2nd Week: Properties of Astrophysical Plasmas B

Before an in-depth analysis of nuclear astrophysics can begin, one must review the basics of nuclear physics. This begins with thermodynamics.

Basics of Thermodynamics

Here are the definitions of some of the basic quantities.

$n=\int _{0}^{\infty }{\omega (p)f(p)}$

Occupation probabilities

The 1st law of Thermodynamics in a system (or subsystem) with variable number of particles is

$dE=TdS-PdV+\mu dN$ ...

Maxwell-Boltzmann

The probability distribution can be found by:

$f(p)=\left(e^{{E-\mu } \over {kt}}\right)^{-1}$

Fermi-Dirac

Suppose that our system has discrete energies and that $n_{k}$ is the number of particles occupying the energy level $\epsilon _{k}$. This two quantities must satisfy

$N=\sum _{k}n_{k}$

$E_{Total}=\sum _{k}n_{k}\epsilon _{k}$

Since we are dealing with fermions, $n_{k}$ can be 0 or 1. The thermodynamic potential for a particular energy sate $\epsilon _{k}$ can be written as

$\Omega _{k}=-kT\log(\sum _{n_{k}=0}^{1}e^{n_{k}(\mu -\epsilon _{k})/kT})$

$\Omega _{k}=-kT\log(1+e^{(\mu -\epsilon _{k})/kT})$

Recall that, the mean particle number in a certain energy state $\epsilon _{k}$ is minus the derivative of the thermodynamic potential $\Omega _{k}$ with respect to the chemical potential $\mu$, at V and T constant. Therefore

$f(\epsilon _{k})=-{\partial \Omega _{k} \over \partial \mu }=...$

$f(\epsilon _{k})={1 \over {e^{(\epsilon _{k}-\mu )/kT}+1}}$