16.5 Diocotron modes 467
Equation (16.42) reduces to the special case of rigid body rotation when the density is
uniform, but in the more general case there is a shear in the angular velocity. It is seen from
Eq.(16.39) that
1
r
d
dr
(
r^2 ω 0 (r)
)
=
1
Br
d
dr
(
r
dφ 0
dr
)
=−
nq
ε 0 B
=−
ω^2 p(r)
ωc
. (16.43)
Furthermore, since
d
dr
(
1
r
dr^2
dr
)
=0 (16.44)
an arbitrary constant can be added toω 0 (r)in the last term in Eq.(16.40), which conse-
quently can be written as
[ω−lω 0 (r)]
(
d
dr
(
r
dφ ̃l
dr
)
−
l^2
r
φ ̃l
)
− ̃φl
d
dr
(
1
r
d
dr
[
r^2 (ω−lω 0 )
]
)
=0; (16.45)
this is called the diocotron wave equation.
The dioctron wave equation can be written in a more symmetric form by defining
Φ = rφ ̃l
G = r^2 (ω−lω 0 (r)) (16.46)
and using
d
dr
(
r
d ̃φl
dr
)
=r
d
dr
(
1
r
dΦ
dr
)
+
Φ
r^2
(16.47)
to obtain
r
d
dr
(
1
r
dΦ
dr
)
+
(
1 −l^2
)
Φ−Φ
r
G
d
dr
(
1
r
dG
dr
)
=0. (16.48)
16.5.1Wall boundary condition
While the wall boundary condition does not affect the equilibrium of a non-neutral plasma,
it plays a critical role for the diocotron modes since these modes are non-axisymmetric. Be-
cause the wall is perfectly conducting, the azimuthal electric field must vanish at the wall,
i.e.,Eθ(a)=0.This is trivially satisfied for the equilibrium because the equilibrium is ax-
isymmetric andEθ=−r−^1 ∂φ/∂θvanishes everywhere for an axisymmetric field. How-
ever, for a non-axisymmetric perturbation, the azimuthal electric field isE ̃θ,l=−ilφ ̃l/r
which is, in general, finite. Thus, in order to have the azimuthal electric field vanish at the
wall, each finitelmode must satisfy the wall boundary condition ̃φl(a)=0.
16.5.2Thel=1diocotron mode: a special case
Thel=1mode is a special case and has the exact solution
Φ=λG (16.49)
whereλis an arbitrary constant. In terms of the original variables, thissolution is (White,
Malmberg and Driscoll 1982)
̃φl=1(r)=ω−ω^0 (r)
2 ω
Sr (16.50)