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The mixed Faraday tensor $F^\mu{}_\nu$ explicitly in natural units is:
$$(F^\mu{}_\nu)=\left(\begin{array}{cccc}0&E_x&E_y&E_z\\E_x&0&B_z&-B_y\\E_y&-B_z&0&B_x\\E_z&B_y&-B_x&0\end{array}\right)$$
and thus has the same form as a general element of the Lorentz algebra $\mathfrak{so}(1,\,3)$. Is there a physical interpretation of this or is it co-incidence?
Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$):
$$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$
Now let's think of a particle's four-velocity evolving on the particle's world line. If the time evolution is linear then $v(\tau) = G(\tau)\,v(0)$ for some proper-time-varying transformation matrix $G(\tau)\in GL(4,\,\mathbb{R})$.
But now recall that a four-velocity's pseudo-norm is always constant: $\langle v,\,v\rangle = v_\nu\,v^\nu = c^2$. Therefore, $G(\tau)$ must conserve this norm; in other words, $G(\tau) \in SO(1,\,3)$ the group of matrices that do so (it can't be in the four-coset $O(1,\,3)$ outside $SO(1,\,3)$ if $G(\tau)$ varies continuously, since $G(0)=\mathrm{id}$, but this is an aside). Therefore, we must have:
$$\frac{\mathrm{d} v}{\mathrm{d}\tau} = \dot{G}(\tau) v(0) = \dot{G}(\tau) \,G(\tau)^{-1} v(\tau) = F(\tau)\, v(\tau)\tag{2}$$
where $F(\tau) = \dot{G}(\tau) \,G(\tau)^{-1}$ belongs to the Lie algebra of $SO(1,\,3)$. (2) then is the form that (1) must take, for any four-force that depends homogeneously and linearly on the four-velocity alone. So the structure of the mixed Faraday tensor is precisely defined by the fact that four velocities have constant norm and the notion is general to any situation of the kind just described, not only the Lorentz force law.
If we now lower the index $\mu$ in (1) to have the covariant velocity on the left hand side of (1), the Faraday tensor is skew-symmetric, so that it can be a two form and thus fulfil $d\,F=0$ to resume the Great Faraday's thought picture that tubes of $F$ never end.