Seebeck Effect - Thermopower

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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

$$\vec{E}= Q\nabla T$$

where the thermopower $Q$ should be $< 0$.

Let's calculate $Q$ in the Drude model.
Strategy: In steady state

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

Diffusion

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Suppose the gradient is along the $x$-direction. Consider the electrons crossing the $x = 0$ plane a time $t$ 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 $x = 0$ for the $x$-component is

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

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

where we used $\vec{\j}= - n e v$. $\vec{\j}_Q$ is the electric current due to the thermal gradient.

E-field current

Recall

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

Steady-State: (Heat current cancels $\vec{E}$-field current)

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

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

$$\vec{E} = Q \nabla T$$

where

In the Kinetic Theory of Gases,

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