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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.

Assume an air-filled or evacuated lossless electromagnetic waveguide, and a pulse bandwidth narrow enough to use its carrier's local group velocity. The longitudinal dispersion relation isIt givesHere $f_c/f=20/40=1/2$, soThe phase velocity exceeds $c$, but the pulse envelope travels at the subluminal group velocity; a moving constant-phase surface does not by itself transport information. Their product is $v_{\mathrm{p}}{v}_{\mathrm{g}}=c^2$. A broad pulse can distort because this waveguide is dispersive.

For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency $f_c=10\mathrm{GHz}$ and carrier frequency $f=30\mathrm{GHz}$, $\sqrt{1-(f_c/f{)}^2}=\sqrt{8}/3$. ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check $v_p{v}_g=c^2$ follows directly from the dispersion relation. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.

Fixed endpoints require $k_n=n\pi /L$, with $n=1,2,\dots$. Writing $c_s=\sqrt{T/\rho }$,For equal amplitudes $A$, the two requested profiles areThe first is asymmetric, with an interior zero at $x=2L/3$; the second has its interior node at $L/2$. For two-mode string recurrence, orthogonality means the whole initial profile recurs only when both cosine coefficients equal one. Since ${\omega }_2=2{\omega }_1$,

The weak-stiffness string frequency shift follows by expanding the supplied dispersion relation:Higher normal modes acquire larger fractional frequency shifts. Let $\eta =\alpha k_1^2/c_s^2\ll 1$ and retain ${\omega }_1^{(0)}=c_s{k}_1$ for the original ideal frequency. At the originally defined time $t_1=\pi /(2{\omega }_1^{(0)})$,Thus the $t=0$ sketch is unchanged, whereas the original quarter-period sketch gains a small negative fundamental component. If $t_1$ is redefined using the new fundamental frequency, its fundamental component is exactly zero and the second-mode change there is only second order. Both conventions agree that generic later shapes change because of relative phase drift.

The new frequency ratio isAn exact recurrence now requires ${\omega }_{1}t=2\pi q$ and ${\omega }_{2}t=2\pi p$ for integers $p,q$. If $r=p/q$ in lowest terms, the first recurrence is $2\pi q/{\omega }_1$; for irrational $r$ there is no exact finite recurrence. Near-recurrences remain possible. Positive stiffness alone does not prove that the ratio is irrational.

Substitution into the wave equation gives ${\partial }_x^2\psi +{\partial }_y^2\psi =-(k_x^2+k_y^2)\psi$ and ${\partial }_t^2\psi =-{\omega }^2\psi$. Therefore ${\omega }^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.

The element's mass is $\rho dxdy$, and its speed is $-\omega A\sin (k_{x}x)\sin (k_{y}y)\sin (\omega t)$. Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are $a/2$ and $b/2$, soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy $\rho c^2|\nabla \psi {|}^2/2$ and use the dispersion relation.

For $b=4a/5$, the two lowest modes are $(p,q)=(1,1)$ and $(2,1)$:The fundamental has one antinode at $(a/2,b/2)$ and no interior nodal line. The next mode has a node along $x=a/2$ and opposite-sign antinodes at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a node.

The square fundamental has $f=c\sqrt{2}/(2\beta a)$. Equating it to $f_{11}$ givesAt the same peak displacement amplitude and frequency, time-averaged kinetic energy is proportional to membrane area. ThereforeThe reciprocal ratio is $41/40$ if the rectangle is placed first.

Use base dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The proposed dimensional analysis givessoEquating the independent dimensional powers requiresConsequentlyThe mass density drops out because there is no other mass-bearing quantity with which to combine it under the stated assumptions.

Dimensional analysis fixes the powers but leaves a dimensionless coefficient undetermined. For a linear deep-water gravity wave, the dispersion relation is ${\omega }^2=gk$, giving phase velocity $\omega /k=g/\omega$ and group velocity $d\omega /dk=g/(2\omega )$. Both obey the inferred scaling. The phrase wave speed alone does not distinguish these two velocities or fix the coefficient.

Work with the explicitly corrected reduced-pressure system in b. Write $d=1/(2H)$ and use the complex wave amplitude, taking real parts at the end. The vertical factor satisfies $({\partial }_z^2-H^{-1}{\partial }_z)w=-(m^2+d^2)w$, so the wave equation givesFor a Kelvin wave, continuity, zonal momentum and the buoyancy equation then determine all polarizations:In particular $u,w,b,\Phi$ have amplitude proportional to $e^{z/(2H)}$, whereas pressure $p^{\prime }$ has amplitude proportional to $e^{-z/(2H)}$. Directly substituting these expressions into vertical momentum gives ${\Phi }_z=-i\omega (m^2+d^2)w/k^2=i(\omega -N^2/\omega )w$, verifying both the sign and the dispersion relation.

Meridional momentum is $\beta yu+{\Phi }_y=0$. Since $\Phi =cu$, it integrates to a Gaussian $\exp [-\beta y^2/(2c)]$. ThusThe eastward intrinsic branch $c>0$ is selected by decay away from the equator; choosing $c<0$ would make the meridional amplitude grow without bound. Although the squared dispersion relation has both frequency signs, a meridionally trapped Kelvin wave has this one phase-propagation direction.

For short vertical wavelengths, $m^2\gg k^2+d^2$, so ${\omega }^2\simeq N^2{k}^2/m^2$, $c\simeq N/|m|$ and $L_D^2\simeq N/(\beta |m|)$. The wave is a slowly oscillating, hydrostatic internal gravity wave with strong vertical variation and relatively narrow trapping. Here $u\simeq -(m/k)w$ and $p^{\prime }/{\rho }_0\simeq -(\omega m/k^2)w$; the buoyancy expression above remains valid.

For long vertical wavelengths, $|m|\ll \sqrt{k^2+d^2}$, the exact limiting speed is $c=N/\sqrt{k^2+d^2}$. To obtain the stated height-controlled limit also requires long horizontal wavelength, $|k|H\ll 1$. With both $|m|H\ll 1$ and $|k|H\ll 1$,The finite density scale height prevents the hydrostatic $1/|m|$ phase speed from diverging as $m\to 0$. The polarizations tend to $u\simeq -iw/(2Hk)$, $\Phi \simeq -iNw/k$ and $b\simeq -iNw/(2Hk)$. Long vertical wavelength alone at fixed large $kH$ would not justify $c=2NH$. In the opposite horizontally short limit the frequency approaches $N$ instead. The growth $e^{z/(2H)}$ compensates decreasing reference density in wave-energy density; a radiating wave train is a local modal description, not a finite-total-energy eigenfunction over an infinite vertical domain.

Apply the intended, corrected dispersion relation in c. A uniform mean zonal wind gives ${\omega }_{\mathrm{ground}}=\hat{\omega }+kU$. Stationarity with $U=-20\mathrm{m}{\mathrm{s}}^{-1}$ therefore requires eastward intrinsic phase speed $\hat{\omega }/k=-U=20\mathrm{m}{\mathrm{s}}^{-1}$. With $k=1/a$,The period determines $\Omega =2\pi /(24.4\times 3600\mathrm{s})$ and $\beta =2\Omega /a\simeq 4.21\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. Hence

A storm within roughly $12^{\circ }$ of the equator has substantial projection on the Gaussian mode. There is no sharp latitude cutoff: $|y|=L_D$ gives amplitude $e^{-1/2}$ and the $1/e$ amplitude latitude is about $16.4^{\circ }$. Forcing outside that core has progressively weaker overlap.

The intrinsic wave is eastward even though its observed ground pattern is stationary; westward advection cancels its phase propagation. Its intrinsic period is about $12.4$ Earth days. The inferred $|m|\gg 1/(2H)$ puts it in the short-vertical-wavelength branch, well below the long-wave speed $2NH=200\mathrm{m}{\mathrm{s}}^{-1}$. For upward energy radiation, $\partial \hat{\omega }/\partial m$ must be positive, which selects $m<0$ for positive $k,\hat{\omega }$: vertical phase propagation is then downward. Dust absorption supplies thermal forcing with a zonally asymmetric component; the response does not imply that dust itself moves eastward at the intrinsic phase speed.

Southern spring to summer, the warmer perihelion-side dust season, is a reasonable climatological window for such a regional storm. This seasonal inference uses observed dust climatology, not the dispersion relation: the specified wind and wave parameters alone do not determine a unique season, and individual regional storms can occur outside that window.^[ref]

For $U=0$ and fixed $\omega >0$, completing the square in the dispersion relation givesThese Rossby-wave isofrequency circles lie in the $k\le 0$ half-plane and touch the origin, where the dispersion relation is undefined. Their radii decrease as frequency increases. The zero-frequency line is $k=0,l\ne 0$.

The group velocity is normal to the constant-frequency curves, toward increasing frequency:At $l=0$, the group velocity is eastward; close to $k=0$ it is westward. For the positive-frequency circles its arrows point inward toward the circle centre. Its meridional component is southward in $k<0,l>0$ and northward in $k<0,l<0$.

For $k\ne 0$, the zonal phase velocity, meaning propagation at fixed $y$, isIt is westward everywhere it is defined. The zonal group velocity is eastward for $|k|>|l|$, westward for $|k|<|l|$, and zero on $|k|=|l|$. There is no region of eastward zonal phase velocity. At $k=0$, $\omega /k$ is undefined, though its continuous limiting expression is westward and $c_{gx}=-\beta /l^2$. Using instead the normal-to-wavecrest phase velocity vector gives ${\mathbf{c}}_p=\omega (k,l)/(k^2+l^2)$; its zonal component is also westward for $k\ne 0$, but is not the same scalar as $\omega /k$.

Seek nonzero time-harmonic waves of the form $\varphi =\mathrm{Re}[F(y)e^{i(kx-\omega t)}]$ and $\eta =\mathrm{Re}[A{e}^{i(kx-\omega t)}]$, taking $k>0$. Laplace's equation reduces to $F^{\prime \prime }-k^{2}F=0$. The impermeable-bed condition $F^{\prime }(-h)=0$ gives $F(y)=C\cosh (k(y+h))$.

The linearized kinematic and dynamic linearized free-surface boundary conditions at zero becomeEliminating $A$ and the nonzero amplitude $C$ givesConversely, if this dispersion relation holds, either amplitude can be chosen nonzero and the other set by the second boundary equation, producing a wave satisfying every condition. The expression also holds for negative $k$ because $k\tanh (kh)$ is even.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

The band effective mass is determined by the second derivative of the dispersion relation:With the stated sign convention and $t>0$, the band minimum is at $k=0$. Expanding the cosine, or differentiating twice, givesComparison with a free electron's kinetic energy ${\hslash }^2{k}^2/(2m_e)$ therefore requiresThe hopping integral has units of energy. A negative $t$ would move the minimum to the edge of the Brillouin zone and give the same positive bottom-of-band effective mass with $|t|$ in this expression.

Different measurements reveal different aspects of electronic band structure; a convincing determination combines their constraints.

Angle-resolved photoemission. In angle-resolved photoemission spectroscopy, photons of known energy eject occupied electrons. Measuring their kinetic energy and emission angle reconstructs binding energy and the component of crystal momentum parallel to the surface:Here $\Phi$ is the sample work function when kinetic energy is referenced to the nearby vacuum; an analyzer energy scale requires its own work-function calibration. Momentum is conserved modulo a surface reciprocal lattice vector, neglecting photon momentum. Tracking the peaks maps occupied dispersion relations and the Fermi surface. The measured intensity also contains the matrix element, the Fermi-Dirac distribution and the spectral function, so an absent bright feature does not establish an absent band. For a three-dimensional solid, the perpendicular momentum requires a final-state model. Inverse photoemission spectroscopy probes unoccupied states by injecting electrons and detecting emitted photons.

Optical spectroscopy. Photon absorption and ellipsometry measure interband transitions, optical selection rules and the complex dielectric response. Because optical photon momentum is small on the scale of the Brillouin zone, direct transitions approximately preserve crystal momentum. Their onset constrains a direct band gap; a threshold involving a phonon can reveal an indirect band gap. Peak positions constrain energy differences between bands, but the response also depends on the joint density of states and optical matrix elements, so it does not uniquely determine every $E_n(\mathbf{k})$.

Quantum oscillations and cyclotron resonance. In clean samples at low temperature, the de Haas-van Alphen effect measures oscillations in magnetization and the Shubnikov-de Haas effect measures oscillations in electrical resistivity as the inverse magnetic field is varied. Their period in inverse magnetic field is $1/F$, where Onsager quantization relates $F$ to an extremal Fermi-surface area perpendicular to the field:Rotating the magnetic field maps the geometry of extremal orbits. The thermal damping determines the cyclotron effective mass,with the appropriate magnitude for a hole orbit. Cyclotron resonance supplies the corresponding dynamical mass through ${\omega }_c=eB/m_c$. These are detailed probes near the Fermi level, rather than measurements of all occupied bands. Diffraction supplies the reciprocal lattice needed to label the measured momenta, but diffraction by itself is not a measurement of electronic dispersion.

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

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.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

Let $A$ be the real-space area and use ${\sum }_{\mathbf{p}}\to C\int d^{2}p$, with $C=A/(2\pi \hslash {)}^2$. There is one state per momentum cell, since the fermions are spinless. An annulus between $p$ and $p+dp$ contains $C2\pi pdp$ states. Near the Fermi surface, the quasiparticle dispersion relation has $dϵ/dp=v_F^{\ast }=p_F/m^{\ast }$, so the total density of states isThe density of states per unit area and the occupied-state count are, respectively,In particular, $D^{\ast }(\mu )=2N{m}^{\ast }/p_F^2$. No factor of two for spin degeneracy is present. This counts Landau quasiparticle levels, so it is not multiplied by the quasiparticle residue of the microscopic particle Green's function.

Within a stable normal Fermi liquid, the low-energy states are described by long-lived Landau quasiparticles and holes in a thin shell about the Fermi surface; the Pauli exclusion principle suppresses their decay. Expanding the energy about the equilibrium occupations gives the one-quasiparticle dispersion relation at first order and the interaction between occupation changes at second order, namely the Landau quasiparticle energy functional. For a smooth interaction, radial deviations in that interaction may be neglected at leading order, leaving its angular dependence; higher occupation terms contribute higher powers of the small shell thickness. Collective occupation deformations are included in the same description. The sufficiency claim is conditional on the normal Fermi liquid regime, and on supplementing this static energy functional with dynamics when required: it is not a description of arbitrary low-energy phases such as a superconductor or non-Fermi liquid.