ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence September 26, 2016, Axel Brandenburg

Magnetorotational Instability for inclined fields

1 Governing equations

In the presence of rotation and shear, the MHD equations for the departure from the shear flow u, the magnetic field B, and the density ρ becomes

\frac{∂u}{∂t} + Sx\frac{∂u}{∂y} + u_{x}Sŷ + u · ∇u + 2Ω × u = −ρ^{−1}∇P + ρ^{−1}J × B,  (1)

\frac{∂B}{∂t} + Sx\frac{∂B}{∂y} + u · ∇B = B_{x}Sŷ + B · ∇u − B∇ · u,  (2)
\frac{∂ρ}{∂t} + Sx\frac{∂ρ}{∂y} + u · ∇ρ = −ρ∇ · u.  (3)

Let us here, for simplicity, consider an isothermal equation of state, i.e., P = ρc_{s}^{2}, where c_{s} = const. The equations can be readily linearized about u = 0, B = B_{0} = const, and ρ = ρ_{0} = const. For the following, we assume B_{0} = (0, B_{0y}, B_{0z}) and ∇ = (0, 0, ∂_{z}). We assume that all perturbations are proportional to e^{σt+ikz}. Thus we have

( \begin{matrix}σ & −2Ω & 0 & 0 & −ik \frac{B_{0z}}{μ_{0}ρ_{0}} & 0 \\ S + 2Ω & σ & 0 & 0 & 0 & −ik \frac{B_{0z}}{μ_{0}ρ_{0}} \\ 0 & 0 & σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & 0 & ik & σ & 0 & 0 \\ −ikB_{0z} & 0 & 0 & 0 & σ & 0 \\ 0 & −ikB_{0z} & ikB_{0y} & 0 & −S & σ \end{matrix} ) ( û_{x1} ; û_{y1} ; û_{z1} ; ρ̂_{1}/ρ_{0} ; B̂_{x1} ; B̂_{y1} ) = 0.  (4)

We can split the determinant into 3 parts, T_{1} + T_{2} + T_{3} = 0 with

T_{1} = σ det ( \begin{matrix}σ & 0 & 0 & 0 & −ik \frac{B_{0z}}{μ_{0}ρ_{0}} \\ 0 & σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & ik & σ & 0 & 0 \\ 0 & 0 & 0 & σ & 0 \\ −ikB_{0z} & ikB_{0y} & 0 & −S & σ \end{matrix} )  (5)

T_{2} = 2Ω det ( \begin{matrix}S + 2Ω & 0 & 0 & 0 & −ik \frac{B_{0z}}{μ_{0}ρ_{0}} \\ & σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & ik & σ & 0 & 0 \\ −ikB_{0z} & 0 & 0 & σ & 0 \\ 0 & ikB_{0y} & 0 & −S & σ \end{matrix} )  (6)

T_{3} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} det ( \begin{matrix}S + 2Ω & σ & 0 & 0 & −ik \frac{B_{0z}}{μ_{0}ρ_{0}} \\ 0 & 0 & σ & ikc_{s}^{2} & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & 0 & ik & σ & 0 \\ −ikB_{0z} & 0 & 0 & 0 & 0 \\ 0 & −ikB_{0z} & ikB_{0y} & 0 & σ \end{matrix} )  (7)

The first part becomes

T_{1} = σ^{2} det ( \begin{matrix}σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ ik & σ & 0 & 0 \\ 0 & 0 & σ & 0 \\ ikB_{0y} & 0 & −S & σ \end{matrix} ) − ik \frac{B_{0z}}{μ_{0}ρ_{0}} σ det ( \begin{matrix}0 & σ & ikc_{s}^{2} & 0 \\ 0 & ik & σ & 0 \\ 0 & 0 & 0 & σ \\ −ikB_{0z} & ikB_{0y} & 0 & −S \end{matrix} ) (8)

The second part becomes

T_{2} = 2Ω(S+2Ω) det ( \begin{matrix}σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ ik & σ & 0 & 0 \\ 0 & 0 & σ & 0 \\ ikB_{0y} & 0 & −S & σ \end{matrix} )−ik \frac{B_{0z}}{μ_{0}ρ_{0}} 2Ω det ( \begin{matrix}0 & σ & ikc_{s}^{2} & 0 \\ 0 & ik & σ & 0 \\ −ikB_{0z} & 0 & 0 & σ \\ 0 & ikB_{0y} & 0 & −S \end{matrix} ) (9)

and the third part is in turn split into 3 parts, T_{3} = T_{3a} + T_{3b} + T_{3c} with

T_{3a} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} (S + 2Ω) det ( \begin{matrix}0 & σ & ikc_{s}^{2} & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & ik & σ & 0 \\ 0 & 0 & 0 & 0 \\ −ikB_{0z} & ikB_{0y} & 0 & σ \end{matrix} )  (10)

T_{3b} = σik \frac{B_{0z}}{μ_{0}ρ_{0}} det ( \begin{matrix}0 & σ & ikc_{s}^{2} & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ 0 & ik & σ & 0 \\ −ikB_{0z} & 0 & 0 & 0 \\ 0 & ikB_{0y} & 0 & σ \end{matrix} )  (11)

T_{3c} = − (k \frac{B_{0z}}{μ_{0}ρ_{0}} )^{2} det ( \begin{matrix}0 & 0 & σ & ikc_{s}^{2} \\ 0 & 0 & ik & σ \\ −ikB_{0z} & 0 & 0 & 0 \\ 0 & −ikB_{0z} & ikB_{0y} & 0 \end{matrix} )  (12)

1.1 T_{1}

Let us start with T_{1} = T_{1a} + T_{1b} so

T_{1a} = σ^{2} det ( \begin{matrix}σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ ik & σ & 0 & 0 \\ 0 & 0 & σ & 0 \\ ikB_{0y} & 0 & −S & σ \end{matrix} )  (13)

T_{1a} = σ^{3} det ( \begin{matrix}σ & 0 & 0 \\ 0 & σ & 0 \\ 0 & −S & σ \end{matrix} ) − ikc_{s}^{2}σ^{2} det ( \begin{matrix}ik & 0 & 0 \\ 0 & σ & 0 \\ ikB_{0y} & −S & σ \end{matrix} ) − ik \frac{B_{0y}}{μ_{0}ρ_{0}} σ^{2} det ( \begin{matrix}ik & σ & 0 \\ 0 & 0 & σ \\ ikB_{0y} & 0 & −S \end{matrix} ) (14)

so

T_{1a} = σ^{6} + k^{2}c_{s}^{2}σ^{4} + k^{2} \frac{B_{0y}^{2}}{μ_{0}ρ_{0}} σ^{4} = σ^{6} + (ω_{c}^{2} + ω_{Ay}^{2})σ^{4}  (15)

and for T_{1b} we have

T_{1b} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} σ det ( \begin{matrix}0 & σ & ikc_{s}^{2} & 0 \\ 0 & ik & σ & 0 \\ 0 & 0 & 0 & σ \\ −ikB_{0z} & ikB_{0y} & 0 & −S \end{matrix} )  (16)

with

T_{1b} = ik \frac{B_{0z}}{μ_{0}ρ_{0}} σ^{2} det ( \begin{matrix}0 & σ & 0 \\ 0 & 0 & σ \\ −ikB_{0z} & 0 & −S \end{matrix} ) + k^{2} \frac{B_{0z}}{μ_{0}ρ_{0}} σc_{s}^{2} det ( \begin{matrix}0 & ik & 0 \\ 0 & 0 & σ \\ −ikB_{0z} & ikB_{0y} & −S \end{matrix} )  (17)

or

T_{1b} = k^{2} \frac{B_{0z}^{2}}{μ_{0}ρ_{0}} σ^{4} + k^{4} \frac{B_{0z}^{2}}{μ_{0}ρ_{0}} σ^{2}c_{s}^{2} = ω_{Az}^{2}σ^{2}(σ^{2} + ω_{c}^{2})  (18)

So in summary

T_{1} = σ^{6} + ω_{s}^{2}σ^{4} + ω_{Ay}^{2}σ^{4} + ω_{Az}^{2}σ^{4} + ω_{Az}^{2}σ^{2}ω_{s}^{2}  (19)

1.2 T_{2}

Next, we work with T_{2} = T_{2a} + T_{2b} so

T_{2a} = 2Ω(S + 2Ω) det ( \begin{matrix}σ & ikc_{s}^{2} & 0 & ik \frac{B_{0y}}{μ_{0}ρ_{0}} \\ ik & σ & 0 & 0 \\ 0 & 0 & σ & 0 \\ ikB_{0y} & 0 & −S & σ \end{matrix} )  (20)

so

T_{2a} = 2Ω(S+2Ω) [σ det ( \begin{matrix}σ & 0 & 0 \\ 0 & σ & 0 \\ 0 & −S & σ \end{matrix} ) − ikc_{s}^{2} det ( \begin{matrix}ik & 0 & 0 \\ 0 & σ & 0 \\ ikB_{0y} & −S & σ \end{matrix} ) − ik \frac{B_{0y}}{μ_{0}ρ_{0}} det ( \begin{matrix}ik & σ & 0 \\ 0 & 0 & σ \\ ikB_{0y} & 0 & −S \end{matrix} )]

(21)

and thus

T_{2a} = 2Ω(S + 2Ω) (σ^{4} + k^{2}c_{s}^{2}σ^{2} + k^{2} \frac{B_{0y}^{2}}{μ_{0}ρ_{0}} σ^{2})  (22)

while

T_{2b} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} 2Ω det ( \begin{matrix}0 & σ & ikc_{s}^{2} & 0 \\ 0 & ik & σ & 0 \\ −ikB_{0z} & 0 & 0 & σ \\ 0 & ikB_{0y} & 0 & −S \end{matrix} )  (23)

so

T_{2b} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} 2Ω [−σ det ( \begin{matrix}0 & σ & 0 \\ −ikB_{0z} & 0 & σ \\ 0 & 0 & −S \end{matrix} ) + ikc_{s}^{2} det ( \begin{matrix}0 & ik & 0 \\ −ikB_{0z} & 0 & σ \\ 0 & ikB_{0y} & −S \end{matrix} )]  (24)

and thus

T_{2b} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} 2Ω (+σ^{2}SikB_{0z} + k^{2}c_{s}^{2}S ikB_{0z})  (25)

or

T_{2b} = k^{2} \frac{B_{0z}^{2}}{μ_{0}ρ_{0}} 2ΩS (σ^{2} + k^{2}c_{s}^{2})  (26)

So in summary.

T_{2} = 2Ω(S + 2Ω) (σ^{4} + ω_{s}^{2}σ^{2} + ω_{Ay}^{2}σ^{2}) + 2ΩSω_{Az}^{2} (σ^{2} + ω_{s}^{2})  (27)

1.3 T_{3}

And finally, we have

T_{3a} = −ik \frac{B_{0z}}{μ_{0}ρ_{0}} (S + 2Ω) [−σ det ( \begin{matrix}0 & σ & 0 \\ 0 & 0 & 0 \\ −ikB_{0z} & 0 & σ \end{matrix} )  (28)
+ikc_{s}^{2} det ( \begin{matrix}0 & ik & 0 \\ 0 & 0 & 0 \\ −ikB_{0z} & ikB_{0y} & σ \end{matrix} ) − ik \frac{B_{0y}}{μ_{0}ρ_{0}} det ( \begin{matrix}0 & ik & σ \\ 0 & 0 & 0 \\ −ikB_{0z} & ikB_{0y} & 0 \end{matrix} )]  (29)

but there are only zeros in the second row of all matrices, so

T_{3a} = 0.  (30)

Next,

T_{3b} = σik \frac{B_{0z}}{μ_{0}ρ_{0}} [−σ det ( \begin{matrix}0 & σ & 0 \\ −ikB_{0z} & 0 & 0 \\ 0 & 0 & σ \end{matrix} )  (31)
+ikc_{s}^{2} det ( \begin{matrix}0 & ik & 0 \\ −ikB_{0z} & 0 & 0 \\ 0 & ikB_{0y} & σ \end{matrix} ) − ik \frac{B_{0y}}{μ_{0}ρ_{0}} det ( \begin{matrix}0 & ik & σ \\ −ikB_{0z} & 0 & 0 \\ 0 & ikB_{0y} & 0 \end{matrix} )] ,  (32)

so

T_{3b} = σik \frac{B_{0z}}{μ_{0}ρ_{0}} [−σ^{3}ikB_{0z} − ikc_{s}^{2}σk^{2}B_{0z} − ik \frac{B_{0y}}{μ_{0}ρ_{0}} σk^{2}B_{0z}B_{0y}]  (33)

or

T_{3b} = σ^{2}k^{2} \frac{B_{0z}^{2}}{μ_{0}ρ_{0}} [σ^{2} + c_{s}^{2}k^{2} + \frac{B_{0y}^{2}}{μ_{0}ρ_{0}} k^{2}]  (34)

And then

T_{3c} = − (k \frac{B_{0z}}{μ_{0}ρ_{0}} )^{2} [σ det ( \begin{matrix}0 & 0 & σ \\ −ikB_{0z} & 0 & 0 \\ 0 & −ikB_{0z} & 0 \end{matrix} ) − ikc_{s}^{2} det ( \begin{matrix}0 & 0 & ik \\ −ikB_{0z} & 0 & 0 \\ 0 & −ikB_{0z} & ikB_{0y} \end{matrix} )] (35)

or

T_{3c} = − (k \frac{B_{0z}}{μ_{0}ρ_{0}} )^{2} [−σ^{2}k^{2}B_{0z}^{2} − k^{2}c_{s}^{2}k^{2}B_{0z}^{2}]  (36)

or

T_{3c} = (k^{2} \frac{B_{0z}^{2}}{μ_{0}ρ_{0}} )^{2} [σ^{2} + k^{2}c_{s}^{2}]  (37)

So in summary.

T_{3} = σ^{2}ω_{Az}^{2}(σ^{2} + ω_{s}^{2} + ω_{Ay}^{2}) + ω_{Az}^{4}(σ^{2} + ω_{s}^{2})  (38)

1.4 T_{1}+T_{2}+T_{3}

T_{1} = σ^{6} + ω_{s}^{2}σ^{4} + ω_{Ay}^{2}σ^{4} + ω_{Az}^{2}σ^{4} + ω_{Az}^{2}σ^{2}ω_{s}^{2}  (39)
T_{2} = 2Ω(S + 2Ω) (σ^{4} + ω_{s}^{2}σ^{2} + ω_{Ay}^{2}σ^{2}) + 2ΩSω_{Az}^{2} (σ^{2} + ω_{s}^{2})  (40)
T_{3} = σ^{2}ω_{Az}^{2}(σ^{2} + ω_{s}^{2} + ω_{Ay}^{2}) + ω_{Az}^{4}(σ^{2} + ω_{s}^{2})  (41)

So

T = σ^{6} + σ^{4}[ω_{s}^{2} + ω_{Ay}^{2} + 2ω_{Az}^{2} + 2Ω(S + 2Ω)]  (42)
+σ^{2} [2Ω(S + 2Ω)(ω_{s}^{2} + ω_{Ay}^{2}) + ω_{Az}^{2}(2ω_{s}^{2} + ω_{Ay}^{2} + ω_{Az}^{2} + 2ΩS)]  (43)
+ω_{s}^{2}ω_{Az}^{2}(ω_{Az}^{2} + 2ΩS)  (44)

References

Balbus, S. A., & Hawley, J. F., “A powerful local shear instability in weakly magnetized disks. I. Linear analysis,” Astrophys. J. 376, 214-222 (1991).

Velikhov, E. P., “Stability of an ideally conducting liquid flowing between cylinders rotating in a magnetic field,” Sov. Phys. JETP 36, 1398-1404 (1959). (Vol. 9, p. 995 in English translation)

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