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# VbvAstE-001

Book Boris V. Vasiliev Astrophysics

## infinity. 1 With this in

infinity. 1 With this in mind, we can write the energy balance equation of electron E kin = eϕ(r). (10.5) The potential energy of an electron at its moving in an electric field of the nucleus can be evaluated on the basis of the Lorentz transformation [14]§24. If in the laboratory frame of reference, where an electric charge placed, it creates an electric potential ϕ 0, the potential in the frame of reference moving with velocity v is ϕ = ϕ 0 √ 1 − v2 c 2 . (10.6) Therefore, the potential energy of the electron in the field of the nucleus can be written as: E pot = − Ze2 ξ r β . (10.7) Where β = v c . (10.8) and ξ ≡ p m , ec (10.9) m e is the mass of electron in the rest. And one can rewrite the energy balance Eq.(10.5) as follows: where Y = 3 8 mec2 ξY = eϕ(r) ξ β . (10.10) [ ξ(2ξ 2 + 1) √ ] ξ 2 + 1 − Arcsinh(ξ) − 8 3 ξ3 . (10.11) ξ 4 Hence ϕ(r) = 3 m ec 2 βY. (10.12) 8 e In according with Poisson’s electrostatic equation ∆ϕ(r) = 4πen e (10.13) or at taking into account that the electron density is depending on momentum (Eq.(10.2)), we obtain ∆ϕ(r) = 4e ( ) 3 ξ , (10.14) 3π ˜λ C 1 In general, if there is an uncompensated electric charge inside the cell, then we would have to include it to the potential ϕ(r). However, we can do not it, because will consider only electro-neutral cell, in which the charge of the nucleus exactly offset by the electronic charge, so the electric field on the cell border is equal to zero. 77

where ˜ λ C = m ec is the Compton radius. At introducing of the new variable we can transform the Laplacian: ϕ(r) = χ(r) r , (10.15) ∆ϕ(r) = 1 r d 2 χ(r) dr 2 . (10.16) As (Eq.(10.12)) χ(r) = 3 m ec 2 Yβr , (10.17) 8 e the differential equation can be rewritten: where L = d 2 χ(r) dr 2 = χ(r) L 2 , (10.18) ( ) 1/2 9π Yβ ˜ λ 32 αξ 3 C , (10.19) α = 1 is the fine structure constant. 137 With taking in to account the boundary condition, this differential equation has the solution: χ(r) = C · exp ( − r L ) . (10.20) Thus, the equation of equilibrium of the electron gas inside a cell (Eq.(10.10)) obtains the form: Ze r · e−r/L = 3 8 mec2 βY . (10.21) 10.2 The Thomas-Fermi screening Let us consider the case when an ion is placed at the center of a cell, the external shells don’t permit the plasma electron to approach to the nucleus on the distances much smaller than the Bohr radius. The electron moving is non-relativistic in this case. At that ξ → 0, the kinetic energy of the electron E kin = 3 8 mec2 ξY → 3 EF , (10.22) 5 and the screening length √ EF L → . (10.23) 6πe 2 n e Thus, we get the Thomas-Fermi screening in the case of the non-relativistic motion of an electron. 78

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