Cococubed.com Self-heating in reaction networks

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Contact: F.X.Timmes
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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 $$Y_i = \frac{n_i}{n_B} = \frac{N_i}{N_B} \label{eq:y}$$ 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 $X_i$, $$X_i = Y_i \ A_i = \frac{n_i \ A_i}{n_B}$$ 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 $$\sum X_i = \frac{n_B}{n_B} = 1 \label{eq:mconserv}$$ is invariant under nuclear reactions. Define the baryon density, in atomic mass units, as $$\rho_B = n_B \ m_u = \frac{n_B}{N_A} \hskip 0.2in {\rm g \ cm}^{-3}$$ 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 first law of thermodynamics is $$dE = \delta Q -\delta W = \delta Q - PdV \hskip 0.2in {\rm erg} \label{eq:first}$$ where $dE$ is the change in internal energy and $\delta Q$ is the heat added to the system. $\delta W = PdV$ is the mechanical work performed in expanding or contracting the system, where P [erg cm$^{-3}$] is the presure and $dV$ is the change in volume [cm$^3$]. If one chooses the entropy S [erg/K], volume $V$, and $N_i$ as the independent thermodynamic variables, then the total differential of the internal energy is \begin{align} dE & = \left ( \partf{E}{S}\right )_{V,N_i} dS + \left ( \partf{E}{V}\right )_{S,N_i} dV + \sum_i \left ( \partf{E}{N_i}\right )_{S,V} dN_i \notag \\[8pt] & = T dS - PdV + \sum_i \left ( \partf{E}{N_i}\right )_{S,V} dN_i \ . \label{eq:svn} \end{align} Comparing equations ($\ref{eq:first}$) and ($\ref{eq:svn}$), gives the familiar equality $$\delta Q \equiv dE + PdV = T dS + \sum_i \left ( \partf{E}{N_i}\right )_{S,V} dN_i \label{eq:heat}$$
On the other hand, one may {\it choose} the temperature T, volume $V$, and $N_i$ as the independent thermodynamic variables. Then the total differential of the internal energy is \begin{align} dE & = \left ( \partf{E}{T}\right )_{V,N_i} dT + \left ( \partf{E}{V}\right )_{T,N_i} dV + \sum_i \left ( \partf{E}{N_i}\right )_{T,V} dN_i \notag \\ & = C_v dT - PdV + \sum_i \left ( \partf{E}{N_i}\right )_{T,V} dN_i \ , \label{eq:tvn} \end{align} where $C_v$ is the specific heat [erg/K] at constant volume. Comparing equations ($\ref{eq:first}$) and ($\ref{eq:tvn}$) gives the following equality $$\delta Q \equiv dE + PdV = T dS + \sum_i \left ( \partf{E}{N_i}\right )_{S,V} dN_i = c_v dT + \sum_i \left ( \partf{E}{N_i}\right )_{T,V} dN_i$$ Note the $\partial E / \partial N_i$ terms are sometimes called the chemical potential $\mu_i$. The heat added to the system from nuclear reactions is $$\delta Q = E_{\rm nuc} - E_{\nu}$$ which gives the equalities \begin{align} \delta Q & \equiv dE + PdV \notag \\ & = T dS + \sum_i \left ( \partf{E}{N_i}\right )_{S,V} dN_i \notag \\ & = C_v dT + \sum_i \left ( \partf{E}{N_i}\right )_{T,V} dN_i \notag \\ & = E_{\rm nuc} - E_{\nu} \end{align} Multiplying by the constant $N_A /N_B$ [1/g] gives the specific form of the first law: \begin{align} \delta q & \equiv de + P d \left(\frac{1}{\rho} \right) \notag \\ & = T ds + \sum_i \left ( \partf{e}{Y_i}\right )_{S,\rho} dY_i \notag \\ & = c_v dT + \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} dY_i \notag \\ & = \epsilon_{\rm nuc} - \epsilon_{\nu} \hskip 1.5in {\rm erg} \ {\rm g}^{-1} \end{align} Which could have been gotten by using the $(T,\rho,Y_i)$ Helmholtz free energy triplet as the independent variables. Operating with the time derivative $d/dt$ gives \begin{align} \drvf{e}{t} - \frac{P}{\rho^2} \drvf{\rho}{t} & = T \drvf{s}{t} + \sum_i \left ( \partf{e}{Y_i}\right )_{S,\rho} \drvf{Y_i}{t} \notag \\ & = c_v \drvf{T}{t} + \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \notag \\ & = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} \hskip 1.5in {\rm erg} \ {\rm g}^{-1} \ {\rm s}^{-1} \label{eq:specific} \end{align} One may solve any combination of the expressions in equation ($\ref{eq:specific}$).

It is interesting to consider equation ($\ref{eq:specific}$) for an idea gas. It has been shown previously that the specific internal energy $$e = \frac{3}{2} \frac{P}{\rho} = \frac{3}{2} \frac{N_A kT}{\overline{{\rm A}}} = \frac{3}{2} \ N_A kT \ \sum_{i} Y_i \hskip 0.2 in {\rm erg \ g}^{-1}$$ has the derivative $$\left ( \partf{e}{Y_i} \right )_{\rho, N_{i \ne j}} = \frac{3}{2} N_A kT \ .$$ Its also been previously shown in these notes that nuclear energy generation rate is $$\dot{\epsilon}_{\rm nuc} = - N_A c^2 \sum_{i} \drvf{Y_i}{t} \ m_i \hskip 0.2 in {\rm erg \ g}^{-1} \ {\rm s}^{-1} \ .$$ Neglecting thermal neutrino losses, the temperature evolution version of equation ($\ref{eq:specific}$) becomes \begin{align} c_v \drvf{T}{t} & = \dot{\epsilon}_{\rm nuc} - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \notag \\[8pt] & = - N_A c^2 \sum_{i} \drvf{Y_i}{t} \ m_i - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \notag \\[8pt] & = - N_A c^2 \sum_{i} \drvf{Y_i}{t} \ m_i - \frac{3}{2} N_A kT \sum_i \drvf{Y_i}{t} \ . \label{eq:evol1} \end{align} Physically equation ($\ref{eq:evol1}$) says the change in the energy is the energy release from nuclear reactions minus the thermal energy of the now heavier (and fewer) gas particles. For example, say there are 4 protons. Each has an energy of 3/2 $kT$. When the 4 protons combine to form 1 helium nucleus, there is an energy gain from the nuclear reaction but now there is only 1 particle with an energy of 3/2 $kT$. The system gains energy because $$\frac{m_{4He} - 4 m_p}{4m_p} \cdot c^2 \simeq 0.007 \cdot c^2 \simeq 7\times 10^{18} \gg N_A k T \simeq 8 \times 10^{14} \left ( \frac{T}{10^7 {\rm K}} \right) \hskip 0.2in {\rm erg} \ {\rm g}^{-1}$$ for typical hydrogen burning temperatures. At the other extreme, the system looses energy when the second term in equation ($\ref{eq:evol1}$) becomes larger than the first term. At temperatures of $\simeq$ 10$^{10}$ K, this occurs when the relative mass change from nuclear burning becomes smaller than $\eta = \Delta m/m = N_A k T / c^2 \simeq 8 \times 10^{16} / 10^{21} \simeq 10^{-4}$, about a factor of 100 smaller than hydrogen burning's 0.007. In general, when $$\frac{N_A k T}{\eta c^2 } \ll 1$$ the second term in equation ($\ref{eq:specific}$) can be safely dropped.

Let's consider a few specific instantiations of equation ($\ref{eq:specific}$).

For self-heating at constant density, $d\rho = 0$, and equation($\ref{eq:specific}$) reduces to $$\drvf{e}{t} = c_v \drvf{T}{t} + \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} \ .$$ is a convenient form. If one evolves the specific energy, one must do a root find on the equation of state to get the temperature for a given $(e, \rho, Y_i)$ triplet. If one evolves the temperature directly, the $\partial e / \partial Y_i$ terms must be included: $$\drvf{T}{t} = {1 \over c_v} \left [ \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \right ] \hskip 0.5in {\rm K} \ {\rm s}^{-1} \ . \label{eq:temp}$$

For self-heating at constant pressure, one approach is to solve equation ($\ref{eq:temp}$) and do a root find on the equation of state to get the density for a given $(T, P, Y_i)$ triplet - a Gibbs free energy basis set. {\it This is the easiest approach}. Alternatively, the hard way, expanding the total specific energy differential of equation ($\ref{eq:specific}$) in the $(T, \rho, Y_i)$ Helmholtz free energy basis gives \begin{align} & \drvf{e}{t} - {P \over \rho^2} \drvf{\rho}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} \notag \\[8pt] & \left ( \partf{e}{T}\right )_{\rho,Y_i} \drvf{T}{t} + \left ( \partf{e}{\rho}\right )_{T,Y_i} \drvf{\rho}{t} + \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} - {P \over \rho^2} \drvf{\rho}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} \notag \\[8pt] & \left ( \partf{e}{T}\right )_{\rho,Y_i} \drvf{T}{t} + \left [ \left ( \partf{e}{\rho}\right )_{T,Y_i} - {P \over \rho^2} \right ] \drvf{\rho}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \ . \label{eq:step1} \end{align}

Now, the first law is an exact differential, which requires $$P = \rho^2 \left ( \partf{e}{\rho} \right )_{T,Y_i} + T \left ( \partf{P}{T} \right )_{\rho, Y_i} \ . \label{eq:max}$$ Using equation ($\ref{eq:max}$) to eliminate $(\partial e / \partial \rho )_{T,Y_i}$ in equation ($\ref{eq:step1}$) gives $$c_v \drvf{T}{t} - {T \over \rho^2} \left ( \partf{P}{T}\right )_{\rho,Y_i} \drvf{\rho}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \label{eq:step2}$$

In addition, for constant pressure, the total differential is $dP = 0$ $$dP = \left ( \partf{P}{T}\right )_{\rho,Y_i} dT + \left ( \partf{P}{\rho}\right )_{T,Y_i} d\rho + \sum_i \left ( \partf{P}{Y_i}\right )_{T,\rho} dY_i = 0 \ .$$ Solving for $d\rho$, $$d\rho = - {1 \over \left ( \partf{P}{\rho}\right )_{T,Y_i} } \left [ \left ( \partf{P}{T}\right )_{\rho,Y_i} dT + \sum_i \left ( \partf{P}{Y_i}\right )_{T,\rho} dY_i \right ] \ . \label{eq:drho}$$ Substituting equation ($\ref{eq:drho}$) into equation ($\ref{eq:step2}$) $$c_v \drvf{T}{t} + {T \over \rho^2} \left ( \partf{P}{T}\right )_{\rho,Y_i} {\left ( \partf{\rho}{P}\right )_{T,Y_i} } \left [ \left ( \partf{P}{T}\right )_{\rho,Y_i} \drvf{T}{t} + \sum_i \left ( \partf{P}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t} \right ] \notag$$ $$= \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left ( \partf{e}{Y_i}\right )_{T,\rho} \drvf{Y_i}{t}$$ Rearranging, $$\left [ c_v + {T \over \rho^2} \left ( \partf{P}{T}\right )^2_{\rho,Y_i} \left ( \partf{\rho}{P}\right )_{T,Y_i} \right ] \drvf{T}{t} \notag$$ $$= \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left [ \left ( \partf{e}{Y_i}\right )_{T,\rho} + {T \over \rho^2} \left ( \partf{P}{T}\right )_{\rho,Y_i} \left ( \partf{\rho}{P}\right )_{T,Y_i} \left ( \partf{P}{Y_i} \right )_{T,\rho} \right ]\drvf{Y_i}{t}$$ using the so-called thermal and density exponents $$\chi_t = {T \over P} \partf{P}{T} \hskip 1.0in \chi_{\rho} = {\rho \over P} \partf{P}{\rho}$$ one has $$\left [ c_v + {P \over \rho T} {\chi_T \over \chi_{\rho} } \right ] \drvf{T}{t} = \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left [ \left ( \partf{e}{Y_i}\right )_{T,\rho} + \left ( \partf{\rho}{P}\right )_{T,Y_i} \left ( \partf{P}{Y_i} \right )_{T,\rho} \right ]\drvf{Y_i}{t}$$ or c_p \drvf{T}{t}$$= \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left [ \left ( \partf{e}{Y_i}\right )_{T,\rho} + {T \over \rho^2} \left ( \partf{P}{T}\right )_{\rho,Y_i} \left ( \partf{\rho}{P}\right )_{T,Y_i} \left ( \partf{P}{Y_i} \right )_{T,\rho} \right ]\drvf{Y_i}{t}$$ where $c_p$ [erg g$^{-1}$ K$^{-1}$] is the specific heat at constant pressure. Hence, the final temperature evolution equation for a fixed pressure and changing composition: $$\drvf{T}{t} = {1 \over c_p} \left [ \dot{\epsilon}_{\rm nuc} - \dot{\epsilon}_{\nu} - \sum_i \left [ \left ( \partf{e}{Y_i}\right )_{T,\rho} + {T \over \rho^2} \left ( \partf{P}{T}\right )_{\rho,Y_i} \left ( \partf{\rho}{P}\right )_{T,Y_i} \left ( \partf{P}{Y_i} \right )_{T,\rho} \right ]\drvf{Y_i}{t} \right ]$$ Whew. Equation ($\ref{eq:temp}$) with a root find is simpler.

For self-heating at entropy, a model for a burning convective region, one approach is to solve equation ($\ref{eq:temp}$) and do a root find on the equation of state to get the density for a given $(T, S, Y_i)$ triplet. Alternatively, the hard way, is to follow the example above to get an explicit evolution equation for the entropy.