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.

The particle density: 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 = \int_{0}^{\infty}{\omega(p)f(p)dp}}

The energy density: 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 u = \int_{0}^{\infty}{E\omega(p)f(p)dp}}

The pressure: 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 1 \over 3 \int_{0}^{\infty}{pv\omega(p)f(p)dp}}

Occupation probabilities

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

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

Maxwell-Boltzmann

The probability distribution can be found by:

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 f(\epsilon_k) = \left( e^{ {\epsilon_k - \mu}/{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 (unknown function "\label"): {\displaystyle N=\sum_{k}n_k \label{eq:N}}

Failed to parse (unknown function "\label"): {\displaystyle E_{Total}=\sum_{k}n_k\epsilon_k \label{eq:E}}

Since we are dealing with fermions, $n_{k}$ can be 0 or 1. The thermodynamic (Landau) 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

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

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

Bose-Einstein

Consider a gas of bosons in which the particles satisfy equations (\ref{eq:N}) and(\ref{eq:E}). Similarly as in the Fermi-Dirac case, we can write the thermodynamic potential for a particular energy $\epsilon _{k}$ as

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

Notice that in this case there is no restriction on the number of particles occupying the same state 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 is because the particles have integer spin, and therefore do not satisfy Pauli exclusion principle.