## The consequences of quasi-neutrality in the cusps

Discuss how polywell fusion works; share theoretical questions and answers.

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Art Carlson
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### Carlson sheath equation

icarus wrote:Seriously though, I would be really interested to look at it, don't know if you have it in Latex or some other symbolic electronic form, since I find these line text equations hard to read and sometimes open to interpretation. Is it possible we can use it to estimate the thickness of the layer, i.e., the "Carlson length",, for the mono-energetic electrons?

I actually like doing my math with latex, but this stuff is all on scratch paper.

electron density (proof still not yet revealed):
n_e = n_0 * ( 1 + e*phi/W_0 )
ion density (same as in the standard derivation, but using my result c_s = sqrt(W_0/m_i) )
n_i = n_0 * ( 1 - 2*e*phi/W_0 )^-0.5
charge density
rho = e*( n_i - n_e )
Poisson equation
-epsilon_0/(e*n_0) * (d/dx)^2 phi = rho / (e*n_0) = ( 1 - 2*e*phi/W_0 )^-0.5 - ( 1 + e*phi/W_0 )

Now we change to normalized variables:
x -> X * lambda_D; phi -> -Phi * W_0/e
where
lambda_D = sqrt(epsilon_0*W_0/(n_0*e^2))
is the equivalent definition of the Debye length, which uses the electron energy W_0 instead of the electron temperature.
(d/dX)^2 Phi = ( 1 + 2 Phi )^-0.5 - ( 1 - Phi )

With the sheath equation in this form it is obvious that the thickness will be on the order of lambda_D. But we can go still farther, still following the standard derivation.
(d/dX) (d Phi / d X )^2 = 2 * (d Phi / d X ) * (d^2 Phi / (d X)^2 ) = 2 * (d Phi / d X ) * [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ]
[Edit: Missing square bracket supplied.] (Ugh! Wiki really isn't made for this!) Integrate a first time over X and then take the square root:
d Phi / d X = sqrt( 2 * integral{d Phi * [ ( 1 + 2 Phi )^-0.5 - ( 1 - Phi ) ] } )
[Removed an extraneous factor of 2.] I'm running out of steam, but this integral can be solved analytically, so we have
d Phi / d X = f(Phi)
which we can write as
x(Phi) = integral{ d Phi * f(Phi) }
From here we have to get numerical, but it's no big. Just integrate f(Phi) from 0 to 1 (which is where the most perpendicular electron finally gets turned around by the field in the sheath). The result is the thickness of the sheath in units of lambda_D. Cool, isn't it?
Last edited by Art Carlson on Mon Feb 16, 2009 9:27 am, edited 2 times in total.

TallDave
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rnebel wrote:3. Energy confinement on the WB-7 exceeds the classical predictions (wiffleball based on the electron gyro-radius) by a large factor.

I'm not 100% clear whether that means it's better than n*kBolt*Te = B**2/(2*mu0), or whether that means better than something else which is called classical here (perhaps WB without recirc?). I'm leaning toward the latter, based on the comment that the WB and recirc appeared to be "reasonable."

Art Carlson
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### proff of electron density self-similarity

icarus wrote:Art said:

Better yet, I'll try this business model: Anyone who wants to see the proof send me 2 Euros and I'll mail it to them.

Okay, I'll take the bait. It sounds interesting enough that I'll take a look at it. I can offer three forms of payment: i) I'll proof read the proof for the discount rate of 2 euros and we call it square, ii) you can call it in lieu for services for being the namer of the name "Carlson sheath" and we call it square or iii) I'll buy you "ein stein bier" next time I'm over your way (preferably at Oktoberfest).

Your prices are too high. But I'll offer you 1 euro for the proofreading and 1 euro for the name.

*** Shrink wrap: Anyone except icarus who reads beyond this point agress to pay me 2 euros! ***

Actually it's easier than I thought. Write the electron energy distribution this way:
f(v_x, v_y, v_z) = n_0 * delta( v_z - sqrt( 2*W_0/m_e - v_x^2 - v_y^2 ) ) / ( pi * 2*W_0/m_e )

The probability density in the v_x and v_y directions is uniform and gets projected down onto the sphere with radius sqrt( 2*(W_0+e*phi)/m_e ). It is obvious that the form of the distribution is the same no matter what phi is, and the integrations (which I was doing before with trig functions) become trivial. The density is essentially the area of the sphere projected onto 2 dimensions, which leads pretty quickly to the expression proportional to phi.

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

icarus
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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 ) ] } )

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.

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

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?

Art Carlson
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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).

TallDave
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For those worried Art is talking about putting on a vinyl suit, here's a quick intro to LaTeX.

We've got an impressive lot of equations here, but of course they're of trivial interest unless they describe reality. So what observations are we predicting here for WB-7?

Art Carlson
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TallDave wrote:We've got an impressive lot of equations here, but of course they're of trivial interest unless they describe reality. So what observations are we predicting here for WB-7?

This particular calculation lends some credence to my model, but I already said it won't convince anybody who doesn't want to be convinced. Given (hopefully by accurate measurements)
plasma density
magrid voltage
magnetic field

my model gives a lower bound on the energy loss rate. (The bound itself is fuzzy, say within a factor of 2 or 3.) Rick Nebel, if I'm reading him right, has said that the WB-7 results are inconsistent with this bound, but nobody has seen the data or any details of the calculations. Applying the model to the reactor regime predicts that net power at a reasonable size is hopeless.

krenshala
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Art Carlson wrote:Applying the model to the reactor regime predicts that net power at a reasonable size is hopeless.

My maths are, unfortunately, not up to the majority of what you posted but does this statement mean you think a Polywell could get net power, but only at a ridiculously large size?

imaginatium
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Art Carlson wrote:
TallDave wrote:We've got an impressive lot of equations here, but of course they're of trivial interest unless they describe reality. So what observations are we predicting here for WB-7?

This particular calculation lends some credence to my model, but I already said it won't convince anybody who doesn't want to be convinced. Given (hopefully by accurate measurements)
plasma density
magrid voltage
magnetic field
my model gives a lower bound on the energy loss rate. (The bound itself is fuzzy, say within a factor of 2 or 3.) Rick Nebel, if I'm reading him right, has said that the WB-7 results are inconsistent with this bound, but nobody has seen the data or any details of the calculations. Applying the model to the reactor regime predicts that net power at a reasonable size is hopeless.

The mere fact that Rick Nebel, has said that the WB-7 results are inconsistent with this bound, should be reason enough to abandon using it in your calculations. So applying a model inconsistent with the data (whether you've seen the data personally or not), to the reactor regime, predicts nothing but false conclusions.

Billy Catringer
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Art Carlson wrote: This particular calculation lends some credence to my model, but I already said it won't convince anybody who doesn't want to be convinced.

Either you have a case or you don't, Art. What anyone wants or does not want is of no consequence whatsoever. Do you have a case?

Betruger
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Unless he is basing himself on what Dr Nebel informed him of privately.

jabowery
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imaginatium wrote:The mere fact that Rick Nebel, has said that the WB-7 results are inconsistent with this bound, should be reason enough to abandon using it in your calculations.

In the absence of published, peer-reviewed and reproducible results, its not unreasonable to speculate about Nebel's interpretation -- that it is right or that it is wrong. Indeed, it is appropriate to have skeptics like Carleson follow their hunch that Nebel is wrong and attempt to flesh out a theory of what might have misled Nebel.

I know in my own case, I was misled by Koloc's supposed multiple video frames of a "plasmoid" that it turns out were individual shots of a CCD artifact assembled after the fact to appear to be a single plasmoid in continuous motion for nearly a hundred milliseconds. I don't think Nebel is committing fraud, and I'm unsure of Koloc -- he seemed to exhibit a multiple personality disorder that may have been brought on by work with his mercury switches or other phenomenon (I'm just glad to have survived working in his garage with those high pulse power instruments of his). But when we're dealing with such golden carrots dangling on such lengthy sticks, there is a lot of room for following up on a lot of theories.

D Tibbets
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I certainly cannot argument against Art Carlson's conclusions based on his assumptions. But, that is the heart of the matter. Are his assumptions correct? He has taken a pessimistic attitude based on his experiance and reasoned arguments. He, or some other equally qualified physist needs to assume an optimistic attidude and develope his arguments on his percieved assumptions. Then there can be a debate on the merrits of each approach. The hints of the results from Dr Nebel suggest that qusi neutrality in the cusps may not exist, so it certainly warrents a model complient with this assumption.
The lack of difinitive data is frustrating, but with time the insiders at least, will know more after focused measurements, and experments. The lack of the aviability of this data has been used as a negative argument. The same arguments could be used for the atomic bomb. It couldn't work based on percieved physics, because there was no open peer review of the supporting data. ie- Dr Heisenberg* during WWII.

* A convieniant over simplification. More details of Heisenberg's atomic bomb work in WWII Germany -
www-personal.umich.edu/~sanders/physics/HeisenbergCP.pdf

Dan Tibbets
To error is human... and I'm very human.

Art Carlson
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1. What I have presented here is, in my opinion, the most complete and consistent theory of polywell physics there is. Explanations by Bussard and others sometimes contradict themselves, sometimes contradict known physics, and resort to handwaving at an early stage.
2. The published experimental basis on polywells is too thin to draw any conclusions.
3. Dr. Nebel's statements about his measurements have not been explicit and extensive enough to draw any firm conclusions, even if we trusted him fully. (I have not received any experimental information from him beyond what he has said here, but I haven't given up hope.)
4. I have no doubts that Rick is an honest and competent physicist, but physics is hard. His statements should be treated with caution until other, independent experts have had a chance to review them.

Art Carlson
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krenshala wrote:
Art Carlson wrote:Applying the model to the reactor regime predicts that net power at a reasonable size is hopeless.

My maths are, unfortunately, not up to the majority of what you posted but does this statement mean you think a Polywell could get net power, but only at a ridiculously large size?

That's what the model in it's current form says. One thing that is missing is any consideration of the power load to the in-vessel components and first wall, and especially to the wall behind the cusps. That will kill you long before you have to face the prospects of organizing the distribution of electricity from a 1000 GW reactor to an entire continent.