VbvAstE-001

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.
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