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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.

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.