Question 4.2.5: A sheared force-free magnetic field B^(0) = (0,B0 sin(θ(x)),......

The external perturbation results in fluid motion, which can be described in terms of the displacement vector \vec{ξ}(\vec{r}, t) = \vec{ξ}(x, y) \exp(−iωt). For the weak perturbation under discussion, (a << l), the linearized ideal MHD equations apply, so that equation of motion takes the form

ρ\frac{∂^{2}\vec{ξ}}{∂t^{2}} = −ω^{2}ρ\vec{ξ} = \frac{1}{c} (\vec{j}^{(0)} ×\vec{b}) + \frac{1}{c}(\vec{j}^{(1)} × \vec{B}^{(0)}),      (4.29)

where \vec{b} is the magnetic field perturbation with the associated current \vec{j}^{(1)}, and

\vec{j}^{(0)} = \frac{c}{4π} \frac{dθ}{dx} \vec{B}^{(0)}

is the initial electric current of the sheared magnetic field (4.27). The magnetic induction equation yields

\vec{b} = \vec{∇} × (\vec{ξ} × \vec{B}^{(0)}),        (4.30)

which, after being inserted into (4.29), results in a single equation for the displacement vector \vec{ξ}. Since the initial magnetic field is force-free and, hence, \vec{j}^{(0)} is parallel to \vec{B} ^{(0)}, it follows from (4.29) that \vec{ξ} is perpendicular to \vec{B}^{(0)}.

Therefore, the displacement vector \vec{ξ} can be represented as

ξ_{x}(x, y) = ξ_{x}(x) \cos(ky) \exp(−iωt),

ξ_{⊥}(x, y) = ξ_{y} \cos θ − ξ_{z} \sin θ = ξ_{⊥}(x) \sin(ky) \exp(−iωt)

Then, the magnetic field perturbation of (4.30) takes the form:

b_{x} = −kB_{0}ξ_{x}(x) \sin(ky) \sin θ \exp(−iωt),

b_{y} = −B_{0} \cos(ky) \frac{d}{dx} [\sin θ ξ_{x}(x)] \exp(−iωt),          (4.31)

b_{z} = −B_{0} \cos(ky) \exp(−iωt) \left\{kξ_{⊥}(x) + \frac{d}{dx} [\cos θ ξ_{x}(x)]\right\}

Finally, the equation of motion (4.29) yields

ξ_{⊥}(x) = \frac{k \cos θ}{(ω^{2}/V^{2}_{A} − k^{2})} \frac{dξ_{x}}{dx},

where ξ_{x}(x) satisfies the following equation:

\frac{d}{dx} \left[δ(x) \frac{dξ_{x}}{dx} \right] + \left(\frac{ω^{2}}{V^{2}_{A}}− k^{2} \right) δ(x)ξ_{x} = 0,        (4.32)

where δ(x) ≡ ω^{2} − k^{2}V^{2}_{A} \sin^{2} θ(x). This equation acquires a singularity, if at some location x = x_{r} δ(x_{r}) = 0, i.e., ω^{2} = k^{2}V^{2}_{A} \sin^{2} θ(x_{r}) = (\vec{k} · \vec{V}_{A})^{2}. The latter equality is nothing but the dispersion relation for shear Alfven waves (see Problem 4.2.1). Thus, in this case the external perturbation resonates at x = x_{r} with the shear Alfven wave, causing the magnetic energy dissipation even in an apparently ideal fluid.

The resulting dissipation power can be derived in the following way (L. Chen and A. Hasegawa, Physics of Fluids, vol.17, 1399, 1974). In the vicinity of the resonant point the frequency shift δ(x) can be approximated as

δ(x) ≈ \left( \frac{dδ}{dx}\right)_{x_{r}} (x − x_{r}) = δ^{′}(x_{r})(x − x_{r}),

and, therefore, the displacement ξ_{x}(x) behaves at this point, according to equation (4.32), as

ξ_{x}(x) ≈ ξ_{r} \log(x − x_{r}),        (4.33)

where the amplitude ξ_{r} is determined by a global solution of (4.32) with the boundary condition specified by (4.28). Consider now the energy balance inside a thin layer enclosing the resonant point, i.e., x_{r} − \epsilon < x < x_{r} + \epsilon, and per unit area in the (y, z) plane:

\frac{dW_{A}}{dt} =< P_{x}(x_{r} − \epsilon) > − < P_{x}(x_{r} + \epsilon) >,      (4.34)

where < P_{x} > is an x-component of the Poynting vector \vec{P} = c(\vec{E} × \vec{B})/4π, averaged over its variation with time and y-coordinate. Since in ideal MHD \vec{E} = −(\vec{v} × \vec{B} /c) ≈ iω(\vec{ξ} × \vec{B}^{(0)})/c, a non-vanishing averaged contribution to < P_{x} > is due to the magnetic field perturbation \vec{b}. Hence, the respective part of P_{x} is P_{x} = c[ℜ(E_{y})ℜ(b_{z})−ℜ(E_{z})ℜ(b_{y})]/4π, which, by using (4.31), can be written as

Therefore, in the vicinity of the resonant point x_{r}, where δ(x) ≈ δ^{′}(x_{r})(x−x_{r}) and dξ_{x}/dx = ξ_{r}(x − x_{r})^{−1}, Equation (4.35) reduces to

P_{x} = \frac{ωB^{2}_{0}}{4π} \cos^{2} (ky)Re \left(iξ_{x} \exp(−iωt)  \frac{δ^{′}(x_{r})ξ_{r} \cos(ωt)}{(ω^{2} − k^{2}V^{2}_{A})} \right)

As seen from this expression, the power (4.34) supplied into the resonant area, if non-zero, should be due to a discontinuity of ξ_{x}(x) at x = x_{r}, i.e., \Delta ξ_{x} = ξ_{x}(x_{r} − \epsilon) − ξ_{x}(x_{r} + \epsilon), so that

\frac{dW_{A}}{dt} = \frac{ωB^{2}_{0}}{8π} \frac{δ^{′}(x_{r})ξ_{r}}{(ω^{2} − k^{2}V^{2}_{A})} ℜ[i\Delta ξ_{x} \exp(−iωt)) \cos(ωt)] >

In order to derive \Delta ξ_{x}, it is convenient to resolve the singularity in equation (4.32) by introducing an infinitesimally small positive imaginary part to the driving frequency, i.e., by replacing ω with ω + iγ, γ → 0, which in physical terms means that the external perturbation is adiabatically “switched-on” at t → −∞. It follows then from the expression (4.32) for δ(x) that such a procedure results in the resonant point being now displaced into the complex plane of variable x by \Delta x_{r} = −2iγω/δ^{′}(x_{r}). Thus, the resonant point becomes located below the real axis if δ^{′}(x_{r}) > 0, and vice versa. In other words, introduction of γ > 0 provides one with a rule of passing the resonant singularity (see also Problem 3.4.8): it should be passed above if δ^{′}(x_{r}) > 0, and below if δ^{′}(x_{r}) < 0. Thus, while passing from x = x_{r} − \epsilon to x = x_{r} + \epsilon the variable (x − x_{r}) acquires additional phase factor \exp(i \Delta \phi) with \Delta \phi = −π signδ^{′}(x_{r}), which, according to expression(4.33), yields \Delta ξ_{x} = iπξ_{r}sign[δ^{′}(x_{r})]. Therefore, the Alfven resonance power supply is equal to

\frac{dW_{A}}{dt} = \frac{ωB^{2}_{0}}{16 (k^{2}V^{2}_{A}− ω^{2})}  |δ^{′}(x_{r})|ξ^{2}

which, by using expression (4.32) for δ(x) and the resonant condition δ(x_{r}) = 0, takes the form

\frac{dW_{A}}{dt} = \frac{ω^2B^{2}_{0}}{8 (k^{2}V^{2}_{A}− ω^{2})^{1/2}} |\frac{dθ}{dx}| ξ^{2}_{r}        (4.36)

Then, always present finite dissipation effects such as fluid resistivity or viscosity result in this power being absorbed inside a narrow layer at the resonant surface x = x_{r}.

As an example, consider a linear force-free magnetic field for which θ(x) = αx, assuming, for simplicity, that the perturbation length-scale k^{−1} and the shear length α^{−1} are comparable to the size of the system, i.e., kl ∼ αl ∼ 1 (as  shown in Problem 4.3.5, such a field becomes MHD unstable when αl > π/2).

In this case there are two Alfven resonances located at x_{r} = ±l \sin^{−1}(ω/kV_{A}), which allow a broad interval of the driving frequency ω, ranging from ω_{max} ∼ kV_{A} ∼ V_{A}/l ∼ τ^{−1}_{A} , the inverse Alfven transit time, down to ω = 0. A general dimensional consideration yields that in this case the resonance amplitude ξ_{r} ∼ a. Thus, for the low driving frequencies, when ωτ_{A} ≪ 1, the dissipated power can be estimated from equation (4.36) as

\frac{dW_{A}}{dt} ∼ V_{A} \frac{B^{2}_{0}}{8π} (ωτ_{A})^{2}(a/l)^{2}.        (4.37)