 Solution

Follow a fluid element with position $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives the material derivativeApply Newton's second law to a material element of volume $dV$ and mass $\rho dV$. The net pressure force is $-\nabla pdV$, and gravity gives $-\rho \nabla \varphi dV$. Dividing by the element mass, neglecting viscosity and using the chain rule componentwise gives the Euler equation

For a static spherical isothermal gas, $p=c_s^2\rho$ and $dp/dr=-GM(r)\rho /r^2$. Trial density $\rho =A/r^2$ gives enclosed mass $M(r)=4\pi Ar$, soThis is the singular isothermal sphere. Its central density diverges, although the enclosed mass tends to zero linearly and its total mass inside a finite radius is finite. It is not the finite-central-density regular isothermal solution.

Truncate at $r_0$ with boundary pressure $p_0$. Then $M=4\pi A{r}_0=2c_s^2{r}_0/G$ and $p_0=c_s^2\rho (r_0)$, givingThe external medium is being treated as a prescribed pressure boundary; its own gravitational field is excluded from the cloud self-energy calculation.

For a monatomic perfect gas, the equilibrium is isothermal but the internal-energy relation still uses its adiabatic index $\gamma =5/3$:The gravitational self-energy follows by assembling shells:Both integrals remain finite at the singular center. ConsequentlyThe pressure-confined singular isothermal gas sphere therefore satisfies the virial theorem including the work of the confining surface pressure. One can see its origin directly by integrating $\mathbf{r}\cdot \nabla p$: integration by parts gives $\oint p\mathbf{r}\cdot d\mathbf{S}-3\int pdV$, whose surface term is $4\pi r_0^3{p}_0$. An isolated vacuum-boundary version would omit that term and could not truncate this same positive-temperature isothermal profile at finite radius. The equality expresses equilibrium force and energy balance, rather than proving dynamical stability.