2nd Week: Properties of Astrophysical Plasmas C

Thermodynamic Quantities

Important thermodynamic quantities for a volume $V$ can be found by integrating over the product of the density of state function $\omega (p)$ and the occupation probability of each state $f(p)$ for all momenta. So for example, the number density is:

$n={\frac {N}{V}}=\int {\omega (p)f(p)dp}$

and the energy density is:

$e={\frac {E}{V}}=\int {E(p)\omega (p)f(p)dp}$

and pressure is:

$P=\int {pv\omega (p)f(p)dp}$

So for example, to find $\omega (p)$ for a particle in a box, we can start with the well known solution for such a particles energy levels:

$E={\frac {\pi ^{2}\hbar ^{2}}{2md^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})$

Then we want the number of states that lead to a particle having an energy of E or less. Take the three n quantum numbers as a vectors within a sphere with radius, then for large energies the number of possible values that these n's can take to give that energy or less is just the area of the sphere $4/3\pi R^{3}$. Of course, we don't know what to do with negative values of n, so we'll just take the 1/8th of the sphere for which all values are positive. This gives us, for number of states:

$n_{states}={\frac {4g\pi }{3\hbar ^{3}}}(2mE)^{3/2}$

And our density of states is just the number per, dE, so taking the derivative with respect to E:

$\omega ={\frac {2\pi g}{\hbar ^{3}}}(2m)^{3/2}E^{1/2}$