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Abundance variables

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Contact: F.X.Timmes
my one page vitae,
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research statement, and
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$ \def\drvop#1{{\frac{d}{d{#1}}}} \def\drvf#1#2{{\frac{d{#1}}{d{#2}}}} \def\ddrvf#1#2{{\frac{d^2{#1}}{d{#2}^2}}} \def\partop#1{{\frac{\partial}{\partial {#1}}}} \def\ppartop#1{{\frac{\partial^2}{\partial {#1}^2}}} \def\partf#1#2{{\frac{\partial{#1}}{\partial{#2}}}} \def\ppartf#1#2{{\frac{\partial^2{#1}}{\partial{#2}^2}}} \def\mpartf#1#2#3{{\frac{\partial^2{#1}}{\partial{#2} \ {\partial{#3}}}}} $ A pdf of this note is avaliable.

Baryon number is an invariant. Define the abundance of species $Y_i$ by \begin{equation} Y_i = \frac{n_i}{n_B} = \frac{N_i}{N_B} \end{equation} where $N_i$ is the number of particles of isotope $i$, $N_B$ is the number of baryons, $n_i$ is the number density [cm$^{-3}$] of isotope $i$ and $n_B$ is baryon number density [cm$^{-3}$]. The number of baryons in isotope $i$ divided by the total number of baryons is the baryon fraction $X_i$, \begin{equation} X_i = Y_i \ A_i = \frac{n_i \ A_i}{n_B} \end{equation} where $A_i$ is the atomic mass number, the number of baryons in an isotope. Usually the baryon fraction is called the ``mass fraction''. Note \begin{equation} \sum X_i = \frac{n_B}{n_B} = 1 \end{equation} is invariant under nuclear reactions. Define the baryon density, in atomic mass units, as \begin{equation} \rho_B = n_B \ m_u = \frac{n_B}{N_A} \hskip 0.2in {\rm g \ cm}^{-3} \end{equation} where $m_u$ is the atomic mass unit [g] and $N_A$ is the Avogadro number [g$^{-1}]$ in a system of units where the atomic mass unit is {\it defined} as 1/12 mass of an unbound atom of $^{12}$C is at rest and in its ground state.

Mean atomic number \begin{equation} \overline{\rm A} = \frac{\sum n_i {\rm A}_i}{\sum n_i} = \frac{n_B}{\sum n_i} = \frac{\sum Y_i {\rm A}_i}{\sum Y_i} = \frac{1}{\sum Y_i} \end{equation} Mean charge \begin{equation} \overline{\rm Z} = \frac{\sum n_i {\rm Z}_i}{\sum n_i} = \frac{\sum Y_i {\rm Z}_i}{\sum Y_i} = \overline{\rm A} \sum Y_i {\rm Z}_i \end{equation} Electron to baryon ratio, where the second equality assumes full ionization \begin{equation} Y_e = \frac{n_e}{n_B} = \frac{\sum n_i Z_i}{n_B} = \sum Y_i Z_i = \frac{\overline{\rm Z}}{\overline{\rm A}} \end{equation} Neutron excess \begin{equation} \eta = \sum ({\rm N}_i - {\rm Z}_i) Y_i = \sum ({\rm A}_i - 2 {\rm Z}_i) Y_i = \sum {\rm A}_i Y_i - 2 Y_e = 1 - 2 Y_e \end{equation} Mean ion molecular weight \begin{equation} \mu_{{\rm ion}} = \overline{\rm A} \end{equation} Mean electron molecular weight \begin{equation} \mu_{{\rm ele}} = \frac{1}{Y_e} = \frac{\overline{\rm A}}{\overline{\rm Z}} \end{equation} Mean molecular weight \begin{equation} \mu = \left [ \frac{1}{\mu_{ion}} + \frac{1}{\mu_{ele}} \right ]^{-1} = \left [ \frac{1}{\overline{\rm A}} + Y_e \right ]^{-1} = \left [ \frac{1}{\overline{\rm A}} + \frac{\overline{\rm Z}}{\overline{\rm A}} \right ]^{-1} = \frac{\overline{\rm A}}{\overline{\rm Z} + 1} = \frac{ n_B}{\sum n_i + n_e} \end{equation}