Thermal Conductivity of a Metal: The Drude Picture
In the Drude model, we treat electrons as a gas of free particles moving through a lattice of ions. Just as these electrons carry electric charge (electrical current), they also carry kinetic energy. When a temperature gradient is applied to a metal, electrons moving from the hot side carry more energy than those from the cold side, resulting in a net flow of heat. This section derives the coefficient of thermal conductivity mathematically.
1. Physical Setup and Fourier's Law
Consider a metallic rod of cross-sectional area \(A\) with a temperature gradient applied along the \(x\)-axis.
- Left End: Temperature \(T(x)\) (Hotter)
- Right End: Temperature \(T(x+dx)\) (Colder)
We assume a steady state where no electrical current flows, only heat. We define the heat current density (thermal current per unit area) \(\mathbf{J}_Q\) as:
This flow is governed by Fourier's Law, which states that heat flows against the temperature gradient (from hot to cold):
Here, \(\kappa\) (kappa) is the thermal conductivity of the metal. Our goal is to find an expression for \(\kappa\) using microscopic physics.
2. Microscopic Picture (1D Model)
Let us simplify the problem to one dimension. Electrons move with a characteristic thermal speed \(v\).
- \(n\): Number density of electrons.
- \(\tau\): Relaxation time (time between collisions).
- \(\ell = v\tau\): Mean free path (distance traveled between collisions).
Consider a plane located at position \(x\).
Statistically, half the electrons at this plane arrived from the left, and half arrived from the right.
3. Energy Transport and Taylor Expansion
The crucial insight is that electrons carry the thermal energy characteristic of the location where they last collided.
- From the Left: These electrons last scattered at \(x - \ell\). Their energy is \(\varepsilon_L = \varepsilon[T(x-\ell)]\).
- From the Right: These electrons last scattered at \(x + \ell\). Their energy is \(\varepsilon_R = \varepsilon[T(x+\ell)]\).
Here, \(\varepsilon(T)\) is the average energy per electron at temperature \(T\).
Taylor Expansion
Since the mean free path \(\ell\) is microscopic, the temperature change over this distance is small. We can expand the energy function around \(x\):
\[ \varepsilon[T(x \pm \ell)] \approx \varepsilon[T(x)] \pm \ell \frac{dT}{dx} \frac{d\varepsilon}{dT} \]4. Calculating Net Heat Current
The net heat current density is the energy carried from left to right minus the energy carried from right to left.
\[ J_Q = J_Q^{L \to R} - J_Q^{R \to L} \]Substituting flux and energy:
\[ J_Q = \left( \frac{n}{2}v \cdot \varepsilon[T(x-\ell)] \right) - \left( \frac{n}{2}v \cdot \varepsilon[T(x+\ell)] \right) \]Factoring out constants:
\[ J_Q = \frac{n}{2}v \left[ \varepsilon(x-\ell) - \varepsilon(x+\ell) \right] \]Now, substitute the Taylor expansions:
\[ J_Q = \frac{n}{2}v \left[ \left( \varepsilon(T) - \ell \frac{d\varepsilon}{dT} \frac{dT}{dx} \right) - \left( \varepsilon(T) + \ell \frac{d\varepsilon}{dT} \frac{dT}{dx} \right) \right] \]The \(\varepsilon(T)\) terms cancel out:
\[ J_Q = \frac{n}{2}v \left[ -2\ell \frac{d\varepsilon}{dT} \frac{dT}{dx} \right] \] \[ J_Q = -n v \ell \frac{d\varepsilon}{dT} \frac{dT}{dx} \]Using \(\ell = v\tau\), we get the final derived equation for heat current:
5. Thermal Conductivity and Specific Heat
We now compare our derived equation with Fourier's Law (\(J_Q = -\kappa \frac{dT}{dx}\)):
\[ -\kappa \frac{dT}{dx} = -n v^2 \tau \frac{d\varepsilon}{dT} \frac{dT}{dx} \]Solving for \(\kappa\):
\[ \kappa = n v^2 \tau \frac{d\varepsilon}{dT} \]We can relate the term \(n \frac{d\varepsilon}{dT}\) to the electronic heat capacity per unit volume, \(C_V\). Since \(\varepsilon\) is energy per particle and \(n\) is particles per volume: \[ C_V = \frac{d(Energy/Volume)}{dT} = n \frac{d\varepsilon}{dT} \]
Substituting this back, we get the fundamental Drude result for thermal conductivity:
Note: In a rigorous 3D analysis accounting for the random direction of velocity components, a geometric factor of \(1/3\) appears, yielding \(\kappa = \frac{1}{3} v^2 \tau C_V\).
6. Classical Drude Evaluation & Wiedemann-Franz Law
Drude used classical kinetic theory (Maxwell-Boltzmann statistics) to evaluate this expression.
Applying Equipartition
For a classical monatomic gas: \[ \varepsilon = \frac{3}{2} k_B T \quad \Rightarrow \quad \frac{d\varepsilon}{dT} = \frac{3}{2} k_B \] \[ v^2 = \frac{3 k_B T}{m} \]
Substituting these into \(\kappa = n v^2 \tau \frac{d\varepsilon}{dT}\):
\[ \kappa = n \left( \frac{3 k_B T}{m} \right) \tau \left( \frac{3}{2} k_B \right) \]The Wiedemann-Franz Ratio
The Drude model also gives us the electrical conductivity: \[ \sigma = \frac{n e^2 \tau}{m} \] If we take the ratio \(\frac{\kappa}{\sigma T}\), the material-specific terms (\(n, \tau, m\)) conveniently cancel out:
\[ \frac{\kappa}{\sigma T} = \frac{\frac{9}{2} \frac{n k_B^2 T \tau}{m}}{\frac{n e^2 \tau}{m} T} = \frac{9}{2} \left( \frac{k_B}{e} \right)^2 \]This predicts that the ratio of thermal to electrical conductivity is a universal constant for all metals.
7. Success and Failure of the Model
The Success: The Wiedemann-Franz law holds experimentally. The prediction that good electrical conductors are also good thermal conductors is correct.
The "Lucky" Failure: The Drude derivation makes two huge errors that cancel each other out:
- It overestimates the heat capacity \(C_V\) (using classical \(\frac{3}{2}nk_B\) instead of the much smaller quantum electronic heat capacity).
- It underestimates the velocity \(v^2\) (using thermal velocity instead of the much higher Fermi velocity \(v_F\)).
The Correction (Sommerfeld Model): Later, Sommerfeld corrected this using Fermi-Dirac statistics. The corrected form uses the Fermi velocity \(v_F\) and the quantum heat capacity proportional to \(T\):