Proof of Thomson's theorem on electrostatics using variational calculus

Physics Asked by Juan Pablo Arcila on February 22, 2021

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

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 $g = Q^{(n)}-intrho dV^{(n)}_{int}-intsigma dS^{(n)}$. 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. $Q^{(n)}=intrho dV^{(n)}_{int}+intsigma dS^{(n)}$. Later in this article the same methode is deployed with constraints as results of Poisson's equation.

Answered by Chen on February 22, 2021