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A Fourier mode advected with a homologous spherical flow has phase $\mathbf{q}\cdot \mathbf{r}/R(t)$. Its material derivative in the background vanishes because $\dot{\mathbf{k}}=-(\dot{R}/R)\mathbf{k}$. Thus contraction shortens its physical wavelength while preserving its phase on each fluid element. Linearizing the continuity equation and pressureless momentum equation gives $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$ and ${\dot{\mathbf{u}}}_1+(\dot{R}/R){\mathbf{u}}_1=-i\mathbf{k}{\Phi }_1$. Combining these with the Poisson equation for Newtonian gravity yields $\ddot{\delta }+2(\dot{R}/R)\dot{\delta }-4\pi G\rho \delta =0$ for nonzero compressive wavenumber. It is a local bulk calculation, without the moving cloud-boundary conditions.

The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

Applying the Poisson equation for Newtonian gravity to a flattened power-law gravitational potential yieldsFor $0<\beta <1$, nonnegative mass density everywhere outside the origin requires and is ensured by $\beta \le {\lambda }^2\le 2/(\beta +1)$. The central density singularity is locally integrable and carries no point mass, although the total mass diverges at large radius in this scale-free model.

The Poisson equation for Newtonian gravity givesFor $V>0$, nonnegative mass density away from the origin is equivalent to $q^2\ge 1/2$. Strict positivity requires $q^2>1/2$. This condition does not by itself guarantee a nonnegative galactic distribution function.

The vertical Jeans equation determines $K_z={\partial }_z\Phi$ from a tracer's density and velocity dispersion, with a correction for the tilted velocity ellipsoid. Integrating the axisymmetric Poisson equation for Newtonian gravity givesfor $z>0$ and reflection symmetry. Subtracting independently estimated stellar and gas surface densities constrains any additional gravitating component; a dark matter halo also contributes, so an excess is not automatically a dark disk.

A sinusoidal surface-density component ${\Sigma }_k\cos (kx)$ on $y=0$, with $k>0$ and translational invariance in the remaining direction, produces the displayed gravitational potential. Laplace's equation away from the sheet separates into a cosine in $x$ and a decaying exponential in $|y|$. Continuity across the sheet excludes a potential jump, while integrating Poisson equation for Newtonian gravity across it gives $[{\Phi }_y{]}_{0^{-}}^{0^{+}}=4\pi G{\Sigma }_k\cos (kx)$. For a potential $C\cos (kx)e^{-k|y|}$ that derivative jump is $-2kC\cos (kx)$, giving $C=-2\pi G{\Sigma }_k/k$.

More generally a specified jump $J\cos (kx)$ gives potential $-J\cos (kx)e^{-k|y|}/(2k)$. Each spatial Fourier mode decays on the length scale $1/k$, so fine density structure has a shorter-range field. The sinusoid changes sign and is naturally interpreted as a density perturbation component. General sheet distributions are built by superposing their Fourier modes; decay alone does not restrict a harmonic potential to a single mode.

For a symmetric self-gravitating slab with polytropic equation of state $p=K{\rho }^2$ and constant $K>0$, hydrostatic equilibrium gives $2K{\rho }^{\prime }=-{\psi }^{\prime }$. The plane-parallel Poisson equation for Newtonian gravity then yields ${\rho }^{\prime \prime }+(2\pi G/K)\rho =0$. Reflection symmetry at the midplane fixes the title profile. For a vacuum surface, retain the nonnegative profile only on $|z|\le z_s=\pi /(2k)$ and put $\rho =0$ outside. The half-thickness is independent of central density in this ideal index-one model, while the surface mass density is $\Sigma =2{\rho }_c/k$.

For an isothermal gas slab, a horizontal Fourier mode has $\delta \rho =-ik\rho {\xi }_x-(\rho {\xi }_z{)}_z$, $W=c_s^2\delta \rho /\rho +\varphi$, ${\omega }^2{\xi }_x=ikW$, ${\omega }^2{\xi }_z=W_z$ and $({\partial }_z^2-k^2)\varphi =4\pi G\delta \rho$. Multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate by parts, then use the Poisson equation for Newtonian gravity. Vanishing pressure-work and gravitational boundary terms give the displayed real quadratic identity, with $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$. Thus every admissible squared eigenfrequency is real; a negative value gives gravitational growth rather than an oscillatory instability.

Without an escape-energy cutoff, a galactic distribution function proportional to $\exp [(\Psi -v^2/2)/{\sigma }^2]$ gives $\rho \propto e^{\Psi /{\sigma }^2}$ and a constant one-dimensional velocity dispersion. Since $\Psi ={\Phi }_0-\Phi$, differentiation gives the displayed hydrostatic relation. An ideal gas at constant temperature has pressure $P=\rho k_{B}T/m_0$, so its hydrostatic equilibrium equation is identical at the displayed dispersion-temperature identification. Coupling either relation to the same Poisson equation for Newtonian gravity, with matched central density and boundary data, therefore gives the same density structure. Truncating the stellar velocities at escape changes this equivalence.

For a spherical, nonstreaming stellar population with isotropic velocity dispersion, each component of the random pressure tensor equals $p=n⟨v_r^2⟩$. The observed second stellar velocity moment satisfies $N(R){\sigma }_{\mathrm{los}}^2(R)=2{\int }_R^{\infty }p(r)r/\sqrt{r^2-R^2}dr$. Composing this Abel transform with itself gives the displayed inverse under sufficient decay and regularity. The Spherical Jeans equation then gives ${\psi }^{\prime }=p^{\prime }/n$, with gravitational acceleration $\nabla \psi$. The Poisson equation for Newtonian gravity recovers total gravitating mass density, not just the tracer density. Without velocity isotropy the same projected moment also contains anisotropic stresses.

A Kuzmin disk is an infinitesimally thin axisymmetric astrophysical disk whose relative potential is $\psi =Gm/\sqrt{R^2+(|z|+b{)}^2}$, with $m,b>0$. Away from the disk this is a point-mass vacuum field. The jump of its vertical derivative is $-2Gmb/(R^2+b^2{)}^{3/2}$, so the Poisson equation for Newtonian gravity, written ${\nabla }^2\psi =-4\pi G\rho$, gives the displayed surface density. Integrating $2\pi R\Sigma$ gives total mass $m$.

The reflection-symmetric Newtonian gravitational potential $\Phi =v_0^2\log [(\sqrt{R^2+z^2}+|z|)/R_{\ast }]$ is harmonic off the plane. Its normal-derivative jump is $2v_0^2/R$, so the Poisson equation for Newtonian gravity gives surface density of a disk $S=v_0^2/(2\pi GR)$ and volume mass density $\rho =S\delta (z)$. The midplane circular speed is constant. The scale-free model has infinite total mass and does not use a finite-mass boundary condition at infinity.

For a Newtonian gravitational potential obeying the Poisson equation for Newtonian gravity, define $T_{ij}=[{\partial }_i\Phi {\partial }_j\Phi -{\delta }_{ij}|\nabla \Phi {|}^2/2]/(4\pi G)$. Commutation of mixed partial derivatives gives ${\partial }_j{T}_{ij}=\rho {\partial }_i\Phi$. Thus the gravitational force density can be put in conservative form as $-{\partial }_j{T}_{ij}$. The sign is chosen so that $T_{ij}$ is added to the momentum flux; reversing the stress convention reverses the divergence sign.

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]

Using $[{\nabla }_c,{\nabla }_d]V^a=R^a{}_{bcd}{V}^b$, the Newton–Cartan gravitational connection has only the displayed Hessian components and their antisymmetric partners in the last two indices. The geodesic deviation equation gives relative acceleration $-{\partial }_i{\partial }_j\Phi \delta x^j$. The Poisson equation for Newtonian gravity is therefore $R_{ab}=4\pi G\rho (dt{)}_a(dt{)}_b$. A spatially uniform gravitational acceleration has zero tidal curvature even if the connection coefficients do not vanish in the original coordinates.

For small isentropic perturbations of a fixed-composition gas, linearizing its equation of state gives $p_1=(\partial p/\partial \rho {)}_s{\rho }_1=c_s^2{\rho }_1$. The adiabatic sound speed is evaluated at the uniform background, so it is spatially constant at this order. If fast heat exchange instead enforces constant temperature, use the isothermal sound speed consistently. A completely arbitrary entropy perturbation would require an additional thermodynamic variable.

The uniform static infinite medium requires the Jeans swindle: its background gravitational field is subtracted or externally balanced, and only the perturbation sources the Poisson equation for Newtonian gravity. Linearizing the continuity equation, Euler equation and gravitational equation then givesTaking one more time derivative of the first equation and substituting the other two yieldsA Fourier mode proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore obeys the Jeans instability relationFor $k>k_J$, the mode oscillates as a sound wave. For $0<k<k_J$, $\omega =\pm i\sqrt{4\pi G{\rho }_0-c_s^2{k}^2}$ gives a growing and a decaying mode; the static Jeans growth-rate spectrum has no preferred finite fastest-growing wavelength in the infinite-medium approximation. The threshold $k=k_J$ is marginal at linear order. The Jeans length is ${\lambda }_J=2\pi /k_J=c_s\sqrt{\pi /(G{\rho }_0)}$. Defining the Jeans mass as a sphere whose diameter is this wavelength givesOther radius conventions alter its numerical coefficient; the mass scale compares self-gravity with pressure communication.

For the pressureless finite cloud, the shell theorem gives an edge acceleration $\ddot{r}=-GM/r^2$, with fixed enclosed $M=4\pi r_0^3{\rho }_0/3$. Multiplication by $\dot{r}$ and integration from rest givesThe negative root describes collapse. Substituting $r=r_0{\sin }^2\theta$ gives the free-fall time of a uniform sphereThe surface and inner shells reach the origin together in pressureless homologous spherical collapse; pressure, rotation or shell crossing would change this ideal solution.

Equating this time to the specified radius-based sound-crossing time $r_0/c_s$ gives $r_0=c_s\sqrt{3\pi /(32G{\rho }_0)}$. The resulting crossing-time cloud support mass isThis is the same dimensional mass scale as $M_J$, with a different coefficient because the adopted length and time criteria differ. In particular, $t_{\mathrm{ff}}=r_0/c_s$ is not the exact marginal dispersion-relation condition.

With the requested isotropic support approximation, replace $c_s$ by the Alfvén speed $v_A=B/\sqrt{{\mu }_0{\rho }_0}$. In the crossing-time convention the Magnetic Jeans mass isUsing the wavelength-based Jeans mass convention instead replaces $C_{\mathrm{cross}}$ by ${\pi }^{5/2}/6$. The scaling, rather than a convention-independent coefficient, is the shared prediction.

For a fixed cloud mass undergoing isothermal homologous contraction, $\rho /{\rho }_0=(r_0/r{)}^3$ and $c_s$ is constant. Its number of thermal Jeans masses therefore grows asA cloud initially close to the support threshold has $N_{J,0}$ of order one, so smaller pieces can subsequently become independently unstable: this is isothermal fragmentation.

For approximately isotropic contraction with magnetic flux freezing, $B{r}^2$ stays constant, so $B\propto {\rho }^{2/3}$. Hence $M_B\propto B^3/{\rho }^2$ stays constant andThis ideal magnetic support therefore does not create progressively more unstable masses in the way thermal isothermal support does. The same conclusion follows from the conserved mass-to-flux ratio. A magnetically subcritical cloud cannot initiate unrestricted cross-field collapse under ideal flux freezing alone; flux loss or initially supercritical conditions are needed. A real magnetic field also gives anisotropic support: motion and fragmentation along its field lines are not excluded by this scalar-speed estimate.

Write the polytropic law as $p=K{\rho }^{1+1/n}$, with constant $K$ and $n>0$. Its pressure potential, or polytropic enthalpy relative to zero density, isHydrostatic balance is $dH/dr=-d\psi /dr$. If the surface has zero density and potential ${\psi }_0$, integration gives $H={\psi }_0-\psi$. At the centre, $\Delta \psi \equiv {\psi }_0-{\psi }_c=(n+1)K{\rho }_c^{1/n}>0$. ThereforeInsert $\psi ={\psi }_0-\Delta \psi \theta$ and $\rho ={\rho }_c{\theta }^n$ into the spherical Poisson equation for Newtonian gravity:The central conditions are $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$; the latter follows from a regular spherically symmetric centre. This reproduces the supplied coordinate and potential scaling rather than introducing an unrelated normalization.

For the sound-speed task, the adiabatic sound speed of a perfect polytropic gas isA diatomic gas with active rotational but frozen vibrational degrees of freedom has $\gamma =7/5$ and $n=5/2$. With the requested surface-potential gauge ${\psi }_0=0$,The interior potential is then negative. Choosing zero potential at the surface is legitimate; for a finite isolated sphere it is a different additive gauge from setting potential to zero at infinity. Both choices cannot generally be imposed simultaneously.

For the mass-radius task, let ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. A finite-radius polytrope hasThusAt fixed central temperature, fixed composition and fixed polytropic index, $M\propto R$. These are homologous polytropes at fixed central temperature. They do not in general have constant $K$; imposing constant entropy instead gives a different polytropic mass-radius sequence. If $n$ or $\mu$ varies between stars, the displayed dimensionless structure factor and composition factor must be retained: equal $T_c$ alone is not a universal exact proportionality for arbitrary stellar models.

For a plane-parallel slab, all equilibrium quantities depend only on height $z$. The vertical hydrostatic equilibrium equation and the Poisson equation for Newtonian gravity areWith the polytropic equation of state $p=K{\rho }^2$ and constant $K$,In the gas interior, where $\rho >0$, divide by $\rho$ and differentiate once:ConsequentlyThe coefficient $2\pi G/K$ has units inverse square metres. The index-one plane-parallel polytrope illustrates the resulting structure: symmetry about $z=0$ gives $\rho (z)={\rho }_c\cos (kz)$, with $k=\sqrt{2\pi G/K}$. A slab bounded by vacuum ends at its first zero, $|z|=\pi /(2k)$; the negative continuation of the cosine is not physical mass density.

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.

Use $c=1$ and a spatially flat background dominated by the stated perfect fluid, so $4\pi G{a}^2{\rho }_0=3{\mathcal{H}}^2/2$. In Newtonian gauge, the scalar energy and momentum constraints from the linearized Einstein field equations combine to eliminate the potential's time derivative and give the stated relativistic Poisson constraint. Its source contains the velocity correction to the coordinate density perturbation. On sub-horizon scales that correction is small for the ordinary growing modes, recovering the Poisson equation for Newtonian gravity; near or outside the horizon the gravitational field cannot be sourced by $\delta$ alone. Equality of the two metric potentials also assumes negligible scalar anisotropic stress.

The fluid equations are the linear energy-continuity and momentum-Euler parts of ${\nabla }_{\mu }{T}^{\mu \nu }=0$. Here $c_s^2=\delta p/\delta \rho$ is the relevant dimensionless sound-speed squared in the adopted fluid closure; for a constant-$w$ adiabatic barotropic fluid it equals $w$. A fluid with entropy perturbations needs additional pressure information. The equations actually use $\theta =\nabla \cdot \mathbf{v}$ in Fourier space, the peculiar-velocity divergence. Calling it simply momentum density is imprecise: it is the divergence of the velocity, with momentum weighted by ${\rho }_0+p_0$ before normalization. In particular $\theta$ must have inverse-conformal-time units for the displayed continuity equation to be dimensionally consistent.