Question

Calculation of average kinetic ebergy from density of states

Answer 1

In general the density of states, related to volume V and N countable energy levels, is defined by:

{\displaystyle D(E)={\frac {1}{V}}\cdot \sum _{i=1}^{N}\delta \left(E-E\left({\vec {k}}_{i}\right)\right).}

Using {\displaystyle (\Delta k)^{d}=\left({\frac {2\pi }{L}}\right)^{d}} (the smallest allowed change of {\displaystyle k} for a particle in a box of dimension {\displaystyle d} and length {\displaystyle L}) under the limit {\displaystyle L\to \infty }, one derives the volume-related density of states for continuous energy levels

{\displaystyle D(E):=\int _{\mathbb {R} ^{d}}{\frac {\mathrm {d} ^{d}k}{(2\pi )^{d}}}\cdot \delta \left(E-E\left({\vec {k}}\right)\right).}

With {\displaystyle d} of the spatial dimension of the considered system and {\displaystyle k} the wave vector.

Equivalently, the density of states can also be understood as the derivative of the microcanonical partition function {\displaystyle Z_{m}(E)}with respect to the energy:

{\displaystyle D(E)={\frac {1}{V}}\cdot {\frac {\mathrm {d} Z_{m}(E)}{\mathrm {d} E}}}

The number of states with energy {\displaystyle E'} (degree of degeneracy) is given by:

{\displaystyle g\left(E'\right)=\lim _{\Delta E\to 0}\int _{E'}^{E'+\Delta E}D(E)\mathrm {d} E=\lim _{\Delta E\to 0}D\left(E'\right)\Delta E,}

where the last equality only applies when the mean value theorem for integrals is valid.