It is well known that electromagnetic field is a 2form and Maxwell's equation can be reformulated in language of differential forms. What is the Poynting vector in this language?

6$\begingroup$ Seems like this would be a better question on Physics SE. $\endgroup$– WojowuDec 9 '20 at 17:06

5$\begingroup$ It seems ontopic here too. $\endgroup$– YCorDec 10 '20 at 10:35
Too long for a comment.
It is worth mentioning the controversy that arose about the expression of the Poynting vector $P$, whether the Abraham form $E\times H$ or the Minkowski form $D\times B$ is valid. The notations are that the electromagnetic field, a closed $2$form in Minkowski space, has coordinates $(E,B)$. The vectors $(D,H)$ are given by $$D=\frac{\partial L}{\partial E},\qquad H=\frac{\partial L}{\partial B}$$ where $L(E,B)$ is the density of the Lagrangian from which the Maxwell's equations derive. For instance, the standard equations follow from the choice $L=\frac12(E^2B^2)$ (here the light speed is set to $1$). Other choices are possible, like that of BornInfeld, which yield nonlinear models.
The amazing resolution of the controversy is that both expressions equal each other if, and only if $L$ is Lorentz invariant. In mathematical terms, this means that $L$ is a function of the quantities $$\frac12(E^2B^2),\qquad E\cdot B$$ only.
Going further, the Maxwell's equations imply that the following tensor is (rowwise) divergence free $$T=\begin{pmatrix} W & E\times H \\ D\times B & W_B\otimes BW_D\otimes D+\sigma I_3 \end{pmatrix}$$ where $$\sigma:=B\cdot W_B+D\cdot W_DW.$$ Hereabove $W:=D\cdot EL(E,B)$ is the energy density (a partial Legendre transform of $L$), and subscripts are differentials. That ${\rm Div}_{t,x}T=0$ expresses the conservation laws of the energy and of the Poynting vector. The important point is that $T$ is a symmetric tensor, and this is ensured by the Lorentz invariant.
Poynting vector is a 3d vector which can be thought of as a part of 4d stressenergy tensor $T$. In invariant language, you fix your reference frame by choosing a timelike 4velocity $u$. Consider $T(u)$: it gives you a flow of energymomentum. $T(u,u)$ is the flow of energy, and the vector $P=T(u)+T(u,u)u$ is the flow of momentum. By construction, $P(u)=0$ so you can think of it as 3d spacelike vector. So far, this applies to any stressenergy $T$. For Maxwell $T$, express it in terms of $F$: $T = F.F  1/4 g (F,F)$, where $g$ is metric and $F.F$ is a partial scalar product of $F$ with itself, with respect to the first index: this $T$ is traceless. You can use this to express $T(u)$ and $T(u,u)$ and $P$ in terms of $F(u)$ (aka the electric field) and $(^*\!F)(u)$ (aka the magnetic field). In the end you should find something like $P=u\wedge F(u)\wedge (^*\!F)(u)$.

$\begingroup$ Indeed, $P=u\wedge F(u)\wedge (*F)(u)$ is an amazing expression through exterior calculus operations only! The particular case when $u=(1,0,0,0)$, that is, $P=(1,0,0,0)\wedge(0,E_x,E_y,E_z)\wedge(0,B_x,B_y,B_z)$ is even clearer. Correcting a possible typo: $T(u,u)$ is energy density, and $T(u)+T(u,u)u$ is energy flux, while the momentum flux is a tensor rather than a vector (see en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor for the particular case $u=(1,0,0,0)$). $\endgroup$ Dec 14 '20 at 20:04