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:

\[\varepsilon\frac{d \bm v}{dt} = \pm \frac{\hat q}{\hat m}(\bm E + \bm v \times \bm B); \quad \frac{d \bm r}{d t} = \bm v\]

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:

\[\tau_c \ll \tau \quad\Rightarrow\quad \frac{\tau_c}{\tau} \sim \varepsilon \ll 1\] \[\bm r(t) \quad \rightarrow \quad \bm r(t,\varphi(t)) \quad \rightarrow \quad \bm r(t,\varphi)\]

Next, we express our solution in the following form:

\[\bm r(t,\varphi) = \bm R(t) + \varepsilon \bm \rho(t,\varphi); \quad \bm v(t,\varphi) = \bm U(t)+ \bm u(t,\varphi)\] \[\left<\bm \rho\right> = \frac{1}{2\pi}\oint \bm \rho d\varphi = \bm 0; \quad \left<\bm u\right> = \frac{1}{2\pi}\oint \bm u d\varphi = \bm 0\]

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:

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

Magnetic field gradient:

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

\[\nabla \bm B = \left[ \begin{array}{cc|c} \partial B_x/\partial x & \partial B_y/\partial x & \partial B_z/\partial x \\ \partial B_x/\partial y & \partial B_y/\partial y & \partial B_z/\partial y \\ \hline \partial B_x/\partial z & \partial B_y/\partial z & \partial B_z/\partial z \\ \end{array} \right]\]

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:

\[\pm q\bm v \times \bm B = \pm q\bm v \times \bm B(\bm R,t) \pm \varepsilon q\bm v \times (\bm \rho\cdot\nabla)\bm B(\bm R,t)\]

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

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

Leading order solution

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 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:

\[\bm U_{1\perp} = \pm \frac{\bm 1_b}{\omega_c} \times \left[ \frac{d \bm U_0}{dt} +\frac{\mu_0}{m}\nabla B \right]\]

Magnetic drift

\[\pm \frac{\mu_0}{m\omega_c} \bm 1_b \times \nabla B\]

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:

\[\pm \frac{U_{\parallel 0}^2}{\omega_c} \bm 1_b \times \bm \kappa\]

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

\[\pm \frac{\bm 1_b}{\omega_c} \times \frac{d \bm U_{\perp 0}}{dt}\]

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

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

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


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


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


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


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

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:

\[\frac{d}{dt} \mu_0 = 0\]

See HAZE18 p 29 to learn more about Poincaré invariants and Adiabatic invariants

Magnetic mirror

bg right:40% 90%

In stationary fields,

\[H = \frac{1}{2} mv_\parallel^2 + \mu B = \text{const}\] \[\mu = \frac{mv_\perp^2}{2B} = \text{const}\]

Further reading