Question 4.1.5: Consider a linear force-free magnetic field with the poloida......

a) From equation (4.14), the magnetic field components of this configuration are equal to:

B_{x} = \frac{∂Ψ}{∂y} = −B_{0} \frac{κ}{k} \cos(kx) exp(−κy),

B_{y} = −\frac{∂Ψ}{∂x} = −B_{0} \sin(kx) \exp(−κy),        (4.15)

B_{z} = αΨ = B_{0} \frac{α}{k} \cos(kx) \exp(−κy)

A straightforward calculation yields the following magnetic energy (per unit length along the z-axis),

W_{M} = \int_{−π/2k}^{π/2k} dx \int_{0}^{∞} dy \frac{B^{2}}{8π} = \frac{B^{2}_{0}}{16k^{2}(1 − α^{2}/k^{2})^{1/2}}        (4.16)

It can be seen from equation (4.16) that the magnetic energy is minimal for α = 0, which corresponds to a potential magnetic field with B_{z} = 0 (shown in Figure 4.2). Note that magnetic flux distribution at the base of the arcade, y = 0, does not depend on α. Therefore, the magnetic energy variation with the force-free parameter α is in full agreement with the general consideration for the excess magnetic energy given in Problem 4.1.3. Hence, for this configuration the excess energy is equal to

\Delta W_{M} = W_{M}(α) −W_{M}(0) = \frac{B^{2}_{0}}{16k^{2}} \left( \frac{1}{\sqrt{1 − α^{2}/k^{2}}} − 1 \right)

b) Each magnetic field line of the arcade, defined by the equation Ψ(x, y) = const, can be more conveniently labeled by x_{0}: the x-coordinate of its footpoints at y = 0 (see Figure 4.3) (they relate to each other by Ψ= \frac{B_{0}}{ k} \cos(kx_{0})). A non-potential field with a non-zero α has these footpoints being shifted in z-direction on ±z(x_{0}). The non-potential arcade as a whole bulges upwards since the characteristic height, κ^{−1} = k^{−1}(1−α^{2}/k^{2})^{−1/2}, increases with α. Quantatively, the shearing of magnetic  field lines in z-direction can be derived as

2z(x_{0}) = \int_{−x_{0}}^{+x_{0}} dz = \int_{−x_{0}}^{+x_{0}} \frac{B_{z}}{B_{x}} dx

with the integral being carried out along the field line Ψ = const = (x_{0}). By making use of expressions for B_{z} and B_{x} in equation (4.15), one gets

z(x_{0}) = − \frac{α}{κ} x_{0} = − \frac{α}{k(1 − α^{2}/k^{2})^{1/2}} x_{0}        (4.17)

Note that this shift is negative for the right-hand side footpoint and positive for the left-hand side one, as shown in Figure 4.3.

Assume now that this shearing is provided by some external force that slowly moves magnetic footpoints, such that their z-coordinates vary with time. Then, the magnetic arcade will evolve through a sequence of the force-free equilibria (4.14), with the parameter α varying with time in accordance with the relation (4.17). Thus, we have

v_{z}(x_{0}) = \frac{dz(x_{0})}{dt} = − \frac{x_{0}}{k} \frac{d}{dt} [ \frac{α}{(1 − α^{2}/k^{2})^{1/2}} ] = − \frac{dα/dt}{k(1 − α^{2}/k^{2})^{3/2}} x_{0}        (4.18)

These flows at the base of the arcade lead to Poynting flux of energy \vec{P} into the arcade, with the relevant component P_{y} equal to P_{y}|_{y=0} = \frac{c}{4π} (E_{z}B_{x} − E_{x}B_{z})_{y=0}. In a perfectly conducting fluid \vec{E} = − \frac{1}{c} (\vec{v} × \vec{B}).

Thus, for the z-velocity of equation (4.18), E_{z}|_{y=0} = 0, E_{x}|_{y =0} = \frac{1}{c}(v_{z}B_{y})|_{y=0}, and hence

P_{y}|_{y=0} = − \frac{v_{z}}{4π} (B_{y}B_{z})|_{y=0} = \frac{B^{2}_{0}}{4πk^{2}} \frac{αdα/dt}{(1 − α^{2}/k^{2})^{3/2}} x_{0} \sin(kx_{0}) \cos(kx_{0})

This yields the total power supply into the arcade

\frac{dW}{dt} = \int_{−π/2k}^{π/2k} P_{y}|_{y=0}dx_{0} = \frac{B^{2}_{0}}{16k^{4}} \frac{αdα/dt}{(1 − α^{2}/k^{2})^{3/2}},

which, according to equation (4.16), is equal to the rate of change of its  magnetic energy.