Particle and energy current

5.3 Particle and energy current

As often used before each volume element \(dk\) contains

\begin{equation*} 2 \frac{d^3k}{(2\pi)^3} \end{equation*}

(5.19)

electronic states which are occupied with

\begin{equation*} dn = \frac{d^3k}{4\pi^3} f(\vec{r},\vec{k}) \end{equation*}

(5.20)

particles. The complete current density is therefor

\begin{equation*} \vec{j} = \frac{q}{4\pi^3}\int_{V_k} \vec{v}(\vec{k}) f(\vec{r},\vec{k}) d^3k = \frac{q}{4\pi^3}\int_{V_k} \vec{v}(\vec{k}) f^{(1)}(\vec{r},\vec{k}) d^3k \qquad . \label{curr_dens_solution} \end{equation*}

(5.21)

Here we used

\begin{equation*} \int_{V_k} \vec{v}(\vec{k}) f_0(\vec{r},\vec{k}) d^3k = 0 \end{equation*}

(5.22)

since \(f_0(\vec{r},\vec{k})\) is a symmetric function in \(\vec{k}\), \(\vec{v}\) an antisymmetric function in \(\vec{k}\) and we integrate over symmetric boundaries (In equilibrium no currents are flowing!). Correspondingly we find for the energy flux density:

\begin{equation*} W = \frac{1}{4\pi^3}\int_{V_k} E(\vec{k})\vec{v}(\vec{k}) f^{(1)}(\vec{r},\vec{k}) d^3k \qquad . \end{equation*}

(5.23)