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Introduce the slow time $T=ϵt$ and expand $y=y_0(t,T)+ϵy_1(t,T)+\cdots$. At leading order writeThe first-order equation isIts resonant sine and cosine components must vanish by the solvability condition in the method of multiple scales. Let angle brackets denote a full-period average. The cosine projection of the nonlinear term is zero, since $⟨{\sin }^n\psi \cos \psi ⟩=0$. The sine projection givesThis is the averaged oscillator damping by a power of velocity.

If $n$ is odd, defineThen $R_T=-C_n{R}^n$, givingFor every fixed $n$, these leading multiple-scale expansions are uniform on bounded slow-time intervals $0\le ϵt\le T_0$. The unforced phase remains $t$ to this order. A bounded first-order correction supplies the small adjustment to the initial derivative and initial displacement; it does not change this leading result.

If $n$ is even, the average of the odd power ${\sin }^{n+1}\psi$ vanishes. Thus $R_T=0$ andThe first-order forcing has only a constant and even harmonics, so it produces no fundamental resonance. Its bounded particular solution can be supplemented by homogeneous sine and cosine terms to satisfy the initial conditions. There can be an $O({ϵ}^2)$ frequency correction, but its phase accumulation is only $O(ϵ)$ on the present time scale. In contrast to the odd case, the even-power force is invariant under velocity reversal and the equation is reversible; calling it positive damping for both signs of velocity would be wrong. The mechanical energy identity makes the distinction clear:This is nonpositive for odd $n$ and changes sign for even $n$; its cycle average reproduces the amplitude equation above.

For large odd $n$, the central binomial coefficient givesThis form reveals several time regimes. For $L_nϵt\ll 1$, $R=1-C_nϵt+\cdots$; at $t=O(1/(ϵ\sqrt{n}))$, the amplitude change is only $O(1/n)$. For $L_nϵt\gg 1$ but $\log (L_nϵt)\ll n$, the leading change is $R\simeq 1-\log (L_nϵt)/n$. In particular, at $t=O(1/ϵ)$ it is only $O(\log n/n)$, despite the long elapsed time. High powers damp mainly near velocity maxima, and even a small amplitude reduction strongly suppresses subsequent damping.

An order-one amplitude reduction in the averaged law requires $\log (L_nϵt)=O(n)$. Formally, $t=e^{(n-1)\sigma }/(ϵL_n)$ gives $R\sim e^{-\sigma }$, and for fixed odd $n>1$ its late envelope is an algebraic power $t^{-1/(n-1)}$. These exponential-in-$n$ time scales are predictions of extending the averaged law; the fixed-$n$, bounded-$ϵt$ multiple-scale error estimate alone does not establish uniform accuracy that far. A joint large-$n$, small-$ϵ$ approximation also needs control of the increasingly sharp nonlinear peaks, rather than assuming fixed-$n$ error constants are uniform. Taking $n$ fixed first and then examining the large-$n$ envelope avoids that unsupported interchange of limits.

Use a linear, lossless arterial pressure wave model with an incompressible fluid of mass density $\rho$, small wall displacement and negligible mean-flow advection compared with the wave speed. Let the parent vessel have equilibrium area $A_1$ and speed $c_1$. Linearizing mass conservation and axial momentum gives $a_t+A_1{u}_x=0$ and $\rho u_t=-p_x$. With wall compliance $p=(dp/dA)a$, these imply $c_1^2=(A_1/\rho )dp/dA$. A stiffer wall therefore has larger wave speed. A downstream travelling pressure wave carries flux $Q_{+}=Y_1{p}_{+}$, where $Y_1=A_1/(\rho c_1)$ is the characteristic admittance of an arterial wave; a returning wave has $Q_{-}=-Y_1{p}_{-}$.

Place the reflecting junction at $x=L$. Treat the daughters as outgoing-wave loads, with combined admittance $Y_d={\sum }_j{A}_j/(\rho c_j)$ and no further returning waves. At the junction, pressure continuity and flux conservation give $p_I+p_R=p_T$ and $Y_1(p_I-p_R)=Y_d{p}_T$. Thus the bifurcation reflection of an arterial pressure wave hasA decrease in total area gives $0<r<1$ if the parent and daughter wave speeds are equal. More generally this requires $Y_d<Y_1$, an assumption about both area and stiffness. To isolate the effect of ageing, assume all relevant wave speeds increase by the same factor, so their ratios and hence $r$ stay fixed. Unchanged geometry alone does not ensure unchanged $r$ if the vessels stiffen by different factors.

At a station $x<L$, put $\varphi =\omega (t-x/c_1)$ and $\vartheta =2\omega (L-x)/c_1$, the round-trip phase to the reflector. The incident pressure harmonic and its reflection giveFor time convention $e^{i\varphi }$, the two phasors are $P=1+r{e}^{-i\vartheta }$ and $V=1-r{e}^{-i\vartheta }$. Their maximum times are shifted by $-\arg P/\omega$ and $-\arg V/\omega$, respectively. Hence the pressure-flow phase lag from one arterial reflection isIndeed $V/P$ has the argument of $1-r^2+2ir\sin \vartheta$. On the branch $0<\vartheta <\pi$, this phase lag is positive: the pressure maximum follows the flow maximum. There is no cycle-wrap ambiguity on this branch because $1-r^2>0$.

The age trend need not have the same sign at every station. Differentiating the phase givesIncreasing $c_1$ decreases $\vartheta$. At a distal station with $0<\vartheta <\pi /2$, the lag therefore decreases. At a more proximal, ascending-aortic station with $\pi /2<\vartheta <\pi$, it increases. Thus one sufficient set of assumptions reproducing both trends isAs long as ageing keeps the stations in these ranges, the distal lag falls while the proximal lag rises. The effect is interference between pressure waves that add and flow waves that subtract, not merely the shortening of one travel time. The lag is largest at a round-trip phase of $\pi /2$.

This is a conditional explanation, with fixed representative harmonic frequency, fixed reflector location, negligible attenuation, negligible additional reflections and common stiffness scaling. The given stiffness and geometry facts alone do not specify the phase ranges, so they do not force either age trend without these extra assumptions. Here the two stations lie on the incident-plus-reflected segment. A station beyond the sole junction in an ideal outgoing daughter has only a transmitted wave and zero pressure-flow lag in this model. Actual peripheral measurements beyond that junction require returning waves from further junctions or terminal loads; a single-reflector calculation cannot establish their lag. Non-sinusoidal waveforms, distributed viscosity, mean flow and differently evolving reflection coefficients can also shift peak times. These limitations matter when applying the harmonic explanation to a real arterial network.

At normal incidence the P wave and the two S wave polarizations decouple. In each uniform layer the relevant displacement satisfies $\rho u_{tt}=C{u}_{zz}$, with $C=\lambda +2\mu$ for a longitudinal wave and $C=\mu$ for a shear wave. Its speed is $v=\sqrt{C/\rho }$ and its seismic impedance is $Z=\rho v$. A welded interface preserves particle velocity and traction. For displacement or velocity amplitudes, incidence from medium 0 onto medium 1 therefore givesReflection reverses the displacement sign when the receiving seismic impedance is larger. The energy-flux fractions areThe impedance factor matters: a displacement transmission coefficient larger than one does not imply energy amplification. Equal impedance eliminates reflection even when the densities and speeds differ.

For a stack, every interface creates reflected waves which can return to all earlier interfaces. A seismic layer transfer matrix incorporates those multiple reflections exactly. Use harmonic time dependence $e^{-i\omega t}$, particle velocity $V$ and compressive traction $P=-{\sigma }_{zz}$ for a P-wave, or the negative relevant shear traction for an S-wave. Through layer $j$ of thickness $h_j$,This follows by solving $V_z=i\omega P/C_j$ and $P_z=i\omega {\rho }_{j}V$. Continuity means that the state passes unchanged across each welded interface. Put $M=M_N\cdots M_1=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, with $ad-bc=1$, and let the exterior impedances be $Z_L,Z_R$. A unit incident velocity gives top state $(1+r,Z_L(1-r))$ and bottom state $(t,Z_{R}t)$. Eliminating these states yieldsFor real moduli and no attenuation, conservation of elastic-wave energy flux gives $|r{|}^2+(Z_R/Z_L)|t{|}^2=1$. Reciprocal incidence gives equal energy transmission in the reverse direction; flux-normalized amplitudes express that reciprocity without unequal exterior impedances obscuring it.

The wave phase accumulated in each layer determines constructive or destructive interference. For one layer between identical exterior media, an integral number of half-wavelengths makes $M_j=\pm I$ and gives perfect transmission, regardless of the layer impedance. At quarter-wave phase thickness, a single matching layer with $Z_j=\sqrt{Z_L{Z}_R}$ also gives zero reflection: the layer transforms the terminal impedance to $Z_j^2/Z_R=Z_L$. Away from these matching conditions there are transmission peaks, reflection maxima and rapid frequency-dependent phase changes. In the time domain these are a direct transmitted pulse followed by delayed internal reverberations and reflected echoes. Long residence in a resonant stack produces narrow peaks and long ringing, without creating energy.

For a periodic two-layer stack, the half-trace of one cell isThis is the periodic seismic multilayer stop band criterion. When the absolute right side exceeds one, the cell eigenvalues are a reciprocal growing/decaying pair and a long finite stack has exponentially small transmission. The missing transmitted energy is reflected, not absorbed. Large impedance contrast and repeated quarter-wave layers strengthen such stop bands. In pass bands the cell modes propagate, and a finite stack superposes transmission resonances on that propagation. An irregular stack need not have equally spaced peaks: the individual travel times and impedance contrasts set its interference pattern.

At sufficiently low frequency a finite stack has $M\to I$, so its leading reflection is that of the two exterior media alone. For many thin repeated layers with wavelength much longer than a cell, their effective density is the volume-weighted density and their normal modulus is the harmonic mean, $C_{\mathrm{eff}}^{-1}=⟨C^{-1}⟩$. This follows by expanding each propagation matrix to first order and averaging its two coefficients. At high frequency, sharp interfaces still reflect: there is no general disappearance of reflection, only increasingly rapid interference fringes. Smooth impedance grading instead suppresses short-wavelength reflection when its variation scale is long compared with the wavelength.

These calculations provide a reproducible investigation: vary the layer travel times, impedances and number of cells; compute the matrix product; plot reflected/transmitted fluxes and check their sum. Adding attenuation makes the flux sum smaller than one and reduces reverberation peaks, while oblique incidence introduces Snell law for elastic waves and P-SV mode conversion absent from this one-dimensional setting. Impedance contrasts generate reflections, propagation phases organize their interference, and energy conservation separates reflection from genuine absorption.

Away from the line source, insert a leading wavefront form $u\sim a(\mathbf{x})W(t-T(\mathbf{x}))$ into the shear-horizontal wave equation and match the two most singular orders. The first gives the eikonal equation $|\nabla T|=1/\beta$. The next givesEquivalently, SH ray-tube amplitude transport conserves $\rho \beta a^{2}J$, where $J$ is the transverse width of a two-dimensional ray tube. A line source has straight radial elastic rays and $J=rd\theta$ for fixed angular width, soThe same geometrical spreading applies to the first nonzero derivative discontinuity or leading wavefront singularity; the source history determines which waveform $W$ and normalization $C$ occur. In three dimensions a point source instead has $J\propto r^2$ and amplitude proportional to $r^{-1}$.

This does not contradict the SH line-source near-field singularity. Ray transport describes an arrival-front limit with $t-r/\beta$ small, or the high-frequency field outside the immediate source neighborhood. The logarithm describes $r\downarrow 0$ at fixed elapsed time and includes the tail behind the two-dimensional wavefront. For example, a causal step force $F_{0}H(t)$ has the exact radial displacementFor $r\ll \beta t$, this is $F_0\log (2\beta t/r)/(2\pi \mu )$, the near-source logarithm. For $t=r/\beta +\tau$ with $0<\tau \ll r/\beta$, it isThe coefficient of the leading arrival singularity therefore has exactly the ray-theory dependence. A force impulse gives the corresponding $r^{-1/2}H(\tau )/\sqrt{\tau }$ leading field. The limits describe different regions of one causal solution; neither is a uniform approximation to the other.

With vertical shear modulus ${\rho }^{\prime }{\beta }_z^{\prime 2}$ and horizontal shear modulus ${\rho }^{\prime }{\beta }_x^{\prime 2}$, force balance therefore gives the anisotropic shear-horizontal wave equationFor $u_y=Y(z)e^{i(kx-\omega t)}$, the layer vertical wavenumber and substrate decay exponent areThe free top still gives $Y=A\cos (qz)$; the substrate still has $Y=B{e}^{-\kappa (z-h)}$. Welded displacement continuity and the now anisotropic traction continuity giveHence the anisotropic Love-wave dispersion relation isEquivalently, writing $c=\omega /k$,This replaces the isotropic speed in the layer equation by separate horizontal and vertical speeds and replaces the layer traction modulus by ${\rho }^{\prime }{\beta }_z^{\prime 2}$. No additional SH/P-SV coupling arises for the aligned symmetry axes and constitutive law here.

A rigid base still imposes $q_n={\chi }_n/h$, but nowThusThe vertical speed sets the resonance frequencies, while the horizontal speed sets the high-frequency limits and $c_n{U}_n={\beta }_x^{\prime 2}$.

For the elastic substrate, put ${ϵ}_x={\beta }_x^{\prime }/\beta$, $\zeta ={\rho }^{\prime }{\beta }_x^{\prime }{\beta }_z^{\prime }/(\rho {\beta }^2)$, $W=\omega h/{\beta }_z^{\prime }$ and $K=kh{\beta }_x^{\prime }/{\beta }_z^{\prime }$. The exact parameterization isIt reproduces the isotropic analysis with ${\beta }^{\prime }$ replaced by ${\beta }_z^{\prime }$ in the frequency scale, by ${\beta }_x^{\prime }$ in the velocities, and by the two effective contrast parameters above. In particular the true cutoffs are ${\omega }_{c,n}=n\pi {\beta }_z^{\prime }/[h\sqrt{1-{ϵ}_x^2}]$, the group minima occur near ${\omega }_{n,z}$, and a sufficient rigid-limit validity range isFor comparable horizontal and vertical layer speeds, a bounded density ratio and fixed mode index, the minimum estimate becomes ${\delta }_{\min }\sim ({\zeta }^2{\chi }_n/2{)}^{1/3}$ and $U_{\min }/{\beta }_x^{\prime }\sim 3(\zeta /{\chi }_n{)}^{1/3}/2^{2/3}$. If their ratio is extreme, that more specific minimum-depth estimate additionally needs $\zeta /{\delta }_{\min }\gg {ϵ}_x$; the exact parameterization and load criterion remain valid without it. As in the isotropic case, neither a slow horizontal nor a slow vertical speed alone justifies the rigid boundary at cutoff or within the strong-dispersion transition.

The full Love wave problem replaces the rigid base by a decaying substrate displacement. With $Y=A\cos (qz)$ in the layer and $Y=B{e}^{-\kappa (z-h)}$ below it,where ${\mu }^{\prime }={\rho }^{\prime }{\beta }^{\prime 2}$ and $\mu =\rho {\beta }^2$. Displacement continuity gives $B=A\cos (qh)$ and traction continuity gives ${\mu }^{\prime }q\sin (qh)=\mu \kappa \cos (qh)$. The rigid base is therefore a good approximation only when $\mu \kappa \gg {\mu }^{\prime }q$: then the substrate displacement is small and $qh$ lies close to ${\chi }_n=(n+1/2)\pi$.

Put $ϵ={\beta }^{\prime }/\beta$, $\eta ={\mu }^{\prime }/\mu$, $y=qh$, $W=\omega h/{\beta }^{\prime }$ and $K=kh$. The complete dispersion branch can be parameterized without root finding:Then $c={\beta }^{\prime }W/K$ and $U={\beta }^{\prime }{W}^{\prime }/K^{\prime }$. At the lower end, $c\to \beta$ and $U\to \beta$. The Love-wave cutoff frequencies are ${\omega }_{c,n}=n\pi {\beta }^{\prime }/[h\sqrt{1-{ϵ}^2}]$, with zero cutoff for the fundamental. Thus the rigid-layer formula cannot describe the fundamental's low-frequency limit or the low-frequency part of an overtone: it would either give no propagating mode or predict $c>\beta$, outside the trapping interval ${\beta }^{\prime }<c<\beta$.

For comparable densities and a fixed mode index, $\eta =({\rho }^{\prime }/\rho ){ϵ}^2\ll 1$. Write $y={\chi }_n-\delta$ near the strong-dispersion transition. In the region controlling the group-velocity minimum of a high-contrast Love wave, $\delta \ll 1$ and $\eta /\delta \gg ϵ$. The leading derivatives giveTheir minimum is atThe frequency estimate follows by inserting ${\delta }_{\min }$ into $W\simeq ({\chi }_n-\delta )[1+{\eta }^2/(2{\delta }^2)+{ϵ}^2/2]$: the two leading terms of size $\delta$ cancel. The real group velocity has a finite minimum and turns upwards near the rigid-layer threshold, whereas the rigid formula has zero group speed at its artificial threshold.

For a useful estimate of the validity range, put $\Delta =W-{\chi }_n>0$. The rigid prediction has $K_R^2=W^2-{\chi }_n^2\simeq 2{\chi }_n\Delta$. The finite-load phase correction is $\delta \simeq \eta {\chi }_n/(\kappa h)$; its changes to both $K$ and $dK/dW$ are small ifThe first condition keeps $K_R$ away from the substrate propagation boundary; the second controls the rapidly changing interface phase. Equivalently, the rigid formulas are reliable sufficiently above the actual group-velocity minimum, beyond its narrow transition interval. At fixed $\omega /{\omega }_n>1$ separated from one, they become asymptotically accurate as the contrast grows. They are particularly simple at $\omega \gg {\omega }_n$, where both velocities approach ${\beta }^{\prime }$. Within the transition and below it the full dispersion relation is required.

The small speed ratio alone is insufficient if the layer density grows so strongly that ${\mu }^{\prime }/\mu$ is not small. For such materials the exact load criterion $\mu \kappa \gg {\mu }^{\prime }q$ and its frequency derivatives should be used; the preceding contrast estimates assume a bounded order-one density ratio and fixed $n$.

Let the layer occupy $0<z<h$, with a free surface at $z=0$ and rigid base at $z=h$. Write an SH-wave as $u_y=Y(z)e^{i(kx-\omega t)}$. The equation and boundary conditions giveThus $Y=A\cos (qz)$ and the guided SH modes between rigid and free planes haveThe propagating branch has $\omega >{\omega }_n$ and $k=\sqrt{{\omega }^2-{\omega }_n^2}/{\beta }^{\prime }$. Hence the rigid-base approximation to Love waves predictsAt each threshold $c_n\to \infty$ and $U_n\to 0$. As frequency rises, the phase velocity decreases towards ${\beta }^{\prime }$ and the group velocity increases towards ${\beta }^{\prime }$. Frequencies below ${\omega }_n$ have imaginary horizontal wavenumber and do not give propagating guided modes. The infinite threshold phase speed does not violate causality: the limiting group speed is zero. These quarter-wave thresholds differ from the actual Love-wave cutoff frequencies of an elastic substrate.

The equilibrium isothermal equation of state is $p=c_s^2\rho$. Hydrostatic equilibrium gives $p_z=-\rho {\Omega }^{2}z$, henceHere $\Omega \ne 0$ is assumed. This is the same local harmonic-gravity balance as isothermal vertical structure in stellar gravity.

For $\gamma >1$, this equilibrium is stably stratified under adiabatic fluid perturbations. Its squared buoyancy frequency isA general perturbation can therefore support acoustic waves restored by compression and internal gravity waves restored by buoyancy, with nonoscillatory entropy or vortical disturbances also possible. The prescribed velocity is purely vertical and horizontally uniform. It describes vertical acoustic waves; it does not contain the horizontally varying circulation required for an oscillatory internal-gravity branch. For zero horizontal wavenumber, that branch has zero frequency. An isothermal equilibrium must not be confused with an isothermal perturbation: the latter would remove the buoyancy restoring force.

Introduce the fluid Lagrangian displacement $\xi (z)e^{i\sigma t}$, so $w=i\sigma \xi$. Use $\delta \rho ,\delta p$ for fluid Eulerian perturbations and $\Delta \rho ,\Delta p$ for fluid Lagrangian perturbations. The continuity equation and the adiabatic relation areThe fixed external gravitational potential is not perturbed. The vertical linearized Euler equations consequently giveSubstituting the preceding relations, and using $p_z=-\rho g$ and $p_{zz}=-{\rho }_{z}g-\rho g_z$, reduces its right-hand side toSince $g={\Omega }^{2}z$ and $g_z={\Omega }^2$, the same equation holds for $w$:The rigid fixed base supplies the essential condition $w(0)=0$. At infinity the condition is finite kinetic energy, not bounded velocity amplitude; a polynomial can grow while the Gaussian mass density makes its energy finite.

Put $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The equation becomes the Hermite differential equation in its probabilists' convention,For $w={\sum }_{j\ge 0}{a}_j{x}^j$, its recurrence isThe base condition makes $a_0=0$ and therefore every even coefficient zero. An odd series terminates at degree $2n+1$ when $\lambda =2n+1$. It remains to prove that a nonterminating odd solution cannot have finite energy; termination alone would not establish the requested exhaustion of modes.

For that purpose, set $y=e^{-x^2/4}w$. The finite-energy condition is $y\in L^2(0,\infty )$, and its equation isBecause the coefficients are even and the solution is odd at the base, extend $y$ oddly to the whole real line. Define lowering and raising operatorsIntegration by parts gives $\Vert ay{\Vert }^2=\lambda \Vert y{\Vert }^2$, so $\lambda \ge 0$. These identities apply to the square-integrable solution: multiplying its equation by a compactly cut off conjugate and taking larger cutoffs bounds $y^{\prime }$ and $xy$ in $L^2$, and removes the endpoint terms. The same argument applies to each lowered eigenfunction.

If $\lambda$ is not a nonnegative integer, repeatedly apply $a$. Each application lowers $\lambda$ by one and remains nonzero whenever its current value is positive, by the norm identity. After finitely many applications it would yield a nonzero square-integrable eigenfunction with negative $\lambda$, contradicting that identity. Thus $\lambda =N$ is a nonnegative integer. At $N=0$, $ay=0$ gives $y\propto e^{-x^2/4}$, which is even. Lowering reverses parity, so an initially odd eigenfunction can reach this even ground state only if $N$ is odd. Consequently $N=2n+1$. The recurrence then terminates and uniquely determines the odd solution from $a_1$.

The complete adiabatic vertical modes of a Gaussian atmosphere are thereforeHere ${\mathrm{He}}_j$ is the Probabilists' Hermite polynomial. For example the first shapes are $x$, $x^3-3x$, and $x^5-10x^3+15x$, with frequencies squared ${\Omega }^2(1+\gamma )$, ${\Omega }^2(1+3\gamma )$, and ${\Omega }^2(1+5\gamma )$. Either sign of each real frequency gives a time-harmonic solution. Gaussian moments make every displayed polynomial have finite kinetic energy. Without the fixed-base condition, finite energy alone would also admit even polynomials, which is why that condition is necessary.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.

In the Maxwell-Boltzmann velocity distribution, each Cartesian component has mean-square velocity $k_B{T}_e/m_e$. Thus $⟨v^2⟩=3k_B{T}_e/m_e$. The original PDF calls this quantity an rms velocity, but the rms speed is its square root. Also, its $n_{\gamma }$ must mean the dimensionless mean occupation number $n$, since its equilibrium value is a Planck photon distribution; it is not the frequency-integrated photon number density.

Take $x=h\nu /(k_B{T}_{\gamma })$ at a fixed local radiation temperature. Because $x$ is proportional to frequency,The hot-electron Kompaneets equation therefore becomesFor traversal length $d\ell =cdt$, the Compton y parameter isIt measures electron pressure integrated along the line of sight, or the Thomson scattering optical depth weighted by $k_B{T}_e/(m_e{c}^2)$. In this hot-electron limit the mean fractional energy gain is approximately $4y$. The derivation assumes nonrelativistic electrons, Thomson-limit photon energies and $T_e\gg T_{\gamma }$. The original PDF uses $h\nu$, as required for ordinary frequency; the converted TeX incorrectly substitutes $\hslash \nu$.

Write $n=n_0+y\mathcal{L}{n}_0+O(y^2)$, with $n_0=(e^x-1{)}^{-1}$ and $\mathcal{L}=x^{-2}{\partial }_x(x^4{\partial }_x)$. Differentiating the Planck photon distribution givesDividing by $n_0$ proves the first-order thermal Sunyaev-Zeldovich effect:In the Rayleigh-Jeans law limit, $x{e}^x/(e^x-1)\to 1$ and $x\coth (x/2)\to 2$, so $\Delta n/n_0\to -2y$. In the Wien approximation, the prefactor approaches $x$ and the bracket approaches $x-4$, so $\Delta n/n_0\sim yx(x-4)\sim y{x}^2$. The last limit is an asymptotic description of the first-order coefficient: a perturbative prediction at large $x$ still requires $y{x}^2\ll 1$.