What is the radiation pattern for an electric dipole rotating in the horizontal plane?

Question

In trying to answer $\sigma$ transition and angular momentum conservation the following question arose.

Suppose we have an electric dipole $\boldsymbol{d}$ which is oriented and rotating within the horizontal ($xy$) plane. i.e. initially $\boldsymbol{d}$ points along $\boldsymbol{x}$, and then a quarter period later it points along $\boldsymbol{y}$.** What will be the resultant radiation pattern? A complete answer would give something like:

$\boldsymbol{E}(r, \theta, \phi) = |E(r, \theta, \phi)|e^{i\Phi(r, \theta, \phi)} \boldsymbol{\epsilon}(r, \theta, \phi)$

That is, it will give the amplitude, phase, and unit polarization vector as a function of the 3 polar coordinates, $(r, \theta, \phi)$.

My guess is that $\Phi$ and $\boldsymbol{\epsilon}$ are independent of $r$, though I'm not actually 100% certain about this. I could imagine the light is emitted such that it seems like it emanates from a point slightly off the z-axis, in which case I think $\Phi$ and $\boldsymbol{\epsilon}$ have a non-trivial dependence on $r$.. I think this would be related to the presence of non-zero orbital angular momentum.

** The case which is usually treated is a dipole oriented along $\boldsymbol{z}$ and oscillating along this axis. This corresponds to a $m=0$ type oscillation while I am interested in $m=\pm1$ type oscillations.

Answer 1

The fields can be treated as the sum of the fields from two oscillating electric dipoles of magnitude $d$ that are $\pi/2$ out of phase, with one dipole pointing along the x-axis and one pointing along the y-axis. i.e. $$\vec{d_{\rm tot}} = d\cos(\omega t) \hat{x} + d \sin(\omega t)\hat{y},$$ where $\omega$ is the angular velocity.

I think then that you have (radiation) E-fields of the form $$\vec{E} = \frac{\omega^2 d}{c^2r}\left(\cos(\theta)\cos(\phi) \hat{\theta} - \sin(\phi) \hat{\phi}\right)\cos[\omega(t - r/c)] - \frac{\omega^2 d}{c^2r}\left(\cos(\theta) \sin(\phi) \hat{\theta} + \cos(\phi) \hat{\phi}\right)\sin[\omega(t- r/c)]\ . $$

The light is circularly polarised along the z-axis and linearly polarised along the x- and y-axes.

Answer 2

The far-field radiation field from an oscillating dipole of the form:

\begin{align} \boldsymbol{d} = \boldsymbol{d}_0 e^{-i\omega t} \end{align}

is given by

\begin{align} \boldsymbol{E}(\boldsymbol{r}, t) = \frac{\omega^2}{4\pi \epsilon_0 c^2} \frac{e^{i\omega r/c}}{r} \left(\hat{\boldsymbol{r}}\times \boldsymbol{d}\right)\times\hat{\boldsymbol{r}} e^{-i\omega t} \end{align} Source: Wikipedia: Dipole Radiation

A dipole rotating in the $xy$ plane as described in the question is given by $\boldsymbol{d_0} = \frac{1}{\sqrt{2}}(\hat{x}+i\hat{y})$.

Plotting the real part of this vector field on a surface of constant amplitude gives:

We can see that light propagating upwards along the $z$ axis is right hand circularly polarized, light traveling downwards is left hand circularly polarized and light propagating horizontally is linearly polarized in the $xy$ plane. Of note is that the light propagating in the $x$ direction is $\pi/2$ out of phase with light propagating in the $y$ direction. This gives the field the sense of rotation at all locations.

In the vertical direction the light has angular momentum due to the rotation of the polarization vector. This is spin angular momentum. In the horizontal direction the light has angular momentum due to the global rotating structure of the field. This is orbital angular momentum.