How do I translate between spectral radiometric quantities and electromagnetic ones? In the study of electricity and magnetism we learn that light is just a series of waves in the electromagnetic field. We learn about the Poynting vector
$$\mathbf{S}=\mathbf{E}\times\mathbf{H},$$ 
the magnitude of which is also identified as the intensity. 
Meanwhile, in radiometry there are many different quantities that are defined as being spectral densities, and I'm having trouble relating them to the fundamental electromagnetic fields. I'm particularly interested in relating spectral radiance and/or spectral irradiance back to the field quantities.
Are there relationships for going back and forth between spectral radiance and fields?
 A: Here is a preliminary partial answer that covers a solvable case that may be expandable to cover the general case.
The key to making the transition from field theoretic to radiometric quantities is to start from a simple case with known answers on both sides and generalize from there. The known case is the linearly polarized plane-wave with
\begin{align}
  \mathbf{E} &= \mathbf{E}_0e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t}\ \mathrm{and}\\
  \mathbf{B} &= \frac{\hat{k}\times\mathbf{E}_0}{c}e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t}.
\end{align}
For that case the time averaged Poynting vector is
\begin{align}
  \langle \mathbf{S}\rangle &= \frac{1}{2\mu_0} \mathbf{E}^*\times\mathbf{B} \\
   &= \frac{\mathbf{E}_0\times\left(\hat{k}\times\mathbf{E}_0\right)}{2\mu_0c}.\tag{1}
\end{align}
For that same linearly polarized plane wave the spectral radiance is given by
$$I_\nu(\nu,\theta,\phi) = I_0 \delta\left(\nu-c\frac{k}{2\pi}\right)\frac{\delta(\theta-\theta_k)\,\delta(\phi-\phi_k)}{\sin\theta} \tag{2}$$
where the delta functions concentrate the spectral radiance on the direction and magnitude of the wave vector $\mathbf{k}= k\left(\sin\theta_k \cos\phi_k \hat{x} + \sin\theta_k\sin\phi_k\hat{y} + \cos\theta_k\hat{z}\right)$. Because the net energy flux is given by $\frac{\operatorname{d}E}{\operatorname{d}A\operatorname{d}t} = \int I_\nu \cos\theta \operatorname{d}\nu\operatorname{d}\Omega$ we can derive that
\begin{align}
  I_0 &= \frac{E_0^2}{2c\mu_0}.
\end{align}
The obvious next step would be to spread out the delta functions in equation (2), but we first have to consider when that makes sense in the field context. See, superposition is done at the level of the fields, not the fluxes. (to be continued...)
