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Let $u=u_r<0$ for accretion. The steady spherical continuity equation and radial Euler equations for an inviscid fluid giveFor a barotropic fluid, $dp/dr=c_s^{2}d\rho /dr$. Dividing continuity equation by $r^2\rho u$ gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Eliminating this derivative yieldsAt a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:These are the sonic and regularity conditions. Otherwise the derivative is singular and the proposed smooth transonic passage fails. The local slope must also be a real root compatible with the desired branch. For example, defining $C=\rho d(c_s^2)/d\rho$ at the point and differentiating both sides givesThis explains why simultaneous vanishing is a necessary regularity condition, rather than an automatic proof that every potential admits a critical crossing.

Define the barotropic enthalpy function by $dh=dp/\rho$. The momentum equation integrates to the Bernoulli equationChanging the reference mass density in $h$ changes only the constant $B$. For an isothermal closure this logarithmic function is a barotropic barotropic pressure potential; it should not be confused with the constant thermodynamic specific enthalpy of an ideal gas held at fixed temperature.

For isothermal transonic accretion in the Paczyński-Wiita potential, assume $M>0$ and $c_s>0$, and define $r_B=GM/(2c_s^2)$ and $\beta =16c_s^2/c^2$. The critical equation iswhose two roots are $r_G+\frac{1}{2}{r}_B(1\pm \sqrt{1+\beta })$. Only the plus root lies outside $r_G$. ThereforeFor this isothermal case $C=0$, and the slope relation reduces toThe positive slope is the inward-accretion solution connecting a small inward speed at large radius to the supersonic inward branch. Thus the exterior critical point is a genuine nondegenerate transonic point.

Choose the reference mass density as ${\rho }_0$, so $h=c_s^2\log (\rho /{\rho }_0)$ and $B=0$ from the conditions at infinity. At $u_c=-c_s$,Substitution into $\dot{M}=4\pi r_c^2{\rho }_c{c}_s$ givesThis is the rate selected by the smooth transonic accretion solution. Arbitrary static or subsonic formal solutions are not assigned this rate merely by specifying conditions at infinity. As $r_G/r_B\to 0$, the result tends to the isothermal Bondi accretion rate $\pi e^{3/2}{\rho }_0(GM{)}^2/c_s^3$, a useful normalization check.

The steady Euler equations for an inviscid fluid with no body force are $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho$. Use the convective acceleration identityto obtainFor the question's streamfunction convention, $\mathbf{u}=({\psi }_y,-{\psi }_x,0)$, so $\mathbf{u}\times \mathbf{\omega }=-\omega \nabla \psi$. Constant $\omega$ therefore gives the Bernoulli function for planar constant-vorticity flowon the connected fluid region.

For circular streamlines, write $\mathbf{u}=w(r)\hat{\mathbf{\varphi }}$. The cylindrical curl gives $(rw{)}^{\prime }/r=\omega$, so $w=\omega r/2+C_1/r$. The outer wall condition gives $C_1=-2\omega a^2$. Choose the positive azimuthal orientation so that $w(a)=V$, as implicit in the requested sign for $\omega$. ThenReversing the circulation reverses $\omega$ but leaves the speed and pressure difference unchanged. For this uniform-vorticity circular flow in an annulus, one convenient streamfunction is $\psi =V(r^2/2-4a^2\log (r/a))/(3a)+C_2$.

The radial component of the Euler equations for an inviscid fluid is $p_r=\rho w^2/r$. Thus pressure increases outwards, andTherefore

Write the radial velocity as $u(r)$, allowing $u<0$ for accretion flow, and let $c^2=\gamma p/\rho$ be the squared adiabatic sound speed. The steady continuity equation, radial Euler equations for an inviscid fluid and entropy advection equation giveOn a nonzero smooth flow branch, the polytropic equation of state is $p=K{\rho }^{\gamma }$ with constant $K$. Thus the two useful first integrals for spherically symmetric adiabatic flow areHere $\mathcal{J}$ is the signed mass flux and $\mathcal{B}$ is the Bernoulli function. The specific enthalpy is $c^2/(\gamma -1)$; the prescribed Newtonian gravitational potential has no contribution from the gas's own gravity. Eliminating the density derivative using mass conservation gives the differential formThis sonic point equation displays both the singular coefficient and the numerator that must vanish for a smooth transonic branch.

Take a static, nonrotating, nonmagnetic spherical equilibrium with $\nabla P=-\rho \nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. Use a fluid displacement $\mathbf{\xi }(\mathbf{r})e^{-i\omega t}$. Write $p_1,{\rho }_1,{\varphi }_1$ for Eulerian fluid perturbations. The Lagrangian pressure perturbation and corresponding mass density change obey $\Delta P=p_1+\mathbf{\xi }\cdot \nabla P$, $\Delta \rho ={\rho }_1+\mathbf{\xi }\cdot \nabla \rho$. Linearizing mass conservation, the Euler equations for an inviscid fluid and Poisson equation for Newtonian gravity givesThere is no equilibrium acceleration to multiply a perturbed mass density. The adiabatic process condition, with composition carried by the parcel, closes the system:These are the complete linear adiabatic stellar oscillation equations, including the perturbation of self-gravity. Neglect of heat exchange is appropriate when oscillation periods are short compared with relevant thermal relaxation times; it does not determine nonadiabatic excitation or damping.

For the pressure and buoyancy modes, separate angular dependence using spherical harmonics:The horizontal amplitude ${\xi }_h$ has dimensions of length. Angular differentiation gives $\nabla \cdot \mathbf{\xi }=[r^{-2}(r^2{\xi }_r{)}^{\prime }-\ell (\ell +1){\xi }_h/r]Y_{\ell }^m$, and tangential momentum gives ${\omega }^{2}r{\xi }_h=p_{\ell }/\rho +{\varphi }_{\ell }$. Define the adiabatic sound speed, stellar buoyancy frequency and Lamb frequency byThe adiabatic mass density relation becomes ${\rho }_{\ell }/\rho =p_{\ell }/(\rho c_s^2)+(N^2/g){\xi }_r$. Substitution produces a radial form of the full oscillation equations for nonzero $\omega$:The apparent $N^2/g$ factor is evaluated through its defining gradient at the centre rather than by dividing two zeros.

Regularity at the centre excludes singular solutions. At a free surface the Lagrangian pressure perturbation vanishes, $\Delta P=0$; outside the star the gravitational perturbation decays as $r^{-\ell -1}$. For a model with mass density tending to zero at its surface, continuity of ${\varphi }_{\ell }$ and its radial derivative gives ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R$. If the equilibrium mass density jumps to vacuum, include the displaced-surface mass sheet: the outward-minus-inward derivative jump is $4\pi G\rho (R){\xi }_r(R)$, so the interior condition is ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R-4\pi G\rho (R){\xi }_r(R)$. An atmospheric boundary condition can replace the ideal free surface.

These conditions make ${\omega }^2$ an eigenvalue, not an arbitrary local sound frequency. With conservative boundary conditions the adiabatic operator is self-adjoint, giving real ${\omega }^2$; negative values describe instability. The frequencies depend on $P(r),\rho (r),{\Gamma }_1(r)$, self-gravity, stratification and boundaries, as well as angular degree $\ell$ and radial order. In a spherical nonrotating star they are degenerate in $m$. For homologous equilibrium structures, their scale isthus the typical oscillation time measures inverse square root of mean mass density, while individual frequencies probe the interior adiabatic sound speed and stellar buoyancy frequency profiles.

For radial modes, write ${\xi }_r=r\eta$. Eliminating the pressure and gravitational perturbations gives the radial stellar pulsation equationThis is a Sturm-Liouville problem. Multiplication by $\eta$ and integration, with vanishing boundary terms, gives its Rayleigh quotientFor constant ${\Gamma }_1>4/3$, $P^{\prime }<0$ makes both numerator contributions nonnegative. At ${\Gamma }_1=4/3$, a homologous displacement is neutral; for constant ${\Gamma }_1<4/3$ the same trial displacement makes the quotient negative. For the uniform-density stellar model, $\eta =$ constant is an exact mode and ${\omega }^2=(3{\Gamma }_1-4)GM/R^3$. With varying ${\Gamma }_1$, the integral criterion, rather than a universal pointwise threshold, controls radial stability.

The Cowling approximation neglects ${\varphi }_{\ell }$ while retaining the equilibrium gravitational field. It is useful for short-wavelength modes, but is not needed for the full derivation above. In a locally slowly varying region, take both remaining amplitudes proportional to $e^{i\int k_{r}dr}$ and retain the leading derivative terms. ThenEliminating either amplitude gives the acoustic-gravity propagation relationPositive $k_r^2$ is oscillatory propagation; negative $k_r^2$ means an evanescent wave. The high-frequency branch, ${\omega }^2>S_{\ell }^2,N^2$, describes stellar acoustic modes, restored chiefly by compressibility and pressure. At frequencies well above $N$, this gives ${\omega }^2\simeq c_s^2[k_r^2+\ell (\ell +1)/r^2]$. The low-frequency propagating branch in stable stratification, ${\omega }^2<N^2,S_{\ell }^2$, instead describes stellar gravity modes, restored by buoyancy. At $\ell =0$ there is no such nonradial gravity-wave cavity.

A stellar acoustic mode is trapped between an inner turning point near $\omega =S_{\ell }$ and an outer reflecting region. Low-$\ell$ modes penetrate deeply; radial modes reach the centre. Higher-degree modes turn farther out. Standing waves require the WKB quantization conditionwhere the phase $\alpha$ depends on the central or turning-point behaviour and surface reflection. For high radial order and small degree, the leading acoustic travel-time result is the large frequency separationThe $\ell /2$ term is the leading central angular phase shift; smaller frequency separations depend on detailed interior gradients. The near-surface mass density stratification sets the acoustic cutoff frequency. In a plane-parallel isothermal atmosphere with mass density scale height $H_{\rho }$, ${\omega }_{\mathrm{ac}}\simeq c_s/(2H_{\rho })$. Modes below this cutoff can reflect and form a cavity; waves sufficiently above it escape and need an outgoing-wave boundary condition. Nonadiabatic damping, driving and rotation alter real-star mode properties, but the adiabatic frequency problem isolates their dependence on the equilibrium structure.

Let $\mathbf{v}$ denote the velocity relative to the uniformly rotating frame, and write $r_{\perp }^2=x^2+y^2$. For rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, the Euler equations for an inviscid fluid in that frame areThe negative quadratic term is the centrifugal potential, whose negative gradient is the outward centrifugal acceleration per unit mass. In steady flow, dotting the equation with $\mathbf{v}$ eliminates the Coriolis force, since $\mathbf{v}\cdot (\mathbf{\Omega }\times \mathbf{v})=0$. The remaining acceleration term is $\mathbf{v}\cdot \nabla (|\mathbf{v}{|}^2/2)$. ThusBy part (b), the steady rotating-frame Bernoulli integral isSteadiness here is measured in the rotating frame, including the body-force potential used in that frame.

In steady flow, the Euler equations for an inviscid fluid and part (a) giveTaking the dot product with $\mathbf{u}$ yields $\mathbf{u}\cdot \nabla H=0$, because a cross product is perpendicular to its first factor. Part (b) therefore gives Bernoulli's quantity along streamlines:For rotational flow these constants can differ between streamlines, since $\mathbf{u}\times \mathbf{\omega }$ need not vanish and therefore $\nabla H$ need not be zero. Special rotational flows can still have one global constant when that cross product vanishes; the general conclusion is only constancy along individual streamlines.

Write the conservative body force per unit mass as $-\nabla \Phi$. For a fluid of constant density $\rho$, the Euler equations for an inviscid fluid and the incompressible flow condition areUsing part (a), these becomeFor irrotational flow, locally choose a velocity potential with $\mathbf{u}=\nabla \varphi$. Then $\nabla ({\partial }_t\varphi +H)=0$, so the unsteady Bernoulli integral isThe constant is spatially uniform on a connected region where the velocity potential is defined; its time dependence can be absorbed into the additive time-dependent gauge of $\varphi$. A global single-valued velocity potential requires an appropriate topological condition, such as a simply connected domain; without it, the integral is locally valid. In a connected steady irrotational region, $\nabla H=0$ directly gives one spatial Bernoulli constant even without choosing a global potential.

For constant density $\rho$ and constant angular velocity $\mathbf{\Omega }$, the Euler equations for an inviscid fluid in the rotating frame isThe last two forces are gravity and centrifugal acceleration; $2\mathbf{\Omega }\times \mathbf{u}$ is the Coriolis force term on the left. Incorporate both conservative forces into the modified pressureso the right side is $-\nabla \Pi$.

At latitude $\lambda$ on a spherical Earth of radius $R$, the outward local vertical component of centrifugal acceleration is ${\Omega }_E^{2}R{\cos }^2\lambda$. Its ratio to gravity is at most ${\Omega }_E^{2}R/g\simeq 3.5\times 10^{-3}$, using ${\Omega }_E\simeq 7.3\times 10^{-5}{\mathrm{s}}^{-1}$, $R\simeq 6.4\times 10^6\mathrm{m}$, and $g\simeq 9.8{\mathrm{m}\mathrm{s}}^{-2}$. Thus this component is small compared with gravity everywhere on Earth.

For a velocity scale $U$ and length scale $\ell$, nonlinear advection scales as $U^2/\ell$, whereas the Coriolis term scales as $2\Omega U$. Their ratio is the Rossby number $U/(2\Omega \ell )$. If it is small, for example by increasing rotation at fixed $U,\ell$, neglect nonlinear advection in perturbations about a fluid at rest. The conservative equilibrium forces have already been absorbed into pressure; with a rotation-timescale unsteady motion the leading equations areThese are the linear equations governing inertial waves.

For constant fluid density $\rho$, steady Euler equations for an inviscid fluid under a conservative force per unit mass $-\nabla \Phi$ read $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho -\nabla \Phi$. Here $\mathbf{u}$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf{u}$ givesThis is the Bernoulli equation. The constants can differ between streamlines. If density varies, the same derivation for a barotropic fluid replaces $p/\rho$ by ${\int }^{p}d{p}^{\prime }/\rho (p^{\prime })$; constant density is the appropriate water approximation here.

For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio ${ϵ}^2/h^2$, so it can be neglected when $h\gg ϵ$. The Bernoulli equation then gives Torricelli's law, $U\simeq \sqrt{2gh}$. Conservation of mass givesHence the leading drainage time of a conical tank isRemoving the tip places the aperture at height of order $ϵ$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(ϵ)$ changes the displayed leading time by relative order $(ϵ/h_0{)}^{5/2}$; replacing the actual head $h-ϵ$ by $h$ introduces a larger, but still vanishing, relative correction of order $ϵ/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.

For constant-density incompressible inviscid flow with conservative body force, Euler equations for an inviscid fluid give ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla (p/\rho +\Phi )$. Using $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$, take the curl to obtain ${\partial }_t\mathbf{\omega }=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. Since both divergences vanish, the vector identity expands this toWithout incompressibility or the constant-density/barotropic hypothesis, extra compression or baroclinic terms are present; the printed equation uses these flow assumptions.

Let $\kappa (t)=\dot{h}/h$, with $h>0$. Axisymmetric mass conservation is $r^{-1}{\partial }_r(r{u}_r)+{\partial }_z{u}_z=0$. If $u_r$ is independent of $z$, then $u_z$ is affine in $z$. The impermeable moving-plate conditions $u_z(r,\pm h,t)=\pm \dot{h}$ force $u_z=\kappa z$, and integrating the radial equation gives $u_r=-\kappa r/2+C(t)/r$. Regularity at the axis excludes $C(t)/r$. The meridional part of the axisymmetric inviscid flow between moving parallel plates is thereforeFor no swirl, $u_{\theta }=0$ completes the field. Continuity alone does not determine azimuthal motion; this must be specified separately. More generally, prescribe a smooth regular initial profile $U(r)$ independent of $z$. The azimuthal part of Euler equations for an inviscid fluid isso conservation of $r{u}_{\theta }$ along radial particle paths gives $u_{\theta }(r,t)=\lambda U(\lambda r)$, where $\lambda =\sqrt{h/h_0}$. Together with the meridional components above, this gives a family of complete fields, with a suitable pressure determined by the radial and vertical equations. The field is therefore not unique without rotational data.

With the stated solid-body rotation, $u_{\theta }=\Omega (t)r$ and $\mathbf{\omega }=2\Omega {\mathbf{e}}_z$. The vorticity equation gives $2\dot{\Omega }=2\Omega {\partial }_z{u}_z=2\Omega \kappa$, henceFor a fluid particle, $\dot{r}=-\kappa r/2$, so $r^{2}h$ is constant and $r^2\Omega$ is constant. This is conservation of angular momentum about the axis: radial spreading when the plates approach decreases the angular velocity. The meridional and rotating fields satisfy the inviscid equations with an appropriate pressure; no tangential no-slip condition is imposed on inviscid plates.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let $v(r)>0$ denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state giveUsing the adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$ and $A\propto r^3$,The Bernoulli equation iswhere the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

Write $\mathbf{x}$ for a comoving coordinate, $\mathbf{u}$ for the peculiar velocity, $H=\dot{a}/a$ for the Hubble parameter, and $\delta =\delta \rho /\bar{\rho }$ for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation areHere $c_s$ is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating $\mathbf{u}$ and $\varphi$ givesA Fourier mode with comoving wavenumber $k$ therefore obeysThe Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is $2\pi /k$, the result isBelow this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.

For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining ${\sigma }^2=⟨u_x^2⟩=⟨|\mathbf{u}{|}^2⟩/3$, the static collisionless marginal-stability calculation givesThe coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance $\int vdt/a$ traveled since decoupling.

For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination, $a\propto t^{1/2}$, to matter domination, $a\propto t^{2/3}$. For the usual schematic dynamical Jeans scale use ${\bar{\rho }}_{\mathrm{tot}}$ in the estimate, soNear the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With $R_b=3{\bar{\rho }}_b/(4{\bar{\rho }}_{\gamma })\propto a$, its photon-baryon sound speed isFor example, this follows from $\delta {\rho }_{\gamma }/{\bar{\rho }}_{\gamma }=(4/3)\delta {\rho }_b/{\bar{\rho }}_b$ and $\delta P=c^2\delta {\rho }_{\gamma }/3$. In the radiation-dominated limit $R_b\ll 1$, $c_s\simeq c/\sqrt{3}$; in the strongly baryon-loaded matter-era limit $R_b\gg 1$, $c_s\propto a^{-1/2}$. Thus the baryon Jeans length across recombination has the asymptotic historyThe transition to the plateau is smooth: retaining $1+R_b$ gives ${\lambda }_{J,b}^{\mathrm{com}}\propto a^{1/2}/\sqrt{1+R_b}$ in matter domination. If baryon loading is still small, this intermediate segment instead rises as $a^{1/2}$ until loading becomes important.

At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas, $c_{s,b}^2=\gamma k_B{T}_b/(\mu m_p)$, with $\gamma =5/3$. The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives $T_b\propto {\bar{\rho }}_b^{\gamma -1}\propto a^{-2}$, so $c_{s,b}\propto a^{-1}$ and ${\lambda }_{J,b}^{\mathrm{com}}\propto a^{-1/2}$. Residual energy exchange through Compton scattering can initially keep $T_b\propto a^{-1}$, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

log(comoving baryon Jeans scale)
  ^
  |                   __________________
  |                 /                   |
  |               /                     |  loss of photon pressure
  |             /                       |
  |           /                         +---\
  |         /                                \
  +-------------------|-----------------|----------> log(time)
                    t_eq              t_rec
        slope 1/2          slope ~0        slope -1/3

For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering $t_{\mathrm{NR}}<t_{\mathrm{dec}}<t_{\mathrm{eq}}<t_{\mathrm{rec}}$. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled, $\sigma \propto \sqrt{T}\propto a^{-1/2}$; after kinetic decoupling, particle momentum redshifts as $a^{-1}$ and $\sigma \propto a^{-1}$. Combining these with the background dynamical density givesThere is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.