Converting drive voltage into collision energy.

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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Converting drive voltage into collision energy.

Post by Chris Bradley » Mon May 05, 2008 10:06 am

Hi.

I've been picked up twice now in other threads for my conversion of drive voltage into collision energy (I say the latter is a half of the former, for unit charges like deuteron ions). If I understand them both, one says I am over by a factor of two, the other says I am under by a factor of 4.

Room for a specific discussion on this topic then!?

The actual collision energy experienced by particles driven through given voltages seems, to me, to be the most basic and important issue in trying to quantitatively enumerate the performance of IEC devices. So I'd like to nail this down well. At the moment, if I'm wrong I'm going to have to correct an awful lot of calculations.

Some of this may be actual scientific understanding, some may be 'just' nomenclatures. So I've broken it down into statements and you, dear reader, can let me know which you think are correct and which are wrong:

1. Let us say we have accelerated a deuteron ion through 100kV in the lab-frame and it approaches along a trajectory that brings it into a sticky collision (viz, fusion) with another, but lab-stationary, deuteron.

2. In actual numbers, this means that the relative velocity between them (as seen in any non-relativistic frame) is SQRT(2x100kVx1.6E19Jx1.6E-27kg) =~ 4,472,000m/s

3. The question is, what is the collision energy between these deuterons. I understand this term to mean the amount of kinetic energy converted into other [excitation] energy in that impact, i.e. it is the resultant kinetic energy loss.

4. If I am an observer at the centre of mass of these two deuterons, I will see them approaching at 2,236,000m/s, that is, each with a KE of 25keV.

5. They come into fused contact at their CofM (viz, where I am sitting) and thereafter neither have KE in my frame.

6. A total of 2 x 25keV has therefore been lost as kinetic energy.

7. The amount of energy released in the impact is 50keV to me as an observer at the CofM.

8. Try another frame; I am now an observer 'along for the ride' with one of the particles, so I see one particle as stationary. The other approaches at 4,472,000m/s with a KE of 100keV in my frame.

9. As they collide in front of me, they career off with a new velocity of 2,236,000m/s away from me (conservation of momentum). The particle that had been coming towards me has now lost 75keV, but the one I was with has just gained 25keV. Total loss = 50keV.

10. or another way, the resultant fused particle now has a mass twice that of a deuteron but is moving at 2,236,000, which again equals 50keV and therefore represents a loss of 100keV-50keV=50keV.

11. I could do this for any frame of reference, and still end up with a total KE loss of 50keV when a deuteron accelerated through 100kV comes into sticky collision with another deuteron not so accelerated.

12. You can repeat this for any drive voltage (V) with like particles and the KE loss is always ½ x V.e

If anyone doesn't agree, please identify which statement is in error. Statements of agreement also welcome!

Best regards,

Chris MB.

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Re: Converting drive voltage into collision energy.

Post by Chris Bradley » Mon May 05, 2008 11:07 am

Sorry, a quick self-correction: The absolute velocities were for protons, not deuterons. (Doesn't affect the final energy figures, though.)

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Re: Converting drive voltage into collision energy.

Post by Carl Willis » Mon May 05, 2008 3:55 pm

Chris,

This looks like a correct treatment.

The trick with this kind of discussion is to be very clear about what quantities are being considered. It's easy to lose track and calculate a wrong number or inaccurately read someone else's numbers. If one is talking about the kinetic energy of individual particles, the oncoming particle in the lab frame has 4 times the energy of a particle in the CM frame. If one is talking about system kinetic energy, then the system in the target frame has 2 times the energy of the system in the CM frame. When one comes to the crucial consideration of cross-sections and yields, it's important to understand that by default they are reported in the target frame. The energy of relevance to a cross-section is the "collision energy" as you put it, a function of the relative velocities between the particles. In fact, as you noticed, calculations are much more intuitive if you use relative velocity as the variable: find relative velocity from the acceleration voltage, then determine the target-frame energy corresponding to this relative velocity, and look up CS or TTY at that energy. On the other hand, the convention is to report differential angular cross sections with angle in the CM frame and energy in the lab frame.

Confusion abounds!

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley » Mon May 05, 2008 4:04 pm

So...if that's right so far, I understood reported fusion cross-section coefficients to be derived equivalent to CofM energy - having been converted from the target energy.

So in the case of DD's peak fusion cross section, at 1250keV and cross-section = 96millibarn, this is the collision/cross-section you'd get if you accelerate an initially lab-stationary deuteron through 2500kV of potential into another lab-stationary deuteron.

Y/N??

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Re: Converting drive voltage into collision energy.

Post by Carl Willis » Mon May 05, 2008 6:24 pm

>reported fusion cross-section coefficients to be derived equivalent to CofM energy - having been converted from the target energy.

It depends whose work you read. It may be that in astrophysics or fusion energy fields (not my fields, by the way), data gets reported in CM by default. I do know that if you look up cross-sections in CSISRS or another nuclear data service, and see energy "EN" without "CM" next to it, you're dealing with a lab-frame energy. If you see "EN-CM," then you have the center-of-momentum collisional energy being used. It's too bad researchers are sometimes not explicitly clear on this point and rely on what may be dubious convention.

>So in the case of DD's peak fusion cross section, at 1250keV and cross-section = 96millibarn, this is the collision/cross-section you'd get if you accelerate an initially lab-stationary deuteron through 2500kV of potential into another lab-stationary deuteron.

>Y/N??

N.

This cross-section is obtained from a nuclear data service like CSISRS, right? There's a nice evaluated cross-section courtesy of Liskien available there which shows the peak in the vicinity of 1.6 MeV(that would be in the target frame) at 106 mb. This 1.6 MeV implies that the deuteron beam is accelerated through the equivalent of a 1.6 MV potential difference.

The deuteron is moving relative to the other one at Vrelative = SQRT(2*1.6 MeV / Mdeuteron). In the CM frame, this deuteron has a velocity relative to the observer of Vrelative / 2. Its kinetic energy is 1/4 that of 1.6 MeV, but the system ("collision") kinetic energy is 1/2 of 1.6 MeV because there are two deuterons with equal energy in this system.

With a fusor, the operator is in the CM frame, a departure from the usual conditions for reported cross-sections. The particles of charge q are accelerated by a cathode voltage of U volts and the system energy for a head-on collision is 2*U*q. But this is not where you look up the cross-section. You'd look that up at 4*U*q. Example would be my fusor "Carl's Jr." running at U = 70 kV. Each deuteron ideally ends up in the center at 70 keV. The system energy in a head-on collision is 2*70 keV = 140 keV, but I look up the cross-section for this in CSISRS or similar at 4*70 keV = 280 keV. This is one of the advantages of arranging head-on, center-of-momentum collisions in the lab--you can get by with lower voltages (and smaller feedthroughs, transformers, etc). As I think you pointed out in another post though, energy is conserved regardless of what frame you do the calculation in.

This horse has been near beat to death in previous discussions on this board, dating way back. I suspect if you search "center of momentum" or similar, you'll dredge up the goods.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley » Mon May 05, 2008 7:19 pm

It is relevant of you to draw out that astrophysical texts differ - the sources I've read are, indeed, generally such texts.

I have also paid particular regard to a freely down-loadable chapter from a book on nuclear fuel reactions, which is also explicitly CM oriented; http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf

Further, you have been talking about head-to-head collisions, both particles moving, whereas I have been talking about a beam into a stationary particle. I suspect we've been discussing convention of notations, then, and nothing more.

May I please test this, then, by setting one final agreed 'datum' of data before closing the thread. By way of example (using differential masses and a specific energy) there is a peak fusion resonance for p-11B at a lower energy than its main peak, at 148keV (CM). To accelerate a proton into a lab-stationary, fixed, 11B target and to get to this resonant cross-section, the proton needs to be moving with a velocity of 5,682,500m/s towards a target 11B nucleus and thus the proton needs to be accelerated though a lab potential of 161.5kV to get it to that velocity in the boron's stationary lab-frame.

Is this right?

best regards,

Chris MB.

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Re: Converting drive voltage into collision energy.

Post by Carl Willis » Mon May 05, 2008 7:46 pm

Chris,

Without going through the paces of doing the whole problem myself, I'm inclined to agree that this sounds right. The boron is a heavy nucleus compared to the proton, and of course in the limit that the total system mass is no different than the target mass, the CM and lab frames are the same. The lab energy you found for the proton in the lab frame is only slightly higher than the CM frame, so it looks right.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley » Mon May 05, 2008 8:27 pm

Thanks, Carl. That one's done now. I think it was just the conventions of notation/terminology that were the issue, not the science.

Please keep a friendly look out for any similar mis-interpretations of mine in other threads - I am working from first principles here and it is quite probable that there are many other commonly held conventions of which I am not remotely aware and need guidance/correction on.

best regards,

Chris MB.

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Re: Converting drive voltage into collision energy.

Post by Richard Hull » Tue May 06, 2008 2:57 pm

Thanks are due to Carl for belaboring this point to the bitter end. I was originally very confused about this myself back on songs and actually went to the old Ask a Scientist at the princeton fusion website. I was given careful tutelage on this very issue and reported it in Songs.

It comes back to what Carl noted. Who is doing the talking? Where you are picking up your cross sections from? What the particles under discussion are doing. Even after all this, many remain confused. (including some pro's trying to cross over to other disciplines.)

For us, it is simple. Using the fusion cross section charts mentioned by Carl that were derived for fusion research, the defaualt in our fusors demands for the most ideal and fortuitous collisonal scenario that the cross section be read at 4*V applied*q.

To amplify on Carl. This is the sole great glory of the fusor, as we use it. Most every other aspect of the fusor is a loser or rife with issues and problems due to its extreme simplicity. So, extreme simplicity and a super cross section achieved per unit voltage applied are the attractors for the amateur community.

Richard Hull
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Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

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Re: Converting drive voltage into collision energy.

Post by Chris Bradley » Fri May 09, 2008 4:16 pm

I am sad that you find this boring, but as you say it is about misunderstanding the context.

I was originally picked up for dividing the fusor drive voltage by 2 to get the collision energy. I was unequivocal with the scenario I was calculating for. It was a lab-fast particle into a lab-stationary particle. I was therefore right to divide the drive voltage by two.

The subsequent issue was then discussed because the presumption was that collisions in the fusor are lab-fast into lab-fast, with the lab at rest with the CofM. This would clearly be a <drive voltage x 4> factor, as was then raised as a correction to my submission.

In point of fact, I cannot see where we actually disagreed with the calculation after that, it was just this misunderstanding of setting, not of science.

A factor of x8 between two results may be here-nor-there with a fusor because it is a case of bunging in a good load of 10's of kV to get the best performance, and it always will be.

Though this may satisfy the practical, experimental interests of fusioneers, it does not satiate my desire to understand exactly what is happening.

I believe it is quite wrong to be so sure that fusions in the fusor are lab-fast into lab-fast (the x4 drive voltage factor). Some may happen this way, but I believe that the majority are lab-fast into lab-stationary (which is therefore the x0.5 factor).

I can provide the basis of a proof either way by means of the very calculation which has been boringly discussed here:

Let us compare the neutron outputs of two otherwise identically operated fusors, one operating at 20kV, the other at 10kV ('10kV!!' you may say... exactly my point, please read on..):

A) If the fusor operates lab-fast into lab-fast particles, the collision energy is 80keV for the higher drive voltage and 40keV for the lower voltage. Using an astrophysical factor of S(0)=56 and Gamow energy of 986kV to calculate cross sections for DD (as the experimental data is a bit sparse in the lower energy ranges), we end up with cross-sections of 10millibarn for the fusor collisions operating at 10kV and 20millibarn for that operating at 20kV. Recall that reaction rate is also related to velocity (which therefore differs by SQRT(2)). This means that whatever the neutron rate you get out of a fusor operating at 20kV, you should also be able to achieve 35% of that rate at 10kV. (I presume this means that 10kV rates should be within the possibility of being measured? - Just run it for (less than) 3 times as long.)

B) If the fusor operates lab-fast into lab-stationary (i.e. only collisions with background particles) then the collision energies are then 10keV and 5keV which gives cross sections of 0.27millibarn and 9microbarn respectively. This means that comparing neutron rates at 10kV drive voltage should be (including velocity factor) about 3% what you would get at 20kV drive voltage, which I suspect would be generally below measurement capability, or a very long run.

Summary: I do not have the practical data to show one way or the other, but if neutron rates at 20kV for a given time period are detectable but neutron rates at 10kV, by the same kit, are not measurable with a run of just 3 times the duration of the first, then this is at least prima facie evidence that the collisions are not lab-fast into lab-fast, and that the lab is not in the CofM frame for the fusion collisions.

I have been digging around to find any one's reported neutron output rates below 10kV but to no avail. I therefore have to suspect that the fusor does, indeed, function by lab-fast into lab-stationary particles as otherwise I'd expect measurable rates in the 5-10kV range because they would be fairly comparable and within the same order of magnitude as that with, say, a 20kV drive voltage. I will tend to think this is correct, unless there is experimental evidence to the contrary.

best regards,

Chris MB.

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