icarus wrote:Art:

That's quick and dirty but correct. I hope you can follow it.

Mostly. I think you have an extra factor of 2 on the LHS of this step

d Phi / d X = sqrt( 2 * integral{d Phi * 2 * [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ] } )

I don't know where you see a problem. Here it is in a little more detail:

- (d/dX) (d Phi / d X )^2 = 2 * (d Phi / d X ) * (d^2 Phi / (d X)^2 )

(d/dX) (d Phi / d X )^2 = 2 * (d Phi / d X ) * [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ]

(d Phi / d X )^2 = integral{ dX * 2 * (d Phi / d X ) * [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ] }

(d Phi / d X )^2 = 2 * integral{ d Phi [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ] }

(d Phi / d X ) = sqrt( 2 * integral{ d Phi [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ] } )

OK, I see it now, too. Quite right. And there were some brackets missing, too.

icarus wrote:and I haven't quite figured out where your electron density comes from, though it is clearly related to the sheath potential energy equation and the electron energy as they are added to the sheet.

In this model, the probability density function for electrons does not depend on v_x and v_y. As an electron rolls up the beach, v_x and v_y do not change, but v_z is reduced so that the total energy (kinetic plus potential) remains constant. When it gets so far that v_z is reduced to zero, it is reflected. That occurs sooner for electrons that have larger values of perpendicular velocity. The number of electrons still in the game at any potential is proportional to the area of the disk bounded by electrons with v_z = 0. For these 0.5*m*v_z^2 = W_0 + e*phi. (Remember that W_0 is the kinetic energy of the electrons upstream, where phi = 0.) If you stare at this a bit, it should become evident that the electron density must be linearly proportional to the potential phi.

icarus wrote:Basically looks good I think, neat, as you say. I could latec it up if you like?

I'm not planning on taking this anywhere, and there's no way to display latex in this wiki, at least not directly. Feel free, but it's probably not worth the trouble.

icarus wrote:And as I ponder the implications of this negative sheath it makes some more aspects of the machine clearer too. The sheath creates the sharp potential drop from the positive region near Magrid to the well in the center. Your equations now show that the sheath will be consistent with a huge drop in potential across a small layer, due to high electron density there.

Inside the sheath it is quasi-neutral, so the potential is not affected by particle charge interior to the sheath, unless there other layers/regions of charge build-up.

Perhaps we could also consider that any/all electron excess in the Polywell will be manifest in the sheath surrounding the neutral spherical core? I.e., all the deviation from neutrality will be taken up by the sheath and the interior will then be totally neutral?

I'm not sure you've got the picture, especially the point that the "quasi" in "quasi-neutrality" means that you can certainly have electric fields in the plasma, even though the electron and ion densities are very nearly equal. In this model, the sheath, where most of the potential is dropped if you go to higher and higher potential differences between magrid and wall, is snugged up against the wall, well outside the magrid. On the other end, you have moderate electric fields somewhere upstream, that accelerate the ions to the speed they need to satisfy the Bohm criterion, sqrt(W_0/m_i).