Mean angular power
= Mean angular power
Since $\mathbf E\cdot\mathbf n=0$, the <Poynting vector> is
$$
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B
=\epsilon_0c|\mathbf E|^2\mathbf n.
$$
The area on a distant sphere corresponding to <solid angle> $d\Omega$ is $r^2d\Omega$. Consequently
$$
\boxed{\frac{d\langle P\rangle}{d\Omega}
=\frac{q^2}{16\pi^2\epsilon_0c^3}
\left\langle |\mathbf a|^2-(\mathbf n\cdot\mathbf a)^2\right\rangle
=\frac{q^2}{16\pi^2\epsilon_0c}
\left\langle|\mathbf n\times\dot{\boldsymbol\beta}|^2\right\rangle.}
$$
The average is over the source motion at <retarded time>. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are <power> per <solid angle>, $\mathrm{W\,sr^{-1}}$.
The <acceleration covariance in nonrelativistic radiation> is another way to organize this result: with $C_{ij}=\langle a_i a_j\rangle$, the angular factor is $\operatorname{tr}C-\mathbf n^TC\mathbf n$. Integrating over all observation directions uses $\int[|\mathbf a|^2-(\mathbf n\cdot\mathbf a)^2]d\Omega=(8\pi/3)|\mathbf a|^2$ and gives the averaged <Larmor formula>, $\langle P\rangle=q^2\langle a^2\rangle/(6\pi\epsilon_0c^3)$.