If we have a planar and harmonic EM wave, with $B$ field:

$$B=A\left(\begin{array}{c} 1\\ i\\0 \end{array} \right)e^{-i(\omega t-\vec k\cdot\vec r)}$$

and with it's corresponding $E$ field. This is a circularly polarised wave, but that field does not have 0 divergence, the three components of it, when taking real part, are:

$$x=\cos (\vec k\cdot\vec r-\omega t)$$ $$y=i\cdot i\sin(\vec k\cdot\vec r-\omega t)=-\sin(\vec k\cdot\vec r-\omega t)$$ $$z=0$$

Tha divergence won't be 0 unless $\vec k=(0,0,a)$, for some $a$. so what's the problem here? Isn't that a wave unless it spread on the $z$ axis? If the $E$ field is just the same but with different phase, I guess the same thing would have to hold if the wave had not matter around: $\nabla\cdot E =0$, right?


$\vec{E}$ has not different phase, but different polarisation : $\vec{B} = \vec{n}\times\vec{E}$

Yes, this solution cannot be a solution of Maxwell's equations for all $\vec{k}$, cause $\nabla{ \vec{E}} =0$ for $\vec{E} = \vec{E_0} e^{-i(\omega t - \vec{k}\vec{r})} $ implies $\vec{k} \vec{E_0} = 0$, so electromagnetic waves in the empty space are transverse. The profound reason of this is the gauge invariance of the EM field and, finally, the masslessness of photons.

  • $\begingroup$ Thank you. And, in this particular case, the conclusion of that is that for $E$ and $B$ to be a solutions of Maxwell's equations, that $k$ must have only $z$ component, because $E_0$ and $B_0$ are in the $x-y$ plane, and they're perpendicular. The fact that $\vec{B} = \vec{n}\times\vec{E}$ means that if $B$ makes circles, $E$ must make the too, right? So if they're asking me if those can be waves in the empty space, I should say that only if $k$ has only $z$ component, because otherwise they wouldn't be a solution to Maxwell's equation.? $\endgroup$ – MyUserIsThis Mar 17 '13 at 18:56
  • $\begingroup$ Yes, you're right. :) $\endgroup$ – const.maslov Mar 18 '13 at 15:48

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