# Can light exists in $2+1$ or $1+1$ spacetime dimensions?

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 equations and their particular solution - electromagnetic radiation (light)?

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Related calculation for arbitrary spacetime dimensions: physics.stackexchange.com/a/20578/2451 – Qmechanic Jan 24 '13 at 12:55

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.

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@RonMainon Thanks for the answer, but I would naively say that in 2d space (see, please, again my question) perpendicular waves can exist. AS regards the second paragraph of your answer I admit that I don't understand how it is related to light. – Leos Ondra Jul 24 '12 at 5:39
@LeosOndra if you mean 2 spatial dimensions we'd normally describe this as 3D or 1+2D. I suspect Ron thought you meant 2 dimensions in total i.e. 1 time and 1 space. – John Rennie Jul 24 '12 at 10:37
@RonRaimon Thanks for the additional answer for 2+1 dimensions. In the meantime I found a note in Wikipedia Spacetime article according to which Weyl in his 1922 book Space, time, and matter ruled out possibility of Maxwell equations in 1+2D, but a cursory look at the relevant pages suggests that this conclusion is based on his own geometrical theory of electromagnetism and is not valid. – Leos Ondra Jul 24 '12 at 13:42
There is a major qualitative difference between 2+1 and 3+1 dimensions : Huygens' principle is only valid for odd numbers of space dimensions, see en.wikipedia.org/wiki/Wave_equation . – jjcale Jul 25 '12 at 0:09
@Ron Maimon: No, Huygen's principle holds only for odd number of space dimensions : See end of section "Solution of a general initial-value problem" and the section "Scalar wave equation in two space dimensions" in the Wikipedia article. – jjcale Jul 25 '12 at 18:32

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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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.

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Thanks for your answer, this makes sense to me. – Leos Ondra Jul 27 '12 at 10:16
I have finally read the answer carefully and I must confess I don't understand it. I have always thought that E and B are perpendicular to each other and both of them to the wave vector. How can this be in 2+1D? If light propagates along x axis and E is in y direction we are left with no axis for B. – Leos Ondra Jul 29 '12 at 14:49
The fact that $\vec{B}$ is perpendicular to $\vec{E}$ and to the wave vector is due to a specific coincidence that occurs in 3+1D space. Namely, that can happen because the magnetic field happens to have the same number of components (3) as a spatial vector, and so in 3+1D space only, you can kind of treat the magnetic field as a vector. But in other numbers of dimensions, the magnetic field is not a rotational vector. In 2+1D specifically, it is a rotational scalar - in other words, it has no direction, only a magnitude. In 4+1D, it would have six components, and so on. – David Z Jul 30 '12 at 2:30
Thank you for the explanation. – Leos Ondra Jul 30 '12 at 6:37