Fundamentals of Plasma Physics

110 Chapter 3. Motion of a single plasma particle

and thezaxis has been chosen to be in the direction ofB 0 .Equation (3.197) can be solved
forvby first dotting withAto obtain

A·v=C·A (3.199)

and then crossing withAto obtain

v×A+AA·v−vA^2 =C×A. (3.200)

Substituting forA·vusing Eq.(3.199) and forv×Ausing Eq.(3.197) gives

v=

C+AA·C−C×A

1 +A^2

=C‖zˆ+

C⊥

1 +A^2

+

A×C

1 +A^2

(3.201)

whereChas been split into parallel and perpendicular parts relative toB 0 andAA·C
=A^2 C‖zˆhas been used. On substituting forAandCthis becomes

v=

iq

ωm

[

E ̃‖(x)ˆz+

E ̃⊥(x)

1 −ω^2 c/ω^2

−

iωc

ω

zˆ×E ̃(x)

1 −ω^2 c/ω^2

]

e−iωt. (3.202)

The third term on the right hand side is a generalization of theE×Bdrift, since for
ω <<ωcthis term reduces to theE×Bdrift. Similarly, the middle term on the right hand
side is a generalization of the polarization drift, since forω <<ωcthis term reduces to the
polarization drift. The first term on the right hand side, the parallel quiver velocity, does
not involve the magnetic fieldB 0. This non-dependence on magnetic field is to be expected
because no magnetic force results from motion parallel to a magnetic field. In fact, if the
magnetic field were zero, then the second term would add to the first and thethird term
would vanish, giving a three dimensional unmagnetized quiver velocityv= iqE ̃(x)/ωm.
If the electric field is in addition decomposed into spatial Fourier modes with depen-
dence∼exp(ik·x),then the velocity for a typical mode will be

v(x,t)=

iq

ωm

[

E ̃‖ˆz+

E ̃⊥

1 −ω^2 c/ω^2

−

iωc

ω

zˆ×E ̃

1 −ω^2 c/ω^2

]

eik·x−iωt. (3.203)

The convention of a negative coefficient forωand a positive coefficient forkhas been
adopted to give waves propagating in the positivexdirection. Equation (3.203) will later
be used as the starting point for calculating wave-generated plasma currents.

3.9 Wave-particle energy transfer

3.9.1 ‘Average velocity’

Anyone who has experienced delay in a traffic jam knows that it is usually impossible to
make up for the delay by going faster after escaping from the traffic jam. To see why,
defineαas the fraction of the total trip length in the traffic jam,vsas the slow (traffic jam)