\hslash \omega /2$ above the potential minimum. The harmonic-trap energy-zero convention subtracts that constant, while the displayed density of states is the leading semiclassical continuum expression, neglecting the finite-level corrections produced by that shift. The discrete ground state must be counted separately.

Condensate population. Below the transition, use ${\mu }_{\mathrm{ch}}\simeq 0$ for the excited states. The saturation of harmonic-trap excited states follows from the supplied integral:where $\zeta$ is the Riemann zeta function and $\zeta (3)\simeq 1.202$. Setting $N_{\mathrm{ex}}(T_0)=N$ gives the harmonic-trap critical-temperature prefactor:Above $T_0$ there is no macroscopic ground-state fraction in this approximation. For completeness, above the transition the quantum gas fugacity $z=e^{{\mu }_{\mathrm{ch}}/(k_{B}T)}<1$ satisfies $N=(k_{B}T/\hslash \omega {)}^3{\mathrm{Li}}_3(z)$; the polylogarithm ${\mathrm{Li}}_3$ therefore defines the plotted negative branch. A finite trapped gas has rounded behaviour, weak interactions shift the threshold, and the continuum thermal population is inaccurate when $k_{B}T$ is comparable to or below $\hslash \omega$. The power law is intended for the large-$N$ condensed regime, not an exact discrete-spectrum formula at arbitrarily small temperature.^[ref]

For the ideal dilute Maxwell-Boltzmann gas implicit in a single-particle partition function, indistinguishability gives $Z_N=Z_1^N/N!$. With Stirling's approximation,Differentiate at fixed $T,V$:This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.

Equilibrium constraints and the Gibbs minimum. The open-system equilibrium contacts allow exchange of heat, volume and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' chemical potentials across the contacts. These conditions follow by maximizing total entropy under conserved total energy, volume and species amounts.

The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir entropy change givesFor candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus the Fixed-particle-number Gibbs minimum selects stable equilibrium at fixed $T,p,N$, with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.

The simple-compressible-system first law givesFor a homogeneous extensive single-component phase, the Single-component Gibbs Euler relation follows from $U=TS-pV+{\mu }_{\mathrm{ch}}N$, yielding $G={\mu }_{\mathrm{ch}}N$. For a mixture, the corresponding identities are $G={\sum }_i{\mu }_i{N}_i$ and $dG=-SdT+Vdp+{\sum }_i{\mu }_{i}d{N}_i$. Surface effects or other nonextensive contributions require additional terms.^[ref]

Reaction equilibrium. Let $d\xi$ be an allowed reaction extent for the forward process, so $d{N}_A=-d\xi$ and $d{N}_B=2d\xi$. At fixed $T,p$,For an interior equilibrium allowing both reaction directions, the coefficient vanishes:At fixed $T,V$, minimizing $F$ gives the same condition. The reaction formula-unit conservation gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This conservation law replaces fixed total particle number for the reacting mixture.

Using the stipulated single-particle functions,The number-based dissociation equilibrium ratio is thereforeThis number-based quantity has a factor of volume; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on temperature alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.

Define the dissociated fraction of original A formula units by $\alpha =N_B/(2M)$. Then $N_A=M(1-\alpha )$ and $N_B=2M\alpha$, soThe dissociation fraction versus conserved density follows because the left side increases monotonically from zero to infinity on $0<\alpha <1$. At fixed $V,T$, increasing $M$ decreases $q$, hence decreases dissociation. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high density the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.

In the grand canonical ensemble, the system exchanges both energy and particles with a reservoir at fixed temperature $T$ and chemical potential $\mu$. The probability of a particular many-body microstate $i$, with energy $E_i$ and particle number $N_i$, is its Boltzmann weight $e^{-\beta (E_i-\mu N_i)}$ divided by the grand canonical partition function $\Xi$, where $\beta =1/(k_{B}T)$. The denominator normalizes ${\sum }_i{p}_i=1$; a degeneracy is accounted for by counting distinct microstates, rather than changing an individual state's probability.

For noninteracting particles, specify a microstate by the occupation numbers $n_r$ of one-particle levels of energy ${ϵ}_r$. Then $E-\mu N={\sum }_r({ϵ}_r-\mu )n_r$, so the Gibbs probability factorizes between levels. Put $x_r=e^{-\beta ({ϵ}_r-\mu )}$.

For fermions, the Pauli exclusion principle permits $n_r=0,1$ in each complete one-particle state, giving ${\Xi }_r=1+x_r$. Its mean occupancy isthe Fermi-Dirac distribution.

For bosons, $n_r=0,1,2,\dots$, so the geometric sum is ${\Xi }_r=1/(1-x_r)$ when $x_r<1$. Differentiating that sum givesthe Bose-Einstein distribution. Finite-system convergence requires $\mu <{ϵ}_{\min }$; condensation is the limiting approach to the lowest energy with macroscopic ground-state occupation. Both quantum distributions reduce to $e^{-\beta ({ϵ}_r-\mu )}$ when occupation is small. The quantities derived here are mean occupancies of states, rather than normalized probabilities over the set of one-particle energies.

Excited-state saturation. Consider first a uniform three-dimensional noninteracting gas of identical spinless bosons in a large box. Set the one-particle ground energy to zero. The Bose-Einstein distribution is $\bar{n}(ϵ)=[z^{-1}{e}^{\beta ϵ}-1{]}^{-1}$, with quantum gas fugacity $z=e^{\beta \mu }\le 1$. For dispersion $ϵ=p^2/(2m)$, the excited density of states isUsing the thermal de Broglie wavelength ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, integration givesThe polylogarithm is ${\mathrm{Li}}_s(z)={\sum }_{\ell \ge 1}{z}^{\ell }/{\ell }^s$. Its finite value at $z=1$ implies a maximum excited population $V\zeta (3/2)/{\lambda }_T^3$, with $\zeta (3/2)\simeq 2.612$. Once this is less than $N$, the excess particles occupy the ground state macroscopically. This is Bose-Einstein condensation, rather than occupation of a state by merely one or a few atoms.

Transition and condensate fraction. Setting the saturation number equal to $N$ gives the homogeneous ideal Bose gas condensation temperatureBelow this temperature, $\mu$ approaches zero from below in the thermodynamic limit, and particle conservation givesAbove $T_c$ there is no macroscopic condensate and $\mu <0$ adjusts to keep $N$ fixed. At finite $N$, the chemical potential remains slightly below the ground energy and the transition is rounded.

Energy and the role of geometry. The excited-state internal energy isThe zero-energy condensate contributes no term to this energy in the ideal model. Below $T_c$,Condensation depends on low-energy state counting: for a uniform quadratic-dispersion gas, a finite-temperature bulk condensate requires spatial dimension greater than two. A trap changes the density of states and therefore the temperature law; a three-dimensional harmonic trap gives a cubic, rather than $3/2$, condensate-fraction exponent. Finally, an ideal-gas condensate should not automatically be called a superfluid: its free quadratic spectrum has zero critical speed by the Landau criterion. Interactions matter for the robust collective-flow properties of real dilute condensates.

Quanta of collective motion. An elementary excitation is a low-energy departure from the many-body ground state that can often be treated as a quasiparticle. Near an ordered equilibrium, small collective motions form normal modes. Quantization turns a harmonic mode of frequency ${\omega }_{\mathbf{k}}$ into bosonic excitations with energy $\hslash {\omega }_{\mathbf{k}}$. Their number is not an independently conserved thermodynamic variable in equilibrium, so their chemical potential is zero andHere $\nu$ labels branches, and $U_0$ contains the ground-state contribution, including phonon zero-point energy where appropriate. The approximation is useful while interactions and damping of the excitations are weak enough that the modes remain identifiable.

Phonons. A phonon is a quantum of an elastic lattice-vibration mode. Long-wavelength acoustic phonon branches have $\omega \simeq c_{\nu }k$, with one longitudinal and two transverse polarizations in a three-dimensional monatomic solid. A multi-atom primitive cell can also support optical phonon branches, generally with nonzero frequency at $k=0$. Linear acoustic dispersion gives $D(ϵ)\propto {ϵ}^2$, so the energy integral has $U-U_0\propto T^4$ and the heat capacity obeys the Debye $T^3$ law. In the Debye model,At temperatures high compared with all phonon energies but below structural changes, the classical harmonic limit gives approximately $C_V=3N{k}_B$. Optical modes are frozen out at temperatures well below their gaps.

Spin waves. A spin wave is a spatially varying collective precession of ordered spins; its quantum is a magnon. In a simple isotropic ferromagnet, a long-wavelength gapless branch has energy ${ϵ}_{\mathbf{k}}\simeq D_s{k}^2$, where the spin stiffness $D_s$ has units $\mathrm{J}{\mathrm{m}}^2$. State counting gives $D(ϵ)\propto {ϵ}^{1/2}$. For one isotropic branch in three dimensions,This exponent is specific to a gapless quadratic ferromagnetic branch. Long-wavelength antiferromagnetic spin waves can be linear and instead give a $T^3$ heat-capacity law in three dimensions. Magnetic anisotropy or an applied field can open a gap, causing activated behaviour at still lower temperatures. Near a magnetic ordering transition, a dilute noninteracting spin-wave description is no longer adequate.

These examples illustrate the low-temperature energy of gapless bosonic excitations: for $ϵ\propto k^s$ in $d$ dimensions, $D(ϵ)\propto {ϵ}^{d/s-1}$, hence $U-U_0\propto T^{1+d/s}$ and $C_V\propto T^{d/s}$. The exponent records both the spatial dimension and the dispersion, making calorimetry a probe of elementary excitations.

Component chemical potentials. Define the per-spin quantum concentration $n_Q={\lambda }_T^{-3}=(m{k}_{B}T/(2\pi {\hslash }^2){)}^{3/2}$. For a classical component with one-particle potential energy $U_{\sigma }(\mathbf{r})$, the local Helmholtz free energy density is $k_{B}T{n}_{\sigma }[\ln (n_{\sigma }/n_Q)-1]+n_{\sigma }{U}_{\sigma }$. Its derivative with respect to number density gives the full local chemical potential, including the magnetic energy. With the up state defined as the lower level,The Bohr magneton ${\mu }_B$ is a magnetic moment; ${\mu }_{B}B$ is an energy. The factor two for the two spin states is not already included in $n_Q$.

Total equilibrium chemical potential. If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-energy cost. Thusnot their sum or zero. The total particle number, or a reservoir, fixes this common value; the supplied information does not specify a numerical chemical potential. A useful local expression is $\mu =k_{B}T\ln [n/(2n_Q\cosh x)]$, with $x={\mu }_{B}B/(k_{B}T)$, whose density dependence is found below.

Why it is uniform and common. Moving a small number $dN$ of one species from ${\mathbf{r}}_1$ to ${\mathbf{r}}_2$ changes the free energy by $[{\mu }_{\sigma }({\mathbf{r}}_2)-{\mu }_{\sigma }({\mathbf{r}}_1)]dN$. If this difference were nonzero, a transfer could lower the free energy. Equilibrium under diffusion therefore requires each full chemical potential to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes free energy by $({\mu }_{\uparrow }-{\mu }_{\downarrow })dN$, proving equality when spin relaxation is permitted. This is chemical equilibrium of a classical spin gas. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying magnetic field is balanced by a spatially varying density, not by a varying full chemical potential.