In the wing frame the unperturbed plasma moves with velocity

- u_{0} e_{x}

. The ideal magnetohydrodynamics condition

E + u \times B / c = 0

therefore produces the motional electric field

E = - \frac{( - u _{0} e _{x} ) \times ( B _{0} e _{z} )}{c} = - \frac{u _{0} B _{0}}{c} e_{y} .

(1)

The wing's surface conductivity then gives the surface current density

J = Σ E = - \frac{Σ u _{0} B _{0}}{c} e_{y} .

(2)

Its direction also follows directly from the Lorentz force on the wing's mobile charges.

b

 0  0

Solution

 0  0

The wing and its forcing are time independent in the co-moving frame, so after transients have propagated away the perturbation is stationary there. Write the background velocity and field as

U = - u_{0} e_{x}, B_{0} = B_{0} e_{z},

(1)

and take every perturbation to be independent of

y

. The cold-plasma linearized ideal magnetohydrodynamic equations are

- ρ_{0} u_{0} \partial_{x} δ u = \frac{1}{4 π} (\nabla \times δ B) \times B_{0},

(2)

- u_{0} \partial_{x} δ B = (B_{0} \cdot \nabla) δ u - B_{0} \nabla \cdot δ u .

(3)

Their relevant components are

- ρ_{0} u_{0} \partial_{x} δ u_{x} = \frac{B _{0}}{4 π} (\partial_{z} δ B_{x} - \partial_{x} δ B_{z}),

(4)

- u_{0} \partial_{x} δ B_{x} = B_{0} \partial_{z} δ u_{x}, - u_{0} \partial_{x} δ B_{z} = - B_{0} \partial_{x} δ u_{x} .

(5)

Differentiate the momentum equation with respect to

x

and use the two induction relations. With the Alfvén speed

u_{A}^{2} = \frac{B _{0}^{2}}{4 π ρ _{0}},

(6)

the result is

(u_{A}^{2} - u_{0}^{2}) \frac{\partial ^{2} δ u _{x}}{\partial x ^{2}} + u_{A}^{2} \frac{\partial ^{2} δ u _{x}}{\partial z ^{2}} = 0 .

(7)

c

 0  0

Solution

 0  0

Integrating the

y

component of Ampère's law,

\partial_{z} B_{x} - \partial_{x} B_{z} = \frac{4 π}{c} j_{y},

(1)

through the current sheet gives the magnetic-field jump across a surface current

δ B_{x} (0^{+}) - δ B_{x} (0^{-}) = \frac{4 π}{c} J_{y} = - \frac{4 π Σ u _{0} B _{0}}{c ^{2}} .

(2)

Reflection in the wing plane reverses the tangential perturbation

δ B_{x}

. If

b = \frac{2 π Σ u _{0} B _{0}}{c ^{2}},

(3)

the upper and lower boundary values on the wing are consequently

δ B_{x} (x, 0^{+}) = - b, δ B_{x} (x, 0^{-}) = b, ∣ x ∣ < \frac{L}{2} .

(4)

Outside the wing, the corresponding boundary value is zero.

d

 0  0

Solution

 0  0

For

u_{0} > u_{A}

, define the Alfvénic Mach-cone slope

m = \frac{u _{0}^{2}}{u _{A}^{2}} - 1 ≃ \frac{u _{0}}{u _{A}} .

(1)

The equation in part (b) is the hyperbolic partial differential equation

\partial_{z}^{2} δ u_{x} - m^{2} \partial_{x}^{2} δ u_{x} = 0,

(2)

whose characteristic curves are

x \pm m z = constant

. Let

W (s)

be one for

∣ s ∣ < L /2

and zero otherwise. Selecting the characteristics that trail downstream, toward negative

x

, and imposing part (c) gives

δ B_{x} = - sgn (z) bW (x + m ∣ z ∣) .

(3)

The stationary induction and mass conservation equations then give

δ u_{x} = \frac{u _{0} b}{m B _{0}} W (x + m ∣ z ∣),

(4)

δ B_{z} = \frac{b}{m} W (x + m ∣ z ∣), δ ρ = \frac{ρ _{0} b}{m B _{0}} W (x + m ∣ z ∣),

(5)

with

δ u_{z} = δ u_{y} = δ B_{y} = 0

. In the strongly super-Alfvénic limit,

δ u_{x} ≃ u_{A} b / B_{0}

and

δ B_{z} ≃ b u_{A} / u_{0}

inside the disturbed region.

The perturbations have support only where

∣ x + m ∣ z ∣∣ < \frac{L}{2} .

(6)

Because the wing is infinite in

y

, this is the union of two inclined slabs bounded by the characteristic planes

x + m ∣ z ∣ = \pm L /2

. These two trailing slabs are the Alfvén wings generated by the conductor.

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 3

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