icarus wrote:Art, you said:
Assume B is in the z-direction, E is in the x-direction, and the magnitude of B and E are functions of x only (no dependence on y, z, or t). Then B = curl A implies A is in the y-direction and B = (dA/dx).
Just so that we're on the same page, we're talking about the interface region between plasma and B-field here, correct?
yes
icarus wrote:
If so, your statement above seems to imply that you think the B-field will penetrate quite some way into the plasma, i.e. to the bottom of this interface region, such that at x=0, B is non-zero?
I don't know where you got that. I made no assumptions about how big the region is. The argument applies whether it is infinitesimal or infinite or anything in between. I also made no assumption about which parts may have a B field, an E field, both, or neither.
icarus wrote:
How would you justify that? I think that in this interfacial region, B will be function of electron density (and therefore also a function of x), so that as you move in towards the plasma there will be some threshold value of electron density that is sufficient to sustain mag-field-repelling loop currents. My contention is that the steep electric field gradient associated with the interface penetrates further than the x position where the B-field disappears. The bunching of electrons at the interface (sheath) ensures it, no?
I made no assumptions about how the fields are produced.
icarus wrote:
In this picture, the electrons will have been aligned (by the electric field) to already have largely only radial velocity components before encountering the B-field (on an outwardly directed path).
That doesn't change anything. I just showed that an electron leaving the sheath will have the same angle as it had entering the sheath. If it goes in at a steep angle, then it bounces back at a steep angle.
icarus wrote:
So we add to your equations above the following;
for x<a, a - being the radius/altitude at which the B-field goes to zero,
x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx)
y: (m/q)*v_x*(dv_y/dx) = 0
z: (m/q)*v_x*(dv_z/dx) = 0
and for x>a, the above become (your original);
x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx) + v_y*(dA/dx)
y: (m/q)*v_x*(dv_y/dx) = -v_x*(dA/dx)
z: (m/q)*v_x*(dv_z/dx) = 0
Solving for x<a;
v_y and v_z are now both constant, and v_x = sqrt(2*q*phi/m) (again quadratic but this time only phi dependent). So as phi increases with increasing x up to a, v_x increases but v_y and v_z remain constant. The angle that the electrons enter the B-field region of the interface, at x=a, given by the ratio of tangential to normal velocity components
alpha =atan(sqrt(v_y^2 + v_z^2)/v_x)
and is equivalent to a loss cone angle. We see then that it is solely dependent upon the value of the voltage in the region just beneath the point where the B-field begins. QED?
Correct, but not relevant to the point I was making.
icarus wrote:
Aside: I don't think that you can assume that the curl of the vector potential is orthogonal to the vector potential generally, as you have done. The result that I'm after doesn't change just that your equations should have begun;
x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx) + v_y*(dA_x/dy- dA_y/dx)
y: (m/q)*v_x*(dv_y/dx) = v_x*(dA_y/dx - dA_x/dy)
z: (m/q)*v_x*(dv_z/dx) = 0
I assumed that the only variation in any quantity was in the x-direction. In that case dA_x/dy vanishes, and B and A are orthogonal.