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Diffstat (limited to 'source/know/concept/guiding-center-theory')
| -rw-r--r-- | source/know/concept/guiding-center-theory/index.md | 178 |
1 files changed, 90 insertions, 88 deletions
diff --git a/source/know/concept/guiding-center-theory/index.md b/source/know/concept/guiding-center-theory/index.md index 6de54fa..5368966 100644 --- a/source/know/concept/guiding-center-theory/index.md +++ b/source/know/concept/guiding-center-theory/index.md @@ -11,14 +11,14 @@ 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/) $\vb{B}$ +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 $\vb{F}$ +combines the Lorentz force $$\vb{F}$$ with Newton's second law: $$\begin{aligned} @@ -28,7 +28,7 @@ $$\begin{aligned} \end{aligned}$$ We now allow the fields vary slowly in time and space. -We thus add deviations $\delta\vb{E}$ and $\delta\vb{B}$: +We thus add deviations $$\delta\vb{E}$$ and $$\delta\vb{B}$$: $$\begin{aligned} \vb{E} @@ -38,10 +38,10 @@ $$\begin{aligned} \to \vb{B} + \delta\vb{B}(\vb{x}, t) \end{aligned}$$ -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 $\vb{u} = \vb{u}_{gc} + \vb{u}_L$. +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: $$\begin{aligned} @@ -49,7 +49,7 @@ $$\begin{aligned} = q \big( \vb{E} + \delta\vb{E} + (\vb{u}_{gc} + \vb{u}_L) \cross (\vb{B} + \delta\vb{B}) \big) \end{aligned}$$ -We already know that $m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$, +We already know that $$m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$$, which we subtract from the total to get: $$\begin{aligned} @@ -59,8 +59,8 @@ $$\begin{aligned} This will be our starting point. Before proceeding, we also define -the average of $\Expval{f}$ of a function $f$ over a single gyroperiod, -where $\omega_c$ is the cyclotron frequency: +the average of $$\Expval{f}$$ of a function $$f$$ over a single gyroperiod, +where $$\omega_c$$ is the cyclotron frequency: $$\begin{aligned} \Expval{f} @@ -74,19 +74,19 @@ 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 $\delta\vb{E} = 0$: +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, +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 $E_\parallel$: +$$u_{gc\parallel}$$ and $$E_\parallel$$: $$\begin{aligned} m \dv{u_{gc\parallel}}{t} @@ -95,8 +95,8 @@ $$\begin{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: +Next, the equation for $$\vb{u}_{gc\perp}$$ is found by +subtracting $$u_{gc\parallel}$$'s equation from the original: $$\begin{aligned} m \dv{\vb{u}_{gc\perp}}{t} @@ -104,11 +104,11 @@ $$\begin{aligned} = q (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}) \end{aligned}$$ -Keep in mind that $\vb{u}_{gc\perp}$ explicitly excludes gyration. -If we try to split $\vb{u}_{gc\perp}$ into a constant and a time-dependent part, +Keep in mind that $$\vb{u}_{gc\perp}$$ explicitly excludes gyration. +If we try to split $$\vb{u}_{gc\perp}$$ into a constant and a time-dependent part, and choose the most convenient constant, we notice that the only way to exclude gyration -is to demand that $\vb{u}_{gc\perp}$ does not depend on time. +is to demand that $$\vb{u}_{gc\perp}$$ does not depend on time. Therefore: $$\begin{aligned} @@ -116,8 +116,8 @@ $$\begin{aligned} = \vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B} \end{aligned}$$ -To find $\vb{u}_{gc\perp}$, we take the cross product with $\vb{B}$, -and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$: +To find $$\vb{u}_{gc\perp}$$, we take the cross product with $$\vb{B}$$, +and use the fact that $$\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$$: $$\begin{aligned} 0 @@ -125,10 +125,10 @@ $$\begin{aligned} = \vb{B} \cross \vb{E} + \vb{u}_{gc\perp} B^2 \end{aligned}$$ -Rearranging this shows that $\vb{u}_{gc\perp}$ is constant. +Rearranging this shows that $$\vb{u}_{gc\perp}$$ is constant. The guiding center drifts sideways at this speed, -hence it is called a **drift velocity** $\vb{v}_E$. -Curiously, $\vb{v}_E$ is independent of $q$: +hence it is called a **drift velocity** $$\vb{v}_E$$. +Curiously, $$\vb{v}_E$$ is independent of $$q$$: $$\begin{aligned} \boxed{ @@ -138,8 +138,8 @@ $$\begin{aligned} \end{aligned}$$ 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 $q$: +$$\vb{E}$$ 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 $$q$$: $$\begin{aligned} \boxed{ @@ -151,18 +151,18 @@ $$\begin{aligned} ## Non-uniform magnetic field -Next, consider a more general case, where $\vb{B}$ is non-uniform, -but $\vb{E}$ is still uniform: +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, $\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$. +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: $$\begin{aligned} @@ -182,7 +182,7 @@ $$\begin{aligned} + \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \big) \end{aligned}$$ -Where we have used that $\Expval{\vb{u}_{gc}} = \vb{u}_{gc}$. +Where we have used that $$\Expval{\vb{u}_{gc}} = \vb{u}_{gc}$$. The two averaged expressions turn out to be: $$\begin{aligned} @@ -198,9 +198,9 @@ $$\begin{aligned} <label for="proof-nonuniform-B-averages">Proof</label> <div class="hidden" markdown="1"> <label for="proof-nonuniform-B-averages">Proof.</label> -We know what $\vb{x}_L$ is, -so we can write out $(\vb{x}_L \cdot \nabla) \vb{B}$ -for $\vb{B} = (B_x, B_y, B_z)$: +We know what $$\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} @@ -212,7 +212,7 @@ $$\begin{aligned} \end{pmatrix} \end{aligned}$$ -Integrating $\sin$ and $\cos$ over their period yields zero, +Integrating $$\sin$$ and $$\cos$$ over their period yields zero, so the average vanishes: $$\begin{aligned} @@ -220,8 +220,8 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Moving on, we write out $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$, -suppressing the arguments of $\sin$ and $\cos$: +Moving on, we write out $$\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} @@ -247,7 +247,7 @@ $$\begin{aligned} \end{pmatrix} \end{aligned}$$ -Integrating products of $\sin$ and $\cos$ over their period gives us the following: +Integrating products of $$\sin$$ and $$\cos$$ over their period gives us the following: $$\begin{aligned} \Expval{\cos^2} = \Expval{\sin^2} = \frac{1}{2} @@ -256,7 +256,7 @@ $$\begin{aligned} \end{aligned}$$ Inserting this tells us that the average -of $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$ is given by: +of $$\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} } @@ -268,11 +268,11 @@ $$\begin{aligned} \end{pmatrix} \end{aligned}$$ -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 $B \equiv |\vb{B}| \approx B_z$: +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 $$B \equiv |\vb{B}| \approx B_z$$: $$\begin{aligned} \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } @@ -290,6 +290,7 @@ $$\begin{aligned} \end{pmatrix} = - \frac{u_L^2}{2 \omega_c} \nabla B \end{aligned}$$ + </div> </div> @@ -301,11 +302,11 @@ $$\begin{aligned} = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}$$ -Let us now split $\vb{u}_{gc}$ into -components $\vb{u}_{gc\perp}$ and $u_{gc\parallel} \vu{b}$, +Let us now split $$\vb{u}_{gc}$$ into +components $$\vb{u}_{gc\perp}$$ and $$u_{gc\parallel} \vu{b}$$, which are respectively perpendicular and parallel -to the magnetic unit vector $\vu{b}$, -such that $\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$. +to the magnetic unit vector $$\vu{b}$$, +such that $$\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$$. Consequently: $$\begin{aligned} @@ -322,9 +323,9 @@ $$\begin{aligned} = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}$$ -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: +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: $$\begin{aligned} \dv{\vu{b}}{t} @@ -336,8 +337,8 @@ $$\begin{aligned} <label for="proof-nonuniform-B-curvature">Proof</label> <div class="hidden" markdown="1"> <label for="proof-nonuniform-B-curvature">Proof.</label> -Assuming that $\vu{b}$ does not explicitly depend on time, -i.e. $\ipdv{\vu{b}}{t} = 0$, +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: $$\begin{aligned} @@ -346,16 +347,16 @@ $$\begin{aligned} = u_{gc\parallel} \dv{\vu{b}}{s} \end{aligned}$$ -Where $\dd{s}$ is the arc length of the magnetic field line, -which is equal to the radius $R_c$ times the infinitesimal subtended angle $\dd{\theta}$: +Where $$\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}$$ -Meanwhile, across this arc, $\vu{b}$ rotates by $\dd{\theta}$, -such that the tip travels a distance $|\dd{\vu{b}}|$: +Meanwhile, across this arc, $$\vu{b}$$ rotates by $$\dd{\theta}$$, +such that the tip travels a distance $$|\dd{\vu{b}}|$$: $$\begin{aligned} |\!\dd{\vu{b}}\!| @@ -363,15 +364,15 @@ $$\begin{aligned} = \dd{\theta} \end{aligned}$$ -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 $\vu{b}$: +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 $\dd{\vu{b}}$, +Combining these expressions for $$\dd{s}$$ and $$\dd{\vu{b}}$$, we find the following derivative: $$\begin{aligned} @@ -380,6 +381,7 @@ $$\begin{aligned} = - \frac{\vu{R}_c}{R_c} = - \frac{\vb{R}_c}{R_c^2} \end{aligned}$$ + </div> </div> @@ -391,8 +393,8 @@ $$\begin{aligned} = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg) \end{aligned}$$ -Since both $\vb{R}_c$ and any cross product with $\vb{B}$ -will always be perpendicular to $\vb{B}$, +Since both $$\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} @@ -405,7 +407,7 @@ $$\begin{aligned} The parallel part simply describes an acceleration. The perpendicular part is more interesting: -we rewrite it as follows, defining an effective force $\vb{F}_{\!\perp}$: +we rewrite it as follows, defining an effective force $$\vb{F}_{\!\perp}$$: $$\begin{aligned} m \dv{\vb{u}_{gc\perp}}{t} @@ -416,7 +418,7 @@ $$\begin{aligned} \end{aligned}$$ 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, +We thus assume that $$\vb{u}_{gc\perp}$$ is constant in time, such that the equation reduces to: $$\begin{aligned} @@ -426,8 +428,8 @@ $$\begin{aligned} \end{aligned}$$ This is analogous to the previous case of a uniform electric field, -with $q \vb{E}$ replaced by $\vb{F}_{\!\perp}$, -so it is also solved by crossing with $\vb{B}$ in front, +with $$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} @@ -436,11 +438,11 @@ $$\begin{aligned} \equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2} \end{aligned}$$ -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** $\vb{v}_{\nabla B}$: +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** $$\vb{v}_{\nabla B}$$: $$\begin{aligned} \boxed{ @@ -454,11 +456,11 @@ $$\begin{aligned} } \end{aligned}$$ -Such that $\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$. +Such that $$\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$$. We are still missing a correction, -since we neglected the time dependence of $\vb{u}_{gc\perp}$ earlier. -This correction is called $\vb{v}_p$, -where $\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$. +since we neglected the time dependence of $$\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} @@ -466,10 +468,10 @@ $$\begin{aligned} = \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B} \end{aligned}$$ -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 $\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$, +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: $$\begin{aligned} @@ -477,11 +479,11 @@ $$\begin{aligned} = q \vb{v}_p \cross \vb{B} \end{aligned}$$ -To isolate this for $\vb{v}_p$, -we take the cross product with $\vb{B}$ in front, +To isolate this for $$\vb{v}_p$$, +we take the cross product with $$\vb{B}$$ in front, like earlier. We thus arrive at the following correction, -known as the **polarization drift** $\vb{v}_p$: +known as the **polarization drift** $$\vb{v}_p$$: $$\begin{aligned} \boxed{ @@ -490,8 +492,8 @@ $$\begin{aligned} } \end{aligned}$$ -In many cases $\vb{v}_E$ dominates $\vb{v}_F$, -so in some literature $\vb{v}_p$ is approximated as follows: +In many cases $$\vb{v}_E$$ dominates $$\vb{v}_F$$, +so in some literature $$\vb{v}_p$$ is approximated as follows: $$\begin{aligned} \vb{v}_p @@ -502,7 +504,7 @@ $$\begin{aligned} The polarization drift stands out from the others: it has the opposite sign, -it is proportional to $m$, +it is proportional to $$m$$, and it is often only temporary. Therefore, it is also called the **inertia drift**. |