Question 4.3.6: Investigate the condition for the current sheet formation at......

Consider a planar magnetic field invariant along the z-coordinate:

\vec{B} = (\vec{∇}ψ(x, y) ×\vec{e}_{z})

If this field is potential with a neutral X-point at the origin of the polar coordinate system (r, θ) as shown in Figure 4.13, its respective flux function is

ψ_{0}(r, θ) = \frac{B_{0}r^{2}}{2R} \cos 2θ,        (4.77)

where R is a radius of the circular domain under consideration. Assume now that this field, while being embedded into a perfectly conducting fluid (ideal MHD), becomes deformed by some displacement of its field line footpoints at the boundary surface r = R, which can be described by the relation θ_{1}(θ), where θ and θ_{1} are, respectively, the initial and the displaced azimuthal coordinates of the footpoints. Since the magnetic field is frozen-in, each field line preserves its flux function magnitude; thus, the latter remains unchanged for the magnetic footpoints at the boundary:

ψ[R, θ_{1}(θ)] = ψ_{0}(R, θ)        (4.78)

(for instance, ψ(R, θ_{s}) = 0, where θ = θ_{s} are locations of the separatrix footpoints S_{1,2,3,4} shown in Figure 4.13(a)). Our interest is in the resulting deformed magnetostatic equilibrium, which in the case of zero-β plasma should be a force-free one.

For a planar magnetic field with B_{z} = 0, the flux function of the deformed force-free equilibrium is, according to the Grad-Shafranov equation (4.10), a solution of the Laplace equation

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

∇^{2}ψ(r, θ) = 0,        (4.79)

with the boundary condition specified by Equation (4.78) (which means that the deformed magnetic field remains a potential one). From the mathematical viewpoint this is a well-known Dirichlet problem (see, e.g., R. Courant and D. Hilbert, Methods of Mathematical Physics, v.2, Ch.4, Interscience, 1989) that always has a regular solution. However, it does not mean that such a solution can be realized under the ideal MHD frozen-in constraint on the magnetic field evolution. The reason lies in the particular structure of the initial magnetic field (4.77), where separatrix lines OS_{i} (i=1,2,3,4)  the whole domain into four quadrants, with the magnetic flux within each of them equal to \Delta ψ_{0} = ψ_{0}(R, 0) − ψ_{s} = B_{0}R/2 (note that ψ_{s} = ψ_{0}(R, π/4) = 0).

Therefore, the appropriate solution of equation (4.79) must not only satisfy the boundary condition (4.78), but it should also preserve the magnetic footpoints’ connectivity inside each of the quadrants and, hence, the given above amount of the magnetic flux \Delta ψ.

Consider, for example, the azimuthal deformation

θ_{1}(θ) = θ + \frac{δ_{0}}{R} \cos θ,

with a small displacement amplitude δ_{0} ≪ R. According to relation (4.78), it yields the following re-distribution of the magnetic flux at the boundary r = R:

ψ^{(R)}(θ) = ψ^{(R)}_{0} (θ − δθ) ≈ ψ^{(R)}_{0} (θ) − \frac{dψ^{(R)}_{0}}{dθ} δθ =

ψ^{(R)}_{0} (θ) + B_{0}δ_{0} \sin 2θ \cos θ =

ψ^{(R)}_{0} (θ) + \frac{B_{0}δ_{0}}{2} \sin θ + \frac{B_{0}δ_{0}}{2} \sin 3θ        (4.80)

As seen from expression (4.80), such a perturbation results in appearance of a dipole (m = 1) and m = 3 azimuthal components in the boundary flux distribution. As far as the latter is concerned, a regular solution of the Laplace equation (4.79), which accounts for the respective contribution in (4.80), reads

ψ_{3}(r, θ) = \frac{B_{0}δ_{0}}{2R^{3}} r^{3} \sin 3θ        (4.81)

By adding perturbation (4.81) to the initial flux function ψ_{0} given by equation (4.77), one concludes that ψ_{0} dominates in the vicinity of the origin r = 0.

Therefore, the neutral X-point is not moved from the center, and the separatrix lines, that originate from the X-point, correspond to ψ = ψ_{s} = 0.

Thus, they end up at the displaced footpoints (S^{′}_{1} , S^{′}_{2} , S^{′}_{3} , S^{′}_{4} ) of the initial separatrix as shown in Figure 4.13(b). Therefore, the deformed field has the same field line connectivity as the initial one of Figure 4.13(a), since the amount of magnetic flux inside each quadrant remains unchanged and equal to \Delta ψ_{0} = B_{0}R/2. It also follows from this consideration that the same is the case for any harmonic of the boundary flux perturbation with the azimuthal number m ≥ 2.

All is different for the dipole perturbation with ψ^{(R)}_{1} (θ) = \frac{B_{0}δ_{0}}{2} \sin θ. Indeed, the respective regular solution inside the circle r = R is

ψ_{1}(r, θ) = \frac{B_{0}δ_{0}}{2R} r \sin θ = \frac{B_{0}δ_{0}}{2R} y,

which corresponds to a uniform magnetic field B_{1x} = ∂ψ_{1}/∂y = B_{0}δ_{0}/2R.

Being combined with the initial magnetic field (4.77), it moves the neutral X-point up from the center to the location (x_{1} = 0, y_{1} = δ_{0}/2). Moreover, this displaced X-point acquires now a magnitude of ψ that is different from the initial one (which is equal to zero). Indeed,

ψ(x_{1}, y_{1}) = ψ_{0}(x_{1}, y_{1}) + ψ_{1}(x_{1}, y_{1}) = − \frac{B_{0}δ^{2}_{0}}{8R} + \frac{B_{0}δ^{2}_{0}}{4R} = \frac{B_{0}δ^{2}_{0}}{8R}          (4.82)

Since the separatrix originates from the X-point, its flux function magnitude ψ_{s} = ψ(x_{1}, y_{1}) ≠ 0; therefore, all separatrix footpoints at the boundary r = R cannot coincide with their displaced initial counterparts (S^{′}_{1} , S^{′}_{2} , S^{′}_{3} , S^{′}_{4} , where ψ^{(R)} = 0 (see Figure 4.13(c)). Consequently, the amount of magnetic flux confined now in each of the four quadrants becomes different from the initial value of \Delta ψ_{0} = B_{0}R/2. Indeed, as seen from Figure 4.13, it is increased in the upper and lower quadrants by \Delta ψ_{s} = B_{0}δ^{2}_{0}/8R, and, respectively, it is reduced by the same amount in the left-hand side and the right-hand side ones. Thus, the regular solution of Figure 4.13(c) is not compatible with the initial one of Figure 4.13(a). Hence, in the framework of ideal MHD such external perturbation leads to a singular deformed magnetic equilibrium with a current sheet (see Problem 4.3.7).