Mass fractions
The mass fraction of H, He and
the metals is a quick way to specify the chemical composition of the
stellar plasma. The mass fraction of hydrogen is defined as ratio of
the mass of hydrogen particles and the total mass of all particles (in a
given volume):
$$
X = \frac{M_\mathrm{H}}{M}= \frac{\rho_\mathrm{H}}{\rho},
$$
where $\rho_\mathrm{H}$ and $\rho$ are the respective (mass) densities.
The hydrogen mass density is equal to the hydrogen number density
($n_\mathrm{H}$) times the mass of one hydrogen particle $m_{\mathrm{H}}
= A_{\mathrm{H}}\,m_{\mathscr{A}}$, thus:
$$
X = \frac{M_\mathrm{H}}{M}= \frac{n_\mathrm{H}\,m_{\mathrm{H}}}{\rho}=
\frac{n_\mathrm{H}\,A_{\mathrm{H}}\,m_{\mathscr{A}}}{\rho}.
$$
Similar to that, for helium and the metals we define:
$$
Y = \frac{M_\mathrm{He}}{M}=
\frac{n_\mathrm{He}\,m_{\mathrm{He}}}{\rho}=
\frac{n_\mathrm{He}\,A_{\mathrm{He}}\,m_{\mathscr{A}}}{\rho},
$$
$$
Z = \frac{M_\mathrm{metals}}{M}= \frac{n_\mathrm{metals}\,m_{\mathrm{metals}}}{\rho} =
\frac{\sum_{i=3} n_\mathrm{i}\,A_{\mathrm{i}}\,m_{\mathscr{A}}}{\rho},.
$$
On the other hand, in the same volume there is the total number of $N$
particles (free and bound atoms and free electrons):
$$
N = N_\mathrm{e} + N_{\mathrm{a}} = N_\mathrm{e} + N_\mathrm{H} +
N_\mathrm{He} + N_\mathrm{metals},
$$
where the indices $\mathrm{e}$ and $\mathrm{a}$ stand for the electrons
and for the atoms.
The same equation for the number densities reads:
$$
n = n_\mathrm{e} + N_{\mathrm{a}} = n_\mathrm{e} + n_\mathrm{H} +
n_\mathrm{He} + n_\mathrm{metals}.
$$
For the mass of one particle of the species $i$ ($i \ge 1$) we have:
$$
m_{i} = A_{i}\,m_{\mathscr{A}},
$$
or expressed relative to hydrogen:
$$
m_{i} =\frac{A_{i}\,m_{\mathrm{H}}}{A_{\mathrm{H}}}.
$$
Since $m_{\mathrm{H}} \approx m_{\mathscr{A}}$, sometimes in the
literature the last equation appears as:
$$
m_{i} =\frac{A_{i}\,m_{\mathscr{A}}}{A_{\mathrm{H}}}.
$$
Abundances
Abundances specify
detailed chemical composition of a stellar atmosphere. There are three
common ways to express them in astrophysics:
1)
$\alpha_i$ - relative number of the atoms (nuclei) of the element Z
relative to the number of hydrogen (nuclei) in the linear scale (for hydrogen, $\alpha_\mathrm{H} \equiv 1$): \begin{equation}\alpha_i = \frac{N_i}{N_\mathrm{H}} = \frac{n_i}{n_\mathrm{H}}\end{equation};
2)
$\varepsilon_{i}$ - abundances relative to the amount of hydrogen in the
logarithmic scale normalized so that the abundance of hydrogen
$\varepsilon_{\mathrm{H}} = 12$:\begin{equation}\varepsilon_i = 12 + \log\alpha_i;\end{equation}
3) $x_i$ - relative number of the nuclei of the element Z relative to the total number of nuclei in the linear scale: \begin{equation}
x_i = \frac{N_i}{N_\mathrm{a}} = \frac{N_i}{\sum N_j} = \frac{10^{\varepsilon_i}}{\sum 10^{\varepsilon_j}}. \end{equation}
For hydrogen for the solar composition, $x_\mathrm{H} \approx 0.9$.
The
relations between $x_i$ and $\alpha_i$ are the following:
$$
x_i = \frac{N_i}{\sum N_j} = \frac{\alpha_i N_{\mathrm{H}}}{\sum
\alpha_j N_{\mathrm{H}}}= \frac{\alpha_i}{\sum \alpha_j},
$$
$$
\alpha_i = \frac{N_i}{N_\mathrm{H}} = \frac{x_i
N_\mathrm{n}}{N_\mathrm{H}}.
$$
Finally, sometimes there is a need to express the individual abundances
of non-hydrogen species relative to their total number:
$$
(x_i)_{\mathrm{nonH}} = \frac{N_i}{\sum_{j\gt 1} N_j} =
\frac{N_i}{-N_\mathrm{H} + \sum_{j\gt 1} N_j} = \frac{\alpha_i
N_\mathrm{H}}{-N_\mathrm{H} + \sum_{j\gt 1} \alpha_j N_\mathrm{H}} =
\frac{\alpha_i }{-1 + \sum_{j\gt 1} \alpha_j }.
$$
Mass fractions in terms of abundances
For
any $X_i$ ($X_i \in [X, Y, Z]$) we can write:
$$
X_i = \frac{n_i m_i}{\rho} = \frac{\alpha_i m_i n_\mathrm{H}}{\rho}=
\frac{\alpha_i m_i n_\mathrm{H}}{\sum \alpha_j m_j n_\mathrm{H}}=
\frac{\alpha_i m_i}{\sum \alpha_j m_j}.
$$
From $m_i = A_i m_\mathrm{H}/ A_\mathrm{H}$ follows:
\begin{equation}
X_i = \frac{\alpha_i A_i}{\sum \alpha_j A_j}.
\label{eq:xyzabund}
\end{equation}
Therefore:
$$
X = \frac{\alpha_\mathrm{H} A_\mathrm{H}}{\sum \alpha_j
A_j}\;\;\;\;\;\;\;Y = \frac{\alpha_\mathrm{He} A_\mathrm{He}}{\sum
\alpha_j A_j}\;\;\;\;\;\;\;Z = \frac{\sum_{j>2} \alpha_j A_j}{\sum
\alpha_j A_j}.
$$
Pressure and density in terms of number densities
The
total gas pressure may be represented as a sum of the partial gas
pressures of the particles of different species (possible species are
now atoms (neutrals and ions), electrons and molecules). It is Dalton's
law:
$$
p \equiv p_{\mathrm{tot}} = p_{\mathrm{e}} +
p_{\mathrm{a}^{\mathrm{free}}} + p_{\mathrm{mol}}.
$$
If the ideal gas approximation is applicable, for each component of the
mixture it holds:
$$
p = n\,kT = (n_{\mathrm{e}} + n_{\mathrm{a}}^{\mathrm{free}} +
n_{\mathrm{mol}})\,kT,
$$
where $n_{\mathrm{a}}^{\mathrm{free}}$ represents the sum over all
chemical elements and
$n_{\mathrm{mol}}$ the sum over all molecules.
If there is no molecules formed ($n_{\mathrm{a}}^{\mathrm{free}}=n_{\mathrm{a}}$):
$$
p = n\,kT = (n_{\mathrm{e}} + n_{\mathrm{a}})\,kT.
$$
Therefore the total number density of the atoms may be found when the total gas pressure and the electron pressure are known:
$$
n_{\mathrm{a}} = (p - p_{\mathrm{e}})/kT.
$$
The total mass density is
$$
\rho = \rho_{\mathrm{e}} + \rho_{\mathrm{a}}.
$$
This definition hold no matter if the molecules are formed or not. Since
$\rho_{\mathrm{e}} \ll \rho_{\mathrm{a}}$ it is commonly taken that
$$
\rho \approx \rho_{\mathrm{a}} = \rho_{\mathrm{H}} + \rho_{\mathrm{He}} +
\rho_{\mathrm{m}}.
$$
Note that pressure is proportional to the number of the particles
while the density is proportional to their mass. As a clear
consequence, the increasing ionization rate (the increasing number of
free electrons) increases the total pressure but not the density.
Mean molecular weight
The mean molecular weight is primarily used in chemistry where it is known as molar mass. In chemistry it is defined as the mass of a given substance divided by its amount of substance:
$$
\mu = \frac{M [\mathrm{g}]}{\widetilde{n} [\mathrm{mol}]}.
$$
If there is, for an example, one mole ($\widetilde{n} = 1 $)
of neutral hydrogen gas, there
is $N_\mathscr{A}$ hydrogen atoms in it. Total mass of these particles
is $
M = N_\mathscr{A} m_{\mathrm{H}}$, where $m_{\mathrm{H}} =
A_{\mathrm{H}} m_\mathscr{A}$ is the mass of one H atom in grams. Numerically,
$N_\mathscr{A}\, m_\mathscr{A}= 1$, and therefore the total mass $M$ is equal
to the value of $A_{\mathrm{H}}$ (but the unit is still gram, not a.m.u.!). The molar mass is simply
that mass divided by the amount of substance, $\widetilde{n} =
1\,\mathrm{mol}$, thus $\mu = 1\, \mathrm{g/mol}$. If the gas is
completely ionized, its mass remains the same, while the amount of
substance (i.e. the number of particles) doubles, so that $\mu = 1/ 2
\,A_{\mathrm{H}}$.
Equivalently, the $\mu$ can be defined as the mean mass of a
particle in atomic mass units. Let's again consider neutral hydrogen gas. There is only one type of particles in that case,
they are all equal and, therefore, the mean molecular weight is equal to
the mass of one particle in a.m.u., $\mu = A_{\mathrm{H}} \approx 1\,
\mathrm{a.m.u.}$. If the gas gets fully ionized, the total mass in unchanged, the
number of particles is doubled and, clearly, $\mu = 1/2
A_{\mathrm{H}}$. Formally this can be written as: $$
\mu = \frac{\sum N_i\,A_i [\mathrm{a.m.u.}]}{N},
$$ where $N_i$ is the number of particles of type $i$, $N$ is the
total number of particles and $A_i$ is the atomic weight of one particle of type $i$
(in a.m.u.).
There is often need to relate $\mu$ to the density. The following relations hold:
\begin{eqnarray}
\mu &=& \frac{M [\mathrm{g}]}{\widetilde{n} [\mathrm{mol}]} =\\
&=& \frac{M [\mathrm{g}]}{N}\,N_\mathscr{A}
[\mathrm{mol^{-1}}] = \\ &=& \frac{M [\mathrm{g}] / V
[\mathrm{cm^3}]}{N / V [\mathrm{cm^3}]}\,N_\mathscr{A}
[\mathrm{mol^{-1}}]= \\&=& \frac{\rho [\mathrm{g\,cm^{-3}}]}{n
[\mathrm{cm^{-3}}]}\,N_\mathscr{A} [\mathrm{mol^{-1}}].\end{eqnarray}☞a.m.u. is (in terms of writing equations and crunching numbers) not much
more special then any imperial unit, pound or ounce for example.
Avogadro's number can be taken as the conversion factor between grams
(or kilograms) and a.m.u.
Mean molecular weight of a mixture
Now
that we understand all about $\mu$ and its units, we take the last of
the the relations from the previous section and apply it to the gas that consists of different
atoms, neutral and ionized, free electrons and molecules. That relation
says that $\mu$ depends upon the density and the number of particles.
The density remains unchanged in the basic micro-processes in the
stellar atmospheres (ionization, recombination, molecular association
and disassociation). On the other hand, the number of the free particles
(atoms, electrons, molecules) changes in these processes and $\mu$
follows that change as well. The actual chemical composition enters
$\mu$ through the density. Therefore, $\mu$ is a single number that
tells us about the chemical composition if the ionization fractions are
known or about the ionization fraction if the chemical composition is
fixed. An obvious question is if we can split the two parts of $\mu$ -
one dependent only on the number of the free electrons and other
dependent only on the chemical composition.
Let's
assume that there is no molecules in the plasma and write the density
as a sum per specie:
$$
\rho = \sum_{i = 1}^{i_\mathrm{max}} \rho_i = \rho_\mathrm{H} +
\rho_\mathrm{He} + \rho_\mathrm{m},
$$
where we neglected the density of the free electrons ($\rho_\mathrm{e}
\ll \rho_i$ for ant $i$). The total number density explicitly written
is:
$$
n= n_\mathrm{e} + \sum_{i = 1}^{i_\mathrm{max}} n_i = n_\mathrm{e} +
n_\mathrm{H} + n_\mathrm{He} + n_\mathrm{metals}.
$$
The mean molecular weight is then:
$$
\mu = \frac{\rho_\mathrm{H} + \rho_\mathrm{He} +
\rho_\mathrm{metals}}{(n_\mathrm{e} + n_\mathrm{H} + n_\mathrm{He} +
n_\mathrm{metals})\,m_\mathscr{A}},
$$
and its inverse:
$$
\frac{1}{\mu} = \frac{n_\mathrm{e}\,m_\mathscr{A}}{\rho} +
\frac{n_\mathrm{H}\,m_\mathscr{A}}{\rho} +
\frac{n_\mathrm{He}\,m_\mathscr{A}}{\rho} +
\frac{n_\mathrm{metals}\,m_\mathscr{A}}{\rho}.
$$
In the three atomic terms we recognize expressions proportional to the
mass fractions defined above:
$$
\frac{1}{\mu} = \frac{n_\mathrm{e}\,m_\mathscr{A}}{\rho} + X
\frac{m_\mathscr{A}}{m_\mathrm{H}} +
Y\frac{m_\mathscr{A}}{m_\mathrm{He}} +
Z\frac{m_\mathscr{A}}{m_\mathrm{m}}.
$$
The first term:
$$
\frac{1}{\mu_\mathrm{e}} = \frac{n_\mathrm{e}\,m_\mathscr{A}}{\rho}
$$
depends only on the number of the free electrons. The second term
depends only on the chemical composition:
$$
\frac{1}{\mu_\mathrm{a}} = X \frac{m_\mathscr{A}}{m_\mathrm{H}} +
Y\frac{m_\mathscr{A}}{m_\mathrm{He}} +
Z\frac{m_\mathscr{A}}{m_\mathrm{m}}.
$$
If we replace $m_\mathrm{H}$ in the first term with
$A_\mathrm{H}\,m_\mathscr{A}$ and approximate $A_\mathrm{H}\approx 1$,
that term reduces to $X$ only. In the other two terms we use $m_i =
A_i\,m_\mathrm{H}/A_\mathrm{H}$ to get:
$$
\frac{1}{\mu_\mathrm{a}} = X + Y\frac{A_\mathrm{H}}{A_\mathrm{He}} +
Z\frac{A_\mathrm{H}}{A_\mathrm{m}}.
$$
A similar expression employing the abundances instead of the mass
fractions can easily be derived using the relation Eq.\ref{eq:xyzabund}
between the mass fractions and the abundances that we can write for
separately for any element under the consideration:
$$
X_i = \frac{\alpha_i A_i}{\sum \alpha_j A_j}, \;\;\;\;\;\;\;\mathrm{1
\le i \le i_\mathrm{max}}.
$$
For $1/\mu_\mathrm{a}$ now we have:
$$
\frac{1}{\mu_\mathrm{a}} = \sum_i X_i \frac{A_\mathrm{H}}{A_i} = \sum_i
\frac{\alpha_i A_i}{\sum \alpha_j A_j} \frac{A_\mathrm{H}}{A_i} =
\frac{A_\mathrm{H} \sum_i \alpha_i }{\sum \alpha_j A_j},
$$
and finally:
$$
\mu_\mathrm{a} = \frac{\sum \alpha_j A_j}{A_\mathrm{H} \sum_j \alpha_j
}.
$$
This expression, for example, appears in the important article by
Mihalas (1967, in Methods in Computational Physics, eds. Alder, Fernbach
& Rotenberg, p.16, eq.54).
Different forms of the ideal gas law
The
ideal gas law is one of the most useful and widely applicable
approximations in astrophysics. I already used it to introduce the total
and the partial pressure:
$$
p = n\,kT = (n_{\mathrm{e}} + n_{\mathrm{a}})\,kT.
$$
This form comes from statistical mechanics. The constant $k$ is the
Boltzmann constant that is equal to the universal gas constant ($R$)
divided by Avogardo's number ($N_\mathscr{A}$). Other useful forms can
be easily derived:
$$
p = n\,kT = \frac{N}{V}\,\frac{R}{N_\mathscr{A}}\, T =
\frac{N}{N_\mathscr{A}}\,\frac{RT}{V} = \frac{\widetilde{n}RT}{V},
$$
or
$$
p = \frac{\widetilde{n}RT}{V} = \frac{M}{\mu}\,\frac{RT}{V} = \frac{\rho
RT}{\mu}.
$$
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