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Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the energy isThe periodic boundary conditions imposeHence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every Ising spin, givingThis is an odd antiferromagnetic Ising ring: geometrical frustration forces one defect, whose location is unrestricted. The ground-state degeneracy includes both its possible locations and global spin inversion symmetry. The formula also includes the self-bond case $N=1$ under the stated Hamiltonian convention.

For $J>0$ each bond minimizes its energy when its two Ising spins are opposite, giving bond energy $-J$. An even ring allows every bond to do this simultaneously, including the closing bond required by the periodic boundary conditions. Its two ground states areThe two configurations are related by spin inversion symmetry. There is no geometrical frustration.

Set $K=\beta J$. The transfer matrix for the one-dimensional Ising model assigns a Boltzmann factor to each bond. In the Ising spin order $(+1,-1)$, it isMultiplying these bond factors and summing every Ising spin implements the canonical ensemble. The periodic boundary conditions close the product into a trace:The eigenvectors $|+⟩=(1,1{)}^T/\sqrt{2}$ and $|-⟩=(1,-1{)}^T/\sqrt{2}$ form an orthonormal basis, with eigenvaluesThe negative second eigenvalue encodes alternating correlations. Defining $q={\lambda }_{-}/{\lambda }_{+}=-\tanh K$, the exact partition function and total Helmholtz free energy areThe parity-dependent term must be retained for finite $N$; in particular it nearly cancels the leading term of $Z_N$ at low temperature when $N$ is odd.

Represent insertion of one Ising spin by $\Sigma =\mathrm{diag}(1,-1)$. In the eigenbasis of $T$, this matrix swaps $|+⟩$ and $|-⟩$. Its diagonal entries in that eigenbasis vanish, giving the exact one-point correlation functionEquivalently, configurations pair with their opposites under spin inversion symmetry, with the same Boltzmann factor and opposite $S_1$.

Two insertions separated by $n$ bonds give the two-point correlation functionThus the exact finite antiferromagnetic Ising-ring correlation isThe two numerator terms represent the two arcs between the sites. The formula has $C_N(0)=1$ and $C_N(n)=C_N(N-n)$, as required by the periodic boundary conditions. Since the one-point correlation function is zero, this is also the connected correlation function. At $K=0$, every Ising spin is independent, so $C_N(n)=0$ for distinct sites and $C_N(0)=1$; powers with exponent zero are understood as one.

Near a spatially extended Hopf bifurcation, a critical oscillatory normal mode has a slowly varying complex amplitude. For a selected travelling-wave branch, a method of multiple scales expansion takes the formHere the distance from onset is of order ${\epsilon }^2$, $v_0$ is the critical eigenvector, and $v_g$ is the group velocity. The comoving coordinate removes the first spatial derivative of the envelope. Projection at the first resonant order gives $A_T=\rho A+D_0{A}_{XX}-G_0|A{|}^{2}A$. Spatial translation and the temporal phase of the Hopf oscillation give a constant-phase symmetry $A\mapsto e^{i\theta }A$, which permits the cubic term $|A{|}^{2}A$ but forbids a generic $A^2$ term. The coefficients are complex because the envelope has both growth and frequency detuning. For a supercritical branch with positive real diffusion, real rescalings and a uniform phase rotation give the complex Ginzburg–Landau equationA subcritical Hopf branch requires higher saturation terms instead. If spatial reflection makes opposite travelling waves critical together, two coupled amplitude equations are required initially; the scalar equation applies to a selected single-wave branch when its competing amplitude is stable.

Distant boundaries can still determine which travelling pattern is observed. They can select its phase and wavenumber, reflect a travelling disturbance, inject a competing wave, or let a growing wave packet leave the domain. In particular, amplification in a comoving frame can be convective wave-packet instability rather than absolute wave-packet instability at a fixed laboratory point. A finite-domain outcome then need not reflect intrinsic bulk modulation dynamics. Periodic boundary conditions are the appropriate idealization when studying bulk instabilities without end selection or reflected-wave forcing: they eliminate physical end layers and make the allowed envelope Fourier modes discrete. They do not assert that real distant boundaries are always negligible. Choose a periodic length compatible with the carrier, fix the phase winding, and compare perturbations within that same periodic problem. On an open physical domain the group velocity and actual boundary conditions must be restored.

The optimum carrier wavenumber corresponds to the spatially uniform envelope $A_f=e^{-icT}$, often called a flat state. Its phase is arbitrary. Write $A=e^{-icT}(1+r)e^{i\varphi }$ with real small amplitude and phase disturbances. To first order,A Fourier mode $e^{\lambda T+ikX}$ therefore has matrix and dispersion relationThe trace is negative. The amplitude mode at $k=0$ has eigenvalue $-2$, while the phase mode is neutral because constant phase shifts preserve the solution. Put $D=1+bc$. For every nonzero real $k$, the determinant is positive if $D\ge 0$, whereas $D<0$ makes it negative for $0<k^2<-2D/(1+b^2)$. ThusThis is the Benjamin-Feir stability condition. At $D=0$, nonzero modes still decay, but the leading long-wave decay is fourth order rather than diffusive. On a periodic envelope interval of length $L$, the smallest nonzero wavenumber is $2\pi /L$, so strict stability of all allowed nonzero modes requiresA sufficiently short periodic domain can therefore exclude the unstable band even when the infinite-domain Benjamin-Feir stability condition fails.

The amplitude is damped while the phase is slow. For a slowly varying phase, the leading slaved amplitude is $r=-\frac{1}{2}\{b{\varphi }_{XX}+({\varphi }_X{)}^2\}+\cdots$. Substitution in the phase equation gives the diffusion term $D{\varphi }_{XX}$ and the nonlinear frequency shift $(c-b)({\varphi }_X{)}^2$. The latter also follows directly from a constant phase gradient: an envelope with gradient $q$ has amplitude squared $1-q^2$ and frequency $c+(b-c)q^2$. Constant phase symmetry rules out undifferentiated $\varphi$; translation invariance gives constant coefficients. The leading comoving Ginzburg-Landau equation also has spatial parity, so odd linear derivatives are absent at this order. Its neutral phase eigenvalue expands asThis coefficient follows by inserting a power series for $\lambda$ in the quadratic dispersion relation. Close to $D=0$, it becomes $E=(1+b^2)/2>0$ at leading order. The long-wave phase reduction of the complex Ginzburg-Landau equation is consequentlyThe fourth derivative regularizes negative phase diffusion. Terms omitted here have higher order in the long-wave expansion. First-order advection has been removed by the group-velocity frame; higher odd derivatives in a more general travelling-wave problem are beyond this leading amplitude approximation.

Take $D=-d$ with $d>0$ small. At the instability boundary $c=-1/b$, so $g=-(1+b^2)/b$ is nonzero. Set $\xi =\sqrt{d/E}X$, $\tau =d^{2}T/E$, and $\varphi =(d/g)\Phi$. At leading order the phase dynamics become the potential Kuramoto-Sivashinsky equation,With $u=-2{\Phi }_{\xi }$ this isThe variable $u$ is a phase-gradient disturbance, not the original physical field. Periodicity of the phase fixes ${\int }_0^{P}ud\xi =0$ for the zero-winding branch; the Kuramoto-Sivashinsky equation conserves this mean. The scaled period is $P=L\sqrt{d/E}$.

At fixed period $P$, the flat phase has growth rates ${\lambda }_n=k_n^2-k_n^4$, $k_n=2\pi n/P$. Its first loss of stability occurs at $P=2\pi$. To find what replaces it, put $k=2\pi /P$ and consider $P$ just above $2\pi$. After choosing an origin by translation, expand a small steady modulation as $u=a\sin (k\xi )+b_2\sin (2k\xi )+\cdots$. Projection of $-u{u}_{\xi }$ onto the first two harmonics yieldsSince ${\lambda }_2=-12$ at onset, this harmonic is slaved to $b_2=k{a}^2/(2{\lambda }_2)+\cdots$. The fundamental amplitude obeysThe primary periodic bifurcation of the Kuramoto-Sivashinsky equation is therefore supercritical. For $0<P-2\pi \ll 1$,The radial eigenvalue of this nonzero branch is $-2{\lambda }_1+O({\lambda }_1^2)<0$; all higher harmonics remain damped and the modulation's spatial phase is neutral. Thus the small finite-amplitude modulation is stable in the sense of orbital stability, at fixed mean and fixed period, despite the flat state's instability. The two signs of $a$ represent translations of the same modulation. In the potential formulation the spatial mean of $\Phi$ advances at $⟨({\Phi }_{\xi }{)}^2⟩$: the original wave acquires a nonlinear frequency correction rather than requiring a time-independent absolute phase.

For larger $P$, further linear modes become active and the small-amplitude argument no longer determines stability. A periodic modulation $\bar{u}$ must then be tested with the operator $v_{\tau }=-v_{\xi \xi }-v_{\xi \xi \xi \xi }-{\partial }_{\xi }(\bar{u}v)$ on mean-zero periodic disturbances. Its translation mode is always neutral, and secondary oscillatory or more complicated modulation can occur. Moreover, stability against perturbations of the same period does not prove stability against disturbances of a larger period; those require a separate sideband or Floquet theory calculation. The controlled conclusion near the first threshold is the existence of a stable small modulation family, not universal stability of all periodic Kuramoto-Sivashinsky solutions.

Because $⟨B⟩=0$, a periodic primitive $H$ exists. Fix its additive constant by $⟨H⟩=0$. Integrating the conserved-field equation once in space gives $H_T=\sigma (H_X+\delta A^2{)}_X$: the possible integration constant vanishes on taking the spatial mean. Treat the spatial average as the inner product defining functional derivatives. Integration by parts, with no boundary terms under periodic boundary conditions, givesThus the conserved-field real amplitude equation is a gradient flow ascending $V$, or descending $-V$, with positive diagonal mobilities. Along a classical solution,For the mixed term put $P=⟨A^4{⟩}^{1/2}$ and $Q=⟨H_X^2{⟩}^{1/2}$. The Cauchy-Schwarz inequality and a completed square giveConsequently, for $\beta =1-\delta >0$,The negative leading quartic coefficient proves the upper bound. The stronger pointwise square completionis also useful: a superlevel set $V\ge V(0)$ bounds the amplitude in $L^4$, its spatial derivative in $L^2$, and $B=H_X$ in $L^2$, because a quartic polynomial with negative leading coefficient dominates its lower-order terms.

The dynamics dissipates toward the steady-state set rather than sustaining nonconstant periodic motion. Indeed $V$ increases to a finite limit, and the time integral of $⟨A_T^2+H_T^2/(2\delta \sigma )⟩$ is finite. For global regular trajectories with the usual parabolic precompactness, every limit point is stationary, by the LaSalle invariance principle. A nonconstant periodic or recurrent orbit would force a strictly increasing $V$ to return to its original value and is impossible. Boundedness of a monotone functional alone does not prove convergence to one particular equilibrium; isolation of the limiting equilibrium or an additional convergence theorem supplies that stronger claim.

For a periodic potential, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting unitary operators, so within each energy eigenspace one can choose simultaneous translation eigenstates. Their eigenvalues have modulus one and can be written $e^{i\mathbf{k}\cdot {\mathbf{a}}_i}$. Translation by any lattice vector then givesThis proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.

The reciprocal lattice consists of vectors $\mathbf{g}$ with $\mathbf{g}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$. Changing $\mathbf{k}$ to $\mathbf{k}+\mathbf{g}$ and $u$ to $e^{-i\mathbf{g}\cdot \mathbf{r}}u$ gives the same physical wavefunction with a periodic new $u$. The eigenvalue problem on one cell has a discrete set of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\mathbf{k}+\mathbf{g})=E_n(\mathbf{k})$. A reciprocal-lattice fundamental domain, conventionally a Brillouin zone, therefore labels inequivalent wavevectors.

Periodic boundary conditions in physical volume $\mathcal{V}$ give one allowed wavevector per reciprocal-space volume $(2\pi {)}^3/\mathcal{V}$. Including the two spin states, there are $2\mathcal{V}{d}^{3}k/(2\pi {)}^3$ states; per unit physical volume the count is the printed $2d^{3}k/(2\pi {)}^3$. Each occupied state carries charge $-e$ and velocity ${\hslash }^{-1}{\nabla }_k{E}_n$, soFor a full band, the integral of ${\nabla }_k{E}_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence a full band carries zero electric current.

Let $h=\Delta x$, $k=\Delta t$ and $\mu =k/h$. For the exact solution $u(x,t)=F(x+t)$ of the advection equation, the three left-hand samples are translations by $(\mu -1)h$, $\mu h$ and $(\mu +1)h$ relative to $(x_m,t_n)$; the right-hand samples are translations by zero and $h$.

Denote their coefficients by $a=\mu (1+\mu )/6$, $b=(2-\mu )(1+\mu )/3$, $c=(2-\mu )(1-\mu )/6$, $d=(2-\mu )/3$ and $e=(1+\mu )/3$. Taylor expansion compares momentswhere $0^0=1$ in the zeroth moment. Direct calculation givesThus the unscaled exact-solution residual isFor fixed nonzero $\mu$, dividing by the time step yieldsThe scheme is therefore third order under fixed-Courant refinement, with an unscaled one-step defect of order four. Stability then gives third-order accumulated error for smooth data with periodic boundary conditions. Calling the unscaled defect fourth order is a different convention, not fourth-order approximation of the differential equation.

At $\mu =-1,1,2$, the amplification factor in (b) becomes exactly $e^{i\mu \theta }$: the numerical step translates the grid data by that integer number of cells and is exact for this constant-speed advection equation. At $\mu =0$ it is the identity, a zero-time-step limit. These are exact special shifts, rather than merely a cancellation of one term of the Taylor expansion. This is the implicit advection scheme with exact integer shifts.

Choose a spatial interval with periodic boundary conditions and a uniform grid, so no separate inflow boundary closure of a difference scheme is required. The discrete Fourier transform is a unitary operator in the discrete L2 norm. Every circulant stencil is diagonal in that transform, and a one-step Fourier multiplier is a contraction exactly when its modulus is at most one for every grid frequency. For a mesh-uniform statement, use all $\theta \in [-\pi ,\pi ]$, the limiting frequency range.

A Fourier mode $e^{im\theta }$ giveswith the coefficients from (a). Squaring moduli and putting $v=\cos \theta$ yieldsThis has the required nonnegative sign for every frequency exactly whenOne must also check that the implicit equation is solvable, rather than divide by a vanishing symbol. HereFor $\mu \ne 1/2$, a zero would have to occur at zero or $\pi$; the latter occurs only at $(1\pm \sqrt{3})/2$, both outside the displayed stable intervals. At $\mu =1/2$, $D=3/4+(1/4)\cos \theta \ge 1/2$. Thus there is no denominator zero in the stable range. For fixed such $\mu$, continuity on the compact frequency interval gives a uniformly bounded inverse, and $|G|\le 1$ proves $\Vert U^n{\Vert }_{2,h}\le \Vert U^0{\Vert }_{2,h}$ for all steps.

Outside these intervals the displayed difference is negative at every nonzero frequency where $D\ne 0$, so $|G|>1$. On periodic refining meshes there are frequencies approaching any selected such value. Their amplification is bounded away from one and is repeated $O(h^{-1})$ times over a fixed physical interval when $k=\mu h$, violating a mesh-uniform stability bound. At the two singular parameters the implicit operator itself fails on meshes containing the Nyquist mode. These observations prove necessity as well as sufficiency.

For the physical convention $k>0$, $h>0$, the result is $0<\mu \le 1$ or $\mu \ge 2$, with zero admitted as the identity limit. The negative intervals describe the algebraic contraction result for signed Courant numbers. In particular, the additional large positive interval must not be discarded by imposing the usual explicit-scheme Courant–Friedrichs–Lewy condition: this scheme is implicit. Stability at large fixed Courant number is not a guarantee of a small error constant.

The homogeneous fixed point must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular spatially homogeneous equilibriumThis is the basally forced quadratic activator-inhibitor model. The linearization of its reaction terms gives the Jacobian matrixPut $a=(r-1)/(r+1)$, so $-1<a<1$. Its trace and determinant are $a-q$ and $q$, and its eigenvalues areSince $q>0$, both eigenvalues have negative real part exactly when $q>a$. HenceFor $r\le 1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with ${\lambda }_{\pm }=\pm i\sqrt{a}$; the nonlinear qualification at this threshold is proved below.

Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The diffusion terms subtract $k^2\mathrm{diag}(1,p)$ from $J$. With $s=k^2$ the mode trace and determinant areFor a strictly stable homogeneous state, $q>a$ makes $\tau (s)<0$ for all $s\ge 0$. A nonzero Fourier mode can therefore grow only if $\Delta (s)<0$, in which case its two real eigenvalues have opposite signs. The determinant parabola has its minimum at $s_{\ast }=(pa-q)/(2p)$, and that minimum is $q-(pa-q{)}^2/(4p)$. The two-species Turing criterion is consequently derived here asThese force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt{p}$ givesThus the inhibitor in a reaction-diffusion system must diffuse sufficiently fast. Inside this region, the growing wavenumbers obeyAt onset, the minimum touches zero and the two determinant roots coalesce. ThereforeHere $p$ is evaluated on the onset surface. These are the Turing threshold of a basally forced quadratic activator-inhibitor model and its continuum critical wavelength. On a finite or domain with periodic boundary conditions an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the diffusion criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.

For completeness, the weak focus at the Hopf threshold of a basal activator-inhibitor model resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and set $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $X=x$, $Y=(hy-ax)/\omega$. Expanding the rational reaction term to cubic order giveswhereWrite $X=\varrho \cos \theta$, $Y=\varrho \sin \theta$ and evaluate these homogeneous polynomials on the unit circle. If $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, and $C=\cos \theta G_2-\sin \theta F_2$, division of the radial and angular equations givesThe angular integral of $A$ is zero because it changes sign under $\theta \mapsto \theta +\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero return map term is cubic. Using $⟨{\cos }^4\theta ⟩=3/8$, $⟨{\cos }^2\theta {\sin }^2\theta ⟩=1/8$ and the corresponding sixth moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map isBecause $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore nonlinearly asymptotically stable, with weak rather than exponential attraction, despite its neutral linear eigenvalues. The strict inequalities above describe the ordinary diffusion-driven instability from a strictly linearly stable homogeneous state.

Since the endpoints represent the same point of the tube, both density and diffusive flux must match. For a positive diffusion coefficient $\lambda$, the periodic boundary conditions areFor separation of variables, periodic spatial modes solve $X^{\prime \prime }+n^{2}X=0$ and have time factors $e^{-\lambda n^{2}t}$. The constant mode has no decay. The periodic diffusion on a circle solution is thereforewhere the Fourier coefficients areThe expression for $a_n$ includes $n=0$.

The specified profile is even, so $b_n=0$. Its mean follows from ${\int }_0^{\pi }(\pi -x{)}^{2}dx={\pi }^3/3$, giving $a_0=2{\pi }^2/3$. For $n\ge 1$, two integrations by parts giveConsequentlyAll nonconstant normal modes decay for $\lambda >0$, so the density becomes uniform:The conserved total amount is ${\int }_{-\pi }^{\pi }udx=2{\pi }^3/3$. The initial cusp at zero is compatible with an L2 space solution and is immediately smoothed by diffusion. At $\lambda =0$ there would be no relaxation, so positivity is needed for this long-time conclusion.

spin normalization in two-dimensional free-electron state counting. Treat the mobile electrons as independent particles in a constant interior potential, with parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m)$, and impose confinement or periodic boundary conditions. electron–electron interactions and the detailed periodic ionic potential are neglected, while Fermi statistics are retained. Each spatial state occupies area $(2\pi {)}^2/A$ in two-dimensional wavevector space. For spin degeneracy $g_s$,The printed coefficient $Am/(2\pi {\hslash }^2)$ is the per-spin density of states. Unpolarized spin-one-half electrons have $g_s=2$, doubling it. Actual graphene near its Dirac points has approximately linear dispersion, so its energy-dependent density of states is not this parabolic-model result; the plane geometry alone does not establish the free-electron dispersion.^[ref]

Occupied disk and Fermi energy. At zero temperature fill the lowest states up to a disk of radius $k_F$. State counting givesThus the physical spin-degenerate result is ${ϵ}_F=\pi {\hslash }^{2}N/(Am)$. If $N$ instead counts one spin species, the per-spin convention gives $2\pi {\hslash }^{2}N/(Am)$. Each occupied wavevector can contain one electron of each allowed spin, never two in the same complete one-particle state.

Occupation probability and chemical potential of a two-dimensional ideal Fermi gas. A single fermion state permits occupation zero or one. In a grand canonical ensemble their relative weights are $1$ and $e^{-\beta (ϵ-\mu )}$. Normalizing givesThe chemical potential is fixed by the total particle number. Here the total density of states $D=g_{s}Am/(2\pi {\hslash }^2)$ is constant above the band bottom. Integrating the occupation gives $N=D{k}_{B}T\ln (1+e^{\mu /(k_{B}T)})$, and thereforeThe approach is exponentially close in this ideal two-dimensional constant-density-of-states model; a three-dimensional $T^2$ correction should not be inserted here. A narrow interval of width of order $k_{B}T$ around $\mu$ is thermally smeared.

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

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.