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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: – Qmechanic Jan 24 '13 at 12:55

3 Answers 3

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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 . – 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

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

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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