# Proof of Thomson's theorem on electrostatics using variational calculus

I'm following a proof of Thomson's theorem but I'm a bit confused when they use a lagrange multiplier to include the charge conservation constraint, could someone explain to me how is this multiplier deducted? I would appreciate it so much. By the way, I took the image from the paper "A variational proof of Thomson's theorem," by Miguel C.N. Fiolhais a,b,c,∗, Hanno Essén d, Tomé M. Gouveia e https://doi.org/10.1016/j.physleta.2016.06.039

• Could you elaborate on what exactly in the paper's derivation is unclear? – Qmechanic Feb 26 '18 at 12:58

Are you familiar with using Lagrange multiplier to find extrema of a system unter constraints? If not, check Wikipedia. In short in 2d, one constraint case, $f(x,y)$ has extrema when $\nabla \mathcal{L(x,y)}=0$, in which $\mathcal{L} = f(x,y)-\lambda \cdot g(x,y)$ with constraint $g(x,y)=0$.
In this case $g = Q^{(n)}-\int\rho dV^{(n)}_{int}-\int\sigma dS^{(n)}$. This comes from charge conservation. The total charge of a conductor Q is equal to all the charges inside the conductor and on the surface, i.e. $Q^{(n)}=\int\rho dV^{(n)}_{int}+\int\sigma dS^{(n)}$. Later in this article the same methode is deployed with constraints as results of Poisson's equation.