# Why do you need weak spin-orbit interaction for $LS$ coupling?

From my lecture notes, the residual electrostatic perturbation causes the orbital angular momentum to couple and also spin $L=l_1+l_2 \ ,\ S=s_1+s_2$ (and it's not clear how residual electrostatic causes the coupling? ).

$L$ and $S$ are then constants of motion (because there's no external torque?). Why do you need weak spin-orbit interaction for the coupling?

From the wording of your question, there might be a confusion between LS "coupling" and spin-orbit coupling; the former is bad jargon (it's just a name for the representation scheme where you treat L and S as 'good' numbers for labelling the quantum states), while the latter is an actual coupling in the sense that you add a perturbation term ($\vec{S}\cdot\vec{L}$) to the Hamiltonian. When the electrons shake one another via the residual (non-central) electrostatic interaction, they may change each other's angular momentum. But since these interactions are internal, they cannot change the total angular momentum $\vec{J} = \vec{L} + \vec{S}$. The LS scheme goes one step further and assumes that the changes in L and S do not mix, and so $\vec{L}$ and $\vec{S}$ are each individually conserved.
But assuming that you understand this, you might be asking why does spin-orbit coupling have to be small in order for the LS representation scheme be valid. The reason is that spin-orbit coupling removes the premise that $\vec{L}$ and $\vec{S}$ are individually conserved. It's very name suggests that: "spin" and "orbital" angular momenta "coupling" (another word for mixing in the quantum world). So while J is still a good quantum number, L and S are not.