366 Chapter 12. Magnetic reconnection
Equation (12.18) can be recast using Eqs.(12.15) and (12.19) as aninduction equation,−
∂A 1 z
∂t−U 1 x∂A 0 z
∂x=ηJ 1 z. (12.20)The current has the form
μ 0 Jz=ˆz·∇×B=ˆz·∇×(∇×Azˆz)=ˆz·∇×(∇Az×zˆ)=−∇^2 ⊥Az (12.21)where the subscript⊥means perpendicular toˆz.Thus, the induction equation becomes
∂A 1 z
∂t+U 1 x∂A 0 z
∂x=
η
μ 0∇^2 ⊥A 1 z. (12.22)To proceed, it is necessary to express the perturbed velocityU 1 xin terms ofA 1 z; this
relation is obtained from the equation of motion.
While we could just plow ahead and manipulate the equation of motion to obtainU 1 x
in terms ofA 1 z, it is more efficient to exploit the incompressibility relation. In two-
dimensional hydrodynamics, incompressibility simplifiesflow dynamics so thatflow is
described by two related scalars, the stream-functionfand the vorticityΩ. For two-
dimensional motion in thex−yplane of interest here, the general incompressible velocity
can be expressed as
U=∇f×zˆ (12.23)
since
∇·U=∇·(∇f׈z)=ˆz·∇×∇f=0. (12.24)
The vorticity is the curl of the velocity and because the velocity lies inthex−yplane, the
vorticity vector is in thezdirection. The vorticity magnitudeΩis given by
Ω=ˆz·∇×U=ˆz·∇×(∇f׈z)=∇·[(∇f×zˆ)׈z]=−∇^2 ⊥f (12.25)where⊥means perpendicular with respect toz. Given the vorticity,fcan be found by
solving the Poisson-like Eq.(12.25), and then knowingf, the velocity can be evaluated
using Eq.(12.23). Appropriate boundary conditions must be specified for bothfandΩ;
these boundary conditions are that the vorticity is antisymmetric inxand is only large in
the vicinity ofx=0as indicated in Fig.12.2(c).
The curl of the equation of motion provides the vorticity evolution and also annihilates
∇P;this elimination ofPfrom consideration is why the vorticity/stream-line method is a
more efficient approach rather than direct solution of the equation of motion.
Let us now solve forU 1 xfollowing this procedure. The linearized equation of motion
is
ρ 0∂U 1
∂t=(J×B) 1 −∇P1; (12.26)
taking the curl and dotting withˆzgives
ρ 0∂Ω 1
∂t= ˆz·∇×(J×B) 1
= ˆz·∇×[Jzˆz×(∇Az×zˆ)] 1
= ˆz·∇×(Jz∇Az) 1
= ˆz·(∇Jz×∇Az) 1. (12.27)