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Model the post-eruption flow as an axisymmetric, shallow, hydrostatic, turbulently mixed layer on a horizontal bed. Neglect entrainment, erosion, heat exchange, rotation and large-scale viscous drag. Particles are dilute, with density ${\rho }_p$, uniform-in-depth particle volume fraction $c(r,t)$ and constant downward settling speed $w_s>0$ relative to the carrier gas. Deposited particles are not resuspended. Their volume is small enough that settling does not significantly change the bulk carrier volume. Let ${\rho }_a$ be ambient air density and ${\rho }_g$ carrier-gas density at its local temperature, and use a leading Boussinesq approximation for the mixture motion. DefineThe hot gas reduces the density excess; the ground-hugging current model requires $g^{\prime }>0$. If density contrasts are too large for Boussinesq dynamics, density-weighted balances would be needed instead. The equations below are a stated shallow-layer model, not a description of the initial explosive vent jet.

For layer height $h$ and radial depth-averaged speed $u$, the appropriate shallow-water balances areThe hydrostatic force gives the nonconservative momentum equation $u_t+u{u}_r=-g^{\prime }{h}_r-(h/2)g_r^{\prime }$. In particular a spatially varying particle fraction must not be treated as constant reduced gravity when differentiating pressure. With no heat exchange or entrainment, a carrier-temperature anomaly $\Theta$ is advected:At this order $b_T$ is proportional to $-\Theta$. At the bed the carrier has no normal penetration and particles have absorbing deposition flux $w_{s}c$; at the top the pressure equals ambient pressure. Axis regularity requires $u(0,t)=0$ and smooth scalar fields, with no further source after the initial pulse. The initial height, particle inventory and temperature anomaly specify the released volume and ash content. Outside the front $r=R(t)$, the current layer is absent. At a resolved advancing head, use the material-front condition and the gravity-current front condition $\dot{R}=\mathsf{F}\sqrt{g_f^{\prime }{h}_f}$, where $\mathsf{F}$ is an order-one front Froude number and $h_f,g_f^{\prime }$ refer to the fluid feeding the head. The front closure replaces unresolved head dynamics; simply prescribing both a zero-depth mathematical edge and a finite-depth box-head speed would be inconsistent.

To obtain a gravity-current box model, assume horizontal as well as vertical mixing, uniform $h(t),c(t),\Theta (t)$ in a disk of radius $R(t)$, and a negligible-volume head whose feeding depth is the box depth. The conserved bulk volume $\mathcal{V}$ givesA bulk velocity $u=(\dot{R}/R)r$ satisfies interior continuity for this box. Rather than impose a uniform pressure on the full shallow-water equations and thereby lose the head force, replace their momentum dynamics by the front closure. Total suspended particle volume is $\mathcal{V}c$, so bed deposition givesThese ordinary differential equations, together with initial $R_0,c_0,\mathcal{V},{\Theta }_0$, are the axisymmetric settling box model. The maintained negative thermal term may eventually cancel particle buoyancy and cause lift-off; beyond $g^{\prime }=0$ this horizontal gravity-current model does not apply.

For the requested particle-only calculation set $b_T=0$ and define the initial reduced gravity $g_0^{\prime }=b_p{c}_0$. The equations reduce toEliminating time gives $d\sqrt{c}/dR=-w_s{R}^3/[2\mathsf{F}\sqrt{b_p}{A}^{3/2}]$. Integrating from the release state yieldsThus the monodisperse circular ash runout has limiting radiusThe maximum outward distance beyond the initial release edge is $R_{\infty }-R_0$. When the vent scale is negligible, set $R_0=0$ in this late-time source idealization. The resulting radius scales as ${\mathcal{V}}^{3/8}(g_0^{\prime }{)}^{1/8}{w}_s^{-1/4}$; it is not obtained by conserving reduced gravity after particles settle out.

To determine the time, put $s=R^2/R_{\infty }^2$, $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. The concentration relation becomes $\sqrt{c/c_0}=(1-s^2)/(1-s_0^2)$, and the front equation reduces toConsequentlyThe furthest reach is a limiting radius, approached asymptotically rather than attained at a finite stopping time. As the residual particle concentration vanishes, the front slows exponentially. A finite observational threshold for remaining ash would give a finite practical stopping time, but no such threshold is part of these balances.

The local ash deposit density is most naturally its mass per horizontal area, $\Sigma (r)$. Once the front has arrived at $r$, its local deposition flux is ${\rho }_p{w}_{s}c(t)$. For $r<R_{\infty }$, define $s_a=\max (s_0,r^2/R_{\infty }^2)$; the arrival time is zero inside the initial footprint and otherwise $t_a=\tau [\mathrm{artanh}{s}_a-\mathrm{artanh}{s}_0]$. Sincewe can integrate the flux exactly. Using $\int {\mathrm{sech}}^4\xi d\xi =\tanh \xi -({\tanh }^3\xi )/3$ gives the circular ash deposit profileIt is constant under the initial disk, then decreases to zero at the limiting edge. Particle solid volume per area is $\Sigma /{\rho }_p$; deposit thickness is $\Sigma /{\rho }_d$ if a bulk deposit density ${\rho }_d$ including voids is supplied. The problem does not determine that packing density.

For a negligible initial radius this simplifies toIts integral $2\pi {\int }_0^{R_{\infty }}r\Sigma (r)dr={\rho }_p{c}_0\mathcal{V}$ confirms that the complete initial ash inventory is deposited. The finite-$R_0$ formula gives the same budget including its central flat portion.

The diagram uses the negligible-vent-radius idealization and time normalized by $\tau$. It illustrates both the asymptotic approach to $R_{\infty }$ and the smoothly vanishing outer deposit. All these predictions depend on the mixing and constant-Froude head closure; they do not claim to resolve the initial blast or the ultimate inertia-dominated motion of almost particle-free gas.

Let $z>0$ measure height from the lower edge and $y>0$ distance from one face of the vertical plate. Use the steady Boussinesq approximation, homogeneous isotropic permeability of a porous medium $K$, reference fluid mass density ${\rho }_0$, dynamic viscosity $\mu$, thermal expansion coefficient $\alpha$, effective thermal conductivity $k_e$ and liquid volumetric heat capacity $C_f={\rho }_0{c}_f$. Define $B={\rho }_{0}g\alpha K/\mu$ and $\kappa =k_e/C_f$, consistently with the thermal convention used above. Write $\theta =T-T_a$, where $T_a$ is the uniform ambient temperature.

Interpret $F$ as uniform wall heat-flux density: a plate segment of height $dz$ inputs heat $Fdz$ per unit horizontal span. The cumulative heat input up to height $z$ is therefore $Fz$. If $F$ instead denotes the sum of flux densities into two identically heated faces, replace it by $F/2$ in each face formula before summing. A specification only of total power per span, without its distribution in height, would not determine this uniformly heated-wall problem. We calculate one heated face; identical heating on both faces doubles the integrated fluxes.

Away from the leading edge, suppose the boundary layer is slender, $\delta (z)\ll z$, and porous inertia and longitudinal heat conduction are negligible. Lateral pressure adjustment leaves the vertical Darcy velocity proportional to buoyancy:At the plate, $v=0$ and $-k_e{\theta }_y=F$; in the far field $w,\theta \to 0$. Darcy law permits tangential discharge at an impermeable plate, so imposing a further no-slip condition $w=0$ would be inappropriate in this leading porous model.

Balance $\theta \sim F\delta /k_e$, $w\sim B\theta$ and $w\theta /z\sim \kappa \theta /{\delta }^2$ to obtainThus the warmed layer thickens and accelerates upward. Define a streamfunction by $w={\psi }_y$, $v=-{\psi }_z$ and setThe last two expressions agree because ${\delta }^3=\kappa k_{e}z/(BF)$. Differentiation gives $v=-(\kappa /3\delta )(2f-\eta f^{\prime })$. Substituting in the thermal equation leaves the single nonlinear equation for the uniformly heated vertical plate in a porous medium:The first condition is impermeability; the second fixes the wall heat flux; the third fixes ambient temperature and vertical velocity. Seek the positive-temperature branch with $f^{\prime }\ge 0$ and finite $f(\infty )$. Decay also gives $f^{\prime \prime }(\infty )=0$. The far-field transverse inflow need not vanish: $v_{\infty }=-(2\kappa /3\delta )f(\infty )$ supplies the increasing discharge.

An approximate solution can be obtained without a numerical shooting calculation. Integrate the differential equation over $0<\eta <\infty$. The boundary conditions and integration by parts giveThus the exact thermal budget is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Assume an exponential thermal shape, $f=C(1-e^{-\eta /d})$. The wall derivative requires $C/d^2=1$, and the integral condition requires $C^2/(2d)=1$. Hence $d=2^{1/3}$, $C=2^{2/3}$ and the integral exponential profile for porous wall convection isIt satisfies all boundary conditions, has positive decaying temperature, and matches the exact integrated heat budget. It is an integral approximation, not an exact pointwise solution of the differential equation; its mass-flux coefficient should be read accordingly.

Define the volume discharge, kinematic Darcy momentum flux and buoyancy flux per horizontal span byWith these standard discharge conventions the mass flux is $\dot{m}={\rho }_{0}Q$. Their similarity expressions areThe integral identity evaluates two of them exactly:Equivalently $C_f\int w\theta dy=Fz$: all the heat supplied below height $z$ is carried upward through that section. Although these fluxes are exact within the boundary-layer equations, porous drag means that the momentum flux itself is not conserved with height.

The approximate mass flux isFor clarity, if “momentum flux” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant porosity $\varphi$ and pore speed $w/\varphi$ give $\dot{P}=({\rho }_0/\varphi )\mathcal{M}$. The mass flux remains ${\rho }_{0}Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit porosity error. Validity also requires $\alpha \theta \ll 1$ and a sufficiently slender, low-inertia porous boundary layer; the approximation is not the immediate leading-edge conduction field.

Assume the present universe is expanding, $\dot{a}(t_0)>0$. The stated energy inequality and the Raychaudhuri equation imply $\ddot{a}\le 0$ wherever $a>0$. Therefore for $t<t_0$ in that positive branch, $\dot{a}(t)\ge \dot{a}(t_0)$, andNonnegativity of $a(t)$ forces $t_0-t\le a(t_0)/\dot{a}(t_0)=H_0^{-1}$. If the solution remains valid towards the past, its positive branch must thus end in $a=0$ within that time; a positive scale factor extending further would contradict the inequality. HenceThis argument presumes an expanding branch and the stated energy condition throughout it, rather than proving either assumption.

There are two relevant failures of the naive numerical bound. The supplied ten-billion-year Hubble time is an approximate observational input: a smaller inferred present expansion rate gives a larger Hubble time, and comparison with stellar ages must include these uncertainties. More decisively, a positive cosmological constant, or other dark energy with sufficiently negative pressure, has $\rho +3P/c^2<0$ and permits accelerated expansion, so the concavity assumption fails. Observations of acceleration therefore remove this particular upper-bound argument. This is a discussion of the supplied historical estimate, not an assertion that its numerical value is a current measurement.

Heat capacity at fixed volume is $C_V=(\partial E/\partial T{)}_{V,N}=T(\partial S/\partial T{)}_{V,N}$. From the Helmholtz free energy differential $dF=-SdT-PdV$ at fixed particle number, equality of mixed derivatives gives $(\partial S/\partial V{)}_T=(\partial P/\partial T{)}_V$. Differentiating once more givesFor the canonical partition function, $F=-k_{B}T\log Z$, $E=k_B{T}^2{\partial }_T\log Z$, and $S=k_B(\log Z+T{\partial }_T\log Z)$. Thus, for $V>aN$,andNo Stirling approximation is needed. The heat capacity is independent of volume, while the pressure is linear in temperature, so both sides of the Maxwell-derived identity vanish.

In the lubrication regime the thin region has longitudinal scale $\ell \sim \sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell /b\sim \mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.

Use a frame translating with the cylinder. The wall moves with velocity $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, soContinuity makes $q$ constant. Equal ambient pressures at both infinities give $\int p_{x}dx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. ThereforeThe vertical velocity, with the wall condition $w(0)=0$, follows explicitly from incompressibility:It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.

The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}s$, its integral isForce balance equates this wall force to the required cylinder force at leading order; the pressure and viscous stresses on remote boundaries make no leading contribution. This is the cylinder translating parallel to a wall in a thin gap result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.

Choose a control volume containing the small impingement region, with an inlet section across the incident jet and an outlet annulus across the conical sheet. The remaining boundary follows free surfaces and the wetted cone. Subtract atmospheric pressure and neglect gravity. Bernoulli's equation gives the same speed $U$ in the outgoing sheet as in the incoming jet, and its total volume flux is $AU$.

The incoming axial momentum rate is $\rho A{U}^2$. The outgoing velocity makes angle $\alpha$ with the cone axis, so its axial momentum rate is $\rho A{U}^2\cos \alpha$; transverse components cancel by axisymmetry. The integral momentum equation says that the cone's axial force on the water is their difference, $-\rho A{U}^2(1-\cos \alpha )$. By equal and opposite reaction, the jet's force on the cone is along the incident jet, with magnitudeThe outlet control surface is across the sheet, not along it, so its entire outgoing volume flux is counted.

The free jet is at atmospheric pressure. Bernoulli's equation from the nozzle to height $z$ gives $U(z{)}^2/2+gz=U_0^2/2$. Conservation of volume flux gives $A(z)U(z)=A_0{U}_0$. HenceFor the prescribed well-below-apex regime, $U(z)=U_0[1-gz/U_0^2+O((gz/U_0^2{)}^2)]$ and $A(z)=A_0[1+gz/U_0^2+O((gz/U_0^2{)}^2)]$. The slowing jet therefore thickens. These expressions use the slender-jet, nearly uniform cross-sectional speed approximation underlying the question; the formal vanishing-speed apex is outside that approximation.

For a steady inviscid incompressible fluid of constant density in uniform gravity $\mathbf{g}=-g{\mathbf{e}}_z$, the Euler equation for fluid motion and continuity equation areTaking the scalar product with $\mathbf{u}$ givesThus Bernoulli's quantity is constant along each streamline. For an irrotational connected flow it is constant throughout the flow; constancy across different streamlines is not assumed in general.

Since incompressibility makes $\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=\rho (\mathbf{u}\cdot \nabla )\mathbf{u}$, integrating over a fixed control volume and applying the divergence theorem yields the integral momentum balanceHere $\mathbf{n}$ is outward and the left side is the net outgoing momentum flux. Uniform atmospheric pressure has zero resultant on a closed control surface, so it may be subtracted from all pressures in the jet calculations.

The velocity divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=\alpha -\alpha =0$, satisfying the constant-density incompressible flow condition. Choose the stream function convention $(u_x,u_y)=({\psi }_y,-{\psi }_x)$. Integration then givesIndeed, ${\psi }_y=\alpha x-\beta y$ and $-{\psi }_x=\beta x-\alpha y$. Streamlines are contours of this function. The vector field is also compatible with steady inviscid momentum balance without body forces: $(\mathbf{u}\cdot \nabla )\mathbf{u}=({\alpha }^2-{\beta }^2)(x,y)$ is balanced by pressure $p=p_0-\rho ({\alpha }^2-{\beta }^2)(x^2+y^2)/2$.

The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.

For constant mass density and incompressible flow, use $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ in the Euler equations for an inviscid fluid. Taking the curl removes the pressure and kinetic energy gradients, giving ${\mathbf{\omega }}_t=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. The supplied vector identity, with $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, yields the vorticity equationIn a two-dimensional flow $\mathbf{u}=(u,v,0)$ is independent of $z$, and $\mathbf{\omega }=\omega {\mathbf{e}}_z$. The right side vanishes, so $D\omega /Dt=0$: each material element retains its vorticity. The vortex lines are parallel to $z$ and are carried by the planar velocity. Also planar incompressibility preserves the area of a material cross-section; integrating its conserved vorticity over that area shows that its circulation, or vortex strength, stays constant. Thus the vortex lines move with the fluid without changing strength.

Place a line vortex of circulation $\Gamma$ at $(0,b)$ above the wall $y=0$. Its velocity field is $\Gamma {\mathbf{e}}_z\times (\mathbf{r}-{\mathbf{r}}_0)/(2\pi |\mathbf{r}-{\mathbf{r}}_0{|}^2)$. The method of images puts a vortex of circulation $-\Gamma$ at $(0,-b)$. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$, the resulting stream function isIt is identically zero on the wall, so $u_y=-{\psi }_x=0$ there. In the fluid half-plane it has just the required vortex singularity, with no other singularities, and its velocity decays at infinity. This is the no-background-flow solution with the wall: the difference from another such solution is a regular irrotational, divergence-free decaying field with zero normal velocity. Reflect its tangential component evenly and its normal component oddly across the wall. The reflected components are bounded harmonic functions on the whole plane, so they are constant by harmonic Liouville theorem, and decay forces both constants to be zero. Thus replacing the wall by the image gives the same flow. The real vortex moves under the image velocity, excluding its own singular velocity, soIts signed velocity parallel to the wall is $\Gamma /(4\pi b)$, and its speed is $|\Gamma |/(4\pi b)$.

For the quarter-plane, place images of circulation $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. The paired stream functions make both walls streamlines. Summing their velocities at the real vortex givesIn polar coordinates,It follows directly thatThus the line-vortex trajectory in a quarter-plane is $r\sin 2\theta =\text{constant}$.

Use the spatial-potential convention of this paper and assume $\bar{\rho }+\bar{P}\ne 0$. A passive time displacement changes the spatial scalar by $\tilde{\Psi }-\Psi =\bar{N}H{\xi }^0$ and changes the energy density byThe two contributions to the uniform-density curvature perturbation therefore cancel:A spatial scalar displacement changes $E$ but not this combination. Hence $\zeta$ is gauge invariant at linear order. Pure cosmological-constant matter, for which $\bar{\rho }+\bar{P}=0$, does not supply a energy density clock and this formula is not defined for that component alone.

Now take the definite constant-$w$ barotropic equation of state and use physical-time differentiation $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$. Put $A=\bar{\rho }+\bar{P}$. Background stress-energy conservation gives $\mathcal{D}A=-3H(1+w)A$. Differentiating $\zeta =-\Psi +\delta \rho /(3A)$, then substituting the two supplied perturbation equations, givesThus the lapse and expansion perturbations cancel without imposing a gauge. In terms of the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-w\delta \rho$,For the stipulated barotropic equation of state, $\delta P=w\delta \rho$, so the first term vanishes. For a Fourier mode, $\Delta =-k^2/a^2$. For regular long-wavelength solutions the remaining gradients are suppressed when $k\ll aH$. Consequently $\dot{\zeta }\simeq 0$ on superhorizon scales, with exact conservation in the leading zero-gradient limit. It is not an exact finite-$k$ identity. Entropy perturbations or a singular gradient expansion can invalidate that conclusion; the supplied adiabatic assumption is what removes the unsuppressed pressure source.

This superhorizon conservation of uniform-density curvature is useful because inflationary fluctuations leave the horizon during cosmic inflation and may re-enter only much later. In an adiabatic single-clock evolution, the conserved uniform-density curvature perturbation transports their initial amplitude through reheating and the later radiation in cosmology and matter eras without solving every short-distance process. It then determines the initial conditions for large-scale structure and Cosmic microwave background anisotropy. Extra fluctuating fields or fluids can introduce isocurvature perturbations and convert them into curvature, so conservation must be checked rather than assumed during such transitions.

Let $w=c_s^2$ be constant and small, and write $\theta =i\mathbf{k}\cdot \mathbf{v}$. The supplied synchronous gauge equations implyDifferentiating the first and eliminating $\theta$ givesFor the dominant single component, the Einstein equation gives $h^{\prime \prime }+\mathcal{H}{h}^{\prime }=-3{\mathcal{H}}^2(1+3w)\delta$. Therefore the exact algebraic intermediate relation isThe requested nonrelativistic density equation in synchronous gauge follows to leading nonrelativistic order: discard pressure corrections to the background, Hubble damping, and metric tensor terms while retaining $c_s^2{k}^2\delta$, which can be comparable to gravity at the Jeans wavenumber. Since $w\ll 1$, that scale lies well inside the Hubble radius, so keeping its enhanced gradient term is consistent. The resulting equation iswhere the Friedmann equation gave $3{\mathcal{H}}^2/2=4\pi G{a}^2{\bar{\rho }}_m$. Now ${\delta }^{\prime }=a\dot{\delta }$ and ${\delta }^{\prime \prime }=a^2(\ddot{\delta }+H\dot{\delta })$, soThis is the specified approximation, not an exact finite-$w$ relativistic identity. The pressure-gradient term stabilizes short wavelengths, while gravity drives the long-wavelength Jeans instability.

Now apply the stipulated equation to $P=K{\rho }^{4/3}$ with $K>0$. In the pressure-negligible matter background,Thus $c_s^2{k}^2/a^2$ is proportional to $t^{-2}$, exactly like the gravitational term. DefineThe comoving Jeans wavenumber and $q$ are constant in this era. The equation becomes the Euler-Cauchy equationSetting $\delta \propto t^p$ gives $p^2+p/3+2(q-1)/3=0$. For distinct real roots the full pair of modes isThese are the Jeans modes of a four-thirds polytropic cosmological fluid. The physical Jeans length isHere ${\lambda }_J\propto a$, so its comoving value is constant. For $\lambda >{\lambda }_J$, $q<1$ and $p_{+}>0$, giving genuine growth; the other solution decays. In the long-wavelength pressure-free limit, $p_{+}=2/3$ and $p_{-}=-1$, the familiar matter-era modes. At $\lambda ={\lambda }_J$, the two powers are $0$ and $-1/3$, so the leading mode is constant rather than growing.

For $1<q<25/24$ both powers are negative: pressure prevents growth, although the modes still decay without oscillation. At $q=25/24$ they coincide and the complete solution isFor $q>25/24$ they become a complex-conjugate pair; a real basis isThese short-wavelength acoustic oscillations have a decaying envelope. The onset of instability is $q=1$, while the threshold for oscillatory time dependence is slightly larger, $q=25/24$; expansion damping explains why these two thresholds are not identical.

Let $\mathcal{H}=a^{\prime }/a$ and ${\theta }_N=i\mathbf{k}\cdot {\mathbf{v}}_N$. Choose the synchronous gauge rest frame of cold dark matter, so ${\theta }_C=0$. Its continuity equation then gives $h^{\prime }=-2{\delta }_C^{\prime }$. Substitution in the trace Einstein equation givesFor radiation in cosmology the continuity and Euler equations are ${\delta }_R^{\prime }+4{\theta }_R/3+2h^{\prime }/3=0$ and ${\theta }_R^{\prime }=k^2{\delta }_R/4$. Differentiate the first, eliminate ${\theta }_R^{\prime }$ and $h^{\prime \prime }$, and obtainradiation in cosmology pressure supplies the acoustic restoring term while its energy density contributes twice as strongly to the trace gravitational source through $1+3w_R=2$.

Deep in radiation domination, $a\propto \tau$, $\mathcal{H}=1/\tau$, ${\Omega }_R\simeq 1$ and ${\Omega }_C\simeq 0$. At long wavelengths $k\tau \ll 1$, the adiabatic initial conditions require ${\delta }_C=3{\delta }_R/4$. The cold-matter equation reduces toA power ${\tau }^p$ gives $p^2-4=0$. The regular growing solution therefore hasThe other adiabatic solution diverges as ${\tau }^{-2}$ and is discarded for the regular primordial growing mode. The relation between the density contrasts fixes their relative entropy to zero, not their subsequent equality inside the acoustic horizon.