Is Electrostatics Local?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

The electrostatic potential is always a little funny because of the gauge freedom to add any constant. It is somewhat "nonlocal" in the sense that the equation $\nabla^2 \phi = -\rho$ must be supplemented by a boundary condition at infinity.

The simple answer to your question is that in principle, we can always bypass the potential entirely and directly use Columb's law ${\bf E}({\bf r}) = \int d^3 {\bf r}' \frac{\rho({\bf r}')}{|{\bf r} - {\bf r}'|^3} ({\bf r} - {\bf r}')$ to calculate the electric field, and this equation is manifestly local in the sense that charges farther away contribute less. Only the electric field has physical reality (philosophical claim alert!), so electrostatics is indeed local.

The more subtle answer is that Coulomb's integral only converges if the charge distribution decays sufficiently quickly at spatial infinity. So in this sense, we do need "manually" add in some sense of locality as an additional postulate. If the charge distribution remains nonnegligible out at spatial infinity, then the electric field becomes ill-defined, as discussed at Gauss's law in a uniform charge distribution extending infinitely in all directions.

Answer 5

Is Electrostatics Local?

No, electrostatics is a non-local theory. This is clear from, say, the interaction term: $$ V_{non\;local} = \int d^3 r \rho(\vec r)\phi(\vec r) = \int d^3r d^3r' \rho(\vec r)\rho(\vec r')\frac{1}{|\vec r - \vec r'|}\;, $$ which is clearly "non-local" as the term is usually used, since there is a kernel that depends on two different points in space and is integrated over all space for both of them.

A "local" interaction term would have all the components evaluated at the same position like: $$ V_{local}=\int d^3r \psi^\dagger(\vec r) A(\vec r) \psi(\vec r) $$

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Of course, it turns out that electrostatics is not a complete theory and that quantum electrodynamics actually is local--the electric field or potential actually does not get modified immediately everywhere in space, but rather those changes propagate at the speed of light via a local mechanism.

But the equations of electrostatics, e.g., $$ -\nabla^2\phi = \rho\;, $$ are non-local.