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?
