P-B11 has 2 high energy alphas, not 1
You are perfectly right. And I feel ashamed that I didn't consider the impossibility of conserving momentum while thinking about that possibility.93143 wrote:That wouldn't conserve momentum. If the high-energy alphas are assumed to have 4 MeV each, as stated, the "low-energy" alpha in that case would have to come off with almost twice the velocity of either of the high-energy alphas, or about 13 MeV. There isn't enough energy in the system to do that. If you constrain the energy to 8.7 MeV, you find that the "high-energy" alphas come off with about 1.6 MeV each, compared to the "low-energy" alpha's 5.4 MeV...
If, on the other hand, you assume the angle between the two high-energy alphas is 155° and constrain the total k.e. in the reaction CoM rest frame to 8.7 MeV, the high-energy alphas come off with just under 4 MeV, and the low-energy alpha is left with about 750 keV. So it looks like that's what they're describing.
Upon reading the paper in a more focused way it was also clearly described that are the two energetic alphas that are incident of 155°.
Well, fortunately wiser heads have prevailed. My total energy yield was obviously wrong. Also the mention of the ~ 700KeV alpha is explained. I had a flawed view.
It does seem though that the reaction was not expected to vary at lower collision energies, and thus that this reaction mechanism might be consistent over the broad crossection range where significant fusion rates can occur.
The total Q ratio would not change, but the direct conversion considerations would. As far as the radiation damage to structures of two ~4 MeV alphas and one 700KeV alpha versus one ~4 MeV alpha and two 2.4 MeV Alphas , the effect would be ~ similar (due to the same total amount of energy ?).
The radiation consideration for people outside the machine is non existent as the alphas will only penetrate a small fraction of an inch of the vacuum vessel wall (this ignores possible secondary radiation effects). The gamma ray dose would be the overriding concern here and presumably this branch reaction probability is unchanged.
One possible consideration is Nebel's mention that the alphas (at 4 and 2.4 MeV) do not heat the plasma because their Coulomb collisionality and dwell time before escape is two small. The ~ 700 KeV alpha may be a different matter though as it has only perhaps 2-3 times the energy of the fuel B11 ions*. These lower energy alphas may have marginally to moderate capacity for heating the plasma. Would this be good or bad? It might reduce the input power needed, but compromise the monoenergetic nature of the plasma, and worsen Bremsstrulung, etc.
* I'm unsure of the desired energy of the B11 ions in a Polywell. It depends on whether the low energy resonance peak is targeted and confusing issues with the contributions of the protons and boron ions on the targeted center of mass KE, proton to boron ratios, confluence ( igher percentage of beam - beam fusions), etc.
Dan Tibbets
It does seem though that the reaction was not expected to vary at lower collision energies, and thus that this reaction mechanism might be consistent over the broad crossection range where significant fusion rates can occur.
The total Q ratio would not change, but the direct conversion considerations would. As far as the radiation damage to structures of two ~4 MeV alphas and one 700KeV alpha versus one ~4 MeV alpha and two 2.4 MeV Alphas , the effect would be ~ similar (due to the same total amount of energy ?).
The radiation consideration for people outside the machine is non existent as the alphas will only penetrate a small fraction of an inch of the vacuum vessel wall (this ignores possible secondary radiation effects). The gamma ray dose would be the overriding concern here and presumably this branch reaction probability is unchanged.
One possible consideration is Nebel's mention that the alphas (at 4 and 2.4 MeV) do not heat the plasma because their Coulomb collisionality and dwell time before escape is two small. The ~ 700 KeV alpha may be a different matter though as it has only perhaps 2-3 times the energy of the fuel B11 ions*. These lower energy alphas may have marginally to moderate capacity for heating the plasma. Would this be good or bad? It might reduce the input power needed, but compromise the monoenergetic nature of the plasma, and worsen Bremsstrulung, etc.
* I'm unsure of the desired energy of the B11 ions in a Polywell. It depends on whether the low energy resonance peak is targeted and confusing issues with the contributions of the protons and boron ions on the targeted center of mass KE, proton to boron ratios, confluence ( igher percentage of beam - beam fusions), etc.
Dan Tibbets
Last edited by D Tibbets on Thu Apr 07, 2011 5:05 pm, edited 1 time in total.
To error is human... and I'm very human.
I can't really buy any of this.
The issues seem simple to me; either there are two binary fissions, or a single, 3-alpha fission. (Are there any other possibilities??)
If there is a 3-alpha fission, then the emitted particles would be at 120 degrees to each other, and have the same energy. There is no way, kinetically, that 3 particles of like mass can emerge from a single CoM 'event' but with different energies.
If there is any asymmetry in the energy or direction, I do not see how kinetics offers any explanation for anything other than two binary fissions.
Now, if there were to be a mix of 120 degree, 3-way alpha fissions and double binary fissions, then the resultant combined distributions may make an appearance of some other awkward mix, but it is clear that it is not [just, nor dominantly] a 3-way fission.
If anyone thinks I have misunderstood the issues, please let me know.
Edit: yes, I did misunderstand/mis-state the kinetics issue. Disregard the bollocks above, see below where I am put right.
The issues seem simple to me; either there are two binary fissions, or a single, 3-alpha fission. (Are there any other possibilities??)
If there is a 3-alpha fission, then the emitted particles would be at 120 degrees to each other, and have the same energy. There is no way, kinetically, that 3 particles of like mass can emerge from a single CoM 'event' but with different energies.
If there is any asymmetry in the energy or direction, I do not see how kinetics offers any explanation for anything other than two binary fissions.
Now, if there were to be a mix of 120 degree, 3-way alpha fissions and double binary fissions, then the resultant combined distributions may make an appearance of some other awkward mix, but it is clear that it is not [just, nor dominantly] a 3-way fission.
If anyone thinks I have misunderstood the issues, please let me know.
Edit: yes, I did misunderstand/mis-state the kinetics issue. Disregard the bollocks above, see below where I am put right.
Last edited by chrismb on Thu Apr 07, 2011 11:22 pm, edited 1 time in total.
You are making the same mistake I made above.chrismb wrote:I can't really buy any of this.
The issues seem simple to me; either there are two binary fissions, or a single, 3-alpha fission. (Are there any other possibilities??)
If there is a 3-alpha fission, then the emitted particles would be at 120 degrees to each other, and have the same energy.
Think of the system as a Newton cradle, where the last 3 balls touch each other and form a triangle. One of the vertex is in line with the other ball of the cradle, thus the other 2 ball will form an angle in respect to the line of the cradle.
When the cradle strikes the 2 end ball will get the biggest part of the momentum and leave the cradle at a split angle.
The third ball of the triangle (that transferred the momentum from the cradle line to the other 2 balls) is left with less momentum.
Total energy and momentum is conserved, and momentum distribution among the balls varies according to the angle at which the 2 end balls leave the cradle.
Hope it makes sense.
Nope. The two systems are quite different. In p11B the energy for the particle's trajectory arise from energy at the CoM. It is a fusion reaction. In Newton's cradle, the energy for the trajectory of the particles comes from kinetic energy external to the particles that are 'trajected'.Giorgio wrote:You are making the same mistake I made above.
Think of the system as a Newton cradle, where the last 3 balls touch each other and form a triangle. One of the vertex is in line with the other ball of the cradle, thus the other 2 ball will form an angle in respect to the line of the cradle.
...
Hope it makes sense.
If you wish to, more thoroughly, account for the KE of the incoming particle, well, that is quite simple. With an addition of a given amount of KE vectored in the direction of the beam [because all that happens is the proton fuses, thereby exactly imparting that KE to the system, prior to the equal fission products emitted from that fusion 12C*], what we would then end up with is the addition of that velocity to the three equal emissions of alphas in all, random, direction. This would give a single mode and a symmetrical distribution around it, the half-width of which would be the KE of the incoming proton.
Incorrect.chrismb wrote:If there is a 3-alpha fission, then the emitted particles would be at 120 degrees to each other, and have the same energy. There is no way, kinetically, that 3 particles of like mass can emerge from a single CoM 'event' but with different energies.
Assuming 2 particles diverge at equal energy with an angle of A <= 180, and the 3rd departs in the opposite direction of that angles bisector, there is a specific ratio of energy between the first 2 particles and the 3rd to conserve momentum.
Aligning our coordinate system with the X axis along the trajectory of the 3rd particle, the first 2 particles have momentums <x1, +/- y1>, and the 3rd has momentum <-x1, 0>. For any given angle A, mass, and total energy, there are values x1 and y1 that satisfy conservation of momentum and energy.
As a special case example, 2 particles departing in opposite directions leaving the 3rd at CoM rest satisfies conservation of momentum.
I was trying to point out is that it is a problem of mass and vectors (i.e. Kinetic energy and momentum), not only of mass (i.e. Kinetic energy).chrismb wrote:Nope. The two systems are quite different. In p11B the energy for the particle's trajectory arise from energy at the CoM. It is a fusion reaction. In Newton's cradle, the energy for the trajectory of the particles comes from kinetic energy external to the particles that are 'trajected'.
To work the way you describe, the system getting the extra KE energy should be already composed by 3 Alpha packed together, the distribution of the extra KE should than be uniform over the 3 particle, and finally the rupture line of the system should also be in the exact center of mass of the system.chrismb wrote:If you wish to, more thoroughly, account for the KE of the incoming particle, well, that is quite simple. With an addition of a given amount of KE vectored in the direction of the beam [because all that happens is the proton fuses, thereby exactly imparting that KE to the system, prior to the equal fission products emitted from that fusion 12C*], what we would then end up with is the addition of that velocity to the three equal emissions of alphas in all, random, direction. This would give a single mode and a symmetrical distribution around it, the half-width of which would be the KE of the incoming proton.
What we have instead is a system composed by two Alpha and a third potential alpha missing a neutron.
When the incoming neutron is captured the third Alpha that is formed is acting like the ball in the line of the cradle, with consequent distribution of energy as hanelyp described.
That is exactly what happens. It's called fusion. The proton fuses with the 11B to make a 12C* - 3 alphas just itching to split up from the intertial frame of that 12C*.Giorgio wrote:To work the way you describe, the system getting the extra KE energy should be already composed by 3 Alpha packed together, the distribution of the extra KE should than be uniform over the 3 particle, and finally the rupture line of the system should also be in the exact center of mass of the system.
The only other possibility, which seems to be what you are edging towards, is that two neutrons and a proton jump out of the 11B towards the proton, leaving a 8Be behind. That would not be p+11B fusion.
Last edited by chrismb on Thu Apr 07, 2011 1:50 pm, edited 1 time in total.
That IMHO is not what experiments are finding.chrismb wrote:That is exactly what happens. It's called fusion. The proton fuses with the 11B to make a 12C* - 3 alphas just itching to split up from the intertial frame of that 12C*.Giorgio wrote:To work the way you describe, the system getting the extra KE energy should be already composed by 3 Alpha packed together, the distribution of the extra KE should than be uniform over the 3 particle, and finally the rupture line of the system should also be in the exact center of mass of the system.
A two fission process, yes that is my very limited understanding. The only difference from the classical view is that the total momentum of the three alphas is rearranged. Instead of a 4 MeV alpha, followed by the residual B8 isotope/ isomer splitting into two 2.4 MeV alphas. Instead the C12* splits into a 700KeV alpha, and the resultant B8* splits into two 4MeV alphas. They then go on to explain how the B8 isomer can deliver this energy (I think, I am unfamiliar with the jargon and details of the possible isomers, their energy and decay processes that is possible for C12* and B8* intermediates).
Dan Tibbets
Dan Tibbets
To error is human... and I'm very human.
But unless there is an direct interaction with the .7MeV α, how are the two 4MeV αs constrained to tha 155°? Wouldn't they average out around the sphere?D Tibbets wrote:A two fission process, yes that is my very limited understanding. The only difference from the classical view is that the total momentum of the three alphas is rearranged. Instead of a 4 MeV alpha, followed by the residual B8 isotope/ isomer splitting into two 2.4 MeV alphas. Instead the C12* splits into a 700KeV alpha, and the resultant B8* splits into two 4MeV alphas.