Seebeck Effect - Thermopower

The fact that higher energy electrons move faster means that setting up a temperature gradient would give rise to an electric current. So, if we could set up a temperature gradient on a closed circuit, a current would flow up the gradient (since electrons are negatively charged). On an open circuit, the flow gives rise to a build-up of charge at the ends, and a steady state is obtained when the field $\vec{E}$ produced is sufficient to cancel the drift. Hence we expect

$\displaystyle \frac{d\vec{\j}}{dt}\bigg\vert _{total} = \frac{d\vec{\j}_Q}{dt}\bigg\vert _{\it diffusion} + \frac{d\vec{\j}_E}{dt}\bigg\vert _{field} = 0$

=2.5 true in $\epsfbox{thermopower2.eps}$

Suppose the gradient is along the

-direction. Consider the electrons crossing the

plane a time

after the start of the operation. An electron with velocity $\vec{v}$ will come from a point $\vec{r} = - \vec{v}t$ away and will have a velocity appropriate to $\vec{r}$ . Let $\nabla v_x \vert\vert\;\nabla T\;\vert\vert +{\hat x}$ . The excess velocity at

for the

-component is

$\displaystyle \delta v_x = - t \vec{v}\cdot \nabla v_x = - tv_x\nabla v_x = - \frac{1}{2}t\nabla v_x^2= - \frac{1}{6} t\nabla \overline{v^2}$

$\displaystyle \frac{\partial v_x}{\partial t} = - \frac{1}{6} \nabla \overline{... ...rline{v^2} = + \frac{1}{6} ne\frac{\partial \overline{v^2}}{\partial T}\nabla T$

$\displaystyle \left( \frac{\partial \vec{\j}}{\partial t}\right)_{field} = \frac{ne^2\vec{E}}{m}$

$\displaystyle \frac{d\vec{\j}}{dt}\bigg\vert _{tot} =\frac{d\vec{\j}_Q}{dt}\bigg\vert _{\it diff} + \frac{d\vec{\j}}{dt}\bigg\vert _{\vec{E}} =0$

$\displaystyle \frac{ne^2 \vec{E}}{m}$	$\displaystyle =$	$\displaystyle - \frac{1}{6}ne \frac{\partial \overline{v^2}}{\partial T}\nabla T$
$\displaystyle \vec{E}$	$\displaystyle =$	$\displaystyle Q \nabla T$

$\fbox{\parbox{2.5in}{$\quad Q = - \frac{\textstyle 1}{\textstyle 6e} \frac{\text... ...) = - \frac{\textstyle 1}{\textstyle 3e} \frac{\textstyle c_v}{\textstyle n}$}}$

$\displaystyle c_v = \frac{3}{2} nk_B \Rightarrow Q = - \frac{k_B}{2e} \sim \left({100 \times \mbox{too} \atop \mbox{\quad large for expt.}}\right)$