328 Chapter 10. Stability of static MHD equilibria
If a configuration is to be stable, any perturbation must generate a restoring force which
pushes the perturbed surface back to its equilibrium location. The competition between
destabilizing and restoring forces is found by integrating Eq.(10.141) across the perturbed
surface. This integration gives Eq.(10.144) evaluated at the perturbed surface, i.e., using
Eq.(10.148) to take into account the effect of the perturbation of the surface.Sinceξwas
assumed positive (outward bulge atθ=0,z=0) the perturbed system will be stable if the
effective pressure on the outside of the perturbed surface exceeds the effective pressure on
the inside, i.e., if
[
B^2 vθ+Bvz^2
2 μ 0]
perturbedsfc>
[
P+
Bpz^2
2 μ 0]
perturbedsfc=⇒stable. (10.149)In this case, the restoring force pushes back the bulge and makes the system revert to its
equilibrium condition. Subtracting the equilibrium pressure balance relation, Eq.(10.144),
this becomes
B 0 vθB 1 vθ+B 0 vzB 1 vz
μ 0+ξ∂
∂r[
B 02 vθ+B^20 vz
2 μ 0]
>P 1 +
B 0 pzB 1 pz
μ 0+ξ∂
∂r[
P 0 +
B 02 pz
2 μ 0]
(10.150)
where all quantities are evaluated atr=a.
Equation (10.150) can be simplified considerably because of the following relation-
ships:
- There are no currents in either the plasma interior or the external vacuum so
∂B 0 pz
∂r=
∂B 0 vz
∂r=0. (10.151)
- The adiabatic pressure equation gives
P 1 +ξ·∇P 0 +γP 0 ∇·ξ=0. (10.152)
Since incompressibility∇·ξ=0has been assumed, the adiabatic relation reduces
to
P 1 +ξ∂P 0 /∂r=0. (10.153)- From Ampere’s law it is seen thatB 02 vθ∼r−^2 so
[
∂B 02 vθ/∂r]
r=a=−^2 B2
0 vθ/a;thus
Eq.(10.150) simplifies toB 0 vθB 1 vθ+B 0 vzB 1 vz−ξ
a
B^20 vθ>B 0 pzB 1 pz=⇒stable (10.154)where again all fields are evaluated atr=a.
To simplify the algebra all fields are now normalized toB 0 vθ(a).The normalized fields
are denoted by a bar on top and are
B ̄ 0 vz= B^0 vz
B 0 vθ(a), B ̄ 0 pz=B 0 pz
B 0 vθ(a),
B ̄ 1 vz= B^1 vz
B 0 vθ(a), B ̄ 1 vθ=B 1 vθ
B 0 vθ(a), B ̄ 1 pz=B 1 pz
B 0 vθ(a)