Initial commit after migration from Hugo

1 files changed, 516 insertions, 0 deletions

diff --git a/source/know/concept/guiding-center-theory/index.md b/source/know/concept/guiding-center-theory/index.md
new file mode 100644
index 0000000..d429a50
--- /dev/null
+++ b/source/know/concept/guiding-center-theory/index.md
@@ -0,0 +1,516 @@
+---
+title: "Guiding center theory"
+date: 2021-09-21
+categories:
+- Physics
+- Electromagnetism
+- Plasma physics
+layout: "concept"
+---
+
+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 $m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$,
+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" markdown="1">
+<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 $\vb{B} = (B_x, B_y, B_z)$:
+
+$$\begin{aligned}
+    (\vb{x}_L \cdot \nabla) \vb{B}
+    = \frac{u_L}{\omega_c}
+    \begin{pmatrix}
+        \displaystyle \sin(\omega_c t) \pdv{B_x}{x} + \cos(\omega_c t) \pdv{B_x}{y} \\
+        \displaystyle \sin(\omega_c t) \pdv{B_y}{x} + \cos(\omega_c t) \pdv{B_y}{y} \\
+        \displaystyle \sin(\omega_c t) \pdv{B_z}{x} + \cos(\omega_c t) \pdv{B_z}{y}
+    \end{pmatrix}
+\end{aligned}$$
+
+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 $\cos$:
+
+$$\begin{aligned}
+    \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}
+    &= \frac{u_L^2}{\omega_c}
+    \begin{pmatrix}
+        \cos \\
+        - \sin \\
+        0
+    \end{pmatrix}
+    \cross
+    \begin{pmatrix}
+        \displaystyle \pdv{B_x}{x} \sin + \pdv{B_x}{y} \cos \\
+        \displaystyle \pdv{B_y}{x} \sin + \pdv{B_y}{y} \cos \\
+        \displaystyle \pdv{B_z}{x} \sin + \pdv{B_z}{y} \cos
+    \end{pmatrix}
+    \\
+    &= \frac{u_L^2}{\omega_c}
+    \begin{pmatrix}
+        \displaystyle - \pdv{B_z}{x} \sin^2 - \pdv{B_z}{y} \sin \cos \\
+        \displaystyle - \pdv{B_z}{x} \sin \cos - \pdv{B_z}{y} \cos^2 \\
+        \displaystyle \pdv{B_y}{x} \sin \cos + \pdv{B_y}{y} \cos^2
+        \displaystyle + \pdv{B_x}{x} \sin^2 + \pdv{B_x}{y} \sin \cos
+    \end{pmatrix}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    &= \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \displaystyle - \pdv{B_z}{x} \\
+        \displaystyle - \pdv{B_z}{y} \\
+        \displaystyle \pdv{B_y}{y} + \pdv{B_x}{x}
+    \end{pmatrix}
+\end{aligned}$$
+
+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 $B \equiv |\vb{B}| \approx B_z$:
+
+$$\begin{aligned}
+    \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    &= - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \displaystyle \pdv{B_z}{x} \\
+        \displaystyle \pdv{B_z}{y} \\
+        \displaystyle \pdv{B_z}{z}
+    \end{pmatrix}
+    \approx - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \displaystyle \pdv{B}{x} \\
+        \displaystyle \pdv{B}{y} \\
+        \displaystyle \pdv{B}{z}
+    \end{pmatrix}
+    = - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+</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. $\ipdv{\vu{b}}{t} = 0$,
+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 $\idv{}{\vb{v}p}{t}$ 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.