# Stellar spectra

### Stellar Spectra

\The spectrum that we see from a star gives us information from the photosphere from the star, and contains all the information that we get from stars, aside from neutrinos. Thus, even though it is a very small fraction of the star, it is important observationally. \\The surface (or photosphere) is where the mean free path is comparable to the scale height of the atmosphere. This is given by

${\displaystyle \ell \sim H\sim {\frac {P}{\rho g}}\,\!}$

Plugging in gas pressure, which dominates the photospheres of stars,

${\displaystyle {\frac {1}{\kappa \rho }}\sim {\frac {kT}{mg}}.\,\!}$

This gives us the pressure and density in the photosphere, with

${\displaystyle P\approx {\frac {g}{\kappa }},\,\!}$

and

${\displaystyle \rho \approx {\frac {mg}{kT\kappa }}.\,\!}$

For the Sun, ${\displaystyle T_{\rm {eff}}=5800}$ K, giving a number density of ${\displaystyle n\approx 10^{17}}$ cm${\displaystyle ^{-3}}$, which is ${\displaystyle 10^{-9}}$ the central number density of the Sun. The pressure is ${\displaystyle P\approx 10^{5}}$ erg cm${\displaystyle ^{-3}\approx 0.1}$ atm. The pressure and density, along with the effective temperature, set the observable properties of stars. To first order, we have assumed that the surface of a star is a perfect blackbody emitter. This is not accurate though, as it ignores the presence of spectral lines, the strength of which depends on the density and pressure as well. The spectra of stars are the basis of a classification system, labeling stars O, B, A, F, G, K, M, ranging from the hottest stars to the coolest stars. There are also L and T spectral types, which apply not to stars, but to brown dwarfs. The nice thing about this is that for main sequence stars, the mass maps to an effective temperature, which maps to a spectral type. Our goal is to understand the second fact (since we already understand the first fact). \\The matter on the surface of a star is in thermodynamic equilibrium. This is fundamentally the reason that the temperature determines the spectral lines in a star. The ratio of number of states is just set by the Boltzmann factor, with

${\displaystyle {\frac {N_{2}}{N_{1}}}={\frac {g_{2}}{g_{1}}}e^{-(E_{2}-E_{1})/kT}.\,\!}$

For neutral atoms and molecules, this describes the number of electrons in each state. For determining the fraction of neutral and ionized atoms, we can generalize the Boltzmann factor to an expression known as the Saha equation. In thermal equilibrium, the temperature, pressure, and chemical potential (${\displaystyle \mu }$) will all come in to equality. These three quantities are the macroscopic properties of a gas. The implication of the chemical potential equilibrium is that for a reaction where

${\displaystyle A+B\rightarrow C+D,\,\!}$

and where

${\displaystyle C+D\rightarrow A+B,\,\!}$

then

${\displaystyle \mu (A)+\mu (B)=\mu (C)+\mu (D).\,\!}$

The fact that the reaction be able to go both ways is important to remember. This fact is the reason that we cannot apply chemical potential equilibrium to describe fusion in the centers of star. It is so energetically difficult to break apart a helium atom that there are many more reactions producing helium then there are reactions breaking helium apart to provide hydrogen. So, instead, we’ll apply this to ionization balance, which is a problem chemical equilibrium can solve, and is what we want to understand to understand the surfaces of stars. We can start by looking at the ionization/recombination of hydrogen:

${\displaystyle e^{-}+p\leftrightarrow H+\gamma .\,\!}$

We can solve for the balance of this reaction by writing down the chemical potential of each side,

${\displaystyle \mu (e^{-})+\mu (p)\leftrightarrow \mu (H)+\mu (\gamma ).\,\!}$

The phase space distribution of particles is

${\displaystyle n(p)={\frac {g}{h^{3}}}{\frac {1}{e^{(E_{p}-\mu )/kT}\pm 1}}.\,\!}$

Remember that the plus or minus depends on whether we are dealing with Fermi-Dirac or Bose-Einstein statistics. In the limit of a classical gas, this cannot matter, so we must have the exponential term dominate, leaving

${\displaystyle n(p)={\frac {g}{h^{3}}}e^{-(E_{p}-\mu )/kT}.\,\!}$

The energy of particles is

${\displaystyle E_{p}=mc^{2}+{\frac {p^{2}}{2m}}.\,\!}$

The number density of particles in real space is the integral over the momentum of the particle,

${\displaystyle n=\int 4\pi p^{2}dpn(p)=\int 4\pi p^{2}dp{\frac {g}{h^{3}}}e^{-(mc^{2}-\mu )/kT}e^{-p^{2}/2mkT}.\,\!}$

Only the last term needs to stay in the integral, and it is a Gaussian integral with a nice solution. The result is

${\displaystyle n=ge^{-(mc^{2}-\mu )/kT}\left({\frac {2\pi mkT}{h^{2}}}\right)^{3/2}.\,\!}$

Or, in terms of the quantum number density,

${\displaystyle n=gn_{Q}e^{-(mc^{2}-\mu )/kT}.\,\!}$

This allows us to finally write down what ${\displaystyle \mu }$ is, with

${\displaystyle \mu =mc^{2}-kT\ln \left({\frac {gn_{Q}}{n}}\right).\,\!}$

Now, we have everything we need to solve for the ionization balance of hydrogen. We can write the chemical potential of the proton, electron, and hydrogen atom with the above expression, and the chemical potential of the photon is zero (we can create this for free, so they cannot have a chemical potential). The difference in the mass between the hydrogen atom and the sum of electon and proton masses is just the binding energy of the hydrogen atom, which we’ll call ${\displaystyle \chi }$. Then

${\displaystyle -{\frac {\chi }{kT}}-\ln \left({\frac {g_{H}n_{Q,H}}{n_{H}}}\right)=-\ln \left({\frac {g_{p}n_{Q,p}}{n_{p}}}\right)-\ln \left({\frac {g_{e}n_{Q,e}}{n_{e}}}\right)\,\!}$

Some rearranging, taking advantage of the log rules for adding/subtraction, and then exponentiating both sides, gives the Saha equation,

${\displaystyle {\frac {n_{e}n_{p}}{n_{H}}}={\frac {g_{e}g_{p}}{g_{H}}}n_{Q,e}e^{-\chi /kT}.\,\!}$