5th Week: Reaction Rates, Nuclear Networks & NSE

So far we have studied the nuclear reactions and the main thermodynamic properties. Now we move on further to use this knowledge to predict the astrophysical plasma processes such as energy release and composition changes.

Thermonuclear reaction rates

From the definition of nuclear cross section we know that the nuclear cross section between target $i$ and projectile $j$ can be written as

$\sigma ={\frac {r/n_{i}}{n_{j}v}}\ ,$

where $v$ is the relative velocity between targets with the number density $n_{i}$ and projectiles with the number density $n_{j}$ and $r$ is the number of reactions per $cm^{3}$ per $sec$. From here one can find

$r=\sigma vn_{i}n_{j}\ ,$

or more generally when targets and projectiles follow the specific distribution then

$r_{i;j}=\int \sigma \cdot \left|{\vec {v_{i}}}-{\vec {v_{j}}}\right|dn_{i}dn_{j}\ .$

For the nuclei in astrophysical plasma that obey the Maxwell-Boltzmann distribution we can write

$dn_{j}=n_{j}\phi \left({\vec {v_{j}}}\right)d^{3}v_{j}\ ,$

where

$\phi \left({\vec {v_{j}}}\right)=\left({\frac {m_{j}}{2\pi kT}}\right)^{3/2}\exp \left[-{\frac {m_{j}v_{j}^{2}}{2kT}}\right]\ .$

Using this relation we find the following expression for the reaction rate

$r_{i;j}=n_{i}n_{j}\langle \sigma v\rangle _{i;j}=n_{i}n_{j}\int \sigma \left(\left|{\vec {v_{i}}}-{\vec {v_{j}}}\right|\right)\left(\left|{\vec {v_{i}}}-{\vec {v_{j}}}\right|\right)\phi \left({\vec {v_{i}}}\right)\phi \left({\vec {v_{j}}}\right)d^{3}v_{i}d^{3}v_{j}\ .$

Nuclear Networks

We saw that there could be two types of expressions for the reaction rates

First: $r_{i,j}={\frac {1}{1+\delta _{ij}}}n_{i}n_{j}<\sigma v>\ ,$ where the delta factor is included to avoid double counting when we have identical particles.

Second:$r_{i,j}=\lambda _{i}n_{i}\ .$

So if we have the reaction $i+j\rightarrow o+m\,$ the change in the densities is given by

$\left({\frac {\partial n_{i}}{\partial t}}\right)_{\rho }=\left({\frac {\partial n_{j}}{\partial t}}\right)_{\rho }=-r_{i,j}$

$\left({\frac {\partial n_{o}}{\partial t}}\right)_{\rho }=\left({\frac {\partial n_{m}}{\partial t}}\right)_{\rho }=+r_{i,j}$

Now recall that

$Y_{i}={\frac {n_{i}}{\rho N_{A}}}$

Then taking the derivative with respect to time we have

${\dot {Y}}_{i}={\frac {1}{\rho N_{A}}}\left({\frac {\partial n_{i}}{\partial t}}\right)_{\rho }$

Using the previous equations we can write it as

${\dot {Y}}_{i}={\frac {r_{i,j}}{\rho N_{A}}}=-{\frac {1}{1+\delta _{ij}}}\rho N_{A}<\sigma v>_{i,j}Y_{i}Y_{j}$

This definition is for the case where we destroy a nucleus $i$ and in the case where we produce a nucleus $m$

${\dot {Y}}_{m}={\frac {1}{1+\delta _{ij}}}\rho N_{A}<\sigma v>_{i,j}Y_{i}Y_{j}$