Scaling

Cross section

The equations for calculating the optical depth in physical units look like

d\tau = n \sigma dl

And in computational units

d\tau = n_c \sigma_c dl_c

Dividing the first equation by the second gives

n_c \sigma_c dl_c = n \sigma dl = n_c N_s \sigma dl_c L_s

where N_s and L_s are the number density and length scales in the code.

This gives

\sigma_c = \sigma N_s L_s

Ionizing Flux

The equation for calculating the ionization rate in physical units is

\frac{dn}{dt} = n \sigma F

and in computational units

\frac{dn_c}{dt_c} = n_c \sigma_c F_c

Dividing the first equation by the second, gives

N_s/T_s = N_s \frac{1}{N_s L_s} \frac{F}{F_c}

which gives

F_c=F \frac{T_s}{L_s N_s}

Recombination Rate

The equation for recombination in physical units is

\frac{d n}{dt} = \alpha n_{HII} n_e

and in computational units

\frac{d n_c}{dt} = \alpha_c n_{c,HII} n_{c,e}

Dividing the two and solving for \alpha_c gives

\alpha_c = \alpha \frac{T_s}{N_s}

Ionization Heating

The equation in physical units is

\frac{dE}{dt}=e_\Gamma \sigma n F

and in computational units

\frac{dE_c}{dt_c}=e_{c,\Gamma} \sigma_c n_c F_c

Dividing the two and solving for e_{c,\Gamma} gives

e_{c,\Gamma} = \frac{\sigma}{\sigma_c} N_s \frac{F}{F_c}\frac{T_s}{E_s} = \frac{N_s}{E_s}

Recombination Cooling

In physical units we have

\frac{dE}{dt}=\Lambda kT T^{-.89} n_e n_{H+}

and in computational units we have

\frac{dE_c}{dt_c}=\Lambda_c T T^{-.89} n_{c,e} n_{c,H+}

and dividing the two gives us

\Lambda_c = \Lambda \frac{k N_s^2 T_s}{E_s}

Lyman Alpha Cooling

In physical units we have

\frac{dE}{dt}=\Lambda e^{-118348 K/T} n_e n_{H}

and in computational units we have

\frac{dE_c}{dt_c}=\Lambda_c e^{-118348 K/T} n_{c,e} n_{c,H}

and dividing the two gives us

\Lambda_c = \Lambda \frac{N_s^2 T_s}{E_s}

Test problem

2D version of Tripathi without atmosphere

\rho_c = 6\times 10^8 cm^{-3}

R_c=1.5\times 10^{10} cm

F=2\times 10^{13} cm^{-2} s^{-1}

T= 10^3 K

embedded in a lower density pressure balanced ambient with a density ratio

\chi=100

With a cross section for photo-ionization of 6.3e-18 and our clump's column density of 9e18, we would expect the tau=1 surface to be 5% of the way in towards the clump center.

And with a flux of 2e13, we would expect photo-ionization of the column to take 5 days or so…