468 Chapter 16. Non-neutral plasmas
whereSis a constant. Since Eq.(16.48) is a second-order ordinary differential equation, it
must have two independent solutions and Eq.(16.50) must be one of these solutions. The
other solution has a singularity atr=0and is discarded as being non-physical (this other
solution can be found by assuming it is of the formΦ =FGand substituting this into
Eq.(16.48) to findF). It is now evident that rigid body rotation corresponds to having
G= 0for allrsinceω 0 (r)is a constant andω=ω 0 (a)for a rigid body;ifG= 0then
Eq.(16.48) shows that the form ofΦis unrestricted and so rigid body rotation can have
any perturbation profile. On the other hand, if the rotation has even the slightest amount of
shear, thenGmust be non-zero and the solution is restricted to be of the form specified by
Eq.(16.50).
The perfectly conducting wall boundary conditionEθ= 0implies thatφ ̃l=1(a) = 0
and so Eq.(16.50) gives thel=1mode frequency to be
ω = ω 0 (a)= −q
ε 0 Ba^2∫rp0n 0 (r′)r′dr′= −
Er(a)
Ba(16.51)
which has the interesting feature of depending only onEr(a),the equilibrium radial field
at the wall. SinceEr(a)depends only on the total charge per length, thel= 1mode
frequency depends only on the total charge per length and not on the radial profile of the
charge density. The coefficientShas been chosen so that theperturbedradial electric field
at the wall is
E ̃r(a)=−Sa
2 ω(
dω 0 (r)
dr)
r=a. (16.52)
Sincen 0 (a)=0by assumption, Eq.(16.42) shows that
(
dω 0 (r)
dr)
r=a=
(
−
2
r
ω 0 (r))
r=a
= −2 ω
a(16.53)
and so
S=E ̃r(a). (16.54)16.5.3Energy analysis of diocotron modes
The linearized current density of a diocotron mode is
J 1 = n 1 qu 0 +n 0 qu 1= n 1 qrω 0 (r)ˆθ+n 0 qE 1 ×B
B^2
(16.55)
and then 1 qrω 0 (r)ˆθequilibriumflow term enables the wave energy to become negative.
The wave is electrostatic so Eq.(7.56) gives the wave energy density to be (Briggs, Daugh-