Maxwell's equations in differential form in 2-space+1-time dimensions

Question

How does one write maxwell's equation in 2+1 dimensions? It becomes particularly interesting as the components of 2 forms and 1 form are 3. Are there any sources for this?

Answer 1

The relativistic version of Maxwell's equations, $$ \partial_\mu \, F^{\mu\nu} = J^\nu$$ (plus the dual equation, see wikipedia if that doesn't ring a bell), works fine with any number of spatial dimensions. Of course, $F^{\mu\nu}$ is the field strength tensor and $J^\mu$ is the relativistic current.

Here's where it gets fun. In 2+1 dimensions $$ F^{\mu\nu} = \left( \begin{array}{ccc} 0 & -E_1 & -E_2 \\ E_1 & 0 & B \\ E_2 & -B & 0 \end{array} \right) . $$ So the electric field is a 2d vector $\vec{E}=(E_1, E_2)$, but there's only one component of the magnetic field: $B$ is a scalar (actually a pseudoscalar)!

That's because fundamentally, $E$ is a spatial vector and $B$ is a spatial "bivector". In 3d a bivector happens to be the same as a "psuedovector", so we tend to incorrectly think of $B$ as a vector quantity. But in other numbers of dimensions that's just not the case. To get a much deeper feel for this, I'd recommend checking out the geometric algebra formulation of electrodynamics (e.g. Ch. 7 of Doran and Lasenby).

Answer 2

The relativistic formulation of electromagnetism repackages Maxwell's equations as

$F^{a} = \int d^{3}x\,j^{b}F^{a}{}_{b}$, where $F_{ab}$ is the Maxwell Tensor given by $F_{ab} = \nabla_{[a}A_{b]}$, and $A$ is the electromagnetic potential. In Four dimensions, it can be shown that the first equation reduces to the normal Lorentz force law, and that the Maxwell Tensor is $F_{0i} = E_{i}$ and $F_{ij} = B^{k}\epsilon_{ijk}$, where $\epsilon_{ijk}$ is the standard Levi-Civita symbol.

Then, generalizing to 2+1 dimensions is easy, because all of these equations just cross over. All that changes is that your vector $A$ only has three components, which will mean that you have a 2-dimensional electric field, and a one-dimensional "magnetic" field.

Answer 3

In The differential geometry behind Maxwell’s equations by M. Climente Chapter 4, Maxwell's equations are derived in the differential geometry formalism. If you don't know differential geometry, skip this part and read the last section. In this formalism, Maxwell's equation are \begin{align} d_SB&=0 & \partial_tB+d_SE&=0\\ *_Sd_S*_SE&=\rho & -\partial_tE+*_Sd_S*_SB&=j \end{align} Here $E=E_xdx+E_ydy+\dots$ is a one form that represents the electric field, $B=B_{xy}dx\wedge dy+B_{xz}dx\wedge dz+\dots$ is a two-form representing the magnetic field, $d_S$ is the exterior derivative with respect to the space variables and $*_S$ is the hodge star with respect to the space variables. Also $\rho$ is a scalar function representing the density and $j$ is a one-form representing the current.

The nice thing about differential geometry is that it works in any dimension. So by working through the math, you get the following expressions in (2+1)D:

\begin{align} 0&=0 & &\partial_t b+\partial_xE_y-\partial _y E_x=0\\ \partial_x E_x+\partial_y E_y&=\rho & &\cases{-\partial_tE_x-\partial_y b=j_x\\-\partial_t E_y+\partial_x b=j_y} \end{align} A couple comments. In 2D there is only one two-form, so the magnetic field has one component $B_{xy}=b$. Gauss's law for magnetism completely disappears and becomes $0=0$.

As a final note, I derived these last expressions myself so there could be errors. If you want to use these, please check them for yourself.