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Transcript
```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    Â be the magnetic field in the vacuum. Since there can be no electric
current in the vacuum,
    Â  0 ,
(21.2)
where Â 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  Â
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  20  dn  0  n̂  B  Bdn .
(21.9)
When   0 , this becomes

2 ¨
™p  B °  1 lim n̂  B  Bdn  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

   Â  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  Â   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
    Â  0 ,
(21.20)
subject to the matching conditions
p0   
1
0
  B0  Q B0 
1
0
  B̂
0

   Â  B0
(21.21)
and
n̂  Â   n̂   B̂0
(21.22)
at the vacuum/fluid interface, and the boundary condition
n̂  Â  0
(21.23)
at the perfectly conducting wall.
4
```