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21. BOUNDARY CONDITIONS FOR LINEARIZED IDEAL MHD Before we prove that F is self-adjoint, we must derive the boundary conditions to be imposed on the solutions of the linearized ideal MHD wave equation 0 & & J 0 Q 1 0 Q B0 p0 p0 , (21.1) where Q B0 . We consider the system to consist of a conducting fluid (a plasma) surrounded by a vacuum, all enclosed within a perfectly conducting boundary, as shown in the figure. The conducting boundary (i.e., wall) is denoted as W , and the surface separating the plasma and the vacuum is denoted as S . The magnetic field is everywhere tangent to this surface, by construction. Let B̂  be the magnetic field in the vacuum. Since there can be no electric current in the vacuum,  0 , (21.2) where  is the vector potential in the vacuum region. At the wall W the tangential component of the electric field must vanish: n̂ Ê 0 , where n̂ is the outward drawn normal. Since Ê Â / t , we require that n̂  W 0 . (21.3) This is the boundary condition to be applied to Equation (21.2) at W . Now consider the plasma/vacuum interface S . Here we define n̂ to point from the plasma into the vacuum. From Maxwell’s equations, we know that, at the interface S , the magnetic field must satisfy the conditions ßn̂ B® 0 , ßn̂ B® 0 K (21.4) , (21.5) 1 and n̂ ßE® 0 , (21.6) where ßf ® fV fP is the jump in a quantity across the interface, and K is a surface current flowing within the interface S . If K 0 , then all components of B are continuous across S . The equilibrium force balance condition is B2 B B p . 2 0 0 (21.7) Now define a variable n that measures distance in the local direction of n̂ , i.e., across the interface at n 0 . Then n̂ / n , and the normal component of Equation (21.7) is B2 1 p n̂ B B . n 2 0 0 (21.8) Integrating this expression across the interface S , we have B2 1 n p 20 dn 0 n̂ B Bdn . (21.9) When 0 , this becomes 2 ¨ © ™p B ° 1 lim n̂ B Bdn 0 , ™ ° ™ 2 0 °Æ 0 0 ´ (21.10) since the integrand on the right hand side is continuous across S . The condition (21.10) is called the pressure balance condition. Let r0 be the equilibrium position of the plasma/vacuum interface S0 . Then, from Equation (21.10), the equilibrium variables must satisfy p0 r0 B02 r0 2 0 B̂02 r0 2 0 (21.11) , since the pressure in the vacuum must vanish. During the displacement of the plasma, the plasma/vacuum interface will also be displaced to a new location r r0 . At this perturbed boundary S the pressure balance condition is p0 r0 p1 r0 B0 r0 B1 r0 2 0 1 2 B̂0 r0 B̂1 r0 2 0 1 2 . (21.12) To lowest order in small quantities, this becomes 2 p0 r p1 r 1 B02 r 2B0 r B1 r 2 0 1 B̂02 r 2 B̂0 r B̂1 r . 2 0 (21.13) All quantities at S can be related to quantities at S0 by f r f r0 f r0 , (21.14) and as a consequence, to lowest order B0 r B1 r B0 r0 B1 r0 . The perturbed pressure is determined from the linearized energy equation p1 p0 p0 . (21.15) Finally, substituting all this into Equation (21.13), and using the equilibrium pressure balance condition, Equation (21.11), we find that the condition p0 1 0 B0 Q B0 1 0 B̂ 0  B0 (21.16) must be satisfied at the equilibrium interface S0 . A second condition to be satisfied at S0 is found from Equation (21.6). Let the boundary be moving with velocity V1 / t , let E be the electric field measured in the stationary frame of reference, and let E* E V1 B0 be the electric field seen in the frame moving with the boundary. In the plasma, E V1 B0 , by Ohm’s law. Therefore E* 0 ; the electric field in the frame moving with the boundary must vanish on the plasma side of the interface. Then Equation (21.6) requires that n̂ Ê* 0 , i.e., the tangential electric field in the frame moving with the boundary must vanish in the vacuum. Since Ê* Ê V1 B̂0 by the (non-relativistic) law of transformation of the electromagnetic fields, we have n̂ Ê B̂0 n̂ V1 V1 n̂ B̂0 . But on S0 , n̂ B0 n̂ B̂0 0 , by construction. V1 / t , we require that (21.17) Then using Ê Â / t n̂  n̂ B̂0 and (21.18) be satisfied on S0 . Then the boundary value problem for determining the normal modes of the system can be stated as follows. In the fluid, solve 0 & & J 0 Q 1 0 Q B0 p0 p0 , (21.19) where Q B0 , and in the vacuum, solve 3  0 , (21.20) subject to the matching conditions p0 1 0 B0 Q B0 1 0 B̂ 0  B0 (21.21) and n̂  n̂ B̂0 (21.22) at the vacuum/fluid interface, and the boundary condition n̂  0 (21.23) at the perfectly conducting wall. 4