# Notes on single particle motion II

## Variable fields

We will consider only cases with:

$\ell \ll L; \quad v_\parallel / L \ll \omega_c; \quad |\partial \bm E/\partial t|/E \ll \omega_c; \quad |\partial \bm B/\partial t|/B \ll \omega_c$

where $L$ is the gradient length of changes in $\bm E$ and $\bm B$ ($L\sim E/|\nabla \bm E|\sim B/|\nabla \bm B|$). This ensures that the particle does not “see” large changes in $\bm E$, $\bm B$ within an orbit.

We introduce the small parameter $\varepsilon = \ell / L$ and rewrite $m/q \to \varepsilon \hat m/\hat q$, where $\hat m,\hat q$ are arbitrary normalization constants:

$\varepsilon\frac{d \bm v}{dt} = \pm \frac{\hat q}{\hat m}(\bm E + \bm v \times \bm B); \quad \frac{d \bm r}{d t} = \bm v$
• We will study the limit $\varepsilon \to 0^+$ and drop the hat symbols for simplicity

## Two-scale averaging method

There are two distinct timescales in the problem: $\tau_c = 1/\omega_c$ for the fast changes of the gyrophase $\varphi$, and $\tau$ for the slow changes of the position of the gyrocenter:

$\tau_c \ll \tau \quad\Rightarrow\quad \frac{\tau_c}{\tau} \sim \varepsilon \ll 1$
• While the gyrophase is a function of time, $\varphi = \varphi(t)$, we will write all variables as functions of $t$ and $\varphi$ as if they were two independent variables:
$\bm r(t) \quad \rightarrow \quad \bm r(t,\varphi(t)) \quad \rightarrow \quad \bm r(t,\varphi)$
• We will compute derivatives as: $\dfrac{d \bm r}{dt} = \dfrac{\partial \bm r}{\partial t} + \dot\varphi \dfrac{\partial \bm r}{\partial \varphi}$, with $\dot\varphi \approx \omega_c = O(1/\varepsilon)$

Next, we express our solution in the following form:

$\bm r(t,\varphi) = \bm R(t) + \varepsilon \bm \rho(t,\varphi); \quad \bm v(t,\varphi) = \bm U(t)+ \bm u(t,\varphi)$
• $\bm R$ and $\bm U = \bm U_\perp + U_\parallel \bm 1_b$ describe the gyrocenter, with $\bm U = d\bm R/dt$
• $\bm \rho,\bm u$ represent the fast gyromotion:
• They are perpendicular to $\bm B(\bm R,t)$, i.e., $\bm \rho \cdot \bm 1_b =0$, $\bm u \cdot \bm 1_b =0$
• They are $2\pi$-periodic in $\varphi$ for fixed $t$
• They have zero $\varphi$-average:
$\left<\bm \rho\right> = \frac{1}{2\pi}\oint \bm \rho d\varphi = \bm 0; \quad \left<\bm u\right> = \frac{1}{2\pi}\oint \bm u d\varphi = \bm 0$

The vector basis ${\bm 1b, \bm 1_r, \bm 1\varphi}$ is defined at $(\bm R,t)$.

Finally, we expand:

$\bm R = \bm R_0 +\varepsilon\bm R_1 +\varepsilon^2\bm R_2 + \ldots \\ \bm U = \bm U_0 +\varepsilon\bm U_1 +\varepsilon^2\bm U_2 + \ldots \\ \bm \rho = \bm \rho_0 +\varepsilon\bm \rho_1 +\varepsilon^2\bm \rho_2 + \ldots \\ \bm u = \bm u_0 +\varepsilon\bm u_1 +\varepsilon^2\bm u_2 + \ldots$

The gyrophase evolves as:

$\frac{d \varphi}{dt} = \frac{1}{\varepsilon}\Omega_{-1} + \Omega_0 + \varepsilon \Omega_1 + \ldots$

Consider $\bm B (\bm R,t) = B_0 \bm 1_z$. The gradient of $\bm B$ has different terms:

$\nabla \bm B = \left[ \begin{array}{cc|c} \partial B_x/\partial x & \partial B_y/\partial x & \partial B_z/\partial x \\ \partial B_x/\partial y & \partial B_y/\partial y & \partial B_z/\partial y \\ \hline \partial B_x/\partial z & \partial B_y/\partial z & \partial B_z/\partial z \\ \end{array} \right]$
• Diagonal terms are the divergence terms. Observe that always $\nabla \cdot \bm B = 0$
• The variation of $B_z$ with $x$ and $y$ are the gradient terms
• Variations of $B_x$, $B_y$ with $z$ are the curvature terms
• Terms $\partial B_x/\partial y$ and $\partial B_y/\partial x$ are shear terms

We write:

$\bm B(\bm r,t)\simeq \bm B(\bm R,t) + \varepsilon(\bm \rho\cdot \nabla) \bm B (\bm R,t) + \ldots \\ \bm E(\bm r,t)\simeq \bm E(\bm R,t) + \varepsilon(\bm \rho\cdot \nabla) \bm E (\bm R,t) + \ldots$

and expand the magnetic force as:

$\pm q\bm v \times \bm B = \pm q\bm v \times \bm B(\bm R,t) \pm \varepsilon q\bm v \times (\bm \rho\cdot\nabla)\bm B(\bm R,t)$
• The second term on the RHS can be thought of as a small “extra force” $\bm F$

Furthermore, we require that $E_\parallel$ be small:

$\bm E(\bm R,t) = \bm E_\perp(\bm R,t) + \varepsilon E_\parallel(\bm R,t)\bm 1_b$

To order $O(1/\varepsilon)$, the equation of motion reads:

$\Omega_{-1}\frac{d \bm u_0}{d\varphi} \mp \frac{q}{m}\bm u_0 \times \bm B(\bm R,t) = \pm\frac{q}{m}\bm E_\perp(\bm R,t) \pm \frac{q}{m}\bm U_{\perp 0} \times \bm B(\bm R,t)$

Averaging in $\varphi$:

$\bm 0 = \bm E_\perp(\bm R,t) + \bm U_{\perp 0} \times \bm B(\bm R,t)$

This yields

$\bm U_{\perp 0} = \frac{\bm E_\perp(\bm R,t)\times \bm B(\bm R,t)}{B^2(\bm R,t)}$

Subtracting this from the equation:

$\Omega_{-1}\frac{d \bm u_0}{d\varphi} \mp \frac{q}{m}\bm u_0 \times \bm B(\bm R,t) = \bm 0; \quad \Omega_{-1}\frac{d \bm \rho_0}{d \varphi} = \bm u_0$

In the local cylindrical vector basis, integrating and imposing periodicity in $\varphi$:

$\bm u_0 = \mp u_{\perp 0} \bm 1_\varphi; \quad \Omega_{-1} = \mp \frac{qB(\bm R,t)}{m} = \mp\omega_c(\bm R,t); \quad \bm \rho_0 = \ell_0 \bm 1_r$

Thus, we recover the solution we already had for uniform, constant fields

## Guiding center drifts

To order $O(1)$ the equation reads:

$\Omega_{-1}\frac{\partial \bm u_1}{\partial \varphi} \mp \frac{q}{m}\bm u_1 \times \bm B(\bm R,t) + \frac{d \bm U_0}{dt} + \frac{\partial \bm u_0}{\partial t} + \Omega_{0}\frac{\partial \bm u_0}{\partial \varphi} \\ = \pm \frac{q}{m}E_\parallel(\bm R,t)\bm 1_b \pm \frac{q}{m}(\bm\rho_0\cdot\nabla)\bm E(\bm R,t) \pm \frac{q}{m}\bm U_{1\perp} \times \bm B(\bm R,t) \\ \pm \frac{q}{m}\bm U_{\perp 0} \times (\bm\rho_0\cdot\nabla)\bm B(\bm R,t) \pm \frac{q}{m}\bm u_0 \times (\bm\rho_0\cdot\nabla)\bm B(\bm R,t)$

Averaging in $\varphi$ eliminates most of the terms:

$\frac{d \bm U_0}{dt} = \pm \frac{q}{m}E_\parallel(\bm R,t)\bm 1_b \pm \frac{q}{m}\bm U_{1\perp} \times \bm B(\bm R,t) \pm \frac{q}{m}\left<\bm u_0 \times (\bm\rho_0\cdot\nabla)\bm B(\bm R,t)\right>$
$\pm \frac{q}{m}\left<\bm u_0 \times (\bm\rho_0\cdot\nabla)\bm B(\bm R,t)\right> = - \frac{q}{m} u_{\perp 0} \ell_0\left<\bm 1_\varphi \times (\bm 1_r\cdot\nabla)\bm B(\bm R,t)\right> \\ = \frac{q}{m} u_{\perp 0} \ell_0 \frac{1}{2\pi} \oint d\varphi \left[ \bm 1_b (\bm 1_r\cdot\nabla)\bm B(\bm R,t) \cdot \bm 1_r - \bm 1_r (\bm 1_r\cdot\nabla)\bm B(\bm R,t) \cdot \bm 1_b \right]$

To work out this expression use a local Cartesian vector basis ${\bm 1_x,\bm 1_y, \bm 1_z}$, with $\bm 1_r = \cos\varphi\bm 1_x + \sin\varphi\bm 1_y; \quad \bm 1_b = \bm 1_z$

and use:

$\nabla\cdot\bm B = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0$

After integrating, and noting that $\bm B(\bm R,t) = B_z(\bm R,t) \bm 1_z = B(\bm R,t) \bm 1_z$:

$\pm \frac{q}{m}\left<\bm u_0 \times (\bm\rho_0\cdot\nabla)\bm B(\bm R,t)\right> = - \frac{q}{m} u_{\perp 0} \ell_0 \frac{1}{2} \nabla B(\bm R,t) = - \frac{u_{\perp 0}^2}{2B(\bm R,t)} \nabla B(\bm R,t)$

Introducing the (unsigned) magnetic moment at order $O(1)$:

$\mu_0(\bm R,t) = \frac{m u_{\perp 0}^ 2}{2B(\bm R,t)}$

Now, we can rewrite the averaged equation as

$m\frac{d \bm U_0}{dt} = \pm q E_\parallel(\bm R,t)\bm 1_b \pm q\bm U_{1\perp} \times \bm B(\bm R,t) -\mu_0(\bm R,t)\nabla B(\bm R,t)$

The parallel projection of this equation gives $U_{\parallel 0}$:

$m\frac{d U_{\parallel 0}}{dt} = \pm q E_\parallel(\bm R,t) -\mu_0(\bm R,t)(\bm 1_b\cdot\nabla) B(\bm R,t) - m\bm 1_b \cdot\frac{d \bm U_{\perp 0}}{dt}$

Dropping the $(\bm R,t)$ dependency for the sake of notation:

$m\frac{d U_{\parallel 0}}{dt} = \pm q E_\parallel -\mu_0(\bm 1_b\cdot\nabla) B - m\bm 1_b \cdot\frac{d \bm U_{\perp 0}}{dt}$
• The second term in the RHS is the magnetic mirror force
• It pushes particles in the direction of decreasing $B$ regardless of their sign

The perpendicular projection gives:

$\bm U_{1\perp} = \pm \frac{\bm 1_b}{\omega_c(\bm R,t)} \times \left[ \frac{d \bm U_0}{dt} +\mu_0(\bm R,t)\nabla B(\bm R,t) \right]$

Dropping the $(\bm R,t)$ dependency for the sake of notation:

$\bm U_{1\perp} = \pm \frac{\bm 1_b}{\omega_c} \times \left[ \frac{d \bm U_0}{dt} +\frac{\mu_0}{m}\nabla B \right]$
• The first term is known as the inertial drift
• The second term is the magnetic drift

## Magnetic drift

$\pm \frac{\mu_0}{m\omega_c} \bm 1_b \times \nabla B$
• It displaces the gyrocenter at right angles with $\partial B/\partial \bm 1_\perp$
• The effect is the opposite for positive and negative particles
• It depends on $m$ and on $u_{\perp 0}$ through $\mu_0$

## Inertial drift

We split it in two parts. The derivative of $U_{\parallel 0}\bm 1_b$ gives:

$\pm \frac{\bm 1_b}{\omega_c} \times \frac{d \bm U_{\parallel 0} \bm 1_b}{dt} = \pm \frac{U_{\parallel 0}}{\omega_c} \bm 1_b \times \frac{d \bm 1_b}{dt} \\ = \pm \frac{U_{\parallel 0}}{\omega_c} \bm 1_b \times \left[ \frac{\partial\bm 1_b}{\partial t} + (\bm U_{\perp 0} \cdot \nabla) \bm 1_b + U_{\parallel 0} (\bm 1_b \cdot \nabla) \bm 1_b \right]$

Using the curvature vector $\bm \kappa = (\bm 1_b \cdot \nabla) \bm 1_b$, the last part is the curvature drift:

$\pm \frac{U_{\parallel 0}^2}{\omega_c} \bm 1_b \times \bm \kappa$
• It displaces the gyrocenter at right angles with $\bm \kappa$
• The effect is the opposite for positive and negative particles
• It depends on $m$ and on $U_{\parallel 0}^2$

The derivative of $\bm U_{\perp 0}$ gives:

$\pm \frac{\bm 1_b}{\omega_c} \times \frac{d \bm U_{\perp 0}}{dt}$
• This is known as the polarization drift
• The effect is the opposite for positive and negative particles

## Conservation of $\mu_0$ to $O(1)$

The equation of the $\varphi$-averaged kinetic energy of the particle reads:

$\frac{m}{2} \frac{d}{dt} (u_{\perp 0}^2 + U_0^2) = \pm q U_{\parallel 0}E_\parallel \pm q \bm U_{\perp 1}\cdot \bm E_\perp \pm q \left<(\bm \rho_0\cdot \nabla)\bm E \cdot \bm u_0 \right>$

But:

$m\bm U_0\cdot\frac{d \bm U_0}{dt} = \pm q U_{\parallel 0}E_\parallel \pm q\bm U_{\perp 0}\cdot (\bm U_{1\perp} \times \bm B) -\mu_0\bm U_0\cdot\nabla B$

And:

$U_{\perp 0} = \frac{\bm E_\perp\times\bm B}{B^2} \quad\Rightarrow\quad \pm q\bm U_{\perp 0}\cdot (\bm U_{1\perp} \times \bm B)= \pm q \bm U_{\perp 1}\cdot \bm E_\perp$

$\pm q \left<(\bm \rho_0\cdot \nabla)\bm E \cdot \bm u_0 \right> = \frac{m u_{\perp 0}^2}{2\pi B}\oint d\varphi (\bm 1_r \cdot \nabla)\bm E \cdot \bm 1_\varphi = - \mu_0 \bm 1_b \cdot \nabla \times \bm E = \mu_0 \frac{\partial B}{\partial t}$

Therefore:

$\frac{m}{2} \frac{d}{dt} u_{\perp 0}^2 - \mu_0 \frac{\partial B}{\partial t} -\mu_0 \bm U_0\cdot \nabla B = 0$

Since $\mu_0 = u_{\perp 0}^2/(2B)$ and $dB/dt = \partial B/\partial t + \bm U_0 \cdot \nabla B$ in the averaged equation:

$\frac{d}{dt} \mu_0 = 0$
• $\mu_0$ is conserved to lowest order and is therefore an adiabatic invariant

## Magnetic mirror In stationary fields,

$H = \frac{1}{2} mv_\parallel^2 + \mu B = \text{const}$ $\mu = \frac{mv_\perp^2}{2B} = \text{const}$
• Particles trade parallel to perpendicular kinetic energy as they move toward a region of maximum $B$
• If $B_{max}$ at the maximum, only particles with $v_\perp^2/v^2 < B/B_{max}$ can escape. This defines a loss cone in phase space