From 4e21ad64df85004b6b68691d871b18bd0f08a6d1 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Fri, 19 Apr 2019 18:45:09 -0400
Subject: [PATCH] Add most of an answer for 9.6

---
 _psets/9.md | 59 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 59 insertions(+)

diff --git a/_psets/9.md b/_psets/9.md
index 902fdf4..a5f6fed 100644
--- a/_psets/9.md
+++ b/_psets/9.md
@@ -8,6 +8,65 @@ title: Problem Set 9
 Solve the periodically forced Lorentz model for the dielectric constant as a function of frequency,
 and plot the real and imaginary parts.
 
+The periodically forced Lorentz model is
+
+$$
+m \left( \ddot{x}(t) + \gamma \dot{x}(t) + \omega_0^2 x(t) \right) = -e E(t)
+$$
+
+It models the motion of a particle of mass $$m$$ and charge $$-e$$ subjected to a time-varying
+electric field $$E(t)$$. Assuming a bulk material composed of such particles, we can use this model
+to find a relation between the dielectric constant and frequency of incoming radiation.
+
+To start, the [polarization density](https://en.wikipedia.org/wiki/Polarization_density) can be
+expressed in terms of the number of particles per unit volume, their charge, and their displacement:
+
+$$
+P = -N e x
+$$
+
+But it can also be expressed using the electric field and dielectric constant:
+
+$$
+P = \epsilon_0 E(\epsilon_r - 1)
+$$
+
+Thus the dielectric constant for this material is
+
+$$
+\epsilon_r = \frac{-N e x}{\epsilon_0 E} + 1
+$$
+
+So now let's solve the model. Let's assume a simple sinusoidal solution.
+
+$$
+\begin{align*}
+x(t) &= A e^{i \omega t} \\
+\dot{x}(t) &= i \omega A e^{i \omega t} \\
+\ddot{x}(t) &= - \omega^2 A e^{i \omega t}
+\end{align*}
+$$
+
+Then the Lorentz model reduces to
+
+$$
+m A e^{i \omega t} \left( - \omega^2  + i \omega \gamma + \omega_0^2 \right) = -e E(t)
+$$
+
+or
+
+$$
+\frac{x(t)}{E(t)} = \frac{-e}{m \left( \omega_0^2 - \omega^2 + i \omega \gamma \right)}
+$$
+
+So this solution is valid for a sinusoidally varying electric field.
+
+Finally we just plug this in to find
+
+$$
+\epsilon_r = \frac{N e^2}{\epsilon_0 m \left( \omega_0^2 - \omega^2 + i \omega \gamma \right)} + 1
+$$
+
 
 ## (12.1)
 
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