Using the numbers we measure for $L(H\alpha)$, I'll re-derive key quantities about SN 2009ip.
This analysis implicitly assumes that H$\alpha$ is optically thick at $10^4$ K. I believe we should make that assumption explicit in the text.
2.9e-14 erg/s/cm^2/A
7.40562619552e+25 cm
1.59045568109e+38 erg/s/A
sanity check: wave * 1e8 = 6562.8 = 6562.8
planck(6562.8,1e4) = 1.22966749446e+15 erg/s/cm^2/cm
planck(6562.8,1e4) = 12296674.9446 erg/s/cm^2/A
We're trying to solve an equation of the form:
$$ L(H\alpha) = \pi R^2 B_\lambda(T) d\lambda $$
Since our luminosity is in erg s$^{-1}$ $\dot{A}^{-1}$, we don't have to worry about the bandwidth.
$$ R = (L(H\alpha) / \pi / B_\lambda(T))^{1/2}$$
Emitting area = 4.11702975893e+30 cm^2
R = 1.144767e+15 cm
R = 76.5229413217 AU
R = 0.000370993688718 pc
Light Crossing time at 76 AU = 0.441960 days
R = 5.723835e+14 cm
R = 38.2614706609 AU
R = 0.000185496844359 pc
Light Crossing time at 38 AU = 0.220980 days
Because the brightening was sudden, and because a uniform disk is somewhat implausible, we should come up with a different measurement for a disk with an inner hole. I'll use fiducial numbers of 13000 $km s^{-1}$ and $\Delta t \sim 40$ days.
Inner disk radius = 4.4928e+15 or 300.325130229 AU
Light crossing time at inner disk radius = 1.73453329503 days
Height needed if this is ALL coming from the inner part of a disk: 1.45843491139e+14 cm or 9.74903522735 AU
The speed required to achieve that height in 2 days is 8440.01684832 km/s
We then want $$\int_{r_{inner}}^{r_{outer}} 2 \pi r dr = A_{emitting}$$ or $$ r_{outer}^2 - r_{inner}^2 = A_{emitting} / \pi$$
Outer radius = 4.6363501932e+15 or 309.920867958 AU
Delta-R = 1.43550193199e+14 or 9.59573772859 AU
The shock crossing time from inner->outer at 13000 km/s = 1.27805 days
I think this effectively leaves us with the conclusion that there is a 30 AU radius torus at 300 AU from the central source. That strikes me as somewhat implausible; 30 AU is enormous, particularly at densities $\sim10^{13}$. I'll try computing the mass in the disk assuming it's quite thin, $\sim0.1AU$.
mass = 1.44223224375e+32 g = 72.5067741062 msun
That's ridiculous. I'll invert the assumption: say the disk/torus is 1 msun; how thin must it be?
height of disk = 20632261267.2 cm = 0.0013791814797 AU
These numbers don't feel right to me. Why such an ultrathin disk? It would HAVE to puff up; the pressures are enormous.
What about the hypothesis that the brightening results from optically thin emission? I'll work under the assumption that H$\alpha$ is "just barely" optically thick (i.e. I'll cheat and use the optically thin approach);
in order to get the high H$\beta$ / H$\alpha$ ratio we actually need it to be pretty optically thick and under the Case B regime (Case C just makes it harder to get the observed $\alpha$/$\beta$ ratio).
Emitting path length for density = 1.0e+06 cm^-3 is 4.90881531249e+19 = 15.9083870006 pc
This is 9475 times larger than the largest allowed scale
Densities must be at least 9.734331e+07 for optically thin emission to be plausible
So at densities $>5\times10^7$, if H$\alpha$ starts becoming optically thick but H$\beta$ remains optically thin, we could explain the brightening by a filled sphere of just barely optically thick emission.
Despite all this, we still don't have a great explanation for why the P Cygni profile disappears. We should revisit that if possible.
Fraction of sphere that must be emitting: 0.0324616032628