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Use the isothermal Lane-Emden equation variables $r=a\xi$, $a^2=c_T^2/(4\pi G{\rho }_c)$ and $\rho ={\rho }_c{e}^{-\Psi }$. Truncation at fixed dimensionless radius ${\xi }_b$ gives $R=a{\xi }_b$ and $M=4\pi {\rho }_c{a}^3{\xi }_b^2{\Psi }^{\prime }({\xi }_b)$. Division proves the displayed relation. A common temperature, composition and dimensionless truncation therefore imply $M\propto R$, while the external pressure generally varies along that homologous sequence. Arbitrary changes of truncation at fixed temperature do not retain a common coefficient. A finite regular cloud needs an outer boundary such as that of a pressure-confined isothermal sphere.

In a spherical self-gravitating equilibrium with positive mass density, $d\psi /dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho (r)$ can be expressed as functions of $\psi$. Hydrostatic equilibrium further givesSince $p$ decreases outward wherever $\rho >0$, eliminating $\psi$ gives a barotropic equation of state along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.

For $p=K{\rho }^{1+1/n}$, integrate the polytropic enthalpy balance $dp/\rho =-d\psi$:For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho (R)=0$ and therefore ${\psi }_T=\psi (R)$. With the gravitational potential zero at infinity, an isolated sphere has ${\psi }_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the gravitational potential.

Put $\theta =(\rho /{\rho }_c{)}^{1/n}$ and $\xi =\alpha r$, with ${\alpha }^2=4\pi G{\rho }_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi ={\psi }_T-(n+1)K{\rho }_c^{1/n}\theta$. Substitution into the spherical gravitational Poisson equation yields the Lane-Emden equationLet ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)>0$. Then the Lane-Emden mass formula and radius areFor a common fixed polytropic index with a finite first zero, their scalings areThe usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.

For an ideal gas of fixed composition, $T_c\propto K{\rho }_c^{1/n}$. Consequently homologous polytropes at fixed central temperature have $K\propto {\rho }_c^{-1/n}$ and both $M,R\propto {\rho }_c^{-1/2}$, recovering $M\propto R$. The common central temperature, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different polytropic mass-radius relation, and no finite index makes its generic fixed-$K$ exponent exactly one.

For adiabatic monatomic hydrogen, $\gamma =5/3$ and $n=3/2$. At a common fixed specific entropy, hence common $K$, the same relations giveHere $\bar{\rho }=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that mass sequence.

Efficient radiative cooling in molecular clouds prevents compression from forcing the temperature upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and temperature. It is therefore possible to assemble dense massive cold clouds while disposing of compressional heat. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.

For a common isothermal sound speed $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the homologous isothermal mass-radius relationFor example, the singular isothermal sphere has $M(R)=2c_T^{2}R/G$. A regular finite cloud requires a pressure-confined isothermal sphere or another outer boundary: an unconfined isothermal sphere does not end at a finite vacuum radius. Equal temperature alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.

Write the initial mass density as $\rho =mn$. The ideal gas law gives pressure $p=n{k}_{B}T$, and the sphere has $M=4\pi \rho R^3/3$. Here $k_B$ is the Boltzmann constant and $G$ is the gravitational constant. First obtain the gravitational potential energy by adding concentric shells. The mass inside radius $r$ is $M(r)=M{r}^3/R^3$, soIn a small contraction $dR<0$ at fixed total mass, the gravitational energy released is $-dW=3G{M}^2|dR|/(5R^2)$. The positive work required against the gas pressure is $-pdV=4\pi R^{2}p|dR|$. Equating these givesUse $R=(3M/(4\pi mn){)}^{1/3}$ to eliminate the radius. The uniform-sphere pressure-work collapse criterion isThe units are kilograms: $(k_{B}T/(Gm){)}^{3/2}$ has units $\mathrm{k}{\mathrm{g}}^{3/2}{\mathrm{m}}^{-3/2}$, and ${\rho }^{-1/2}$ supplies $\mathrm{k}{\mathrm{g}}^{-1/2}{\mathrm{m}}^{3/2}$. At fixed initial $n$ and $T$, $M>M_{\mathrm{crit}}$ makes gravity stronger than the pressure-work requirement in this model, allowing contraction.

This is a Jeans mass estimate for homologous contraction of an initially uniform sphere. It is not an exact stability limit for a hydrostatic gas sphere: uniform mass density and uniform temperature would give uniform pressure, rather than the gradient required by hydrostatic equilibrium. The numerical coefficient therefore differs from conventions based on a Jeans instability wavelength or a pressure-confined isothermal sphere. Subsequent contraction also depends on heat exchange; isothermal pressure support during gravitational collapse weakens relative to gravity as the radius decreases.

Let $M(r)$ be the enclosed mass. For spherical symmetry, hydrostatic equilibrium and mass conservation giveThe isothermal equation of state $p=K\rho$ therefore impliesMultiply by $r^2$, differentiate and eliminate $M$:or equivalently $r^{-2}d[r^2{\rho }^{-1}{\rho }^{\prime }]/dr=-4\pi G\rho /K$. This is the density equation for an isothermal sphere. Here $K$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$ and equals the squared isothermal sound speed.

A regular isothermal sphere does not reach zero density at a finite radius. The finite body discussed later must therefore be truncated by a nonzero external pressure, as in a pressure-confined isothermal sphere. A finite vacuum boundary would not be compatible with this equation of state.

Using the supplied first critical value ${\xi }_c\simeq 6.5$, the critical radius of a pressure-confined isothermal sphere isNo radius in metres can be specified without $K$ and a central mass density. Here $c_s$ is the isothermal sound speed, consistent with the fixed-temperature variations in part (d). At a given central mass density, equilibria truncated at $\xi <{\xi }_c$ lie on the first stable branch, while increasing the dimensionless outer radius beyond this first threshold gives an unstable configuration. Numerical integration gives ${\xi }_c\simeq 6.451$, consistent with the paper's rounded $6.5$.

For a transparent comparison with the given Jeans length, set the uniform comparison density ${\rho }_0={\rho }_c$ and use the same speed of sound. ThenBoth scales express the competition between pressure support and self-gravity, with characteristic length $c_s/\sqrt{G\rho }$. Their numerical closeness under this density choice does not mean that the models impose the same threshold: $r_c$ is a boundary radius, whereas ${\lambda }_J$ is the wavelength of a perturbation. Comparing $r_c$ with ${\lambda }_J/2$ would instead give a factor about $2.07$. Using the sphere's mean or boundary density instead of its central density changes the coefficient again, so the reference density must always be stated.

The pressure-confined isothermal sphere gives an actual finite, centrally concentrated hydrostatic equilibrium and an explicit confining pressure, making its radial stability threshold more self-consistent for a roughly spherical isothermal cloud. The uniform Jeans instability model gives a simpler local criterion, includes perturbations at different wavelengths, and does not require constructing a spherical equilibrium. Its ideal uniform static infinite background requires the Jeans swindle or an explicitly specified treatment of background gravity. Neither model includes rotation, magnetic fields, turbulent support or a realistic temperature evolution, so neither is a universal collapse criterion.

A pressure-confined isothermal sphere is a regular isothermal sphere truncated at a boundary where its internal pressure equals an imposed external pressure. The finite central density and the boundary density generally differ. Its radial stability depends on the dimensionless truncation radius as well as the fixed mass and temperature. The first fixed-mass turning point of an isothermal sphere occurs at dimensionless radius about $6.451$; the modestly concentrated branch is stable to the radial isothermal perturbations in this model.

For a uniform ideal gas sphere, set $\rho =mn$, $p=n{k}_{B}T$ and $M=4\pi \rho R^3/3$. Building the sphere by concentric shells gives gravitational potential energy $W=-3G{M}^2/(5R)$. During a small homologous contraction, released gravitational energy balances the work against gas pressure when $3G{M}^2|dR|/(5R^2)=4\pi R^{2}p|dR|$, or $GMm/R=5k_{B}T$. Eliminating $R$ gives the displayed mass. At the same initial mass density and temperature, larger masses have stronger gravitational contraction than pressure support in this estimate. Its numerical coefficient belongs to the uniform-sphere pressure-work model; the infinite-medium Jeans instability and a pressure-confined isothermal sphere use different equilibrium and boundary assumptions.