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The ordinary wave's electric field is perpendicular to the plane containing the wavevector and optic axis. Its refractive index $n_o$ is direction independent at fixed frequency.

Fluid added mass increases inertia but need not alter the same restoring force. A submerged spring's constant buoyancy shifts equilibrium, whereas a pendulum's buoyancy reduces its restoring torque. Thus an unbounded-fluid sphere has spring frequency factor $\sqrt{\rho /(\rho +{\rho }_f/2)}$ and downward-pendulum factor $\sqrt{(\rho -{\rho }_f)/(\rho +{\rho }_f/2)}$. The latter stable configuration requires $\rho >{\rho }_f$.

Measurements at fixed frequency and varying oscillation amplitude. Amplitude-independent moduli identify a possible linear regime; dependence on amplitude signals nonlinear response or an experimental artifact.

Measurements at different frequencies while retaining an amplitude in the linear regime. They compare observation time with material relaxation times.

An oscilloscope displays an electrical signal versus time. In an acoustic wavelength measurement, the microphone trace supplies the tone's frequency and oscillation amplitude as the detector changes position.

In the classical thermal model, a coordinate in a harmonic potential obeys $\gamma \dot{x}+kx=\xi$, where $\gamma$ is the drag coefficient and $⟨\xi (t)\xi (t^{\prime })⟩=2\gamma k_{B}T\delta (t-t^{\prime })$. Its stationary one-sided spectral density per hertz is the displayed Lorentzian line shape. The integral over positive frequencies equals $k_{B}T/k$, consistent with the equipartition theorem. On a log-log plot the spectrum is flat for $f\ll f_c$ and has slope $-2$ for $f\gg f_c$. The crossover is determined by the restoring stiffness and drag; an added detector response multiplies the mechanical spectrum by the squared magnitude of its transfer function.

An overtone is an oscillatory mode above the fundamental mode in the same mode family. Waveguide branches are compared at fixed horizontal wavenumber. Their frequencies need not be integer multiples of the fundamental frequency. In a scalar layered guide, additional vertical nodes distinguish successive overtones.

For a one-electron P-to-S transition with nuclear spin zero, ten electric-dipole sublevel channels have finite strength at weak nonzero field. Six retain finite strength in the complete orbital-spin decoupling limit. Channel count is distinct from spectral frequency count: retaining first-order residual spin-orbit interaction energy gives five distinct frequencies, with two central channels degenerate.

Let $\hat{\mathbf{u}}$ and $\hat{p}$ be the complex plane wave amplitudes and set $k_x=k_{\perp }$, $k_z=k_{\parallel }$. The linearized Euler equations in the rotating frame areHere $p$ is pressure per unit mass density, including the centrifugal potential. The last line is incompressibility, not an additional equation of state.

Taking the curl eliminates pressure. For constant $\mathbf{\Omega }$ and a divergence-free vector field, $\nabla \times (2\mathbf{u}\times \mathbf{\Omega })=2(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. Thus the inertial-wave vorticity equation givesApply $\mathbf{k}\times$ once more and use $\mathbf{k}\times (\mathbf{k}\times \hat{\mathbf{u}})=-k^2\hat{\mathbf{u}}$. For a nonzero oscillatory amplitude,so the inertial-wave dispersion relation isThe restoring influence is the Coriolis acceleration. In particular $|\omega |\le 2|\Omega |$. For $k_{\parallel }=0$, the projected linear dynamics have zero frequency; these are steady balanced disturbances rather than the nonzero-frequency branch used in the division above. The spatially uniform $k=0$ mode must be treated separately from this dispersion relation.

Let $L_{\perp }=k_{\perp }^{-1}$ and let $U$ be the common size of horizontal and vertical velocity perturbations, as found in part b. The linear inertial wave time and nonlinear advection time areThe critical balance in a rapidly rotating fluid is thereforeThe Rossby number is of order $ϵ$. Horizontal geostrophic flow balances the leading pressure gradient against Coriolis acceleration, soThese are dimensionally comparable statements, rather than comparisons of a dimensional frequency with a dimensional velocity. Equivalently, use $L_{\perp }$ and $(2\Omega {)}^{-1}$ as units. ThenThus rotation stays $O(1)$ in these units, while the motions evolve on the slower $O({ϵ}^{-1})$ time. Vertical advection $u_{\parallel }{\partial }_z$ is smaller than horizontal advection by $ϵ$.

Write ${\mathbf{u}}_{\perp }={\mathbf{u}}_{\perp }^{(0)}+{\mathbf{u}}_{\perp }^{(1)}+\cdots$, with ${\mathbf{u}}_{\perp }^{(1)}=O(ϵU)$, and $p=p_0+p_1+\cdots$. At leading horizontal order,Define the stream function $\Phi =p_0/(2\Omega )$, including its horizontally uniform part in this convention. ThenThe next-order incompressibility condition is not zero horizontal divergence:This term has size $ϵk_{\perp }U$ and is essential: its product with the large rotation rate is a leading contribution to the vertical vorticity equation.

To see that contribution explicitly, write $D_{\perp }={\partial }_t+{\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }$. At the first horizontal inertial order,Taking its vertical curl eliminates $p_1$. The two-dimensional advection identity and the two divergence orders giveEquivalently the ageostrophic flow satisfieswhose divergence is $-D_{\perp }\zeta /(2\Omega )$. This explicitly verifies that the next-order divergence has the required sign.

The vertical momentum equation, at the same slow order, isFinally ${\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }f=\{\Phi ,f\}$, where the canonical two-dimensional Poisson bracket is $\{f,g\}=f_x{g}_y-f_y{g}_x$. The reduced equations for an anisotropic rotating fluid areThe omitted vertical-advection, ageostrophic-advection and relative-vorticity stretching terms are smaller by $O(ϵ)$ in the equations retained here. Linearizing these equations recovers ${\omega }^2=4{\Omega }^2{k}_{\parallel }^2/k_{\perp }^2$, the anisotropic limit of part a.

For a uniform axial magnetic field, the anisotropic Shear Alfvén wave is predominantly horizontally polarized, whereas the pseudo-Alfvén wave is predominantly vertically polarized. Without rotation both have characteristic frequency ${\omega }_A\sim |k_{\parallel }|v_A$. Pressure projection makes the effective Coriolis acceleration couple these two transverse wave polarization states at the inertial-wave rateThus the useful comparison is not $\Omega$ alone, nor anisotropy alone, butIn this limit rotational coupling of shear and pseudo-Alfvén waves is negligible on an Alfvén propagation time, so the two familiar linear polarizations are decoupled to leading order. The factor $k_{\parallel }$ cancels: simply making both disturbances more anisotropic does not by itself remove their relative rotational coupling.

For a nonlinear reduction one must also make the rotational mixing slow compared with the nonlinear evolution being retained. Under Alfvénic critical balance, $|k_{\parallel }|v_A\sim k_{\perp }U$, the same inequality gives ${\omega }_I\ll k_{\perp }U$. Rapid rotation relative to the horizontal nonlinear rate can therefore coexist with leading Alfvénic decoupling in the windowHigh electrical conductivity makes magnetic diffusion slow, and the corresponding inviscid approximation is also assumed. If $2\Omega \gtrsim k_{\perp }{v}_A$, rotational and magnetic restoring effects are comparable and the modes are mixed magneto-Coriolis disturbances; in the rotation-dominated limit they separate into fast inertial-like and slow magnetostrophic branches, rather than into independently evolving shear and pseudo-Alfvén polarizations.

At finite amplitude, independence is generally one-way in the usual anisotropic ideal magnetohydrodynamics reduction: the Alfvénic subsystem has no leading feedback from pseudo-Alfvénic perturbations, but it can still advect them. The small rotational-coupling condition removes an additional two-way coupling, rather than all nonlinear advection.

This is an asymptotic finite-timescale statement. The two nonrotating Alfvén polarizations are degenerate, so even a weak nonzero rotational coupling can rotate their superposition over the much longer time ${\omega }_I^{-1}$. Exact decoupling for arbitrarily long times does not follow from a small ratio alone. The strictly $k_{\parallel }=0$ nonpropagating limit is separate.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

In this question, the coefficient multiplying ${\nabla }^2\mathbf{B}$ is the magnetic diffusivity. Denote it ${\eta }_m$; it is the paper's $\eta$, with dimensions of length squared per time. If ${\eta }_{\mathrm{Ohm}}$ denotes electrical resistivity, then ${\eta }_m={\eta }_{\mathrm{Ohm}}/{\mu }_0$. The two conventions must not be interchanged.

For an imposed real positive $\omega$, retain the momentum equation and useElimination gives the exact Resistive damping of an Alfvén wave relationHere $k$ is complex because the driving frequency is real. Introduce the dimensionless diffusion parameter $ϵ={\eta }_m\omega /v_A^2\ll 1$. For propagation and attenuation along $+z$, choose the square-root branch with $\mathrm{Re}k,\mathrm{Im}k>0$. The spatial damping of a driven Alfvén wave isWith the convention $e^{i(kz-\omega t)}$, the amplitude decays as $e^{-\mathrm{Im}kz}$. Define its e-folding length $L_A=1/\mathrm{Im}k$ and the wavelength $\lambda =2\pi /\mathrm{Re}k$. ThenIf damping length instead refers to wave energy, the squared amplitude decays twice as fast: $L_E=L_A/2$ and $L_E/\lambda \sim 1/(2\pi ϵ)$. Stating the convention avoids a factor-of-two ambiguity. The temporal damping rate ${\eta }_m{k}^2/2$ at fixed real $k$ is a different specification from this imposed-frequency problem.

Linearized magnetic flux freezing gives the Eulerian magnetic field perturbationThe background is uniform, and incompressibility gives $\nabla \cdot \mathbf{\xi }=0$. For a horizontal displacement proportional to $e^{ikz}$, $\delta \mathbf{B}=ik{B}_0\mathbf{\xi }$ and $\delta B_z=0$. Thus the first-order magnetic pressure perturbation vanishes, and the magnetic-tension response to an incompressible displacement isBending a field line produces a restoring force proportional to displacement, like a spring obeying Hooke's law. Per unit mass, its squared spring frequency is $k^2{v}_A^2$, where $v_A=B_0/\sqrt{{\mu }_0\rho }$ is the Alfvén speed. The equilibrium can be independent of $z$ while the perturbation varies with $z$; neglect of background vertical structure does not set this perturbation's wavenumber to zero.

Work locally at $R_0$ with $\Omega =\Omega (R_0)$, and defineHere $\kappa$ is the radial epicyclic frequency. Combining the radial imbalance, the Coriolis force and the magnetic spring gives the spring model of magnetorotational instability:For a normal mode proportional to $e^{-i\omega t}$, the coefficient matrix isSetting its determinant to zero yields the ideal magnetorotational dispersion relation,Its two values of ${\omega }^2$ areso they are real. For $a>0$, their product is negative exactly when $a+D<0$, giving an exponentially growing normal mode. Conversely, $a+D>0$ also makes their sum positive and therefore gives two positive squared frequencies. A fixed nonzero magnetic mode is exponentially unstable for $k^2{v}_A^2+D<0$, oscillatory for $k^2{v}_A^2+D>0$, and marginal at equality. Stability against every allowed nonzero wavenumber in the unbounded ideal model requires $D\ge 0$; finite-domain restrictions can remove the unstable band.

For $\Phi =-GM/R$, the equilibrium relation gives ${\Omega }^2=GM/R^3$, so $D=-3{\Omega }^2$ and ${\kappa }^2={\Omega }^2$. Hence the Keplerian disk has the magnetorotational instability bandWithout the magnetic field, the polynomial becomes ${\omega }^2({\omega }^2-{\kappa }^2)=0$. The Rayleigh centrifugal stability criterion instead requires ${\kappa }^2>0$, equivalently $d(R^4{\Omega }^2)/dR>0$. A Keplerian disk satisfies that hydrodynamic criterion. Magnetic coupling therefore destabilizes a disk that has a restoring hydrodynamic epicyclic oscillation; the field couples angular-momentum exchange between elements, rather than acting only as an isolated restoring spring.

The plotted maximum is the maximum growth rate of ideal magnetorotational instability, $3\Omega /4$ at $k^2{v}_A^2=15{\Omega }^2/16$.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

Insert the interaction-picture expansion into the Schrödinger equation. The phases multiplying the basis states already solve the unperturbed evolution, leavingHermiticity requires $H_{21}^{\prime }=\hslash ({\omega }^{\prime }{)}^{\ast }$, not $\hslash {\omega }^{\prime }$ unless the coupling is real. Since the diagonal matrix elements vanish and ${\omega }_0={\omega }_2-{\omega }_1$, expanding the cosine givesThe conjugate pair of couplings conserves $|c_1{|}^2+|c_2{|}^2$. Near resonance, the difference frequency is slow while the sum frequency is approximately $2{\omega }_0$. Rapid terms alternate sign before an appreciable slow transfer accumulates. The rotating-wave approximation retains the slow terms when $|{\omega }^{\prime }|,|\omega -{\omega }_0|\ll \omega +{\omega }_0$; resonance alone is insufficient for a strong drive. Counter-rotating corrections include small shifts of the resonance and need not remain negligible at arbitrary strength or arbitrarily long times.

For the spin preparation, use $H=-\mathbf{\mu }\cdot \mathbf{B}=-\gamma \mathbf{s}\cdot \mathbf{B}$ and $\mathbf{s}=\hslash \mathbf{\sigma }/2$. Choose $B>0$ and $\gamma >0$ to fix helicity, with $|\uparrow ⟩=(1,0{)}^T$, $|\downarrow ⟩=(0,1{)}^T$ and ${\omega }_0=\gamma B$. The initial $|\uparrow ⟩$ is the lower state. For negative $\gamma$, reverse the RF rotation sense; pulse areas use $|\gamma |$, and signed Larmor precession determines the corresponding phase adjustment.

A circularly rotating magnetic drive uses two transverse fields in quadrature during $0<t<\tau$:With $U_0(t)=e^{+i{\omega }_{0}t{\sigma }_z/2}$, their interaction Hamiltonian is exactly constant,The minus sign in the second RF component selects co-rotation for the specified positive gyromagnetic ratio. Exponentiating this Pauli matrix givesThe laboratory phase of a resonant spin pulse includes free precession. Returning to the laboratory frame and removing a common phase,Unlike a linearly polarized RF field, this exactly co-rotating field has no counter-rotating component. A single linearly polarized field $B_{x,\mathrm{peak}}\cos ({\omega }_{0}t+\phi )$ gives the same slow rotation in the weak-drive rotating-wave approximation with $\Omega =|\gamma |B_{x,\mathrm{peak}}/2$. For the circular field above, $\Omega =|\gamma |B_1$. Keeping these amplitude conventions distinct avoids a factor-of-two pulse-duration error.

For a single harmonic pulse detuned by $\Delta ={\omega }_0-\omega$, the effective rotating-frame field tilts out of the transverse plane. Its rotation rate becomes $\sqrt{{\Omega }^2+{\Delta }^2}$, and its maximum population transfer is ${\Omega }^2/({\Omega }^2+{\Delta }^2)$. Exact inversion to $-z$ is impossible with one fixed nonzero-detuning pulse. Half transfer, needed for a transverse eigenstate, gives the equatorial reachability of a detuned spin pulse criterion $|\Delta |\le \Omega$. With that condition, changing the duration and RF phase can still prepare $+x$ or $+y$ at the chosen laboratory stopping time. The mere statement $\omega \approx {\omega }_0$ does not ensure this condition, because the drive can be much weaker than either frequency. If $|\Delta |>\Omega$, neither transverse target is reached by that single pulse. These restrictions concern fixed-frequency, fixed-amplitude single-pulse control; retuning, chirping or phase-controlled composite pulses change the available controls and can remove the restriction. No dissipative spin relaxation is included in this coherent two-state model.

Use the electronic weak-field Zeeman splitting model, neglecting hyperfine structure. The Landé g-factor is $g_g=2$ for $3S_{1/2}$ and $g_e=2/3$ for $3P_{1/2}$. The shift of an absorption frequency is thereforeHere the magnetic field is weak relative to spin-orbit coupling, so $J$ remains a useful quantum number, but the spectral line widths are small enough to resolve its magnetic substates.

For linear polarization parallel to $z$, the optical electric field has only the field-axis $\pi$ component. The electric-dipole selection rule is $\Delta m=0$. Its two allowed transitions are $m_g=m_e=\pm 1/2$, givingFor propagation perpendicular to $z$ with linear polarization also perpendicular to $z$, the optical electric field is an equal-amplitude combination of the two field-axis circular components. It drives ${\sigma }^{\pm }$ transitions with $\Delta m=\pm 1$. These connect $-1/2\to +1/2$ and $+1/2\to -1/2$, givingThus each orientation selects two lines, with twice the separation in the perpendicular-polarization case. Neither pair has an unshifted central component. This is polarization-selective Zeeman splitting of an alkali D1 line, an anomalous Zeeman effect rather than a normal orbital triplet.

The equation describes small plane-wave perturbations of a uniform, stationary plasma in ideal magnetohydrodynamics, with propagation parallel to the background magnetic field. The nonzero field defines $\hat{\mathbf{b}}={\mathbf{v}}_A/v_A$; $c_s$ is the sound speed for the assumed thermodynamic closure. Neglect viscosity, electrical resistivity, gravity and microscopic dispersive effects. A uniform background flow would instead introduce its Doppler-shifted frequency.

Decompose $\mathbf{u}={\mathbf{u}}_{\perp }+u_{\parallel }\hat{\mathbf{b}}$. The displayed matrix equation then becomesThus there are two independent transverse Alfvén wave polarizations with ${\omega }^2=k^2{v}_A^2$, restored by magnetic tension, and a longitudinal acoustic mode with ${\omega }^2=k^2{c}_s^2$, restored by pressure. The transverse modes have no first-order density compression; a strictly longitudinal displacement along the field does not bend it. These are the parallel-propagating ideal magnetohydrodynamic wave branches. In the general oblique problem, fast and slow magnetosonic modes mix compression and magnetic response; their parallel limits coincide with the faster and slower of these speeds. If $c_s=v_A$, all three velocity polarizations share the same frequency magnitude. The velocity equation alone does not enumerate nonpropagating entropy perturbations.