116 Chapter 3. Motion of a single plasma particle
which corresponds to considering small amplitude waves sinceα∼ω−b^1 andωb∼
√
φ 0.
We wish to determine how these untrapped particles exchange energy with the wave.
To accomplish this the lab-frame kinetic energy must be expressed in terms of wave-frame
quantities. From Eqs.(3.220) and (3.224) the lab-frame velocity is
v=1
k(
ω+dθ
dt)
=
ωb
k(
ω
ωb+
dθ
dτ)
(3.232)
so that the lab-frame kinetic energy can be expressed as
W=
1
2
mv^2 =mω^2 b
2 k^2[(
ω
ωb) 2
+ 2
ω
ωbdθ
dτ+
(
dθ
dτ) 2 ]
. (3.233)
Substituting for(dθ/dτ)^2 using Eq.(3.227) gives
W=
mω^2 b
2 k^2[(
ω
ωb) 2
+ 2
ω
ωbdθ
dτ+λ+ 2cosθ]
. (3.234)
Since wave-particle energy transfer is of interest, attention is nowfocused on thechanges
in the lab-frame particle kinetic energy and so we consider
dW
dt=
mω^3 b
k^2[
ω
ωbd^2 θ
dτ^2−
dθ
dτsinθ]
= −
mω^3 b
k^2sinθ[
ω
ωb+
dθ
dτ]
(3.235)
where Eq.(3.225) has been used. To proceed further, it is necessary to obtain the time
dependence of bothsinθanddθ/dt.
Solving Eq.(3.227) fordθ/dτand assumingα >> 1 (corresponding to untrapped par-
ticles) gives
dθ
dτ
= ±
√
λ+ 2cosθ= ±
√
α^2 + 2(cosθ−cosθ 0 )= α(
1 +
2(cosθ−cosθ 0 )
α^2) 1 / 2
≃ α+
cosθ−cosθ 0
α.
(3.236)
This expression is valid for both positive and negativeα, i.e. for particles going in either
direction in the wave frame. The first term in the last line of Eq. (3.236) gives the velocity
the particle would have if there werenowave (unperturbed orbit) while the second term
gives theperturbationdue to the small amplitude wave. The particle orbitθ(τ)is now
solved for iteratively. To lowest order (i.e., dropping terms of orderα−^2 ) the particle
velocity is
dθ
dτ
=α (3.237)