3
$\begingroup$

My qualitative understanding is that the mathematical form of the interaction between particles is constrained by their gauge symmetry, so that, for example, the U(1) gauge symmetry in QED gives rise to the $\frac{1}{r}$ coulomb potential.

My question is:

If we lived in a "topologically nontrivial" space, e.g. a ring or a sphere or a three-sphere or mobius strip or whatever, what form would the coulomb potential take?

On a ring, for example, how could one even have field lines emitting from a point source, since the line leaving the charge to the charge's left would continue around the ring and run counter the line that leaves to its right?

$\endgroup$
0
3
$\begingroup$

An exactly solvable example

As a warm-up, consider Maxwell's equations in one-dimensional space. In that case, there is no magnetic field, and we have only one equation (with two components) governing the electric field: $$ \partial_1 E\propto J^0 \hskip2cm \partial_0 E\propto -J^1 \tag{1} $$ where $E$ is the electric field, $J^0$ is the charge density, $J^1$ is the current density, and $0,1$ are the indices of the time,space coordinates respectively.

Forget about the potential and focus instead on the electric field $E$. Consider a static situation with no current $J^1$, so both sides of the second equation are zero. Let $x$ denote the spatial coordinate, and consider a situation in which $J^0(x)\propto \delta(x)$, corresponding to a single point charge. Then we can write $J^0(x)\propto\partial_1\theta(x)$ where $\theta(x)$ is the usual step-function, so equations (1) should be solved by $E(x)=a + b\theta(x)$ for constants $a,b$.

If the one-dimensional space is non-compact (I mean, if it is the whole real line), then this solution works fine. It says that the electric field has one value on one side of the charge and a different value on the other side. The charge determines the difference between these two values.

But if the one-dimensional space is compact (I mean, if it is a circle), then equations (1) do not have a solution for a single static charge, because the would-be solution $E(x)a+b\theta(x)$ is not consistent with the periodic boundary condition that defines the circle. A single charged particle could not exist if space were a circle.

By the way, two particles of opposite charge can exist on the circle. In that case, the electric field $E$ has one constant value in one of the two charge-free intervals and has a different constant value in the other charge-free interval.

A general result

The conclusion that an isolated charge is impossible on a circle can be generalized to higher-dimensional compact spaces, at least some of them. This is mentioned in the question

Electric charges on compact four-manifolds

which is expressed using differential forms. In case that formalism is unfamiliar, here's the idea:

Consider a $D$-dimensional space that can be cut into two parts by deleting a lower-dimensional sphere. For example, a $D$-dimensional sphere can be cut in half by deleting the equator, which is a $(D-1)$-dimensional sphere. Let $M$ be the compact manifold, let $S$ be a lower-dimensional sphere that cuts it into two parts, and let $M_1$ and $M_2$ denote those two parts, each with boundary $S$.

Suppose that $M_1$ contains a net charge. We will prove that $M_2$ must contain an equal and opposite net charge.

One of Maxwell's equations is $\nabla\cdot\mathbf{E}\propto J^0$. If we integrate this over $M_1$, we get $$ \int_{M_1} \nabla\cdot\mathbf{E}\propto Q $$ where $Q$ is the total charge contained in $M_1$. We can use integration-by-parts on the left-hand side to rewrite the volume integral as a surface integral over the components of $\mathbf{E}$ normal to $S$. If we instead consider the volume integral over the other part $M_2$, then we again get the same surface integral except for an overall sign (because the normal component of the electric field points in the opposite direction). This shows that if the net charge in $M_1$ is $Q$, then the net charge in $M_2$ is $-Q$, as claimed. In particular, a single charged particle cannot exist in a compact space of this type.

This argument applies to a large class of compact manifolds, but not to all of them. The question linked above asks if there are at least some compact manifolds in which an isolated charge can exist. An answer to that other question hasn't been posted yet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.