bjschaeffer
Thinker
- Joined
- Apr 13, 2012
- Messages
- 148
And you have the sign wrong.
The sign is not wrong, is only omitted for conveneince. Every body knows that a binding energy has to be negative.
And you have the sign wrong.
Helium-4 is unbound because it has less binding energy per nucleon than 3H + n.
You made a mistake, you make a confusion between binding energy per nucleon and total binding energy.
Helium-4 is the α particle very strongly bound.
Nuclear Physics A Volume 747, Issues 2–4, 24 January 2005, Pages 362–424. Inputs: QCD, chiral field theory, high energy nucleon scattering data. Outputs: deuteron bound state energy, quadrupole moment, radius, D/S ratio. And it gets them right.
Here's the funny thing, bjschaeffer. You're standing here rejecting QCD and saying (a) it doesn't seem to apply at low energy, and (b) it's fit to the data.
Do you remember what happened, earlier in this thread, when people pointed out problems with your theory? Your theory is, of course, fit to the data, and your theory doesn't seem to apply---worse than that, it's disproven---in any basic scattering experiment.
Funny that you should be perfectly happy with this state for your own theory (do vague fits at low energies, and flatly refuse to consider other energies), and reject the far-superior state of QCD (very precise at high energy, applied at other energies with the help of data-driven effective theories). Huh.
Who is ths Professor Brodsky?I have the proof from Professor Brodsky that the binding energy of the simplest nucleus beyond the proton has never ...
The two-nucleon system at next-to-next-to-next-to-leading orderNuclear Physics A Volume 747, Issues 2–4, 24 January 2005, Pages 362–424. Inputs: QCD, chiral field theory, high energy nucleon scattering data. Outputs: deuteron bound state energy, quadrupole moment, radius, D/S ratio. And it gets them right.
We consider the two-nucleon system at next-to-next-to-next-to-leading order (N3LO) in chiral effective field theory. The two-nucleon potential at N3LO consists of one-, two- and three-pion exchanges and a set of contact interactions with zero, two and four derivatives. In addition, one has to take into account various isospin-breaking and relativistic corrections. We employ spectral function regularization for the multi-pion exchanges. Within this framework, it is shown that the three-pion exchange contribution is negligibly small. The low-energy constants (LECs) related to pion–nucleon vertices are taken consistently from studies of pion–nucleon scattering in chiral perturbation theory. The total of 26 four-nucleon LECs has been determined by a combined fit to some np and pp phase shifts from the Nijmegen analysis together with the nn scattering length. The description of nucleon–nucleon scattering and the deuteron observables at N3LO is improved compared to the one at NLO and NNLO. The theoretical uncertainties in observables are estimated based on the variation of the cut-offs in the spectral function representation of the potential and in the regulator utilized in the Lippmann–Schwinger equation.
Who is ths Professor Brodsky?
Why have you not included the email that he relied to to provide context?
He is not one of the authors of the paper that ben m cited
The two-nucleon system at next-to-next-to-next-to-leading order
E. Epelbaum, W. Glöckle, Ulf-G. Meißner
Professor Brodsky is here
http://www.thp.univie.ac.at/english/research/particle/part-home.htm
FYI, bjschaeffer:
So you remain wrong.
- "a two-nucleon system" is the simplest nucleus beyond the proton.
- "chiral effective field theory" is QCD.
- "deuteron observables" includes the binding energy according to ben m.
This has nothing to do with the abysmal failures of your idea.
Stan Brodsky is a former head of the theory group at the Stanford Linear Accelerator Center, which he joined in 1966.Who is ths Professor Brodsky?
He like you believe that a neutron is a nucleon, the same thing as a proton which is wrong because their electric and magnetic properties are different.

Wikipedia said:In chemistry and physics, a nucleon is one of the particles that makes up the atomic nucleus....There are two kinds of nucleon: the neutron and the proton.
...
I didn't obtain this standards, presumably inexistent.
...
Stan Brodsky is a former head of the theory group at the Stanford Linear Accelerator Center, which he joined in 1966.
Complaining about his paper's rejection by the European Physical Journal A, bjschaeffer wrote:
From Wikipedia's current article on nucleons:
Disputing the accepted definition of nucleon would be one way to avoid meeting the scientific standards of any reputable journal.
It is not Brodsky but Meissner who rejected my paper. He doesn't distinguish between protons and neutrons :
"We consider the two-nucleon system at next-to-next-to-next-to-leading order (N3LO) in chiral effective field theory." N means nucleon. see here:
http://www.sciencedirect.com/science/article/pii/S0375947404010747
It is not Brodsky but Meissner who rejected my paper. He doesn't distinguish between protons and neutrons
Your inability to understand "nucleons" is striking.
The proton and neutron are identical in their strong interactions. They differ in electromagnetic and weak interactions. Therefore, it makes sense to describe a single strong-interacting object ("the nucleon") which has two subtypes (an isospin doublet) of different charges---the same as we do for the pion (three subtypes, an isospin triplet), the delta baryons (a quadruplet), etc.
I understand this is not how your model works. But it is long known, and works just fine, in the mainstream model.

in the context of n3lo, n stands for next to. Hence n3 stands for next-to-next-to-next-to and n3lo stands for next-to-next-to-next-to leading order.
your inability to understand "nucleons" is striking.
The proton and neutron are identical in their strong interactions. They differ in electromagnetic and weak interactions. Therefore, it makes sense to describe a single strong-interacting object ("the nucleon") which has two subtypes (an isospin doublet) of different charges---the same as we do for the pion (three subtypes, an isospin triplet), the delta baryons (a quadruplet), etc.
I understand this is not how your model works. But it is long known, and works just fine, in the mainstream model.
My model is very simple its potential is on the picture below I have shown already :
You've posted this repeatedly, and it is not and cannot be the correct model of nuclear binding. Do you want to talk about anything about it, other than reposting your impression of "success" at predicting the deuteron binding energy?
Disputing the accepted definition of nucleon would be one way to avoid meeting the scientific standards of any reputable journal.
This means that new ideas are censored as in the time of Copernicus and Galileo…
No, it means that in order to communicate with people you need to understand the meanings of the terms that they use