Particle and energy current

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5.3 Particle and energy current

As often used before each volume element $d k$ contains

$2 \frac{d^{3} k}{{(2 π)}^{3}}$

(5.19)

electronic states which are occupied with

$d n = \frac{d^{3} k}{4 π^{3}} f (\vec{r}, \vec{k})$

(5.20)

particles. The complete current density is therefor

$\vec{j} = \frac{q}{4 π^{3}} \int_{V_{k}} \vec{v} (\vec{k}) f (\vec{r}, \vec{k}) d^{3} k = \frac{q}{4 π^{3}} \int_{V_{k}} \vec{v} (\vec{k}) f^{(1)} (\vec{r}, \vec{k}) d^{3} k .$

(5.21)

Here we used

$\int_{V_{k}} \vec{v} (\vec{k}) f_{0} (\vec{r}, \vec{k}) d^{3} k = 0$

(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:

$W = \frac{1}{4 π^{3}} \int_{V_{k}} E (\vec{k}) \vec{v} (\vec{k}) f^{(1)} (\vec{r}, \vec{k}) d^{3} k .$

(5.23)

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