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Contact: F.X.Timmes
my one page vitae,
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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} \label{eq:y} \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 \label{eq:mconserv} \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.

The rest-mass energy is \begin{equation} E = - M c^2 \hskip 0.2 in {\rm erg} \, \end{equation} where $M$ is the total baryonic mass [g]. The minus sign indicates that creating mass reduces the energy reservoir of a closed system. For $M$ being composed of $i$ isotopes \begin{equation} E = - \sum_{i=1}^{k} N_i \ m_i \ c^2 \hskip 0.2 in {\rm erg} \ . \end{equation} Multiplying by the constant $N_A / N_B$ and using equation ($\ref{eq:y}$) gives gives the specific nuclear energy of the baryons \begin{equation} \epsilon = - N_A c^2 \sum_{i=1}^{k} Y_i \ m_i \hskip 0.2 in {\rm erg \ g}^{-1} \ . \end{equation} Taking the time derivative yields the specific nuclear energy generation rate \begin{equation} \dot{\epsilon} = - N_A c^2 \sum_{i=1}^{k} \dot{Y_i} \ m_i \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \label{eq:epsnuc} \end{equation} Note one only needs to evaluate, not integrate, the right-hand sides of the $\dot{Y_i}$ ODEs defining the nuclear reaction network to obtain the instantaneous $\dot{\epsilon}$. Nice.

The atomic mass $m_i$ in equation ($\ref{eq:epsnuc}$) can be defined as \begin{equation} m_i = {\rm A}_i m_u + \Delta_i \hskip 0.2in {\rm g} \label{eq:mass} \end{equation} where $\Delta_i$ is the mass excess of each species in atomic mass units [g]. This expression neglects the electronic binding energy, and $\Delta_i$ is thus independent of the ionization state of a given species. However, the electron rest masses are included in this definition since the $m_i$ are atomic masses - this is important for accurately tracking weak reactions. The mass excess is related to the binding energy $B_i$, mass excess of the proton $\Delta_p$ and mass excess of the neutron $\Delta_n$ by \begin{equation} \Delta_i = {\rm Z}_i \Delta_p + N_i \Delta_n - \frac{B_i}{c^2} \hskip 0.2in {\rm g} \label{eq:massex} \end{equation} Substituting equations ($\ref{eq:mass}$) and ($\ref{eq:massex}$) into equation ($\ref{eq:epsnuc}$) gives \begin{equation} \dot{\epsilon} = - N_A c^2 \sum_{i=1}^{k} \dot{Y_i} \ \left ( A_i m_u + {\rm Z}_i \Delta_p + N_i \Delta_n - \frac{B_i}{c^2} \right ) \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \end{equation} From equation ($\ref{eq:mconserv}$), $\sum \dot{X_i} = \sum \dot{Y_i} {\rm A}_i = 0$, hence the above reduces to \begin{equation} \dot{\epsilon} = N_A \sum_{i=1}^{k} \dot{Y_i} B_i - \sum_{i=1}^{k} \dot{Y_i} ({\rm Z}_i \Delta_p + N_i \Delta_n) \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \end{equation} This is a handy form as changes due to weak reactions are isolated by the second term on the right-hand side. If weak reactions are not important, \begin{equation} \dot{\epsilon} = N_A \sum_{i=1}^{k} \dot{Y_i} B_i \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \end{equation}

Equation ($\ref{eq:epsnuc}$) is the {\it instantaneous} energy generation rate. In this form it can be added as an ODE to the nuclear reaction network equations, or added to the system of PDEs in a fully coupled hydrocode. In an operator-split hydrocode, over a timestep $\Delta t$ one usually uses the finite difference approximation \begin{equation} \dot{Y_i} = \frac{Y_{i,{\rm end}} - Y_{i,{\rm start}}}{\Delta t} \hskip 0.2in {\rm s}^{-1} \end{equation} and equation ($\ref{eq:epsnuc}$) becomes \begin{equation} \left < \dot{\epsilon} \right > = - N_A c^2 \sum_{i=1}^{k} \left [ \frac{Y_{i,{\rm end}} - Y_{i,{\rm start}}}{\Delta t} \right ] \ m_i \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \ , \end{equation} and represents the average energy generation rate over a finite timestep.

A note on the mass of isotope $i$. The intuitive mass counting of nucleons minus the binding energy is \begin{equation} m_i = {\rm Z}_i m_p + N_i m_n - \frac{B_i}{c^2} \hskip 0.3in {\rm g} \end{equation} Eliminating the binding energy $B_i$ using equation ($\ref{eq:massex}$) \begin{align} m_i & = {\rm Z}_i m_p + N_i m_n - \Delta_i - {\rm Z}_i \Delta_p - N_i \Delta_m \notag \\ & = {\rm Z}_i (m_p - \Delta_p) + N_i (m_n - \Delta_n) - \Delta_i \end{align} By definition, $m_p = m_u + \Delta_p \simeq (1 + 0.007276) m_u$ and $m_n = m_u + \Delta_n \simeq (1 + .008664) m_u$. Substituting, \begin{align} m_i & = {\rm Z}_i m_u + N_i m_u - \Delta_i \notag \\ & = ({\rm Z}_i + N_i) m_u - \Delta_i \notag \\ & = {\rm A}_i m_u - \Delta_i \end{align} which is equation ($\ref{eq:mass}$) for the atomic mass.