Notes on single particle motion I
Charged particle motion
m\frac{d \bm v}{dt} = \pm q(\bm E + \bm v \times \bm B)- m: particle mass; q: (unsigned) particle charge
- If fields are stationary, mechanical energy is conserved
- Other exact conservation laws may exist in each problem
Constant and uniform \bm B field
Take \bm B = B \bm 1_z:
\frac{d v_x}{dt} = \pm \frac{qB}{m}v_y; \frac{d v_y}{dt} = \mp \frac{qB}{m}v_x; \frac{d v_z}{dt} = 0
Last one is trivial:
v_z = v_\parallelThe other two in matrix form:
\frac{d }{dt} \left[ \begin{array}{c} v_x \\ v_y \\ \end{array} \right] = \pm\omega_c \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right] \cdot \left[ \begin{array}{c} v_x \\ v_y \\ \end{array} \right]- Where \omega_c = qB/m is the (unsigned) gyrofrequency
Homogeneous ODE with solution:
\varphi = \mp \omega_c t + \varphi_0; \quad v_x = \pm v_\perp \sin\varphi; \quad v_y = \mp v_\perp \cos\varphi- The initial phase angle \varphi_0 is irrelevant here
Integrating again we get:
x = x_{gc} + \ell \cos \varphi; \quad y = y_{gc} + \ell \sin \varphi- \ell = v_\perp/\omega_c = m v_\perp / (qB) is the (unsigned) gyroradius or Larmor radius
Constant and uniform \bm B and \bm E fields
Take $\bm B = B \bm 1z and \bm E = E\parallel \bm 1z + E\perp\bm 1_y$:
The parallel part is trivial again
\frac{d v_z}{dt} = \pm \frac{q}{m}E_\parallel \quad \Rightarrow \quad v_z = v_{\parallel 0} \pm \frac{q}{m}E_\parallel tThe perpendicular part is a inhomogeneous ODE:
\frac{d }{dt} \left[ \begin{array}{c} v_x \\ v_y \\ \end{array} \right] = \pm\omega_c \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right] \cdot \left[ \begin{array}{c} v_x \\ v_y \\ \end{array} \right] \pm \frac{q}{m} \left[ \begin{array}{c} 0 \\ E_\perp \\ \end{array} \right] v_x = \frac{E_\perp}{B} \pm v_\perp \sin \varphi; \quad v_y = \mp v_\perp \cos \varphiThe motion is the sum of the following. In a local basis ${\bm 1b, \bm 1_r(\varphi), \bm 1\varphi(\varphi)}$:
- A fast gyromotion about the gyrocenter with gyrofrequency \omega_c and gyroradius \ell:
- A slow drift, known as the ExB drift:
- The magnitude is E_\perp/B
- The ExB drift is independent of charge sign, q, m
- An accelerated parallel motion, according to the E_\parallel field:
By the way: the ExB drift can also be understood transforming into a moving frame.
- Define a frame S^\prime that displaces with velocity \bm v_d with respect to the original one S
- In this frame the new \bm E^\prime and \bm B^\prime fields are as follows (non-relativistic limit):
- This means that in S^\prime we have E_\perp^\prime = 0, i.e. the gyromotion-only case
Final solution in constant, uniform fields
Using the local vector basis ${\bm 1b, \bm 1_r(\varphi), \bm 1\varphi(\varphi)}$:
\bm v = \bm v_d \mp v_{\perp 0} \bm 1_\varphi + v_{\parallel 0} \bm 1_b; \quad \varphi = \mp\omega_c t;An extra force
Just like the \pm q\bm E force gives rise to the ExB drift, an extra constant force \bm F gives rise to a similar drift:
m\frac{d \bm v}{dt} = \pm q\bm v \times \bm B + \bm F \quad\Rightarrow\quad \bm v_d = \pm \frac{\bm F\times \bm B}{qB^2}- In this case, ions and electrons have opposite drift directions and a net current develops