What Does a Supernova Look Like?
As a supernova blastwave collides with the interstellar medium, it accelerates charged particles behind the shock which then release synchrotron radiation as they move through a magnetic field. This emission $j_\nu$ occurs at all frequencies ($\nu$, or 'nu') according to a power law.
As this light (radiation) travels toward us and passes through the surrounding interstellar medium (ISM), some is absorbed and re-emitted in a different direction. This absorption process is very frequency dependent, and so light from the supernova may look very different depending on which frequency (and angle!) it is viewed.
The below interactive figure calculates this process in real time, so you can adjust both frequency and viewing angle to see how the emission we receive changes. By default, both a thick circumstellar disk and the interior of the supernova are shown in blue. Clicking the 'Toggle ISM' button will remove this, showing how the light is obscured by the ISM.
Frequency Adjuster : nu = 10^9.5
The above figure displays total intensity $I$ at each point as a function of frequency $\nu$, found by integrating along the line of site: $$ I(\nu) = \int e^{-\tau_\nu} j_\nu\ d\ell $$ Each point behind the shock has an emission spectrum $j_\nu$ of $$ j_\nu = c_j(s) K B^{(s+1)/2} \nu^{(1-s)/2}, $$ and optical depth as a function of distance is $$ \tau_\nu (\ell) = \int_0^\ell \kappa_\nu\ dr, $$ where $\kappa_\nu$ is given by $$ \kappa_\nu = 0.212 n_e^2 T^{-1.35} \nu^{-2.1}\ \text{cm}^{-1}, $$ and $K$ (from $j_\nu$ equation) is given by $$ K = K_0 (E/E_0)^{s-2} (\rho/\rho_0)^{4/3}. $$ We currently assume $E/E_0$ is constant, but that can easily be added in.
The actual calculation being performed for each pixel is $$ I(\nu) = \sum_\ell \Delta I_\nu = \sum_\ell e^{-\tau_\nu(\ell-\Delta \ell)} j_\nu(\ell)\ \Delta \ell, $$ where $$ \tau_\nu(\ell) = \sum_\ell \kappa_\nu(\ell)\ \Delta \ell. $$
