*
Cococubed.com

Abundance variable
first derivative examples

Home

Astronomy research
Software instruments
Presentations
Illustrations
cococubed YouTube
Bicycle adventures
Public Outreach
Education materials
     Solar Systems Astronomy
     Energy in Everyday Life
     Geometry of Art and Nature
     Calculus
2022 ASU Solar Systems Astronomy
2022 ASU Energy in Everyday Life

AAS Journals
AAS YouTube
2022 Earendel, A Highly Magnified Star
2022 TV Columbae, Micronova
2022 White Dwarfs and 12C(α,γ)16O
2022 Black Hole mass spectrum
2022 MESA VI
2022 MESA in Don't Look Up
2022 MESA Marketplace
2012-2023 MESA Schools
2022 MESA Classroom
2021 Bill Paxton, Tinsley Prize


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.

We've previously shown the mean atomic number \begin{equation} \overline{\rm{A}} = \frac{\sum n_i {\rm A}_i}{\sum n_i} = \frac{\sum Y_i {\rm A}_i}{\sum Y_i} = \frac{1}{\sum Y_i} \label{eq:abar} \end{equation} and 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 \label{eq:zbar} \end{equation} have derivatives \begin{equation} \drvf{\overline{{\rm A}}}{Y_i} = - \overline{{\rm A}}^2 \end{equation} and \begin{equation} \partf{\overline{\rm Z}}{Y_i} = \overline{\rm{A}} \ ( Z_i - {\overline{\rm Z}} ) \end{equation} respectively. For an arbitrary quantity $\beta$ that is written in terms of $\overline{\rm{A}}$ and $\overline{\rm{Z}}$, the derivative with respect to $Y_i$ is \begin{equation} \drvf{\beta}{Y_i} \ = \ \partf{\beta}{\overline{\rm Z}} \ \partf{\overline{\rm Z}}{Y_i} + \partf{\beta}{\overline{\rm A}} \ \partf{\overline{\rm A}}{Y_i} \ = \ \partf{\beta}{\overline{\rm Z}} \ \overline{\rm{A}} \ (Z_i - {\overline{\rm Z}}) - \partf{\beta}{\overline{\rm A}} \ \overline{\rm{A}}^2 \ . \label{eq:dbdy} \end{equation} One assumes all partials of $\beta$ with respect to $\overline{\rm{A}}$ and $\overline{\rm{Z}}$ are available from the physics is at hand (e.g., from an eos).



One simple but relevant example is the specific energy of an ideal gas \begin{equation} e = \frac{3}{2} \frac{P}{\rho} = \frac{3}{2} \frac{N_A kT}{\overline{{\rm A}}} \hskip 0.2 in {\rm erg \ g}^{-1} \label{eq:egas} \end{equation} has the partial derivative of with respect to ${\overline{\rm Z}}$ \begin{equation} \partf{e}{\overline{\rm Z}} = 0 \end{equation} and the partial derivative of with respect to ${\overline{\rm A}}$ \begin{equation} \partf{e}{\overline{\rm A}} = - \frac{3}{2} \frac{N_A kT}{\overline{{\rm A}}^2} \ . \end{equation} Hence by equation ($\ref{eq:dbdy}$) the derivative with respect to $Y_i$ is \begin{equation} \drvf{e}{Y_i} = \partf{e}{\overline{\rm Z}} \ \partf{\overline{\rm Z}}{Y_i} + \partf{e}{\overline{\rm A}} \ \partf{\overline{\rm A}}{Y_i} = \left [ - \frac{3}{2} \frac{N_A kT}{\overline{{\rm A}}^2} \right ] \left [ - \overline{\rm A}^2 \right] = \frac{3}{2} N_A kT \hskip 0.2 in {\rm erg \ g}^{-1} \end{equation} as expected when recasting equation ($\ref{eq:egas}$) as \begin{equation} e = \frac{3}{2} \frac{P}{\rho_B} = \frac{3}{2} \ N_A kT \ \sum Y_i \hskip 0.2 in {\rm erg \ g}^{-1} \ . \end{equation} Note $\partial e / \partial Y_i$ is usually known as the chemical potential in the first law of thermodynamics when composition changes are taken into account.



Another example is the 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} The partial derivative of $\mu$ with respect to ${\overline{\rm Z}}$ is \begin{equation} \partf{\mu}{\overline{\rm Z}} = \partf{}{\overline{\rm Z}} \left ( \frac{\overline{\rm{A}}}{\overline{\rm{Z}} + 1 } \right ) = - \frac{\overline{\rm A}}{\left ( \overline{\rm Z} + 1 \right )^2} = - \frac{\mu}{\overline{\rm Z} + 1} = - \frac{\mu^2}{\overline{\rm A} } \ . \end{equation} The partial derivative of $\mu$ with respect to ${\overline{\rm A}}$ is \begin{equation} \partf{\mu}{\overline{\rm A}} = \partf{}{\overline{\rm A}} \left ( \frac{ \overline{\rm{A}}}{\overline{\rm{Z}} + 1 } \right ) = \frac{1}{\left ( \overline{\rm Z} + 1 \right )} = \frac{\mu}{\overline{\rm A}} \end{equation} Applying equation ($\ref{eq:dbdy}$), the derivative of $\mu$ with respect to $Y_i$ is \begin{align} \drvf{\mu}{Y_i} & = \partf{\mu}{\overline{\rm Z}} \ \partf{\overline{\rm Z}}{Y_i} + \partf{\mu}{\overline{\rm A}} \ \partf{\overline{\rm A}}{Y_i} = \left [ -\frac{\mu^2}{\overline{\rm A} } \right ] \left [ \overline{\rm{A}} (Z_i - {\overline{\rm Z}}) \right] + \left [ \frac{\mu}{\overline{\rm A} } \right ] \left [ - \overline{\rm A}^2 \right] \notag \\[8pt] & = \mu^2 ({\overline{\rm Z}} - Z_i) - \mu {\overline{\rm A}} \end{align}


Sometimes the composition is defined in terms of the hydrogen baryon (mass) fraction $\mathcal{X}$, helium baryon (mass) $\mathcal{Y}$, and metal baryon (mass) $\mathcal{Z}$. Bayron number conservation is then expressed as \begin{equation} \mathcal{X}+ \mathcal{Y} + \mathcal{Z} = 1 \ . \label{eq:xyz} \end{equation} These abundance variables are related to the usual $Y_i = X_i \ A_i$ abundance variables by \begin{equation} \mathcal{X} = Y_{\rm H} \hskip 0.5in \mathcal{Y} = 4Y_{\rm He} \hskip 0.5in \mathcal{Z} = A_{\rm z} Y_{\rm Z} \ , \end{equation} which mostly explains the use of calligraphic font to avoid the same symbols for different quantities. Applying equation ($\ref{eq:abar}$) \begin{equation} \overline{\rm{A}} = \frac{1}{\mathcal{X} + \mathcal{Y}/4 + \mathcal{Z}/A_{\rm z}} = \frac{4 A_{\rm z}}{A_{\rm z} \mathcal{X} + A_{\rm z} \mathcal{Y} + 4 \mathcal{Z} } = \frac{4 A_{\rm z}}{A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z}} \ , \label{eq:fabar} \end{equation} where the last expression eliminated $\mathcal{Y}$ using equation ($\ref{eq:xyz}$). Applying equation ($\ref{eq:zbar}$) \begin{align} \overline{\rm{Z}} & = \frac{\mathcal{X} + 1/2 \mathcal{Y} + Z_{\rm z}/A_{\rm z} \mathcal{Z}}{\mathcal{X} + \mathcal{Y}/4 + \mathcal{Z}/A_{\rm z}} \notag \\[8pt] & = (\mathcal{X} + 1/2 \mathcal{Y} + Z_{\rm z}/A_{\rm z} \mathcal{Z}) \cdot \overline{\rm{A}} \notag \\[8pt] & = \frac{2 A_{\rm z}(1 + \mathcal{X} - \mathcal{Z}) + 4 Z_{\rm z} \mathcal{Z}}{A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z}} \ , \label{eq:fzbar} \end{align} where again $\mathcal{Y}$ was eliminated using equation ($\ref{eq:xyz}$). Then the first derivatives of $\overline{\rm{A}}$ are \begin{align} \partf{\overline{\rm{A}}}{\mathcal{X}} & = - \frac{12 A_{\rm z}^2}{(A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z})^2} \notag \\[8pt] \partf{\overline{\rm{A}}}{\mathcal{Z}} & = \frac{4 A_{\rm z}^2 - 16 A_{\rm z}}{(A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z})^2} \label{eq:fdabar} \end{align} The first derivatives of $\overline{\rm{Z}}$ are \begin{align} \partf{\overline{\rm{Z}}}{\mathcal{X}} & = \frac{4 A_{\rm z} [ A_{\rm z}(\mathcal{Z} - 1) + \mathcal{Z}(2 - 3 Z_{\rm z}) ] }{(A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z})^2} \notag \\[8pt] \partf{\overline{\rm{Z}}}{\mathcal{Z}} & = - \frac{4 A_{\rm z} [ 2 + \mathcal{X}(2 (A_{\rm z} - 3 Z_{\rm z}) - Z_{\rm z}] }{(A_{\rm z} + 3 A_{\rm z} \mathcal{X} + 4 \mathcal{Z} - A_{\rm z} \mathcal{Z})^2} \label{eq:fdzbar} \end{align} For an arbitrary quantity $\beta$ that is written in terms of $\overline{\rm{A}}$ and $\overline{\rm{Z}}$, the derivatives with respect to $\mathcal{X}$ and $\mathcal{Z}$ are \begin{align} \drvf{\beta}{\mathcal{X}} & = \partf{\beta}{\overline{\rm Z}} \ \partf{\overline{\rm Z}}{\mathcal{X}} + \partf{\beta}{\overline{\rm A}} \ \partf{\overline{\rm A}}{\mathcal{X}} \notag \\[8pt] \drvf{\beta}{\mathcal{Z}} & = \partf{\beta}{\overline{\rm Z}} \ \partf{\overline{\rm Z}}{\mathcal{Z}} + \partf{\beta}{\overline{\rm A}} \ \partf{\overline{\rm A}}{\mathcal{Z}} \label{eq:chain} \end{align}


Note that equations ($\ref{eq:fabar}$) and ($\ref{eq:fzbar}$) can be solved for the either unknown or inconvenient variables $A_{\rm z}$ and $Z_{\rm z}$: \begin{align} A_{\rm z} & = \frac{4 \overline{\rm A}\mathcal{Z}}{4 - \overline{\rm A} - 3 \overline{\rm A} \mathcal{X} - \overline{\rm A} \mathcal{Z}} \notag \\[8pt] Z_{\rm z} & = \frac{4 \overline{\rm Z} - 2\overline{\rm A}(1 + \mathcal{X} - \mathcal{Z})}{4 - \overline{\rm A} - 3 \overline{\rm A} \mathcal{X} - \overline{\rm A} \mathcal{Z}} \ . \end{align} Eliminating $A_{\rm z}$ and $Z_{\rm z}$ from equation (\ref{eq:fdabar}) gives \begin{align} \partf{\overline{\rm A}}{X} & = - \frac{3 \overline{\rm A}^2}{4} \notag \\[8pt] \partf{\overline{\rm A}}{Z} & = \frac{\overline{\rm A} \ (\overline{\rm A} + 3 \overline{\rm A} \mathcal{X} -4)}{4 \mathcal{Z}} \ . \end{align} Eliminating $A_{\rm z}$ and $Z_{\rm z}$ from equation (\ref{eq:fdzbar}) gives \begin{align} \partf{\overline{\rm Z}}{X} & = \frac{\overline{\rm A}\ (2 - 3 \overline{\rm Z})}{4} \notag \\[8pt] \partf{\overline{\rm Z}}{Z} & = \frac{\overline{\rm A} \ [\overline{\rm Z} + (3 \overline{\rm Z} - 2) \mathcal{X} - 2]}{4 \mathcal{Z}} \ . \end{align} Equation ($\ref{eq:chain}$) still holds.