why Nuclear Physics cannot be entirelly correct

[Belz...]

Thanks. But do I sense a bit of sarcasm? lol.

No. Just read the words in the post, and don't add anything you might imagine is there, please.

 

[Dancing David]

Distribution of nucleons into the nuclei​

In Atomic Physics, the electron is considered as a cloud of probability about the nucleus.

But in Nuclear Physics this is not possible. The nucleons have to be considered as particles, because if we consider them as a cloud of probability there is no way to explain some properties of the nuclei.

For example, you don't believe in Bose_Einstein condensate?

Nucleons have waveforms, you can't just wish them away. The waveforms are constrained but still wave forms.

 

[Dancing David]

I am not speaking about any theory.

I am speaking about the position of nucleons within a nucleus.

The position of two deuterons, in a given instant, have to be like shown in the figure:

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_489414da28450e8cf7.jpg[/qimg]​

If this position does not occur, it's impossible for the nucleus 8O16 to have:
- null nuclear spin
- and null magnetic moment,
no matter what is the theory considered

If you shove the cat into a box smaller than the cat, the cat is no longer a cat but mush.

David's Cat

Actually I like cats quite a bit and would never mush one, no cats were harmed in the making of this analogy.

 

[Dancing David]

Well obviously they aren't considered as a cloud of probability about the nucelus since they make up the nucleus.


False. In general, one describes the nucleon orbitals using wavefunctions. If you look at a shell model level scheme you'll find that the nomenclature for each of the levels is very similar to those of the atomic model (though the ordering is different, largely due to the good old spin-orbit interaction).


Not true. Deuteron like pairing is a very rare occurrence. Most nuclei can be though of as having proton spin-0 proton and neutron pairs. Some excited states of 16O do look a bit like a collection of alpha particles though.


I have no idea what you are talking about. If like nucleons are arranged in spin-0 pairs then they have a large overlap of wavefunctions which means they effectively feel more of the strong force and are more tightly bound. Accordingly, the ground state of all even-even nuclei are spin 0.


You're talking crap.


Trepidation? What is it scared of?

ThanksTubbythin!




Aside:
This is the issue, Pedrone will try to force the language of physics to meet his defintions, it is not the english language barrier, it is the physics concepts barrier.

 

[Nihilianth]

What I see is indistinct images from a book in a language that I cannot read (Italian?).

No, not Italian. I know a little Italian. However, the little c character with the little currly-q thing below it sort of gives it away as Portuguese. That, and he said it was Portuguese earlier.

What I don't understand, is how someone can take a book and translate it into a language they don't have full control over, all by themselves, and expect the translation to be taken seriously. That's what I don't understand.

 

[dafydd]

Nobody is obliged to read what I write.

Quite right. We just do it for the giggles. Where did you study and what are your qualifications?

 

[slingblade]

I THOUGHT you seemed to have a pretty good grasp on the language. But you mentioned it wasn't your first language

I don't think I did, because it's my only language. I think I said that we try to cut ESLers a break. But it's all good--go forth, young scientist!

 

[blackcream]

What about the new particle that may have been discovered.

 

[Tubbythin]

ThanksTubbythin!

Thanks, though I should have said
"having spin-0, proton-proton and neutron-neutron pairs"
or something like that.

 

[pedrone]

Quote:
But in Nuclear Physics this is not possible. The nucleons have to be considered as particles, because if we consider them as a cloud of probability there is no way to explain some properties of the nuclei.

False. In general, one describes the nucleon orbitals using wavefunctions. If you look at a shell model level scheme you'll find that the nomenclature for each of the levels is very similar to those of the atomic model (though the ordering is different, largely due to the good old spin-orbit interaction).

False what you state.
A nucleus as 8O16 cannot have null nuclear spin and null nuclear magnetic moment if each proton do not have another proton symmetrically with regard to the center , with contrary spin and contrary direction of the magnetic moment vector.
Just because the nucleus has a spin. So, the following occurs:
- The rotation of a proton A regarding the center of the nucleus yields a nuclear magnetic moment.
- There is need other proton B, gyrating about the center of the nucleus, so that to yield a nuclear magnetic moment which cancells that produced by the proton A.
- If this do not occur, it's impossible to have a nucleus 8O16 with null nuclear magnetic moment.

Quote:
For instance, let's consider the oxygen nucleus 8O16.
The physicists discovered that into the nuclei the protons and neutrons are linked together (the nuclei are filled by deuterons).

Not true. Deuteron like pairing is a very rare occurrence. Most nuclei can be though of as having proton spin-0 proton and neutron pairs. Some excited states of 16O do look a bit like a collection of alpha particles though.

So, again you accuse Eisberg and Resnick to be liars.

I have not their book at hand here now. But in the end of the week I will put it here.

And what you say is very strange, since neutron pairs cannot exist, because in short distance they have repulsion.
From what you say, we had to expect the existence of dineutrons outside the nuclei, and they do not exist.

And worst is proton pairs, since they have strong repulsion

Quote:
The 8O16 has null nuclear spin and also null magnetic moment.
If the nucleons into the 8O16 should be a cloud of probability in disorder, its nuclear spin and magnetic moment could not be null.

I have no idea what you are talking about. If like nucleons are arranged in spin-0 pairs then they have a large overlap of wavefunctions which means they effectively feel more of the strong force and are more tightly bound. Accordingly, the ground state of all even-even nuclei are spin 0.

You have no idea what I'm talking about, and you have no idea what you're talking about

Trepidation? What is it scared of?

Trepidation
Vibration
Shake
wheel imbalance
etc., any word you wish, due to unbalance of masses.

Or do you think that a nucleus with unbalanced masses gyrates without any trepidation ?
If you do, then you believe that Nature is magic

 

[pedrone]

For example, you don't believe in Bose_Einstein condensate?

Nucleons have waveforms, you can't just wish them away. The waveforms are constrained but still wave forms.

No, nucleons have NOT waveforms.

What they have is ZITTERBWEGUNG (helical trajectory) wrongly interpreted in Quantum Mechanics as waveforms.

The Zitterbewegung Interpretation of Quantum Mechanics:
http://geocalc.clas.asu.edu/pdf-preAdobe8/ZBW_I_QM.pdf

But into nuclei the nucleons are constrained to lose their zitterbewegung, because of the confinement into short distance.

 

[Reality Check]

A nucleus as 8O16 cannot have null nuclear spin and null nuclear magnetic moment if each proton do not have another proton symmetrically with regard to the center , with contrary spin and contrary direction of the magnetic moment vector.

You are showing your ignorance again, pedrone
A nucleus as 8O16 has zero nuclear spin and zero nuclear magnetic moment for the same reason athat an atom can have zero atomic spin and zero atomic magnetic: The sum of the QM spins is zero.

These are QM systems not classical systems.

 

[Tubbythin]

False what you state.
A nucleus as 8O16 cannot have null nuclear spin and null nuclear magnetic moment if each proton do not have another proton symmetrically with regard to the center , with contrary spin and contrary direction of the magnetic moment vector.
Just because the nucleus has a spin. So, the following occurs:
- The rotation of a proton A regarding the center of the nucleus yields a nuclear magnetic moment.
- There is need other proton B, gyrating about the center of the nucleus, so that to yield a nuclear magnetic moment which cancells that produced by the proton A.
- If this do not occur, it's impossible to have a nucleus 8O16 with null nuclear magnetic moment.

Pardon?

So, again you accuse Eisberg and Resnick to be liars.

Nope.

Quote]
I have not their book at hand here now. But in the end of the week I will put it here.

And what you say is very strange, since neutron pairs cannot exist, because in short distance they have repulsion.[/Quote]
They're bound by the potential of all the nucleons. But those in time-reversed coplanar orbits feel stronger residual interactions.

From what you say, we had to expect the existence of dineutrons outside the nuclei, and they do not exist.

No you wouldn't, since dineutrons do not feel the potential of the rest of the nucleus.

And worst is proton pairs, since they have strong repulsion

As above.

You have no idea what I'm talking about, and you have no idea what you're talking about

Yes and no, respectively.

Trepidation
Vibration
Shake
wheel imbalance
etc., any word you wish, due to unbalance of masses.

Still not the faintest clue what you're talking about. Nuclei may have vibrational modes but not in the ground state. And they bare no resemblance to your previous diagrams.

Or do you think that a nucleus with unbalanced masses gyrates without any trepidation ?

Trepidation means fear. As nucleons do not feel emotions I can't imagine how they could feel fear. So yes, no nucleus feel trepidation, regardless of whether it is unbalanced or not.

If you do, then you believe that Nature is magic

Nope. I think believing nuclei can feel fear is far more of a belief in magic.

 

[Reality Check]

No, nucleons have NOT waveforms.

What they have is ZITTERBWEGUNG (helical trajectory) wrongly interpreted in Quantum Mechanics as waveforms.

The Zitterbewegung Interpretation of Quantum Mechanics:
http://geocalc.clas.asu.edu/pdf-preAdobe8/ZBW_I_QM.pdf

Try reading at least the title of what you cite:
The Zitterbewegung Interpretation of Quantum Mechanics:

The zitterbewegung is a local circulatory motion of the electron presumed to be the basis of the electron spin and magnetic moment. A reformulation of the Dirac theory shows that the zitterbewegung need not
be attributed to interference between positive and negative energy states as originally proposed by Schroedinger. Rather, it provides a physical interpretation for the complex phase factor in the Dirac wave function generally. Moreover, it extends to a coherent physical interpretation of the entire Dirac theory, and it implies a zitterbewegung interpretation for the Schroedinger theory as well.

This is a possible interpretation of the Schroedinger theory. That means that starts with wave functions ("wave forms") and may give a physical explanation for them. Zitterbewegung explains the complex phase factor in the Dirac wave function, it does not replace it.

But into nuclei the nucleons are constrained to lose their zitterbewegung, because of the confinement into short distance.

Nucleons are not described by the above paper at all. The Dirac equation is for particles like electrons. The theory that describes protons and neutrons is QCD.


Citations for your assertions that

1. nucleons have zitterbewegung
2. that zitterbewegung is "lost" at the short distances in nuclei.

 

[Tubbythin]

No, nucleons have NOT waveforms.

What they have is ZITTERBWEGUNG (helical trajectory) wrongly interpreted in Quantum Mechanics as waveforms.

The Zitterbewegung Interpretation of Quantum Mechanics:
http://geocalc.clas.asu.edu/pdf-preAdobe8/ZBW_I_QM.pdf

But into nuclei the nucleons are constrained to lose their zitterbewegung, because of the confinement into short distance.

They have wavefunctions. As Eisberg and Resnick say.

 

[Daylightstar]

...
Trepidation means fear.
...

Trepidation can also mean a shivering, tremulous (trembling) motion....of the body, as an emotional response.
It does not, afaik, refer to motion of things.

Finding (by Pedrone) interesting sounding words is one thing, understanding them is something else.
Pedrone appears to be very lax in his thinking- and judgement skills.

 

[Tubbythin]

Neutron decay

According to Maxwell theory, when an electric particle (as the electron) is submitted to acceleration, it emits energy (photons).

So, consider the beta decay of a free neutron: n -> p + e + v

At once the neutron becomes a proton, electron, and antineutrino, the electron moves away the proton. As there is Coulombic attraction between the proton and the electron, this one has attraction with the proton, and so it is submitted to deceleration.
Therefore, according to Maxwell theory, the electron would have to emit photons, in any neutron beta decay.
But the emission of photons is not observed experimentally. And we realize that something is wrong with the neutron model composed by quarks.

Err...

 

[pedrone]

Originally Posted by pedrone
Neutron decay

According to Maxwell theory, when an electric particle (as the electron) is submitted to acceleration, it emits energy (photons).

So, consider the beta decay of a free neutron: n -> p + e + v

At once the neutron becomes a proton, electron, and antineutrino, the electron moves away the proton. As there is Coulombic attraction between the proton and the electron, this one has attraction with the proton, and so it is submitted to deceleration.
Therefore, according to Maxwell theory, the electron would have to emit photons, in any neutron beta decay.
But the emission of photons is not observed experimentally. And we realize that something is wrong with the neutron model composed by quarks.

Err...

Tubbythin,
why did you hide the link you posted with that "Err..." ?

Well, let's look what he have there:
url='http://prola.aps.org/abstract/PR/v93/i3/p518_1'


Wow !!!

This is not beta decay of neutron:
Internal bremsstrahlung spectra from S35 and Pm147 have been investigated with the NaI scintillation spectrometer

Tubbythin,
dont you know the difference between the beta decay of a free neutron and the beta decay of nuclei ? (as S35 and Pm147)


Tubbythin,
I suspect that you're acting dishonestly on purpose

 

[Reality Check]

Tubbythin,
why did you hide the link you posted with that "Err..." ?

Well, let's look what he have there:
url='http://prola.aps.org/abstract/PR/v93/i3/p518_1'


Wow !!!

This is not beta decay of neutron:
Internal bremsstrahlung spectra from S35 and Pm147 have been investigated with the NaI scintillation spectrometer

Tubbythin,
dont you know the difference between the beta decay of a free neutron and the beta decay of nuclei ? (as S35 and Pm147)


Tubbythin,
I suspect that you're acting dishonestly on purpose

I can answer this, Tubbythin:

pedrone,
why did you not know that the standard way to have hyperlinks is with the text as in neutron (stability and beta decay)?

Free neutrons decay by emission of an electron and an electron antineutrino to become a proton, a process known as beta decay

Wow !!!

This is beta decay of neutron:
Internal bremsstrahlung spectra from S35 and Pm147 have been investigated with the NaI scintillation spectrometer

pedrone,
dont you know there is no difference between the beta decay of a free neutron and the beta decay of a neutron in a nuclei ? (as S35 and Pm147)


pedrone,
I suspect that you're acting ignorant on purpose


By hiding the link a bit, Tubbythin was trying to not embarass you. Pity that you insisted in doing it for yourself!

 

[pedrone]

Not true. Deuteron like pairing is a very rare occurrence. Most nuclei can be though of as having proton spin-0 proton and neutron pairs. Some excited states of 16O do look a bit like a collection of alpha particles though.

Tubbythin,
I seriously suspect you are lying.

The most probable occurrence within the nuclei is the deuteron, as mentioned by Eisberg and Resnick (in the end of the week I'll post what they state in here).

Proton-proton pairing occurrence is very improbable, because there is a strong Coulombic repulsion between them.

Neutron-neutron pairing occurrence is also very improbable, because there is repulsion between neutrons in short distance (that's why there no exist dineutrons in nature).

Unlike, deuterons exist in Nature. They are stable, with a considerable binding energy (about 2MeV).
So, as deuterons exist in free nature, then more reason we have to suppose that they exist within the nuclei.
Therefore its occurrence within the nuclei is very probable.

We cannot say the same about proton-proton pairing, or about neutron-neutron pairing. The occurrence of these two pairing is very, very improbable.

So, I suspect you're lying.