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Expand a single-electron wavefunction as $|\psi ⟩={\sum }_n{\psi }_n|n⟩$. Project the Schrödinger equation onto $⟨n|$ to obtainTranslation symmetry suggests the Bloch state ${\psi }_n=N^{-1/2}{e}^{ikna}$ on a periodic chain of $N$ sites. Substitution givesTake the conventional positive hopping $t>0$. The first Brillouin zone is $-\pi /a\le k<\pi /a$; the band has minimum ${ϵ}_0-2t$ at $k=0$ and maximum ${ϵ}_0+2t$ at the zone edges. Its width is $4t$. A negative hopping interchanges the extrema but does not change the density of states after replacing $t$ by $|t|$.

There are $N$ allowed wavevectors and two spin states per wavevector. For the total spin-summed density of states $D(E)$, each interior band energy has two roots, $\pm k$. Using the spacing $2\pi /(Na)$ givesThus the density of states per atom isIt is symmetric about ${ϵ}_0$ and has integrable inverse-square-root Van Hove singularities at both band edges. Its integral over the band is two states per atom. The plot labels the dispersion extrema and both singular edges.

The first Brillouin zone is the squareAt half filling, translation of the wavevector by $\mathbf{Q}=(\pi /a,\pi /a)$ exchanges the two band energies:Thus the same energy-pair argument as in part c givesAt zero temperature, the Fermi surface satisfies $\cos (k_{x}a)+\cos (k_{y}a)=0$. Within the square Brillouin zone this is the diamondwhose vertices are $(\pm \pi /a,0)$ and $(0,\pm \pi /a)$. The occupied region is the diamond's interior, containing the band minimum at the zone centre. It has half the Brillouin zone area, consistent with one electron per atom and two spin states per wavevector.

A phonon is a quantum of a normal mode of Bravais lattice vibration. A harmonic mode of angular frequency $\omega$ has energy $\hslash \omega (n+1/2)$, and adding one phonon increases its energy by $\hslash \omega$. The phonon carries crystal momentum $\hslash \mathbf{q}$, defined modulo a reciprocal lattice vector.

The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice about the origin: the set of wavevectors at least as close to the origin as to any other reciprocal lattice point. Its boundaries are reciprocal-space perpendicular bisectors. It gives one representative of each wavevector modulo reciprocal lattice translations.

A consistent mechanism is a transition to G-type antiferromagnetism at the Néel temperature. On the manganese primitive cubic lattice, take the magnetic moments to beEvery nearest neighbour has the opposite magnetic moment. The nuclear crystal structure can retain its original lattice constant, while the magnetic order introduces the propagation wavevectorSumming the scattering amplitude from the magnetic moments over sites gives peaks at $\mathbf{q}=\mathbf{G}+\mathbf{Q}$, where $\mathbf{G}$ is a nuclear reciprocal lattice vector. Thus all three components of the new Miller indices, expressed relative to the original unit cell, are half-integers. The two smallest squared magnitudes in units of $(2\pi /a{)}^2$ areThe first predicts $2\theta \simeq 21.29^{\circ }$, agreeing with the new ring. The next squared magnitude is $19/4$, outside the measured angular range. Taking the observed first magnetic ring as the calibration, Bragg's law predictsUsing the independently fitted lattice constant gives $41.42^{\circ }$, within the input rounding. One ring alone does not uniquely determine a magnetic order; this proposed nearest-neighbour antiferromagnetic pattern explains both the observed ring and the count of two new rings in the specified interval.

Given independent primitive lattice translation vectors ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$, defineThen ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. The reciprocal lattice consists ofequivalently all wavevectors for which $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct lattice vector $\mathbf{R}$.

For the face-centered cubic lattice, chooseTheir scalar triple product is $a^3/4$, and direct evaluation of the cross products givesEvery integer combination has coordinates $(2\pi /a)(H,K,L)$ with $H,K,L$ all even or all odd. Conversely, for such coordinates the coefficients $(K+L)/2,(H+L)/2,(H+K)/2$ are integers, so every such point occurs. The even-coordinate points are the corners of cubes of edge $4\pi /a$, and the odd-coordinate points are their body centres. ThereforeThis is a property of the Bravais lattice; the two-atom atomic basis changes structure factors rather than these reciprocal lattice translation vectors.

Subtracting the three acoustic phonon branches from the six displacement normal modes leavesAt small wavevector, these involve relative motion of the two atomic basis sites rather than a rigid translation of the primitive unit cell. Their angular frequencies remain nonzero at the zone centre. The count uses the primitive unit cell; using all eight chemical atoms in the conventional unit cell would spuriously multiply the number of distinct branches in the first Brillouin zone.

There are two chemical atoms in the primitive unit cell, hence six displacement coordinates and six phonon branches at each wavevector. Three rigid translations cost no restoring potential energy in the long-wavelength limit. ConsequentlyFor propagation along $[100]$, one acoustic phonon has longitudinal polarization along $x$ and two have transverse polarization along $y,z$. The cubic crystal symmetry makes those two transverse directions equivalent. In terms of the elastic stiffness tensor and the mass density,Thus there are two distinct acoustic velocities in general, one longitudinal and a doubly degenerate transverse value. This degeneracy of normal modes concerns the two transverse polarizations; it does not imply equality with the longitudinal speed of sound.

Use the long-wavelength linear dispersion relation $\omega \simeq v_L|\mathbf{k}|$ as an estimate. The phonon energy is thenThis extrapolates the measured low-wavevector speed of sound to the $X$ point, the first Brillouin zone boundary along $[100]$. It is not an exact phonon dispersion relation there: the discrete crystal structure generally bends the branch, and the initial slope alone cannot determine its boundary energy. In particular, $(2\pi /a,0,0)$ is not a reciprocal lattice vector of the face-centered cubic lattice; the smallest such axial vector is $(4\pi /a,0,0)$. There is therefore no reason for this acoustic phonon energy to vanish.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

At the zone centre, the dynamical matrix commutes with the point-group representation on displacements. An irreducible representation of dimension two enforces a doublet; distinct one-dimensional representations need not have equal frequencies. The tetragonal point group $D_{4h}$ has two-dimensional $E_g$ and $E_u$ representations, so optical modes transforming that way are doubly degenerate at the centre. Atomic positions and the displacement representation would be needed to assign every optical branch to a particular representation; the point-group label alone is not a complete mode decomposition.

For propagation along the tetragonal fourfold axis, the two transverse acoustic polarizations transform together and have equal sound speeds and frequencies. For example ${\omega }_{\mathrm{TA},x}={\omega }_{\mathrm{TA},y}\simeq v_T|q_z|$, whereas the longitudinal speed is independent. An optical doublet may split away from the centre along a direction whose little group has only one-dimensional representations. It need not remain degenerate at every nearby wavevector.

All irreducible representations of $D_{2h}$ are one-dimensional, so the orthorhombic point group enforces no analogous optical doublet or equality of transverse sound speeds. The three acoustic frequencies are nevertheless all zero at the exact centre in both crystals, because of translations; that common zero is not evidence that their slopes are equal. Accidental equalities remain possible. For a polar crystal, a direction-dependent macroscopic electric field also distinguishes longitudinal from transverse infrared-active limits; an analytic zone-centre symmetry classification alone does not remove LO–TO splitting.

Each atom supplies three displacement coordinates. Translational invariance gives three acoustic branches for either structure, leavingThese count branches at a general wavevector; a doubly degenerate pair still counts as two modes. The stated atom counts refer to primitive cells, so no additional centring factor is needed.

For neutron energy loss, $E_f=30-20=10\mathrm{meV}$. Write $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. The neutron dispersion $E={\hslash }^2{k}^2/(2m_n)$, or $E[\mathrm{meV}]=2.0721k[{\mathring{\mathrm{A}}}^{-1}{]}^2$, givesA triangle of these wavevectors exists ifWith $c=4.50\mathring{\mathrm{A}}$, the requested momentum transfers have $Q=(2\pi /c)|l|=1.3963|l|{\mathring{\mathrm{A}}}^{-1}$. ConsequentlyFor example $|l|=1$ is too small even for forward scattering at these unequal neutron energies, and $|l|=5$ exceeds the backscattering maximum. This identifies accessible reciprocal lattice points, not guaranteed nonzero intensity: a one-phonon structure factor contains the atomic scattering lengths, eigenvectors and factors $\mathbf{Q}\cdot {\mathbf{e}}_j$. A transverse eigenvector perpendicular to $\mathbf{Q}$ may be invisible, and basis interference can cause extinctions. The supplied geometry and cell constants do not specify that full intensity selection rule.

For a transverse electromagnetic plane wave in a nonmagnetic medium, Maxwell's equations give $c^2{k}^2={\omega }^2{ϵ}_r(\omega )$. Set damping to zero and define $u=c^2{k}^2/{ϵ}_{\infty }$. Substitution yieldsso ${\omega }^4-({\omega }_{\mathrm{LO}}^2+u){\omega }^2+u{\omega }_{\mathrm{TO}}^2=0$. Therefore the two phonon polariton branches areTake the nonnegative square root for the positive physical frequencies. At $k=0$, ${\omega }_{-}=0$ and ${\omega }_{+}={\omega }_{\mathrm{LO}}$. At small wavevector the lower branch is photon-like with velocity $c/\sqrt{{ϵ}_r(0)}$; at large wavevector it approaches ${\omega }_{\mathrm{TO}}$ from below. The upper branch ultimately becomes photon-like with velocity $c/\sqrt{{ϵ}_{\infty }}$. There is no real propagating transverse branch with frequency strictly between ${\omega }_{\mathrm{TO}}$ and ${\omega }_{\mathrm{LO}}$ in this lossless bulk model. Coupling changes the eigenmodes from a bare photon crossing a flat oscillator into mixed normal modes.

In Raman scattering, the photon transfers $\hslash \omega =\hslash ({\omega }_i-{\omega }_s)$ and $\mathbf{q}={\mathbf{k}}_i-{\mathbf{k}}_s$ to a collective excitation. Changing scattering geometry selects different $q$ and probes its dispersion.

The open circles rise and then flatten near $45\mathrm{meV}$: they trace the lower transverse phonon polariton branch. It is strongly dispersive on its photon-like side, then becomes predominantly TO-phonon-like and approaches the TO energy at larger wavevector. The open triangles stay near $50\mathrm{meV}$ and are the nearly nondispersive longitudinal optical phonon. Longitudinal motion creates its own polarization field and is shifted above the transverse resonance. It does not hybridize with a propagating transverse photon in the same way, and it should not be misidentified as the upper transverse polariton branch, which rises with wavevector. Raman tensor and polarization selection rules determine which excitation is visible.

Scale qualification: the source horizontal label appears to print $10^{-4}\mathrm{c}{\mathrm{m}}^{-1}$. Taken literally, one plotted unit would have vacuum photon energy only $\hslash c\times 0.01{\mathrm{m}}^{-1}\simeq 2\times 10^{-6}\mathrm{meV}$, inconsistent with the observed tens of meV and the dispersive polariton. The physically compatible scale is of order $10^4\mathrm{c}{\mathrm{m}}^{-1}$; the qualitative branch identification does not require silently treating the printed negative exponent as correct. The original model figure in solution explicitly labels its calculated wavevector scale.

Decomposing a plane displacement into components parallel and perpendicular to its wavevector gives compressional and shear speedsThe incident polarization is longitudinal. Its reflected longitudinal wave reverses the vertical wavevector component while retaining its horizontal component. The converted shear wave has wavevector $k_s(\sin \beta ,-\cos \beta )$ perpendicular to polarization $(\cos \beta ,\sin \beta )$.

Every term must have the same phase along the rigid boundary. ThusChoose $\cos \beta >0$ so the reflected shear wave propagates downward into $y<0$. At $y=0$, both displacement components must vanish for every boundary phase:Solving this two-by-two system yieldsFor a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha <\pi$; choosing $\beta \in (-\pi /2,\pi /2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha \le \pi /2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no critical angle for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At normal incidence $\alpha =\pi /2$, $\beta =0$, $R_1=1$ and $R_2=0$, cancelling the incident displacement with the reversed longitudinal polarization. This is compressional-wave reflection from a rigid elastic boundary.

Write $e_{xx}=u_{x,x}$, $e_{yy}=u_{y,y}$ and $2e_{xy}=u_{x,y}+u_{y,x}$. Differentiating twice and cancelling the mixed partial derivatives givesA convenient set of six three-dimensional Saint-Venant compatibility equations isThey follow by inserting $e_{ij}=(u_{i,j}+u_{j,i})/2$ and commuting derivatives. They form the six symmetric components of the incompatibility tensor $C_{ij}={ϵ}_{ik\ell }{ϵ}_{jmn}{e}_{\ell n,km}$, up to the indicated signs and permutation of components.

They obey the three differential identities ${\partial }_i{C}_{ij}=0$, because a symmetric pair of derivatives contracts with an antisymmetric Levi-Civita symbol. Thus there are three independent differential constraints, represented symmetrically by six component equations; they are not six unrelated constraints on the six strain functions. For example, at a nonzero wavevector directed along the third axis, compatibility imposes $e_{11}=e_{22}=e_{12}=0$ while $e_{13},e_{23},e_{33}$ can arise from the three displacement components. This confirms rank three. One cannot simply discard the same arbitrary three component equations in every direction; their independence is differential, with the stated identities. On a simply connected domain these are the local Saint-Venant compatibility equations for strain to be a symmetric displacement gradient, while domains that are multiply connected regions require global closure conditions as in the screw example. This is Saint-Venant strain compatibility and differential identities.

Growing and decaying modes. Neglect pressure and insert the Einstein-de Sitter background. A trial $\Delta \propto t^p$ givesThereforeThe quoted growth law selects the growing mode; arbitrary initial data can also contain the decaying one. In a static homogeneous Jeans model, $\ddot{\Delta }=(4\pi G\bar{\rho }-k^2{c}_s^2)\Delta$, so an unstable mode grows exponentially, $e^{\sqrt{4\pi G\bar{\rho }-k^2{c}_s^2}t}$. Expansion introduces the $2H\dot{\Delta }$ damping term and a falling background density, yielding power-law growth instead. If the wavevector in the supplied pressure term is comoving, that term would need $R^{-2}$; the pressureless result does not depend on this convention.

Amplitude problem. Using the stated recombination epoch, matter-era growth from then until today is only about $1501$. A baryonic density perturbation of order $10^{-5}$ would reach merely $1.5\times 10^{-2}$, far below the order-unity regime needed for nonlinear collapse. photon coupling also prevents baryons from freely collapsing before recombination. The observed microwave temperature contrast is not an exact direct measurement of baryon density contrast at every scale: acoustic transfer and gravitational redshift matter. The estimate expresses the insufficient growth of tiny photon-coupled baryonic seeds, not an exact conversion between the two observables.

Coupled gravitational response. Both pressureless components respond to the same perturbed potential. The equations areTheir difference obeys $\ddot{D}+2H\dot{D}=0$, where $D={\Delta }_D-{\Delta }_B$. Thus $\dot{D}\propto R^{-2}$ and, in Einstein-de Sitter, $D=C_0+C_1{t}^{-1/3}=C_0+{\tilde{C}}_1{R}^{-1/2}$.

Catch-up branch and its condition. With dark matter dominating the source and in its growing mode, write ${\Delta }_D=AR$. The full baryon solution isThe requested baryon catch-up approximation after recombination selects the branch with negligible relative decaying mode, ${\tilde{C}}_1=0$. Defining $C_0=AR(z_0)$ then givesThis branch applies at and after $z_0$ and has ${\Delta }_B(z_0)=0$ and matched baryon/dark-matter growth velocities at that epoch. It does not follow uniquely from vanishing initial baryon density contrast alone. If both ${\Delta }_B$ and its initial time derivative vanish at $R=R(z_0)$, the zero-initial-baryon-velocity catch-up transient instead givesBoth approach unity for $x\gg 1$. The displayed target therefore needs its stated no-relative-decaying-mode assumption; a finite baryon fraction modifies the transient further.

Physical consequence. cold dark matter is not held to the radiation by Thomson scattering, and can build gravitational wells before baryon decoupling, with appreciable matter-era growth after equality. Its growing perturbations then attract baryons once pressure and radiation drag become unimportant. The relative offset stays constant or decays while the common mode grows, so baryons catch up. Nonlinear structure need not grow solely from the tiny baryonic perturbation inferred at decoupling. cold dark matter growth during radiation domination is slower than the matter-era law, and the stated formula is neither a radiation-era nor a nonlinear-collapse solution.

The inverse of the $(v,\lambda )$ metric tensor block isThe normal covector $d\lambda$ has squared norm ${\lambda }^2/m^2$. ThereforeMore explicitly, the induced metric tensor is $-{\lambda }_0^{2}d{v}^2+m^{2}d{\Omega }_2^2$; at zero its sole zero direction is ${\partial }_v$. The coefficients are independent of $v$, so $K={\partial }_v$ is a Killing vector field. On $\lambda =0$,Hence $K$ is both tangent and normal to this null hypersurface, proving it is a Killing horizon. The definition of surface gravity is ${\nabla }_{K}K=\kappa K$ on the Killing horizon, or equivalently $d(K^2)=-2\kappa K^{♭}$. Since $d(K^2)=-2\lambda d\lambda$ vanishes there while $K^{♭}\ne 0$,Thus this is a degenerate Killing horizon, and $v$ is an affine parameter on each of its null geodesics.

For a concrete horizon frequency comparison, fix the same angular position for a narrow high-frequency light beam and two freely falling radial observers. Work in geometric optics, with the beam's affinely transported wavevector tangent to the Killing horizon, $k=qK$. Here $q>0$ is constant because ${\nabla }_{K}K=0$. Choose the observers as translates of one timelike geodesic by the isometry generated by $K$, crossing at $v_1$ and $v_2$. Give them the same positive Killing energy per unit rest mass, $E=-K\cdot W$. At their crossings their normalized four-velocities havewhich satisfy $g(W,W)=2m{W}^v{W}^{\lambda }=-1$. They can be realized by local geodesics with precisely these initial data. The measured angular frequency isTo connect this measurement to surface gravity, if ${\nabla }_{K}K=\kappa K$ with constant $\kappa$, affine transport of $k=q(v)K$ would give $dq/dv=-\kappa q$. For observers related by the same Killing vector field translation, $-K\cdot W$ agrees at the crossings, giving ${\omega }_2/{\omega }_1=e^{-\kappa (v_2-v_1)}$. Here there is no accumulated redshift along the Killing horizon for these equally prepared observers. Arbitrary observers with different four-velocities can still measure different frequencies through their different Killing energies.

Use a plane wave with phase $kx+mz-\omega t$, and take $N>0$, $k,m\ne 0$. The positive-frequency branch of the given hydrostatic dispersion relation is $\omega =N|k|/|m|$. The wavevector is $\mathbf{K}=(k,m)$. Differentiating with respect to its components gives the group velocity, which carries the wave packet and its energy:The normal phase velocity, which follows a crest in its normal direction, isConsequently,Thus the group velocity lies along the constant-phase lines $z=-(k/m)x+\mathrm{constant}$, perpendicular to the phase velocity. For $k,m>0$, crests move upward and rightward, whereas energy moves downward and rightward. Choosing $m<0$ gives upward and rightward energy propagation and downward and rightward normal phase propagation. Conjugate or oppositely directed components supply the other outgoing directions from a localized source.

The speeds areThe hydrostatic limit is $|k/m|\ll 1$. In that limit energy propagates almost horizontally, with $|c_{g,x}|=N/|m|$ and $|c_{g,z}|=N|k|/m^2$, while normal phase propagation is almost vertical and has speed approximately $N|k|/m^2$. The horizontal speed at which a crest intersects a fixed height is $\omega /k=c_{g,x}$; this is not the horizontal component of the normal phase velocity vector. For example, $N=10^{-2}{\mathrm{s}}^{-1}$, vertical wavelength $2\pi /|m|=10\mathrm{km}$ and horizontal wavelength $2\pi /|k|=1000\mathrm{km}$ give $|c_{g,x}|\simeq 15.9\mathrm{m}{\mathrm{s}}^{-1}$ and $|c_{g,z}|\simeq 0.159\mathrm{m}{\mathrm{s}}^{-1}$.

These are local speeds in a uniform medium, without a background current. The non-rotating approximation requires $|f|\ll \omega$, while the hydrostatic approximation requires $\omega \ll N$. At lower frequencies, rotation changes the dispersion relation to that of inertia-gravity waves.

Write the velocity field as $u=U+u\prime$, $v=v\prime$. The uniform background has zero relative vorticity, and the perturbation has $\xi \prime =v{\prime }_x-u{\prime }_y$. Keeping only terms linear in the perturbations givesThe incompressibility condition permits a streamfunction $\psi$ with $u\prime =-{\psi }_y$, $v\prime ={\psi }_x$, hence $\xi \prime ={\nabla }_h^2\psi$. Thus the Linear Rossby-wave equation here isChoose a plane wave $\psi =\mathrm{Re}\{\hat{\psi }{e}^{i(kx+ly-\omega t)}\}$, with $K^2=k^2+l^2>0$. Substitution givesand therefore the local Rossby-wave dispersion relation in a zonal jetThe $Uk$ term is the background Doppler shift; the intrinsic phase velocity is westward. A real physical wave can always be represented using the conjugate wavevector if necessary to choose $\omega \ge 0$.

The relevant ground-frame zonal phase velocity and group velocity areTime increases downward in the source diagram. The individual phase contours lean towards increasing longitude as time increases, so their phase velocity is eastward. The envelope moves eastward faster: successive crests grow at the downstream edge and decay upstream. This illustrates Rossby-wave downstream development.

For the intended purely zonal wave packet, $l=0$, andConsequently, with $U,\beta >0$ and the positive-frequency choice $k>0$,There is a small interpretation qualification: if “zonally propagating” specified only the plotted direction while allowing an unknown nonzero meridional wavenumber, eastward phase velocity alone would imply $k^2+l^2>\beta /U$, not the stronger bound on $k$ alone. In this uniform model a strictly zonal group velocity with $\beta k\ne 0$ does imply $l=0$, which supplies the intended interpretation.