Notes on single particle motion I

                Equipo de Propulsión Espacial y Plasmas, Mario Merino, 2021

Charged particle motion

mdvdt=±q(E+v×B)m\frac{d \bm v}{dt} = \pm q(\bm E + \bm v \times \bm B)

  • mm: particle mass; qq: (unsigned) particle charge
  • If fields are stationary, mechanical energy is conserved
  • Other exact conservation laws may exist in each problem

Constant and uniform B\bm B field

Take B=B1z\bm B = B \bm 1_z:

dvxdt=±qBmvy;\frac{d v_x}{dt} = \pm \frac{qB}{m}v_y;

dvydt=∓qBmvx;\frac{d v_y}{dt} = \mp \frac{qB}{m}v_x;

dvzdt=0\frac{d v_z}{dt} = 0

Last one is trivial:

vz=v∥v_z = v_\parallel

The other two in matrix form:

ddt[vxvy]=±ωc[01−10]⋅[vxvy]\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 ωc=qB/m\omega_c = qB/m is the (unsigned) gyrofrequency

Homogeneous ODE with solution:

φ=∓ωct+φ0;vx=±v⊥sin⁡φ;vy=∓v⊥cos⁡φ\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 φ0\varphi_0 is irrelevant here

Integrating again we get:

x=xgc+ℓcos⁡φ;y=ygc+ℓsin⁡φx = x_{gc} + \ell \cos \varphi; \quad y = y_{gc} + \ell \sin \varphi

  • ℓ=v⊥/ωc=mv⊥/(qB)\ell = v_\perp/\omega_c = m v_\perp / (qB) is the (unsigned) gyroradius or Larmor radius

Constant and uniform B\bm B and E\bm E fields

Take B=B1z\bm B = B \bm 1_z and E=E∥1z+E⊥1y\bm E = E_\parallel \bm 1_z + E_\perp\bm 1_y:

The parallel part is trivial again

dvzdt=±qmE∥⇒vz=v∥0±qmE∥t\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 t

The perpendicular part is a inhomogeneous ODE:

ddt[vxvy]=±ωc[01−10]⋅[vxvy]±qm[0E⊥]\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]

vx=E⊥B±v⊥sin⁡φ;vy=∓v⊥cos⁡φv_x = \frac{E_\perp}{B} \pm v_\perp \sin \varphi; \quad v_y = \mp v_\perp \cos \varphi

The motion is the sum of the following. In a local basis {1b,1r(φ),1φ(φ)}\{\bm 1_b, \bm 1_r(\varphi), \bm 1_\varphi(\varphi)\}:

  • A fast gyromotion about the gyrocenter with gyrofrequency ωc\omega_c and gyroradius ℓ\ell:

vc=∓v⊥1φ\bm v_c = \mp v_\perp \bm 1_\varphi

  • A slow drift, known as the ExB drift:
    • The magnitude is E⊥/BE_\perp/B
    • The ExB drift is independent of charge sign, qq, mm

vd=E×BB2\bm v_d = \frac{\bm E\times \bm B}{B^2}

  • An accelerated parallel motion, according to the E∥E_\parallel field:

v∥=[v∥0±qmE∥t]1b\bm v_\parallel = \left[v_{\parallel 0} \pm \frac{q}{m}E_\parallel t\right]\bm 1_b

By the way: the ExB drift can also be understood transforming into a moving frame.

  • Define a frame S′S^\prime that displaces with velocity vd\bm v_d with respect to the original one SS
  • In this frame the new E′\bm E^\prime and B′\bm B^\prime fields are as follows (non-relativistic limit):

E′=E+vd×B=E∥1b;B′=B\bm E^\prime = \bm E + \bm v_d \times \bm B = E_\parallel \bm 1_b; \quad \bm B^\prime = \bm B

  • This means that in S′S^\prime we have E⊥′=0E_\perp^\prime = 0, i.e. the gyromotion-only case

Final solution in constant, uniform fields

Using the local vector basis {1b,1r(φ),1φ(φ)}\{\bm 1_b, \bm 1_r(\varphi), \bm 1_\varphi(\varphi)\}:

v=vd∓v⊥01φ+v∥01b;φ=∓ωct;\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 ±qE\pm q\bm E force gives rise to the ExB drift, an extra constant force
F\bm F gives rise to a similar drift:

mdvdt=±qv×B+F⇒vd=±F×BqB2m\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