Final failure of the nuclear models of the Standard Nuclear Physics

[wlad]

Ahead is a discussion between Mr. Joe, Mr. JR, and Guglinski, in the blog Journal of Nuclear Physics.

Physics of rotating and expanding black hole universe « Journal of Nuclear Physics



From the discussion we get the following indisputable conclusion:

"the light nuclei with equal pair quantity of protons and neutrons violate the monopolar nature of the electric charges"

Joe
April 10th, 2014 at 1:59 AM

Wladimir,

I repeat what I answered in (1). Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation. That is what we know empirically. The other variables (a, c, d) that you mention in your list have a dipolar nature that can not possibly logically have an effect to change a property of a monopolar nature. Switching spins, or reversing fluxes n(o), will not affect the electric charge of a particle. It could possibly logically only affect the dipolar properties of that particle. The way that you would have it, the deuteron would be treated no different from the neutron, even though the deuteron has something extra: an electric charge. What you are actually doing is ignoring the electric charge when considering induced magnetic moments from rotation of nucleons.

All the best,
Joe

 

Wladimir Guglinski
April 11th, 2014 at 6:01 PM

Joe,
I dont see any problem, and let me tell you why, as follows.

1) Consider a nucleus 2He4.
As its two deuterons have contrary spins, the 2He4 has a total magnetic moment zero. And therefore the two charges of the two deuterons cannot induce a magnetic field, since the two charges are gyrating within a space without any magnetic field.

2) Now let us consider the 3Li6 as example for explaining how occurs the magnetic force Fm between the deuteron and the central 2He4, as follows.

3) The deuteron of the 3Li6 is captured by the flux n(o) of the 2He4, and it is dragged by the rotation of the flux n(o), in order that the rotation of the deuteron induces a magnetic moment in the 3Li6.

4) The magnetic moment of the 3Li6 is +0,822, measured in experiments.

5) Suppose that the rotation of the deuteron induces a magnetic moment +0,820

6) The two deuterons of the 2He4 have a very short orbit radius of rotation about the center of the 3Li6 nucleus. As the two deuterons are gyrating wihin a magnetic moment +0,820 induced by the deuteron captured by the flux n(o), then the two positives charges of the 2He4 (gyrating within a magnetic moment +0,820)_ induce an additional very weak magnetic moment (because the radius of their orbits is very short).

7) Suppose that the additional magnetic moment induced by the two deuterons of the central 2He4 is +0,002.

8) Then the total magnetic moment of the 3Li6 becomes +0,820 + 0,002 = +0,822

Note that the two charges of the deuterons of the central 2He4 induce a weak magnetic moment only when a particle with charge is captured by the flux n(o) of the 2He4, as occurs with a proton or a deuteron.
In the case of a free 2He4 the two charges of the deuterons do not induce magnetic field, since the two charges rotate in a space with no magnetic field.

As the neutron is not captured by the flux n(o), the phenomenon does not occur with the neutron.
A neutron influences the magnetic moment of a nucleus (for instance the 3Li7) only when the neutron is captured by a deuteron via spin-interaction, and then the neutron also contributes for the magnetic moment of the 3Li7.

I hope to have responded your question

regards
wlad

 

Joe
April 12th, 2014 at 12:21 AM

Wladimir,

You state the following:

“In the case of a free 2He4 the two charges of the deuterons do not induce magnetic field, since the two charges rotate in a space with no magnetic field.”

This idea is false since rotating electric charges create their own magnetic field independent of any pre-existent magnetic field. Magnetic fields from various sources are simply summed. A null magnetic field from a combination of sources does not prevent another source from exhibiting its own magnetic field within that same space, and yielding a non-null net magnetic field for that space.

All the best,
Joe

 

Wladimir Guglinski
April 12th, 2014 at 8:35 AM

Joe,
consider the following:

1- suppose the body of the electron gyrates in the clockwise direction (spin-up)

2- consider one electron moving in an orbit with radius R about a vertical axis in the clockwise direction

3- suppose a particle X with negative charge (equal to the charge of the electron) which body gyrates in the anti clockwise direction (spin-down, so contrary of the spin of the electron).

4- consider the particle X moving in an orbit with radius R about a vertical axis in the clockwise direction, as happens with the electron.

.

Then I ask you:

1- Will the electron and the particle X create the same magnetic field ?

.

You said:
“I repeat what I answered in (1). Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation. That is what we know empirically.”

However, all the experiments with a body electrically charged are made with a body which has its electric charge thanks to the electrons in its electrosphere. And the spin of the electron has always the same direction.

But suppose we are be able to create a particle X with the same charge of the electron, but its spin being contrary of the spin of the electron.
And suppose we create a body, which electrosphere is composed by the particle X.

Then I ask you:

2- Would electric charge to have a monopolar nature?

3- Or, in another words:
would a body formed by electrons in its electrosphere induce the same magnetic field of a body formed by particles X in its electrosphere?

regards
wlad

 

Wladimir Guglinski
April 12th, 2014 at 5:51 PM

Joe wrote in April 10th, 2014 at 1:59 AM

Wladimir,

I repeat what I answered in (1). Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation.
———————————-

Joe,
we had a good discussion here.
And we arrived to a very interesting conclusion:

It’s IMPOSSIBLE to explain the magnetic moment zero of the 2He4 by considering the current nuclear models of the Standard Nuclear Physics

Indeed, let us see why:

1- All the nuclei have rotation. The Nobel Laureate Hans Bette estimated that 10% of the magnetic moment of the nuclei is due to their rotation.
For instance, consider the 3Li6. Suppose that 2 protons and 2 neutrons cancell each other their magnetic moment. The third proton has magnetic moment +2,793, while the third neutron has magnetic moment -1,913.
If the 3Li6 had no rotation, its magnetic moment would have to be +2,793-1,913 = +0,880. But the magnetic moment of 3Li6 is +0,835 , and the difference +0,045 is due to the rotation of the 3Li6.

2- Consider the 2He4. Consider that, if the 2He4 had no rotation, its magnetic moment would be zero, since the two protons cancell each other their magnetic moment, and the two neutrons also cancell each other.

3- However the 2He4 has rotation. And therefore the two protons move in the same direction.

4- But “Electric charge has a monopolar nature. And the direction of the magnetic moment that is induced by its rotation is dependent only on the direction of that rotation”.
And so by considering the current nuclear models the two protons of the 2He4 have to induce a magnetic moment.

5- Therefore the 2He4 cannot have magnetic moment zero, according to the current nuclear models of the Standard Nuclear Physics.

.

I wonder if Mr. JR would come here to show us how such puzzle is solved in the Standard Nuclear Physics.

regards
wlad

 

 

Wladimir Guglinski
April 12th, 2014 at 5:56 PM

Dear Mr JR

The readers of the JoNP will feel themselves very happy if you come here to explain us how the nucleus 2He4 can have magnetic moment zero, according to the current models of the Standard Nuclear Physics.

All of us will be very thankfull to your explanation

regards
wlad

 

Joe
April 12th, 2014 at 11:46 PM

Wladimir,

I do not know of any experiment that has switched the intrinsic spin to observe if the same electric charge with the same direction of rotation would produce a magnetic moment in the opposite direction. Logically there should be no difference since electric charge and intrinsic spin are independent of one another. For example, a neutrino has no electric charge but does have an intrinsic spin.

To have the behavior that you want, a switched intrinsic spin would have to effect a corresponding switch in the electric charge. But the electric charge is monopolar: it can not be flipped to an alternate state that does not exist. And flipping the spin is not going to change the sign of the charge (which would give you the behavior that you want). For example, there is no relationship such as, electron implies negative spin, and positron implies positive spin. Proof of this is in the fact that a maximum of two electrons (same sign) can occupy one orbital, and each electron must have a spin that is the opposite of the other. This is the Pauli Exclusion Principle.

All the best,
Joe

 

Wladimir Guglinski
April 13th, 2014 at 5:58 AM

Dear Joe,
you and me have proven here, together, the following:

1- It is impossible to explain the magnetic moment zero of the 2He4 by keeping the current dogma of the monopolar nature of the electric charge.

And as consequence:

2- Therefore it is impossible to explain the magnetic moment zero of the 2He4 by taking as starting point any nuclear model which do not work by considering a structure of the aether formed by particles and antiparticles as gravitons, magnetons, electricitons, etc.

3- Any nuclear model developed by no considering the participation of the aether within its structure is according to the monopolar nature of the electric charge, and therefore cannot be correct.

4- All the current nuclear models of the Standard Nuclear Physics are wrong, because they do not consider the aether within their structure, and so they cannot explain the magnetic moment zero of the 2He4.

5- Nowadays all the nuclear theories of the authors trying to explain the cold fusion cannot be entirely successful, because they are developed by considering wrong nuclear models, since all they are based on the monopolar nature of the electric charge.

.

I and the readers of the JoNP hope you have the honesty to admit it.

regards
wlad

 

 

Wladimir Guglinski
April 13th, 2014 at 10:57 AM

Joe wrote in April 12th, 2014 at 11:46 PM

Wladimir,

1- I do not know of any experiment that has switched the intrinsic spin to observe if the same electric charge with the same direction of rotation would produce a magnetic moment in the opposite direction.
Logically there should be no difference since electric charge and intrinsic spin are independent of one another. For example, a neutrino has no electric charge but does have an intrinsic spin.
———————————

COMMENT

Consequence:

Therefore it is IMPOSSIBLE the existence of the nucleus 2He4, because the experiments measured that it has magnetic moment zero, however due to fact that charge is monopolar, it is IMPOSSIBLE to exist the nucleus 2He4 with magnetic moment zero.

——————————————
To have the behavior that you want, a switched intrinsic spin would have to effect a corresponding switch in the electric charge. But the electric charge is monopolar: it can not be flipped to an alternate state that does not exist.
——————————————

COMMENT

I dont want nothing.

As the electric charge is monopolar, therefore the 2He4 nucleus cannot exist, since it has magnetic moment zero as measured by experiments, but due to its rotation and the fact that electric charge is monopolar the 2He4 cannot have magnetic moment zero.

The same we can say about all the other light even-even nuclei with Z=N pairs: 4Be8, 6C12, 8O16, 10Ne20, etc.

Therefore we have to eliminate all those nuclei from the Periodic Table of Elements, because from the known principles of Physics they cannot exist.

So, I propose a New Periodic Table, where the elements with nuclei 2He4, 4Be8, 6C12, 8O16, 10Ne30, 12Mg24, etc. will be eliminated, because it is IMPOSSIBLE to explain their magnetic moment zero from any nuclear model based on the monopolar nature of the electric charge.
And the sequence of the elements will become the following:

1H

2He2, 2He3, ….(eliminated 2He4)… , 2He5, 2He6 , etc

3Li

4Be5, 4Be6, 4Be7, …(eliminated 4Be8)… , 4Be9 , 4Be10, 4Be11, etc

5B

6C8, 6C9, 6C10, 6C11, ….(eliminated 6C12)…. , 6C13, 6C14 , etc

7N

8O13, 8O14, 8O15, …. (eliminated 8O16)… , 8O17, 8O18, etc

9F

etc, etc.

Unless you propose, dear Joe, that the rotation of the nuclei does not exist.
But unfortunatelly, by this sort of solution is impossible to explain the magnetic moments of the other nuclei.

Or perhaps you will claim that only the light even-even nuclei with Z=N do not have rotation.

Dear Joe,
please ask the help of Mr. JR so that to decide what is the best solution:

1- To eliminate the nuclei 2He4, 4Be8, 6C12, 8O16, etc. from the Periodic Table

2- To reject the hipothesis of rotation of the nuclei

3- To propose that only the nuclei 2He4, 4Be8, 6C12, etc. do not have rotation

I will be waiting the decision of yours and Mr. JR, telling me what is the best solution.

regards
wlad

 

 

 

JR
April 13th, 2014 at 11:11 AM

Wladimir said:

The readers of the JoNP will feel themselves very happy if you come here to explain us how the nucleus 2He4 can have magnetic moment zero, according to the current models of the Standard Nuclear Physics.

All of us will be very thankfull to your explanation

I doubt that you will be thankful, since you didn’t like the explanation very much the last time I answered this question (for 12Be), but here it goes:

4He is a spin-0 nucleus and so, by definition, has no magnetic moment.

In conclusion, I don’t believe that your misunderstanding of how magnetic moments are defined is enough to throw out all of classical electromagnetism or quantum electrodynamics. But keep trying.

-John

 

 

 

Wladimir Guglinski
April 13th, 2014 at 4:59 PM

JR wrote in April 13th, 2014 at 11:11 AM

4He is a spin-0 nucleus and so, by definition, has no magnetic moment.
———————————–

COMMENT

Dear JR
nuclear properties of nuclei cannot be established by definition

The nuclei have nuclear properties, which have to be explained by any nuclear theory taking in consideration the known Laws of Physics

If a nuclear model violates a known Law of Physics, as the current nuclear models are violating the monopolar nature of the electric charges, the theorists have two alternatives:

1- To reject the nuclear model, and to look for another model able to be suit to the law (in the present case the monopolar nature of the electric charges).

2- To discover the reason why the Law is being seemingly violated

.

Dear JR
such a solution of yours (claiming that “4He is a spin-0 nucleus and so, by definition, has no magnetic moment”) is actually proposed according to the phantasmagoric Heisenberg’s method.

The physicists who do not have interest to solve the questions regarding the Fundamental Physics, as happens to you, may be satisfied with your solution.

But any sincere physicists who has respect to his own honesty cannot accept such sort of explanation.
Because your explanation is actually a way of running away of the theoretical puzzles which defy the foundations of Modern Physics.

regards
wlad

 

 

 

Eric Ashworth
April 13th, 2014 at 11:35 AM

Dear Wladimir,

I do not profess to know nuclear physics as you do and I do not think many other people do also, other than to days modern day physicists. What you wrote April 13 at 5.58 am are, I beleive, some of the clearest statements you have made so far, as far as I am concerned, with regards an explanation of this anomaly. From what I gather it is to do with a monopolar nature and a zero reading. From my own understanding of nature and investigations of natural phenomena you cannat have a mono situation, it has to be binary i.e. a duality to create,exist and destroy at its most primary principle. Having spent many years developing a technology based upon this binary understanding of nature and considerable outside pressure in specific areas against its development I have discovered that some mysteries must remain mysteries for a specific reason. Wladimir do you wonder why your QR theory is rejected so strongly by certain individuals or have you a good idea why? because there is one. The reason is well worth finding out. All the best Eric Ashworth.

 

 

Wladimir Guglinski
April 13th, 2014 at 5:28 PM

Eric Ashworth wrote in April 13th, 2014 at 11:35 AM

Dear Wladimir,

do you wonder why your QR theory is rejected so strongly by certain individuals or have you a good idea why? because there is one.
———————————–

COMMENT

DEar Eric,
by looking at the explanation given by Mr. JR we realize why.

Look at his “brilliant” solution:

“4He is a spin-0 nucleus and so, by definition, has no magnetic moment”

The reason is because the theorists have not respect to the fundamental questions in Physics.

When the experimental data prove that current nuclear models violate some fundamental law and they cannot give a satisfactory explanation (satisfactory for a honest scientist), they give explanations which disagree to what we expect from a honest scientist:

they give orders to the nucleus, telling him the rules he has to follow, in order do not defy their nuclear models.

regards
wlad

 

 

Wladimir Guglinski
April 13th, 2014 at 5:10 PM

Dears Joe and Mr. JR

there is a 4th sort of solution which you may propose, for the violation of the monopolar nature of the electric charges by the nuclei.

The 4th solution you can propose I explain ahead.

As we know, when Bohr discovered his hydrogen model, he realized that his model violates some fundamental known Laws of Physics.

Then Bohr proposed a solution based on postulates, by claiming that the atom is able to violate some fundamental laws.

So, dears Joe and Mr. JR, you can propose the following postulate:

The light even-even nuclei with Z=N can violate the monopolar nature of the electric charges.

.

Therefore, my dears friends, you have now 4 alternatives for chosing what is the best solution.

regards
wlad

 

 

Joe
April 14th, 2014 at 12:20 AM

Wladimir,

It may be that the ultimate test of veracity in QRT is in explaining how 4Be8 is unstable while 6C12 and other nuclei with a symmetrical distribution of only deuterons, are stable.

All the best,
Joe

 

 

 

Wladimir Guglinski
April 14th, 2014 at 8:30 AM

Joe,
the reason why 4Be8 is not stable is shown in the page page 17, item 3.13.5, Fig. 14, of the paper Stability of Light Nuclei
http://www.journal-of-nuclear-physic...t%20nuclei.pdf

As we see in the Fig. 14, the 4Be8 is the unique nucleus in which there are two deuterons occupying opposite perfectly symmetric positions regarding to the central 2He4.

A perfect symmetry occurs when:

1- one deuteron is in the side ANA, and the other in the side DOUGLAS

2- one deuteron is in the inferior part of ANA, and the other in the superior part of DOUGLAS (or vice-versa).

A partial symmetry occurs when one deuteron is in the superior part of ANA and the other deuteron is also in the superior part, but in the side of DOUGLAS.
A partial symmetry between a deuteron and two neutrons can be seen in the superior part of the Fig. 7.

Looking at the Fig. 14 you realize that the spin-interaction Fsi(green arrows) promotes a force of attraction between the two deuterons (the red arrows show only the direction of their spins)

In the Fig. 14 I supposed that the spin-interaction Fsi is stronger than the repulsion force Fr (pink arrow) because the 4Be8 decays in two alpha particles , and so such sort of decay requires to suppose that the two deuterons are captured by the central 2He4, and the 4Be8 decays emitting two nucleons 2He4:
4Be8 -> 2He4 + 2He4

If in the Fig. 14 the repulsion force Fr should be stronger than Fsi, then the two deuterons would be expelled in contrary direction, and the decay of 4Be8 would be:
4Be8 -> 2He4 + D + D

.

The reason why 6C12 is stable is shown in the item 3.13.6, Fig. 15 and 16.

.

With the 8O16 the first hexagonal floor is complete, and so the nuclei with Z > 8 (as 10Ne20, 12Mg24, 14Si28, etc) are stable thanks to the spin-interactions between their deuterons.

regards
wlad

 

 

JR
April 14th, 2014 at 8:40 AM

Wladimir,

I’m not sure how you propose to explain the properties of nuclei if you aren’t willing to define what those properties are and then stick to those definitions. What you asked is how conventional models explain the fact that 4He has a zero magnetic moment. The magnetic moment of 4He is zero because the magnetic moment is defined in such a way that it is zero for spin-0 particles; nothing else is needed.

If you wish to define some new quantity that is similar to the magnetic moment but different, that’s fine. But you should (1) provide a definition (2) call it something different (3) not confuse it with the already-defined magnetic moment and (4) not assume it’s zero because the magnetic moment is zero.

 

 

Wladimir Guglinski
April 14th, 2014 at 11:48 AM

JR wrote in April 14th, 2014 at 8:40 AM

Wladimir,

1)——————————
The magnetic moment of 4He is zero because the magnetic moment is defined in such a way that it is zero for spin-0 particles; nothing else is needed.
——————————–

COMMENT

Show us here where did you find such definition of the magnetic moment

2) —————————-
If you wish to define some new quantity that is similar to the magnetic moment but different, that’s fine. But you should (1) provide a definition (2) call it something different (3) not confuse it with the already-defined magnetic moment and (4) not assume it’s zero because the magnetic moment is zero.
———————————–

COMMENT:

The definition of magnetic moment is independent of the spin of the nuclei.

Definition by wikipedia:
Magnetic moment - Wikipedia, the free encyclopedia
“The magnetic moment of a magnet is a quantity that determines the torque it will experience in an external magnetic field. A loop of electric current, a bar magnet, an electron, a molecule, and a planet all have magnetic moments“.

And magnetic moment of a nucleus:
“Since the electromagnetic moments of the nucleus depend on the spin of the individual nucleons, one can look at these properties with measurements of nuclear moments, and more specifically the nuclear magnetic dipole moment.”

Therefore, although the magnetic moment of a nucleus depends on the spin of individual nucleons, however it does not means that it depeneds ONLY on the spin of the INDIVIDUAL nucleons.

The nuclei have also rotation:
Phys. Rev. 53, 778 (1938) - On the Rotation of the Atomic Nucleus

And therefore the magnetic moment depends also on the rotation of the nucleus.

The total spin due to the individual nucleons in the 2He4 is zero.
However, as the two protons of the 2He4 gyrate in the same direction, then (by considering the monopolar nature of the charge) the two protons have to induce a magnetic field, which will be responsible for a magnetic moment for the 2He4.

But as the experiments detect that 2He4 has no magnetic moment, this means that 2He4 violates the monopolar nature of the charge, by considering the current nuclear models.

There is no need to be a genius to understand it.
And you, Mr. JR, you are actually showing to everybody that your understanding of Physics is very poor.

regards
wlad

[Sensei]

1- It is impossible to explain the magnetic moment zero of the 2He4 by keeping the current dogma of the monopolar nature of the electric charge.

 

And as consequence:

 

2- Therefore it is impossible to explain the magnetic moment zero of the 2He4 by taking as starting point any nuclear model which do not work by considering a structure of the aether formed by particles and antiparticles as gravitons, magnetons, electricitons, etc.

 

3- Any nuclear model developed by no considering the participation of the aether within its structure is according to the monopolar nature of the electric charge, and therefore cannot be correct.

 

4- All the current nuclear models of the Standard Nuclear Physics are wrong, because they do not consider the aether within their structure, and so they cannot explain the magnetic moment zero of the 2He4.

 

Your conclusions are quite ridiculous.

 

 

5- Nowadays all the nuclear theories of the authors trying to explain the cold fusion cannot be entirely successful,

 

How can scientists explain something that never have been detected?

 

Fleischmann detected unexpected heat. At least that's what they said. We have just their words for it. That's it.

Their claim that it must be cold fusion, have as much value as your claim that magnetic moment 0 in Helium leads to existence of aether..

 

Fleischmann did electrolysis of heavy water in vacuum flask.

And instead of 30 C he measured 50 C.

I have many times reached 100 C (and have to stop) during electrolysis of normal water, but never thought about it as cold fusion.

 

First question that I would ask- what material is vacuum flask made of.

Usually they're made of aluminum. We would have to ask Fleischmann (he is dead now) or his team..

We can easily see that if it's made of aluminum, there will be Al(OH)₃ floating in it after a while of electrolysis.

The more of it floating, the more current is flowing.

I have literally kg of Al(OH)₃ (and after removing water Al₂O₃) created during electrolysis.

 

Another question- what electrolyte was used.

 

because they are developed by considering wrong nuclear models, since all they are based on the monopolar nature of the electric charge.

 

Are you saying that f.e. proton doesn't have 938.272 MeV etc. with other particles ?

 

If not, which part of nuclear models you don't agree?

 

Edited April 16, 2014 by Sensei

[Sensei]

 

The same we can say about all the other light even-even nuclei with Z=N pairs: 4Be8, 6C12, 8O16, 10Ne20, etc.

 

Beryllium-8 is very unstable isotope.

It's decaying to two alpha particles.

Therefore we have to eliminate all those nuclei from the Periodic Table of Elements, because from the known principles of Physics they cannot exist.

 

Or, what is much more probable, you have no idea why they are existing.

Existence of Helium-4 is doubtless.

You can just ask question- why is it stable.

 

I have calculated hypothetical Decay Energy of a few lighter elements, and every time isotope is stable Decay Energy is negative value.

Proton, Deuterium, Helium-3, Helium-4 don't have any hypothetical daughter element which is conserving both energy and electric charge.

When element is unstable you have Decay Energy which is higher than 0.

Helium-5 has Decay Energy > 0 only for neutron emission (DE<0 for beta decay-, beta decay+ and proton emission). And that's the way this isotope is decaying.

[swansont]

 

The nuclei have also rotation:

Phys. Rev. 53, 778 (1938) - On the Rotation of the Atomic Nucleus

 

 

I'm not going to address the semi-infinite number of problems with your post, nor address the questionable practice of copying an argument from one forum into another.

 

The above article is from 1938. Nineteen-frikkin'-thirty-eight. About five years after the discovery of the neutron. How mature was nuclear theory at that point?

 

If your point is that the nuclear theory of 1938 was not correct/complete, I don't think that there's going to be much argument. But if you're basing ANY argument of current nuclear theory on models that date back to then, I have to ask if you think anyone is going to take you seriously. If you want to critique the current state of nuclear physics, you'll need to do a much better job of demonstrating that you understand physics.

 

You might start with the fact that the abstract of that article (emphasis added)

The spacing of the levels in the fine structure of alpha- and beta-ray processes and the existence of metastable nuclear states (isomers, isobars) are in contradiction with the existence of low lying levels corresponding to the rotation of the nucleus as a whole.

 

 

implies that nuclei do not, in fact, rotate. (The article itself is paywalled.)