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My question is: should any of the field lines in this depiction have components along the line of sight between the radiator and the field point? As they do.

I was inspired by this question Electromagnetic Fields of light waves. In his question, student509 shared this depiction of near field EM radiation from a dipole radiator.

Felder um Dipol

Cribbed from wikipedia:Near and far field.

The equation which I am familiar with for the electric portion of EM radiation is this:

In the Feynman Lectures, Volume I, 28-1 Feynman promulgates this expression for the electric field due to a moving charge. My notation differs from the original.

$$\mathfrak{E}=-\frac{q}{4\pi\varepsilon_{o}}\left(\frac{\hat{\rho}}{\rho^{2}}+\frac{\rho}{c}\frac{d}{dt}\left[\frac{\hat{\rho}}{\rho^{2}}\right]+\frac{1}{c^{2}}\frac{d^{2}\hat{\rho}}{dt^{2}}\right).$$

Here $\rho$ represents the retarded position of the source; and $\hat{\rho}$ is the corresponding unit vector pointing from the field point toward the source's previous location.

The equation Feynman provides has three terms. The first two are "near-field" terms which fall off as $\frac{1}{\rho^2}$. They are both Coulombic. The first is the standard Coulombic field strength produced by the charge when it was located at $\vec{\rho}$.

$$\mathfrak{E}_{1}=-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}.$$

The second is the time derivative of the first term, multiplied by the time delay $\frac{\rho}{c}=\Delta t$,

$$\mathfrak{E}_{2}=\frac{d}{dt}\left[-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}\right]\Delta t.$$

In the case of a neutrally charged unexcited radiator, the Coulombic field will be $\vec{0}$. That suggests to me that the first term will be of zero magnitude. And since the second term is the time derivative of the first, it too should be zero.

The contribution of the first term points radially along the line of sight between the source and the field point. The second term has two components:

$$\mathfrak{E}_{2}=-\frac{q}{4\pi\varepsilon_{o}c}\rho\left(\frac{1}{\rho^{2}}\frac{d\hat{\rho}}{dt}-\frac{2}{\rho^{3}}\hat{\rho}\right)$$ One is parallel to the line of sight. The other is perpendicular to the line of sight, and lies in the plane spanned by the axle of the transmitter and $\hat{\rho}$.

The third term lies in the plane spanned by the axle of the transmitter and $\hat{\rho}$.

If the depiction is correct, then how should that be understood in terms of Feynman's formula?

Edit to add a depiction of the curl of the velocity field of the laminar flow of a Newtonian fluid, in order to show that the curl of a vector field can be non-vanishing, and yet all field vectors are parallel.

enter image description here

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  • $\begingroup$ Note one of the Maxwell's equations: $\nabla\times \vec E=-\partial_t \vec B$. You do know that $\vec B$ changes in time, thus it shouldn't surprise you that $\vec E$ field lines form closed loops: $\vec E$ has nonzero curl. $\endgroup$ – Ruslan Jan 25 '18 at 19:06
  • $\begingroup$ Is this correct? "In previous chapters we have not been concerned with the direction of oscillation of the electric field, except to note that the electric vector lies in a plane perpendicular to the direction of propagation." feynmanlectures.caltech.edu/I_33.html#Ch33-S1 $\endgroup$ – Steven Thomas Hatton Jan 25 '18 at 19:17
  • $\begingroup$ A non-vanishing curl does not necessarily imply the field has vectors pointing in all directions in the plane normal to the curl. Consider the example of laminar flow of a Newtonian fluid. The curl man not vanish, and yet all vectors of the velocity field are parallel. $\endgroup$ – Steven Thomas Hatton Jan 25 '18 at 19:46

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