Drude model

Proof using the equation of motion^{[Ashcroft & Mermin 15]}

Given $${\displaystyle {\begin{aligned}\mathbf {p} (t)&=\Re {\left(\mathbf {p} (\omega )e^{-i\omega t}\right)}\\\mathbf {E} (t)&=\Re {\left(\mathbf {E} (\omega )e^{-i\omega t}\right)}\end{aligned}}}$$ And the equation of motion above $${\displaystyle {\frac {d}{dt}}\mathbf {p} (t)=-e\mathbf {E} -{\frac {\mathbf {p} (t)}{\tau }}}$$ substituting $${\displaystyle -i\omega \mathbf {p} (\omega )=-e\mathbf {E} (\omega )-{\frac {\mathbf {p} (\omega )}{\tau }}}$$ Given $${\displaystyle {\begin{aligned}\mathbf {j} &=-ne{\frac {\mathbf {p} }{m}}\\\mathbf {j} (t)&=\Re {\left(\mathbf {j} (\omega )e^{-i\omega t}\right)}\\\mathbf {j} (\omega )&=-ne{\frac {\mathbf {p} (\omega )}{m}}={\frac {(ne^{2}/m)\mathbf {E} (\omega )}{1/\tau -i\omega }}\end{aligned}}}$$ defining the complex conductivity from: $${\displaystyle \mathbf {j} (\omega )=\sigma (\omega )\mathbf {E} (\omega )}$$ We have: $${\displaystyle \sigma (\omega )={\frac {\sigma _{0}}{1-i\omega \tau }};\sigma _{0}={\frac {ne^{2}\tau }{m}}}$$

Proof using Maxwell's equations^{[Ashcroft & Mermin 16]}

Given the approximations for the ${\displaystyle \sigma (\omega )}$ included above

• we assumed no electromagnetic field: this is always smaller by a factor v/c given the additional Lorentz term ${\displaystyle -{\frac {e\mathbf {p} }{mc}}\times \mathbf {B} }$ in the equation of motion
• we assumed spatially uniform field: this is true if the field does not oscillate considerably across a few mean free paths of electrons. This is typically not the case: the mean free path is of the order of Angstroms corresponding to wavelengths typical of X rays.

The following are Maxwell's equations without sources (which are treated separately in the scope of plasma oscillations), in Gaussian units: $${\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &=0;&\nabla \cdot \mathbf {B} &=0;\\\nabla \times \mathbf {E} &=-{\frac {1}{c}}{\frac {\partial \mathbf {B} }{\partial t}};&\nabla \times \mathbf {B} &={\frac {4\pi }{c}}\mathbf {j} +{\frac {1}{c}}{\frac {\partial \mathbf {E} }{\partial t}}.\end{aligned}}}$$ Then $${\displaystyle \nabla \times \nabla \times \mathbf {E} =-\nabla ^{2}\mathbf {E} ={\frac {i\omega }{c}}\nabla \times \mathbf {B} ={\frac {i\omega }{c}}\left({\frac {4\pi \sigma }{c}}\mathbf {E} -{\frac {i\omega }{c}}\mathbf {E} \right)}$$ or $${\displaystyle -\nabla ^{2}\mathbf {E} ={\frac {\omega ^{2}}{c^{2}}}\left(1+{\frac {4\pi i\sigma }{\omega }}\right)\mathbf {E} }$$ which is an electromagnetic wave equation for a continuous homogeneous medium with dielectric constant ${\displaystyle \varepsilon (\omega )}$ in the Helmoltz form $${\displaystyle -\nabla ^{2}\mathbf {E} ={\frac {\omega ^{2}}{c^{2}}}\varepsilon (\omega )\mathbf {E} }$$ where the refractive index is ${\textstyle n(\omega )={\sqrt {\varepsilon (\omega )}}}$ and the phase velocity is ${\displaystyle v_{\text{p}}={\frac {c}{n(\omega )}}}$ therefore the complex dielectric constant is $${\displaystyle \varepsilon (\omega )=\left(1+{\frac {4\pi i\sigma }{\omega }}\right)}$$ which in the case ${\displaystyle \omega \tau \gg 1}$ can be approximated to: $${\displaystyle \varepsilon (\omega )=\left(1-{\frac {\omega _{\rm {p}}^{2}}{\omega ^{2}}}\right);\omega _{\rm {p}}^{2}={\frac {4\pi ne^{2}}{m}}{\text{(Gaussian units)}}.}$$ In SI units the ${\displaystyle 4\pi }$ in the numerator is replaced by ${\displaystyle \varepsilon _{0}}$ in the denominator.

Derivation and Drude's errors^{[Ashcroft & Mermin 16]}

Solids can conduct heat through the motion of electrons, atoms, and ions. Conductors have a large density of free electrons whereas insulators do not; ions may be present in either. Given the good electrical and thermal conductivity in metals and the poor electrical and thermal conductivity in insulators, a natural starting point to estimate the thermal conductivity is to calculate the contribution of the conduction electrons.

The thermal current density is the flux per unit time of thermal energy across a unit area perpendicular to the flow. It is proportional to the temperature gradient. $${\displaystyle \mathbf {j} _{q}=-\kappa \nabla T}$$ where ${\displaystyle \kappa }$ is the thermal conductivity. In a one-dimensional wire, the energy of electrons depends on the local temperature ${\displaystyle \varepsilon [T(x)]}$ If we imagine a temperature gradient in which the temperature decreases in the positive x-direction, the average electron velocity is zero (but not the average speed). The electrons arriving at location x from the higher-energy side will arrive with energies ${\displaystyle \varepsilon [T(x-v\tau )]}$, while those from the lower-energy side will arrive with energies ${\displaystyle \varepsilon [T(x+v\tau )]}$. Here, ${\displaystyle v}$ is the average speed of electrons and ${\displaystyle \tau }$ is the average time since the last collision.

The net flux of thermal energy at location x is the difference between what passes from left to right and from right to left: $${\displaystyle \mathbf {j} _{q}={\frac {1}{2}}nv{\big (}\varepsilon [T(x-v\tau )]-\varepsilon [T(x+v\tau )]{\big )}}$$ The factor of ⁠1/2⁠ accounts for the fact that electrons are equally likely to be moving in either direction. Only half contribute to the flux at x.

When the mean free path ${\displaystyle \ell =v\tau }$ is small, the quantity ${\displaystyle {\big (}\varepsilon [T(x-v\tau )]-\varepsilon [T(x+v\tau )]{\big )}/2v\tau }$ can be approximated by a derivative with respect to x. This gives $${\displaystyle \mathbf {j} _{q}=nv^{2}\tau {\frac {d\varepsilon }{dT}}\cdot \left(-{\frac {dT}{dx}}\right)}$$ Since the electron moves in the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ directions, the mean square velocity in the ${\displaystyle x}$ direction is ${\displaystyle \langle v_{x}^{2}\rangle ={\tfrac {1}{3}}\langle v^{2}\rangle }$. We also have ${\displaystyle n{\frac {d\varepsilon }{dT}}={\frac {N}{V}}{\frac {d\varepsilon }{dT}}={\frac {1}{V}}{\frac {dE}{dT}}=c_{v}}$, where ${\displaystyle c_{v}}$ is the specific heat capacity of the material.

Putting all of this together, the thermal energy current density is $${\displaystyle \mathbf {j} _{q}=-{\frac {1}{3}}v^{2}\tau c_{v}\nabla T}$$ This determines the thermal conductivity: $${\displaystyle \kappa ={\frac {1}{3}}v^{2}\tau c_{v}}$$ (This derivation ignores the temperature-dependence, and hence the position-dependence, of the speed v. This will not introduce a significant error unless the temperature changes rapidly over a distance comparable to the mean free path.)

Dividing the thermal conductivity ${\displaystyle \kappa }$ by the electrical conductivity ${\displaystyle \sigma ={\frac {ne^{2}\tau }{m}}}$ eliminates the scattering time ${\displaystyle \tau }$ and gives $${\displaystyle {\frac {\kappa }{\sigma }}={\frac {c_{v}mv^{2}}{3ne^{2}}}}$$

At this point of the calculation, Drude made two assumptions now known to be errors. First, he used the classical result for the specific heat capacity of the conduction electrons: ${\displaystyle c_{v}={\tfrac {3}{2}}nk_{\rm {B}}}$. This overestimates the electronic contribution to the specific heat capacity by a factor of roughly 100. Second, Drude used the classical mean square velocity for electrons, ${\displaystyle {\tfrac {1}{2}}mv^{2}={\tfrac {3}{2}}k_{\rm {B}}T}$. This underestimates the energy of the electrons by a factor of roughly 100. The cancellation of these two errors results in a good approximation to the conductivity of metals. In addition to these two estimates, Drude also made a statistical error and overestimated the mean time between collisions by a factor of 2. This confluence of errors gave a value for the Lorenz number that was remarkably close to experimental values.

The correct value of the Lorenz number as estimated from the Drude model is^{[Ashcroft & Mermin 19]} $${\displaystyle {\frac {\kappa }{\sigma T}}={\frac {3}{2}}\left({\frac {k_{\rm {B}}}{e}}\right)^{2}=1.11\times 10^{-8}\,{\text{W}}\Omega /{\text{K}}^{2}.}$$

Proof together with the Drude errors^{[Ashcroft & Mermin 21]}

From the simple one dimensional model $${\displaystyle v_{Q}={\frac {1}{2}}[v(x-v\tau )-v(x+v\tau )]=-v\tau {\frac {dv}{dx}}=-\tau {\frac {d}{dx}}\left({\frac {v^{2}}{2}}\right)}$$ Expanding to 3 degrees of freedom ${\displaystyle \langle v_{x}^{2}\rangle ={\frac {1}{3}}\langle v^{2}\rangle }$ $${\displaystyle \mathbf {v_{Q}} =-{\frac {\tau }{6}}{\frac {dv^{2}}{dT}}(\nabla T)}$$ The mean velocity due to the Electric field (given the equation of motion above at equilibrium) $${\displaystyle \mathbf {v_{E}} =-{\frac {e\mathbf {E} \tau }{m}}}$$ To have a total current null ${\displaystyle \mathbf {v_{E}} +\mathbf {v_{Q}} =0}$ we have $${\displaystyle Q=-{\frac {1}{3e}}{\frac {d}{dT}}\left({\frac {mv^{2}}{2}}\right)=-{\frac {c_{v}}{3ne}}}$$ And as usual in the Drude case ${\displaystyle c_{v}={\frac {3}{2}}nk_{\rm {B}}}$ $${\displaystyle Q=-{\frac {k_{\rm {B}}}{2e}}=0.43\times 10^{-4}~\mathrm {V/K} }$$ where the typical thermopowers at room temperature are 100 times smaller of the order of microvolts.^{[Ashcroft & Mermin 20]}