Nuclear Strong Force is a Fiction

bjschaeffer said:
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.
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But if you subtract the mass of the electron then you get the right answer.

There are at least 7 H isotopes including the proton and 2H.
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Well... sort of.
 
bjschaeffer said:
The excess neutrons are not unbound but slightly bound; otherwise the binding energies of 4H, 5H, 6H, and 7H would be the same.
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Huh? 4H and 6H decay via neutron emission, therefore they are not bound.

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.
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How do you know it isn't just numerology?

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?
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Achieving order of magnitude by randomly adding some constants together until you get the right answer means nothing.
The nuclear energy is much greater because the strong force is much stronger than the electromagnetic force (hence the name).

I have the formula you not. This is the most important thing to know in nuclear physics.
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No it isn't. You can't do anything with it. It's numerology and it's meaningless.

It is the equivalent of Bohr formula for the hydrogen atom, characterizing the chemical energy. The Bohr formula is 1/2α2mec2. The nuclear energy formula is αmpc2. 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.
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No, ignoring the fundamental laws of nuclear interaction would be using numerology to find one number that vaguely matches another and then claiming victory, whilst getting everything else completely wrong.

You are unable to explain clearly the physical basis of your theory. Mine is simple: application of Coulomb's laws with no angular momentum.
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It may be simple, but it is also wrong.

It is easy to criticize but you are unable to calculate the binding energy of the deuteron.
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So are you.
 
bjschaeffer said:
The excess neutrons are not unbound but slightly bound; otherwise the binding energies of 4H, 5H, 6H, and 7H would be the same.
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The National Nuclear Data Center says you're wrong. 4H, 5H, 6H, and 7H are unbound. Look it up.

It is easy to criticize but you are unable to calculate the binding energy of the deuteron.
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My colleagues are able to calculate the binding energy of the deuteron, remember? You just don't like the way they did it. They did it using a complicated theory, and you don't like that because you guessed that all such theories should be simple.

You are also not able to calculate the binding energy of the deuteron. You obtained the number 1.6 MeV---not actually the binding energy of the deuteron---by manipulating the constants associated with an electromagnetically-bound object of your own invention which has virtually no resemblance to actual deuterons.
 
ben m said:
The National Nuclear Data Center says you're wrong. 4H, 5H, 6H, and 7H are unbound. Look it up.



My colleagues are able to calculate the binding energy of the deuteron, remember? You just don't like the way they did it. They did it using a complicated theory, and you don't like that because you guessed that all such theories should be simple.

You are also not able to calculate the binding energy of the deuteron. You obtained the number 1.6 MeV---not actually the binding energy of the deuteron---by manipulating the constants associated with an electromagnetically-bound object of your own invention which has virtually no resemblance to actual deuterons.
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Now I know that you are unable to think by yourself. You don't know the difference between unbound and slightly bound. You can see on my post #414 the difference between unbound and slightly bound for 4He.

I still have not seen any (except mine of course) detailed calculation of the deuteron binding energy from fundamental laws, simply because it doesn't exist.
 
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bjschaeffer said:
Now I know that you are unable to think by yourself. You don't know the difference between unbound and slightly bound.
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He sure seems to, the difference is of the sign of the binding energy:
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.
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Tubbythin said:
Huh? 4H and 6H decay via neutron emission, therefore they are not bound.


How do you know it isn't just numerology?


Achieving order of magnitude by randomly adding some constants together until you get the right answer means nothing.
The nuclear energy is much greater because the strong force is much stronger than the electromagnetic force (hence the name).


No it isn't. You can't do anything with it. It's numerology and it's meaningless.


No, ignoring the fundamental laws of nuclear interaction would be using numerology to find one number that vaguely matches another and then claiming victory, whilst getting everything else completely wrong.


It may be simple, but it is also wrong.


So are you.
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4H and 6H are unstable but they have a binding energy that I have calculated :




The application of the Coulomb's electric and magnetic laws to the nucleus is not randomly adding some constants together.

The strong force is a mysterious and unknown force. Its fundamental laws and constants are unknown
 
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ben m said:
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.
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You are an ignorant, look at recent nuclear tables. Specialists will be able to obtain a good precision when the electromagnetic theory of the nuclear interaction will be recognized.

The sign is the same as for the deuteron, that is, negative, thus attractive. It is positive on the graph for practical reasons:
 
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bjschaeffer said:
Now I know that you are unable to think by yourself. You don't know the difference between unbound and slightly bound. You can see on my post #414 the difference between unbound and slightly bound for 4He.
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4He is very bound for its mass. That's because of its shell structure, something that real nuclear physics explains quite nicely.
 
bjschaeffer said:
4H and 6H are unstable but they have a binding energy that I have calculated :
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Then your calculations are wrong. See here and here.

The application of the Coulomb's electric and magnetic laws to the nucleus is not randomly adding some constants together.
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Doing it properly isn't. In fact it is crucial when trying to compare neutron-neutron, proton-proton and neutron-proton interactions.

The strong force is a mysterious and unknown force. Its fundamental laws and constants are unknown
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It is a complex force. It is not unknown. We have been studying it for a century (whether we knew it or not).
 
bjschaeffer said:
You are an ignorant, look at recent nuclear tables.
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I just did. They told me they were unbound. Here is a recent experimental paper on 6H. Note the final sentence:
Paper said:
These results show the availability of nuclear structure information well outside the bounding limits, resulting in an extraordinary input to improve the present models and understanding of nuclear matter.
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Tubbythin said:
I just did. They told me they were unbound. Here is a recent experimental paper on 6H. Note the final sentence:
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Originally Posted by bjschaeffer
You are an ignorant, look at recent nuclear tables.
I just did. They told me they were unbound. Here is a recent experimental paper on 6H. Note the final sentence:
Originally Posted by Paper
These results show the availability of nuclear structure information well outside the bounding limits, resulting in an extraordinary input to improve the present models and understanding of nuclear matter.

This is literature, not physics
 
Tubbythin said:
Then your calculations are wrong. See here and here.


Doing it properly isn't. In fact it is crucial when trying to compare neutron-neutron, proton-proton and neutron-proton interactions.


It is a complex force. It is not unknown. We have been studying it for a century (whether we knew it or not).
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If you are unable to show its fundamental laws and constants it means that it is only speculative.

The Coulomb's laws are well known the strong force laws are inexistent.

np, nn and pp interactions are different because the neutron and the proton have different electric and magnetic properties.
 
bjschaeffer said:
If you are unable to show its fundamental laws and constants it means that it is only speculative.
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No it doesn't. Something that is based on hundreds of thousands of pieces of experimental evidence is not speculative.

The Coulomb's laws are well known the strong force laws are inexistent.
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We might not have a nice simple set of equations to use but that is irrelevant. We don't have a nice simple set of equations to model a human being either. It doesn't mean human beings don't exist.

np, nn and pp interactions are different because the neutron and the proton have different electric and magnetic properties.
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Actually nn and pp interactions are almost exactly the same when you remove Coulomb effects. np interactions are also very similar.
 
You're being misled by plotting binding energy "per nucleon". Make a plot of total binding energy (B/A x A). On this curve, it will be clear that 4H-6H are unstable in the real world (6H has lower binding energy than 5H+n, or 4H+2n, or 3H+3n). And it is clear that your "prediction" is the opposite; you predict increasing binding energy for each added neutron.

You predict that additional neutrons are attracted to 3H, 4H, 5H, and 6H, whereas in reality they're repelled. You have the wrong sign.

Same problem on the helium curve. In reality, 5He decays to 4He+n; 7He decays to 6He+n; again, your prediction is wrong.
 
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