# Why is the Mixed Faraday Tensor a matrix in the algebra so(1,3)?

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?

• It is antisymmetric in its lower indices, therefore the mixed tensor lies in the adjoint representation of the Lorentz group. – Prof. Legolasov Sep 1 '15 at 16:46
• @Hindsight Indeed, the "algebra" of mixed Faraday tensors has to be invariant under the adjoint representation, and this explains how the symmetry, once gotten, is true in all frames. But is this not a little circular? What stops the algebra being a superset of $\mathfrak{so}(1,\,3)$? So I guess the question becomes what is the physical meaning of the skew symmetry of $F_{\mu\,\nu}$? Of course the way it is defined in terms of the potential makes it so, but I think (as I show in my answer) we can explain this right at the outset when we think of $F$ as an encoding for the effect of the .... – Selene Routley Sep 1 '15 at 23:45
• @Hindsight ... the EM field on charge, even before we propose field equations and their implied symmetry. Or am I missing something more basic that you're driving at? – Selene Routley Sep 1 '15 at 23:46
• What stops the algebra from being a superset of so(1, 3) - did you mean a subset? because otherwise the answer is component count. – Prof. Legolasov Sep 2 '15 at 1:17
• @Hindsight Thanks for perservering. I have this uneasy feeling I am missing something really trivial here. What I meant was $\mathfrak{gl}(4,\,\mathbb{R})$ with $\mathfrak{so}(1,\,3)$ as a subspace invariant under the adjoint representation. As far as I can tell, number of components comes from $F$'s nature as a two form, which in turn comes from $F_{\mu\,\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. Is there another way to see this? – Selene Routley Sep 2 '15 at 3:49

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.