rework drude lecture

src/3_drude_model.md

@@ -26,22 +26,25 @@ _(based on chapter 3 of the book)_
     - Use Drude theory to analyze transport of electrons through conductors in electric and magnetic fields.
     - Describe concepts such as electron mobility and the Hall resistance.
 
-## Drude theory
+## The starting point
 
 The Ohm's law, familiar to most from the high school, states that voltage is proportional to current $V=IR$.
 Since we are dealing with *material properties*, let us rewrite this into a relation that does not depend on the material geometry:
-$$V = I ρ \frac{l}{A} ⇒ E = ρ j,$$
+$$
+V = I ρ \frac{l}{A} ⇒ E = ρ j,
+$$
 where $E≡V/l$ is the electric field, $ρ$ the material resistivity, and $j≡I/A$ the current through a unit cross-section.
 Our goal is to understand how this law may arise microscopically, starting from reasonable (but definitely incomplete) assumptions.
 
 - Electrons fly freely, and scatter at random uncorrelated times, with an average scattering time $τ$.
 - After each scattering event, the electron's momentum randomizes with a zero average $⟨\mathbf{p}⟩=0$.
-- The Lorentz force ${\bf F}_{\rm L}=-e\left({\bf E}+{\bf v}\times{\bf B}\right)$ acts on the electrons.
+- The Lorentz force $\mathbf{F}_L=-e\left(\mathbf{E}+\mathbf{v}×\mathbf{B}\right)$ acts on the electrons.
 
 The first assumption here is the least obvious: why the time between scattering events not depend on e.g. electron velocity?
 Also observe that for now we forget that electrons also have fermionic statistics—this will come up in the next lecture, and turns out also helps to justify the first assumption.
 
-Even under these minimal assumptions, our problem seems hard. This is how electron motion looks like under these assumptions:
+Even under these minimal assumptions, our problem seems hard.
+This is how electron motion looks like under these assumptions:
 
 ```python
 %matplotlib inline
@@ -105,7 +108,7 @@ def animate(i):
         line.set_data(trace[0][:i, lnum], trace[1][:i, lnum])
         data = np.stack((scatter_pts[0][:i,lnum], scatter_pts[1][:i, lnum])).T
         scatterers[lnum].set_offsets(data)
-#         
+
 anim = animation.FuncAnimation(fig, animate, interval=100)
 plt.axis('off');
 plt.close();
@@ -117,64 +120,103 @@ Stop here for a second, and ask yourself how you would deal with this problem?
 
 ---
 
-We start by considering only an electric field (_i.e._ ${\bf B}=0$). What velocity do electrons acquire in-between collisions?
+### Key idea
+
+The answer follows from two important principles:
+
+**Always write down what you want to compute.**
 
+Our goal is finding the *electric current density* $j$.
+Each electron with charge $e$ and velocity $\mathbf{v}$ carries current $e\mathbf{v}$.
+Therefore if the electron density is $n$, the *average* current they carry is $ne⟨\mathbf{v}⟩$.
+Our goal is then to compute the *average* velocity.
+
+**Figuring out what happens on the average is simpler than figuring out what
+happens with each individual element.**
+
+Let us compute how the average velocity changes with time.
+The equation with the Lorentz force we just average right away:
 $$
-{\bf v}=-\int_0^\tau\frac{e{\bf E}}{m_{\rm e}}{\rm d}t=-\frac{e\tau}{m_{\rm e}}{\bf E}=-\mu{\bf E},
+m\frac{d⟨\mathbf{v}⟩}{dt} = -e\left(\mathbf{E}+⟨\mathbf{v}⟩×\mathbf{B}\right).
+$$
+Almost there, but we still need to do something with the change of the average velocity due to scattering.
+Consider the effect that scattering has over a small time $dt$.
+Most electrons continue with the same velocity, however a fraction $dt/τ$ will scatter, and that their average velocity becomes zero.
+Therefore we get
+$$
+⟨\mathbf{v}(t+dt)⟩ = ⟨\mathbf{v}(t)⟩(1 - dt/τ) + 0⋅(dt/τ) ⇒ \frac{d⟨\mathbf{v}⟩}{dt} = -\frac{⟨\mathbf{v}⟩}{τ}.
 $$
 
-where we have defined the _mobility_ $\mu\equiv e\tau/m_{\rm e}$. The current density ${\bf j}$ [A/m$^2$] is given by:
-
+That's it!
+We now combine both contributions into a single equation and get
 $$
-{\bf j}=-en{\bf v}=\frac{n e^2\tau}{m_{\rm e}}{\bf E}=\sigma{\bf E}\ ,\ \ \sigma=\frac{ne^2\tau}{m_{\rm e}}=ne\mu
+m\frac{d⟨\mathbf{v}⟩}{dt} = -m\frac{⟨\mathbf{v}⟩}{τ} -e\left(\mathbf{E}+⟨\mathbf{v}⟩×\mathbf{B}\right).
 $$
+Observe that the first term on the right hand side has the same form as the viscous friction force—it tries to stop each particle at place.
 
-where $n$ is the density of electrons, and $\sigma$ is the conductivity, which is the inverse of resistivity $\rho=\frac{1}{\sigma}$.
+We have now derived the necessary equation, the rest is merely applying it.
 
-If we now take $j=\frac{I}{A}$ and $E=\frac{V}{l}$, we retrieve Ohm's Law: $\frac{I}{A}=\frac{V}{\rho l}$.
+### Consequences of the Drude model
 
-Scattering is caused by collisions with:
+For convenience from now on we will omit the average signs, and write $\mathbf{v}$ instead of $⟨\mathbf{v}⟩$
+For a warm-up consider $\mathbf{B} = 0$, and a constant $\mathbf{E}$.
+After we wait long enough, we expect the average electron velocity to become constant, $d\mathbf{v}/dt = 0$, and we immediately get
+$$
+\mathbf{v}=-\frac{eτ}{m}\mathbf{E}=-μ\mathbf{E},
+$$
+where we have defined the *mobility* $μ\equiv eτ/m$—the ratio between the electron *drift velocity* and the electric field.
+Substituting this in the definition of the current density we arrive to
+$$
+\mathbf{j}=-en\mathbf{v}=\frac{n e^2\tau}{m}\mathbf{E}=\sigma\mathbf{E},\quad \sigma=\frac{ne^2\tau}{m}=ne\mu,
+$$
+where $\sigma$ is the conductivity, so that $\rho=\frac{1}{\sigma}$.
+
+#### Origins of scattering
 
-- Phonons: $\tau_{\rm ph}(T)$ ($\tau_{\rm ph}\rightarrow\infty$ as $T\rightarrow 0$)
-- Impurities/vacancies: $\tau_0$
+Something that Drude could not guess is that electrons do not scatter off of every atom (more on that in later weeks).
+Instead scattering always happens due to some deviation from a perfect crystal:
 
-Scattering rate $\frac{1}{\tau}$:
+- Phonons: $\tau_\mathrm{ph}(T)$ ($\tau_\mathrm{ph}\rightarrow\infty$ as $T\rightarrow 0$)
+- Impurities/vacancies or other crystalline defects: $\tau_0$
 
+The scattering rates $1/\tau$ due to different mechanisms add up:
 $$
-\frac{1}{\tau}=\frac{1}{\tau_{\rm ph}(T)}+\frac{1}{\tau_0}\ \Rightarrow\ \rho=\frac{1}{\sigma}=\frac{m}{ne^2}\left( \frac{1}{\tau_{\rm ph}(T)}+\frac{1}{\tau_0} \right)\equiv \rho_{\rm ph}(T)+\rho_0
+\frac{1}{\tau}=\frac{1}{\tau_\mathrm{ph}(T)}+\frac{1}{\tau_0}\ \Rightarrow\ \rho=\frac{1}{\sigma}=\frac{m}{ne^2}\left( \frac{1}{\tau_\mathrm{ph}(T)}+\frac{1}{\tau_0} \right)\equiv \rho_\mathrm{ph}(T)+\rho_0
 $$
 
-![](figures/matthiessen.svg)
+This explains the empirical *Matthiessen's Rule* (1864).
 
-_Matthiessen's Rule_ (1864). Solid (dashed) curve: $\rho(T)$ for a pure (impure) crystal.
+![](figures/matthiessen.svg)
 
-How fast do electrons travel through a copper wire? Let's take $E$ = 1 volt/m, $\tau$ ~ 25 fs (Cu, $T=$ 300 K).
+Here the solid is $\rho(T)$ of a pure crystal, and the dashed of an impure one.
 
-$\rightarrow v=\mu E=\frac{e\tau}{m_{\rm e}}E=2.5$ mm/s ! (= 50 $\mu$m @ 50 Hz AC)
+How fast do electrons travel through a copper wire?
+Let's take $E = 1$ volt/m, $τ∼25$ fs (Cu, $T=300$ K).  
+$⇒ v=\mu E=\frac{eτ}{m}E=2.5$ mm/s ! (= 50 μm @ 50 Hz AC)
 
 ### Hall effect
-We now consider a conductive wire in a magnetic field ${\bf B} \rightarrow$ electrons are deflected in a direction perpendicular to ${\bf B}$ and ${\bf j}$ by the Lorentz force.
+We now consider a conductive wire in a magnetic field $\mathbf{B} \rightarrow$ electrons are deflected in a direction perpendicular to $\mathbf{B}$ and $\mathbf{j}$ by the Lorentz force.
 
 ![](figures/hall_effect.svg)
 
-${\bf E}_{\rm H}$ is the electric field caused by the Lorentz force, leading to a _Hall voltage_ in the direction perpendicular to ${\bf B}$ and ${\bf j}$.
+$\mathbf{E}_\mathrm{H}$ is the electric field caused by the Lorentz force, leading to a _Hall voltage_ in the direction perpendicular to $\mathbf{B}$ and $\mathbf{j}$.
 
-In steady state, there is no current flow in the $y$-direction because 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:
+In steady state, there is no current flow in the $y$-direction because the $y$-component of the Lorentz force $-e\mathbf{v}_x\times\mathbf{B}$ is being compensated by the Hall electric field $\mathbf{E}_\mathrm{H}=\mathbf{v}_x\times\mathbf{B}=\frac{1}{ne}\mathbf{j}\times\mathbf{B}$. The total electric field then becomes:
 
 $$
-{\bf E}=\left(\frac{1}{ne}{\bf j}\times{\bf B}+\frac{m}{ne^2\tau}{\bf j}\right).
+\mathbf{E}=\left(\frac{1}{ne}\mathbf{j}\times\mathbf{B}+\frac{m}{ne^2\tau}\mathbf{j}\right).
 $$
 where the second term is associated with the Drude resistivity derived above.
 
-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:
+We now introduce the _resistivity matrix_ $\tilde{\rho}$ as $\mathbf{E}=\tilde{\rho}\mathbf{j}$, 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:
 
 $$
-\rho_{xy}=\frac{B}{ne}\equiv -R_{\rm H}B
+\rho_{xy}=\frac{B}{ne}\equiv -R_\mathrm{H}B
 $$
 
-where $R_{\rm H}=-\frac{1}{ne}$ is the *Hall coefficient*. So by measuring the Hall coefficient, we can obtain $n$, the density of free electrons in a material.
+where $R_\mathrm{H}=-\frac{1}{ne}$ is the *Hall coefficient*. So by measuring the Hall coefficient, we can obtain $n$, the density of free electrons in a material.
 
-While most materials have $R_{\rm H}<0$, interestingly some materials are found to have $R_{\rm H}>0$. This would imply that the charge of the carriers is positive. We will see later (chapter 17) how to interpret this.
+While most materials have $R_\mathrm{H}<0$, interestingly some materials are found to have $R_\mathrm{H}>0$. This would imply that the charge of the carriers is positive. We will see later (chapter 17) how to interpret this.
 ## Conclusions
 
   1. Drude theory leads to Ohm's law. Resistivity is caused by electrons that scatter with some characteristic time $\tau$.
@@ -182,6 +224,13 @@ While most materials have $R_{\rm H}<0$, interestingly some materials are found
 
 
 ## Exercises
+
+### Warm-up questions
+
+1. How does the resistance of a purely 2D material depend on its size?
+2. Check that the units of mobility and the Hall coefficient are correct. (As you should always do!)
+3. Explain why the scattering rates due to different types of scattering events add up.
+
 ### Exercise 1: Extracting quantities from basic Hall measurements
 We apply a magnetic field $\bf B$ perpendicular to a planar (two-dimensional) sample that sits in the $xy$ plane. The sample has width $W$ in the $y$-direction, length $L$ in the $x$-direction and we apply a current $I$ along $x$.
 
@@ -190,11 +239,11 @@ We apply a magnetic field $\bf B$ perpendicular to a planar (two-dimensional) sa
   3. Express the longitudinal resistance $R=V/I$, where $V$ is the voltage difference over the sample along the $x$ direction, in terms of the longitudinal resistivity $\rho_{xx}$. Suppose we extracted $n$ from a measurement of the Hall resistance, what quantity can we extract from a measurement of the longitudinal resistance? Does the result depend on the geometry of the sample?
 
 ### Exercise 2: Motion of an electron in a magnetic and an electric field.
-We first consider an electron in free space, moving in a plane perpendicular to a magnetic field ${\bf B}$ with velocity ${\bf v}$.
+We first consider an electron in free space, moving in a plane perpendicular to a magnetic field $\mathbf{B}$ with velocity $\mathbf{v}$.
 
   1. Write down the Newton's equation of motion for the electron, compute $\frac{d\mathbf{v}}{{dt}}$.
   2. What is the shape of the motion of the electron? Calculate the characteristic frequency and time-period $T_c$ of this motion for $B=1$ Tesla.
-  3. Now we accelerate the electron by adding an electric field $\mathbf{E}$ that is perpendicular to ${\bf B}$. Adjust the differential equation for $\frac{d\mathbf{v}}{{dt}}$ found in (1) to include ${\bf E}$. Sketch the motion of the electron.
+  3. Now we accelerate the electron by adding an electric field $\mathbf{E}$ that is perpendicular to $\mathbf{B}$. Adjust the differential equation for $\frac{d\mathbf{v}}{{dt}}$ found in (1) to include $\mathbf{E}$. Sketch the motion of the electron.
   4. We now consider an electron in a metal. Include the Drude scattering time $\tau$ into the differential equation for the velocity you formulated in 4.