Nuclear Strong Force is a Fiction

[ben m]

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.

 

[Farsight]

...In short, it is impossible to replicate gravity by electromagnetic means...

I think it's amazing how you can skim through a thread and things like this just jump right out at you.

 

[bjschaeffer]

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.

It says that "There are no figures or tables for this document."
I don't have the full document but in the abstract it is not written that they have calculated the binding energy of the deuteron.

I calculate the binding energies; I cannot do everything. More over diffusion has a very low precision and there is no standard method as for the binding energies.

 

[Ziggurat]

It says that "There are no figures or tables for this document."

There may not be any associated with the web page, but there are plenty of both within the article itself.

I don't have the full document but in the abstract it is not written that they have calculated the binding energy of the deuteron.

They do. They calculate a range from -2.216 to -2.223 MeV. The listed experimental value is −2.224575(9) MeV, and they state explicitly that the binding energy was NOT used as an input for any of their calculations, making it a true prediction of their model. There is a discrepancy, but that is to be expected when you truncate the approximation series at a finite number of terms.

 

[bjschaeffer]

If you have a model with, say, two parameters, then you could reasonably expect it to fit any two data points by adjusting those parameters. But if you're trying to fit ten different data points with just two parameters, then no, you cannot get anything you want.

General relativity, for example, has only two parameters: G and c. But we've got hundreds of independent data points that we can fit with GR, and they all fit. You really cannot do that with just any two-parameter model.

There's a lot about nuclear physics that I don't know. But one of the things that I do know is that the number of parameters available for fitting is a LOT less than the number of data points. Which means that the model really is well-constrained by the data, and we really do have good reason for confidence in its accuracy.

What is better is the use of only fundamental constants to calculate the binding energy of the nuclides. But it seems that they (the "Argonne V18" cited by ben m on post #320) use 40 parameters for very few results of binding energies if any. It would be acceptable if they had calculated thousands of nuclei, but I have never seen that, except with the semi-empirical mass formula where there are half a dozen fitted parameters.

 

[bjschaeffer]

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.

I begin with the simplest cases.
There no fits in my theory because I use only fundamental constants and a theory known since two centuries. And it works.

 

[ben m]

What is better is the use of only fundamental constants to calculate the binding energy of the nuclides. But it seems that they (the "Argonne V18" cited by ben m on post #320) use 40 parameters for very few results of binding energies if any. It would be acceptable if they had calculated thousands of nuclei, but I have never seen that, except with the semi-empirical mass formula where there are half a dozen fitted parameters.

What's the obsession with binding energies? What makes you think that binding energies are a good place to test a new nuclear theory? In reality, we have a vast amount of data on scattering cross sections, and this data is extremely easy to interpret in terms of a potential. You start with the stuff with a clear data/theory connection, then you learn how to do more complicated stuff.

Are you aware, bjschaeffer, that no one can use "fundamental constants" to calculate the atomic energy levels of large atoms, like gold? Nope. You have to use mean-field approximations, and you calibrate those approximations with fits to data. Not because quantum mechanics is wrong, not because we don't know the laws governing atoms, but because many-body physics is messy and intractable.

 

[Ziggurat]

Are you aware, bjschaeffer, that no one can use "fundamental constants" to calculate the atomic energy levels of large atoms, like gold? Nope. You have to use mean-field approximations, and you calibrate those approximations with fits to data. Not because quantum mechanics is wrong, not because we don't know the laws governing atoms, but because many-body physics is messy and intractable.

Hell, you don't even need to go to gold. The Helium atom isn't even exactly solvable. In that case, you don't need mean field theory, you can use the variational principle to calculate approximate results. But to get a very good approximation, you need a test function, and the more parameters the test function has to adjust, the closer you can get it to the real solution (and hence the real energy). So you could have a test function with 40 parameters, and it would give you a pretty good estimate of the ground state of the Helium atom. But those 40 parameters are not parameters of the theory of quantum mechanics. They are parameters of a specific wave function. Because of the way that the variational principle works, adding more parameters will make it MORE accurate, not less accurate. The only limit is computational difficulty. I'm not going to bother trying to dig through the references to find out what the 40 parameters he refers to are, but it wouldn't surprise me in the least if they were similar in nature to the parameters one uses with the variational principle.

 

[Dancing David]

Thank you I just noticed some strange things in my posts. I will take care now and verify that I have it correct like that
{quote}
bjschafer

says this
Quote:by you
{quote}
dancing david

replied this

{/quote}
{/quote}

Cool!

 

[Reality Check]

What are, precisely, the fundamental laws and constants of the nuclear interaction?

See Standard Model.

 

[Reality Check]

You are unable to give me the proof that ²H has bean calculated from fundamental laws and constants. This doesn't exist because the fundamental laws of the nuclear interaction are unknown.

Calculations for simple elements do exist because the fundamental laws of the nuclear interaction are known .

Calculations for complex elements do not exist because the fundamental laws of the nuclear interaction are known and doing calculations from then is hard !

 

[ben m]

I begin with the simplest cases.
There no fits in my theory because I use only fundamental constants and a theory known since two centuries. And it works.

No it doesn't. You multiplied a bunch of masses and constants together and got 1.6 MeV, a number which is not in the "it worked" ballpark. Not by nuclear standards.

When a complex, many-body calculation gets the wrong answer by 30%, you say, "this may be due to the many approximations we made during the calculation." When a simple, tightly-constrained calculation, with precise inputs, gets the wrong answer by 30%, you say, "well, that disproves the theory".

How are you able to say, on one hand, "my theory starts with the simplest calculation", and on the other hand "this calculation is so complex, I had to ignore factors of order 1.3 in order to carry it out"?

Your paper goes on to get more things wrong---by much worse than 30%. You predict that 3H is more tightly bound than 2H---very wrong. You predict that 4H, 5H, 6H, and 7H are bound---wrong, these are unbound. They're just resonances. The experimental binding energy has the opposite sign from what you "predict", and you apparently misread the experimental data in thinking that 7H is an actual nuclear bound state. (If you squeeze 6n and 1p together, what happens is that 4 neutrons immediately leap back out---they're repelled, not attracted.)

I congratulate you on one point: your theory is doing the right thing---it's making clear predictions of experimentally-measurable quantities. That makes it at least plausibly scientific (in nice contrast to DHamilton's and Farsight's work that I've seen) rather than crackpotty. Unfortunately, your theory is in fact wrong, and it's easy to tell that it's wrong.

If you have some objection to my nuclear analysis, let me point out that your theory is also easy to test at higher energies---in electron-proton scattering, in electron-deuteron scattering, in neutron decay, etc.---and again it's very easy to see that it is experimentally wrong. If you're interested, I'm happy to explain.

 

[Farsight]

Who are you calling crackpot? I've demonstrated that I know more physics than others here, and there's nothing crackpot about about careful explanations, supported by robust evidence and papers, that people can actually understand. If you can point to something I've said that's incorrect, please do so, and I'll take it on the chin and say "Sorry, I was wrong". I've done it before, and I will be doing it again. But do please resist the urge to be abusive.

 

[bjschaeffer]

Hell, you don't even need to go to gold. The Helium atom isn't even exactly solvable. In that case, you don't need mean field theory, you can use the variational principle to calculate approximate results. But to get a very good approximation, you need a test function, and the more parameters the test function has to adjust, the closer you can get it to the real solution (and hence the real energy). So you could have a test function with 40 parameters, and it would give you a pretty good estimate of the ground state of the Helium atom. But those 40 parameters are not parameters of the theory of quantum mechanics. They are parameters of a specific wave function. Because of the way that the variational principle works, adding more parameters will make it MORE accurate, not less accurate. The only limit is computational difficulty. I'm not going to bother trying to dig through the references to find out what the 40 parameters he refers to are, but it wouldn't surprise me in the least if they were similar in nature to the parameters one uses with the variational principle.

I have calculated (to be published) the binding energy of ⁴He with different approximations. The figure below showis the Coulomb electromagnetic potential and, at its minimum, the binding energies of ²H and ⁴He .



proving that it is electromagnetically solvable, e.g. by trial and error with two parameters, the distance between the neutron and the proton and between the electric charges of the neutron (to be published soon). The values of the binding energies are per neutron-proton bond!

You can see my older paper on the ²H isotopes here
http://link.springer.com/article/10.1007%2Fs10894-010-9365-0?LI=true

 

[Tubbythin]

Who are you calling crackpot? I've demonstrated that I know more physics than others here,

No, no you haven't.

and there's nothing crackpot about about careful explanations, supported by robust evidence and papers, that people can actually understand.

There is nothing crackpot about careful explanations, supported by robust evidence and papers, that people can actually understand. When you provide some people will stop calling you a crackpot.

If you can point to something I've said that's incorrect, please do so, and I'll take it on the chin and say "Sorry, I was wrong". I've done it before, and I will be doing it again. But do please resist the urge to be abusive.

You have repeatedly claimed to be the person that knows what Einstein actually meant and you have repeatedly been shown to be wrong, for example.
You are the abusive one. Constantly going around telling everyone they're not worthy to lick your boots.

 

[Tubbythin]

I have calculated (to be published) the binding energy of 4He with different approximations. The figure below showis the Coulomb electromagnetic potential and, at its minimum, the binding energies of 2H and 4He

I think you must have forgotten to subtract the mass of the extra electron in the calculation for ⁴He.

ETA:

You can see my older paper on the ²H isotopes here
http://link.springer.com/article/10.1007%2Fs10894-010-9365-0?LI=true

How many ²H isotopes do you think there are?

 

[dafydd]

I've demonstrated that I know more physics than others here,

No.

 

[bjschaeffer]

I think you must have forgotten to subtract the mass of the extra electron in the calculation for ⁴He.

ETA:

How many ²H isotopes do you think there are?

It is only when my calculations will be very very precise that I will have to take into account the electron. It would be ridiculous to do it now.
There are at least 7 H isotopes including the proton and ²H.

 

[bjschaeffer]

No it doesn't. You multiplied a bunch of masses and constants together and got 1.6 MeV, a number which is not in the "it worked" ballpark. Not by nuclear standards.

When a complex, many-body calculation gets the wrong answer by 30%, you say, "this may be due to the many approximations we made during the calculation." When a simple, tightly-constrained calculation, with precise inputs, gets the wrong answer by 30%, you say, "well, that disproves the theory".

How are you able to say, on one hand, "my theory starts with the simplest calculation", and on the other hand "this calculation is so complex, I had to ignore factors of order 1.3 in order to carry it out"?

Your paper goes on to get more things wrong---by much worse than 30%. You predict that 3H is more tightly bound than 2H---very wrong. You predict that 4H, 5H, 6H, and 7H are bound---wrong, these are unbound. They're just resonances. The experimental binding energy has the opposite sign from what you "predict", and you apparently misread the experimental data in thinking that 7H is an actual nuclear bound state. (If you squeeze 6n and 1p together, what happens is that 4 neutrons immediately leap back out---they're repelled, not attracted.)

I congratulate you on one point: your theory is doing the right thing---it's making clear predictions of experimentally-measurable quantities. That makes it at least plausibly scientific (in nice contrast to DHamilton's and Farsight's work that I've seen) rather than crackpotty. Unfortunately, your theory is in fact wrong, and it's easy to tell that it's wrong.

If you have some objection to my nuclear analysis, let me point out that your theory is also easy to test at higher energies---in electron-proton scattering, in electron-deuteron scattering, in neutron decay, etc.---and again it's very easy to see that it is experimentally wrong. If you're interested, I'm happy to explain.

The excess neutrons are not unbound but slightly bound; otherwise the binding energies of 4H, 5H, 6H, and 7H would be the same.

This calculation is perhaps not very precise but there is no fit and moreover it is analytical. It is only to obtain a better precision that numerical calculations are necessary.
There are many people who forget the main thing, the order of magnitude and search the figures after the comma. How do you explain why the nuclear energy is around one million times the chemical energy?

I have the formula you not. This is the most important thing to know in nuclear physics. It is the equivalent of Bohr formula for the hydrogen atom, characterizing the chemical energy. The Bohr formula is 1/2α²m_ec². The nuclear energy formula is αm_pc². If you cannot calculate it you ignore the fundamental laws of the nuclear interaction. Therefore the ratio between nuclear and chemical energy is 1836 x 137.

You are unable to explain clearly the physical basis of your theory. Mine is simple: application of Coulomb's laws with no angular momentum.

It is easy to criticize but you are unable to calculate the binding energy of the deuteron.

 

[ben m]

There are at least 7 H isotopes including the proton and ²H.

No, there are 3. Hydrogen, deuterium, tritium.

You've got the wrong binding energy for 2H.

You've got the wrong binding energy for 3H.

You've got the wrong sign of the "binding energy" for 4H, 5H, and 6H. You think they're bound states with positive binding energy. They're not. Experiments show that these are unbound resonances with negative binding energy.