Does an em-wave in far field approximation always have to be linearily polarized if the source is approximately one-dimensional? For an em-wave generated by an arbitrary source the fields in far field approximation and compact source, are approximately (following Griffiths):
$$
\vec{E}(\vec{r},t) \approx \frac{\mu_0}{4\pi r} \left[\hat{r} \times \left(\hat{r} \times \ddot{\vec{p}}\left(t - \frac{r}{c} \right)\right) \right ]
$$
and
$$
\vec{B}(\vec{r},t) \approx -\frac{\mu_0}{4\pi r c} \left [\hat{r} \times \ddot{\vec{p}}\left(t - \frac{r}{c}\right) \right]
$$
Where 
$$
\vec{p}(t_0) = \int \vec{r}' \rho(\vec{r}',t_0) \mathrm{d Vol}'
$$
From this it seems to be clear, that $\vec{E}$ and $\vec{B}$ are in phase and perpendicular do each other. 
However it seems that both may not have a particular polarization. Is it correct that from those formulas follows that the fields have to be linearily polarized if the source is approximately 1 dimensional. 
Then the only missing parts to be a simple plain wave are that it doesn't need to be harmonic since the time dependency of the charge $\rho$ density is arbitrary and that the amplitude is not constant because of the $\frac{1}{r}$. So one may further say that in far field approximation and over a small region in space an EM-wave generated by an approximately one dimensional source is approximately a plane wave if any only if the source is harmonic.
 A: The polarization of an EM wave can be elucidated a bit by applying a quick vector identity:
$$ 
\vec{E}(\vec{r},t) \approx \frac{\mu_0}{4\pi r} \left[\hat{r} \times \left(\hat{r} \times \ddot{\vec{p}} \right) \right ] = \frac{\mu_0}{4\pi r} \left[ \ddot{\vec{p}} - \hat{r} (\ddot{\vec{p}} \cdot \hat{r}) \right]
$$
(I have omitted the explicit time-dependence of $\ddot{p}$ here, but it should still be understood to be time-dependent.)
Now, think about what the vector in square brackets is:  


*

*$\ddot{\vec{p}} \cdot{r}$ is the radial component of $\ddot{\vec{p}}$,

*which means $\hat{r} (\ddot{\vec{p}} \cdot{r})$ is the projection of $\ddot{\vec{p}}$ in the radial direction,

*which means that $\ddot{\vec{p}} - \hat{r} (\ddot{\vec{p}} \cdot{r})$ is just $\ddot{\vec{p}}$ minus its component in the radial direction—i.e., the projection of $\ddot{\vec{p}}$ orthogonal to $\hat{r}$.


In other words, the polarization of the EM waves from a compact electric dipole source is the projection of $\ddot{\vec{p}}$ perpendicular to the line of sight to the source.
From this, we can see that if $\vec{p}$ always points in the same direction and only varies in magnitude (i.e., $\vec{p} = p(t) \hat{n}$, where $\hat{n}$ is a constant unit vector), then the electric field in the far-field radiation will always be parallel to the projection of $\hat{n}$ perpendicular to $\hat{r}$.  Since this projection vector would be constant at a given location, the resulting EM wave will be linearly polarized along this axis at this location.  In particular, if the source is effectively 1-D, the oscillating charges will only move along one axis, $\vec{p}$ will be of the above form, and the resulting far-field wave will be linearly polarized.
Note that the converse is not true:  one can have 2-D motion and still get linear polarization in a particular direction, since only the acceleration of $\vec{p}$ matters and since other components might get "projected out".  An example of this would be an undulator, which produces highly polarized synchrotron radiation along the beam direction.
