 Cococubed.com Abundance variable first derivative nunaces

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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.

The composition derivatives are perhaps more nuanced than one might initially imagine.

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$^{-3}$]. Define the average of any quantity $\overline{\beta}$ by the number density $n_i$ weighted average \begin{equation} \overline{\beta} = \frac{\sum \beta_i n_i}{\sum n_i} \end{equation} with equation ($\ref{eq:y}$) becomes \begin{equation} \overline{\beta} = \frac{\sum \beta_i Y_i}{\sum Y_i} \ . \label{eq:betabar} \end{equation} The partial derivative with respect to abundance $Y_i$ is \begin{equation} \frac{ \partial \overline{\beta}}{\partial Y_i} = \frac{\beta_i}{\sum Y_i} - \frac{\sum \beta_i Y_i}{\left ( \sum Y_i \right )^2} = \frac{\beta_i}{\sum Y_i} - \frac{\overline{\beta}}{\sum Y_i} = \frac{\beta_i - \overline{\beta}}{\sum Y_i} \ . \end{equation} In this case, \begin{align} \partf{{\overline{\rm A}}}{Y_i} & = \frac{A_i - {\overline{\rm A}}}{\sum Y_i} \notag \\[8pt] \partf{{\overline{\rm Z}}}{Y_i} & = \frac{Z_i - {\overline{\rm Z}}}{\sum Y_i} \label{eq:azbar1} \end{align}

However, reconsider the special case when the arbitrary quantity $\beta_i$ is the number of nucleons ${\rm A}_i$. As shown earlier, baryon conservation requires $\sum X_i = \sum A_i Y_i = n_B / n_B = 1$. Then ${\overline{\rm A}} = 1/{\sum Y_i}$ and another valid expression for $\partial \overline{{\rm A}} / \partial Y_i$ is \begin{align} \partf{{\overline{\rm A}}}{Y_i} & = - \overline{{\rm A}}^2 \notag \\[8pt] \partf{{\overline{\rm Z}}}{Y_i} & = {\overline{\rm A}} \ (Z_i - {\overline{\rm Z}}) \ . \label{eq:abar2} \end{align} Despite looking very different, equations ($\ref{eq:azbar1}$) and ($\ref{eq:abar2}$) are equivalent, provided one consistently uses either set of expressions. Mixing the two formalisms, essentially when to set $\sum X_i = 1$, causes confusion and incorrect results. Let's show the equivalence for the two forms of ${\rm d}{\overline{\rm A}}/{\rm d}Y_i$.

The ideal gas \begin{equation} P = \left ( \sum n_i \right ) kT \ = \ \left ( \sum Y_i \right ) n_B kT \ , \end{equation} has the trivial first derivative with respect to abundance \begin{equation} \partf{P}{Y_i} = n_B kT \label{eq:py} \end{equation}
Consider the formalism of equation ($\ref{eq:azbar1}$) where $\sum X_i = 1$ was not explicitly invoked. The ideal gas law, using equation ($\ref{eq:azbar1}$) becomes \begin{equation} P = n_B kT \left ( \sum Y_i \right ) \ = \ n_B kT \left ( \frac{ \sum {\rm A}_i Y_i}{\overline{{\rm A}}} \right ) \ . \label{eq:pgas1} \end{equation} The derivative with respect to abundance is \begin{align} \partf{P}{Y_i} & \ = \ n_B kT \left[ \frac{{\rm A}_i}{\overline{{\rm A}}} - \frac{ \sum {\rm A}_i Y_i}{\overline{{\rm A}}^2} \ \partf{\overline{{\rm A}}}{Y_i} \right ] \notag \\[8pt] & \ = \ n_B kT \left[ \frac{{\rm A}_i}{\overline{{\rm A}}} - \frac{ \sum {\rm A}_i Y_i}{\overline{{\rm A}}^2} \ \left ( \frac{{\rm A}_i - \overline{{\rm A}}}{\sum Y_i} \right ) \right ] \notag \\[8pt] & \ = \ n_B kT \left[ \frac{{\rm A}_i}{\overline{{\rm A}}} - \frac{{\rm A}_i}{\overline{{\rm A}}^2} \frac{ \sum {\rm A}_i Y_i}{\sum Y_i} + \frac{1}{\overline{{\rm A}}} \frac{ \sum {\rm A}_i Y_i}{\sum Y_i} \right ] \notag \\[8pt] & \ = \ n_B kT \left[ \frac{{\rm A}_i}{\overline{{\rm A}}} - \frac{{\rm A}_i}{\overline{{\rm A}}} + 1 \right ] \notag \\[8pt] & \ = \ n_B kT \end{align} which is identical to equation ($\ref{eq:py}$).

Consider the formalism of equation ($\ref{eq:abar2}$) where $\sum X_i = 1$ is used. The ideal gas law becomes \begin{equation} P = n_B kT \left ( \sum Y_i \right ) \ = \ \frac{n_B kT}{\overline{{\rm A}}} \ . \label{eq:pgas2} \end{equation} The derivative with respect to abundance is \begin{equation} \partf{P}{Y_i} \ = \ \partf{P}{ \overline{{\rm A}}} \ \partf{ \overline{{\rm A}}}{Y_i} \ = \ \left[ -\frac{n_B kT}{\overline{{\rm A}}^2 } \right ] \left [ - \overline{{\rm A}}^2 \right ] \ = \ n_B kT \end{equation} which is identical to equation ($\ref{eq:py}$).

This shows either formalism may be used, provided it is used consistently. Since many expressions that use $\overline{{\rm A}}$ are written in the form of equation ($\ref{eq:pgas2}$) rather than equation ($\ref{eq:pgas1}$), the first derivative form of equation ($\ref{eq:abar2}$) is preferred.