2002 I 1
From QED
The magnetic field is:
The equilibrium momentum equation gives:
Using:
Plugging in:
Using vector identities:
Plugging this in:
Or:
Expanding:
Since the only vector on the left is , p is a function only of ψ, and we can write , and dot with :
Or:
In slab coordinates:
We choose the equilibrium:
We are given:
From the definition:
So:
The current is:
So:
The equilibrium equation becomes for p = 0:
If we take the curl (to get rid of the unknown Bz term):
Writing this out:
for any scalar function:
This gives the first order equation:
Since ψ1 and ψ0 are only functions of x:
Dotting with (the only nonzero component):
Or:
Using :
We write:
at the boundary:
So:
And:
At the other boundary:
So:
The boundary condition:
At the upper boundary:
Putting in the definition for the B:
Using :
Simplifying:
At the lower boundary, we get by analogy:
So:
For the equation:
Near x = 0:
Which can be written as:
We get:
The jump condition across the boundary will be:
If b2 = 0:
Using the boundary conditions gives:
Which gives the exterior solution:
So that the total flux function is:
Near x = 0:
O- and X-points then occur where :
So points occur at ky = nπ, x = 0. O-points are minima, and X-points are maxima:
Failed to parse (syntax error): \partial^{2}\psi/\partial y^{2}>0
at ky = 2πn, so these are
O-points. , so these points are X-points.
This page was recovered in October 2009 from the Plasmagicians page on Generals_2002_I_1 dated 19:55, 27 April 2007.