By the way, I had in mind that this discussion happened before and I found it:
viewtopic.php?p=13981
P-B11 has 2 high energy alphas, not 1
You're assuming something about the internal process that leads to the three-body fission; namely, that it is perfectly symmetric with respect to all three daughter particles. This is not a given. Certainly the event that kicks it off exhibits no such symmetry...chrismb wrote:Then it HAS to be two binary fissions. Which is what the conventional wisdom says. What other outcome can there possibly be, other than a 3-body fission, or two binary fissions!?!?Giorgio wrote:What I mean is that experiments are not finding 3 Alphas splitting with the same energy, hence, just by looking at experimental data is clear that the process cannot be as the one you described.
Momentum and energy can be simultaneously conserved without all three alphas having equal CoM energies. This is because, even if you assume the fission takes place on a plane (for just three particles, anything else does violate conservation of momentum) and concern yourself only with the relative angles between particles, there remain five free variables and only three constraints. (In other words, the "high-energy" alphas don't even need to have equal energies.) Therefore you have to invoke an intranuclear particle physics argument in order to maintain your point. You have not done this.
It is possible, as far as I can tell, that the distributions as observed could be produced by two binary fissions. I haven't studied the problem closely enough to determine whether or not the excited beryllium would have to fission perpendicular to the axis of the initial split or whether it could be in a random direction, but the former at least would yield the observed results. However, without some intranuclear explanation of why you believe a trinary fission would necessarily be symmetric, you have no demonstrated basis for claiming that dual binary fission is the only way to explain the results.
No further explanations are required. This is basic mechanics. A three element 'explosion' of equal masses of any object will result in a uniform, symmetrical 120deg ejection of such parts in a 2D plane.93143 wrote:However, without some intranuclear explanation of why you believe a trinary fission would necessarily be symmetric, you have no demonstrated basis for claiming that dual binary fission is the only way to explain the results.
It's just conservation of momentum at work. Resolve in any axis and zero net momentum must remain. This means symmetry of the projected parts.
In two binary fissions, the geometry and dynamics of the second are independent of the first (excepting for the relative velocity of the CoM of the second fission).
I can not take away much from that previous thread except for two points.
Apparently the experimental results for P-B11 fusion and subsequent fission processes is lean and there is wiggle room for data reliability and interpretation. I don't know whether this later experiment and discussion has any more validity or precision.
The thread also discuses some concern about alpha ash buildup. This was before Nebel's revelation of the subsequently obvious situation of the fusion derived alphas having too much kinetic energy to be retained by the potential well, so they will only be contained by the Wiffleball cusp confinement of a few thousand passes. The electrostatic potential well confinement (helped by annealing) can apparently confine fuel ions indefinitely- or at least long enough for almost all to fuse). The fuel ions may dwell inside the magrid for millions of passes or time frames of 10's of milliseconds. Compare this with perhaps a few thousand passes or ~ <0.1 milliseconds (without recirculation) for the electrons. The fusion alphas would have similar containment times (in terms of transits) as the electrons. And that these alphas would not hit the magrids provided that the B field strength between the Wiffleball border and the magnet casing was great enough so that the alphas gyroradii was less than the distance to the magrid casings ( my recolection is that this would be ~ >3.5 T in a production sized Polywell). The alphas would only exit via the cusps, and do so without interacting much with the fuel ions.
If one of the alphas have ~ 700KeV of energy, this situation may be more uncertain, mostly because of the increased Coulomb collisionality of these lower energy alphas.
Dan Tibbets
Apparently the experimental results for P-B11 fusion and subsequent fission processes is lean and there is wiggle room for data reliability and interpretation. I don't know whether this later experiment and discussion has any more validity or precision.
The thread also discuses some concern about alpha ash buildup. This was before Nebel's revelation of the subsequently obvious situation of the fusion derived alphas having too much kinetic energy to be retained by the potential well, so they will only be contained by the Wiffleball cusp confinement of a few thousand passes. The electrostatic potential well confinement (helped by annealing) can apparently confine fuel ions indefinitely- or at least long enough for almost all to fuse). The fuel ions may dwell inside the magrid for millions of passes or time frames of 10's of milliseconds. Compare this with perhaps a few thousand passes or ~ <0.1 milliseconds (without recirculation) for the electrons. The fusion alphas would have similar containment times (in terms of transits) as the electrons. And that these alphas would not hit the magrids provided that the B field strength between the Wiffleball border and the magnet casing was great enough so that the alphas gyroradii was less than the distance to the magrid casings ( my recolection is that this would be ~ >3.5 T in a production sized Polywell). The alphas would only exit via the cusps, and do so without interacting much with the fuel ions.
If one of the alphas have ~ 700KeV of energy, this situation may be more uncertain, mostly because of the increased Coulomb collisionality of these lower energy alphas.
Dan Tibbets
To error is human... and I'm very human.
That is not true; that only works for a two-element explosion of equal masses. Watch this:chrismb wrote:A three element 'explosion' of equal masses of any object will result in a uniform, symmetrical 120deg ejection of such parts...
Particles: alpha_1, alpha_2, alpha_3
Fission energy: E = 8.68 MeV
Free variables: v_1, v_2, v_3, theta_12, theta_13
Constraints:
v_1 + v_2*cos(theta_12) + v_3*cos(theta_13) = 0 (conservation of momentum in the axis of motion of alpha_1)
v_2*sin(theta_12) + v_3*sin(theta_13) = 0 (conservation of momentum perpendicular to said axis)
v_1^2 + v_2^2 + v_3^2 = 2*E/m_alpha (conservation of energy)
Let theta_12 = 150° and theta_13 = 250°, to make the number of unknowns equal to the number of constraints.
Then
v_1 - 0.866*v_2 - 0.342*v_3 = 0
-> v_1 = 0.866*v_2 + 0.342*v_3
0.5*v_2 - 0.940*v_3 = 0
-> v_2 = 1.879*v_3
(1.970*v_3)^2 + (1.879*v_3)^2 + v_3^2 = 418.6
-> v_3 = 7.054 Mm/s (1.032 MeV)
-> v_2 = 1.879*7.054 = 13.26 Mm/s (3.645 MeV)
-> v_1 = 1.970*7.054 = 13.89 Mm/s (4.003 MeV)
Check:
13.89 + 13.26*cos(150) + 7.054*cos(250) = 0
13.26*sin(150) + 7.054*sin(250) = 0
13.89^2 + 13.26^2 + 7.054^2 = 418.6
or
1.032 + 3.645 + 4.003 = 8.68 MeV
See? Conservation of momentum and energy, just by picking two angles at random. I could have picked a speed and an angle, or two speeds, and the system would still be solvable. Or I could pick an angle and assume two of the speeds are equal, which is what I did to come up with my approximate picture of what the paper describes. So long as I add two constraints before trying to solve the system, it will work out.
You can't justify 120° symmetry without making further assumptions about the mechanism of the explosion. Conservation of momentum and energy of the fragments won't do it.
This may well be true, and I have assumed it in the past when attempting to calculate the energy spectrum from the one-high, two-low model. However, it inherently requires the assumption that the internal processes in the excited nucleus that result in its fission are, or become, geometrically uncorrelated with the axis of fission with the recently-ejected alpha. This is a particle-physics consideration, and should be supported by citation or analysis.In two binary fissions, the geometry and dynamics of the second are independent of the first (excepting for the relative velocity of the CoM of the second fission).
Yes. And it is quite obvious as well because [all that was necessary to say] whether you had a black box with two asymmetric binary fissions or one three-body fission, you'd also see the resultant as three particles with different directions and speeds.93143 wrote:See? Conservation of momentum and energy, just by picking two angles at random.
So, I admit I was quite wrong to suggest the 120 deg even scatter was a consequence of 3 body fissions. I will edit and withdraw that statement. I wasn't thinking straight. I was thinking of the stochastic description of such a scenario, that you'd get an even distribution, then did a brain-blip and turned that into meaning it was a 120 scatter.
If it is any sort of mitigation, as I wrote that last one, I thought 'hmm... not sure about that because I have used pusher fans with asymmetric, but balanced, blades on before' and that situation is numerically (torque == momentum) similar. But I got a call for beer down the pub and so didn't get to do the check. I trust that is a satisfactory excuse!!!
However (!)... I'll post the 'however' later... bed time beckons...
Sorry for the brain-blip.
But ONLY if you assume perfect symmetry to begin with. Since the starting momenta are a-symmetrical, the resulting momenta will be.chrismb wrote: No further explanations are required. This is basic mechanics. A three element 'explosion' of equal masses of any object will result in a uniform, symmetrical 120deg ejection of such parts in a 2D plane.
It all depends on who called you for that beer.chrismb wrote:If it is any sort of mitigation, as I wrote that last one, I thought 'hmm... not sure about that because I have used pusher fans with asymmetric, but balanced, blades on before' and that situation is numerically (torque == momentum) similar. But I got a call for beer down the pub and so didn't get to do the check. I trust that is a satisfactory excuse!!!
Post a pic of her and we will see if it is a satisfactory excuse!
It happens. I've argued on the basis of Gauss' Law, only to have it pointed out that Gauss' Law didn't apply to the geometry in question the way I was trying to apply it - not once, but twice, on this very website...
It's hard for me to tell sometimes; my Internet kyuushojutsu is pretty weak. I figured I'd let the math speak for me, rather than risk a shouting match over what had to be a simple misunderstanding.[all that was necessary to say]
Gauss is a slippery bastard sometimes, even in my limited experience.
At 750kEv I'm starting to wonder about ash. If the high-energy alphas are making about 1,000 transits (iirc) then how long will a low-energy alpha get stuck in the Magrid? That should be solvable. But then again this doesn't apply to PWs anyway.
At 750kEv I'm starting to wonder about ash. If the high-energy alphas are making about 1,000 transits (iirc) then how long will a low-energy alpha get stuck in the Magrid? That should be solvable. But then again this doesn't apply to PWs anyway.
Heh, maybe they're planning to operate an FRC at 675KeV.It is interesting that the work was partly sponsored by Tri Alpha Energy
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
I think that ~750 KeV alphas should exit about as quickly (in terms of passes/ total distance traveled) as any higher energy alpha, with two conditions. The KE must be higher than the potential well so that the only containment is the cusp leaking magnetic field. Secondly, the KE must not be so much that the particle penetrates the magnetic field and hits the magnet surface (the gyroradii must be less than the distance from the Wiffleball border to the magrid surface at a given magnetic field strength).
What will be different is the mean free path of the alphas. If the previously accepted lower alpha energy was ~ 2.4 MeV, this new low energy alpha would be ~3.3 times less than this. Coulomb collisionality crossections scales at ~ the inverse square of the energy (or is that velocity?) so if the MFP of the 2.4 MeV alpha is 1000 meters in a Polywell (perhaps only ~ 1 collision before escape), the MFP of a ~750 KeV alpha will be ~ 100 meters. This would mean it might have ~ 10 collisions before escape. Is that enough to cause some thermalization with other particles within the plasma? Is it significantly good or bad? Could the alpha be down scattered enough that it would present an ash problem?
Because of the relative masses and the Z of the alpha the interaction with other protons, borons and alphas would be different. Collisions with protons would not slow the alpha as much and would increase the speed of the protons (possibly good?). Collisions with borons would tend to slow the alphas more and accelerate the borons (possibly bad from a bremsstrulung standpoint), and both heat and cool other low energy alphas (not much interaction with the 4 MeV alphas).
This will change the the optimal knob positions on the machine (drive voltage, etc.) at the least and may lower the predicted optimal performance. You may need less electron potential and the ions may be proportionately hotter than the electrons at the same net temperature. This may be good. But the plasma may be more thermalized and this may be bad. Instead of being a purely power amplifying device , it may have some self heating baby ignition properties.
The electrons may or may not be accelerated over their lifetimes. The saving grace may be the relatively high loss rate of electrons, especially upscattered electrons. This, plus recirculation may allow for tolorable energy losses, and increasing more important, the inhibition of electron energy runaway.
Dan Tibbets
What will be different is the mean free path of the alphas. If the previously accepted lower alpha energy was ~ 2.4 MeV, this new low energy alpha would be ~3.3 times less than this. Coulomb collisionality crossections scales at ~ the inverse square of the energy (or is that velocity?) so if the MFP of the 2.4 MeV alpha is 1000 meters in a Polywell (perhaps only ~ 1 collision before escape), the MFP of a ~750 KeV alpha will be ~ 100 meters. This would mean it might have ~ 10 collisions before escape. Is that enough to cause some thermalization with other particles within the plasma? Is it significantly good or bad? Could the alpha be down scattered enough that it would present an ash problem?
Because of the relative masses and the Z of the alpha the interaction with other protons, borons and alphas would be different. Collisions with protons would not slow the alpha as much and would increase the speed of the protons (possibly good?). Collisions with borons would tend to slow the alphas more and accelerate the borons (possibly bad from a bremsstrulung standpoint), and both heat and cool other low energy alphas (not much interaction with the 4 MeV alphas).
This will change the the optimal knob positions on the machine (drive voltage, etc.) at the least and may lower the predicted optimal performance. You may need less electron potential and the ions may be proportionately hotter than the electrons at the same net temperature. This may be good. But the plasma may be more thermalized and this may be bad. Instead of being a purely power amplifying device , it may have some self heating baby ignition properties.
The electrons may or may not be accelerated over their lifetimes. The saving grace may be the relatively high loss rate of electrons, especially upscattered electrons. This, plus recirculation may allow for tolorable energy losses, and increasing more important, the inhibition of electron energy runaway.
Dan Tibbets
To error is human... and I'm very human.