A Microscopic View of Ohm's Law — Question 8

Question 8

The drift energy balance identifies where steady electrical work goes

Use the relaxation model $m\,d\vec v_d/dt=-e\vec E-m\vec v_d/\tau$, with electron charge $-e$, mass m and constant positive relaxation time tau. Let $u_d=nm|\vec v_d|^2/2$ be the energy density of the mean drift, and $\vec J=-ne\vec v_d$.

1. Derive an energy balance by dotting the equation with $n\vec v_d$.

2. In steady drift, express the dissipative term in terms of conductivity and field, and explain why steady drift energy does not increase.

Question 8 - Solution

The drift energy balance identifies where steady electrical work goes

(a) Method and calculation

$$nm\vec v_d\cdot\frac{d\vec v_d}{dt}=-ne\vec E\cdot\vec v_d-\frac{nm}{\tau}|\vec v_d|^2.$$$$\frac{du_d}{dt}=\vec J\cdot\vec E-\frac{nm}{\tau}|\vec v_d|^2.$$

(b) Interpretation and further calculation

$$\frac{nm}{\tau}|\vec v_d|^2=\frac{ne^2\tau}{m}E^2=\sigma E^2=\vec J\cdot\vec E.$$$$du_d/dt=0.$$The phenomenological friction term transfers drift energy to other degrees of freedom.

Physical check

The energy of the mean drift is not the full thermal or quantum kinetic energy of the electron gas.