add solutions to Drude exercises

src/2_drude_model_solutions.md

+# Solutions for Drude model exercises
+
+### Exercise 1: Extracting quantities from basic Hall measurements
+
+1.
+
+Hall voltage is measured across the sample width. Hence,
+
+$$
+V_H = -\int_{0}^{W} E_ydy
+$$
+
+where $E_y = -\nu_xB$.
+
+$R_{xy}$ = $-\frac{B}{ne}$, so it does not depend on the sample geometry.
+
+
+2.
+
+If hall resistance and charge density are known, magnetic field is calculated from $R_{xy} = -\frac{B}{ne}$.
+
+As $V_x = -\frac{I_x}{ne}B$, smaller n ensures large hall voltages easy to measure.
+
+3.
+
+$$
+R_{xx} = \frac{\rho_{xx}L}{W}
+$$ 
+
+where $ \rho_{xx} = \frac{m_e}{ne^2\tau}$. Therefore, scattering time ($\tau$) is known and $R_{xx}$ does not depend upon sample geometry.
+
+### Exercise 2: Motion of an electron in a magnetic and an electric field
+
+1.
+
+$$
+m\frac{d\bf v}{dt} = -e(\bf v \times \bf B)
+$$
+
+Magnetic field affects only the velocities along x and y, i.e., $v_x(t)$ and $v_y(t)$ as they are perpendicular to it. Therefore, the equations of motion for the electron are
+
+$$
+\frac{dv_x}{dt} = \frac{ev_yB_z}{m}
+$$
+
+$$
+\frac{dv_y}{dt} = -\frac{ev_xB_z}{m}
+$$
+
+
+
+2.
+
+Compute $v_x(t)$ and $v_y(t)$ by solving the differential equations in 1. Calculate the positions $x(t)$ and $y(t)$ from $v_x(t)$ and $v_y(t)$ respectively with the initial conditions: $v_x=u_x$ and $v_y=0$. It results in
+
+$$
+(x(t) - x_0)^2 + (y(t) - y_0)^2 = \frac{u_x^2}{\omega_c^2}
+$$
+
+where $\omega_c = \frac{e\bf B}{m}$. This represents the circular motion.
+
+$\omega_c$ is the characteristic frequency also called as *cyclotron* frequency. Intuition:    $\frac{mv^2}{r} = evB$ (centripetal force = Lorentz force due to magnetic field).
+
+3.
+
+Due to applied electric field $\bf E$ the equations of motion has an extra term,
+$$
+m\frac{d\bf v}{dt} = -e(\bf E + \bf v \times \bf B)
+$$
+
+Thus the differential equation for velocity differs,
+$$
+\frac{dv_x}{dt} = -\frac{eBv_y}{m}
+$$
+
+$$
+\frac{dv_y}{dt} = \frac{eBv_x}{m} + \frac{eE}{m}
+$$
+
+Repeat steps followed in 2. to compute x(t) and y(t) after which the particle motion is given by,
+
+$$
+\left(x(t) - \frac{Et}{B}\right)^2 + \left(y(t) - \frac{E}{B\omega_c}\right)^2 = \left(\frac{E}{B\omega_c}\right)^2
+$$
+
+This represents the [cycloid](https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif) where the motion is along along x with velocity $\frac{E}{B}$.
+
+4.
+
+$$
+m\left(\frac{d\bf v}{dt} + \frac{\bf v}{\tau}\right) = -e(\bf E + \bf v \times \bf B)
+$$
+
+### Exercise 3: Temperature dependence of resistance in the Drude model
+
+1.
+
+Find electron density from $n_e =   \frac{Z\times n \times N_A}{W} $ 
+
+where *Z* is valence of copper atom, *n* is density, $N_A$ is Avogadro constant and *W* is atomic weight. Use $\rho$ from the lecture notes to calculate scattering time.
+
+2.
+
+$\lambda = \langle v \rangle\tau$
+
+3.
+
+Scattering time $\tau \propto \frac{1}{\sqrt{T}}$; $\rho \propto \sqrt{T}$
+
+4.
+
+In general, $\rho \propto T$ as the phonons in the system scales linearly with T (remember high temperature limit of Bose-Einstein factor becomes $\frac{kT}{\hbar\omega}$ leading to $\rho \propto T$). Inability to explain this linear dependence is a failure of the Drude model.
+
+### Exercise 4: The Hall conductivity matrix and the Hall coefficient
+
+1.
+
+$\rho_{xx}$ is independent of B and
+
+$\rho_{xy} \propto B$
+
+2, 4.
+
+Refer to the lecture notes
+
+3.
+
+$$
+\sigma_{xx} \approx \frac{1}{1 + \rho_{xy}^2}
+$$
+
+$$
+\sigma_{xy} \approx \frac{-\rho_{xy}}{1 + \rho_{xy}^2}
+$$
+
+This describes [Lorentzian](https://en.wikipedia.org/wiki/Spectral_line_shape#Lorentzian).
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