Estimate ionized density of photoevaporation flow from LDN 1616

[will-henney]

Once we know the ionizing flux that is incident on the globule, then we can estimate the ionized density by balancing the photon flux against two terms: the advection of new atoms through the ionization front, plus the recombinations in the ionized flow

The effective thickness, h, can be assumed to be roughly 0.1 times the globule radius.

For the recombination coefficient, use 2.3e-13 cm^3 / s

For the gas velocity, use 10 km/s

[will-henney]

The next step will then be to calculate the predicted H alpha surface brightness.

[RobeReyes]

We can compute the ionized density by balancing the photon flux using two terms: the advection of new atoms through the ionization front, $n_0 u0$, and the re-combinations in the ionized flow, $\int{r0}n^{2} \alpha{\beta} dr=n0^2 h \alpha{\beta}$. So we use $F_0=n_0 u_0 + n0^2 h \alpha\beta$ to compute the ionized density $n_0$, where $u_0$ is the gas velocity, we use $10 km/s$, $h$ is the effective thickness, we take $h=0.1r_0$ where $r0$ is the globule radius that we compute, and $\alpha\beta$ is the re-combination coefficient, we use $2.3\times 10^{-13} cm^3/s$, so we have

h = 0.1 * globule_radius.to('cm') #The effective thickness
alpha_beta = 2.3E-13 * u.cm**3 / (u.second) #The recombination coefficient
u_0 = 10 * u.kilometer / (u.second) #The gas velocity
cm_s = u.cm / u.second #convert the gas velocity to cgs
F_0 = flux_io_eps_ori_cms
a = h*alpha_beta
b = u_0.to(cm_s)
c = -F_0
n_0 = (-b+(b**2-(4*a*c))**(1/2))/(2*a)

To compute the ionized density we resolved our quadratic function $n0^2(h\alpha\beta)+n_0(u_0)-F_0=0$ and we take only the positive solution, so the solution $n_0=\frac{-u_0+\sqrt{u0^2+4h\alpha\beta F0}}{2h\alpha\beta}$ is

9.3998478 cm^{-3}

If we only take the advection term then the ionized density is $n_0=\frac{F_0}{u_0}$ and we have

F_0/(u_0.to(cm_s))

17.805235 cm^{-3}

similarly if we only take the re-combinations in the ionized flow then the ionized density is $n_0=\sqrt{\frac{F0}{h\alpha\beta}}$ and we have

(F_0/(h*alpha_beta))**(1/2)

13.680936 cm^{-3}

[will-henney]

After looking at H alpha emission, we have decide that a globule radius of 0.2 degrees would be more appropriate.

@RobeReyes please recalculate the density using this new radius, and also correcting the factor of 2 that was missing from your calculation in #1