# Maxwell's Equations using Differential Forms

Maxwell's Equations written with usual vector calculus are

$$\nabla \cdot E=\rho/\epsilon_0 \qquad \nabla \cdot B=0$$ $$\nabla\times E=-\dfrac{\partial B}{\partial t} \qquad\nabla\times B=\mu_0j+\dfrac{1}{c^2}\dfrac{\partial E}{\partial t}$$

now, if we are to translate into differential forms we notice something: from the first two equations, it seems that $E$ and $B$ should be $2$-forms. The reason is simple: we are taking divergence, and divergence of a vector field is equivalent to the exterior derivative of a $2$-form, so this is the first point.

The second two equations, though, suggests $E$ and $B$ should be $1$-forms, because we are taking curl. Thinking of integrals, the first two we integrate over surfaces, so the integrands should be $2$-forms and the second two we integrate over paths and so the integrands should be $1$-forms.

In that case, how do we represent $E$ and $B$ with differential forms, if in each equation they should be a different kind of form?

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you're missing that we're also changing dimensions from 3 to 4; you can of course show that it all works out by computing $dF$ and $d\star F$ in a local basis of the exterior algebra, but there's probably a nicer way to show this... –  Christoph Nov 14 '13 at 11:15
In fact, $\vec{E}$ is a 1-form, while $\vec{B}$ is a 2-form. –  Danu Nov 14 '13 at 11:35
probably the most economic way to see what's going on is to write the field tensor as $F=E_idx^i\wedge dt+\star(B_idx^i\wedge dt)$ (sign?) and figure out how $d$ and $d\star$ act on 'fake 3-vectors' $f_idx^i\wedge dt$; $d$ should turn out to be the curl, $d\star$ should end up with the time derivatives in the spatial components and the divergence in the time component –  Christoph Nov 14 '13 at 14:37
@Danu: only from a 3-space perspective –  Christoph Nov 14 '13 at 14:39
see also this answer, which would be appropriate here as well –  Christoph Nov 15 '13 at 22:10

Your problem is that you did not take relativity into account:

In Minkowski space, the relation between exterior derivatives and classical vector operators is different from the one in Euclidean 3-space, and $E$ and $B$ actually turn out to be components of a single 2-form $F$ (which is necessary to get the correct transformation laws under boosts).

Because I'm lazy, I'm going to work backwards from ${\rm d}F$ and ${\rm d}\star F$.

First, the electromagnetic tensor can be decomposed into $$F = \sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i$$ I'm assuming a $(+---)$ convention for the Minkowski metric. Please note that the sign above might be incorrect - I know I messed up somewhere (I started out with a $+$ in the formula above, and 'fixed' it after I got the wrong result), so it might be a good idea for someone to check these calculations and correct my answer if they are wrong.

The exterior derivative on 2-forms can be written as \begin{align*} {\rm d}\sum_i A_i\,{\rm d}t\wedge{\rm d}x^i &= \star\sum_i (\nabla\times A)_i\,{\rm d}x^i \\ {\rm d}\star\sum_i A_i\,{\rm d}t\wedge{\rm d}x^i &= -\star(\nabla\cdot A\,{\rm d}t + \sum_i \frac{\partial A_i}{\partial t}\,{\rm d}x^i) \end{align*} and we arrive at \begin{align*} {\rm d}F &= \star\sum_i (\nabla\times E)_i\,{\rm d}x^i + \star(\nabla\cdot B\,{\rm d}t + \sum_i \frac{\partial B_i}{\partial t}\,{\rm d}x^i) \\&= \star\sum_i ( \nabla\times E + \frac{\partial B}{\partial t} )_i\,{\rm d}x^i + \star\nabla\cdot B\,{\rm d}t \\ {\rm d}\star F &= {\rm d}\left( \star\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i + \sum_i B_i\,{\rm d}t\wedge{\rm d}x^i \right) \\&= -\star(\nabla\cdot E\,{\rm d}t + \sum_i \frac{\partial E_i}{\partial t}\,{\rm d}x^i) + \star\sum_i (\nabla\times B)_i\,{\rm d}x^i \\&= \star\sum_i ( \nabla\times B - \frac{\partial E}{\partial t} )_i\,{\rm d}x^i - \star\nabla\cdot E\,{\rm d}t \end{align*} from which we get the left-hand sides of the Maxwell equations by looking and space and time components separately.

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Yes. The last equation could be written $d*F = *J$, where $J$ is the one form $J = J_\mu dx^\mu$ –  Trimok Nov 14 '13 at 19:01

Define a 4-potential $A_\mu$. Then you can form the 1-form $A = A_\mu dx^\mu$. The field-strength is then $F = dA = \frac{1}{2} F_{\mu\nu} dx^\mu \wedge dx^\nu$. So in fact, $E$ and $B$ are the components of a 2-tensor.

Note that this implies $dF = d^2A = 0$, which give things like $\nabla \cdot B = 0$.

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Since I hate (for reasons I can not make objective I confess) the $\star$ symbol, I propose below to reconstruct the Maxwell's equations from the definition of the 2-form $$F=E_{i}dx^{i}\wedge dt+B_{\left\vert i\right\vert }dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }$$ where I introduce the modulus notation $\left\vert i\right\vert$ such that $\left\vert i+3\right\vert =\left\vert i\right\vert =i$. Greek indices are space-time ones, latin's range only over space indices. I will do nothing else than Christoph in his answer, except perhaps for the definition of the current, see below.

Then one has

$$\text{d}F=-\dfrac{\partial E_{i}}{\partial x^{j}}dx^{i}\wedge dx^{j}\wedge dt+\dfrac{\partial B_{\left\vert i\right\vert }}{\partial x^{\left\vert i\right\vert }}dx^{\left\vert i\right\vert }\wedge dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }+\dfrac{\partial B_{\left\vert i\right\vert }}{\partial t}dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }\wedge dt$$

and thus

$$\text{d}F = \left(\dfrac{\partial E_{y}}{\partial x}-\dfrac{\partial E_{x}}{\partial y}+\dfrac{\partial B_{z}}{\partial t}\right)dx\wedge dy\wedge dt+\left(\dfrac{\partial E_{z}}{\partial y}-\dfrac{\partial E_{y}}{\partial z}+\dfrac{\partial B_{x}}{\partial t}\right)dy\wedge dz\wedge dt + \left(\dfrac{\partial E_{x}}{\partial z}-\dfrac{\partial E_{z}}{\partial x}+\dfrac{\partial B_{y}}{\partial t}\right)dz\wedge dx\wedge dt+\mathbf{\nabla\cdot B}\text{vol}^{3}$$

with obvious change of notations and $\text{vol}^{3}=dx^{\left\vert i\right\vert }\wedge dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }=dx\wedge dy\wedge dz$ for the volume 3-form.

Then, imposing $dF=0$ leads to the two Maxwell equations $$\mathbf{\nabla\cdot B}=0\text{ and }\mathbf{0}=\mathbf{\nabla\times E}+\dfrac{\partial\mathbf{B}}{\partial t}$$ which can thus be seen as a single continuity equation for the $F$ 2-form.

Since we impose $dF=0$, one has locally $F=dA$, with the 1-form $A=A_{\alpha}dx^{\alpha}$ verifying

$$\mathbf{E}=\mathbf{\nabla}\varphi-\dfrac{\partial\mathbf{A}}{\partial t}\text{ and }\mathbf{B}=\mathbf{\nabla\times A}$$

by direct identification of the $dx^{i}\wedge dt$ and the $dx^{\left\vert i\right\vert }\wedge dx^{\left\vert i+1\right\vert }$ in both $F$ and $dA$. Now, one can add any exact form to $A\rightarrow A +d\chi$ ($\chi$ is a 0-form by construction) without changing $F=dA\rightarrow d\left(A+d\chi\right) =dA$

Then of course neither $E$ nor $B$ is a form, but $F$ is a 2-form.

One can also form a 3-form

$$j=J_{\left\vert i\right\vert }dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }\wedge dt-\rho\text{vol}^{3}$$

which is conserved $dj=0$ and thus locally $j=dG$, with $G$ a 2-form, by construction. We recognise $j$ as the usual charge current, and we want to construct $G$. Let us define

$$\gamma=\gamma_{\left\vert i\right\vert }dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }+\gamma_{i}^{0}dx^{i}\wedge dt$$

as the most general 2-form in space-time. The strange space / time separation is necessary since one want just one index for $\gamma$. Otherwise we could have just written $\gamma_{\mu \nu}dx^{\mu}\wedge dx^{\nu}$. Note it was the same for $F$. Then we calculate

$$d\gamma=\dfrac{\partial\gamma_{\left\vert i\right\vert }}{\partial x^{\left\vert i\right\vert }}dx^{\left\vert i\right\vert }\wedge dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }+\dfrac{\partial\gamma_{\left\vert i\right\vert }}{\partial t}dt\wedge dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }+\dfrac{\partial\gamma_{i}^{0}}{\partial x^{j}}dx^{j}\wedge dx^{i}\wedge dt$$

or, separating space and time one more time, one has

$$\mathbf{J}=\dfrac{\partial\mathbf{\gamma}}{\partial t}+\mathbf{\nabla\times\gamma}^{0}\text{ and }-\rho=\mathbf{\nabla\cdot\gamma}$$

where we recognise the second set of Maxwell's equations if we define

$$\mathbf{\gamma}=-\mathbf{D}\text{ and }\mathbf{\gamma}^{0}=\mathbf{H}$$

and -- one more time -- neither $H$ nor $D$ is a form, but $\gamma$ is a proper 2-form, call it $$-G=D_{\left\vert i\right\vert }dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }+H_{i}dt\wedge dx^{i}$$ if you wish. The funny part is that of course any substitution $G\rightarrow G+d\phi$ ($\phi$ is a 2-form) will not redefine the charge $j=dG\rightarrow d\left(G+d\phi\right)=dG$.

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