Advanced fluid dynamics course of the University of Oxford (MMathPhys/MSc paper)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15271h1.pdf

Solutions to both questions and all nine parts. The two-sphere calculation uses the leading free-draining approximation, with negligible connector drag; the limitations of that approximation are made explicit.

To first order in fluid velocity divided by the speed of light, the Lorentz transformation of electromagnetic fields gives ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{u}\times \mathbf{B}$. Applying Ohm's law in the local fluid frame gives $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\eta \mathbf{J}$. Neglecting the displacement current in Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. Substitution in Faraday's law givesHere ${\eta }_m$ is the magnetic diffusivity; the paper's $\eta$ is electrical resistivity, rather than magnetic diffusivity. Expanding the curl with the solenoidal magnetic-field constraint gives the resistive induction equation in its compressible form:The left side is the material derivative of the magnetic field. The three right-hand terms respectively describe field-line stretching and reorientation, amplification or dilution by compression or expansion, and magnetic diffusion. In particular, compression has $\nabla \cdot \mathbf{u}<0$ and amplifies a field component perpendicular to the compression.

The assumptions are nonrelativistic bulk motion, the low-frequency approximation permitting omission of the displacement current, a conducting-fluid continuum, isotropic local Ohm's law with spatially constant scalar electrical resistivity, constant magnetic permeability ${\mu }_0$, and $\nabla \cdot \mathbf{B}=0$. Hall, electron-pressure and other non-Ohmic corrections to the generalized Ohm's law are excluded. Incompressible flow is not required: imposing it would remove the compression term.

Let $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ be the total magnetic energy. Multiply the curl form of the resistive induction equation by $\mathbf{B}/{\mu }_0$ and integrate over the fixed volume. The identityfollows from the divergence theorem. The stated zero boundary fields remove its surface term. Applying it to the motional and resistive contributions yields the magnetic energy balance in resistive magnetohydrodynamics:The first term is the negative of the mechanical power delivered to the fluid by the Lorentz force density. It transfers energy between fluid motion and the magnetic field, with either sign. The second is the nonpositive magnetic contribution to Joule heating. A nonzero positive electrical resistivity therefore dissipates magnetic energy at rate $\eta {\int }_V{J}^{2}dV$; it does not force the total magnetic energy to decrease when fluid motion supplies a larger positive power.

Choose a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$. The resistive induction equation permits the evolution equationA gauge transformation changes $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ by a surface integral of the normal magnetic field, which is zero here. Thus the integrated magnetic helicity has gauge invariance under the stated boundary conditions.

Differentiate both factors in $H$. Use the stated vanishing-boundary assumption to remove the helicity surface flux; this can be implemented in the usual closed-boundary gauge $\mathbf{n}\times \mathbf{A}=0$, or by extending the localized field and integrating over all space with sufficient decay. The curl integration by parts identity then givesThe motional term is zero pointwise because $(\mathbf{u}\times \mathbf{B})\cdot \mathbf{B}=0$. The scalar-potential term integrates to zero by $\nabla \cdot \mathbf{B}=0$ and the boundary condition. The resistive magnetic-helicity balance is thereforeUnlike the resistive magnetic energy contribution, this signed integral need not have a fixed sign.

A magnetic field varying over a length $\ell$ has $|\nabla \times \mathbf{B}|$ of order $B/\ell$. Consequently, the resistive magnetic energy loss contains two inverse powers of $\ell$, while the resistive magnetic-helicity balance contains only one. Small-scale gradients can therefore dissipate substantial magnetic energy while retaining an approximately conserved large-scale magnetic helicity.

A dimensionally consistent version is the selective decay bound for magnetic helicity. With $D_B={\eta }_m{\int }_V|\nabla \times \mathbf{B}{|}^{2}dV/{\mu }_0$, the Cauchy-Schwarz inequality givesFor bounded $E_B$ and bounded, possibly nonvanishing $D_B$ as ${\eta }_m\to 0$, the resistive change in magnetic helicity tends to zero, even though resistive magnetic energy loss may remain finite. Magnetic energy is preferentially dissipated, while magnetic helicity is approximately conserved in this bounded-energy, bounded-dissipation limit.

This is not a comparison of quantities with different physical units or a universal ordering of fractional decay rates. For a single Beltrami field satisfying $\nabla \times \mathbf{B}=k\mathbf{B}$, both integrated quantities have fractional decay rate $2{\eta }_m{k}^2$ without fluid motion. The boundedness assumptions and distribution of magnetic helicity across spatial scales matter.

The stress tensor of the incompressible Newtonian fluid is $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ is the rate-of-strain tensor. The homogeneous Stokes equations give $\nabla \cdot \mathbf{\sigma }=0$. Apply the divergence theorem to the fluid between the body and a surrounding sphere $S_R$. If ${\mathbf{n}}_B$ points from the body into the fluid, the fluid-domain normal on the inner boundary is $-{\mathbf{n}}_B$. HenceThis is the force flux at infinity in Stokes flow. A nonzero resultant force must be supported by a stress of order $R^{-2}$ on a surface of area of order $R^2$. More precisely, the leading exterior solution is a Stokeslet with force on the fluid $\mathbf{f}=-\mathbf{F}$:Thus $\mathbf{u}=O(R^{-1})$, $p-p_{\infty }=O(R^{-2})$ and $e=O(R^{-2})$. These are leading generic orders for a translating body with nonzero drag; the integral alone does not determine every angular component, and force-free disturbances may decay faster.

For two Stokes flows in the same fluid domain with the same dynamic viscosity, symmetry of the Newtonian fluid stress tensor and incompressibility give the local reciprocal identityIn index notation, its remaining cross-gradient terms are both $2\mu e:e^{\prime }$. On the remote sphere the integrated reciprocal flux tends to zero: the velocity is $O(R^{-1})$, the stress tensor is $O(R^{-2})$ and the area is $O(R^2)$. On the body, the no-slip boundary condition makes the two velocities the constants $\mathbf{U},{\mathbf{U}}^{\prime }$. The Lorentz reciprocal theorem therefore givesBy the Linearity of Stokes flow, the sum of the two velocity-pressure fields satisfies the boundary conditions for body velocity $\mathbf{U}+{\mathbf{U}}^{\prime }$. Uniqueness of the exterior Stokes flow identifies it with the third solution. Since the stress tensor and its surface integral are linear too, $\mathbf{F}(\mathbf{U}+{\mathbf{U}}^{\prime })=\mathbf{F}(\mathbf{U})+\mathbf{F}({\mathbf{U}}^{\prime })$; scalar multiplication follows in the same way. Thus $\mathbf{F}=-A\mathbf{U}$ for a fixed-orientation hydrodynamic resistance matrix $A$. Substituting arbitrary $\mathbf{U},{\mathbf{U}}^{\prime }$ in the reciprocal identity yieldsThe minus sign is natural because viscous dissipation gives $-\mathbf{F}\cdot \mathbf{U}=2\mu \int e:edV>0$ for nonzero rigid translation in an otherwise stationary fluid; $A$ is a positive-definite matrix.

Use the free-draining approximation: neglect the hydrodynamic interaction between the widely separated spheres and neglect the connector's viscous drag. Stokes law gives ${\mathbf{f}}_i=-6\pi \mu a_i\mathbf{U}$ for each sphere. Their resultant isThis is the leading two-sphere approximation, not an exact finite-separation hydrodynamic resistance matrix. Writing $d=|{\mathbf{r}}_2-{\mathbf{r}}_1|$ and $\mathbf{n}=({\mathbf{r}}_2-{\mathbf{r}}_1)/d$, the first Stokeslet reflection gives, if the connector remains negligible,The negative correction expresses reduced drag from fluid entrainment by the other sphere. A very thin but arbitrarily long rod need not have negligible total drag: its typical transverse drag scales as $\mu d/\log (d/b)$ for rod radius $b$. Thus the simple boxed result uses the intended negligibly dragging connector limit as well as $a_i/d\ll 1$; the two printed size inequalities alone do not specify a uniform limit in which rod drag vanishes.

Hold the two-sphere object fixed in a uniform stream $\mathbf{V}$. In the free-draining approximation, its drag forces are ${\mathbf{f}}_i={\zeta }_i\mathbf{V}$, with ${\zeta }_i=6\pi \mu a_i$. The hydrodynamic torque about ${\mathbf{r}}_c$ isTo make it zero for every incoming-stream direction, choose the hydrodynamic centre of a two-sphere dumbbell:It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin torque enters this leading approximation.

Finite-separation hydrodynamic interactions and connector drag can change the displayed location; the calculation has the same approximation as part c.

For rotation about the hydrodynamic centre of a two-sphere dumbbell, the sphere centres have velocities ${\mathbf{v}}_i=\mathbf{\Omega }\times ({\mathbf{r}}_i-{\mathbf{r}}_c)$. Their resultant drag force in the free-draining approximation isEach sphere also experiences a spin drag torque, but its isolated rotation causes no net force. Those spin contributions therefore do not alter the cancellation.

This illustrates translation-rotation reciprocal coupling. About any fixed origin, write the fluid's rigid-body force and torque asThe transposed cross-blocks follow from the Lorentz reciprocal theorem. If pure translation causes zero torque about an origin for every $\mathbf{U}$, then $B^T=0$, hence pure rotation about that origin causes zero force. This implication is general for linear Stokes flow; the existence and leading location of such an origin here use the two-sphere geometry and the approximation in part c.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/advanced_fluid_dynamics_2024.pdf

Solutions to both questions and all nine parts. Part 2(d) contains a false identity in the source; its counterexample and the consistent free-energy balance are supplied.

Start in an inertial frame with the inviscid magnetohydrodynamic momentum equation,Here $\rho$ is mass density, $P$ is the isotropic fluid pressure, $\mathbf{v}$ is inertial-frame velocity, $\mathbf{J}$ is electric current density, $\mathbf{B}$ is the magnetic field, and $\Phi$ is gravitational potential per unit mass. In the low-frequency approximation to Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. The vector calculus identityseparates the Lorentz force density into magnetic tension and a magnetic pressure gradient. Transform the material acceleration to a rigid frame rotating with constant angular velocity ${\mathbf{\Omega }}_0$, using the stated frame transformation. The rotating-frame magnetohydrodynamic momentum equation isThe final terms are the Coriolis force and centrifugal force densities; $\mathbf{u}$ is the velocity relative to the rotating frame, and $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ uses that frame.

This requires a smooth conducting-fluid continuum with nonrelativistic motion, a constant magnetic permeability ${\mu }_0$, the solenoidal magnetic-field constraint, a negligible displacement current and negligible net electric-charge force. Viscous forces and anisotropic fluid stresses are omitted. A time-dependent frame would add an Euler force. Incompressible flow is not needed for this force equation. Finite electrical resistivity does not change its displayed instantaneous magnetic force; the later frozen-field calculation additionally uses ideal magnetohydrodynamics. A differentially rotating fluid does not define a single rigid frame with spatially varying ${\mathbf{\Omega }}_0$: choose a fixed local reference rate and retain the fluid's shear in $\mathbf{u}$.

In an inertial frame, a circular orbit at radius $R$ has radial acceleration $-R\Omega (R{)}^2{\mathbf{e}}_R$. Neglecting fluid and magnetic forces givesIn a local rotating frame with ${\Omega }_0=\Omega (R_0)$, the fluid is instantaneously at rest at the reference radius $R_0$. The radial centrifugal force $+\rho R_0{\Omega }_0^2$ then balances the inward gravitational force. Away from $R_0$, the fluid's nonzero rotating-frame velocity must also be retained; the entire differentially rotating disk is not at rest in that one frame.

Take outward radial force as positive and keep the displaced element's angular rate fixed at ${\Omega }_0=\Omega (R_0)$. Its imbalance isUse the equilibrium relation from part b, ${\Phi }^{\prime }(R)=R\Omega (R{)}^2$, and a first-order Taylor expansion:The absolute values are needed for a magnitude. As printed, the proposed magnitude has a signed right side; it would be negative for outward displacement in a Keplerian disk. For decreasing ${\Omega }^2$, the signed imbalance above points outward and therefore amplifies that fixed-angular-rate displacement. This calculation fixes the angular rate, rather than the specific angular momentum; the latter is the conserved quantity for an uncoupled inviscid radial displacement and leads to the Rayleigh centrifugal stability criterion.

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 $k^2{v}_A^2=15{\Omega }^2/16$.

Using the given linear momentum balance, multiply by the velocity and integrate over the fixed volume:This is integration by parts component by component, or the divergence theorem. The first term is boundary stress power. With component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the second is the double contraction $\Phi =\sigma :K$.

For a Newtonian fluid, $\sigma =-pI+2\mu e$, where $e=(K+K^T)/2$ is the rate-of-strain tensor and $\mu \ge 0$ is dynamic viscosity. Incompressibility removes the pressure contribution, and the symmetric tensor $e$ is orthogonal to the antisymmetric part of $K$. ThereforeThe displayed energy identity follows for the stated linear balance. A full advective momentum equation in a fixed control volume would also contribute an advective kinetic energy flux. The nonnegativity statement here is for a purely Newtonian viscous stress; a viscoelastic stress can exchange stored free energy with the flow.

Let $K_{ij}={\partial }_j{u}_i$, $C_{ij}=⟨R_i{R}_j⟩$, $\alpha =2H/\zeta$ and $D_R=2k_{B}T/\zeta$. The configuration equation of the Hookean dumbbell model becomesHere the paper's $k$ denotes the Boltzmann constant $k_B$. Integrating the second moment equation against $R_i{R}_j$ and discarding boundary terms at infinite separation givesFor example the spring drift differentiates each factor $R_i,R_j$, while the Laplacian gives ${\nabla }_{\mathbf{R}}^2(R_i{R}_j)=2{\delta }_{ij}$. The Hookean dumbbell conformation equation isIn the row-vector convention of $\mathbf{R}\cdot \nabla \mathbf{u}$, the displayed gradient tensor is $K^T$, giving exactly the transposes in the requested equation. The dimensionless conformation tensor $A=HC/(k_{B}T)$ consequently has upper-convected derivative $-(A-I)/\lambda$, with polymer relaxation time $\lambda =\zeta /(4H)$. This normalization matters: the unscaled $C$ has equilibrium value $(k_{B}T/H)I$, rather than $I$.

Write the centre position as $\mathbf{x}$ and the bead positions as ${\mathbf{r}}_{1,2}=\mathbf{x}\mp \mathbf{R}/2$. In the affine velocity field, $\mathbf{u}({\mathbf{r}}_{1,2})=\mathbf{u}(\mathbf{x})\mp K\mathbf{R}/2$. The midpoint follows the flow, so its two slip velocities areEach bead loses mechanical power $\zeta$ times its squared slip speed. Adding the two powers proves the drag dissipation of a bead-spring pair:The factor one half comes from two equal-and-opposite bead slips, not from assigning only one bead a drag force.

The claimed equality does not hold for the printed joint distribution. The original PDF specifies a centered Maxwell-Boltzmann distribution of $\mathbf{v}=\dot{\mathbf{R}}$ independent of $\mathbf{R}$, withSince the double average includes number density $n$, evaluating the Gaussian integrals givesThe cross term vanishes because the conditional mean of $\mathbf{v}$ is zero. In particular, set $K=0$ and take the equilibrium configuration probability density ${\psi }_{\mathrm{eq}}\propto e^{-H{R}^2/(2k_{B}T)}$. The left side is $3n\zeta k_{B}T/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier conformation tensor equation can prove the printed statement.

The discrepancy is thermal bead drag and heat-bath power: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf{R}}$ in the earlier configuration continuity equation is a probability current velocity, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.

A consistent overdamped result is the free-energy balance of a Hookean dumbbell. Define $\mathbf{q}=H\mathbf{R}+k_{B}T{\nabla }_{\mathbf{R}}\log \psi$ andDifferentiate the relative entropy and use the configuration Fokker-Planck probability current $\psi \mathbf{V}$. After integration by parts,The affine osmotic term is $n{k}_{B}T\int \nabla \psi \cdot K\mathbf{R}d\mathbf{R}=-n{k}_{B}T\mathrm{tr}K=0$. HenceThe polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has ${\sigma }_p:K=nHC:K$ in an incompressible flow. It stores or releases free energy as well as dissipating it, so the total stress-work term in part a need not itself be a purely dissipative quantity for this viscoelastic material. Stress power equals $\mathcal{D}$ only for stationary free energy, and $\mathcal{D}$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/adavanced-fluid-dynamics.pdf

Solutions to both questions and all eight parts. The diffusion coefficient and the amplitude-versus-energy damping convention are stated explicitly; the final rotation calculation uses the supplied isolated-sphere incident flow.

Take the inviscid limit of the Navier-Stokes equation, neglect gravity, and use ideal magnetohydrodynamics. The governing equations areThe continuity equation also gives $D_t\rho =0$. Write the first-order magnetic perturbation as $\mathbf{b}={\mathbf{B}}^{\prime }$ and linearize about the uniform stationary background:For a nonzero-wavenumber plane wave proportional to $e^{i(kz-\omega t)}$, the two divergence constraints imply $v_z=b_z=0$. The longitudinal momentum equation then gives $P^{\prime }=0$ for that mode, and ${\partial }_t{\rho }^{\prime }=0$ gives ${\rho }^{\prime }=0$ for nonzero $\omega$. Thus the propagating Alfvén wave is transverse. Its two independent transverse directions have the same dispersion relation.

The Linear Alfvén wave equations reduce toEliminating either amplitude givesFor $B_0>0$, the branch traveling along $+z$ has $\omega =k{v}_A$ and $\mathbf{b}=-\sqrt{{\mu }_0{\rho }_0}\mathbf{v}$. The opposite branch reverses that amplitude sign. Finite viscosity would additionally damp the motion; it is neglected in obtaining the real ideal dispersion relation.

Subtract the constant equilibrium magnetic energy density. Since ${\mathbf{B}}_0\cdot \mathbf{b}=0$, the perturbation densities areThe traveling-wave amplitude relation from part a givesThis is pointwise Alfvén wave energy transport for one traveling wave, not a claim that every superposition or standing wave has pointwise equipartition.

For the real fields written as $\mathbf{v}=\mathrm{Re}(\hat{\mathbf{v}}{e}^{i(kz-\omega t)})$ and similarly for $\mathbf{b}$, the period averages areThe factors one half distinguish a sinusoid's peak amplitude from its mean squared value; the same complex-vector convention also covers circular polarization of waves.

In ideal magnetohydrodynamics, the first-order electric field is ${\mathbf{E}}_1=-\mathbf{v}\times B_0{\mathbf{e}}_z$. Retaining its product with both the background and first-order magnetic field gives the Poynting vectorThe first term is transverse and oscillatory, with zero period average. The second is quadratic and transports wave energy. A potential quadratic ideal electric-field term $-\mathbf{v}\times \mathbf{b}$ vanishes here because the real traveling-wave fields are parallel or antiparallel.

For the $+z$ branch, $\mathbf{b}=-\sqrt{{\mu }_0{\rho }_0}\mathbf{v}$, soFor the opposite branch, its sign reverses. The averaged Poynting flux is the total wave energy density, including equal kinetic and magnetic shares, multiplied by the group velocity. This agrees with Alfvén wave energy transport. Although the flux in this expression is electromagnetic, it supplies the transport of both components of the coupled fluid-field wave.

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.