Virial Analysis

Hydrodynamic equation of motion using the Lagrangean time derivative [eq.(A.3)] is

\rho\left(\frac{d \mbox{\boldmath${v}$}}{d t}\right)=-\mbox{\boldmath${\nabla}$} p -\rho \mbox{\boldmath${\nabla}$} \Phi.
\end{displaymath} (4.17)

Multiplying the position vector r and integrate over a volume of a cloud, we obtain the Virial relation as
\frac{1}{2}\frac{d^2 I}{d t^2}=2(T-T_0) + W,
\end{displaymath} (4.18)

I=\int \rho r^2 dV,
\end{displaymath} (4.19)

is an inertia of the cloud,
T=\int\left(\frac{3}{2}p_{\rm th}+\frac{1}{2}\rho v^2\right)dV=\frac{3}{2}\bar{p}{V_{\rm cl}},
\end{displaymath} (4.20)

is a term corresponding to the thermal pressure plus turbulent pressure,
T_0=\int_S P_{\rm th}{\bf r}\cdot {\bf n} dS=\frac{3}{2}P_0 V_{\rm cl}
\end{displaymath} (4.21)

comes from a surface pressure, and
W=-\int \rho {\bf r}\cdot \nabla \Phi dV=-\frac{3}{5}\frac{GM^2}{R}
\end{displaymath} (4.22)

is a gravitational energy. To derive the last expression in each equation, we have assumed the cloud is spherical and uniform. Here we use a standard notation as the radius $R$, the volume $V_{\rm cl}=4\pi R^3/3$, the average pressure $\bar{P}$, and the mass $M$.

To obtain a condition of mechanical equilibrium, we assume $d^2 I/dt^2=0$. Equation (4.18) becomes

4\pi \bar{p}R^3 - 4\pi p_0 R^3-\frac{3}{5}\frac{GM^2}{R}=0.
\end{displaymath} (4.23)

Assuming the gas is isothermal $p=c_{\rm is}^2 \rho$, the average pressure is written as
\bar{p}=c_{\rm is}^2 \bar{\rho}=c_{\rm is}^2\frac{3M}{4\pi R^3}.
\end{displaymath} (4.24)

Using equation (4.24) to eliminate $\bar{p}$ from equation (4.18), the external pressure is related to the mass and the radius as
p_0=\frac{3c_{\rm is}^2 M}{4\pi R^3}-\frac{3GM^2}{20\pi R^4}.
\end{displaymath} (4.25)

Keeping $M$ constant and increasing $R$ from zero, $p_0$ increases first, but it takes a maximum, $p_{0,{\rm max}}=3.15 c_{\rm is}^8/(G^3M^2)$, and finally declines. This indicates that the surface pressure must be smaller than $p_{0,{\rm max}}$ for a cloud to be in the equilibrium. In other words, keeping $p_0$ and changing $R$, it is shown that $M$ has a maximum value to have a solution. The maximum mass is equal to
M_{\rm max}=1.77 \frac{c_{\rm is}^4}{G^{3/2}p_0^{1/2}}.
\end{displaymath} (4.26)

The cloud massive than $M_{\rm max}$ cannot be supported against the self-gravity. This corresponds to the Bonnor-Ebert mass [eq.(4.9)], although the numerical factors are slightly different.

Kohji Tomisaka 2012-10-03