Virial Analysis

For a system to achive a mechanical equlibrium, a relation must be satisfied between energies such as potential, thermal and kinetic energies. This is called the Virial relation. For example, a harmonic oscillator has a potetial energy of and a kinetic energy . Averaging these two energies over one period, both energies give the same absolute value proportional to the oscillation amplitude squared as and . Another example is a Kepler problem. For simplicity, consider mass running on a circular orbit with a radius from a body with a mass . The gravitational and kinetic energies are equal to and , where we used the centrifugal balance as . As a result, for the harmonic oscillator while for the circular Kepler problem . This ratio is known to be related to the power of the potetial as . Important nature of the self-gravity is understood only with this relation without solving the hydrostatic balance equations. In the following, we describe the Virial relation satisfied with isolated systems such as stars.

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

(2.104) |

Multiplying radius to the equation and integrating by the volume over a volume from to , we obtain the Virial relation as

where

is an inertia of this body, and and are, respectively, the kinetic and thermal energies as

Further,

is a gravitational energy, where and

Since

(2.111) |

(2.112) |

using equations (2.107) and (2.108).

On the other hand, the first term of the rhs of equation (2.105)
becomes

(2.113) |

To derive this equation, we have assumed the pressure diminishes at a radius and the surface pressure term does not appear in the final expression. This is valid for an isolated system such as a star.

The last term of the rhs of equation (2.105) is written as

(2.114) |

(2.115) |

The energy per unit mass is necessary for a gas element to move from the radius , inside which mass is contained, to the infinity. Adding the energy for all the gas, the potential energy is obtained. In the case of a star composed of uniform density ,

(2.116) |

To obtain a condition for the mechanical equilibrium, we assume .
Equation (2.106) becomes

In the case of this reduces to . The total energy is expressed as

For the gas with , equation (2.119) gives a negative total energy and the system is in a confined state. However, if , the gravity can not confine the gas.

For
, equation (2.118) gives