It is well known that electromagnetic field is a 2-form 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. $\endgroup$
    – Wojowu
    Dec 9 '20 at 17:06
  • 5
    $\begingroup$ It seems on-topic here too. $\endgroup$
    – YCor
    Dec 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 electro-magnetic 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 $L=\frac12(|E|^2-|B|^2)$ (here the light speed is set to $1$). Other choices are possible, like that of Born-Infeld, which yield non-linear 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 $$\frac12(|E|^2-|B|^2),\qquad E\cdot B$$ only.

Going further, the Maxwell's equations imply that the following tensor is (row-wise) divergence free $$T=\begin{pmatrix} W & E\times H \\ D\times B & -W_B\otimes B-W_D\otimes D+\sigma I_3 \end{pmatrix}$$ where $$\sigma:=B\cdot W_B+D\cdot W_D-W.$$ Hereabove $W:=D\cdot E-L(E,B)$ 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 stress-energy tensor $T$. In invariant language, you fix your reference frame by choosing a time-like 4-velocity $u$. Consider $T(u)$: it gives you a flow of energy-momentum. $T(u,u)$ 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 space-like vector. So far, this applies to any stress-energy $T$. 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

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .