10.2 MHD Rayleigh-Taylor instability 303onyand thus rotates as a function ofy.In this latter case, the magnetic field lines are said
to be sheared, since adjacenty-layers of field lines are not parallel to each other.
The magnetofluid is assumed incompressible and so using Eq.(9.10) the linearized
equation of motion becomes
ρ 0
∂v 1
∂t=−∇P ̄ 1 +
B 0 ·∇B 1 +B 1 ·∇B 0
μ 0−ρ 1 gyˆ (10.23)where
P ̄ 1 =P 1 +B^0 ·B^1
μ 0
is the perturbation of the combined hydrodynamic and magnetic pressure, i.e., the perturba-
tion ofP+B^2 / 2 μ 0 .It is again assumed that all quantities vary in the manner of Eq.(10.9)
so Eq.(10.23) has the respectiveyand⊥components
γρ 0 v 1 y=−∂P ̄ 1
∂y+
i(k·B 0 )B 1 y
μ 0
−ρ 1 g (10.24)γρ 0 v 1 ⊥=−ikP ̄ 1 +1
μ 0[
i(k·B 0 )B 1 ⊥+B 1 y∂B 0
∂y]
. (10.25)
In analogy to the glass of water problem, Eq.(10.25) is dotted withikand Eq.(10.10) is
invoked to obtain
−γρ 0∂v 1 y
∂y=k^2 P ̄ 1 +1
μ 0[
−(k·B 0 )k·B 1 ⊥+iB 1 y∂(k·B 0 )
∂y]
. (10.26)
Because∇·B 1 =0, the perturbed perpendicular field is
ik·B 1 ⊥=−∂B 1 y
∂y(10.27)
so that Eq.(10.26) can be recast as
k^2 P ̄ 1 =−γρ 0∂v 1 y
∂y−
1
μ 0[
−i(k·B 0 )∂B 1 y
∂y+iB 1 y∂(k·B 0 )
∂y]
. (10.28)
Following a procedure analogous to that used in the inverted glass of waterproblem,P ̄ 1 is
eliminated in Eq.(10.24) by substitution of Eq. (10.28) to obtain
γρ 0 v 1 y=−1
k^2∂
∂y{
−γρ 0∂v 1 y
∂y−
1
μ 0[
−(ik·B 0 )∂B 1 y
∂y
+B 1 y∂(ik·B 0 )
∂y]}
+
i(k·B 0 )B 1 y
μ 0−ρ 1 g.
(10.29)
To proceed further, it is necessary to knowB 1 y.The complete vectorB 1 is found by first
linearizing the MHD Ohm’s law to obtain
E 1 +v 1 ×B 0 =0, (10.30)then taking the curl, and finally using Faraday’s law to obtain
γB 1 =∇×[v 1 ×B 0 ]. (10.31)