So I've been trying to convert from Maxwell's equations in terms of differential forms to the integral versions of Maxwell's equations that we know from vector calculus.

We have, in vector calculus

$$\left\{\begin{align} \nabla\cdot\mathbf{E}&=~\rho\\ \nabla\times\mathbf B~&=~\mathbf J+\frac{\partial\mathbf E}{\partial t}\\ \nabla\times\mathbf E~&=-\frac{\partial\mathbf B}{\partial t}\\ \nabla\cdot\mathbf{B}~&=~0 \end{align}\right.$$

Which equates to

$$ F = E_i\,{\rm d}t\wedge{\rm d}x^i + \star B_i\,{\rm d}t\wedge{\rm d}x^i\\ J = \star \left( \rho \, {\rm d} t + J_i \,{\rm d} x^i \right)\\ d\star F = J \\ dF = 0 $$

in differential geometry, with implied sum over the $i$ index.

Stoke's theorem in differential geometry tells us that

$$ \int_M \mathrm{d} \omega = \int_{\partial M} \omega $$

For a given $n-1$ form $\omega$ on manifold $M$ of dimension $n$, with boundary $\partial M$.

So our integral Maxwell's equations in differential geometry are

$$ \int_{\partial M} \star F = \int_M J \\ \int_{\partial M} F = 0 $$

Reducing this expression back to vector calculus for the divergence equations is then an exercise in choosing the correct surfaces. Take the domain

$$ D_{t_0} = \left\{ \left(\begin{array}{c} t_0 \\ x \\ y \\ z\end{array}\right) \in \mathbb{R}^4 \;\;\Bigg| \;\;\left(\begin{array}{c} x \\ y \\ z\end{array}\right) \in D \subset \mathbb{R}^3 \right\} $$

ie. a domain $D$ in $\mathbb{R}^3$, at a chosen time $t_0$. This is a 3 dimensional manifold.

Let $\imath$ be the inclusion mapping of $D_{t_0}$ in Minkowski spacetime, and then $\imath^*F$ and $\imath^* \star F$ are both 2 forms on a 3 dimensional manifold, and $\imath^*J$ is a 3 form on a 3 dimensional manifold. Now we can apply our integral versions of Maxwell's equations.

Calculating the pull-backs gives us that $\imath^*{\rm d} t = 0$, $\imath^*{\rm d} x^i = {\rm d} x^i$, so

$$ \imath^*F = \star B_i\,{\rm d}x^i\\ \imath^* \star F = \star E_i\,{\rm d}x^i\\ \imath^* J = \rho \; {\rm d} x \wedge {\rm d} y \wedge {\rm d} z $$

Then the homogeneous divergence equation gives us (with an implied pull-back by the inclusion mapping of $\partial D$ in $D$)

$$ \int_{\partial D} \imath^* F = \int_{\partial D} \star B_i\,{\rm d}x^i = \int_{\partial D} \mathbf{B} \cdot {\rm d} \mathbf{A} = 0 $$

And the inhomogeneous divergence equation;

$$ \int_{\partial D} \imath^* \star F = \int_{\partial D} \star E_i\,{\rm d}x^i = \int_{\partial D} \mathbf{E} \cdot {\rm d} \mathbf{A} = \int_D \imath^* J = \int_D \rho \; {\rm d} x \wedge {\rm d} y \wedge {\rm d} z = \int_D \rho \,\rm{d} V = Q $$

And so we recover two of the integral versions of Maxwell's equations.

I am, however, a bit lost recovering the two remaining integral versions of Maxwell's equations; those that involve the line integrals around closed loops. Because these equations still involve time derivatives, we've only actually applied Stokes' theorem in 3D and not in 4D space. I'm unsure about how to deal with this, I was thinking about maybe taking a domain of integration something like

$$ D_{x_0} = \left\{ \left(\begin{array}{c} t \\ x_0 \\ y \\ z\end{array}\right) \in \mathbb{R}^4 \;\; \Bigg| \;\;\left(\begin{array}{c} y \\ z\end{array}\right) \in A \subset \mathbb{R}^2, t \in [t_0, t_0 + h] \right\} $$

(Obviously this could be done for a surface oriented in a general direction rather than in the $y-z$ plane, but let's make the maths more simple by Lorentz invariance) And then taking a limit as $h \rightarrow 0$ to recover the time derivative, but it seems a bit of a messy way to do it really, and I also can't work out how the boundaries of this set match up with the 4D Stokes' theorem and things. If anyone could show me how to do it like this that would be great?

My other thought was about pulling back to three dimensional space before applying Stokes' theorem, but I feel like this is treating the equations a bit un-symmetrically and so there should be some way to keep the 4D Stokes theorem and not add another time derivative, and still reproduce the integral forms of the equations. So does anyone have any better ideas?

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    $\begingroup$ If you like this question you may also enjoy reading this article. $\endgroup$
    – Qmechanic
    Commented Sep 27, 2017 at 12:40
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    $\begingroup$ @Qmechanic that's a lovely article. Thanks so much for alerting us to it. Also, a little older and not so polished is this one or this one (I've only looked over the latter one). $\endgroup$ Commented Oct 10, 2017 at 7:31

1 Answer 1


I think your limiting procedure should work.

For the homogeneous Maxwell equations, when we restrict to a surface of constant $x$, the pull back of the field strength looks something like

$$\iota^* F = E^\perp_i dt \wedge dx^i + B_x dy \wedge dz,$$

where $E^\perp$ is the component of $E$ in the $yz$-plane. Now your timelike pillbox integral will have three contributions, one from the sides of the cylinder, and two from the top and bottom. The tangent vectors of the sides are $\partial_t$ and $v^\mu$, the tangent vector of the curve bounding the surface in the $yz$-plane. So the side contribution gives

$$\int dt\oint \mathbf{E}\cdot\mathbf{v}ds,$$

where $s$ is a parameter on the curve. The top and bottom of the integral comes entirely from the $B_x dy \wedge dz$ term, and is just the usual flux of $B$ through the surface. Since the total contribution is zero, we have

$$\int_t^{t+h} dt \oint \mathbf{E}\cdot\mathbf{v}ds + \int_{t+h} \mathbf B\cdot d\mathbf{A} - \int_{t} \mathbf B\cdot d\mathbf{A} = 0$$

Now if you want to take the limit as $h\rightarrow 0$, you can Taylor expand each term and look at the first order in $h$ terms. The first integral just becomes the line integral by the Fundamental Theorem of Calculus. The $\mathbf{B}$ flux integrals can also be seen straightforwardly to approach the time derivative of the flux at first order in $h$. Thus the result is

$$\oint \mathbf{E}\cdot \mathbf{v} ds + \frac{\partial}{\partial t}\int_t \mathbf{B}\cdot d\mathbf{A} = 0$$

Applying basically the same procedure to the dual field strength gives

$$\oint \mathbf{B}\cdot \mathbf{v} ds -\frac{\partial}{\partial t}\int_t \mathbf{E}\cdot d \mathbf{A} = \int_t \mathbf{j}\cdot d\mathbf{A}.$$


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