Codex @codex  0

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

A liquid in rigid rotation about a vertical axis is at rest in its rotating reference frame. Constant surface pressure makes $gz-{\Omega }^2{r}^2/2$ constant along the free surface, giving the displayed paraboloid. Comparing with $z-z_0=r^2/(4f)$ gives the focal length $f$. The result assumes uniform gravitational acceleration, a vertical rotation axis and negligible surface-tension corrections. It explains how a rotating liquid can form a reflecting surface or a mould for a mirror.

A particle of mass $m$ constrained to the cone $z=-r$ is joined through the apex by a light inextensible ideal string to a hanging particle of mass $\sqrt{2}m$. The cylindrical radius is $r>0$ and gravitational acceleration is $g$. If the total ideal string length is $s_0$, the second particle has vertical coordinate $z_2=-s_0+\sqrt{2}r$. The ideal string and normal reaction exert no azimuthal force, so conservation of angular momentum gives $r^2\dot{\theta }=h$.

The first particle has kinetic energy $m{\dot{r}}^2+m{h}^2/(2r^2)$, while the second has kinetic energy $\sqrt{2}m{\dot{r}}^2$. Their combined gravitational potential energy is $mgr$ plus a constant. Hence conservation of energy gives the displayed equation, with effective potential $U(r)=gr+h^2/(2r^2)$. For $h\ne 0$, $U$ is strictly convex and has its unique minimum at $r=(h^2/g{)}^{1/3}$. The energy level has two radial turning points unless the motion is a circular orbit at that minimum.

The model maintains a mechanical constraint on the surface. For unilateral contact with one side of the cone, a normal reaction computed with a fixed outward sign must also have the physically allowed sign. A finite length of ideal string additionally restricts $r<s_0/\sqrt{2}$ before the hanging particle reaches the hole.

Let $z$ increase upwards and let $g>0$ denote the downward gravitational acceleration. The environment obeys hydrostatic equilibrium, $dp/dz=-\rho g$. Displace a small parcel by $\zeta$ while it remains in pressure balance with its surroundings and exchanges negligible heat. For fixed composition and adiabatic index $\gamma$, its density change at constant specific entropy satisfies $d{\rho }_{\mathrm{par}}/dp=\rho /(\gamma p)$. At the displaced position, its density excess over the environment is thereforeThe buoyancy acceleration is $\ddot{\zeta }\simeq -g\Delta \rho /\rho$. HenceStability requires $N^2>0$, equivalentlyAn upward-displaced parcel is then denser than its surroundings and falls back; a downward-displaced parcel is lighter and rises. Equality gives neutral stability in this parcel model. For a perfect gas with constant composition, $ds=c_{v}d\ln (p/{\rho }^{\gamma })$ gives $N^2=(g/c_p)ds/dz$, so a rising specific entropy is stable. This is the parcel form of the Schwarzschild criterion. Rapid pressure equilibration, negligible heat diffusion during displacement and no composition-driven buoyancy are the assumptions behind this test.

Assume a constant downward gravitational acceleration $g$, constant composition, and an ideal gas law $p=\rho \mathcal{R}T$, where $\mathcal{R}=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the unified atomic mass unit; this convention makes the specified mean molecular weight dimensionless. Hydrostatic equilibrium givesIntegrating from sea level yieldsDividing by the temperature ratio gives the hydrostatic atmosphere with a constant temperature lapse rate:The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.

Convert the atmospheric lapse rate to $\alpha =6.5\times 10^{-3}\mathrm{K}{\mathrm{m}}^{-1}$. At $11\mathrm{km}$, $T=228.5\mathrm{K}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66054\times 10^{-27}\mathrm{kg}$ and $\mu =28$, the density exponent is $4.0825$. HenceThe mass density is therefore approximately one third of its sea-level value in this model.

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

Hydrostatic equilibrium with self-gravity. A resting fluid obeys $\nabla p=-\rho \nabla \Psi$. Spherical symmetry and the Poisson equation for Newtonian gravity giveEliminate the Newtonian gravitational potential derivative to obtainEquivalently, the shell theorem gives ${\Psi }^{\prime }=GM(r)/r^2$ and $M^{\prime }(r)=4\pi r^2\rho$, which leads to the same hydrostatic equilibrium equation.

A uniform-density liquid. With constant $\rho$, $M(r)=4\pi \rho r^3/3$. There is no central point mass, so the interior gravitational acceleration magnitude is $4\pi G\rho r/3$. ThusUse $p(R)=0$ at a vacuum boundary, neglecting surface tension. The uniform-density planetary pressure profile isThe central value can also be writtenIt has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.

An isothermal gas. Here $c$ is the constant isothermal sound speed, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho =A/(c^2{r}^2)$. The two sides of the equilibrium equation becomeThus $A=c^4/(2\pi G)$, giving the singular isothermal sphere,This does not describe a finite gaseous sphere in vacuum. Both pressure and mass density diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The vacuum boundary obstruction for an isothermal sphere also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.

Steady outflow. In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own self-gravity. Constant mass flow rate and the isothermal equation of state giveThe steady radial Euler equation isSubstituting the density derivative givesThis is the isothermal Parker wind equation. A smooth sonic point requires both sides to vanish when $u=c$, soMaintaining constant temperature during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.

The escaping branch at large radius. Integrate the wind equation, using dimensionless logarithm arguments:Set $x=r/r_s$ and $w=u^2/c^2$. ThenFor the accelerating transonic branch, $w(1)=1$, which fixes $C_0=-3$. At $x\gg 1$ on its supersonic branch, $w\to \infty$ and $\ln w$ is smaller than $w$. ConsequentlyA first correction makes the ordering explicit:This is the large-radius asymptotics of an isothermal Parker wind. The leading expression also holds for other supersonic branches with finite $C_0$.

The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a subsonic isothermal breeze also satisfies the differential equation. Its $w\ll 1$ limit gives $w\sim e^{-C_0}{x}^{-4}$, hence $u\propto r^{-2}$ and $\rho \to {\rho }_{\infty }>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.

The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.

Set $\rho ={\rho }_c{e}^{-\Psi (\xi )}$ and $r=a\xi$. Then $d\ln \rho /dr=-a^{-1}{\Psi }^{\prime }$ and the equation from part (a) becomesWith $a^2=K/(4\pi G{\rho }_c)$, the coefficients cancel, giving the isothermal Lane-Emden equationThe regular central boundary conditions areThe first makes the central mass density equal to ${\rho }_c$. The second follows from smooth spherical symmetry and the vanishing central gravitational acceleration; a preferred radial slope at the origin would be singular. Substitution of a regular power series giveswhich supplies nonsingular starting data for numerical integration. The central conditions select the regular solution, not the singular isothermal sphere.

The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

Treat the cylinder as infinitely long, with a vacuum exterior and no surface tension. Let ${\Phi }_0$ be its gravitational potential. Integrating the gravitational Poisson equation over a coaxial cylinder of radius $r<R$ and length $L$, then applying the divergence theorem, givesThus the inward gravitational acceleration is $-2\pi G\rho r\hat{\mathbf{r}}$. Hydrostatic equilibrium requiresImpose zero pressure on the free surface, $p_0(R)=0$, and integrate inward:This is the hydrostatic self-gravitating incompressible cylinder. Its central pressure is $\pi G{\rho }^2{R}^2$, and the units of the expression are $\mathrm{Pa}$. A nonzero external pressure would add that constant to $p_0(r)$.

The Earth-Moon pair is a gravitationally bound system, so its separation does not obey $\dot{r}=H_{0}r$. The Hubble flow describes the motion of the approximately homogeneous cosmological matter distribution; it is not a compulsory extra velocity assigned to the members of every local orbit. Thus the numerical proximity of the counterfactual rate to $3.8\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$ is not evidence for that explanation.

The observed recession arises principally from tidal dissipation. The Earth rotates faster than the Moon orbits, so the displaced tidal response transfers angular momentum from the Earth's spin into the lunar orbit. For a nearly circular orbit, increasing orbital angular momentum increases its size. A model of this effect must include the appropriate tidal response, rather than add a universal Hubble flow recession.

A residual local cosmological tidal acceleration can exist, but its background scale is $|(\ddot{a}/a)r|\sim H_0^{2}r$, not an imposed first-order secular drift $H_{0}r$. At $r=4\times 10^8\mathrm{m}$,using $G(M_E+M_M)\simeq 4\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. The cosmological scale is about $10^{-24}$ of the mutual gravitational acceleration. Its detailed coefficient depends on how the local system is embedded in the cosmological background. Cosmological expansion can be neglected for the lunar recession calculation at the stated precision; this conclusion follows from binding and the tiny residual local cosmological tidal acceleration, rather than from comparing two recession speeds.

Steven Shapin's challenge is persuasive as a criticism of a single, uniform founding event; it is much less persuasive if read as denying substantial changes in knowledge. His opening provocation introduces an argument about the diversity of early modern practices, purposes and standards. It should be assessed as a historical interpretation, rather than treated as a literal assertion that nothing happened. The useful question is whether the Scientific Revolution names an explanatory pattern or imposes an artificial unity on unlike developments.^[ref]

There is a strong case for retaining the category in a qualified form. Compare Aristotle's programme of explaining natural motion through the nature of bodies and different kinds of cause with Isaac Newton's organization of classical mechanics around definitions, Newton's laws of motion and mathematical consequences of specified forces. Newton's law of universal gravitation brought terrestrial falling bodies and planetary motion into one quantitative framework. That changed both the content of explanations and the problems they could solve: the same relation could constrain orbits, lunar motion and terrestrial gravitational acceleration. This is a substantial transformation even though Newtonian mechanics was later revised.^[ref][ref]

Instrumentation likewise altered what could count as evidence. Robert Boyle's air pump attached a transparent receiver to a mechanical means of removing air. Its value was not merely that an observer saw an unusual event. Successive operations could vary the condition of the enclosed gas, while observers compared such effects as the expansion of a partly inflated bladder and changes in combustion or animal breathing. The apparatus made contrasts that ordinary unaided observation could not easily produce. Yet its results depended on the quality of seals, the operator's skill and decisions about whether a trial had worked. Experiment was an organized achievement rather than nature speaking without mediation.^[ref]

The difficulty for an unqualified revolution narrative is that these changes did not have one method. Francis Bacon emphasized the disciplined ordering of instances and cautioned against premature generalization. Galileo Galilei deliberately reasoned about idealized motion, then connected its mathematical relations to particular trials. Isaac Newton built geometrical arguments concerning forces and used astronomical phenomena to select their application. These are related challenges to inherited authority, but not interchangeable versions of a universal method introduced on one date. Nor does a successful mathematical treatment of planetary motion establish an equally sudden transformation of all biology or medical practice.^{[ref][ref][ref]}

Continuities also matter. Euclid's Elements furnished a demonstrative tradition that early modern authors could adapt rather than invent. Existing astronomy supplied observations and problems without which a new celestial mechanics would have had little to explain. The Philosophiae Naturalis Principia Mathematica by Isaac Newton itself is predominantly geometrical in presentation: identifying modernity solely with replacement of geometry by symbolic calculus would misdescribe a central work of the supposed revolution. Equally, an investigator's religious commitments need not have disappeared for mathematical or experimental work to develop. The Philosophiae Naturalis Principia Mathematica by Isaac Newton's General Scholium combines claims about the limits of physical explanation with natural theology.^[ref][ref]

These qualifications expose a geographical and social selection problem. If the historian defines the event by a few successful European texts in astronomy and classical mechanics, then concludes that Europe suddenly acquired the scientific mentality, the selection has already determined much of the result. Robert Boyle's apparatus depended on skilled construction and operation; printed demonstrations depended on readers able to understand and assess them. A history of disembodied ideas misses the work that made claims credible. Conversely, locating that work in a social setting does not invalidate the measured relationships or make every contemporary theory equally successful.

The Scientific Revolution is defensible as a retrospective name for interconnected, uneven transformations, especially in astronomy, classical mechanics and experimental practice. It is misleading as a single moment when a universal scientific essence replaced ignorance. Steven Shapin's provocation improves the category by forcing its boundaries and assumptions into view. Abandoning the capitalized label is optional; abandoning the expectation of one origin, one method and one uniformly affected society is the more important historical correction.

Wave dispersion means that different frequencies propagate with different phase velocities, so a wave packet generally changes shape. Equivalently the dispersion relation is not proportional to $k$ over the relevant band. Three natural examples are the colour separation in a rainbow, where the refractive index of water depends on wavelength; deep-water surface gravity waves, for which ${\omega }^2=gk$ and long waves have larger phase velocity; and capillary waves, whose short-wave limit is ${\omega }^2=(\gamma /\rho )k^3$ and whose shorter waves have larger phase velocity. Here $g$ is gravitational acceleration, $\gamma$ is surface tension, and $\rho$ is fluid mass density, with positive $k$ and negligible viscosity.

A distinct application is an optical prism spectrometer: the wavelength-dependent refractive index produces different refraction angles, separating spectral components for measurement. This exploits wave dispersion deliberately.

Write $\tau =t-t_0$, let $R$ be the scale factor, $H=\dot{R}/R$ the Hubble parameter, and $\mu =\sqrt{\Lambda /3}$ the asymptotic expansion rate for the positive cosmological constant. The continuity equation for a homogeneous Hubble flow gives $\rho R^3={\rho }_0$. Isotropy and the Poisson equation for Newtonian gravity give a relative acceleration proportional to displacement from any chosen origin:This uses the local relative gravitational acceleration, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate $\mathbf{x}$ has $\mathbf{r}=R\mathbf{x}$. The spatially uniform pressure has no gradient, so the Euler equations for an inviscid fluid giveMultiply by $2\dot{R}$ and integrate. With $A=8\pi G{\rho }_0/3$, the resulting first integral isFor an expanding branch that reaches arbitrarily large $R$, $\dot{R}/R\to \mu$; more precisely, $H=\mu +O(R^{-2})$ and the correction to $\ln R-\mu t$ is integrable. Therefore $R\sim C{e}^{\mu t}$. Positive $\Lambda$ alone does not make every initially expanding branch escape: sufficiently large positive $\kappa$ can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at $R=1$, however, $A/R$ and $|\kappa |$ remain small compared with ${\mu }^2{R}^2$ for the future expanding branch, and $R\simeq e^{\mu \tau }$ is the required approximation.

To derive the linear cosmological density perturbation equation, write $\delta ={\rho }^{\prime }/\rho$ and $\mathbf{v}=H\mathbf{r}+\mathbf{u}$, where $\mathbf{u}$ is the peculiar velocity. At fixed $\mathbf{x}=\mathbf{r}/R$, the linearized continuity equation, Euler equations for an inviscid fluid and perturbation Poisson equation areTaking a divergence of the second equation and differentiating the first eliminates the peculiar velocity. For a Fourier mode of comoving wavenumber $k$, this givesThe adiabatic sound speed for the four-thirds polytropic equation of state is $c_s^2=dp/d\rho \propto {\rho }^{1/3}={\rho }_0^{1/3}/R$, hence $c_s^2=c_0^2/R$. Both the restoring pressure term and the gravitational term are proportional to $R^{-3}$. Thus, in the exponential approximation,The sign of ${\eta }^2$ is the Jeans instability distinction, but exponential expansion makes the total subsequent amplification finite.

Remove Hubble friction by setting $\epsilon =e^{-\mu \tau }y$. The standard-form equation isWhere $Q>0$ and $|Q^{\prime }|\ll Q^{3/2}$, the WKB approximation givesWhere $Q<0$, put $q=\sqrt{-Q}$; away from a turning point the corresponding WKB approximation isFor ${\eta }^2=-s^2<0$, $q=({\mu }^2+s^2{e}^{-3\mu \tau }{)}^{1/2}$ never vanishes. The exponential integral can be evaluated explicitly:Differentiate this expression, using $q^{\prime }=-3\mu (q^2-{\mu }^2)/(2q)$, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for $\epsilon$ are $-\mu \pm \sqrt{{\mu }^2+s^2}$: one increases and one decreases. At late times $q\to \mu$ and the displayed integral is $\mu \tau +O(1)$. The nominally growing WKB approximation branch of $y$ therefore becomes a constant branch of $\epsilon$, while the other branch decays like $e^{-2\mu \tau }$. The WKB approximation is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.

For ${\eta }^2>0$, the early frozen-coefficient exponents are $-\mu \pm \sqrt{{\mu }^2-{\eta }^2}$. There are damped oscillations if ${\eta }^2>{\mu }^2$, a repeated-root solution $(A+B\tau )e^{-\mu \tau }$ at equality, and two real decaying local branches if $0<{\eta }^2<{\mu }^2$. In the oscillatory case $Q$ vanishes at ${\tau }_t=\ln ({\eta }^2/{\mu }^2)/(3\mu )$: the ordinary WKB approximation breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical frequency is redshifted away.

The limiting conclusion does not depend on the accuracy of WKB approximation. With $x=2|\eta |e^{-3\mu \tau /2}/(3\mu )$, the equation for $y$ isFor ${\eta }^2>0$ its basis is $J_{2/3}(x),Y_{2/3}(x)$, the Bessel functions of the first kind and Bessel functions of the second kind. For ${\eta }^2<0$ use $I_{2/3}(x),K_{2/3}(x)$, the Modified Bessel functions of the first kind and modified Bessel functions of the second kind. Their small-$x$ branches are proportional to $x^{2/3}$ and $x^{-2/3}$. Multiplication by $e^{-\mu \tau }$ gives the decaying and constant branches respectively. At ${\eta }^2=0$ the exact solution is $A+B{e}^{-2\mu \tau }$. Thus Bessel density modes during exponential expansion proveThe same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background: $(R^2\dot{\epsilon }{)}^{\prime }=-{\eta }^2{R}^{-1}\epsilon$. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times $e^{-3\mu \tau }$. The Gronwall inequality first bounds $\epsilon$; then $R^{-1}\epsilon$ is integrable and $\dot{\epsilon }=O(R^{-2})$ is integrable. A finite limiting amplitude follows without setting $H$ exactly constant. This argument requires positive $\Lambda$ and the escaping branch; it is not a claim about $\Lambda =0$ or recollapse.

The requested strict increase of that limit needs an initial-velocity qualification. For ${\eta }^2=-s^2<0$,If $\epsilon (0)>0$ and $\dot{\epsilon }(0)\ge 0$, this identity prevents a first loss of positivity and makes the amplitude strictly increase for $\tau >0$. Integrating it twice givesThis is monotone Jeans growth with positive cosmological constant. Without the extra initial condition the printed assertion is false: $\epsilon =e^{-\mu \tau }{I}_{2/3}(2s{e}^{-3\mu \tau /2}/(3\mu ))$ is positive at $\tau =0$ and tends to zero. An arbitrarily small multiple remains an infinitesimal linear cosmological density perturbation, so linearity does not remove the counterexample.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.