374 Chapter 12. Magnetic reconnection
Continuing this discussion of the ramifications of the antisymmetry ofk·B 0 about
x=0,we now define an artificial reference magnetic fieldB ̄that is parallel to the real
field atx=0,but has no shear (i.e. has noxdependence). The reference field therefore
has the form
B ̄=By 0 (0)ˆy+Bz 0 zˆ=B 0 (0)ˆζforallvaluesofx. (12.70)
We now defineb 0 as the difference between the real field andB ̄so that
B 0 (x)=B ̄+b 0 (x) (12.71)andb 0 (x)has the samex-dependence as theBy 0 field used for the sheet current instability
in the previous section in that (i) it is antisymmetric aboutxand (ii) is in theydirection.
One way of thinking about this is to realize thatb 0 (x)is the component ofB 0 (x)which is
antisymmetric aboutx=0.With this definition
k·B 0 =k·b 0 (x)=kyb 0 (12.72)and so Eq.(12.69) becomes
γρΩ 1 ζ=ikyb 0 J 1 ζ. (12.73)
This equation provides the essence of the dynamics. It shows how the componentof the
magnetic field that is antisymmetric aboutx=0createsfluid vortices that are antisymmet-
ric with respect tox.In particular, sinceb 0 (x)is an odd function ofxandJ 1 ζis an even
function ofx,Ω 1 ζis an odd function ofx.
The complete self-consistent description is obtained by in addition taking into account
the induction equation which shows howfluid motion acts to create perturbations of the
electromagnetic field.
Since we are working with vector potentials, Ohm’s law provides the induction equa-
tion. In the sheet current case the component of Ohm’s law in the direction ofsymmetry
of the perturbation was considered, i.e., the component in thezdirection. Here, the cor-
responding symmetry direction for the perturbation is theζdirection and so the relevant
component of Ohm’s law is theζcomponent,
ˆζ·[E 1 +U 1 ×B 0 (x)]=ˆζ·ηJ 1 (12.74)which becomes
−∂A 1 ζ
∂t+
(
yˆ×ˆζ·U 1)
b 0 (x)=ηJ 1 ζ. (12.75)Since the vorticity vector lies alongˆζthe incompressible velocity must be orthogonal toˆζ
and so has the form
U 1 =∇f 1 ׈ζ. (12.76)
Thus
ˆy׈ζ·U 1 =
(
ˆy׈ζ)
·
(
∇f 1 ׈ζ)
=ikyf 1. (12.77)Again, taking into account the fact that the vorticity points in theζdirection it is seen that
Ω 1 =Ω 1 ζˆζwhere
Ω 1 ζ=ˆζ·∇×U 1 =−∇^2 ⊥f 1 (12.78)
and now⊥means perpendicular toˆζ.
Substituting forΩ 1 ζin Eq.(12.73) using Eq. (12.78) gives
∇^2 ⊥f 1 =−ikyb 0 (x)J 1 ζ
γρ 0