How can I express the Electromagnetic tensor in terms of multilinear map action?

I know that if $F$ is the Electromagnetic Tensor then we can write F as:

$$ \textbf{F} = F_{\alpha \beta} \textbf{e}^{\alpha}\otimes\textbf{e}^{\beta} $$

But I want to know that notion here: $$ [\textbf{F}](?,?) = [F_{\alpha \beta} \textbf{e}^{\alpha}\otimes\textbf{e}^{\beta}](?,?) $$

Ok. If we have $[F_{\alpha \beta} \textbf{e}^{\alpha}\otimes\textbf{e}^{\beta}](\textbf{E},\textbf{B})$, then:

$$[F_{\alpha \beta} \textbf{e}^{\alpha}\otimes\textbf{e}^{\beta}](\textbf{E},\textbf{B}) = [F_{\alpha \beta} \textbf{e}^{\alpha}\otimes\textbf{e}^{\beta}](E^{i}\textbf{e}_{i},B^{j}\textbf{e}_{j}) = F_{\alpha \beta}E^{i}B^{j} [\textbf{e}^{\alpha}\otimes\textbf{e}^{\beta}](\textbf{e}_{i},\textbf{e}_{j}) = F_{\alpha \beta}E^{i}B^{j}\textbf{e}^{\alpha}(\textbf{e}_{i})\textbf{e}^{\beta}(\textbf{e}_{j}) = F_{\alpha \beta}E^{i}B^{j}\delta^{\alpha} _{i}\delta^{\beta}_ {j} = F_{\alpha \beta}E^{\alpha}B^{\beta}$$


1 Answer 1


Yes, the electromagnetic tensor $F$ is a functional which is bilinear and antisymmetrical in its arguments. Moreover this functional is defined pointwise on a manifold (can be $\mathbb{R}^3$ or $\mathbb{R}^4$) and is also shortly called 2-form. The arguments are actually tangent vectors (more see below) of the tangent space of the considered point (The tangent space changes at each point). If it did not depend on the point of the considered manifold, it would only be a bilinear form, a rather simple notion of linear algebra. Typically one would evaluate (apply the EM tensor) F on tangent vectors of a surface, and eventually add up the results along all points of the surface, in that case one gets a well-known integral $\int \bf{B} dA$ (the magnetic flux) or for instance if added up along a line in $\mathbb{R}^3$: $\int \bf{E} d\bf{r}\cdot d t $, i.e. the abstract notion of pointwise defined bilinear form (shortly 2-form) gets a rather intuitive sense.

To make it even more clear in the notation you used one would consider $e^{\alpha}$ and $e^{\beta}$ as co-vectors as elements of the cotangential spaces dual to the corresponding tangential spaces (for each point of the manifold there is another one). The co-vector property is expressed in the relations $e^\alpha(e_\gamma)=\delta^\alpha_\gamma$, where $\delta^i_j$ is the Kronecker symbol.

In the formalism of differential forms one would write these co-vectors as: $e^\alpha = du^\alpha$ ($\alpha =0,1,2,3$ in Minkowski space) with $u=(t,x,y,z)$ for instance (other defintions are possible). Then the whole EM tensor $F$ would be writtten in 4-dim. space at a point $u=(t,x,y,z)$):

$F(u) = E_x(u) dt \wedge dx + E_y(u) dt \wedge dy + E_z(u) dt \wedge dz + B_x(u) dy \wedge dz + B_y(u) dz \wedge dx + B_z(u) dx \wedge dy$

In this formula the antisymmetric $\wedge$ product was used with is defined like $u \wedge v = \frac{1}{2} (u\otimes v - v\otimes u)$ (I would not consider the product defined by $\otimes$ as symmetrical, just as without any symmetry).

If this expression gets applied on a flat surface (for simplicity) in the x-y plane one would use $(e_x, e_y) = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$ are the tangent vectors of the x-y plane. Actually on manifolds tangential vectors are defined as directional derivatives. A directional derivative along the x-vector is just the partial derivative in x-direction, shortly $\frac{\partial}{\partial x}$. The more general case should be intuitively rather clear.

Then the flux given by $F$ through a little surface element $dx dy$ is $F$ applied on the tangential vectors of this surface element:

$F (u)(e_x, e_y) = F(u)(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) = E_x(u) dt \wedge dx (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) + E_y(u) dt \wedge dy (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})+ E_z(u) dt \wedge dz (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) + B_x(u)dy \wedge dz (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) + B_y(u) dz \wedge dx (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) + B_z(u) dx \wedge dy(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$

Using the properties of vector with respect to co-vectors: $ dx (\frac{\partial}{\partial x})=1$ but $dt (\frac{\partial}{\partial x})=0$ and $dy (\frac{\partial}{\partial x})=0$ and $dz (\frac{\partial}{\partial x})=0$ and $dt (\frac{\partial}{\partial y})=0$ and $dx (\frac{\partial}{\partial y})=0$ and $dy (\frac{\partial}{\partial y})=1$ and $dz (\frac{\partial}{\partial y})=0$ one would finally get: $F_u (e_x, e_y) = B_z(u=(t,x,y,z))$.

Keep in mind that the components of $F$ already represent components of $\bf{E}$ and $\bf{B}$, the evaluation is carried out on tangential vectors of a surface, or on tangential vectors of a line, but not on $\bf{E}$ and $\bf{B}$, i.e. $\bf{E}$ and $\bf{B}$ cannot be used (it would not make sense) as arguments in the 2-form $F$ as apparently suggested.

Actually, if the concepts around differential forms appear too abstract, you can still imagine the tangential vectors of each point of the surface as little arrows and imagine $e^{\alpha}$ and $e^{\beta}$ as vectors of the dual space, but sooner or later it is useful to also embrace the analytical aspect of $F$ as a differential form which serves above all to be integrated over. To keep it simple I will not add the formula how to integrate differential forms, but it can be found easily on wikipedia.


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