Timeline for Do Maxwell's Equations overdetermine the electric and magnetic fields?

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Jan 13, 2017 at 17:21	comment	added	Qmechanic♦		1. No, there is not. 2. I assume you are referring to Section III: The massless case $k^2=0$ is one possibility. There's also a massive case $k^2\neq 0$.
Jan 13, 2017 at 17:12	comment	added	image357		1. But there is no actual constraint that tells me that they have to be transversal? 2. Also: What is the argument for $k^2 = 0$ if not the Lorenz-gauge Maxwell equs. $\Box A^\mu = 0$?
Jan 13, 2017 at 17:10	comment	added	Qmechanic♦		It may be viewed as part of the definition of what transversal means.
Jan 13, 2017 at 17:06	comment	added	image357		Why do the transversal modes satisfy $k_{\mu}^{\eta}~\tilde{A}^{\mu}_{T}(k)~=~0$? Maxwell equs. demand $k_{\mu}~\tilde{A}^{\mu}_{T}(k)~=~0$. Where does this extra constraint come from?
Mar 1, 2012 at 15:55	history	edited	Qmechanic♦	CC BY-SA 3.0	corrected minor typos.
Feb 7, 2012 at 11:45	history	edited	Qmechanic♦	CC BY-SA 3.0	Added a bit more explanation.
Feb 5, 2012 at 17:43	history	answered	Qmechanic♦	CC BY-SA 3.0