Is Electrostatics Local? We can solve uniquely for the electrostatic potential $\phi(x)$ of some given charge distribution if we set the boundary condition that $$\lim_{|x|\to\infty}\phi(x) = 0$$ (or whatever boundary condition you want). However, from my experience, the result of an electrostatics experiment doesn't seem to depend on what is happening at the edge of the universe. I find it disturbing that the local physics should depend on boundary conditions arbitrarily far away. Is there a more physical way of posing the problem in terms of purely local conditions?
 A: The edge of the universe doesn't affect a local experiment in the same way that the arbitrary constant you can change to $\phi(x)$ doesn't affect a local experiment. Boundary conditions really just constrain the constant in $\phi(x)$, the formula:
$\phi(x) = \int \frac{\rho(x)}{|x-x'|}d^3x'$
will always give you a correct answer even if you don't treat boundary conditions as anything other than a charge distribution.
A: You are asking about electrostatics first of all, so once you freeze time, there is no notion of causality. Without that notion you seem to be forgetting physics is about models. We accept the Poisson equation based on certain theoretical principles and observations and is successful. If you admit that your field is governed by a differential equation, we know we need to impose boundary conditions in order to choose a single solution out of a family of solutions. The boundary conditions come from the specifics of your problem, does that sound non-local?? No.
Locality only has to do with dependence on neighborhoods as it is the case for any differential equation, they only have derivatives at a single point which in turn only know about what happens in a vicinity of such a point. So the ''dynamics'' if you will are not changing, the boundary condition is not modifying the relation between neighboring points at all. On the contrary the job of the differential equation can be seen as propagating the information of the boundary in a local way, gradually, from the boundary to any point.
P.S. Under your argument, any differential equation is non local, which is just wrong, the definition of locality is not that.
A: Please forgive me for posting an answer to my own question.
But I think physically, the problem we are solving is the solution to the inhomogeneous wave equation:
$$\square \phi(x,t) = \rho(x,t)$$
with Cauchy data $\phi(x,0)$ and $\dot\phi(x,0)$. I think one should be able to prove that if $\rho(x,t)\to \rho_0(x)$, i.e. constant in time, and in the large volume limit, that the function $\phi(x,t)\to \phi_0(x)$, where $\phi_0(x)$ is the solution to the Poisson equation with Dirichlet or Neumann b.c. at infinity. You probably also have to assume that $\rho_0(x)$ decays fast enough for large $x$.
In short, what I'm trying to say is that the boundary conditions are not a cause of the solution, they are just a consequence of the solution to the wave equation in large volume.
