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# VbvAstE-001

Book Boris V. Vasiliev Astrophysics

## In the absence of the

In the absence of the gravitation, the equilibrium state of ideal gas in some volume comes with the pressure equalization, i.e. with the equalization of its temperature T and its density n. This equilibrium state is characterized by the equalization of the chemical potential of the gas µ (Eq.(2.8)). 4.4 The radial dependence of density and temperature of substance inside a star atmosphere For the equilibrium system, where different parts have different temperatures, the following relation of the chemical potential of particles to its temperature holds ([12],25): µ kT = const (4.9) As thermodynamic (statistical) part of chemical potential of monoatomic ideal gas is [12],45: [ ( ) n 2π 2 3/2 ] µ T = kT ln , (4.10) 2 mkT we can conclude that at the equilibrium n ∼ T 3/2 . (4.11) In external fields the chemical potential of a gas [12]25 is equal to µ = µ T + E potential (4.12) where E potential is the potential energy of particles in the external field. Therefore in addition to fulfillment of condition Eq. (4.11), in a field with Coulomb potential, the equilibrium needs a fulfillment of the condition − GMrγ rkT r + P2 r 2kT r = const (4.13) (where m is the particle mass, M r is the mass of a star inside a sphere with radius r, P r and T r are the polarization and the temperature on its surface. As on the core surface, the left part of Eq.(4.13) vanishes, in the atmosphere M r ∼ rkT r. (4.14) Supposing that a decreasing of temperature inside the atmosphere is a power function with the exponent x, its value on a radius r can be written as ( ) x R⋆ T r = T ⋆ (4.15) r 27

and in accordance with Eq.(4.11), the density ( ) 3x/2 R⋆ n r = n ⋆ . (4.16) r Setting the powers of r in the right and the left parts of the condition Eq.(4.14) equal, one can obtain x = 4. Thus, at using power dependencies for the description of radial dependencies of density and temperature, we obtain ( ) 6 R⋆ n r = n ⋆ (4.17) r and ( ) 4 R⋆ T r = T ⋆ . (4.18) r 4.5 The mass of the star atmosphere and the full mass of a star After integration of Eq.(4.17), we can obtain the mass of the star atmosphere ∫ R0 ( ) [ 6 R⋆ M A = 4π (A/Z)m pn ⋆ r 2 dr = 4π ( ) ] 3 R⋆ r 3 (A/Z)mpn⋆R⋆3 1 − R 0 R ⋆ (4.19) It is equal to its core mass (to R3 ⋆ ≈ 10 −3 ), where R R 3 0 is radius of a star. 0 Thus, the full mass of a star M = M A + M ⋆ ≈ 2M ⋆ (4.20) 28

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