Strömgren sphere

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In astrophysics, a Strömgren sphere is the sphere of ionized hydrogen (H II) around a young star of the spectral classes O or B. The most prominent example is the Rosette Nebula. It was derived by and later named after Bengt Strömgren^[1].

An idealized calculation is simple, let's suppose the region is exactly spherical and fully ionized (x=1) and composed only of hydrogen so the numerical density of protons equals the density of electrons ( $n e = n p$ ), then the Strömgren radius will be the region where the recombination rate equals the ionization rate. We will consider the recombination rate $N R$ of all energy levels which is

$N_R = \sum_{n=2}^{\infty}N_n$

$N n$ is the recombination rate of the n-th energy level. The reason we have excluded n=1 is that if a photon with enough energy recombines in to the ground level the hydrogen atom will release another photon capable of ionizing up to the ground level. This is important as electric dipole mechanism always make the ionization up to the ground level so we exclude n=1 to add these ionizing field effect. Now, the recombination rate of a particular energy level $N n$ is (with $n e = n p$ ):

$N_n=n_e n_p \beta_{n}(T_e)=n_e^2 \beta_{n}(T_e)$

where $β n (T e)$ is the recombination coefficient of the nth energy level in a unitary volume at a temperature $T e$ which is the temperature of the electrons and is usually the same of the sphere. So after doing the sum we arrive to:

$N_R=n_e^2 \beta_2(T_e)$

Where $β 2 (T e)$ is the total recombination rate and has an approximate value of:

$\beta_2(T_e) \approx 2 \times 10^{-22} T_e^{3/4} \ \mathrm{[m^{3} s^{-1}]}$ .

Using $n$ as the number of nucleons (in this case, protons), we can introduce the degree of ionization $0\leq x \leq1$ so $n e = x n$ , and the numerical density of neutral hydrogen is $n h = (1 - x) n$ . With a cross section $α 0$ (which has units of area) and the number of ionizing photons per area per second $J$ the ionization rate $N I$ is:

$N I = α 0 n h J$

For simplicity we will consider only the geometric effects on $J$ as we get further from the ionizing source flux $S *$ , so we have an inverse square law:

$J(r)=\frac{3 S_*}{4 \pi r^3}$