Direct Energy, 2018a

304 13.3 Derivation of the Lagrangian
ρ ch =
[ (−5mq
3 2
) 3/2 ( −q
3π 2 ) ] · V 3/2 (13.60)
Finally, we can write E kinetic e
as a function of V .
V
[ (−5mq ) 3/2 ( ) ] E kinetic e
−q
=
V 5/2 (13.61)
V 3 2 3π 2
Notice that the quantity in brackets above is constant. The coecient c 0
is dened from the term in brackets.
( −5mq
c 0 =
3 2
) 3/2 ( −q
3π 2 )
(13.62)
E kinetic e
= c 0 V 5/2 (13.63)
V
We now can describe all of the terms of the Lagrangian in terms of our
generalized path.
E Coulomb e nucl
V
+ E e e interact
= 1 ∣ ∣∣
V 2 ɛ −→
∣ ∣∣
2
∇V
E kinetic e
(13.64)
= c 0 V 5/2 (13.65)
V
The Hamiltonian represents the total energy density, and the Lagrangian
represents the energy density dierence of these forms of energy. The
Hamiltonian and Lagrangian have the form H = H ( )
r, V, dV
dr and L =
L ( )
r, V, dV
dr where r is position in spherical coordinates. There is no θ or φ
dependence of H or L. Everything is spherically symmetric.
H = 1 2 ɛ|−→ ∇V | 2 + c 0 V 5/2 (13.66)
L = 1 2 ɛ|−→ ∇V | 2 − c 0 V 5/2 (13.67)
As an aside, let us consider the Fermi energy E f = μ chem once again.
With some algebra, we can write it as a function of voltage. Use Eqs.
13.37, 13.52, and 13.62.
E f = 2 kf
2 ( )
2m = 2 −3π 2 2/3
ρ ch
(13.68)
2m q
⎡
( ) (
E f = 2 −3π
2 2/3 ( ) ) ⎤
⎣
−5mq −3π
2 −2/3 3/2
2/3
·
V 3/2 ⎦ (13.69)
2m q
3 2 q