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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    $\begingroup$ 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... $\endgroup$
    – Christoph
    Commented Nov 14, 2013 at 11:15
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    $\begingroup$ In fact, $\vec{E}$ is a 1-form, while $\vec{B}$ is a 2-form. $\endgroup$
    – Danu
    Commented Nov 14, 2013 at 11:35
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    $\begingroup$ 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 $\endgroup$
    – Christoph
    Commented Nov 14, 2013 at 14:37
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    $\begingroup$ @Danu: only from a 3-space perspective $\endgroup$
    – Christoph
    Commented Nov 14, 2013 at 14:39
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    $\begingroup$ see also this answer, which would be appropriate here as well $\endgroup$
    – Christoph
    Commented Nov 15, 2013 at 22:10

5 Answers 5


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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    $\begingroup$ Yes. The last equation could be written $d*F = *J$, where $J$ is the one form $J = J_\mu dx^\mu$ $\endgroup$
    – Trimok
    Commented Nov 14, 2013 at 19:01

The Maxwell equations come from 1. equations of motion of electromagnetic action 2. Second Bianchi identity.

Precisely, you cal find elsewhere that

1) The solution of maxwell Lagrangian is $$-\partial _\mu F^{\mu \nu} = J^{\nu} \quad (1)$$ Which implies $$ \mapsto \begin{cases} \nabla \times B - \frac{\partial{E}}{\partial{t}}=\vec j \\ \nabla \cdot E =q \end{cases}$$ 2) The second Bianchi identity $$ \partial_{\alpha} F_{\beta \gamma} + \partial_{\beta} F_{\gamma \alpha} + \partial_{\gamma} F_{\alpha \beta} =0 .\quad (2)$$ implies $$\mapsto \begin{cases} \nabla \cdot B = 0 \\ \nabla \times E + \frac{\partial{B}}{\partial{t}}=0 \end{cases} $$ You can translate (1) and (2) to the differential-form. But I will try more formal method by starting with the action.

The EM action is $${\mathcal{L}}_{\mathtt{Maxwell}} \equiv -(1/4)F_{\mu \nu}F^{\mu \nu} +A_{\mu} J^{\mu}.\quad\quad (3)$$ translate to the differential-form reads $$S_{\mathrm{Maxw}}=\int_{\mathcal{M}}\left(- \frac{1}{2} F \wedge \star F -A \wedge \star J\right),\quad \quad (4)$$ So $$\begin{split} \delta_{A} S_{\mathrm{Maxw}}=\int_{\mathcal{M}}\left(- \mathrm{d}(\delta A) \wedge \star F -\delta A \wedge \star J\right),\\ 0=\int_{\mathcal{M}}\left(- (\delta A) \wedge \star\bar {\mathrm{d}} F -\delta A \wedge \star J\right),\\ \Longrightarrow -\bar {\mathrm{d}}F =J. \\ \boxed{ -\star \mathrm{d} \star F = J }\\ - \mathrm{d} \star F = \star J \Longrightarrow -\partial_{\mu} F^{\mu \nu} = J^{\nu}\end{split}$$

where $\bar {\mathrm{d}}$ is co-differentiation in this case $\bar{\mathrm{d}}= \star \mathrm{d} \star$

Next, Faraday tensor in differential-form reads $$F=\mathrm d A$$ So, obviously $$\mathrm d F =\mathrm d ^2 A =0.$$ This is also a Bianchi second identity and obviosly expand it in local coordinate form gives us equation (2) and so another maxmell equation.

In conclusion the Maxwell equation in the form of differential-form reads $$ \boxed{ - \mathrm{d} \star F = \star J,\quad\quad \mathrm{d}F=0 } $$

Note:Useful for checking your calculations $$ F^{\alpha \beta} = \left( \begin{matrix} {0} & { E_x} & {E_y} &{ E_z} \\ {-E_x} & {0} &{ B_z} &{ -B_y} \\ {-E_y} & {-B_z} &{ 0} &{ B_x} \\ {-E_z} & {B_y }& {-B_x }& {0} \end{matrix} \right)$$

$$F=\frac 1 2 F_{\mu\nu}dx^\mu \wedge dx^\nu$$ $$A^\mu=(q,\vec j)$$ $$F \wedge \star F =\frac{1}{2}F_{\mu \nu} F^{\mu \nu} {\mathrm{d}}^{4}x (=-(E^2 -B^2)).$$ $$ A \wedge \star J =-A_{\mu}J^{\mu}{\mathrm{d}}^{4}x,$$ $$\begin{split} \mathrm{d} \left( \alpha \wedge \star \beta \right) =: \mathrm{d}\alpha \wedge \star \beta - \alpha \wedge \star \bar{\mathrm{d}} \beta \\ \bar{\mathrm{d}} = -(-1)^{np-n+s}\star \mathrm{d} \star \\ (-1)^{s} = \begin{cases} -1 &\text{if Lorentzian } \\ 1 &\text{if Euclidean} \end{cases} \\ \text{ $\beta$ is a p-form}\\ sgn = \text{sign (of $(-1)^N$)} \\ n = N+P\\ \text{Signature} = P-N \\ \text{indeed} &(-1)^{s} := (-1)^{N} = sgn(g)&\\ \end{split}$$ To check your basic skills on exterior algebra, you must understand that $$J=J_\mu dx^\mu \longrightarrow \star J = \frac{1}{3!}(\frac 1 1 \epsilon_{\mu\nu\alpha\beta}J^\mu)dx^\nu \wedge dx^\alpha \wedge dx^\beta$$

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    $\begingroup$ I don't get the minus sign in $ A \wedge \star J =-A_{\mu}J^{\mu}{\mathrm{d}}^{4}x$ when doing the calculation myself, are you sure of this result? $\endgroup$
    – Slayer147
    Commented Aug 19, 2019 at 3:07
  • $\begingroup$ Maybe, you are right. But I have no time to check it now. $\endgroup$ Commented Aug 25, 2019 at 4:20

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 $\nabla \cdot B = 0$.

  • $\begingroup$ Your sum for $F_{\mu\nu}dx^\mu\wedge dx^\nu$ is over all indices or only $\mu<\nu$? $\endgroup$
    – Bellem
    Commented May 1, 2019 at 10:44
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    $\begingroup$ Using normal summation notation it's over all indices. The factor of 1/2 is meant to take care of double counting but people aren't always very consistent. $\endgroup$ Commented May 3, 2019 at 0:11
  • $\begingroup$ Ok then it makes sense to me. Thanks. $\endgroup$
    – Bellem
    Commented May 10, 2019 at 13:39

There is an answer to your question that seems overlooked and worth mentioning: Your intuition is exactly CORRECT.

The electric field intensity E is a 1-form and magnetic flux density B is a 2-form giving you

$\nabla\times E=-\dfrac{\partial B}{\partial t}$ and $\nabla \cdot B=0$

The excitation fields,displacement field D and magnetic field intensity H, constitute a 2-form and a 1-form respectively, rendering the remaining Maxwell's Equations:

$\nabla\times H=j+\dfrac{\partial D}{\partial t}$ and $\nabla \cdot D=0$

The four field descriptions D,H,E,B are geometrically distinct with the D and E being the Hodge dual to each other (same for H and B).

Though your question seems to demand a non-relativistic answer I recommend at some point understanding the fields in spacetime as described excellently by Cristoph et al. as it is quite satisfying seeing how it all comes together. Anyway, in answering an unrelated question for someone earlier I came across this link and I think you may find it useful: https://em.groups.et.byu.net/pdfs/ftext.pdf [Link: Electromagnetics: Richard H. Selfridge, David V. Arnold, and Karl F. Warnick, January 3, 2002]

  • $\begingroup$ Since space is 3d, one-forms and two-forms can be interchanged with no ambiguity, so we could just as well use a two-form to represent $\mathbf{E}$, for example. (Not recommended, but possible.) But that breaks down when we consider the generalization to $D$-dimensional space. For example, when $D=2$, the magnetic field $\mathbf{B}$ has only one component (because it's a two-form), but a one-form has two components. So if $\mathbf{H}$ is a one-form, where does its extra component come from? $\endgroup$ Commented Aug 16, 2020 at 18:11

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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