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diff --git a/content/know/concept/guiding-center-theory/index.pdc b/content/know/concept/guiding-center-theory/index.pdc new file mode 100644 index 0000000..12abac0 --- /dev/null +++ b/content/know/concept/guiding-center-theory/index.pdc @@ -0,0 +1,522 @@ +--- +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. |