Velocity at \tau = 1

If everything up to \tau = 1 is ionized, there must be a constant flow of neutrals from the center of the planet to be ionized in equilibrium. With a planet profile as in Tripathi et al​, the sound speed at any particular point can be calculated and compared to the required outflow velocity.

Planet profile (scaled by values at planet surface)

R_{crit} = \frac{(\gamma-1)G M_p}{\gamma R_p cs_p^2}$ $R_{zero} = \frac{R_{crit}}{R_{crit}-1}$ (radius at which atmosphere ends) $\rho(r) = (R_{crit}(\frac1r-\frac1{R_{zero}}))^\frac1{\gamma-1}$ $P(r) = \rho(r)^\gamma$ $T(r) = \frac{P(r)}{\rho(r)}$ $cs(r) = \sqrt{\frac{k_B T(r)}{m_H}}$

Assuming that everything up to the theoretical \tau = 1 surface is completely ionized, that there is no recombination, and that the ionization front is infinitesimally thin, we can calculate the velocity at the \tau=1 surface required to completely capture the ionizing flux:

v = \frac{J_{phot}}{\rho(R_{\tau=1})n_p}

where n_p is the number density at the planet's surface. For J_phot = 2x10¹³ photons/cm²/s and n_p = 8x10⁷ /cm³, the required velocity is

v = 261,176 \mbox{cm/s}\text{, or } 0.531 cs(R_{\tau=1}).

We can also assume that everything prior to some radius is ionized, and calculate the proportion of the sound speed at that location required to completely replenish the neutral density (and vice-versa, of course). For example, for simulations as of 9/8/17, the velocity required at the steady-state \tau=1 surface is one-fifth the sound speed at that location.

See attached Mathematica notebook​.

Equilibrium Ionization Front

A number of simplifying assumptions make this significantly easier, the most important being a constant temperature (no heating or cooling terms from line transfer, gamma ~ 1). I may return to this and add the heating and cooling terms, which affect the recombination rate, and include the dependence on temperature in the ionization fraction. In addition, hydrodynamics were neglected.

Since the line transfer done by the code is inherently 1D, ionization equilibrium was solved for​ by setting the recombination rate equal to the ionization rate in each successive cell:

$\tau(1) = 0$ $RecomRate(c) = (X(c) n_H)^2 \alpha_{rec}$ $IonRate(c) = (1-X(c))n_H J_{phot} \sigma_H e^{-\tau(c)}$ $\tau(c+1) = \tau(c) + \sigma_H (1-X(c)) n_H dx$

where dx is the cell size in cm, and we are solving for X©, the ionization fraction in the current cell. A modified logistic curve was then fit approximately (i.e. by hand) to the ionization fraction solution and used as the initial conditions for a simulation, which was allowed to relax to true equilibrium. A better fit for initial conditions could be obtained with a minimization routine, but I think it is more instructive to see the simulation have some significant relaxation.

One thing to note is that the mean free path of a photon in the ionization front is not particularly comparable to the actual width of the ionization front in this case, since the ionization front cannot be considered thin on our scale.