Question 4.3.1: A linear force-free magnetic field B^(0) = [0,B0 sin(αx),B0 ......

Since the system remains invariant with respect to z-coordinate, the deformed magnetic field can be represented in the form (4.10), with the initial field (4.38) corresponding to

\vec{B}(x, y) = [ \vec{∇} Ψ(x, y) ×\vec{e}_{z}] + B_{z}(x, y)\vec{e}_{z},        (4.10)

ψ_{0}(x) = \frac{B_{0}}{α} \cos(αx), B^{(0)}_{z} = F_{0}(ψ) = αψ_{0}

By using the perturbation approach, one can write for the deformed field:

ψ = ψ_{0}(x) + δψ(x, y), F(ψ) = F_{0}(ψ) + δF(ψ).        (4.40)

Since the deformed boundary surface (4.39) should remain a magnetic surface, the poloidal flux function ψ is constant there; thus its perturbation must be of the form δψ(x, y) = ψ_{1}(x) \cos ky. Furthermore, the new equilibrium is forcefree and, hence, by applying the Grad-Shafranov equation (4.12), one finds that in the linear approximation the functional dependence F(ψ) remains unchanged, i.e., δF(ψ) = 0 and F(ψ) = F_{0}(ψ) = αψ, so that ψ_{1}(x) satisfies the following equation:

\vec{∇}^{2} Ψ= − F \frac{dF}{dΨ}         (4.12)

ψ^{′′}_{1} + (α^{2} − k^{2})ψ_{1} = 0        (4.41)

As already mentioned, the boundary conditions for the proper solution of equation (4.41) come from the requirement that ψ(x, y) in (4.40) is constant at x = x^{(−)}_{b} = −l and x = x^{(+)}_{b} = l  + a \cos ky. The former condition implies that ψ_{1}(−l) = 0, while the latter yields ψ^{′}_{0} (l)a \cos ky + ψ_{1}(l) \cos ky = 0, i.e., ψ_{1}(l) = aB_{0} \sin(αl). The respective regular solution of equation (4.41) is

ψ^{(r)}_{1} (x) = \frac{aB_{0} \sin(αl)}{\sin(2κl)} \sin[κ(x + l)],        (4.42)

where κ^{2} = (α^{2} − k^{2}) is assumed to be positive, and αl < π/2 (the latter ensures MHD stability of the initial magnetic configuration (see Problem 4.3.5)).

At first glance, there is nothing special about this solution shown in Figure 4.6. However, it turns out that since ψ^{(r)}_{1} (0) ≠ 0, the respective deformed magnetic field acquires a topology which is different from that of the initial configuration of (4.38). Indeed, in the vicinity of x = 0 the poloidal flux function ψ(x, y) takes the form

ψ(x, y) ≈ \frac{B_{0}}{α}(1 − α^{2}x^{2}/2) + ψ^{(r)}_{1} (0) \cos(ky),

which indicates formation there of a periodic chain of identical magnetic islands as shown in Figure 4.7. The O-point with the coordinates x = 0, y = 0 at the center of the island corresponds to ψ = ψ^{(0)} = (B_{0}/α)+ψ^{(r)}_{1} (0), while the two X-points located at x = 0, y = ±π/k have ψ = ψ_{S} = (B_{0}/α) − ψ^{(r)}_{1} (0), the separatrix flux function. Therefore, the total amount of magnetic flux confined inside each island is equal to \Delta ψ = ψ^{(0)} − ψ_{S} = 2ψ^{(r)}_{1} (0). The width of islands is \Delta  = 2[ψ^{(r)}_{1}(0)/B_{0}α]^{1/2}.

Clearly, such a cofiguration cannot be formed in a perfectly conducting fluid with a frozen-in magnetic field. Therefore, another, ideal MHD solution, ψ^{(i)}_{1} , which preserves the magnetic field topology, should exist. It then follows from the consideration given above, that this solution must satisfy the additional requirement, namely that ψ^{(i)}_{1} (0) = 0, which makes it equal to zero at −l ≤ x ≤ 0, and

ψ^{(i)}_{1} (x) = \frac{aB_{0} \sin(αl)}{\sin(κl)} \sin(κx),      (4.43)

at 0 ≤ x ≤ l (see Figure 4.6). As seen from (4.43), this solution is singular: a current sheet with a discontinuity of B_{y} = −∂ψ/∂x is formed at x = 0. The latter is the so-called resonance surface, location of which, in a the perturbation with \vec{k} = (0, k_{y}, k_{z}), is defined by the condition \vec{k}  · \vec{B}^{(0)} = 0.

The poloidal magnetic field pattern of the ideal MHD solution is depicted in Figure 4.8.

It is shown below (see next Problem) that in a highly conducting medium magnetic reconnection acts as a mechanism of magnetic relaxation from the state of higher magnetic energy associated with ψ^{(i)}_{1} , to the lower-energy configuration of ψ^{(r)}_{1} .