|
Cococubed.com
|
|
Abundance variable first derivative nunaces |
Home
Astronomy Research
2025 Neutrinos From De-excitation
2024 Radiative Opacity
2024 Neutrino Emission from Stars
2023 White Dwarfs & 12C(α,γ)16O
2023 MESA VI
2022 Earendel, A Highly Magnified Star
2022 Black Hole Mass Spectrum
2021 Skye Equation of State
2021 White Dwarf Pulsations & 22Ne
Software Instruments
AAS Journals
2025 AAS YouTube
2025 AAS Peer Review Workshops
2025 ASU Energy in Everyday Life
2025 MESA Classroom
Outreach and Education Materials
Solar Systems Astronomy
Energy in Everyday Life
Geometry of Art and Nature
Calculus
Other Stuff:
Bicycle Adventures
Illustrations
Presentations
Contact: F.X.Timmes
my one page vitae,
full vitae,
research statement, and
teaching statement.
$ \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.