Another (simple) FAQ - DONE

[KitemanSA]

I'm not sure how to calculate an ionizing MFP in an electron cascade, though I'm pretty sure they don't travel 10m in a few usecs.

If a boron neutral is gonna ionise to a 5+ in the first few cms of the outer edge, then why does that mean anything other than indicating the electrons there will have a potential of over 600eV and a density of over 10^21/m3 [which is definitely not the picture of 'low energy' nor 'low density' electrons that reside in the outer edge of a Polywell that folks have alluded to]?

First, I am not sure why folks think the B needs to go to +5 at one instant. If a neutral goes to +1 inside the well, it will drop and oscillate. Certainly it will ionize to +5 thereafter. The lower in the well it ionizes, the lower its energy which makes it a bigger target to speed up thru coulomb collisions, no?

Second, I don't know anyone who says that the edge of the well has low energy electrons! Quite the contrary, that is where the electrons are at their HIGHEST energy. I am not sure about the local density, but I am not sure it matters much.

By the bye, even if neutrals DO drift 10m, ON AVERAGE, before they ionize, it just means that some leave the MaGrid and become a Paschen problem. That is what the vacume pump is for!

[TallDave]

They bounce around.

From what I'm reading in Valencia, they don't have to. The KV injected electrons bang the electrons off of them, and those electrons cascade into others, etc. The e-folding dominates. Meanwhile, new hot electrons are banging into those liberated from the ions and heating them. Pretty quickly you get a deep well of hot electrons.

I'm not sure how to calculate an ionizing MFP in an electron cascade, though I'm pretty sure they don't travel 10m in a few usecs.

If a boron neutral is gonna ionise to a 5+ in the first few cms of the outer edge, then why does that mean anything other than indicating the electrons there will have a potential of over 600eV and a density of over 10^21/m3 [which is definitely not the picture of 'low energy' nor 'low density' electrons that reside in the outer edge of a Polywell that folks have alluded to]?

Well, at startup, before the well forms, the electrons are being injected hot. The electrons then smear over the volume of the WB interior, with their average position being in the core where they are slowest (high potential energy, low velocity), so at any given time the density is highest at the core (thus creating a virtual cathode). They will naturally have their highest average velocity near the magrid (the bottom of their inverted well)... but electrons that lose their potential energy and velocity will end up stuck at the bottom of their well near the casings, and leave the system (according to Rick). That's the only place I can see that an ion might occasionally briefly acquire an electron (via a collision with total energy < 671 eV) before a hot electron knocks it off again (ions at low energy are a fat target). The first time it hits anything not at the very lowest end of the energy distribution, it's going to be fully ionized again.

So: generally low density, high energy edges, with cold electrons accumulating at coils and cusps. For electrons, it's an inverted well. The ions get focused at high energy because they converge at the center, the electrons get defocused (spread around the edge) because their "bottom" diverges.

Let me see if I can explain it. Near the most positive potential in the system the positively charged ions will be the slowest.

Yes. They will have low radial velocity there, and some small transverse velocity (which will tend to maxwellianize due to the high collision cross-section, leaving a relatively monoenergetic total energy (large radial potential + small maxwellian transverse velocity)).

Kite -- Yeah, that's odd, I don't know about that.

Did Rick ever give us an optimum p-B11 well depth? I can't remember.

[D Tibbets]

Heh, well, I gave you Bussard's numbers. Feel free to derive an MFP from Bussard's numbers if you want to know what it is.

I was asking you to do your own homework, but sure, easily done;

So he says "The cascade time e-folds at a rate of
1/(no)(sigmaizn)(veo), where (no) is neutral density,
(sigmaizn) is ionization cross-section for low energy
electrons at speed (veo). Typically, for no = 1E13 /cm3 (i.e.
ptorr = 3E-4 torr), veo = 1E9 cm/sec (Ee = 100 eV), and
sigmaizn = 1E-16 cm2, the cascade e- folds with a time
constant of about 1E-6 sec (one usec)."

The MFP is therefore 1/(no)(sigmaizn), in his parlance, giving an MFP of 10 metres.

Can you, please, now explain how these ions get formed if their MFP to ionisation is 10m, whilst in a device of sub 1 metre scale? Taking the discussion into 'time' and using a linear MFP-type calculation to derive timescale is highly disingenuous because the ions do not undergo linear motion during the usec timescales that derives from the answer to an equation that presumes linear motion.

Remember, the ionization collisions are presumably mostly electron- ion collisions. As everything is relative, I'll assume that an electron speed of ~ 10,000 eV, or ~ 10,000,000 meters per second, hitting a stationary neutral fits the MFP figure you gave. Electron speed of 10,000,000 M/s divided by MFP of 10 M = ionizing collision every microsecond. The numbers seem most agreeable.

Also, a later post by ChrisMB implies that the electrons are slow and at lowest density near the edge. Where did that come from? The electrons are hot (fast) on the edge and cold (slow) in the center. Initially, a pure electron injection creates a square potential well with electron density greatest near the Wiffleball border (edge). As ions are injected / created from neutrals they are accelerated towards the center by the potential well, dragging some electrons with them, thus creating the elliptical well with central vertual anode.

Dan Tibbets

[MSimon]

Bussard studied the ionization issue extensively (it was one reason he built PZLx-1). If you have a paper that suggests Bussard's number is way off, I'd be curious, but it seems safe to ignore this in the FAQ. It seems unlikely neutrals can survive more than usecs.

You really just don't seem to understand, and I am clearly very bad at explaining some very basic nuclear collision theory:

Just because a 1+ ion has accelerated into a 65kV well, it doesn't mean that all the electrons simply jump off the damned thing! ions have to be in collision with something for them to loose any more electrons.

Why would electrons jump off an ion just because it is travelling fast????

Fear?

[MSimon]

Sorry Kite, got a little sidetracked there.

What's the parenthetical term in your chart there? I started to make a version with a drive power column added in but I was confused by that.

In the cross-section graph I have the p11B reaction has a sharp peak before the main hump, and the graph flattening out after it. When I used the "45 degree" rule, I ran into the front peak at 130keV. With the -45 rule, I couldn't reach a max til past the end of the graph at 10MeV.

Since the 130keV seems so different than other values mentioned, I am wondering if I have used a bogus data set.

I believe that is correct given the formularies I have studied over the past 3 or so years. I'd put it at 140 KeV but that is a mere quibble.

[KitemanSA]

A variation of the answer last edited by TallDave is now inserted in the FAQ. At this point, this topic is done. If you want to make further comments, either PM me directly or start a new topic please.

If you wish to carry on this otherwise fasinating discussion, would y'all start a new topic please

[chrismb]

What an absolute friggin' joke this whole nonsense has become.

The question is plain and is phrased 'per ion', independent of what device it is in, magnetic or electrostatic confinement. The question is inherently device and method-irrelevant, all that matters is the ion's energy at the moment of fusion, and that is well known, and already well answered by me. It is not remotely difficult to state. The best fusion rate is the peak σv for that ion and reaction, which is slightly above the cross-section peak, not below it as those numbers put up seem to indicate and that has taken no information whatsoever off the nndc database.

You may well have had a different question in mind, seeing as the bizarre non-sequitur nonsense [wrt ion energy] TD has brought to the question, of what the ionisation state is. That is totally irrelevant to what the ion's best energy is at the moment it fuses. So I guess you have confused yourselves because the question you wanted to answer seems to have been; "what is the best drive potential for a Polywell?". That is a question in which ionisation state would play a part, and is, indeed, complex due to the increase in losses with increase of global energy.

[KitemanSA]

Chris,

You were asked REPEATEDLY to edit the answer to make it better. You did not. Two people did, but not you. You merely chose to bitch at their answer.

Next time you are asked to help develop a FAQ answer, please put-up, or shut-up. Edit, don't bitch.

If you think I should change the FAQ answer, tell me SPECIFICALLY how that answer should read.

Thanks

[TallDave]

Thanks Kite, great work on the FAQ.

I stopped by and added a couple minor edits to the main page, but I should probably be doing more. That can be a great general resource for helping out PW newbies who want to know more -- and there may be a flood of those over the next year or two.

[chrismb]

The answer to the original question is, as near as darn it 'the energy of peak cross-section for that particular reaction'. And that is that.

That's all you need say for that question (with inclusion of the numbers that had already been stated, which are close enough.)

I'm sorry the answer wasn't long-winded enough for you to notice it.

[TallDave]

seeing as the bizarre non-sequitur nonsense [wrt ion energy] TD has brought to the question, of

What an odd remark. You brought this up this subject. It was the third post in this thread, long before I showed up.

I'm glad we've apparently convinced you it is nonsense, though.

[chrismb]

What an odd remark. You brought this up this subject. It was the third post in this thread, long before I showed up.

OK. Fair enough. I'll take that one on the chin - but it was not intended to 'introduce' the subject of higher charge states, I was [clearly very badly, in hindsight] trying to reconcile whether it was drive voltage that was intended in the question because the conversation had just plopped into discussing thermal plasmas. mea culpa.

[TallDave]

Easy enough to do. It's hard to think nonthermally!

I had this problem a lot when I was trying to find references. Everyone thinks in terms of thermal distributions.

[KitemanSA]

The answer to the original question is, as near as darn it 'the energy of peak cross-section for that particular reaction'. And that is that.

That's all you need say for that question (with inclusion of the numbers that had already been stated, which are close enough.)

I'm sorry the answer wasn't long-winded enough for you to notice it.

Now I am truly puzzled. I included reference to optimizing to max (sigma x velocity) SPECIFICALLY because you mentioned it. Should I remove it? Guide me oh grumbly guru!

[chrismb]

That bit is OK, and you put that. That's fine. But the comments regarding density are misleading and you've mixed the two concepts. The optimum ion energy for a fusion event is irrespective of density. If you double your density then does the optimum ion energy change? Your rate might change, but not optimum ion energy. <cs.v> is the term for "averaged-equivalent" cross-section across all ion velocities and relates to thermal plasmas. You could put something like "this is how it is for thermal plasmas, but for mono-energetic non-thermal; it's just cs.v because we're assuming all the ions are the same energy, and it is only the rate of reaction that depends on density for non-thermal beam-type processes".

For a given energy over a fusion peak, cs drops off much quicker than v increases, so 'peak cs' will do.