Coulomb's force law on a Riemannian manifold

Question

Suppose we were $2$d beings living on $ S^2 $. How would the Coulomb's force law look like for us?

More generally, given a Riemannian manifold $ (M,g) $, a source charge $ q $ at the point $ p\in M $ and a test charge $ Q $ at the point $ P\in M $, I would say that the force that $ q $ exerts on $ Q $ should be proportional to the normalized velocity vector of the geodesic curve joining $ p $ with $ P $, divided by the square of the geodesic distance.

(If the manifold is closed, let's assume the net charge is zero for consistency, cf. below comment by Michael Seifert.)

But what about the $ 1 / 4\pi $ factor? And the $ 1/\epsilon_0 $ one?

Answer 1

On a Riemannian manifold, you can define the Laplace-Beltrami operator, which allows you to generalise the Poisson equation on arbitrary manifold resulting in Coulomb's law. As already pointed out, you will need to take into account topology and enforce charge neutrality for a closed manifold. For highly symmetric manifolds (sphere, plane, hyperbolic plane), you can still use the same method with Gauss' law.

For example, on a sphere with the usual induced Euclidean metric, in spherical coordinates, the Laplace-Beltrami operator is: $$ \Delta V = \frac1{\sin\theta}\partial_\theta(\sin\theta\partial_\theta V)+\frac1{\sin^2\theta}\partial_\phi^2V $$ You can solve the Poisson problem for two antipodal charges: $$ V = -\frac1{2\pi}\ln[\tan(\theta/2)] \\ E = \frac1{2\pi\sin\theta}e_\theta $$ with the sanity check: $$ V\sim -\frac1{2\pi}\ln\theta \quad \theta\to0\\ V\sim \frac1{2\pi}\ln(\theta-\pi) \quad \theta\to\pi $$ Recall that $\theta$ is basically the geodesic distance to one of the charges, so it is not just the logarithm of geodesic distance as for the plane. The $\frac1{2\pi}$ is the 2D analogue of the $\frac1{4\pi}$ in 3D. The most direct generalisation of Coulomb's law is to add a uniform background neutralising charge. In the case of the sphere, you would get: $$ V = -\frac1{4\pi}\ln[\sin(\theta/2)] \\ E = \frac1{4\pi}\cot(\theta/2)e_\theta $$ and by superposing two opposite, antipodal fields, you recover the first one.

Another classic example is the Coulomb law on a flat torus (with the same background uniform neutralising charge density caveat). There is no simple closed form formula, but there are rapidly converging series using Ewald summation.

I will end with a third example on the hyperbolic plane. It is actually the same as the sphere, just replace the circular functions by hyperbolic ones without the compact caveat: $$ \Delta V = \frac1{\sinh\theta}\partial_\theta(\sinh\theta\partial_\theta V)+\frac1{\sinh^2\theta}\partial_\phi^2V \\ V = -\frac1{2\pi}\ln[\tanh(\theta/2)] \\ E = \frac1{2\pi\sinh\theta}e_\theta $$ Remarkably, in hyperbolic space, the potential does decay to zero at infinity unlike in the flat plane.

Hope this helps.

Answer 2

On the two sphere you cannot solve the Coulomb potential problem for a single point charge, as the Laplacian has a zeo mode.

The spherical-harmonic addition formula $$ \frac{2l+1}{4\pi} P_l(\cos\theta) = \sum_{m=-l}^{l} [Y^m_l({\bf n})]^* Y^m_l({\bf n}'), \quad \cos \theta = {\bf n}\cdot {\bf n}' =x $$ suggests that a Laplace Green function for the Coulomb problem on the two sphere can be written as $$ G_+(\cos\theta) = \sum_{l=1}^\infty\sum_{m=-l}^{l} \frac{1}{l(l+1)}[Y^m_l({\bf n})]^* Y^m_l({\bf n}')\\=\frac 1 {4\pi} \sum_{l=1}^\infty (2l+1) \frac{1}{l(l+1)} P_l(\cos \theta). $$ As we have had to omit the $l=0$ mode this is a modified Green function, {\it i.e} a solution to $$ -\nabla^2 \Phi = \delta^2({\bf n},{\bf n}') - \frac 1 {4\pi} $$ which has a unit positive point charge at ${\bf n}'$ and a unit negative background charge spread uniformly over the sphere.

Another possible Green function is the potential of a unit point charge at the north pole, and a negative point charge at the south pole. This is $$ G_{\pm}(\cos \theta)=-\frac 1{2\pi} \ln \tan(\theta/2)/2\pi= \frac 1{2\pi} Q_0(\cos\theta)= \frac 1 {4\pi} \ln\left(\frac{1+\cos\theta }{1-\cos\theta}\right) , $$ which is zero at the equator. Clearly $$ G_\pm(\theta)= G_+(\cos\theta)- G_+(-\cos\theta), $$ so we have $$ \frac 1{2\pi} Q_0(\cos\theta)= \frac 1 {2\pi} \sum_{l={\rm odd}} (2l+1) \frac{1}{l(l+1)} P_l(\cos \theta). $$ This formula does not appear in the usual tables, but checks numerically.

Answer 3

This is a partial answer, because I'm not opening up the calculations. However, I'll give the main ideas you'd need for a complete calculation. Given your username and the tone of the question, I assume you are familiar with differential geometry (and I'll assume differential forms in particular).

The key remark is that Maxwell's equations can be written in differential form language. Namely, the equations become $$\mathrm{d}F = 0 \quad \text{and} \quad \mathrm{d}\star F = \mu_0 J,$$ where $F$ is the Faraday tensor, and $J$ is the four-current. This form of the equations holds for any Lorentzian manifold. You can then just consider the case of a spacetime of the form $M = \mathbb{R} \times \mathbb{S}^2$ and work out the split between the electric and magnetic fields. In $2+1$ spacetime dimensions, you'll have two components for the electric field and one component for the magnetic field.

The details of Coulomb's law can be complicated, because they will depend on a choice of Green's function that can become messy on curved manifolds. For a spatially compact spacetime, there will be the issue pointed out by Michael Seifert in the comments about it being impossible for the total charge on space to be nonzero. It may be possible to work around that if you remove one point from the sphere (say the south pole) to make a noncompact spatial manifold, and then exploit this behavior. Physically, you would hide whichever charges you need in the south pole while considering point charges elsewhere, similarly to the method of images.