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+---
+title: "Plasma physics"
+firstLetter: "P"
+date: 2021-09-18T14:33:19+02:00
+draft: false
+layout: "category"
+---
+
+This page will fill itself.
diff --git a/content/know/concept/guiding-center-theory/index.pdc b/content/know/concept/guiding-center-theory/index.pdc
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+---
+title: "Guiding center theory"
+firstLetter: "G"
+publishDate: 2021-09-21
+categories:
+- Physics
+- Electromagnetism
+- Plasma physics
+
+date: 2021-09-18T13:47:41+02:00
+draft: false
+markup: pandoc
+---
+
+# Guiding center theory
+
+When discussing the [Lorentz force](/know/concept/lorentz-force/),
+we introduced the concept of *gyration*:
+a particle in a uniform [magnetic field](/know/concept/magnetic-field/) $\vb{B}$
+*gyrates* in a circular orbit around a **guiding center**.
+Here, we will generalize this result
+to more complicated situations,
+for example involving [electric fields](/know/concept/electric-field/).
+
+The particle's equation of motion
+combines the Lorentz force $\vb{F}$
+with Newton's second law:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \big( \vb{E} + \vb{u} \cross \vb{B} \big)
+\end{aligned}$$
+
+We now allow the fields vary slowly in time and space.
+We thus add deviations $\delta\vb{E}$ and $\delta\vb{B}$:
+
+$$\begin{aligned}
+    \vb{E}
+    \to \vb{E} + \delta\vb{E}(\vb{x}, t)
+    \qquad \quad
+    \vb{B}
+    \to \vb{B} + \delta\vb{B}(\vb{x}, t)
+\end{aligned}$$
+
+Meanwhile, the velocity $\vb{u}$ can be split into
+the guiding center's motion $\vb{u}_{gc}$
+and the *known* Larmor gyration $\vb{u}_L$ around the guiding center,
+such that $\vb{u} = \vb{u}_{gc} + \vb{u}_L$.
+Inserting:
+
+$$\begin{aligned}
+    m \dv{t} \big( \vb{u}_{gc} + \vb{u}_L \big)
+    = q \big( \vb{E} + \delta\vb{E} + (\vb{u}_{gc} + \vb{u}_L) \cross (\vb{B} + \delta\vb{B}) \big)
+\end{aligned}$$
+
+We already know that $m \: \dv*{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$,
+which we subtract from the total to get:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \delta\vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big)
+\end{aligned}$$
+
+This will be our starting point.
+Before proceeding, we also define
+the average of $\expval{f}$ of a function $f$ over a single gyroperiod,
+where $\omega_c$ is the cyclotron frequency:
+
+$$\begin{aligned}
+    \expval{f}
+    \equiv \int_0^{2 \pi / \omega_c} f(t) \dd{t}
+\end{aligned}$$
+
+Assuming that gyration is much faster than the guiding center's motion,
+we can use this average to approximately remove the finer dynamics,
+and focus only on the guiding center.
+
+
+## Uniform electric and magnetic field
+
+Consider the case where $\vb{E}$ and $\vb{B}$ are both uniform,
+such that $\delta\vb{B} = 0$ and $\delta\vb{E} = 0$:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} \big)
+\end{aligned}$$
+
+Dotting this with the unit vector $\vu{b} \equiv \vb{B} / |\vb{B}|$
+makes all components perpendicular to $\vb{B}$ vanish,
+including the cross product,
+leaving only the (scalar) parallel components
+$u_{gc\parallel}$ and $E_\parallel$:
+
+$$\begin{aligned}
+    m \dv{u_{gc\parallel}}{t}
+    = \frac{q}{m} E_{\parallel}
+\end{aligned}$$
+
+This simply describes a constant acceleration,
+and is easy to integrate.
+Next, the equation for $\vb{u}_{gc\perp}$ is found by
+subtracting $u_{gc\parallel}$'s equation from the original:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    = q (\vb{E} + \vb{u}_{gc} \cross \vb{B}) - q E_\parallel \vu{b}
+    = q (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B})
+\end{aligned}$$
+
+Keep in mind that $\vb{u}_{gc\perp}$ explicitly excludes gyration.
+If we try to split $\vb{u}_{gc\perp}$ into a constant and a time-dependent part,
+and choose the most convenient constant,
+we notice that the only way to exclude gyration
+is to demand that $\vb{u}_{gc\perp}$ does not depend on time.
+Therefore:
+
+$$\begin{aligned}
+    0
+    = \vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}
+\end{aligned}$$
+
+To find $\vb{u}_{gc\perp}$, we take the cross product with $\vb{B}$,
+and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
+
+$$\begin{aligned}
+    0
+    = \vb{B} \cross (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B})
+    = \vb{B} \cross \vb{E} + \vb{u}_{gc\perp} B^2
+\end{aligned}$$
+
+Rearranging this shows that $\vb{u}_{gc\perp}$ is constant.
+The guiding center drifts sideways at this speed,
+hence it is called a **drift velocity** $\vb{v}_E$.
+Curiously, $\vb{v}_E$ is independent of $q$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_E
+        = \frac{\vb{E} \cross \vb{B}}{B^2}
+    }
+\end{aligned}$$
+
+Drift is not specific to an electric field:
+$\vb{E}$ can be replaced by a general force $\vb{F}/q$ without issues.
+In that case, the resulting drift velocity $\vb{v}_F$ does depend on $q$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_F
+        = \frac{\vb{F} \cross \vb{B}}{q B^2}
+    }
+\end{aligned}$$
+
+
+## Non-uniform magnetic field
+
+Next, consider a more general case, where $\vb{B}$ is non-uniform,
+but $\vb{E}$ is still uniform:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big)
+\end{aligned}$$
+
+Assuming the gyroradius $r_L$ is small compared to the variation of $\vb{B}$,
+we set $\delta\vb{B}$ to the first-order term
+of a Taylor expansion of $\vb{B}$ around $\vb{x}_{gc}$,
+that is, $\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$.
+We thus have:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B}
+    + \vb{u}_{gc} \cross (\vb{x}_L \cdot \nabla) \vb{B}
+    + + \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} \big)
+\end{aligned}$$
+
+We approximate this by taking the average over a single gyration,
+as defined earlier:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B}
+    + \vb{u}_{gc} \cross \expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+    + \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \big)
+\end{aligned}$$
+
+Where we have used that $\expval{\vb{u}_{gc}} = \vb{u}_{gc}$.
+The two averaged expressions turn out to be:
+
+$$\begin{aligned}
+    \expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+    = 0
+    \qquad \quad
+    \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    \approx - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-nonuniform-B-averages"/>
+<label for="proof-nonuniform-B-averages">Proof</label>
+<div class="hidden">
+<label for="proof-nonuniform-B-averages">Proof.</label>
+We know what $\vb{x}_L$ is,
+so we can write out $(\vb{x}_L \cdot \nabla) \vb{B}$
+for $\vb{B} = (B_x, B_y, B_z)$:
+
+$$\begin{aligned}
+    (\vb{x}_L \cdot \nabla) \vb{B}
+    = \frac{u_L}{\omega_c}
+    \begin{pmatrix}
+        \sin\!(\omega_c t) \pdv{B_x}{x} + \cos\!(\omega_c t) \pdv{B_x}{y} \\
+        \sin\!(\omega_c t) \pdv{B_y}{x} + \cos\!(\omega_c t) \pdv{B_y}{y} \\
+        \sin\!(\omega_c t) \pdv{B_z}{x} + \cos\!(\omega_c t) \pdv{B_z}{y}
+    \end{pmatrix}
+\end{aligned}$$
+
+Integrating $\sin$ and $\cos$ over their period yields zero,
+so the average vanishes:
+
+$$\begin{aligned}
+    \expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+    = 0
+\end{aligned}$$
+
+Moving on, we write out $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$,
+suppressing the arguments of $\sin$ and $\cos$:
+
+$$\begin{aligned}
+    \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}
+    &= \frac{u_L^2}{\omega_c}
+    \begin{pmatrix}
+        \cos \\
+        - \sin \\
+        0
+    \end{pmatrix}
+    \cross
+    \begin{pmatrix}
+        \pdv{B_x}{x} \sin + \pdv{B_x}{y} \cos \\
+        \pdv{B_y}{x} \sin + \pdv{B_y}{y} \cos \\
+        \pdv{B_z}{x} \sin + \pdv{B_z}{y} \cos
+    \end{pmatrix}
+    \\
+    &= \frac{u_L^2}{\omega_c}
+    \begin{pmatrix}
+        - \pdv{B_z}{x} \sin^2 - \pdv{B_z}{y} \sin \cos \\
+        - \pdv{B_z}{x} \sin \cos - \pdv{B_z}{y} \cos^2 \\
+        \pdv{B_y}{x} \sin \cos + \pdv{B_y}{y} \cos^2
+        + \pdv{B_x}{x} \sin^2 + \pdv{B_x}{y} \sin \cos
+    \end{pmatrix}
+\end{aligned}$$
+
+Integrating products of $\sin$ and $\cos$ over their period gives us the following:
+
+$$\begin{aligned}
+    \expval{\cos^2} = \expval{\sin^2} = \frac{1}{2}
+    \qquad \quad
+    \expval{\sin \cos} = 0
+\end{aligned}$$
+
+Inserting this tells us that the average
+of $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$ is given by:
+
+$$\begin{aligned}
+    \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    &= \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        - \pdv{B_z}{x} \\
+        - \pdv{B_z}{y} \\
+        \pdv{B_y}{y} + \pdv{B_x}{x}
+    \end{pmatrix}
+\end{aligned}$$
+
+We use [Maxwell's equation](/know/concept/maxwells-equations/) $\nabla \cdot \vb{B} = 0$
+to rewrite the $z$-component,
+and follow the convention that $\vb{B}$
+points mostly in the $z$-direction,
+such that $B \equiv |\vb{B}| \approx B_z$:
+
+$$\begin{aligned}
+    \expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    &= - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \pdv{B_z}{x} \\
+        \pdv{B_z}{y} \\
+        \pdv{B_z}{z}
+    \end{pmatrix}
+    \approx - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \pdv{B}{x} \\
+        \pdv{B}{y} \\
+        \pdv{B}{z}
+    \end{pmatrix}
+    = - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+</div>
+</div>
+
+With this, the guiding center's equation of motion
+is reduced to the following:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+Let us now split $\vb{u}_{gc}$ into
+components $\vb{u}_{gc\perp}$ and $u_{gc\parallel} \vu{b}$,
+which are respectively perpendicular and parallel
+to the magnetic unit vector $\vu{b}$,
+such that $\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$.
+Consequently:
+
+$$\begin{aligned}
+    \dv{\vb{u}_{gc}}{t}
+    = \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t}
+\end{aligned}$$
+
+Inserting this into the guiding center's equation of motion,
+we now have:
+
+$$\begin{aligned}
+    \dv{\vb{u}_{gc}}{t}
+    = m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t} \bigg)
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+The derivative of $\vu{b}$ can be rewritten as follows,
+where $R_c$ is the radius of the field's [curvature](/know/concept/curvature/),
+and $\vb{R}_c$ is the corresponding vector from the center of curvature:
+
+$$\begin{aligned}
+    \dv{\vu{b}}{t}
+    \approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-nonuniform-B-curvature"/>
+<label for="proof-nonuniform-B-curvature">Proof</label>
+<div class="hidden">
+<label for="proof-nonuniform-B-curvature">Proof.</label>
+Assuming that $\vu{b}$ does not explicitly depend on time,
+i.e. $\pdv*{\vu{b}}{t} = 0$,
+we can rewrite the derivative using the chain rule:
+
+$$\begin{aligned}
+    \dv{\vu{b}}{t}
+    = \pdv{\vu{b}}{s} \dv{s}{t}
+    = u_{gc\parallel} \dv{\vu{b}}{s}
+\end{aligned}$$
+
+Where $\dd{s}$ is the arc length of the magnetic field line,
+which is equal to the radius $R_c$ times the infinitesimal subtended angle $\dd{\theta}$:
+
+$$\begin{aligned}
+    \dd{s}
+    = R_c \dd{\theta}
+\end{aligned}$$
+
+Meanwhile, across this arc, $\vu{b}$ rotates by $\dd{\theta}$,
+such that the tip travels a distance $|\dd{\vu{b}}|$:
+
+$$\begin{aligned}
+    |\dd{\vu{b}}\!|
+    = |\vu{b}| \dd{\theta}
+    = \dd{\theta}
+\end{aligned}$$
+
+Furthermore, the direction $\dd{\vu{b}}$ is always opposite to $\vu{R}_c$,
+which is defined as the unit vector from the center of curvature to the base of $\vu{b}$:
+
+$$\begin{aligned}
+    \dd{\vu{b}}
+    = - \vu{R}_c \dd{\theta}
+\end{aligned}$$
+
+Combining these expressions for $\dd{s}$ and $\dd{\vu{b}}$,
+we find the following derivative:
+
+$$\begin{aligned}
+    \dv{\vu{b}}{s}
+    = - \frac{\vu{R}_c \dd{\theta}}{R_c \dd{\theta}}
+    = - \frac{\vu{R}_c}{R_c}
+    = - \frac{\vb{R}_c}{R_c^2}
+\end{aligned}$$
+</div>
+</div>
+
+With this, we arrive at the following equation of motion
+for the guiding center:
+
+$$\begin{aligned}
+    m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} - u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} \bigg)
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+Since both $\vb{R}_c$ and any cross product with $\vb{B}$
+will always be perpendicular to $\vb{B}$,
+we can split this equation into perpendicular and parallel components like so:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    &= q \vb{E}_{\perp} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} + q \vb{u}_{gc} \cross \vb{B}
+    \\
+    m \dv{u_{gc\parallel}}{t}
+    &= q E_{\parallel} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\parallel} B
+\end{aligned}$$
+
+The parallel part simply describes an acceleration.
+The perpendicular part is more interesting:
+we rewrite it as follows, defining an effective force $\vb{F}_{\!\perp}$:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    = \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
+    \qquad \quad
+    \vb{F}_{\!\perp}
+    \equiv q \vb{E}_\perp + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B
+\end{aligned}$$
+
+To solve this, we make a crude approximation now, and improve it later.
+We thus assume that $\vb{u}_{gc\perp}$ is constant in time,
+such that the equation reduces to:
+
+$$\begin{aligned}
+     0
+     \approx \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
+     = \vb{F}_{\!\perp} + q \vb{u}_{gc\perp} \cross \vb{B}
+\end{aligned}$$
+
+This is analogous to the previous case of a uniform electric field,
+with $q \vb{E}$ replaced by $\vb{F}_{\!\perp}$,
+so it is also solved by crossing with $\vb{B}$ in front,
+yielding a drift:
+
+$$\begin{aligned}
+    \vb{u}_{gc\perp}
+    \approx \vb{v}_F
+    \equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2}
+\end{aligned}$$
+
+From the definition of $\vb{F}_{\!\perp}$,
+this total $\vb{v}_F$ can be split into three drifts:
+the previously seen electric field drift $\vb{v}_E$,
+the **curvature drift** $\vb{v}_c$,
+and the **grad-$\vb{B}$ drift** $\vb{v}_{\nabla B}$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_c
+        = \frac{m u_{gc\parallel}^2}{q} \frac{\vb{R}_c \cross \vb{B}}{R_c^2 B^2}
+    }
+    \qquad \quad
+    \boxed{
+        \vb{v}_{\nabla B}
+        = \frac{u_L^2}{2 \omega_c} \frac{\vb{B} \cross \nabla B}{B^2}
+    }
+\end{aligned}$$
+
+Such that $\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$.
+We are still missing a correction,
+since we neglected the time dependence of $\vb{u}_{gc\perp}$ earlier.
+This correction is called $\vb{v}_p$,
+where $\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$.
+We revisit the perpendicular equation, which now reads:
+
+$$\begin{aligned}
+    m \dv{t} \big( \vb{v}_F + \vb{v}_p \big)
+    = \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B}
+\end{aligned}$$
+
+We assume that $\vb{v}_F$ varies much faster than $\vb{v}_p$,
+such that $\dv*{\vb{v}_p}{t}$ is negligible.
+In addition, from the derivation of $\vb{v}_F$,
+we know that $\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$,
+leaving only:
+
+$$\begin{aligned}
+    m \dv{\vb{v}_F}{t}
+    = q \vb{v}_p \cross \vb{B}
+\end{aligned}$$
+
+To isolate this for $\vb{v}_p$,
+we take the cross product with $\vb{B}$ in front,
+like earlier.
+We thus arrive at the following correction,
+known as the **polarization drift** $\vb{v}_p$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_p
+        = - \frac{m}{q B^2} \dv{\vb{v}_F}{t} \cross \vb{B}
+    }
+\end{aligned}$$
+
+In many cases $\vb{v}_E$ dominates $\vb{v}_F$,
+so in some literature $\vb{v}_p$ is approximated as follows:
+
+$$\begin{aligned}
+    \vb{v}_p
+    \approx - \frac{m}{q B^2} \dv{\vb{v}_E}{t} \cross \vb{B}
+    = - \frac{m}{q B^2} \Big( \dv{t} (\vb{E}_\perp \cross \vb{B}) \Big) \cross \vb{B}
+    = - \frac{m}{q B^2} \dv{\vb{E}_\perp}{t}
+\end{aligned}$$
+
+The polarization drift stands out from the others:
+it has the opposite sign,
+it is proportional to $m$,
+and it is often only temporary.
+Therefore, it is also called the **inertia drift**.
+
+
+
+## References
+1.  F.F. Chen,
+    *Introduction to plasma physics and controlled fusion*,
+    3rd edition, Springer.
+2.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/lorentz-force/index.pdc b/content/know/concept/lorentz-force/index.pdc
index 2362766..83357d2 100644
--- a/content/know/concept/lorentz-force/index.pdc
+++ b/content/know/concept/lorentz-force/index.pdc
@@ -5,6 +5,7 @@ publishDate: 2021-09-08
 categories:
 - Physics
 - Electromagnetism
+- Plasma physics
 
 date: 2021-09-08T17:00:32+02:00
 draft: false
@@ -27,6 +28,44 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+## Uniform electric field
+
+Consider the simple case of an electric field $\vb{E}$
+that is uniform in all of space.
+In the absence of a magnetic field $\vb{B} = 0$
+and any other forces,
+Newton's second law states:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \vb{E}
+\end{aligned}$$
+
+This is straightforward to integrate in time,
+for a given initial velocity vector $\vb{u}_0$:
+
+$$\begin{aligned}
+    \vb{u}(t)
+    = \frac{q}{m} \vb{E} t + \vb{u}_0
+\end{aligned}$$
+
+And then the particle's position $\vb{x}(t)$
+is found be integrating once more,
+with $\vb{x}(0) = \vb{x}_0$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{x}(t)
+        = \frac{q}{2 m} \vb{E} t^2 + \vb{u}_0 t + \vb{x}_0
+    }
+\end{aligned}$$
+
+In summary, unsurprisingly, a uniform electric field $\vb{E}$
+accelerates the particle with a constant force $\vb{F} = q \vb{E}$.
+Note that the direction depends on the sign of $q$.
+
+
 ## Uniform magnetic field
 
 Consider the simple case of a uniform magnetic field
@@ -64,12 +103,12 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
-which is always positive:
+which may be negative:
 
 $$\begin{aligned}
     \boxed{
         \omega_c
-        \equiv \frac{|q| B}{m}
+        \equiv \frac{q B}{m}
     }
 \end{aligned}$$
 
@@ -78,7 +117,7 @@ the solution for $u_x(t)$ is given by:
 
 $$\begin{aligned}
     u_x(t)
-    = - u_\perp \sin\!(\omega_c t)
+    = u_\perp \cos\!(\omega_c t)
 \end{aligned}$$
 
 Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
@@ -87,13 +126,12 @@ Then $u_y(t)$ is found to be:
 $$\begin{aligned}
     u_y(t)
     = \frac{m}{q B} \dv{u_x}{t}
-    = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t)
-    = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t)
+    = - \frac{m \omega_c}{q B} u_\perp \sin\!(\omega_c t)
+    = - u_\perp \sin\!(\omega_c t)
 \end{aligned}$$
 
-Where $\mathrm{sgn}$ is the signum function.
-This tells us that the particle moves in a circular orbit,
-and that the direction of rotation is determined by $q$.
+This means that the particle moves in a circle,
+in a direction determined by the sign of $\omega_c$.
 
 Integrating the velocity yields the position,
 where we refer to the integration constants $x_{gc}$ and $y_{gc}$
@@ -101,10 +139,10 @@ as the **guiding center**, around which the particle orbits or **gyrates**:
 
 $$\begin{aligned}
     x(t)
-    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc}
+    = \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + x_{gc}
     \qquad \quad
     y(t)
-    = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc}
+    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + y_{gc}
 \end{aligned}$$
 
 The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
@@ -112,16 +150,18 @@ The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_
 $$\begin{aligned}
     \boxed{
         r_L
-        \equiv \frac{u_\perp}{\omega_c}
+        \equiv \frac{u_\perp}{|\omega_c|}
         = \frac{m u_\perp}{|q| B}
     }
 \end{aligned}$$
 
-Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$:
+Finally, it is easy to integrate the equation
+for the $z$-axis velocity $u_z$, which is conserved:
 
 $$\begin{aligned}
     z(t)
-    = u_z t + z_{gc}
+    = z_{gc}
+    = u_z t + z_0
 \end{aligned}$$
 
 In conclusion, the particle's motion parallel to $\vb{B}$
@@ -129,116 +169,24 @@ is not affected by the magnetic field,
 while its motion perpendicular to $\vb{B}$
 is circular around an imaginary guiding center.
 The end result is that particles follow a helical path
-when moving through a uniform magnetic field.
-
-
-## Uniform electric and magnetic field
-
-Let us now consider a more general case,
-with constant uniform electric and magnetic fields $\vb{E}$ and $\vb{B}$,
-which may or may not be perpendicular.
-The equation of motion is then:
-
-$$\begin{aligned}
-    \vb{F}
-    = m \dv{\vb{u}}{t}
-    = q (\vb{E} + \vb{u} \cross \vb{B})
-\end{aligned}$$
-
-If we take the dot product with the unit vector $\vu{B}$,
-the cross product vanishes, leaving:
+when moving through a uniform magnetic field:
 
 $$\begin{aligned}
-    \dv{\vb{u}_\parallel}{t}
-    = \frac{q}{m} \vb{E}_\parallel
-\end{aligned}$$
-
-Where $\vb{u}_\parallel$ and $\vb{E}_\parallel$ are
-the components of $\vb{u}$ and $\vb{E}$
-that are parallel to $\vb{B}$.
-This equation is easy to integrate:
-the guiding center accelerates according to $(q/m) \vb{E}_\parallel$.
-
-Next, let us define the perpendicular component $\vb{u}_\perp$
-such that $\vb{u} = \vb{u}_\parallel \vu{B} + \vb{u}_\perp$.
-Its equation of motion is found by
-subtracting $\vb{u}_\parallel$'s equation from the original:
-
-$$\begin{aligned}
-    m \dv{\vb{u}_\perp}{t}
-    = q (\vb{E} + \vb{u} \cross \vb{B}) - q \vb{E}_\parallel
-    = q (\vb{E}_\perp + \vb{u}_\perp \cross \vb{B})
-\end{aligned}$$
-
-To solve this, we go to a moving coordinate system
-by defining $\vb{u}_\perp = \vb{v}_\perp + \vb{w}_\perp$,
-where $\vb{v}_\perp$ is a constant of our choice.
-The equation is now as follows:
-
-$$\begin{aligned}
-    m \dv{t} (\vb{v}_\perp + \vb{w}_\perp)
-    = m \dv{\vb{w}_\perp}{t}
-    = q (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B} + \vb{w}_\perp \cross \vb{B})
-\end{aligned}$$
-
-We want to choose $\vb{v}_\perp$ such that the first two terms vanish,
-or in other words:
-
-$$\begin{aligned}
-    0
-    = \vb{E}_\perp + \vb{v}_\perp \cross \vb{B}
-\end{aligned}$$
-
-To find $\vb{v}_\perp$, we take the cross product with $\vb{B}$,
-and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
-
-$$\begin{aligned}
-    0
-    = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B})
-    = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2
-    \quad \implies \quad
     \boxed{
-        \vb{v}_\perp
-        = \frac{\vb{E} \cross \vb{B}}{B^2}
+        \vb{x}(t)
+        = \frac{u_\perp}{\omega_c}
+        \begin{pmatrix}
+            \sin\!(\omega_c t) \\ \cos\!(\omega_c t) \\ 0
+        \end{pmatrix}
+        + \vb{x}_{gc}(t)
     }
 \end{aligned}$$
 
-When $\vb{v}_\perp$ is chosen like this,
-the perpendicular equation of motion is reduced to:
-
-$$\begin{aligned}
-    m \dv{\vb{w}_\perp}{t}
-    = q \vb{w}_\perp \cross \vb{B}
-\end{aligned}$$
-
-Which is simply the case we treated previously with $\vb{E} = 0$,
-with a known solution
-(assuming $\vb{B}$ still points in the positive $z$-direction):
-
-$$\begin{aligned}
-    w_x(t)
-    = - w_\perp \sin\!(\omega_c t)
-    \qquad
-    w_y(t)
-    = - \mathrm{sgn}(q) \: w_\perp \cos\!(\omega_c t)
-\end{aligned}$$
-
-However, this result is shifted by a constant $\vb{v}_\perp$,
-often called the **drift velocity** $\vb{v}_d$,
-at which the guiding center moves transversely.
-Curiously, $\vb{v}_d$ is independent of $q$.
-
-Such a drift is not specific to an electric field.
-In the equations above, $\vb{E}$ can be replaced
-by a general force $\vb{F}/q$ (e.g. gravity) without issues.
-In that case, $\vb{v}_d$ does depend on $q$:
-
-$$\begin{aligned}
-    \boxed{
-        \vb{v}_d
-        = \frac{\vb{F} \cross \vb{B}}{q B^2}
-    }
-\end{aligned}$$
+Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$
+is the position of the guiding center.
+For a detailed look at how $\vb{B}$ and $\vb{E}$
+can affect the guiding center's motion,
+see [guiding center theory](/know/concept/guiding-center-theory/).
 
 
 
diff --git a/content/know/concept/rabi-oscillation/index.pdc b/content/know/concept/rabi-oscillation/index.pdc
new file mode 100644
index 0000000..cf393a4
--- /dev/null
+++ b/content/know/concept/rabi-oscillation/index.pdc
@@ -0,0 +1,218 @@
+---
+title: "Rabi oscillation"
+firstLetter: "R"
+publishDate: 2021-09-22
+categories:
+- Physics
+- Quantum mechanics
+- Optics
+
+date: 2021-09-18T00:41:43+02:00
+draft: false
+markup: pandoc
+---
+
+# Rabi oscillation
+
+In quantum mechanics, from the derivation of
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+we know that a time-dependent term $\hat{H}_1$ in the Hamiltonian
+affects the state as follows,
+where $c_n(t)$ are the coefficients of the linear combination
+of basis states $\ket{n} \exp\!(-i E_n t / \hbar)$:
+
+$$\begin{aligned}
+    i \hbar \dv{c_m}{t}
+    = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp\!(i \omega_{mn} t)
+\end{aligned}$$
+
+Where $\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$
+for energies $E_m$ and $E_n$.
+Note that this equation is exact,
+despite being used for deriving perturbation theory.
+Consider a two-level system where $n \in \{a, b\}$,
+in which case the above equation can be expanded to the following:
+
+$$\begin{aligned}
+    \dv{c_a}{t}
+    &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp\!(- i \omega_0 t) \: c_b - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{a} \: c_a
+    \\
+    \dv{c_b}{t}
+    &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp\!(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b
+\end{aligned}$$
+
+Where $\omega_0 \equiv \omega_{ba}$ is positive.
+We assume that $\hat{H}_1$ has odd spatial parity,
+in which case [Laporte's selection rule](/know/concept/selection-rules/)
+states that the diagonal matrix elements vanish, leaving:
+
+$$\begin{aligned}
+    \dv{c_a}{t}
+    &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp\!(- i \omega_0 t) \: c_b
+    \\
+    \dv{c_b}{t}
+    &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp\!(i \omega_0 t) \: c_a
+\end{aligned}$$
+
+We now choose $\hat{H}_1$ to be as follows,
+sinusoidally oscillating with a spatially odd $V(\vec{r})$:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = V \cos\!(\omega t)
+    = \frac{V}{2} \Big( \exp\!(i \omega t) + \exp\!(-i \omega t) \Big)
+\end{aligned}$$
+
+We insert this into the equations for $c_a$ and $c_b$,
+and define $V_{ab} \equiv \matrixel{a}{V}{b}$, leading us to:
+
+$$\begin{aligned}
+    \dv{c_a}{t}
+    &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!-\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!+\! \omega_0) t\big) \Big) \: c_b
+    \\
+    \dv{c_b}{t}
+    &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!+\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t\big) \Big) \: c_a
+\end{aligned}$$
+
+Here, we make the *rotating wave approximation*:
+assuming we are close to resonance $\omega \approx \omega_0$,
+we decide that $\exp\!(i (\omega \!+\! \omega_0) t)$
+oscillates so much faster than $\exp\!(i (\omega \!-\! \omega_0) t)$,
+that its effect turns out negligible
+when the system is observed over a reasonable time interval.
+
+In other words, over this reasonably-sized time interval,
+$\exp\!(i (\omega \!+\! \omega_0) t)$ averages to zero,
+while $\exp\!(i (\omega \!-\! \omega_0) t)$ does not.
+Dropping the respective terms thus leaves us with:
+
+$$\begin{aligned}
+    \dv{c_a}{t}
+    = - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b
+    \qquad \quad
+    \dv{c_b}{t}
+    = - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a
+\end{aligned}$$
+
+Now we can solve this system of coupled equations exactly.
+We differentiate the first equation with respect to $t$,
+and then substitute $\dv*{c_b}{t}$ for the second equation:
+
+$$\begin{aligned}
+    \dv[2]{c_a}{t}
+    &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + \dv{c_b}{t} \bigg) \exp\!\big(i (\omega \!-\! \omega_0) t \big)
+    \\
+    &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b 
+    - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a \bigg)
+    \exp\!\big(i (\omega \!-\! \omega_0) t \big)
+    \\
+    &= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a
+\end{aligned}$$
+
+In the first term, we recognize $\dv*{c_a}{t}$,
+which we insert to arrive at an equation for $c_a(t)$:
+
+$$\begin{aligned}
+    0
+    = \dv[2]{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a
+\end{aligned}$$
+
+To solve this, we make the ansatz $c_a(t) = \exp\!(\lambda t)$,
+which, upon insertion, gives us:
+
+$$\begin{aligned}
+    0
+    = \lambda^2 - i (\omega - \omega_0) \lambda + \frac{|V_{ab}|^2}{(2 \hbar)^2}
+\end{aligned}$$
+
+This quadratic equation has two complex roots $\lambda_1$ and $\lambda_2$,
+which are found to be:
+
+$$\begin{aligned}
+    \lambda_1
+    = i \frac{\omega - \omega_0 + \tilde{\Omega}}{2}
+    \qquad \quad
+    \lambda_2
+    = i \frac{\omega - \omega_0 - \tilde{\Omega}}{2}
+\end{aligned}$$
+
+Where we have defined the **generalized Rabi frequency** $\tilde{\Omega}$ to be given by:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{\Omega}
+        \equiv \sqrt{(\omega - \omega_0)^2 + \frac{|V_{ab}|^2}{\hbar^2}}
+    }
+\end{aligned}$$
+
+So that the general solution $c_a(t)$ is as follows,
+where $A$ and $B$ are arbitrary constants,
+to be determined from initial conditions (and normalization):
+
+$$\begin{aligned}
+    \boxed{
+        c_a(t)
+        = \Big( A \sin\!(\tilde{\Omega} t / 2) + B \cos\!(\tilde{\Omega} t / 2) \Big) \exp\!\big(i (\omega \!-\! \omega_0) t / 2 \big) 
+    }
+\end{aligned}$$
+
+And then the corresponding $c_b(t)$ can be found
+from the coupled equation we started at,
+or, if we only care about the probability density $|c_a|^2$,
+we can use $|c_b|^2 = 1 - |c_a|^2$.
+For example, if $A = 0$ and $B = 1$,
+we get the following probabilities
+
+$$\begin{aligned}
+    |c_a(t)|^2
+    &= \cos^2(\tilde{\Omega} t / 2)
+    = \frac{1}{2} \Big( 1 + \cos\!(\tilde{\Omega} t) \Big)
+    \\
+    |c_b(t)|^2
+    &= \sin^2(\tilde{\Omega} t / 2)
+    = \frac{1}{2} \Big( 1 - \cos\!(\tilde{\Omega} t) \Big)
+\end{aligned}$$
+
+Note that the period was halved by squaring.
+This periodic "flopping" of the particle between $\ket{a}$ and $\ket{b}$
+is known as **Rabi oscillation**, **Rabi flopping** or the **Rabi cycle**.
+This is a more accurate treatment
+of the flopping found from first-order perturbation theory.
+
+The name **generalized Rabi frequency** suggests
+that there is a non-general version.
+Indeed, the **Rabi frequency** $\Omega$ is based on
+the special case of exact resonance $\omega = \omega_0$:
+
+$$\begin{aligned}
+    \Omega
+    \equiv \frac{V_{ab}}{\hbar}
+\end{aligned}$$
+
+As an example, Rabi oscillation arises
+in the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+where $\hat{H}_1$ is:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - q \vec{r} \cdot \vec{E}_0 \cos(\omega t)
+\end{aligned}$$
+
+After making the rotating wave approximation,
+the resulting Rabi frequency is given by:
+
+$$\begin{aligned}
+    \Omega
+    = \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
+\end{aligned}$$
+
+Where $\vec{E}_0$ is the [electric field](/know/concept/electric-field/) amplitude,
+and $\vec{d} \equiv q \matrixel{a}{\vec{r}}{b}$ is the transition dipole moment
+of the electron between orbitals $\ket{a}$ and $\ket{b}$.
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/content/know/concept/time-dependent-perturbation-theory/index.pdc b/content/know/concept/time-dependent-perturbation-theory/index.pdc
index b16e3ee..1fbd9ce 100644
--- a/content/know/concept/time-dependent-perturbation-theory/index.pdc
+++ b/content/know/concept/time-dependent-perturbation-theory/index.pdc
@@ -181,7 +181,10 @@ which can cause confusion when comparing literature.
 
 In any case, the probability oscillates as a function of $t$
 with period $