Direct Energy, 2018a

298 13.3 Derivation of the Lagrangian
If we have a large atom with many electrons around it, the Coulomb
interaction between any one electron and the nucleus is shielded by the
Coulomb interaction from all other electrons. More specically, suppose
we have an isolated atom with N protons in the nucleus and N electrons
around it. If we pick one of the electrons, E Coulomb e nucl for that electron
describes the energy stored in the electric eld due to the charge separation
between the nucleus of positive charge Nq and that electron. However,
there are also N − 1 other electrons which have a negative charge. The
term E e e interact describes the energy stored in the electric eld due to
the charge separation between the N − 1 other electrons and the electron
under consideration. These terms somewhat cancel each other out because
the electron under consideration interacts with N protons each of positive
charge q and N − 1 electrons each of negative charge −q. However, the
terms do not go away completely. Calculating
E Coulomb e nucl + E e e interact (13.20)
is complicated because the electrons are in motion, and we do not really
know where they are or even where they are most likely to be found. In
fact, we are trying to solve for where they are likely to be found.
As we move the electron under consideration in and out radially, energy
is transferred between (E Coulomb e nucl + E e e interact ) and E kinetic e . The
Hamiltonian is the sum of these two forms of energy per unit volume,
and the Lagrangian is the dierence of these two forms of energy per unit
volume. Both quantities have the units J m
3 . Choose voltage V (r) as the
generalized path and charge density ρ ch (r) as the generalized potential.
The independent variable of these quantities is radial position r, not time.
We can now write the Hamiltonian and Lagrangian.
(
H r, V, dV ) (
ECoulomb e nucl
=
+ E )
e e interact
+ E kinetic e
(13.21)
dr
V
V
V
(
L r, V, dV ) (
ECoulomb e nucl
=
+ E )
e e interact
− E kinetic e
(13.22)
dr
V
V
V
The next step is to write
E Coulomb e nucl
V
+ E e e interact
V
(13.23)
in terms of the path V . As detailed in Table 12.3, the energy density due
to an electric eld −→ E is given by
E
V = 1 2 ɛ|−→ E | 2 . (13.24)