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I read this answer a while ago, and while thinking about $\nabla$, I realized something. Since the cross product can be written as a determinant, in higher dimensions we require extra vector inputs. IIRC it's called the "wedge product" in higher dimensions.

Alright, how does this work when we generalize Maxwell to higher dimensions? Curl can be written (abuse of notation, yes) as a cross product with $\nabla$. But, to generalise it to higher dimensions, we need multiple inputs. We need something like $\nabla_4(\mathbf{B_1},\mathbf{B_2})$ in four dimensions, and so on. So we have two ways to get out of this: We can either use a different way to write curl in multiple dimensions (The wikipedia page has stuff on this which I don't understand), or there are more than one $E$ and $B$ fields in higher dimensions.

So which is it? How are Maxwell's laws generalised to higher dimensions?

Just a note: I never understood the linked answer after the first sentence (didn't know enough), so if there's something obvious there that answers this question, I missed it. I know nothing of higher-dimensional analysis, so if complex notation is going to be unavoidable for the multiple dimensions, I'd be fine if you showed me what happens in four dimensions.

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For more on cross products in dimensions different from 3, see also this question. –  Qmechanic Mar 1 '12 at 16:09
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3 Answers 3

up vote 11 down vote accepted

Qmechanic's answer has the full details, but I figure it might be useful to break this down in a little more detail.

The proper way to generalize Maxwell's equations to higher-dimensional spaces is to use the field tensor $F^{\mu\nu}$. In our normal 3+1D space, it basically looks like this:

$$F = \begin{pmatrix}0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & B_{12} & B_{13} \\ -E_2 & -B_{12} & 0 & B_{23} \\ -E_3 & -B_{13} & -B_{23} & 0\end{pmatrix}$$

If you apply a rotation matrix corresponding to a 3D spatial rotation to this tensor, you will find that it mixes up the three components $E_1$, $E_2$, and $E_3$, and it separately mixes up the three components $B_{12}$, $B_{13}$, and $B_{23}$. The way in which these components mix is consistent with $E$ being a vector in 3D space and with $B$ being a separate vector in 3D space; for example, if you rotate by $\phi$ around the $z$ axis, you'll find that $E'_1 = E_1\cos\phi - E_2\sin\phi$ and $B_{23}' = B_{23}\cos\phi + B_{13}\sin\phi$. Doing this for a few rotations allows you to conclude that $(E_1,E_2,E_3)$ are respectively the $(x,y,z)$ components of $\vec{E}$, and that $(B_{23},-B_{13},B_{12})$ are the $(x,y,z)$ components of $\vec{B}$.

However, if you apply a Lorentz boost (a change in velocity), which is just a rotation that incorporates the time dimension as well as the three spatial dimensions, you will find that it mixes the components of $\vec{E}$ with the components of $\vec{B}$. Nothing in the theory of normal 3D vectors allows this to happen. So if you didn't know about $F$, this would be your first clue that $\vec{E}$ and $\vec{B}$ are not really vectors, but instead are part of some more complex structure. This is in fact exactly what happened in the 1830s-ish, when Michael Faraday discovered that moving a magnet past a wire would induce an electrical current in the wire. The Lorentz boost corresponds to moving the magnet, and the rotation of components of $\vec{B}$ into components of $\vec{E}$ corresponds to the induction of an electric field by the moving magnet. Faraday and the others who capitalized on his work later in that century knew enough to recognize that $\vec{E}$ and $\vec{B}$ were related somehow, in a way that ordinary vectors weren't, but it wasn't until the introduction of special relativity that anyone figured out the actual mathematical structure that both $\vec{E}$ and $\vec{B}$ are part of (namely, the antisymmetric tensor $F$).

Anyway, once you know what $F$ really looks like, it's straightforward to generalize it to higher dimensions. In 4+1D, for example,

$$F = \begin{pmatrix}0 & E_1 & E_2 & E_3 & E_4 \\ -E_1 & 0 & B_{12} & B_{13} & B_{14} \\ -E_2 & -B_{12} & 0 & B_{23} & B_{24} \\ -E_3 & -B_{13} & -B_{23} & 0 & B_{34} \\ -E_4 & -B_{14} & -B_{24} & -B_{34} & 0\end{pmatrix}$$

This has 4 elements of an electric nature, but 6 of a magnetic nature. If you apply a 4D spatial rotation matrix to this, you will still find that the 4 components $E_1, E_2, E_3, E_4$ mix in the way you expect of a 4D vector, but the 6 magnetic components don't. So if we lived in a 4+1D universe, it would be obvious that the magnetic field is not a vector, if nothing else because it has more elements than a 4+1D spatial vector does.

OK, so what about Maxwell's equations? Well, it turns out that you can express the cross product in three dimensions using the antisymmetric tensor $\epsilon^{\alpha\beta\gamma}$, which is defined component-wise as follows:

  • +1 if $\alpha\beta\gamma$ is a cyclic permutation of $123$
  • -1 if it's a cyclic permutation of $321$
  • 0 if any two of the indices are equal

Same goes for the curl; you can write the curl operator in tensor form as

$$\bigl(\vec\nabla\times\vec{V}\bigr)^\alpha = \epsilon^{\alpha\beta\gamma}\frac{\partial V_\beta}{\partial x^\gamma}$$

using the Einstein summation convention. Basically, when you take the curl of a vector, you're constructing all possible antisymmetric combinations of the vector, the derivative operator, and the directional unit vectors:

$$\begin{matrix} \hat{x}_1 \frac{\partial V_2}{\partial x^3} & -\hat{x}_1 \frac{\partial V_3}{\partial x^2} & \hat{x}_2 \frac{\partial V_3}{\partial x^1} & -\hat{x}_2 \frac{\partial V_1}{\partial x^3} & \hat{x}_3 \frac{\partial V_1}{\partial x^2} & -\hat{x}_3 \frac{\partial V_2}{\partial x^1}\end{matrix}$$

The antisymmetric tensor $\epsilon$ is easy to generalize to additional dimensions; you just add on an additional index per dimension, and keep the same rule for assigning components in terms of permutations of indices. However, Maxwell's equations actually involve two different curls, $\vec\nabla\times\vec{E}$ and $\vec\nabla\times\vec{B}$. Since the electric and magnetic fields don't generalize to higher-dimensional spaces in the same way, it stands to reason that their curls may not either.

Let's look at the "magnetic curl" first. The magnetic field generalizes to higher dimensions as an antisymmetric piece of a tensor, so we should write its curl as an operation on that antisymmetric piece of a tensor. Start with the tensor-notation cross product rule,

$$\bigl(\vec\nabla\times\vec{B}\bigr)^\alpha = \epsilon^{\alpha\beta\gamma}\frac{\partial B_\beta}{\partial x^\gamma}$$

and put in the following identity which expresses the components of $\vec{B}$ in terms of components of $F$,

$$B_\beta = -\frac{1}{2}\epsilon_{\beta\mu\nu}F^{\mu\nu}$$

(here the indices range over values 1 to 3), and after some simplifications you get

$$\bigl(\vec\nabla\times\vec{B}\bigr)^\alpha = \frac{\partial F^{\alpha\mu}}{\partial x^\mu}$$

So the curl of something that can be expressed an antisymmetric piece of a tensor is really not a curl at all! That makes it very easy to generalize: the equation

$$\frac{\partial F^{\alpha\beta}}{\partial x^{\beta}} = \mu J^{\alpha}$$

gives you Ampère's law in any number of dimensions $N$, if you just let $\alpha$ and $\beta$ range from $1$ to $N$. (Conveniently, if you let $\alpha$ be equal to zero, you get Gauss's law.)

Now what about the "electric curl"? Well, the electric field generalizes to higher dimensions as a vector (as long as you ignore Lorentz boosts), so it's not really an antisymmetric tensor - at least, the components of $\vec{E}$ don't form a square block of $F$ the way $\vec{B}$ did. But remember, we do have that equation that related $B_\beta$ to $F^{\mu\nu}$. You can actually flip that around and use it to define an antisymmetric tensor that will contain the components of $\vec{E}$. We call this new tensor $G$, the dual tensor to $F$.

$$G^{\mu\nu} = g_{\alpha\beta}\epsilon^{\beta\mu\nu}E^{\alpha}$$

This is the piece of it that contains $E$, anyway. (I might be off by a numerical factor or a sign or something, but that's the gist of it.) If you write out the components of $G$ in 3+1D space, it looks just like $F$ except that the positions of $E$ and $B$ are switched. Using this new dual tensor, you can do the same thing we did with $\vec{B}$ to $\vec{E}$, namely write its curl as

$$\bigl(\vec\nabla\times\vec{E}\bigr)^\alpha = \frac{\partial G^{\alpha\mu}}{\partial x^\mu}$$

and thus Maxwell's other equations are given by

$$\frac{\partial G^{\alpha\beta}}{\partial x^\beta} = 0$$

This can also be easily generalized to higher dimensions, but there is a trick to how you define $G$. The thing is, since it is a dual tensor, it doesn't always have 2 indices. Remember that when you go to higher dimensions, you have to put extra indices on $\epsilon$, and so the definition of $G$ changes. For example, the definition above was for a 3D subset of G. The real $G$ in 3+1D is defined like this:

$$G^{\mu\nu} = \frac{1}{2}g_{\kappa\alpha}g_{\lambda\beta}\epsilon^{\kappa\lambda\mu\nu}F^{\alpha\beta}$$

In 4+1D, it's defined like this

$$G^{\mu\nu\rho} = \frac{1}{6}g_{\kappa\alpha}g_{\lambda\beta}\epsilon^{\kappa\lambda\mu\nu\rho}F^{\alpha\beta}$$

and so on. Notice that $G$ always has $N - 2$ indices, so that the total number of indices on $F$ and $G$ is $N$. This is one key property of dual tensors: in a rough sense, the basis of one is kind of orthogonal to the basis of the other. This is where the exterior calculus that Qmechanic mentioned comes into play: it shows a sort of equivalence between tensors and their duals, and it has ways of neatly dealing with dual tensors which make it possible to write Maxwell's equations very compactly.

$$\begin{align}\mathbf{d}F &= 0 & \mathbf{d}G &= 0\end{align}$$

The exterior derivative $\mathbf{d}$ is an operation that applies to both a tensor field and its dual in similar ways, such that in both cases it reproduces Maxwell's equations.

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Awesome answer!. I didn't understand what you meant by "The curl is not a curl at all". Is it because we have a much simpler expression than the normal curl expression? And in the same expression, is $\mu$ being Einstein-summed (It looks like $\alpha$ isn't)? I doubt we'd get our 3D curl unless it was being Einstein summed. I usually get confused in the summation notation between tensor indices and summation. (My knowledge of tensors is half-baked) –  Manishearth Mar 2 '12 at 12:57
    
You know the identity $\vec\nabla\times(\vec{v}\times\vec{F})=(\vec\nabla\cdot\vec{F} + \vec{F}\cdot\vec\nabla)\vec{v} - (\vec\nabla\cdot\vec{v} + \vec{v}\cdot\vec\nabla)\vec{F}$? It's basically something like that. $\vec{B}$ is already cross-product-like, so when you take the curl of it, you get something that is more like a divergence. –  David Z Mar 2 '12 at 18:33
    
Aah, got it. Thanks! –  Manishearth Mar 3 '12 at 3:15
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The higher-dimensional version of Maxwell's equations is actually written explicitly in the very beginning of the linked answer. However, if you are only familiar with the traditional formulation of Maxwell equations, you will need to study two new subjects to appreciate this.

I) Special relativistic formulation. Even for standard $n=4$ spacetime dimensions, there exist two equivalent formulations of Maxwell's equations, say for simplicity, in vacuum.

  1. The traditional formulation of Maxwell's equations, which uses $\vec{E}$ and $\vec{B}$ fields. (This formulation is, as Einstein famously showed, relativistic covariant, but the vector notation somewhat obscures this important fact.)

  2. The manifestly relativistic covariant formulation, which uses the electromagnetic field tensor $F_{\mu\nu}$.

II) Exterior calculus. As OP correctly observes, the cross product and curl are only defined for 3 spatial directions. To generalize for arbitrary spacetime dimension $n$, we use the mathematical machinery of exterior differential forms, exterior wedge product $\wedge$, and exterior de Rham differential $d$.

In the linked answer, it is also mentioned how many electric and magnetic field components, and how many Maxwell's equations, there are in various spacetime dimensions $n$. Here is a table for small spacetime dimension $n$:

  # of spacetime dimensions n | 1 2 3 4  5  6  7
  ----------------------------------------------
  # of electric fields        | 0 1 2 3  4  5  6
  # of magnetic fields        | 0 0 1 3  6 10 15
  # of Maxwell's eqs.         | 1 2 4 8 15 26 42

For instance, in $n=0+1$ spacetime dimension, there are no electric and magnetic fields, and there is just a single Maxwell equation (Gauss' law), which says that $\rho=0$. Not a very interesting theory!

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So it can't even be written in a simple form in four dimensions? And are there more than 2 fields? I realise that its a tensor, but if we try breaking it into fields what happens? –  Manishearth Mar 1 '12 at 15:43
    
In $n=4+1$ spacetime dimensions, the electric field has $4$ components, and the magnetic field has $6$ components. –  Qmechanic Mar 1 '12 at 15:48
    
Aah. Looking at the other link above it seems that we HAVE to use tensors and there is no vector field per sé, right? Which seems to be spported by the fact that B has six components, which makes no sense if B is a vector.. –  Manishearth Mar 1 '12 at 16:39
    
So in 4+1 dimensions, there is no way to write the laws in terms of vector/scalar fields? Not even in a nonrelativistic case? –  Manishearth Mar 1 '12 at 16:44
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The magnetic field is a two-form, an object with two spatial indices $M_{ij}$ which is antisymmetric on exchange of the indices $M_{ij} = - M_{ji}$. In 3+1 there are 3 independent component of M. There are also 3 components in a vector so we can pretend that field is a vector $\vec{B} =\{M_{yz}, M_{zx},M_{xy}\}$. But note the magnetic field is really 2-form, not a vector. This is why $B$ is not really a vector, its a pseudo-vector (it behaves differently from a vector under reflections). In 4+1 dimensions there are six components of $M$ so we can't pretend its a vector anymore. –  BebopButUnsteady Mar 1 '12 at 16:58
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in n+1 dimensions, E is an n-vector field, and B an antisymmetric tensor field with n(n-1)/2 components. The cross product $\nabla \times E$ becomes the antisymmetric tensor with components $\nabla_j E_k - \nabla_k E_j$, and the cross product $\nabla \times B$ becomes the vector with components $\sum _j \nabla_j B_{jk}$. (or perhaps minus this sum; you could figure out the correct sign from the Lagrangian).

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