4 ms·
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
by orbifold 11y ago
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 11y agoAnd 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 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 11y agoThat'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!