User:SunderB/Draft:Spectroscopy/HII Regions/Ionisation and Temperature

Assumptions

 * Ionisation always occurs from the ground state, also known as Case B Recombination - unlikely for ionisation from higher states as they have shorter lifetimes
 * Ionisation occurs as soon as an electron reaches the ground state - constant supply of high energy electrons
 * Energy is distributed across all components of the plasma quickly, so gas temperature is approximately equal to electron temperature - due to large number of particle interactions

Recombination and Ionisation Rates
To model recombination, we consider that:


 * The overall recombination rate is proportional to the number densities of protons and electrons, and that $$n_pn_e \approx (n_e)^2$$
 * The rate in which electrons recombine into different energy levels varies. Each energy level has its own recombination rate coefficient $$\beta_n$$, which is dependent on temperature.

Combining these two points gives:

$$\frac{dN_R}{dt}=(n_e)^2\times\beta_n(T_e)$$

To get the total recombination rate, we can sum up the number of recombinations into all the energy levels. Assuming Case B recombination:

$$\beta_B(T_e)=(2\times10^{-16)})T^{-\frac{3}{4}}$$

To model ionisation, we consider that:


 * Ionisation rate is proportional to the number density of neutral hydrogen atoms $$n_H$$
 * Ionisation rate is also proportional to the number of ionising photons $$J$$

Therefore, the ionisation rate can be modelled as:

$$\frac{dN_I}{dt}=\alpha_0n_HJ$$

where $$\alpha_0\approx 6.8\times10^{-22}$$m2 is the photoionisation cross-section.

In equillibrium, these two rates are the same:

$$(n_e)^2\times\beta_N(T_e)=\alpha_0n_HJ$$

Fractional Ionisation
To determine the degree of ionisation in the region, we can look at the fractional ionisation $$x = \frac{n_e}{n}$$.

By using $$n=n_p + n_H$$ and assuming $$n_e\approx n_p$$ we get:

$$\begin{align} x &\approx \frac{n_e}{n_e + n_H}\\ xn_e + xn_H&=n_e\\ xn_H &= (1-x)n_e\\ n_H &= \frac{1-x}{x}n_e \end{align}$$

Therefore:

$$\begin{align} (n_e)^2 \beta_n(T_e)&=\alpha_0\frac{x-1}{x}n_eJ\\ x^2n^2\beta_n(T_e)&=\alpha_0(x-1)nJ\\ \therefore \frac{x^2}{x-1} &= \frac{\alpha_0J}{n\beta_B(T_e)}=\frac{\alpha_0Q}{4\pi r^2 n\beta_B(T_e)} \end{align}$$

Using typical values, it can be found that $$(1-x)$$ is approximately $$3.4 \times 10^{-5}$$.

This suggests that H!! regions are very strongly ionised throughout the whole region, and that they have quite hard edges.