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In natural units, the equilibrium phase-space distribution function iswhere the minus sign gives the Bose-Einstein distribution for bosons and the plus sign gives the Fermi-Dirac distribution for fermions.

For this calculation define the net wavevector gained by the neutron as $\mathbf{Q}={\mathbf{k}}_f-{\mathbf{k}}_i$, the opposite of a common sample-transfer convention. Since $\mathbf{Q}\perp {\mathbf{k}}_i$,The perpendicular neutron recoil and phonon absorption condition therefore requires $\omega (Q)=\hslash Q^2/(2m_n)$. In the supplied periodic dispersion, interpret its $q$ as this transfer $Q$ in the extended-zone representation of a phonon dispersion. Put $x=Qa/2$ and $C=m_{n}av/\hslash \simeq 15.84$. Nonzero intersections obeyThere is no root in $0<x\le \pi /2$: $\sin x\ge 2x/\pi$ there and $C>{\pi }^2/4$. On $(\pi /2,\pi )$ the ratio $x^2/\sin x$ increases strictly from ${\pi }^2/4$ to infinity, so exactly one root occurs. Nor can there be a higher root: $x\ge \sqrt{C}$ gives $x^2\ge C\ge C|\sin x|$, with strict inequality at the possible boundary. For $\pi \le x\le \sqrt{C}$, use $|\sin x|\le x-\pi$; $x^2-C(x-\pi )>0$ throughout this short interval, since its minimum there is $C(1+\pi -\sqrt{C})>0$. Thus the only nonzero magnitude satisfiesSolving the exact dispersion, which is permitted instead of the linear approximation, givesA straight chord between the zone edge and the next zone centre gives $\omega \simeq 4v/a-2vQ/\pi$. Its intersection solves $(\hslash /2m_n)Q^2+(2v/\pi )Q-4v/a=0$, yielding $Q\simeq 1.006\times 10^{10}{\mathrm{m}}^{-1}$ and $f\simeq 0.508\mathrm{THz}$ as a less accurate estimate.

The reduced-zone phonon wavevector is $q_{\mathrm{red}}=Q-2\pi /a=-1.873\times 10^9{\mathrm{m}}^{-1}$ along the chosen positive transfer direction. The difference is a reciprocal-lattice vector, not an additional independent phonon outside the first zone. At low temperature the absorption probability contains the Bose-Einstein distribution factor $n_B(\omega )=[e^{\hslash \omega /k_{B}T}-1{]}^{-1}$, which tends to zero. Here $\hslash \omega /k_B\simeq 27.6\mathrm{K}$, so this absorption channel is strongly suppressed well below that scale. elastic scattering and kinematically allowed spontaneous phonon creation need not disappear.

In the harmonic approximation, sum over wavevector $\mathbf{q}$ and phonon branch $s$:The occupation factor is the Bose-Einstein distribution with zero phonon chemical potential; $\frac{1}{2}\hslash \omega$ is the zero-point energy. A temperature-independent zero-point term does not contribute to the heat capacity in this fixed-frequency approximation.

The Debye model treats a solid's low-frequency lattice vibrations as acoustic phonons in an elastic continuum. There are three polarizations per wavevector, one longitudinal and two transverse, with approximately linear dispersion relations. A cutoff is chosen to preserve the total $3N$ vibrational modes of $N$ atoms. The phonon number itself is not fixed; mode occupations follow the Bose-Einstein distribution with zero chemical potential.

For one polarization of speed $v_s$, counting wavevector states in a spherical shell gives $g_s(\omega )=V{\omega }^2/(2{\pi }^2{v}_s^3)$. Define an average Debye speed by $3/v_D^3=1/v_L^3+2/v_T^3$. The total density of states can then be writtenThe Debye wavevector is $k_D=(6{\pi }^{2}N/V{)}^{1/3}$, the Debye frequency is ${\omega }_D=v_D{k}_D$, and the Debye temperature is ${\Theta }_D=\hslash {\omega }_D/k_B$.

Omitting the temperature-independent zero-point energy, the vibrational energy and resulting Debye heat capacity areAt low temperature the upper limit can be extended to infinity. The integral is $4{\pi }^4/15$, givingOnly the long-wavelength acoustic modes are appreciably populated, explaining the Debye T-cubed test. At high temperature every mode contributes $k_B$ by equipartition theorem, so $\boxed{C_V\to 3N{k}_B}$, the Dulong-Petit law.

The model succeeds especially at low temperature, where long-wavelength elasticity is reliable. It replaces real anisotropic branches by a simple continuum and cannot reproduce detailed optical phonon branches or every intermediate-temperature feature. Metals also have an electronic heat capacity approximately proportional to $T$; defects, magnetic excitations or a phase transition can add other terms. Experiments often measure heat capacity at constant pressure, which need not equal the harmonic model's $C_V$ exactly.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

Put $z=e^{\beta \mu }$, with $\beta =1/(k_{B}T)$. For a single quantum state of energy $ϵ$, dilute Maxwell–Boltzmann statistics gives $\ln {\Xi }_{ϵ}\simeq z{e}^{-\beta ϵ}$. For bosons this is the leading expansion of $-\ln (1-z{e}^{-\beta ϵ})$; the occupation must be small for that expansion. Thus the grand potential isWith the ground-state energy set to zero, a dilute classical gas requires $e^{\beta \mu }\ll 1$, giving negative $\mu$. This sign is relative to the chosen energy zero. For an ideal Bose gas, $\mu$ approaches the ground energy from below at condensation, so it approaches zero with this convention.

The printed degeneracy formula requires $n_x,n_y,n_z=0,1,2,\dots$, rather than strictly positive integers. With zero-point energy omitted these are the appropriate oscillator quantum numbers. The stars and bars count of three nonnegative integers summing to $n$ givesThere is one ground state at $n=0$. If the integers were strictly positive, the count would instead be $\left(\genfrac{}{}{0ex}{}{n-1}{2}\right)$ for $n\ge 3$, inconsistent with both the printed formula and its energy reference.

For large $n$, $d_n\simeq n^2/2$. Consecutive energy shells are separated by $\hslash \omega$, giving the continuum density of statesUse this when $k_{B}T\gg \hslash \omega$, and integrate the single-state grand potential over energy:because the integral is $2/{\beta }^3$. Particle number follows by differentiating at fixed temperature and trap frequency:These are the high-temperature expressions for the classical grand potential in a harmonic trap.

Extrapolating to $\mu =0$ gives the requested transition scaleIt is an order-one estimate because Maxwell-Boltzmann occupations cease to be accurate near condensation. The Bose-Einstein distribution gives the more precise ideal-trap result $N_{\mathrm{ex}}^{\max }=\zeta (3)(k_{B}T/\hslash \omega {)}^3$, and hence $T_c=(\hslash \omega /k_B)[N/\zeta (3){]}^{1/3}$, about $0.94$ times the estimate. Excess particles populate the ground state after the excited-state capacity saturates: Bose condensation in a three-dimensional harmonic trap.

The observation of Bose-Einstein condensation in rubidium in 1995 implemented laser precooling followed by radio-frequency evaporation from a magnetic trap. Imaging the released cloud revealed a narrow anisotropic component within a broader thermal population. This provides an experimental connection to the population-counting argument; the measured expansion also reflects interactions and imaging resolution, not just the ideal-trap density of states.^[ref]

An optical phonon often has approximately constant energy $\hslash {\omega }_0$ over the relevant wavevector range. Its absorption or emission therefore moves the electron between energy surfaces separated by this finite amount, with the phonon supplying the associated crystal momentum modulo a reciprocal vector. The last panel shows emission between two points on a parabolic conduction band.

Emission from a band with a lower edge requires at least $\hslash {\omega }_0$ of available kinetic energy and an unoccupied final state. For nondegenerate hot carriers this produces a threshold and can lead to rapid energy loss in successive phonon emissions. Absorption is weighted by the Bose-Einstein distribution $n_B({\omega }_0)$, while emission has the factor $n_B+1$; absorption is strongly suppressed at $k_{B}T\ll \hslash {\omega }_0$. In a metal, Pauli blocking also prevents an electron near the Fermi energy from simply dropping into the occupied Fermi sea.

A periodic harmonic chain with $N$ masses $m_0$, spacing $a$ and nearest-neighbour spring constant $K$ hasHamilton's equations give $m_0{\ddot{u}}_j=K(u_{j+1}+u_{j-1}-2u_j)$. The discrete Fourier transform $u_j=N^{-1/2}{\sum }_k{Q}_k{e}^{ikja}$, $p_j=N^{-1/2}{\sum }_k{P}_k{e}^{ikja}$, with $k=2\pi l/(Na)$ in one Brillouin zone, diagonalizes the quadratic form:The harmonic-chain normal-mode reality condition means $Q_k^{†}=Q_{-k}$, not that $Q_k$ and $Q_{-k}$ are independent real coordinates. For canonical quantization, impose $[u_j,p_l]=i\hslash {\delta }_{jl}$, equivalently $[Q_k,P_{k^{\prime }}]=i\hslash {\delta }_{k,-k^{\prime }}$. For $k\ne 0$ setThen the quantum harmonic chain is a set of independent phonon oscillators:The centre-of-mass zero mode of a harmonic chain is the free coordinate $X=N^{-1}{\sum }_j{u}_j$, with $[X,P_{\mathrm{tot}}]=i\hslash$; it cannot be normalized by dividing by $\sqrt{{\omega }_0}$. Retaining it giveswith the corresponding momentum expansion and uniform term $P_{\mathrm{tot}}/N$. The nonzero modes alone give the site commutator $i\hslash ({\delta }_{jl}-1/N)$; the uniform term supplies the missing $1/N$. Each oscillator occupation is a nonnegative integer, and adding one phonon costs $\hslash {\omega }_k$ and crystal momentum $\hslash k$ modulo a reciprocal-lattice vector. ^[ref]

In the continuum limit of a harmonic chain, $\rho =m_0/a$, $Y=Ka$ andThe wave equation and linear low-$k$ dispersion relation follow. The lattice gives an ultraviolet cutoff; the acoustic approximation does not reproduce the zone-edge flattening.

With the centre of mass fixed, the oscillator partition function and Bose-Einstein distribution areFor a long chain, above its finite-size level spacing and below the zone-edge scale, integrating the two acoustic propagation directions gives $C/(Na)=\pi k_{\mathrm{B}}^{2}T/(3\hslash c)$. At high temperature each retained oscillator instead contributes $k_{\mathrm{B}}$ to the heat capacity.

Measure single-particle energy from the quantum harmonic oscillator ground state energy $\hslash \omega$, and write $\tilde{\mu }={\mu }_{\mathrm{physical}}-\hslash \omega \le 0$. The Bose-Einstein distribution and the supplied density of states giveThis excited-state population increases with the chemical potential and has the finite saturation valueAny additional particles must occupy the ground state: this is two-dimensional harmonic-trap Bose condensation. The critical temperature and the condensate fraction are thereforeHere $N$ is the total particle number, the integral of the inhomogeneous number density over the trap; there is no fixed uniform area in which $N$ itself is a number density. Equivalently $N\kappa ={\pi }^{2}m(k_B{T}_c{)}^2/(6{\hslash }^2)$ relates the trap population to the critical temperature. Harmonic confinement changes the low-energy density of states and permits this saturation in two spatial dimensions.

These formulas are leading large-$N$ results with $k_B{T}_c\gg \hslash \omega$. A sharp transition belongs to the trapped-gas thermodynamic limit $N\to \infty$, $\omega \to 0$ at fixed $N{\omega }^2$; a finite trap has finite-size rounding. In particular, finite $N_0$ requires $\tilde{\mu }=-k_{B}T\log (1+1/N_0)<0$, approaching zero in the condensation limit.

In three dimensions an ideal Bose gas can saturate its excited-state population at chemical potential zero, with the remaining particles forming a Bose-Einstein condensate. The transition follows from the Bose-Einstein distribution integrated against its density of states; below the ideal-gas transition, $n_0/n=1-(T/T_c{)}^{3/2}$. Condensation alone does not supply the finite Landau critical velocity of an interacting superfluid: the ideal quadratic spectrum has ${\inf }_{p}E(p)/p=0$.

By contrast, a weak attractive interaction near a Fermi surface produces the Cooper instability and the BCS gap equation. Its logarithmic enhancement permits pairing at arbitrarily weak attraction in the ideal model. The low-temperature excitations have $E=\sqrt{{\xi }^2+{\Delta }^2}$, and in weak coupling $k_B{T}_c=(e^{{\gamma }_E}/\pi )\Delta (0)\simeq 0.567\Delta (0)$. This transition is controlled by pair formation when phase fluctuations are small; deeply bound pairs instead motivate a Bose-condensation description with their own density and effective mass.

For the given ellipsoidal dispersion, set $q_x=k_x$, $q_y=k_y$ and $q_z=k_z/\sqrt{\gamma }$. Then $\epsilon ={\hslash }^2{q}^2/(2m)$ while the integration measure is $d^{3}k=\sqrt{\gamma }{d}^{3}q$. Thus the anisotropic ideal-gas density of states per unit volume and per internal state isThe full continuum momentum range is assumed; this is an anisotropic three-dimensional gas, not yet a discrete stack of two-dimensional bands.

At the Bose transition, substitute $x=\epsilon /(k_{B}T)$:The integral is $\Gamma (3/2)\zeta (3/2)\simeq 2.315$. Solving for the Bose-Einstein condensation temperature for an anisotropic effective mass givesThe heavier $z$-direction effective mass increases the available excited-state count and lowers the condensation temperature at fixed number density.

For a balanced two-spin Fermi gas, let $n_F$ denote total fermion number density and take $N_s$ to be the single-spin density of states used in the conventional gap equation. Counting the filled ellipsoid givesIf a spin-summed density of states is used instead, the coupling convention in the supplied integral must change accordingly; inserting a factor two into only one side would change the exponent incorrectly.

For ${\epsilon }_c\ll {\epsilon }_F$, replace the density of states across the interaction window by $N_s({\epsilon }_F,\gamma )$. Writing ${\lambda }_0=|g|N_{s0}$, the zero-temperature integral becomesand thereforeThe second expression additionally requires ${\lambda }_0{\gamma }^{1/3}\ll 1$. The explicit density dependence is ${\lambda }_0=|g|m(3{\pi }^2{n}_F{)}^{1/3}/(2{\pi }^2{\hslash }^2)$. This mass-anisotropy enhancement of a weak-coupling BCS gap initially raises the pairing scale, while the ideal Bose condensation scale decreases as ${\gamma }^{-1/3}$.

Large anisotropy weakens transport and coherence in the heavy direction. The separation of pair formation and phase ordering then matters, so a large mean-field gap need not imply a large actual $T_c$. A qualitative sketch has $T_c$ initially rising with the pairing scale and eventually turning down when coherence is the smaller scale. If preformed fermion pairs behave as bosons, use $n_B=n_F/2$ and pair mass about $2m$; in this continuum model this gives $k_B{T}^{\mathrm{BEC}}\simeq 0.218{\epsilon }_{F0}{\gamma }^{-1/3}$.

The turnover cannot be located reliably by taking the weak-coupling formulas to $\gamma \to \infty$: ${\epsilon }_F$ falls, ${\lambda }_0{\gamma }^{1/3}$ grows, the narrow-window assumption fails and the pair chemical potential and mass require a self-consistent treatment. The smooth curve in the figure is explicitly an illustrative interpolation of two extrapolated scales, not such a calculation. If actual decoupled two-dimensional layers are approached, ordinary ideal-gas finite-temperature condensation is absent and interacting phase order requires a different analysis, including the Berezinskii–Kosterlitz–Thouless transition when appropriate.

The cuprates make the distinction between pairing and phase coherence relevant: they are layered and have anisotropic motion, but their strong correlations, lattice band structure and predominantly d-wave superconductivity are outside this attractive continuum $s$-wave model. Anisotropy by itself is therefore not an explanation or quantitative fit of cuprate transition temperatures.

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.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.