Fundamentals of Plasma Physics

12.4 Semi-quantitative estimate of the tearing process 367

Using Eq.(12.21) this can be written as

∂Ω 1

∂t

=

1

μ 0 ρ 0

ˆz·

[

∇Az×∇

(

∇^2 ⊥Az

)]

1. (12.28)

From Fig.12.2(c) it is expected that the vortices have significant amplitude only in the
vicinity of where the current bars are deforming and that at large|x|there will be negli-
gible vorticity. Thus, it is assumed that the vorticity evolution equation has the following
behavior:

1. Inner (tearing/reconnection) region: Here it is assumed that the perturbation has much
   steeper gradients than the equilibrium so

|∇

(

∇^2 ⊥Az 0

)

|

|∇Az 0 |

<<

|∇

(

∇^2 ⊥Az 1

)

|

|∇Az 1 |

. (12.29)

This allows Eq.(12.28) to be approximated as

∂Ω 1

∂t

≃

1

μ 0 ρ 0

ˆz·

[

∇Az 0 ×∇

(

∇^2 ⊥Az 1

)]

=

1

μ 0 ρ 0

dAz 0

dx

∂

∂y

(

∇^2 ⊥Az 1

)

(12.30)

which shows thatJ 1 zcrossed withBy 0 generates vorticity. SinceBy 0 is antisym-

metric with respect tox,the vortices have the assumed antisymmetry. Furthermore,

becauseJ 1 zis symmetric with respect toxand localized in the vicinity ofx=0the

vortices are localized to the vicinity ofx=0.

2. Outer (ideal) region: Here it is assumed thatΩ 1 ≃ 0 ,so Eq.(12.28) becomes

dAz 0

dx

∂

∂y

(

∇^2 ⊥Az 1

)

−

∂Az 1

∂y

d^3 Az 0

dx^3

=0 (12.31)

which is a specification forAz 1 in the outer region for a givenAz 0 .Thus, it is effec-

tively assumed that the outer perturbed field is force-free, i.e.,(J×B) 1 =0so that

no vorticity is generated in the outer region.

The perturbed quantities will now be assumed to have the space-time dependence

Az 1 =Az 1 (x)eiky+γt

Ω 1 =Ω 1 (x)eiky+γt

(12.32)

so that Eq.(12.30) gives the inner region vorticity as

Ω 1 =

1

μ 0 γρ 0

dAz 0

dx

ik

(

∇^2 ⊥Az 1

)

=−

1

γρ 0

dAz 0

dx

ikJz 1. (12.33)

This satisfies all the geometric conditions noted earlier, namely the antisymmetric depen-
dence onx, the localization nearx=0and, consistent with Fig.12.2(c), a periodicity iny
that is 90^0 out of phase with the periodicity ofJ 1 z.
Using Eq. (12.23), it is seen that

U 1 x=

∂f 1

∂y

=ikf 1. (12.34)