# Are / Why are the H II regions positively charged?

One of the questions in our recent astrophysics course homework was to find the general opacity $$\kappa$$ of H II regions. We know that the H II regions are almost entirely ionized hydrogen and I instinctively assumed quasineutrality, which I believed to be quite usual. If we take the individual opacities of the electron and ion gas (proton gas for this case) using the Rosseland mean opacity formula,

$$\kappa_e\cdot\rho\approx n_e\cdot\sigma_e, \quad \text{or} \quad \kappa_p\cdot\rho \approx n_p\cdot\sigma_p,$$

where $$\rho$$ is the mean gas density and $$n$$ and $$\sigma$$ are the number density and Thomson scattering cross-section for electrons or protons respectively, we can simplify the equations with

$$n_j \approx \frac{\rho}{m_j} \quad \Rightarrow \quad \kappa_j = \frac{n_j\sigma_j}{\rho} \approx \frac{\sigma_j}{m_j}.$$

Since $$I(x) = I_0\cdot e^{-\kappa\rho x}$$ for multiple gases, say, gas A and gas B, we can take the intensity of the light that went through one gas,

$$I_A \equiv I_0\cdot e^{-\kappa_A\rho x},$$

as the "initial" intensity of the second gas,

$$I = I_A\cdot e^{-\kappa_B\rho x} = \bigl(I_0\cdot e^{-\kappa_A\rho x}\bigr)\cdot e^{-\kappa_A\rho x} = I_0\cdot e^{-(\kappa_A+\kappa_B)\rho x},$$

so the opacities of individual gas components should just be additive, that is in my case $$\kappa_{tot} = \kappa_e+\kappa_p$$. However, we know that the Thomson cross-section depends explicitly on the reciprocal of the squared mass, and because of the above equation for $$\kappa$$ it follows that:

$$\sigma \sim \frac{1}{m^2} \quad \Rightarrow \quad \kappa \sim \frac{1}{m^3},$$

and knowing the ratio of proton to electron mass,

$$\frac{m_p}{m_e} \approx 1800 \quad \Rightarrow \quad \kappa_p = \frac{\kappa_e}{(m_p/m_e)^3} \approx \frac{\kappa_e}{1800^3},$$

which means the proton opacity is so small compared to the electron one so as to be practically irrelevant for this question, and so I've given the answer in the terms of $$\kappa = \sigma_e/m_e$$. The faculty told me this is not the case though, and that protons dominate the H II regions and the correct answer would be $$\kappa = \sigma_p/m_p$$. The only reasonable conclusion I can draw from that is that the electron density is much smaller and so the H II regions are positively charged, however I can find no mention of this through extensive google-ing and don't have a clue as to where those massive amounts of negative charge go. Do electrons get carried away with the ionization front or are there other processes that I'm not aware of?

• An HII region is neutral. – Rob Jeffries May 11 at 20:56
• @RobJeffries would you mind elaborating a bit? I thought we can assume full ionization with negligible recombination and I'm not seeing how inverse bremsstrahlung contributes to protons dominating the opacity – Andrii Kozytskyi May 11 at 21:16
• It’s ionized, but the aggregate is neutral. Otherwise, the EM force would tear it apart. – Paul May 11 at 21:20
• @Paul that's what I thought too, but I've also seen that the regions are expanding. And then there's still the question of how is the proton domination possible is the cloud is neutral, could you please explain it a bit further? – Andrii Kozytskyi May 11 at 21:22
• I didn't say it did. The answer to your question is no, HII regions are not positively charged. – Rob Jeffries May 11 at 22:08