Question on boundary condition for Maxwell's Equations and Coulomb's law

Question

When deriving Coulomb's law using the differential forms of Maxwell's equation, the boundary condition that $\phi = 0 $ at infinity is also used.

From $\nabla × E = 0, E = \nabla \phi$ for some $V$, plugging this in to Gauss's law we get poisson's equation $\nabla^2 \phi = \rho/e_0$, or $\nabla^2 \phi = q\delta\lvert r-r_0\rvert/e_0$ for a point charge. The general solution is $\phi = \frac{1}{4\pi \epsilon_0} \frac{q}{\lvert r-r_0\rvert} + F$ where $F$ is a harmonic function which satisfies laplace's equation. We have to invoke the condition that at infinity $\phi = 0$ for $F(x)$ to disappear and thus for Coulomb's law to follow.

Maxwell's Equations and the Lorentz force law summarize all of electrodynamics, and that boundary conditions come from the constraints in the problem of consideration. The situation in consideration is a point charge in empty space, so what physical principle motivates this boundary condition, that $\phi = 0$?

Answer 1

Maxwell's equations and Lorentz force law do indeed summarize all electrodynamics. However there are physical situations in which you do not know the charge distribution a priori, but instead you specify some surfaces on which the electric field is always normal, which happens when you have metals around.

In such situations you can in principle solve the three partial differential equations for $E_x, E_y$ and $E_z$ in three variables $x, y, $ and $z$, with the specified Dirichlet boundary conditions. All that introducing the potential did to you was convert this horrible system of PDEs with Dirichlet BCs to a single PDE with Neumann BCs. So it's just a mathematical trick. All physical meaning is still in $E$ alone (classically).

Now once you agreed to introduce $\phi$ to make your life easier, you have to impose the same boundary conditions that you would have otherwise imposed on $E$ itself. For a point particle without introducing $\phi$, to solve the PDEs you had to impose $E\rightarrow 0$ as $r\rightarrow \infty$. This implies $\phi\rightarrow $ constant as $r\rightarrow\infty$. But then we can choose this constant at our disposal because it is physically meaningless (only difference in potential is meaningful).

Answer 2

Causality requires any physical effect outside the light cone of a charge to vanish. So whether or not you believe that you can add an arbitrary constant, or a gauge field, $\phi=0$ at infinity makes sense.