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.

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

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_k} is the number of particles occupying the energy level Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} . This two quantities must satisfy

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N=\sum_{k}n_k}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{Total}=\sum_{k}n_k\epsilon_k}

Since we are dealing with fermions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

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

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