Flux Rope Coupling

Our main objective is to calculate the global equilibrium, in which the flux rope is treated as an infinitessimally thin line currentI(t). However, in order to find the functional formI(t)we must consider the internal structure of the flux rope itself. The current will then be determined by coupling the amount of magnetic flux in the rope with that cancelling at the photospheric PIL below the rope.

Flux Rope Model

In the interior of the flux rope, whereBz andjz may be non-zero, a magnetic field of the form

(6.26) in force-free equilibrium,j×B = 0, satisfies the Grad-Shafranov equation. We use the well-known solution for a uniformly twisted and cylindrically symmetric flux tube given byGold and Hoyle(1960). In cylindrical coordinates(r, θ, z), with origin at(x, y) = (0, h)andz-axis aligned with ourz-axis, this solution may be written (Forbes and Priest,1995, Appendix B) in the form Bz = R₀2Bi R2 0+n2r2 , (6.41) Bθ = nR0rBi R2₀+n2_r2. (6.42)

HereBi is the peak axial fieldBz at the rope centre, and the twist parameternis the number of

turns per lengthR0 along the rope. We setR0to be the radius of the flux rope, which is assumed

small. This solution is shown in Figure6.5for several values ofn. Notice that increasing the twist causes lessBzflux and moreBθflux inside the tube (r/R0 <1). Also, fornlarger than about 1,

6.3 Model with Distributed Sources 137

Figure 6.5: TheBz(a) andBθ (b) components of the Gold-Hoyle solution as a function ofr/R0,

for different values of the twist parameternas given in the legend. HereBi= 1.

Flux Rope Current

Note that our external solution derived in Section6.3.3required us to strictly setBz = 0forr >

R0, so that there will be a discontinuity inBzat the surface of the tube. This discontinuity causes

a surface current (Forbes and Priest,1995), which we need to take into account in determining the total current. There will also be a corresponding discontinuity inBθ, so that inside the tube

Bθ =

nR0rBi

R₀2+n2_r2, Bz =

R2₀Bi

R2₀+n2_r2 forr < R0, (6.43)

and outside the tube

Bθ=

nR0rBo

R2

0+n2r2

, Bz = 0 forr > R0. (6.44)

The total current will then be given by

I = Z A j·dA= 1 µ I r=R+₀ B·dl= 2πnR0Bo µ(1 +n2₎. (6.45)

To determineBoin terms ofBiwe impose magnetic pressure balance at the tube surfacer=R0,

namely

B_θ2(R0+) → Bθ2(R0−) +Bz2(R0−) as→0. (6.46)

Expanding to leading order ingives

B_θ2(R0+) =

n2B_o2

6.3 Model with Distributed Sources 138 B_θ2(R0−) = n2B_i2 (1 +n2₎2 +O(), (6.48) B_z2(R0−) = B_i2 (1 +n2₎2 +O(). (6.49)

Substituting into equation (6.46) gives

Bo=

√ 1 +n2

n Bi (6.50)

so that the total current is

I = 2πR0Bi

µ√1 +n2. (6.51)

We need to know howI depends on the axial magnetic fluxΦzin the tube. This flux is given by

Φz = 2π Z R0 0 dr rBz(r) = πR2₀ln(1 +n2)Bi n2 (6.52)

and hence substituting forBifrom (6.51) gives

I(t) = 2n

2

µR0

√

1 +n2_{ln(1 +n}2₎Φz(t). (6.53)

This is the important relation which determines the currentI(t)in the flux rope, as a function of its axial flux. Notice that the current also depends inversely on the flux rope radiusR0, and increases

with increased twistn, as expected.

Axial Magnetic Flux

The axial magnetic fluxΦz(t)will be coupled to the flux cancelling in the photosphere by setting

Φz(t) = Φ0+δΦc(t), (6.54)

whereΦ0is the initial magnetic flux in the rope att= 0,Φc(t)is the amount of magnetic flux that

has cancelled at the PIL, andδis a constant of proportionality. The choice ofδis arbitrary in our model, but should be between 0 and 1. In thevan Ballegooijen and Martens(1989) scenario for creation of flux ropes by cancellation (Section5.2), a sensible choice forδ is taken to be around 0.5, since the flux from two reconnecting loops is divided between two new field lines: one along the PIL which corresponds to our flux rope, and one transverse to the PIL which submerges. The amount of magnetic fluxΦc(t) cancelled at the underlying photospheric PIL is fixed by our

6.3 Model with Distributed Sources 139 thezdirection is Z ∞ x0 By(x, t)dx=Az(0, t)−Az(∞, t) = B0e1/2ρ20 q(t) , (6.55)

so the required cancelled flux is

Φc(t) =B0e1/2ρ20 1 ρ0 − 1 q(t) lz, (6.56)

since cancellation can occur only at the central PIL. Herelz is the length of the flux rope in the

z-direction where cancellation occurs. This is another undetermined parameter in our model. It should really be a function of time as differential rotation tends to lengthen PILs, although it will be assumed constant for simplicity.