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diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
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--- a/source/know/concept/drude-model/index.md
+++ b/source/know/concept/drude-model/index.md
@@ -9,124 +9,117 @@ categories:
 layout: "concept"
 ---
 
-The **Drude model** classically predicts
-the dielectric function and electric conductivity of a gas of free charge carriers,
+The **Drude model**, also known as
+the **Drude-Lorentz model** due to its analogy
+to the *Lorentz oscillator model*
+classically predicts the [dielectric function](/know/concept/dielectric-function/)
+and electric conductivity of a gas of free charges,
 as found in metals and doped semiconductors.
 
 
+
 ## Metals
 
-An [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
-has an oscillating [electric field](/know/concept/electric-field/)
-$$E(t) = E_0 \exp(- i \omega t)$$
-that exerts a force on the charge carriers,
-which have mass $$m$$ and charge $$q$$.
-They thus obey the following equation of motion,
-where $$\gamma$$ is a frictional damping coefficient:
+In a metal, the conduction electrons can roam freely.
+When an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+passes by, its oscillating [electric field](/know/concept/electric-field/)
+$$\vb{E}(t) = \vb{E}_0 e^{- i \omega t}$$ exerts a force on those electrons,
+so the displacement $$\vb{x}(t)$$ of a particle from its initial position
+obeys this equation of motion:
 
 $$\begin{aligned}
-    m \dvn{2}{x}{t} + m \gamma \dv{x}{t}
-    = q E_0 \exp(- i \omega t)
+    m \dvn{2}{\vb{x}}{t}
+    = q \vb{E} - \gamma m \dv{\vb{x}}{t}
 \end{aligned}$$
 
-Inserting the ansatz $$x(t) = x_0 \exp(- i \omega t)$$
-and isolating for the displacement $$x_0$$ yields:
+Where $$m$$ and $$q < 0$$ are the mass and charge of the electron.
+The first term is Newton's third law,
+and the last term represents a damping force
+slowing down the electrons at rate $$\gamma$$.
 
-$$\begin{aligned}
-    - x_0 m \omega^2 - i x_0 m \gamma \omega
-    = q E_0
-    \quad \implies \quad
-    x_0
-    = - \frac{q E_0}{m (\omega^2 + i \gamma \omega)}
-\end{aligned}$$
-
-The polarization density $$P(t)$$ is therefore as shown below.
-Note that the dipole moment $$p$$ goes from negative to positive,
-and the electric field $$E$$ from positive to negative.
-Let $$N$$ be the density of carriers in the gas, then:
+Inserting the ansatz $$\vb{x}(t) = \vb{x}_0 e^{- i \omega t}$$
+and isolating for the displacement $$\vb{x}$$, we find:
 
 $$\begin{aligned}
-    P(t)
-    = N p(t)
-    = N q x(t)
-    = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} E(t)
+    \vb{x}(t)
+    = \vb{x}_0 e^{- i \omega t}
+    = - \frac{q \vb{E}}{m (\omega^2 + i \gamma \omega)}
 \end{aligned}$$
 
-The electric displacement field $$D$$ is thus as follows,
-where $$\varepsilon_r$$ is the unknown relative permittivity of the gas,
-which we will find shortly:
+The polarization density $$\vb{P}(t)$$ is therefore as shown below.
+Note that the dipole moment vector $$\vb{p}$$ is defined
+as pointing from negative to positive,
+whereas the electric field $$\vb{E}$$ goes from positive to negative.
+Let $$N$$ be the metal's electron density, then:
 
 $$\begin{aligned}
-    D
-    = \varepsilon_0 \varepsilon_r E
-    = \varepsilon_0 E + P
-    = \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) E
+    \vb{P}(t)
+    = N \vb{p}(t)
+    = N q \vb{x}(t)
+    = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} \vb{E}(t)
 \end{aligned}$$
 
-The parenthesized expression is the desired dielectric function $$\varepsilon_r$$,
-which depends on $$\omega$$:
+The electric displacement field $$\vb{D}$$ is then as follows,
+where the parenthesized expression is the dielectric function
+$$\varepsilon_r$$ of the material:
 
 $$\begin{aligned}
-    \boxed{
-        \varepsilon_r(\omega)
-        = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
-    }
+    \vb{D}
+    = \varepsilon_0 \vb{E} + \vb{P}
+    = \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) \vb{E}
+    = \varepsilon_0 \varepsilon_r \vb{E}
 \end{aligned}$$
 
-Where we have defined the important so-called **plasma frequency** like so:
+From this, we define the **plasma frequency** $$\omega_p$$
+at which the conductor "resonates",
+leading to so-called **plasma oscillations** of the electron density
+(see also [Langmuir waves](/know/concept/langmuir-waves/)):
 
 $$\begin{aligned}
+    \varepsilon_r(\omega)
+    = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
+    \qquad\qquad
     \boxed{
         \omega_p
         \equiv \sqrt{\frac{N q^2}{\varepsilon_0 m}}
     }
 \end{aligned}$$
 
-If $$\gamma = 0$$, then $$\varepsilon_r$$ is
-negative $$\omega < \omega_p$$,
-positive for $$\omega > \omega_p$$,
-and zero for $$\omega = \omega_p$$.
-Respectively, this leads to
-an imaginary index $$\sqrt{\varepsilon_r}$$ (high absorption),
-a real index tending to $$1$$ (transparency),
-and the possibility of self-sustained plasma oscillations.
-For metals, $$\omega_p$$ lies in the UV.
-
-We can refine this result for $$\varepsilon_r$$,
-by recognizing the (mean) velocity $$v = \idv{x}{t}$$,
-and rewriting the equation of motion accordingly:
-
-$$\begin{aligned}
-    m \dv{v}{t} + m \gamma v = q E(t)
-\end{aligned}$$
+Suppose that $$\gamma = 0$$,
+then we can identify three distinct scenarios for $$\varepsilon_r$$ here:
 
-Note that $$m v$$ is simply the momentum $$p$$.
-We define the **momentum scattering time** $$\tau \equiv 1 / \gamma$$,
-which represents the average time between collisions,
-where each collision resets the involved particles' momentums to zero.
-Or, more formally:
+*   $$\omega < \omega_p$$, so $$\varepsilon_r < 0$$,
+    so the refractive index $$\sqrt{\varepsilon_r}$$ is imaginary,
+    meaning high absorption and high reflectivity
+    (due to the large complex index difference between media).
+*   $$\omega = \omega_p$$, so $$\varepsilon = 0$$,
+    allowing for self-sustained plasma oscillations.
+*   $$\omega > \omega_p$$, so $$\varepsilon_r > 0$$,
+    so the index $$\sqrt{\varepsilon}$$ is real and asymptotically goes to $$1$$,
+    leading to high transparency and low reflectivity from air.
 
-$$\begin{aligned}
-    \dv{p}{t}
-    = - \frac{p}{\tau} + q E
-\end{aligned}$$
+For most metals $$\omega_p$$ is ultraviolet,
+which explains why they typically appear shiny to us.
+In reality $$\gamma > 0$$, reducing the reflectivity somewhat when $$\omega < \omega_p$$.
 
-Returning to the equation for the mean velocity $$v$$,
-we insert the ansatz $$v(t) = v_0 \exp(- i \omega t)$$,
-for the same electric field $$E(t) = E_0 \exp(-i \omega t)$$ as before:
+The Drude model also lets us calculate the metal's conductivity.
+We already have an expression for $$\vb{x}(t)$$,
+which we differentiate to get the velocity $$\vb{v}(t)$$:
 
 $$\begin{aligned}
-    - i m \omega v_0 + \frac{m}{\tau} v_0 = q E_0
-    \quad \implies \quad
-    v_0 = \frac{q \tau}{m (1 - i \omega \tau)} E_0
+    \vb{v}(t)
+    = \dv{\vb{x}}{t}
+    = - i \omega \vb{x}
+    = \frac{i \omega q \vb{E}}{m (\omega^2 + i \gamma \omega)}
+    = \frac{q \vb{E}}{m (\gamma - i \omega)}
 \end{aligned}$$
 
-From $$v(t)$$, we find the resulting average current density $$J(t)$$ to be as follows:
+Consequently the average current density $$\vb{J}(t)$$ is found to be:
 
 $$\begin{aligned}
-    J(t)
-    = - N q v(t)
-    = \sigma E(t)
+    \vb{J}(t)
+    = N q \vb{v}(t)
+    = \sigma \vb{E}(t)
 \end{aligned}$$
 
 Where $$\sigma(\omega)$$ is the **AC conductivity**,
@@ -134,57 +127,76 @@ which depends on the **DC conductivity** $$\sigma_0$$:
 
 $$\begin{aligned}
     \boxed{
-        \sigma
-        = \frac{\sigma_0}{1 - i \omega \tau}
+        \sigma(\omega)
+        = \frac{\gamma \sigma_0}{\gamma - i \omega}
     }
-    \qquad \quad
+    \qquad\qquad
     \boxed{
         \sigma_0
-        = \frac{N q^2 \tau}{m}
+        \equiv \frac{N q^2}{\gamma m}
     }
 \end{aligned}$$
 
-We can use these quantities to rewrite
-the dielectric function $$\varepsilon_r$$ from earlier:
+Recall that $$\gamma$$ measures friction.
+Specifically, Drude assumed that the electrons often collide with obstacles,
+each time resetting their momentum to zero;
+in that case $$\vb{v}$$ should be interpreted as the average "drift"
+of many electrons in an ensemble.
+The mean time between those collisions is
+the **momentum scattering time** $$\tau \equiv 1 / \gamma$$, so:
+
+$$\begin{aligned}
+    \sigma(\omega)
+    = \frac{\sigma_0}{1 - i \omega \tau}
+    \qquad\qquad
+    \sigma_0
+    = \frac{N q^2 \tau}{m}
+\end{aligned}$$
+
+After defining all those quantities,
+the dielectric function $$\varepsilon_r(\omega)$$ can be written as:
 
 $$\begin{aligned}
     \boxed{
-        \varepsilon_r(\omega)
-        = 1 + \frac{i \sigma(\omega)}{\varepsilon_0 \omega}
+        \begin{aligned}
+            \varepsilon_r(\omega)
+            &= 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
+            \\
+            &= 1 + \frac{i \sigma(\omega)}{\varepsilon_0 \omega}
+        \end{aligned}
     }
 \end{aligned}$$
 
 
+
 ## Doped semiconductors
 
 Doping a semiconductor introduces
-free electrons (n-type)
-or free holes (p-type),
-which can be treated as free particles
-moving in the bands of the material.
-
-The Drude model can also be used in this case,
-by replacing the actual carrier mass $$m$$
-by the effective mass $$m^*$$.
+free electrons (n-type doping) or free holes (p-type doping),
+which can be treated as free charge carriers moving through the material,
+so the Drude model is also relevant in this case.
+
+We must replace the carriers' true mass $$m$$ with their *effective mass* $$m^*$$
+found from the material's electronic band structure.
 Furthermore, semiconductors already have
-a high intrinsic permittivity $$\varepsilon_{\mathrm{int}}$$
-before the dopant is added,
-so the diplacement field $$D$$ is:
+a high intrinsic dielectric function $$\varepsilon_{\mathrm{int}}$$
+before being doped, so the displacement field $$\vb{D}$$ becomes:
 
 $$\begin{aligned}
-    D
-    = \varepsilon_0 E + P_{\mathrm{int}} + P_{\mathrm{free}}
-    = \varepsilon_{\mathrm{int}} \varepsilon_0 E - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} E
+    \vb{D}
+    = \varepsilon_0 \vb{E} + \vb{P}_{\mathrm{int}} + \vb{P}_{\mathrm{free}}
+    = \varepsilon_0 \varepsilon_{\mathrm{int}} \vb{E} - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} \vb{E}
+    = \varepsilon_0 \varepsilon_r \vb{E}
 \end{aligned}$$
 
-Where $$P_{\mathrm{int}}$$ is the intrinsic undoped polarization,
-and $$P_{\mathrm{free}}$$ is the contribution of the free carriers.
+Where $$\vb{P}_{\mathrm{int}}$$ is the intrinsic polarization before doping,
+and $$\vb{P}_{\mathrm{free}}$$ is the expression we calculated above for metals.
 The dielectric function $$\varepsilon_r(\omega)$$ is therefore given by:
 
 $$\begin{aligned}
     \boxed{
         \varepsilon_r(\omega)
-        = \varepsilon_{\mathrm{int}} \Big( 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega} \Big)
+        = \varepsilon_{\mathrm{int}} \bigg( 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega} \bigg)
     }
 \end{aligned}$$
 
@@ -194,29 +206,28 @@ to include $$\varepsilon_\mathrm{int}$$:
 $$\begin{aligned}
     \boxed{
         \omega_p
-        = \sqrt{\frac{N q^2}{\varepsilon_{\mathrm{int}} \varepsilon_0 m^*}}
+        \equiv \sqrt{\frac{N q^2}{\varepsilon_0 \varepsilon_{\mathrm{int}} m^*}}
     }
 \end{aligned}$$
 
 The meaning of $$\omega_p$$ is the same as for metals,
-with high absorption for $$\omega < \omega_p$$.
-However, due to the lower carrier density $$N$$ in a semiconductor,
-$$\omega_p$$ lies in the IR rather than UV.
+but the free carrier density $$N$$ is typically lower in this case,
+so $$\omega_p$$ is usually infrared rather than ultraviolet.
 
-However, instead of asymptotically going to $$1$$ for $$\omega > \omega_p$$ like a metal,
-$$\varepsilon_r$$ tends to $$\varepsilon_\mathrm{int}$$ instead,
-and crosses $$1$$ along the way,
-at which point the reflectivity is zero.
-This occurs at:
+Furthermore, instead of $$\varepsilon_r \to 1$$
+for $$\omega \to \infty$$ like a metal,
+now $$\varepsilon_r \to \varepsilon_\mathrm{int}$$.
+Along the way, there is a point where $$\varepsilon_r = 1$$
+and the reflectivity becomes zero. This occurs at:
 
 $$\begin{aligned}
     \omega^2
     = \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
 \end{aligned}$$
 
-This is used to experimentally determine the effective mass $$m^*$$
-of the doped semiconductor,
-by finding which value of $$m^*$$ gives the measured $$\omega$$.
+If $$N$$ and $$\varepsilon_\mathrm{int}$$ are known,
+this can be used to experimentally determine $$m^*$$
+by finding which value of $$\omega_p$$ would lead to the measured zero-reflectivity point.