How does electrostatic repulsion between electrons in "many electron atom" lead to coupling of individual orbital angular momentum vectors? I just started studying $LS$ coupling scheme, book has described $LS$ coupling in following order
 1. Firstly it mentions due to "spin-spin" correlation individual spin angular momentum vectors couples to from resultant spin angular momentum vector. i.e $\vec{S}$. 
And quantum number $S$ takes values from $|\vec{s_1}+\vec{s_2}+\vec{s_3}.....|_{min}$  to $(\vec{s_1}+\vec{s_2}+\vec{s_3}......)$
 2. Then it says As a result of Residual Electrostatic Interaction individual orbital angular momentum vectors of "optical" electrons are strongly coupled with one another to form a resultant orbital angular momentum vector $\vec{L}$ of magnitude $\sqrt{L(L+1)} \hbar$ which is constant of motion.

My question arises here is how does residual electrostatics interaction which is repulsive electric potential between electrons in an atom, leads to the coupling of individual orbital angular momentum vectors ?
Reference :- Page 144 of the PDF or 140 of the book.
 A: Thanks to @lineage 's comment about seeing section 10.3 of book Quantum Physics of Atoms,Molecules Solids. for insights.
The Coulomb interaction doesn't result in coupling of $\vec{l_1},\vec{l_2}....$ to form $\vec{L}$.
Instead it makes the coupling to happen in such a way that the $\vec{L}$ remains constant.
This happens simply because in most quantum states the charge distributions
of the electrons are not spherically symmetrical, and so they exert torques on each
other. Since the space orientation of the charge distribution of an electron is related
to the space orientation of its orbital angular momentum vector, there are torques
acting between the angular momentum vectors. The torques do not tend to change
the magnitude of the individual orbital angular momentum vectors, but only tend to
make them precess about the total orbital angular momentum vector in such a way
that its magnitude L' remains constant.
The question then arises: Which of the possible values of L' corresponds to the
state of lowest energy?
There are opposing tendencies, but the basis of the one which
usually dominates can be understood even from classical physics by considering two
electrons in a Bohr atom.
Two optically active electrons mov-
ing in the same Bohr orbit tend to remain at op-
posite ends of a diameter so as to minimize their
Coulomb repułsion. As a result, their orbital angu-
lar momenta tend to couple in such a way as to
yield a maximum total orbital angular momentum.
repulsion between the electrons, the most stable arrangement is obtained when the
electrons stay at the opposite ends of a diameter. In this state of lowest energy, the
electrons rotate together with individual orbital angular momentum vcctors parallel,
and therefore with the magnitude L' of the total angular momentum vector a max-
imum. This conclusion is confirmed by an analysis of the spectra produced by atoms
with several optically active electrons. That is, for such atoms the residual Coulomb
interaction produces a tendency for the orbital angular momenta of the optically active
electrons to couple in such a way that the magnitude of the total orbital angular momen-
tum L' is constant, and the energy is usually lowest for the state in which L' is largest.
