Spacetime of special relativity is frequently illustrated with its spatial part reduced to one or two spatial dimension (with light sector or cone, respectively). Taken literally, is it possible for $2+1$ or $1+1$ (flat) spacetime dimensions to accommodate Maxwell's equations and their particular solution - electromagnetic radiation (light)?


4 Answers 4


According to our best understanding of how Maxwell's equations would generalize to other dimensions, then yes, it is. The electromagnetic field is represented by an antisymmetric tensor,

$$F^{\mu\nu} = \begin{pmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{pmatrix}$$

To reduce this into 2+1D, you just chop off one of the spatial rows and columns:

$$F^{\mu\nu} = \begin{pmatrix}0 & -E_x & -E_y \\ E_x & 0 & -B \\ E_y & B & 0\end{pmatrix}$$

This is mathematically equivalent to setting those components to zero, so an electromagnetic wave in 2+1D space would be equivalent to an EM wave in 3+1D which has its electric field oriented in the $xy$ plane and its magnetic field oriented in the $z$ plane, which is indeed possible.

If you wanted to do a proper analysis in 2+1D space itself, you could go straight to the wave equation,

$$\partial_\alpha\partial^\alpha F^{\mu\nu} = 0$$

or, explicitly,

$$-\frac{\partial^2 F^{\mu\nu}}{\partial^2 t} + \frac{\partial^2 F^{\mu\nu}}{\partial^2 x} + \frac{\partial^2 F^{\mu\nu}}{\partial^2 y} = 0$$

for each component, $\nu\in\{t,x,y\}$.

Alternatively you could write Maxwell's equations in a vacuum (assuming no spacetime curvature) as

$$\begin{align}\partial_\mu F^{\mu\nu} &= 0 & \partial_\lambda F_{\mu\nu} + \partial_\nu F_{\lambda\mu} + \partial_\mu F_{\nu\lambda} &= 0\end{align}$$

the first of which becomes the following set of equations

$$\begin{align} -\frac{\partial E_x}{\partial x} - \frac{\partial E_y}{\partial y} &= 0 & \frac{\partial E_x}{\partial t} - \frac{\partial B}{\partial y} &= 0 & \frac{\partial E_y}{\partial t} + \frac{\partial B}{\partial x} &= 0\end{align}$$

and the second of which becomes

$$\frac{\partial B}{\partial t} - \frac{\partial E_x}{\partial y} + \frac{\partial E_y}{\partial x} = 0$$

You can then put these together and derive the wave equation for each of the EM field components.

It's worth mentioning that in 1+1D, the field tensor has only a single component,

$$F^{\mu\nu} = \begin{pmatrix}0 & -E \\ E & 0\end{pmatrix}$$

and the lone Maxwell's equation becomes

$$\frac{\partial E}{\partial x} = 0$$

You could write the wave equation as

$$-\frac{\partial^2 E}{\partial t^2} + \frac{\partial^2 E}{\partial x^2} = 0$$

which makes it seem like EM waves would exist. But Maxwell's equation in 1+1D tells you that the electric field is constant in a vacuum. That doesn't allow for the spatial variation you need to create a wave, which is why EM waves don't exist in that space.

You can also see this by noting that you can't have an EM wave in 3+1D with only an electric field and no magnetic field.


No, because the polarization of the electromagnetic field must be perpendicular to the direction of motion of the light, and there aren't enough directions to enforce this condition. So in 1d, a gauge theory becomes nonpropagating, there are no photons, you just get a long range Coulomb force that is constant with distance.

In the 1960s, Schwinger analyzed QED in 1+1 d (Schwinger model) and showed that electrons are confined with positrons to make positronium mesons. A much more elaborate model was solved by t'Hooft (the t'Hooft model, the nonabelian Schwinger model) which is a model of a confining meson spectrum.

EDIT: 2+1 Dimensions

Yes, light exists in 2+1 dimensions, and there is no major qualitative difference with 3+1 dimensions. I thought you wanted 1+1, where it's interesting.

  • 2
    $\begingroup$ Isn't 2+1D quite different in the sense Huygens principle doesn't apply? $\endgroup$
    – jinawee
    Sep 2, 2017 at 16:48
  • $\begingroup$ Can there be radiation due to scalar field in 1+1 dimensions instead of gauge field? $\endgroup$ Nov 24, 2017 at 6:25

I agree. Light can not exist in 2D spacetime as well as a $\vec{B}$ component that has to be perpendicular to $\vec{E}$. Also, the Gauss law requires $$E \propto \frac{q}{r^{D-1}}$$ where D is the number of spatial dimensions. Therefore it is absent in 2D.

However, 2D or 3D spacetimes still can have the speed of light and Lorentz invariance! $$ ds^2= - c^2 dt^2 + dx^2$$


Light may exist in a 2 dimensional space, but it wouldn't appear the same as ours. With 1 less dimension it would become a point. Add an additional dimension to the point and ,as all dimensions should be perpendicular to each other, you would end up with an electromagnetic wave in 3 dimensional space. This wave would intuitively appear strange due to not all of its components existing in our Time/Space.


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