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

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

The pressure: $P={1 \over 3}\int _{0}^{\infty }{pv\omega (p)f(p)dp}$

where v is the velocity, p is the momentum, E is the energy, $f(p)$ is the probability function, and $\omega (p)$ is the state density.

Wave State Derivation

A brief derivation of a wave state density $\omega (E)$.

One can begin with the stationary Schroedinger equation.

$H\psi _{1,2,...n}=\left(T+V\right)\psi _{1,2,...n}=E\psi _{1,2,...n}$

$T=\sum _{i=1}^{n}t_{i}=\sum _{i=1}^{n}-{\hbar ^{2} \over 2m}\nabla _{i}^{2}$

$V=\sum _{i=1}^{n}\sum _{j>i}^{n}v_{ij}$

However, for noninteracting particles: $t\left(i,j\right)=v\left(i,j\right)=0$

Also since these are stationary states, H describes a particle in a box of size d. So then breaking $\psi$ into $\psi _{x}\psi _{y}\psi _{z}$ one gets the solution:

$\psi _{x}={\sqrt {2 \over d}}sin\left({{\sqrt {2mE}} \over \hbar }x\right)$ and that ${{\sqrt {2mE}} \over \hbar }=n_{x}\pi$

This has a corresponding energy $E_{x}={\pi ^{2}\hbar ^{2} \over 2md^{2}}n_{x}^{2}$

One will get similar solutions for $\psi _{y}$ and $\psi _{z}$.

This means that the total energy of the particle can be found by: $E_{i}={\pi ^{2}\hbar ^{2} \over 2md^{2}}\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)={\pi ^{2}\hbar ^{2} \over 2md^{2}}R_{\rho }^{2}$

One does not want all of the eigenstates but only those within a certain range. These can be described as:

$\phi (E)={1 \over 8}{4\pi \over 3}R_{\rho }^{3}={4\pi \over 3}{V(2m)^{3/2} \over \hbar ^{3}}E^{3/2}$

Then taking the derivative with respect to E and dividing by the volume will give the wave state density.

$\omega (E)={1 \over V}{\partial \phi \over \partial E}={2\pi (2m)^{3/2} \over \hbar ^{3}}E^{1/2}$

Occupation probabilities

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

$dE=TdS-PdV+\mu dN$

Other thermodynamical identities can be found in the lecture notes on Blackboard.

Maxwell-Boltzmann

The probability distribution can be found by:

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

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

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 $\epsilon _{k}$. This is because the particles have integer spin, and therefore do not satisfy Pauli exclusion principle.