Scattering theory.

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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Chris Bradley
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Scattering theory.

Post by Chris Bradley »

It seems to be a matter not often discussed in the detail it deserves, but the issue in all prospective energy producing fusion machines is collision efficiency. Just getting deuterium to go head-to-head with another deuterium is not enough. The coulomb scattering cross-section is much larger than the fusion cross-section. Penetration of one nucleus at a few 10's of keV through the 10's of MeV coulomb repulsion energy of the other is a random, quantum, affair. Electrostatic repulsion is a very, very, likely outcome by comparison with the relatively rare fusion event.

In an accelerator experiment, where beams of, e.g., deuterons, are brought into intersection with another beam then the fact that most of them scatter off is of no concern – there is no energy recovery being attempted. Huge water-colling systems are required, even though the output is a few measly microamps. In a 'thermonuclear' confinement, scattering is irrelevant because it is that very process of scattering that distributes thermal energy to the volume of that reactor. The energy is never 'lost' except when the confinement edges are breached by nucleii whose energies are at the lower or upper limits of the energy distribution, so the only discussion to be had there is simply one of getting the confinement right. So 'scattering' has no real point of discussion in these experiments.

In electrostatic systems, the issue of scattering is not only important, but completely dominant in the dynamics of the machine. There is no essence of 'confinement', instead the principle is to get nucleii hi-tailing it back and forth until there is an interaction. In my analysis of things, the only time you truly get a beam-type behaviour in a fusor is when you first switch it on. Ionised particles in the chamber begin to accelerate in a reciprocating fashion through the centre of the focus across the potential difference. So you can't see this on start-up? Of course not! Why would you see particles jumping around the chamber at millions of m/s? Within a short period of time, a central plasma forms. This is the energy of all the scattered particles that can't get back up to their original potential energy and so their energy begins to become distributed into the background neutral gas that has scattered them. That central plasma ball is of no interest – it represents all the wasted energy of scattered interactions. I would hazard a guess that it is at no more than a few 10's of eV. It can't be much higher than 13eV (~150,000 kelvin) else it would further ionise the background neutrals and its boundary would spread out and just heat up the grid. In fact, that's probably exactly what happens, and some equilibrium is achieved between the heat dissipation of the grid and the thermal gradient to the boundary of the centroid.

Now comes the interesting bit – the beams in the fusor's star mode are where the 'real' energy is. These are 'beam-like', but again if it were a particle beam I don't think it would actually be so visible because to be visible is lossy and a moving particle, unless it is interacting, is not lossy. If it is interacting, then it would decay back into that central ball of waste energy.

I would speculate that it is, like the central ball, a plasma of a few 10k's or 100k's of Kelvin (i.e. Units or 10's of eV) but that it is moving in bulk with an equivalent energy of the drive voltage (30keV, or whatever it is) relative to the neutral gas.

It also seems possible that the particles are not going back-and-forth at the same time. If they did that, they would generate magnetic fields that would then create instabilities. It is quite possible that they do exactly this, of course, in which case they may well execute a very tight ellipse rather than a 'two-way' beam in a single line and it's just that we can't see closely [or quickly] enough what is actually happening.

But I would've tended to think that if this were so then that elliptical orbit could diffuse out to a visible extent. So instead I speculate that those beams could be 'packets' of plasma, at a similar temperature to the centroid, but are moving at very high velocity to the background medium. They would tend to stay in line because of the physical geometry of the inner grid and the direction of the electric fields those form (but if the electric fields are not well aligned around the grids, or if the field is not strong enough and the period of oscillations is too slow, then those plasma packets would spread out under their own internal electrostatic repulsion after exiting the grid). A tiny 'lump' of plasma might then 'sweep up' nucleii that interact with it along its trajectory, hence it maintains a high temperature due to those lossy interactions and therefore continues to emit a black-body spectrum off into the UV, just like the centroid.

I would therefore refer to the fusor as a 'bulk-beam' reactor rather than 'beam'. The device I am patenting is also intended to be a 'bulk-beam' reactor. The mechanistic difference to the fusor is that it is configured such that the remaining energy of any scattered particles can be brought back up to the operational energy. In a fusor, once an individual nucleus is scattered by the background nucleii, its energy is lost because to gain the full energy again it must work its way back to a position at the outer grid (the higher positive potential) and then be accelerated back into the centre through the full potential difference. This is the only way it can regain that full energy potential. Unfortunately, it can't do this as its thermal energy is lost to the centroid and that is that!

Now we can examine if this mechanism can ever release more energy than it takes. Call the cross-section (a function which defines the probability of an interaction event) for fusion as Sf(E) and that for scattering as Sc(E), at some collision energy E. When a nucleii is at a maximum E, within the inner grid, Sf(E)/Sc(E) is at its maximum but it is still considerably smaller than unity. As a nucleus follows its trajectory from the inner grid its velocity, and E, drops off and in doing so Sf drops and Sc increases. Sf(0)/Sc(0) now diminishes to, effectively, zero. So we'll focus on the maximum on Sf(E)/Sc(E).

Let's say that Sf(E)/Sc(E) is 1/1000. That is, the probability of a scattering event is 1000 times more likely than a fusion event, and that that fusion event would release 4MeV (DD reactions). That means that if the nucleii are accelerated to anything more than 4keV to get to this 1/1000 ratio of Sf/Sc, then there can be no energy pay-back.

I regret to say that to get to an Sf/Sc of 1/1000 for D-D, I am lead to believe that the energy required is about 200keV (which means accelerating to 400keV relative to stationary media, due to reduced masses) and therefore this can never pay-back. At 200keV, Sf=~200barn, Sc=~0.2barn. (This is based on a scatter of less than 5 degrees, which itself is not without some energy loss but may be considered a relatively diminutive loss. What the actual degree of tolerable scatter is in this mechanism is something I'm still working on.)

However, there is a small glimmer of hope. Once scattered, the nucleii doesn't loose all of its energy but it's just that fusion becomes significantly less likely. The probability for fusion is a function of e^-(k/SQRT(E)) and so it drops off very very quickly!

If there is no chance to recover that nucleus back up to its original energy then if the statistically averaged energy loss per coulomb collision is j then the probability becomes Sf(E)/Sc(E) + Sf(E-j)/Sc(E-j) + Sf(E-2.j)/Sc(E-2.j) +... (Obviously, j isn't constant for each subsequent collision and I am making an approximation to keep it simple.) As you ramp up the input energy, the Sf/Sc ratio improves, but the j also increases and so that sum decreases. Infact, it improves really well because higher energy means the scattering cross section comes down whilst the fusion cross section goes up. But the ratio NEEDS to be bigger because the ratio of input energy to output energy tends to unity as you use higher energies. You end up 'chasing your tail' all the way up to MeVs of energy, at which point the input energies become greater than the output energies (per collision) and therefore it turns into a big heat-generating machine. (Remember, just generating more heat than you put in isn't enough, it has to be 'useable heat'. If you generate twice as much 'heat' energy as you put in as energy, then Carnot says you can't get anything back. In this case, 1-Tlo/Thi = 50%, so generating 1kW of heat for a 500W input can never be used to generate useable power. A point often overlooked in discussions of fusion energy.)

[DD has a peak cross section at 1.25MeV of 0.1barn, and the scattering cross section for <5 degrees at that energy is, according to the conventional equations, 5.5 barn, so you can see that the DD output energy of 4MeV is never going to pay off because you have to put in 5.5/0.1 x 1.25MeV = 69MeV to get it! So there has to be some other mechanism employed instead of just these direct, lossy collisions if we want to have a hope of energy output.]

Regrettably for the fusor it is even worse than this. The Sf/Sc ratio examined here is only that for when the beam is passing through the inner grid. For most of the time it is accelerating towards the inner grid at velocities lower than its maximum, i.e. The Sf/Sc ratio is, on average, much much smaller than that discussed.

To overcome these limitations I have devised a configuration of grids which dynamically manipulates the track that the nucleii are likely to take and that, after scattering, will accelerate the lower energy nucleii back up to their initial energy. By 'recovering' lost energy, the formulae above are changed to something that looks, potentially, viable. The values are the same but the mechanism differs. When a nucleus suffers an energy loss of j in a coulomb collision, that energy is put back into that particle by accelerating it on a different trajectory until it comes back into alignment with the main beam. Therefore, whereas the condition for more energy out (Eout) than energy in (Ein) was Ein/Eout<Sf(E)/Sc(E) + Sf(E-j)/Sc(E-j) + Sf(E-2.j)/Sc(E-2.j) +... with the series having as many steps as there are collisions, now it becomes (Ein+j+j+j..)/Eout<Sf(E)/Sc(E). That is, with each collision instead of that energy loss being manifested by a reduction in Sf and increase in Sc, it means there is a greater demand on the energy input to boost that nucleus back up to its original energy level. As Sf is a -ve root-exponential function, this brings the prospect of 'energy pay-back' from collision energies, of 10s MeVs [heat generators!] for the fusor, down to 100s of keV, and therefore with the prospect of achieving an output something like 10 times the input power as well as being a bit more practical in terms of power supplies.

This is done in my device by synchronizing a pulse modulated (in the order of MHz), relatively low (500V), e-field with the motion of the particles, and if the particle looses energy and drops out of that synchronization then it gains a big single input of energy from a high voltage DC e-field. Once it gains enough energy to re-synchronize, it's back to its energy. The energy gain that needs to be achieved out of a fusion event therefore needs to be factored in to the equation with extra energy being put in, as in the LHS of the equation above.

I expect that there may be a few ways of doing this. I have focused my efforts on manipulating elliptical and circular orbits, as touched upon briefly in my discussion above. For those with good imagination, I don't think I'm giving anything away (which could count as 'disclosure' in patent terms) by saying that if a fusor and a cyclotron got married, my invention would be one of their kids!
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Re: Scattering theory.

Post by Steven Sesselmann »

Chris,

And I thought you were a man of few words ...

I have a fair idea what you are trying to achieve, but my question to you is, can you retain any of the fusion energy and make it contribute to further fusion reactions?

Without positive feedback I don't believe we can break even...

..not good enough to just capture the pawns, you have to put them back on the board.

Steven
http://www.gammaspectacular.com - Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG
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Re: Scattering theory.

Post by Chris Bradley »

(Sorry for the word count. yes - 25 years of info to get out!)

That's the very issue. Electrostatic fusion is not a 'containment' process, it's a beam process, and therefore it cannot 'contain' any energy produced. It is a 'one-hit' process and you have to be able to get that energy straight out. Otherwise, you are talking about a thermonuclear containment if you are trying to recirculate that thermal energy and there's a great big magnetic bucket being built in France to try to do that.

By definition, if you contain thermal particles then you also contain thermal energy (this is one of the dichotomies of thermonuclear - how do you keep particles in but let energy out). The fusor does not attempt to contain 'thermal' particles, it accelerates individual particles to 'fusible' energies, and therefore it cannot contain thermal energy which might then contribute to a thermonuclear process.

best regards,

Chris MB.
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Re: Scattering theory.

Post by Frank Sanns »

I think what you just said in over 2,000 words, was higher voltage increases the probability of fusion events (up to a point) relative to scattering (and I said that in 12 words). There is nothing new here.

Correction: The beams only show where less than or equal to 13.8 ev of energy was. They show nothing about where fusion is or was.

Cyclotrons or sychrotrons still loose energy because any non linear path of a charged particle results in energy being radiated away. E and M or thermal losses are still losses.

Frank S.
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We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: Scattering theory.

Post by Chris Bradley »

I should've also said, in response to;

"Without positive feedback I don't believe we can break even..."

I rather suspect that you may be correct on this. Will my device give break-even, even if it doesn't have positive feedback? Probably not.

It's not a panacea, none of the devices discussed on this forum are, but I'd like to think it may be a next generation that helps further improvements, as perhaps your device is also.

The act and art of science is the unconstrained progression, and love of, of new ideas, however unrealistic their 'business plan' - a fact that I trust all subscribers to this forum also see as the limitation of the politically directed 'modern research industry'.

best regards,

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Re: Scattering theory.

Post by Chris Bradley »

Higher voltages HUGELY increase the probability of fusion event over scattering! The purpose of the discussion was to examine what that probability actually is, in absolute terms, in respect of input energy and whether that level of input energy is feasible. I've not seen this discussed before.

I don't think there is any disagreement on the beams - I think I said these are the visible consequence of plasma in motion about the centrum, but all the same they must have a kinetic energy equivalent to the ratio of the distance that that beam reaches to the outer grid, else it wouldn't have reached that point within the potential gradient whilst ionised. Particles passing into the grid at their maximum velocity, whether part of that beam or otherwise 'free' particles that are unseen, are the particles most likely to fuse. Is this what you were raising?

Of course, sychrotronic losses become increasingly significant for high velocities and tight radii. There are obviously limitations in this regard, so this must be calculated through to ensure the designed maximum radii do not lead to an onerous consumption of power. It is not too onerous at energies of 10's of kV and unit cms of radius. The coulomb collision losses are greater by orders of magnitude, even at sub-micron pressures, under these conditions, but, agreed, it does become more significant in high 100's kV energies. We're not talking about the MeV/TeV beam accelerators that certainly do have kW worth of sychrotronic emissions.

best regards,

Chris MB.
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Re: Scattering theory.

Post by Frank Sanns »

We will use your word HUGELY then but the net result is still the same. With optimum voltage the best ratio of fusion to scattering still gives you a net gain of 1.00000001 (should be more zeros but I am trying not to be too bleak) ? Huge still does not give you anything worth the effort.

Frank S.
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We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: Scattering theory.

Post by Chris Bradley »

There you go again! Summarising my 'dissertation' in just a few words!!!

And the point of that discussion was to see if it is possible to overcome that gain limitation. I think it can be better than that. As I quantified, at 1.25MeV, for DD in beam collisions it is ~1.018 (in your notation). The fusor will be slightly better than this (some repeat collisions at lower energies due to ion recycling will occur). Still not good enough but maybe a different mechanism for ion control can help improve it to reach the necessary gain.

best regards,

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Re: Scattering theory.

Post by Richard Hull »

Tom Ligon and I used an ocean optics spectrometer way back in 1999 to look at the doppler braodening of the centroid (poissor). The result was that the ions observed in the H beta lines each side of the main peak showed both aproach and recessional velocities EQUAL to the applied volatge energy. Obviously the ball is a mix but it not a wad of pure 13.ev ions.

Fusion in the fusor has long been know to occur mostly outside of the inner grid and involve a lot of neutral-fast collisions.

There can be no real magnetic fields here in our CW devices, just on first principles based on a simple amp-turn basis.

The fusor is and always will be a high loss, scatering limited, recirculating, electrostatic, ion accelerator-collider

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Re: Scattering theory.

Post by Chris Bradley »

Thanks for the info Richard. This is interesting. Are those results somewhere in the forum? I have run a quick search, but to no avail.

I should say that I did not mean to cover a lot of old ground as a primary objective, but did so in an attempt to pick out in detail what mechanisms could be improved. I am interested in understanding if the idea of recovering scattered ions back up to full potential, rather than just accepting the high loss rate of scattered particles, is something of interest/has been considered before/has no hope (with reasons?) etc..

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Re: Scattering theory.

Post by Mike Beauford »

Hi Chris,

I think Steve Sesselmann's S.T.A.R. design was attempting to do just what you are talking about.
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Re: Scattering theory.

Post by Chris Bradley »

I think so too, and I suspect it could have the same outcome. However, I rather suspect that it can achieve this goal with the transient behaviour it exhibits, rather than the steady state condition Steven has been trying for. It appears to go into a fluctuating mode which, to my might, is indicative of a field reversal and, just maybe, it's picking up all those scattered ions and pulling them back up to the full potential so that they can 'have another go'. For this reason I'm quite interested in Steven's design, but I don't really go with his view that there is a build-up of thermal energy in it, for the reasons above.

best regards,

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Re: Scattering theory.

Post by Todd Massure »

Hi Chris,
It sounds like we might be working on the same thing here. I'm going to send you an email.

Todd Massure
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Re: Scattering theory.

Post by Carl Willis »

Hi Chris,

Your idea appears to involve actively kicking particles that fall out of synchronicity with an energetic bunched beam (in this case due to scattering) so that those particles get back in phase and in line with the beam. Having a gaseous fusion target that scatters the beam, and then wanting to recover your unreacted beam, is a severe instance of a rather classic fundamental problem ("emittance growth") in beam physics. Techniques for combating emittance growth in accelerators and colliders are well-studied, up to and including splitting up the beam and re-forming it later with necessarily very complicated non-linear beam optics, which is in essence probably the kernel of your idea. I guarantee there's no merit, from an energy efficiency standpoint, to doing such a thing on a 200-400 keV fusion beam!

That said though, don't let one guy's skepticism discourage you from exploring the idea. Interesting things could certainly be learned from it.

>I regret to say that to get to an Sf/Sc of 1/1000 for D-D, I am lead to believe that the energy >required is about 200keV (which means accelerating to 400keV relative to stationary media, >due to reduced masses) and therefore this can never pay-back. At 200keV, Sf=~200barn, >Sc=~0.2barn.

We should note a couple corrections here: 200 keV in the CM frame translates into 4*200 keV = 800 keV in the stationary-target frame (the relative velocity between beam and target is twice the velocity of the beam relative to the center of momentum). Also, these cross-sections aren't right. At 200 keV CM (= 800 keV), the D(d,n) cross-section is about 90 millibarn. You can look these things up on CSISRS here: <www.nndc.bnl.gov>. The scattering cross-section is more than just Rutherford (elastic) scattering, which I suspect is what you calculated by formula (even though I get about 3 b when I do it). There are also inelastic, energy-sapping atomic processes that occur with some frequency. According to ENDF, the cross-section for inelastic scattering at 800 keV is also about 90 millibarn.

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Re: Scattering theory.

Post by Chris Bradley »

Thanks for this. Well, this is more of the discussion I was hoping to get into so as to get my numbers right.

[Obviously!?] I put Sf and Sc the wrong way around in the clip you show, so I was putting down 200 millibarn for 200keV CM (which was a rough point picked off a graph as I was being a bit lazy to work it out, so will certainly go with your numbers).

Not sure about the x 4 to get to the acceleration required in the stationary frame? Surely one D accelerated through 400keV approaching another D not so accelerated, they will see a relative velocity equivalent to 400keV. So they'll each see the other moving with 200keV relative to their CofM?

The issue of the coulomb scattering: I have, actually, 3 ways of calculating this and I would welcome the input. First off is the way that the Rutherford scattering formula is interpreted, as per the website: http://hyperphysics.phy-astr.gsu.edu/hb ... rosec.html . Second is the way Thomas Smid derives it on his website: http://www.plasmaphysics.org.uk/coulomb.htm , which looks quite convincing to me but comes up with quite small values. And finally I have run a quick step-by-step discretised numerical solution that comes up with a value somewhere in between. (In fact, my calc comes out at per your suggestion.) Can I get a consensus here on a formula for scattering cross-section?

I recognise your point regarding 'other' scattering processes peculiar to particular interactions, but I'm not sure this can be accounted for in generalised equations so I think I will have to run with just the analytical forms for the time being to get somewhere close to understanding what energy levels are likely to get anywhere close to a pay-back scenario.

many thanks,

Chris MB.
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Re: Scattering theory.

Post by Richard Hull »

Chris asked about where the concept of more fusion outside the grid was long known.

This is presented in a number of papers from both U of Illonois and U of Wisconsin.

They used eclipsing or occulting discs infront of an internal pips proton sensor to detemine where the bulk of the fusion was coming from. Poking around in some of the older 1998 and later papers on line at their sites would locate the paper and give the complete explanation and data tables.

The upshot is that the assumption was made that most fusion occurs by volume fusion of lesser energy deuterons and very fast neutrals and not by the idealized focus fusion that satisfies the mind and theory. It is a shotgun approach! There is just more shot in the air in the large volume than in the supposed poissor reaction zone. The shot is not traveling as fast and slam fortuitous fast neutrals in volumetrically more probablistic head ons. Less than elegant, but it works for me.

The far ranging discussions around this are most often found in the theory forum here.

Richard Hull
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Re: Scattering theory.

Post by Chris Bradley »

I might just be a bit thick on this, or something, but on that CSISRS website, I picked H-1 for deuterium and D,n for the reaction (D in, n out, I presume?) and CS for cross section, and got no hits. So trying H-2 instead I got the cross-section curve I'd expect for DD. Fine.

But the reason I entered H-1 to start with is because it has a 'H-0' option. I therefore presumed the numerical value might be the number of neutrons. But, no, it appears to call deuterium H-2.

So - what the HECK is H-0!!! Is this a 'virtual' idea of a hydrogen nucleus?

Is this just a glitch, or am I missing something and I need a guide for this website for people like me who are really very green to this kind of information? It seems clear it's more for people in the field already. The 'help' doesn't quite do it for me.

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Re: Scattering theory.

Post by Carl Willis »

Chris,

The "-0" suffix denotes the element in its natural state, often a mixture of isotopes. This is not so applicable to experiments done with natural hydrogen, which is almost entirely H-1, so I'm not surprised that there is no data for "H-0".

To address your previous question about the kinematics of a reaction in the CM frame versus the target frame, kinetic energy KE = m*v^2 / 2 where m is the particle mass and v is the magnitude of the velocity. Velocity depends on your reference frame. In the CM frame, a head-on DD collision involves particles in which each has velocity v(CM) of the same magnitude relative to reference, and a corresponding KE(CM) of m*v(CM)^2 / 2. Considering the same problem from a particle's perspective now, the other particle has a velocity of 2v(CM) while the reference particle is stationary. So the other particle has a KE(target) = m*[2v(CM)]^2 / 2, or 4*KE(CM). Cross-sections and yields are conventionally reported in the "lab" or target frame.

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Re: Scattering theory.

Post by Chris Bradley »

Thanks for clearing up the 'H-0' thing. Though, following the question through, how could I - a non-industry person - find that out for myself. I did look for some time before asking, for fear of asking a daft question.

To the CM question: Again, perhaps it's really another question of nomenclature rather than science - if you accelerate one particle up to 400keV, it will have 100keV kinetic energy (your x4 factor) in respect of a CM shared between itself and another similar lab-stationary particle. But that 'stationary lab particle' must also have 100keV kinetic energy in respect of the CM because the CM stays dead centre between them and moves at half the lab-framed velocity of the first particle, so both particles are moving towards the CM. So each has 100keV with respect to the CM. Hence, 200keV energy must be available in that collision if it is fully sticky and inelastic (viz. fusion), and they come to stop at the CM in the collision.

Now, I interpreted collision data as being converted FROM the original acceleration lab-derived data (in this example, 400keV) and turned into an equivalent collision energy (200keV) because that seems to be the most eminently sensible measure to use, and subsequent calculations of mine using that have come up with the same figures calculated out in other publications and materials I've read (that is, I only get confident about something when I've used the numbers myself). I think you are saying it's not converted into an equivalent collision energy, but infact is converted into an equivalent thermal bulk temperature (100keV) in which 200keV collisions would occur. That may be the case, I have no good reason to know otherwise, but it doesn't seem very sensible to talk about anything else other than the actual energy in the collision?

Is this correct? And if it is the equivalent bulk temperature quoted rather than collision energy - is there some way I might've been able to find this out for myself, without asking for it - or more importantly still, verify this is how the data has been handled?

best regards,

Chris MB.
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