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Commit 39d645f2 authored by T. van der Sar's avatar T. van der Sar
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Update 3_drude_model.md - polishing

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......@@ -14,7 +14,7 @@ _(based on chapter 3 of the book)_
After this lecture you will be able to:
- discuss the basics of 'Drude theory', which describes electron motion in metals.
- discuss the basics of Drude theory, which describes electron motion in metals.
- use Drude theory to analyze transport of electrons through conductors in electric and magnetic fields.
- describe central terms such as the mobility and the Hall resistance.
......@@ -25,19 +25,21 @@ Ohm's law states that $V=IR=I\rho\frac{l}{A}$. In this lecture we will investiga
- At each scattering event an electron returns to momentum ${\bf p}=0$.
- In-between scattering events electrons respond to the Lorentz force ${\bf F}_{\rm L}=-e\left({\bf E}+{\bf v}\times{\bf B}\right)$.
For now we will consider only an electric field (_i.e._ ${\bf B}=0$). What velocity do electrons acquire in-between collisions?
We start by considering only an electric field (_i.e._ ${\bf B}=0$). What velocity do electrons acquire in-between collisions?
$$
{\bf v}=-\int_0^\tau\frac{e{\bf E}}{m_{\rm e}}{\rm d}t=-\frac{e\tau}{m_{\rm e}}{\bf E}=-\mu{\bf E}
{\bf v}=-\int_0^\tau\frac{e{\bf E}}{m_{\rm e}}{\rm d}t=-\frac{e\tau}{m_{\rm e}}{\bf E}=-\mu{\bf E},
$$
Here we have defined the quantity $\mu\equiv e\tau/m_{\rm e}$, which is the _mobility_. If we have a density $n$ of electrons in our solid, the current density ${\bf j}$ [A/m$^2$] then becomes:
where we have defined the _mobility_ $\mu\equiv e\tau/m_{\rm e}$. The current density ${\bf j}$ [A/m$^2$] is given by:
$$
{\bf j}=-en{\bf v}=\frac{n e^2\tau}{m_{\rm e}}{\bf E}=\sigma{\bf E}\ ,\ \ \sigma=\frac{ne^2\tau}{m_{\rm e}}=ne\mu
$$
$\sigma$ is the conductivity, which is the inverse of resistivity: $\rho=\frac{1}{\sigma}$. If we now take $j=\frac{I}{A}$ and $E=\frac{V}{l}$, we retrieve Ohm's Law: $\frac{I}{A}=\frac{V}{\rho l}$.
where $n$ is the density of electrons in our solid, and $\sigma$ is the conductivity, which is the inverse of resistivity: $\sigma=\frac{1}{\rho}$.
If we now take $j=\frac{I}{A}$ and $E=\frac{V}{l}$, we retrieve Ohm's Law: $\frac{I}{A}=\frac{V}{\rho l}$.
Scattering is caused by collisions with:
......@@ -56,21 +58,21 @@ _Matthiessen's Rule_ (1864). Solid (dashed) curve: $\rho(T)$ for a pure (impure)
How fast do electrons travel through a copper wire? Let's take $E$ = 1 volt/m, $\tau$ ~ 25 fs (Cu, $T=$ 300 K).
$\rightarrow v=\mu E=\frac{e\tau}{m_{\rm e}}E=\frac{10^{-19}\times 2.5\times 10^{-14}}{10^{-30}}=2.5\times10^{-3}=2.5$ mm/s ! (= 50 $\mu$m @ 50 Hz AC)
$\rightarrow v=\mu E=\frac{e\tau}{m_{\rm e}}E=2.5$ mm/s ! (= 50 $\mu$m @ 50 Hz AC)
### Hall effect
Consider a conductive wire in a magnetic field ${\bf B} \rightarrow$ electrons are deflected in a direction perpendicular to ${\bf B}$ and ${\bf j}$.
![](figures/hall_effect.svg)
${\bf E}_{\rm H}$ = _Hall voltage_, caused by the Lorentz force.
${\bf E}_{\rm H}$ is the electric field caused by the Lorentz force, leading to a _Hall voltage_ in the direction perpendicular to ${\bf B}$ and ${\bf j}$.
In equilibrium, assuming that the average velocity becomes zero after every collision: $\frac{mv_x}{\tau}=-eE$
The $y$-component of the Lorentz force $-e{\bf v}_x\times{\bf B}$ is being compensated by the Hall voltage ${\bf E}_{\rm H}={\bf v}_x\times{\bf B}=\frac{1}{ne}{\bf j}\times{\bf B}$. The total electric field then becomes
The $y$-component of the Lorentz force $-e{\bf v}_x\times{\bf B}$ is being compensated by the Hall electric field ${\bf E}_{\rm H}={\bf v}_x\times{\bf B}=\frac{1}{ne}{\bf j}\times{\bf B}$. The total electric field then becomes:
$$
{\bf E}=\left(\frac{1}{ne}{\bf j}\times{\bf B}+\frac{m}{ne^2\tau}{\bf j}\right)
{\bf E}=\left(\frac{1}{ne}{\bf j}\times{\bf B}+\frac{m}{ne^2\tau}{\bf j}\right).
$$
We now introduce the _resistivity matrix_ $\tilde{\rho}$ as ${\bf E}=\tilde{\rho}{\bf j}$, where the diagonal elements are simply $\rho_{xx}=\rho_{yy}=\rho_{zz}=\frac{m}{ne^2\tau}$. The off-diagonal element $\rho_{xy}$ gives us:
......
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