10.8 Analysis of free-boundary instabilities 331If we introduce the normalized pressure
P ̄ 0 =^2 μ^0 P^0
B 02 vθ(a)(10.172)
the MHD equilibrium, Eq.(10.144), can be written in normalized form as
P ̄ 0 +B ̄ 02 pz=1+B ̄^20 vz. (10.173)Substitution forB ̄^20 pzinto Eq.(10.171) and rearranging the order of the second term gives
|k|a[
1+B ̄ 02 vz−P ̄ 0]
[
I|m|
I|′m|]
+
(m+kaB ̄ 0 vz)^2
|k|a[
−I|m|Kˆ|′m|+Iˆ|′m|K|m|
I′|m|Kˆ|′m|−Iˆ|′m|K|′m|]
> 1
(10.174)
which is the general requirement for stability of a configuration having a specified internal
pressure, vacuum axial field, plasma radius, and wall radius.
Conditions for sausage and kink instability Let us now examine the effect of the
various factors in Eq.(10.174). For large argument, the modified Bessel functions have the
asymptotic form
lim
s→∞
I|m|(s)→es; lim
s→∞
K|m|(s)→e−s (10.175)so if the wall radius goes to infinity, the factor
[
−I|m|Kˆ|′m|+Iˆ|′m|K|m|
I′|m|Kˆ|′m|−Iˆ|′m|K|′m|]
→−
K|m|
K|′m|=positivedefinite. (10.176)Since bringing the wall closer is stabilizing, the factor
−I|m|Kˆ′|m|+Iˆ|′m|K|m|
I|′m|Kˆ′|m|−Iˆ′|m|K|′m|will always be positive-definite. In particular, ifb→athenI|′m|Kˆ|′m|→Iˆ|′m|K′|m|so that
the wall stabilization becomes arbitrarily large.
As|k|a→∞the left hand side of Eq.(10.174) becomes infinite. Thus, the configura-
tion is stable with respect to modes having short axial wavelengths. The reason for this is
that short wavelength perturbations cause more bending of the magnetic fieldthan a long
wavelength perturbation and so require more energy.
We therefore focus attention on modes with alongaxial wavelength (i.e., modes with a
smallk) since these offer the only possibility of instability. The analysis can besubdivided
into specific cases, such asm=0,|m|=1,|m|> 1 , close-fitting wall, no wall, low
pressure, high pressure, etc. Before getting into the details let us take a broader look at the
effect of the various terms in Eq. (10.174). SinceI|m|/I|′m|> 0 we see that increasingP ̄ 0
is destabilizing, while increasingB ̄^20 vzis stabilizing. Also, ifm+kaB ̄ 0 vz=0the second
term vanishes, leading toreducedstability;by defining the wavevector ask=(m/a)ˆθ+kzˆ
it is seen thatm+kaB ̄ 0 vz=0corresponds to havingk·B 0 =0at the surface.