6 ms·
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, t
by ngoldbaum 11y ago
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 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!