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.
from astropy import units, constants
pc = units.pc.in_units('cm')
Iha = 2.9e-14 # erg s^-1 cm^-2 A^-1
D = 24e6 * pc
Lha = Iha * D**2
print Iha," erg/s/cm^2/A"
print D," cm"
print Lha," erg/s/A"
# Use the wavelength form of the Planck function
h = constants.cgs.h
k = constants.cgs.k_B
c = constants.cgs.c
def planck(lam, temp):
""" erg s^-1 cm^-2 A^-1 sr^-1 """
return 2*h*c**2 / lam**5 * (exp(h*c/(k*temperature*lam)) - 1)**-1
angstrom = units.angstrom.in_units('cm') # angstroms
wave = 6562.8*angstrom
temperature = 1e4 # K
print "sanity check: wave * 1e8 = 6562.8 = ",wave*1e8
print "planck(6562.8,1e4) = ",planck(wave,temperature)," erg/s/cm^2/cm"
print "planck(6562.8,1e4) = ",planck(wave,temperature)/1e8," 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.
$$ R = (L(H\alpha) / \pi / B_\lambda(T))^{1/2}$$
Aemitting = Lha / (planck(wave,temperature)*pi/1e8)
print "Emitting area = ",Aemitting," cm^2"
R = (Aemitting/pi)**0.5
print "R = %e" % R, " cm"
au = units.AU.in_units('cm')
print "R = ",R/au," AU"
print "R = ",R/pc, " pc"
# Light crossing time
tc = R / constants.cgs.c
day = units.day.in_units('s')
print "Light Crossing time at %i AU = %f days" % (R/au, tc/day)
# What if we assume it's a sphere instead of a disk?
# area -> 4 * pi * r^2
R = (Aemitting/(4*pi))**0.5
print "R = %e" % R, " cm"
au = units.AU.in_units('cm')
print "R = ",R/au," AU"
print "R = ",R/pc, " pc"
tc = R / constants.cgs.c
print "Light Crossing time at %i AU = %f days" % (R/au, tc/day)
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.
kms = (units.km/units.s).in_units('cm/s')
vejecta = 13e3 * kms
deltat = 40 * day
r_collision = vejecta*deltat
print "Inner disk radius = ",r_collision," or ",r_collision/au," AU"
print "Light crossing time at inner disk radius = ",r_collision / constants.cgs.c / day, "days"
H = Aemitting/r_collision/2/pi
print "Height needed if this is ALL coming from the inner part of a disk: ",H," cm or ",H/au," AU"
print "The speed required to achieve that height in 2 days is ", H / (2*day) / 1e5, "km/s"
We then want $$\int_{r_{inner}}^{r_{outer}} 2 \pi r dr = A_{emitting}$$ or $$ r_{outer}^2 - r_{inner}^2 = A_{emitting} / \pi$$
router = (Aemitting/pi+r_collision**2)**0.5
print "Outer radius = ",router," or ",router/au," AU"
print "Delta-R = ",router-r_collision," or ",(router-r_collision)/au," AU"
tcross = (router-r_collision) / vejecta
print "The shock crossing time from inner->outer at %g km/s = %g days" % (vejecta/kms, tcross/day)
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, $\sim0.1AU$.
height = 0.1 * au
volume = Aemitting*height
density = 1e13
nparticles = volume * density
msun = units.msolMass.in_units('g')
mass = nparticles * constants.cgs.m_p * 1.4
print "mass = ",mass,"g = ",mass/msun,"msun"
That's ridiculous. I'll invert the assumption: say the disk/torus is 1 msun; how thin must it be?
mass = 1 * msun
nparticles = mass / (constants.cgs.m_p*1.4)
volume = nparticles/density
height = volume/Aemitting
print "height of disk = ",height,"cm = ",height/au," 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).
density = 1e6 # assume pretty dense, but still plausibly case B
alphaB = 2.6e-13 # cm^3 s^-1
Eha = h * c / wave
nphot_perang = Lha / Eha # phot/s/A
nphot = nphot_perang * 1 # phot/s
# solve for path length
# nphot = density**2 * alphaB * area * path_length
path_length = nphot / (density**2*alphaB*Aemitting)
print "Emitting path length for density = %0.1e" % density,"cm^-3 is ",path_length," = ",path_length/pc," pc"
# use the rise time of ~2 days as an upper limit on the emitting size scale
maxsize = 2*day*c
print "This is %i times larger than the largest allowed scale" % (path_length/maxsize)
print "Densities must be at least %e for optically thin emission to be plausible" % (density*sqrt(path_length/maxsize))
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.
# limb-brightened shell model
half_sph_area = 2*pi*r_collision**2
print "Fraction of sphere that must be emitting: ",Aemitting/half_sph_area
from IPython.core.display import HTML
def blackbg():
"""Black Background code mirror theme."""
html = """
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div.highlight {background:black;} /* needed for nbviewer */
</style>"""
return HTML(html)
blackbg()
