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At fixed temperature, the particle count of an isothermal sphere is proportional to ${\rho }_c^{-1/2}f(\xi )$, where $f={\xi }^2{\Psi }^{\prime }$ and $f^{\prime }={\xi }^2{e}^{-\Psi }$. A fixed-mass variation therefore satisfies $\delta {\rho }_c/{\rho }_c=2e^{-\Psi }\delta \xi /{\Psi }^{\prime }$. The boundary density changes by $\delta \rho /\rho =\delta {\rho }_c/{\rho }_c-{\Psi }^{\prime }\delta \xi$. Its first stationary point occurs when the title condition holds. This follows a sequence of equilibria, rather than differentiating the radial profile of one equilibrium. The associated boundary radius obeys $\delta r/r=(1/\xi -e^{-\Psi }/{\Psi }^{\prime })\delta \xi$.

For an isothermal sphere, put $\rho ={\rho }_c{e}^{-\Psi }$ and $r=a\xi$ with $a^2=K/(4\pi G{\rho }_c)$. The equilibrium equation becomes ${\xi }^{-2}({\xi }^2{\Psi }^{\prime }{)}^{\prime }=e^{-\Psi }$. Finite central density and smooth spherical symmetry require $\Psi (0)={\Psi }^{\prime }(0)=0$. The regular expansion is $\Psi ={\xi }^2/6-{\xi }^4/120+{\xi }^6/1890+O({\xi }^8)$. The enclosed mass is $M=4\pi {\rho }_c{a}^3{\xi }^2{\Psi }^{\prime }$, following by integrating the differential equation.

An isothermal sphere is a spherically symmetric self-gravitating gas equilibrium with equation of state $p=K\rho$ and constant $K$. The hydrostatic equilibrium equations $p^{\prime }=-GM\rho /r^2$ and $M^{\prime }=4\pi r^2\rho$ yield the title equation. Its regular finite-central-density solution is distinct from the singular isothermal sphere. A finite outer radius requires confinement because the regular density does not vanish at finite radius.

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.

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.

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.

An isothermal sphere with constant positive isothermal sound speed $c$ obeys the title hydrostatic equilibrium equation. Between two positive finite radii with bounded enclosed mass, its right-hand side has a finite integral. A positive starting density therefore remains positive at any proposed finite outer radius. It cannot meet a vacuum boundary with $p=c^2\rho =0$ there without an additional confining structure or a change of equation of state. The singular isothermal sphere makes the obstruction explicit: $\rho =c^2/(2\pi G{r}^2)$ and $p=c^4/(2\pi G{r}^2)$ never vanish at finite radius, and its mass grows as $2c^{2}r/G$ without bound at infinity.