Codex @codex  0

Put $U=Q/{\alpha }_0$, $R_0=R({\alpha }_0)$, $c_0^2={\alpha }_0{\tilde{P}}^{\prime }({\alpha }_0)$ and $\nu =\omega -kU$. Let the area and velocity perturbations have amplitudes $a$ and $w$. Linearizing mass conservation and momentum, and using $R_0^{\prime }=-n{R}_0/{\alpha }_0$, givesFor nonzero $k$, eliminate $w=\nu a/(k{\alpha }_0)$ to obtain the friction-induced instability of a collapsible-tube flow relationAt the stated threshold $nU=c_0$, it factors asTherefore one mode has real $\boxed{\omega =k(U+c_0)}$, while the other has $\omega =k(U-c_0)-i{R}_0$ and decays. The PDF's real-frequency statement is valid for the neutral branch, not for both roots. The Fourier convention $e^{i(kx-\omega t)}$ makes positive imaginary frequency mean growth.

Use a time-harmonic wave with time dependence $e^{-i\omega t}$ and let the same symbols denote phasors. The wave amplitude vector obeysAt fixed material contrast, the stated low-frequency condition permits neglecting the diagonal frequency term to zeroth order. All remaining coefficient matrices are scalar multiples of the same exchange matrix, so they commute. Withthe transfer relation isSolve the second row for the outgoing left wave amplitude and substitute in the first. Since $C^2-H^2=1$, this gives the zero-frequency scattering through an elastic layer coefficientsThus the zero-frequency limit depends only on the endpoint seismic impedances. Monotonicity supports the given scale comparison but is not needed to integrate the zero-frequency equation. If $Z_a=Z_b$ in a monotone profile, seismic impedance is constant: there is no directional coupling, with only wave phase accumulation at nonzero frequency.

At a sharp interface, continuity of velocity and traction gives, for unit incident velocity from the left,Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted velocity by its elastic-wave energy flux gives $T_{ba}=\sqrt{Z_b/Z_a}{t}_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A stress reflection coefficient has the opposite sign to the velocity reflection coefficient; the convention matters.

Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing matrix is $S_0=\left(\begin{array}{cc}r& t\\ t& -r\end{array}\right)$, with $r^2+t^2=1$. Therefore $S_0^{†}{S}_0=I$, proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident wave amplitudes, not just incidence from one side.

For all real frequencies the harmonic equations instead giveTheir right sides are equal; the diagonal terms $\pm i\omega /\beta$ have zero real contribution. ConsequentlyThe mean flux for real-part phasors is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real frequency. This relies on real lossless moduli and mass density; absorption would change the identity.

At $c_L=\beta$, the half-space decay exponent $p$ tends to zero and the dispersion relation requires $qh=n\pi$. These are the Love-wave cutoff frequencies,An overtone exists strictly above its cutoff. At the cutoff its half-space field is constant in depth rather than decaying, so it is a limiting nonlocalized state, not a trapped finite-energy eigenfunction. The $n=0$ fundamental has cutoff zero and exists for every positive frequency. Each fixed branch approaches $c_L={\beta }_0$ from above at high frequency. The half-space penetration depth $1/p$ diverges on approach to cutoff.

Choose depth $z$ positive downwards, the free surface at $z=0$ and the interface at $z=h$. A Love wave has only transverse displacement field $u_y=U(z)e^{i(\kappa x-\omega t)}$. Take $\omega >0$, $\kappa =\omega /c_L$, and defineA trapped mode requires ${\beta }_0<c_L<\beta$, so both $q,p$ are positive. The SH equations are $U_0^{\prime \prime }+q^2{U}_0=0$ in the layer and $U^{\prime \prime }-p^{2}U=0$ in the half-space. Zero shear traction at the free surface and decay at large depth selectContinuity of displacement field and shear traction at $h$ gives $C=A\cos (qh)$ and $-{\mu }_{0}qA\sin (qh)=-\mu pC$. ConsequentlySubstituting the definitions of $q,p$ yields the requested speed-frequency dispersion relation. For negative frequency use $|\omega |$ in both vertical wavenumbers; the real displacement field has conjugate frequency partners. Neither equality in the trapped-speed interval gives a strictly localized finite-frequency Love mode.

In the bilinear identity choose field $a$ to be the complex conjugate of the field at frequency $\omega$, equivalently its $-\omega$ partner, and field $b$ to be the field itself. The squared frequencies are equal, so the mass density terms cancel:Multiply by $i\omega$ and use ${\hat{v}}_i=-i\omega {\hat{u}}_i$, ${\hat{v}}_i^{\ast }=i\omega {\hat{u}}_i^{\ast }$ to obtainFor the real fields defined in this question as the sum of positive and negative frequency components, time averaging cancels the $e^{\pm 2i\omega t}$ terms. Their mean body-force work mass density is exactly ${\hat{f}}_i^{\ast }{\hat{v}}_i+{\hat{v}}_i^{\ast }{\hat{f}}_i$, and their outward mean elastic-wave energy flux is $-t_i^{\ast }{\hat{v}}_i-{\hat{v}}_i^{\ast }{t}_i$. Thus mean body-force power equals mean outward boundary power. Mean stored energy does not accumulate in a periodic steady field.

There is no extra factor $1/2$ here: the specified real field is $\hat{f}{e}^{-i\omega t}+{\hat{f}}^{\ast }{e}^{i\omega t}$, not $\mathrm{Re}(\hat{f}{e}^{-i\omega t})$. For the latter phasor convention, both averages instead have the usual factor $1/2$. At $\omega =0$ velocity and both powers vanish.

Choose field $a$ to be $({\hat{\mathbf{u}}}^a{)}^{\ast }$ at frequency $-{\omega }_a$, and field $b$ to be ${\hat{\mathbf{u}}}^b$ at ${\omega }_b$. Zero body forces and zero boundary tractions make all source and surface terms vanish. ThereforeFor distinct squared frequencies this proves weighted orthogonality,Assume positive mass density bounded above and below and finite-energy fields in the ordinary bounded domain. The weighted space $H=L^2(\mathcal{D},\rho dV{)}^3$ is a separable Hilbert space. Select one unit eigenfunction for each distinct squared frequency. They form an orthonormal set. Let $\{d_j\}$ be a countable dense subset of $H$; assign to each eigenfunction a point $d_j$ within distance $1/3$. Two orthogonal unit eigenfunctions have distance $\sqrt{2}$, so the same $d_j$ cannot be assigned to both. This gives an injection into a countable set.

Hence there are at most countably many squared eigenfrequencies, and at most twice that many signed frequencies. Multiplicity at one frequency does not change the argument. A zero frequency associated with rigid translations or rotations is included. This proves the countability of elastic eigenfrequencies directly; proving discreteness or finite multiplicity would require additional spectral regularity arguments.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

For a photon, $p^i{p}_i=0$ and the null displacement along the ray is parallel to $p^i$. Consequently $p_{j}d{x}^j=0$, and the first term of the momentum law vanishes. Write $d{x}^i=p^{i}d\lambda$ with an appropriate path parameter. ThenFor a static potential, a photon emitted and received by laboratories at rest in the background frame has energy proportional to $p^0$. Using $E=h_{\mathrm{P}}\nu$, the scalar-force gravitational redshift isRestoring $c$, the leading fractional shift is $\Delta \nu /\nu =-\Delta {\Phi }_N/c^2$, where ${\Phi }_N$ is the dimensional Newtonian gravitational potential. Near Earth's surface, an upward displacement $H$ has $\Delta {\Phi }_N=gH>0$, hence $\Delta \nu /\nu =-gH/c^2$: the receiver sees a redshift.

Thus yes: the predicted weak-field frequency shift agrees with the Pound-Rebka experiment. The calculation concerns stationary emitter and receiver, with identical local transition-energy standards; it does not add a Doppler shift due to their relative motion. Agreement with this gravitational redshift measurement alone does not establish the tensor theory of general relativity.

Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.

For a plane mode with phase $\theta =kx+mz-\omega t$, let $U,W,P,B$ be its complex amplitudes. The linear equations giveSubstituting into $-i\omega W=-imP+B$ gives the dispersion relationNo waves enter from above. The appropriate radiation condition therefore selects upward vertical group velocity, not upward phase propagation:The boundary-forced internal gravity wave satisfying the boundary condition isThe displacement is $ϵ\sin \theta$, consistent with $b=-N^2$ times the displacement. For a right-moving boundary choose $k>0$, $\omega >0$ without loss of generality. Then $m<0$, and the constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m>0$: the crests rise toward the right. Their acute angle with the horizontal isThe group velocity is along these crests in the upward-right direction, whereas the wavevector and phase propagation point downward-right.

The physical pressure perturbation is ${\rho }_0{p}^{\prime }$. Pressure work at the lower boundary, to second order in amplitude, is its period-mean pressure times vertical boundary velocity. Static hydrostatic terms average to zero over a cycle. ThereforeAs an independent energy check, the wave's mean kinetic-plus-buoyancy energy density is ${\rho }_0{ϵ}^2{N}^2/2$, and multiplication by $c_{gz}$ gives exactly this upward energy flux.

For the exact horizontal mean-momentum equation, incompressibility writes the horizontal momentum equation asAverage over a horizontal period. Require the pressure and velocity fluxes to be periodic, or more generally their horizontal derivative averages to vanish; in particular there must be no imposed mean pressure gradient. ThenHorizontal averaging also makes ${\partial }_z\overline{w}=0$; the zero mean boundary displacement rate gives $\overline{w}=0$. These conditions exclude an independent mean mass flux or horizontal force. The mean equation applies in the fluid, with flattening of the small moving boundary understood at the order used above.

For slow switch-on, define $A(z,t)=ϵ(t-z/c_{gz})$, using $ϵ(s)=0$ for $s<0$. The outgoing internal-wave envelope radiation condition is $A_t+c_{gz}{A}_z=0$. Replace $ϵ$ in the oscillatory fields by $A$. A modulation time $T$ must satisfy $|\omega |T\gg 1$ and $|m|c_{gz}T\gg 1$, with a frequency bandwidth sufficiently narrow to linearize the dispersion relation. The exact boundary derivative also contains $\dot{ϵ}\sin (kx-\omega t)$; it is smaller by the slow-modulation parameter and is beyond this leading envelope approximation.

To leading slow-modulation order, the wave momentum flux isSince ${\partial }_z{A}^2=-c_{gz}^{-1}{\partial }_t{A}^2$, integrating the mean-momentum equation from rest gives the internal-wave momentum depositionThis is second order in wave amplitude, with smaller relative slow-modulation corrections. Mean-flow feedback on the leading wave is higher order. The mean current is created as the wave envelope arrives; an already uniform wavetrain has no local flux divergence.

Finally move horizontally with speed $c_p=\omega /k$. The initial uniform relative flow is $-c_p$, with infinite total energy in the unbounded column. Its finite mean-energy change is insteadAt finite time the envelope is absent sufficiently far above, so integrating the momentum equation givesThis is the boundary work and mean-flow energy of an internal wave relation. In the moving frame the boundary pattern is stationary to leading modulation order. It extracts energy from the incident mean flow and converts it to outgoing internal-wave energy; in the laboratory frame the same energy is supplied as boundary work.

Here write $\sigma$ for the buoyancy perturbation about the resting constant-$N$ stratification; the total buoyancy also includes its background gradient $N^2$. Use kinematic perturbation pressure $p$. The weak-forcing Linearized Boussinesq equations areWeakness means that perturbation advection is small compared with the retained linear terms. With the specified streamfunction convention, the transverse vorticity isDifferentiate the horizontal momentum equation in $z$ and subtract the $x$ derivative of vertical momentum. The pressure derivatives cancel, givingThe second equation uses ${\sigma }_t=N^2{\psi }_x$ and is the forced internal-wave equation.

For free waves at frequency $|\omega |\ll N$, the internal gravity wave dispersion relation gives $|k|/|m|\simeq |\omega |/N\ll 1$. Thus the horizontal response scale is much larger than the vertical scale. Vertical acceleration is small in the vertical momentum equation, leaving the hydrostatic approximation $p_z=\sigma$. Correspondingly the transverse vorticity is predominantly $u_z$: neglect $-w_x={\psi }_{xx}$ relative to ${\psi }_{zz}$ in the inertial vorticity term. The restoring term $N^2{\psi }_{xx}$ must still be retained, since it balances the small time derivatives.

For $m\ne 0$ put $\psi =e^{imz}P(x,t)$, $\sigma =e^{imz}B(x,t)$ and interpret physical fields as real parts. The reduced equation isOne can solve it while obtaining the buoyancy simultaneously. Define $U=imP$ and $H=iB/(cm)$. The first-order hydrostatic equations becomeHence $R_{+}=U+H$ and $R_{-}=U-H$ obeyApply the method of characteristics with zero fields in the past. Withthe solution is $R_{\pm }=I_{\pm }$. Consequently the hydrostatic response to a localized horizontal force isThese are exactly the required two characteristic integrals, derived with their relative sign fixed by the buoyancy equation. Quiet-past initial conditions exclude additional freely propagating solutions.

For the consistency check, set $\tau =s-t$, so both integrals run over the fixed range $(-\infty ,0)$. Differentiation at fixed $x,z$ now givesThis change of variable is important: it includes the time dependence of both the sampled forcing and the moving integration geometry without omitting a term. Each straight path intersects $|x|\le a$ for a time interval of length at most $2a/c$. If $L=\sup |f_{xt}|$, thenThe characteristic size of the hydrostatic buoyancy and pressure-gradient balance is $|m|a{f}_{\max }$, as follows from the analogous bound on $B$ in its integral formula. Thus the omitted acceleration has relative scale at most a constant timesTogether with the low-frequency assumption, this verifies the requested hydrostatic consistency. It is a comparison of characteristic magnitudes, rather than division by a buoyancy value at a node where both terms may vanish. The capital $N$ here is the buoyancy frequency, as printed in the PDF.

For $t\ge 0$, only $0\le s\le t$ contributes. If $|x|>ct+a$, every point $x\mp c(s-t)$ lies outside the source interval, so both integrals vanish:This finite propagation cone belongs to the reduced hydrostatic wave equation.

The hypotheses do not require an even forcing, so no single symmetric graph follows for every permitted $f$. For an illustrative smooth even nonnegative profile, $U$ is even and $H$ is odd. At late time and fixed $x$, if $A_f={\int }_{-a}^{a}f(r,\infty )dr$, thenThus the streamfunction develops a broad nearly level interior response with outgoing fronts, while the buoyancy changes sign across the source and approaches opposite interior plateaus to its left and right. Both vanish beyond the fronts. For $m>0$, a vertical phase with $\sin (mz)=1$ gives real fields $\psi =U/m$ and $\sigma =cmH$. The figure plots those profiles up to positive normalization factors, for an even compact forcing increasing slowly from zero.

Put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and take a compactly supported Lighthill stress tensor. The isotropic term in that tensor is $(p-c_0^2\rho ){\delta }_{ij}$: the Kronecker delta is implicit in the printed shorthand. Convolving the Lighthill acoustic analogy with the retarded acoustic Green function and integrating twice by parts in the source coordinates givesThe boundary terms vanish because of compact support. For source size $\ell$, the acoustic compact-source approximation requires $\omega \ell /c_0\ll 1$. At $r\gg \ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf{y},t)d^{3}y$. In the radiation region $\omega r/c_0\gg 1$, the leading two spatial derivatives act on the retarded time rather than on $1/r$ or $\mathbf{n}$:Thus the leading acoustic quadrupole density isFor the compact acoustic quadrupole Mach-number scaling, let $U$ be the source fluctuation speed, $m=U/c_0\ll 1$, and use the advective source time $\ell /U$. With $T_{ij}=O({\rho }_0{U}^2)$, one has $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{U}^4\ell )$. ConsequentlyThe fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source frequency scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that Mach number estimate.

For the reflection of a P-wave from a rigid plane, use the displacement convention $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$, where $\varphi$ is the P potential and $\psi$ the SV potential. Put $\omega =k{c}_p$, $\kappa =k\sin \theta$, $a=k\cos \theta$, and $b=\sqrt{{\omega }^2/c_s^2-{\kappa }^2}$. The reflected waves must share the tangential wavenumber and frequency and propagate towards negative $y$, so writeThe rigid barrier requires both displacement components to vanish. On $y=0$ this gives $\kappa (1+A)-bB=0$ and $a(1-A)-\kappa B=0$. Solving,The total P potential is the given incident wave plus ${\varphi }_R$, and the S potential is ${\psi }_R$. Reversing the sign convention for $\psi$ reverses $B$ but leaves the displacement unchanged. At normal incidence $A=1,B=0$.

An evanescent reflected S wave would require $b^2<0$, or $\sin \theta >c_p/c_s$. In a stable isotropic elastic solid $c_p>c_s$, so this is impossible for real incidence angles: $b^2=k^2[(c_p/c_s{)}^2-{\sin }^2\theta ]>0$. The reflected S wave is never evanescent in the physical isotropic case. The inequality identifies the hypothetical critical-angle case if the speed ordering were reversed; it is S-to-P conversion, not this P-to-S conversion, that can become evanescent in an ordinary solid.

Use geometrized units, and denote the beacon's proper time by $s$ to distinguish it from the short emitted duration $\tau$. Take $s=0$ at release. The period $M$, the duration $\tau$, and the wavelength $\lambda$ are measured by the beacon's own clock. Its radial timelike geodesic has conserved Killing energy per unit rest mass $E=fdt/ds$, where $f=1-2M/r$. Initially $dr/ds=0$ at $r=8M$, so $E=\sqrt{f(8M)}=\sqrt{3}/2$. Normalization of the four-velocity givesFor the Schwarzschild radial fall from rest at a finite radius, introduceIndeed $ds/d\eta =8M(1+\cos \eta )$ and $dr/d\eta =-4M\sin \eta$ reproduce the inward radial equation. At the Schwarzschild event horizon, $r=2M$ and $\cos {\eta }_H=-1/2$. HenceThis is the finite falling-clock time to the event horizon. By contrast, Schwarzschild time obeys $dt/ds=E/f$ and diverges at crossing. Near the horizon $r-2M\simeq E(s_H-s)$, so $dt/ds\simeq 2M/(s_H-s)$ and $t\simeq -2M\log ((s_H-s)/M)+\text{constant}$.

Every outward optical pulse emitted while $r>2M$ can reach a distant observer, and no pulse emitted after entry into the future black-hole region can do so. The infinite coordinate crossing time does not produce infinite emissions, because the pulse clock runs in proper time. If there is a pulse at release, the escape times are $s_n=nM$ for $n=0,\dots ,23$. Thus 24 pulses are received, counting the pulse at release; 23 are emitted subsequently during the fall. More generally the PDF does not specify the phase of the pulse clock. For a first emission at $s=\delta$ with $0\le \delta <M$, the finite pulse count for a falling Schwarzschild beacon isThe ceiling handles a pulse scheduled exactly at crossing correctly: it cannot escape. This count concerns pulses associated with the fall, not any pulses emitted before release, and assumes an ideal observer able to detect arbitrarily redshifted signals.

To find the stretching of an optical pulse, use the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log (r/(2M)-1)$. An outward radial null geodesic has constant Schwarzschild retarded null coordinate $u=t-r_{\ast }$. At an observer's fixed large radius $R$, its arrival time is $t_R=u+r_{\ast }(R)$; therefore only the varying emission value of $u$ contributes to the interval between arriving wavefronts. Differentiating along the falling source givesThe outgoing pulse redshift from a radial Schwarzschild infaller includes both gravitational redshift and the emitter's inward Doppler shift. A short pulse emitted at radius $r_e$ has, to leading order in its duration,For a finite receiver radius, replace $\mathcal{D}$ by $\sqrt{f(R)}\mathcal{D}$ to measure the receiver's proper time and frequency; this correction is negligible for $R\gg M$. The inverse scaling of frequency follows because the number of carrier cycles is unchanged when the pulse is stretched. Here $\nu =1/\lambda$ is ordinary frequency, not angular frequency, in units with light speed one.

Under the release-phase convention, the last emission is at $s_e=23M$. Solve $8({\eta }_e+\sin {\eta }_e)=23$ to obtain ${\eta }_e\simeq 1.94374184$, $r_e\simeq 2.54255983M$, and $\mathcal{D}(r_e)\simeq 7.49122492$. Because $\tau \ll M$ and this pulse begins $0.683364M$ before crossing, the local short-pulse approximation applies. Thus its approximate observed duration and frequency areWithout fixing that phase, these numerical coefficients are not unique: put $s_e=\delta +(N(\delta )-1)M$, find $r_e$ from the parameterization, and use the boxed general formulas. For a last pulse close to crossing, with $\Delta =s_H-s_e\ll M$ but $\tau \ll \Delta$, the horizon redshift of a Schwarzschild infaller gives the useful estimatesIf a phase instead places the pulse within one pulse-duration of crossing, $\Delta \lesssim \tau$, a single constant stretching factor is inappropriate. The received extent of a complete escaping pulse is exactly $u(s_e+\tau )-u(s_e)$; a pulse straddling crossing has an escaping tail extending to arbitrarily late arrival times, whose instantaneous frequency tends to zero. The finite count remains valid, while its last-pulse duration then has no finite constant multiple of $\tau$. This identifies precisely what the unspecified pulse phase can change.

For the strong-collision ion-neutral Alfvén modes, put $a=k_{\parallel }{v}_A$ and assume $0<|a|/\mu \ll 1$. Rewrite the cubic asThe two slow roots obey $\omega =O(a)$, so the first approximation is ${\omega }_0^2=a^2/2$. Set $\omega ={\omega }_0+\delta \omega$ and retain the next order, of size $a^3$:ConsequentlyThese are weakly damped Alfvén waves. Their phase speed is $v_A/\sqrt{2}=B_0/\sqrt{4\pi ({\rho }_i+{\rho }_n)}$, since collisions make the neutral mass participate in the oscillation. Their ambipolar damping rate is ${\Gamma }_A=a^2/(8\mu )$, with ${\Gamma }_A/|\mathrm{Re}{\omega }_{\pm }|\ll 1$.

For the remaining root, at $a=0$ the cubic is ${\omega }^2(\omega +2i\mu )$, so start from ${\omega }_{d0}=-2i\mu$. Denote the cubic by $F(\omega ,a)$. Its value and derivative at this root areThe first correction is therefore $\delta {\omega }_d=i{a}^2/(4\mu )$, givingThis branch is purely damped. Indeed, the real growth-rate polynomial for $s=-i\omega$ is $s^3+2\mu s^2+a^{2}s+\mu a^2$, and its simple real root near $-2\mu$ stays real for sufficiently small real $a^2$. It represents rapid decay of ion-neutral relative motion. In the $a\to 0$ limit, the other two frequencies tend to zero; all three branches are accounted for.