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Assuming the presence of a dipole consisting of two oscillating point-like charged particles on the z axis. Lets say whose motion follows $q_{\pm}=\pm 1\pm \frac{1}{2}\cos(\omega t)$, it is a well known fact that there is no radiation emitted in the z direction. However, it is clear that there will be an oscillating potential in a far away source even in the z-direction as can be determined by Coulomb's law.

Am I right to interpret this as the longitudinal polarization of the Electric field? Is it not radiating because there's no magnetic counterpart? How should I wrap my mind around the fact that even near infinity there is a longitudinal electric field even though as is well known the polarizations of light are transverse?

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  • $\begingroup$ Is a single field line radiating? No. I don't believe so. Even if it had a meaningful energy density it would have zero volume and hence it couldn't transfer any energy. Does something that can't transfer any energy qualify as a wave? Probably not. Does something that is not a wave have a polarization in a meaningful sense? Is that what you are asking? $\endgroup$ Commented May 19, 2023 at 8:46

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If the oscillating dipole $\mathbf p = p\hat z$ it is not only $E_z=0$ but $\mathbf E = 0$ on the $z$ axis. This is because the two ends of the dipole are infinitesimally close to each other thus in the direction of the dipole moment there is no field, radiated or otherwise.

In general the asymptotic behavior of a radiated field of sources all residing in a finite volume satisfies Sommerfeld's radiation conditions according to which $$ \displaystyle \lim_{r \to \infty} r\left(\sqrt{\frac{\mu_0}{\epsilon_0}} \mathbf {\hat r \times H} +\mathbf E\right) =0\\ \displaystyle \lim_{r \to \infty} r\left(\sqrt{\frac{\epsilon_0}{\mu_0}} \mathbf {\hat r \times E} -\mathbf H\right) =0$$ which means that the longitudinal components that are along the direction $\mathbf {\hat r} $ go to zero faster than $\mathcal O(r^{-1})$ in the far field of the radiation while the transversal components must be at least $\mathcal O(r^{-1})$ asymptotically.

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  • $\begingroup$ I am not sure I agree with your first statement. $E_z$ is not zero on the z axis. I never assumed that the two particles that make the dipole are infinitesimally close to each other. There will always be some non-zero potential that is of course screened and therefore fall as ${\cal O}(r^{-2})$. Regarding your second paragraph, I agree but I am not sure how the longitudinal component (even if it falls as ${\cal O}(r^{-2})$) doesn't carry any radiation. $\endgroup$
    – pip
    Commented May 24, 2023 at 10:07
  • $\begingroup$ An ideal dipole is not two equal and opposite, one positive and one negative, charges at some finite distance. An ideal dipole is the limit of $\lim_{d\to 0, q \to \infty} qD = p$, this is just as an idealized elementary electrodynamic concept is as the concept of and idealized metal is whose conductivity $\sigma = \infty$. $\endgroup$
    – hyportnex
    Commented May 24, 2023 at 12:34
  • $\begingroup$ Without this interpretation, if you strictly follow your own question in which you wrote "Assuming the presence of a dipole consisting of two oscillating point-like charged particles on the z axis. Lets say whose motion follows $q_{\pm}=\pm 1\pm \frac{1}{2}\cos(\omega t)$" it could not be answered because there cannot be such charges oscillating. For starters, it is not a motion, and, more importantly unlike an oscillating ideal dipole would, it does not conserve charge locally, $\endgroup$
    – hyportnex
    Commented May 24, 2023 at 12:35
  • $\begingroup$ Of course, terms that asymptotically are $\mathcal r^{-2}$ or even $\mathcal r^{-3}$ do carry oscillating energy all the way to $\infty$ but they are insignificant and go to zero much faster compared to terms that are $\mathcal r^{-1}$, and $E_r \propto (1/r^2 +jk/r^3)\cos(\theta)p_0$, where $\mathbf p(t)=p_0\cos(\omega t)\hat z $ is the oscillating dipole. $\endgroup$
    – hyportnex
    Commented May 24, 2023 at 12:46

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