Lecture 2

Drude Theory of Thermal Conductivity

The thermal conductivity $k$ is defined as the proportionality constant in Fourier's law:

$$\vec{\j}_Q = - \kappa \nabla T$$

where $\vec{\j}_Q =$ heat current = heat transported across a unit area per unit time.

In Drude's model, electrons transport heat because the electrons thermalized at the ``local" temperature of the last collision. (Assume thermalization at each collision.) Thus, the electrons leaving the hotter regions carry more energy. Let's derive Fourier's law and derive an expression for $\kappa$. We will use the same approach that we used for deriving $\sigma_0$. Look at $\frac{\textstyle d\vec{\j}_Q}{\textstyle dt}\bigg\vert _{total}$. There are 2 contributions - diffusion and collisions (which knock electrons out of the right direction):

$$\frac{d\vec{\j}_Q}{dt}\bigg\vert _{total} = \left( \frac{d\vec{\j... ...dt}\right)_{\it diffusion} + \left( \frac{d\vec{\j}_{Q}}{dt}\right)_{collision}$$

What is $\left( \frac{\textstyle d\vec{\j}}{\textstyle dt}\right)_{\it diffusion}$? Suppose the system has a temperature gradient but no collisions (save those for the other term). Consider the electrons which cross the plane $x = 0$ at a time $t$ after the start of the operation. An electron with velocity $\vec{v}$ will have come from a point $\vec{r} = - \vec{v}t$ away and will therefore have the average energy appropriate for that point, i.e., $\varepsilon (\vec{r})$. The extra energy will be

$$\delta \varepsilon _i = \vec{r}_i \cdot \nabla\varepsilon = \vec{... ...T}\right) = - t \vec{v}_i \cdot \nabla T \left( \frac{d\varepsilon }{dT}\right)$$

The ``extra" energy current is

$$\vec{\j}_Q(t)\bigg\vert _{\it diffusion} = \sum\limits_{i} \vec{v}_i \delta\varepsilon _i = - t\sum\limits_{i}\vec{v}_i (\vec{v}_i\cdot\nabla T)\left( \frac{d\varepsilon }{dT}\right)$$

$$\left( \frac{d\vec{\j}_Q}{dt}\right)_{\it diffusion} = -\left( \frac{d\varepsilon }{dT}\right)\sum\limits_{i}\vec{v}_i (\vec{v}_i\cdot\nabla T)$$

Since the velocity distribution is approximately isotropic (nothing makes the electrons go in the $\nabla T$ direction), the energy diffusion is along $\nabla T$. $\frac{\textstyle d\vec{\j}_Q}{\textstyle dt}$ has a component only along $\nabla T$ and we can replace the $v_i$'s with $\frac{1}{3}\overline{v^2}$. This gives

$$\left( \frac{d\vec{\j}_Q}{dt}\right)_{\it diffusion} = - \left( \frac{dE}{dT}\right)\frac{1}{3} \overline{v^2} \nabla T$$

where the energy per unit volume $E=n\varepsilon$, and $\varepsilon$ is the energy per particle.

Collisions Now put in collisions. We assume that the electrons thermalize at every collision. Thus the extra heat current carried by an electron is annihilated at every collision. $\vec{\j}_Q = \sum\limits_{i}\vec{v}_i\delta \varepsilon _i$ is also lost and we can simply write

$$\left( \frac{d\vec{\j}_Q}{dt}\right)_{collision} = - \frac{\vec{\j}_Q}{\tau}$$

Thus

$$\left( \frac{d\vec{\j}_Q}{dt}\right)_{total} = - \left( \frac{dE}{dT}\right) \frac{1}{3} \overline{v^2}\nabla T - \frac{\vec{\j}_Q}{\tau}$$

In steady state, $(\frac{\textstyle d\vec{\j}_Q}{\textstyle dt})_{total} = 0$. Using $c_v = (\frac{\textstyle dE}{\textstyle dt})\bigg\vert _{V}$ gives the standard Kinetic Theory of Gases result:

To the extent that $\overline{v^2} \tau = v\ell$,
Wiedemann-Franz Law

It is an empirical fact that

$$\frac{\kappa}{\sigma T} \simeq {\rm constant}$$

$$\sim 2\cdot 10^{-8} {\rm watt-ohm}/K^2$$

(See Table 1.6 in AM). $(\kappa /\sigma T)$ is called the Lorenz number. Can the Drude theory explain this? We have

$$\sigma = \frac{ne^2\tau}{m} \hspace*{1in} \kappa = \frac{1}{3} c_v \overline{v^2} \tau$$

Thus

$$\frac{\kappa}{\sigma T} = \frac{1}{3} \quad \frac{c_v \overline{v... ...u}{ne^2\tau T} = \frac{1}{3} \frac{c_vm\overline{v^2}}{ne^2T} \hspace*{1in} (*)$$

Now if one uses a simple Kinetic Theory of Gases picture for a monatomic gas:

$$\frac{1}{2} m\overline{v^2} = \frac{3}{2} k_BT \qquad {\rm and}\quad c_v = \frac{3}{2} nk_B$$

$$\Rightarrow \mbox{\fbox{$\frac{\textstyle \kappa}{\textstyle \sigma T} = \fra... ...extstyle 3}{\textstyle 2}\left( \frac{\textstyle k_B}{\textstyle e}\right)^2$}}$$

This is $1.11 \times 10^{-8} \;\frac{\textstyle W\Omega}{\textstyle k^2}$, i.e., almost exactly half the experimental value. We will see later that Eq. ($*$) is in fact approximately right: it is the evaluation of $c_v$ and $m\overline{v^2}$ that is wrong. Fortuitously, the mistakes cancel except for a factor of 2. Incidentally, there was no evidence in Drude's time for an electronic specific heat of $\frac{\textstyle 3}{\textstyle 2} nk_B$. $(\frac{\textstyle 3}{\textstyle 2} nk_B$ is a lot of specific heat.) So it was puzzling as to why the Lorenz number came out correctly when it appeared that there was no $c_{el}$ at room temperature.