[orbifold]

Actually you can unify E and B into a closed two-form F and get in vacuum

dF = 0

d★F = ★j

Where j = rho dt + jx dx + jy dy + jz dz is the charge and current density, ★ is the Hodgestar operator and d is the exterior derivative. The equations in a medium are slighly less elegant.

[ngoldbaum]

And if you write the electromagnetic field in terms of the 4-vector potential, it's just \Box{A_μ} = 0 (where Box is the D'Alembertian operator). Then again, this doesn't really mean anything without the context of what the notation means...

Feynman has a discussion about this in Volume 2 of the Feynman Lectures: http://www.feynmanlectures.caltech.edu/II_25.html

(I still can't get over how awesome it is that I can deep-link into the Feynman Lectures!!)

[viola11]

That's a charge that applies equally well to the standard formulation and I made this comment just to say that it IS cool you can deeplink the Feynman lectures!