.

*by*

*U.V.S.Seshavatharam*

*Honorary faculty, I-SERVE, Alakapuri, Hyderabad-35, AP, India*

*QA-Spun division, LANCO Industries Ltd, Srikalahasti-517641, AP, India*

*E-mail: seshavatharam.uvs@gmail.com*

.

.

**Abstract**

Based on the big bang concepts- in the expanding universe, ‘rate of decrease in CMBR temperature’ is a measure of the cosmic ‘rate of expansion’. Modern standard cosmology is based on two contradictory statements. They are – present CMBR temperature is isotropic and the present universe is accelerating. In particle physics also, till today laboratory evidence for the existence of ‘dark matter’ and ‘dark energy’ is very poor. Recent observations and thoughts supports the existence of the ‘cosmic axis of evil’. In this connection an attempt is made to study the universe with a closed and growing model of cosmology. If the primordial universe is a natural setting for the creation of black holes and other non-perturbative gravitational entities, it is also possible to assume that throughout its journey, the whole universe is a primordial (growing and rotating) cosmic black hole. Instead of the Planck scale, initial conditions can be represented with the Coulomb or Stoney scale. Obtained value of the present Hubble constant is close to 71 Km/sec/Mpc.

.

.

ERRATA

In my last post the correct is:

1b- So, the fact that proton-neutron interaction is different of

neutron-neutron interaction is not relevantNew challenges for the Standard Model

http://www.livescience.com/38248-no-antineutrino-interaction-detected.html

New challenges for the Standard Model

JR wrote in July 18th, 2013 at 10:08 AM

Dear Wladimir,

1-

——————————————–

However, you still have yet to provide an argument to support this that isn’t based on data that doesn’t exist or using a cartoon-like version of nuclear physics.

——————————————–

COMMENT

I have showed you many inconsistencies of current Nuclear Physics.

For instance, the instability of 8Be,

not explained yet by the most important theorists, among them some Noble Laureate, as Hans Bethe.However, dear JR,as you are no able to understand what is the difference between CAUSE and EFFECT,

then it is easy to understand why you are not able to understand fundamental questions in Physics.2-

———————————————-

You claim that the fact that 2 neutrons don’t bind means that the nuclear force does not bind nucleons. I explained that this is because 2 neutrons interact differently from a neutron and a proton in well understood ways (and gave an example of why this is natural).

Your response was that the interaction between two particles has nothing to do with whether or not they bind.———————————————

COMMENT

No, I did not say what you claim.

What I said is the following:

1- Your explanation why two neutron do not bound is UNACCEPTABLE, because:

1a- Two neutrons interact via strong force.

1b- So, the fact that proton-neutron interaction is different of proton-neutron interaction is not relevant

1c- Because what is relevant is the following:

two neutrons interact via strong force. Therefore they must be bound via strong force.3-

———————————————-

From this, I concluded that you had given up trying to make meaningful arguments on this point and were basically admitting that you were wrong, so I left that discussion alone.

———————————————–

COMMENTNo, dear JR.

Actually the problem is with you,

since you cannot make distinction between CAUSE and EFFECT4-

————————————————

While it’s clear that your understanding of physics is limited, I’m now questioning whether you even understand math and numbers.

————————————————

COMMENT

No, actually your understanding is limited not only in Physics, but it is limited in general.

You dont know to make difference between CAUSE and EFFECT.

You cannot explain a phenomenon by asserting that it is caused by EFFECT produced by the phenomenon5-

————————————————

You claim that the proton radius is 0.26 fm. You claim that this will be proven by an experiment that’s trying to figure out if the radius is 0.88fm (the previously measured value) or 0.84fm, the value determined in the recent muonic hydrogen measurements. So are you claiming that if the new measurement supports the smaller (0.84fm) value, it will confirm that the radius is really 0.26fm? Or are you just claiming that every measurement made for the last 20 years has been wrong, including the new very precise measurement that raised the question of a small adjustment to the radius, and you’re predicting that the next measurement will get 0.26fm because some kind of pixie dust you pulled out of your ass??? You say something about it being related to the mass of the muon, but a muon measurement has been done as says you’re radius is off by a factor of 3.2 (compared to the older measurements that claim you’re wrong by a factor of 3.4). This is vindication for your model??? Being just slightly less way off from all existing measurements???

————————————————-

COMMENT

I am claiming nothing. Forget it.

Let us wait the results of the experiments.

regards

wlad

JR wrote in July 18th, 2013 at 9:50 AM

1-

Why isn’t 8Be bound?

In a very simple picture, a nucleus is bound if it’s mass is below that of all combinations of constituents (and it’s binding energy relative to those states is this mass difference – mass of constituents minus mass of bound state).

COMMENT

ha ha ha ha ha ha

JR, you are joker.

You are making confusion between the

CAUSEand theEFFECT.Also, you are calling stupid all the scientists who tried along decades to explain why 8Be is no stable, as for instance the Nobel Laureate Hans Bethe.

As the answer is so easy as you state, then you also are calling stupid the authors of the paper quoted by me:

http://adsabs.harvard.edu/abs/1971NuPhA.168..438B

The packing loss, quoted by you, is a

CONSEQUENCEof the fact that there are not strong interaction between the nucleons in the 4Be8.The packing loss is not the

CAUSENow I understand your misunderstanding in the fundamental problems of Physics:

You are not able to make distinction between what is CAUSE, and what is CONSEQUENCE.

2=

4He is bound because it’s mass is less than that of 2 proton plus 2 neutrons AND below that of 2 deuterons. If it were not bound with respect to 2p+2n or 2 deuterons, it would just fall apart into those constituents.

COMMENT

ha ha ha ha

4He is bound because there is strong interaction between its nucleons.

This is the CAUSE of why 4He is very stableThe packing loss is not the

CAUSE. It is theCONSEQUENCEThe

CAUSEof the 4He be bound is thestrong interaction between nucleons.If there were no strong interaction between nucleons in the 4He, it`s packing loss would be zero, as happens in the case of 8Be.

The question, which you are not able to understand, is the following:

Why there is strong interaction in 4He, but there is not in 8Be ???????????????????????????????This is just the question faced by Hans Bethe and many other who did never succced to solve the problem.

Well, dear JR,

I will recomend your name to the Nobel Laureate Commission, so that to give you the Nobel Prize in Physics in 2014, since you have solved a puzzle not solved by the most great theorists of Physics.

I think that one of the most serious problem with the current theorists like you is the missing of knowledge in the branch of Philosophy.

I suggest you to read the philosopher Kant, so that to learn the difference between

CAUSEandCONSEQUENCEregards

wlad

Dear Wladimir,

In response to your long treatise on the state of nuclear physics:

1) You assert that there is not a working model of nuclear physics. However, you still have yet to provide an argument to support this that isn’t based on data that doesn’t exist or using a cartoon-like version of nuclear physics.

2) You claim that the fact that 2 neutrons don’t bind means that the nuclear force does not bind nucleons. I explained that this is because 2 neutrons interact differently from a neutron and a proton in well understood ways (and gave an example of why this is natural). Your response was that the interaction between two particles has nothing to do with whether or not they bind. From this, I concluded that you had given up trying to make meaningful arguments on this point and were basically admitting that you were wrong, so I left that discussion alone.

3) While it’s clear that your understanding of physics is limited, I’m now questioning whether you even understand math and numbers. You claim that the proton radius is 0.26 fm. You claim that this will be proven by an experiment that’s trying to figure out if the radius is 0.88fm (the previously measured value) or 0.84fm, the value determined in the recent muonic hydrogen measurements. So are you claiming that if the new measurement supports the smaller (0.84fm) value, it will confirm that the radius is really 0.26fm? Or are you just claiming that every measurement made for the last 20 years has been wrong, including the new very precise measurement that raised the question of a small adjustment to the radius, and you’re predicting that the next measurement will get 0.26fm because some kind of pixie dust you pulled out of your ass??? You say something about it being related to the mass of the muon, but a muon measurement has been done as says you’re radius is off by a factor of 3.2 (compared to the older measurements that claim you’re wrong by a factor of 3.4). This is vindication for your model??? Being just slightly less way off from all existing measurements???

Steven N. Karels

I am totally impressed. Congratulations on your water mission. Hopefully in the near future the 8 watts can be supplied by an LENR E-Cat!

Dear Wladimir,

Why isn’t 8Be bound?

In a very simple picture, a nucleus is bound if it’s mass is below that of all combinations of constituents (and it’s binding energy relative to those states is this mass difference – mass of constituents minus mass of bound state). 4He is bound because it’s mass is less than that of 2 proton plus 2 neutrons AND below that of 2 deuterons. If it were not bound with respect to 2p+2n or 2 deuterons, it would just fall apart into those constituents.

8Be is not bound with respect to two 4He nuclei, so it has a ‘negative binding energy’ if you want to think of it that way, and it just falls apart.

Steven N Karels:

Congratulations,

Warm Regards,

A.R.

Dear Andrea Rossi,

Two months from now I will be installing a prototype village water purification system in Guatemala in the mountains where the Mayan Indians live. It is designed to provide potable water for 100 – 200 and uses only 8W of electrical power during the purification process. The parts cost is under $200USD per system.

To the readers of Rossi`s JNP

THE HISTORY OF NUCLEAR PHYSICSIn the end of the 19th Century the theorists faced a big puzzle: to explain the Balmer scale.

The physicist Voigt proposed a theory developed via mathematics. So, it was missing in his theory any

physical principlefrom which would be possible to explain the Balmer scale.Then Bohr had discovered his hydrogen model of atom. His theory works with physical principles, as for instance an electron moving about a proton, the electron jumping about levels in the proton`s electrosphere and emitting photons, etc.

Thanks to Bohr`s discovery, Schrodinger was able to discover his famous equation, and so thanks to Bohr discovery it was possible to develop Quantum Mechanics.

If Bohr had not discovered his hydrogen model, the Quantum Mechanics would not be developed by having physical principles. Actually Quantum Mechanics would be developed via mathematics, via Voigt theory.

Nowadays we live in Nuclear Physics that same pre-Bohr situation in Atomic Physics. There is not a nuclear model working physical principles.

What the nuclear theorists make today is similar to that mathematical attempt used by Voigt , before Bohr`s discovery of his atom model.

Some principles were established, but they are wrong, and so they deviate the theorists from the correct way.

For instance, the nuclear theorists believe that the nucleons are bound within the nuclei via strong nuclear force. But if that should be the case, two neutrons would have to be bound via strong force, and the dineutron would have to exist, but it does not exist.

So, the fact that dineutrons do not exist suggest that nucleons are not bound within nuclei via strong force.

Why have the nuclear theorists supposed that nucleons are bound via strong force?

The reason is because there are only positive charges within the nuclei (protons), and there is Coulomb repulsions between them. So the theorists supposed that there is need an attractive force capable to keep them within the nuclei. They called such hypothetical force as strong nuclear force.

Then they undertook scattering experiments proton-proton and neutron-proton, so that to verify if strong force really exists. And the experiments showed that strong force indeed exists.

However the strong force is not a fundamental force of nature. Because the experiments have shown that the intensity of the interactions proton-proton and proton-neutron in the experiments depends on the relative velocity between the nucleons that collide. If the relative motion is small, the strong interaction is weaker.

If strong force should be fundamental, it could not change its intensity with the change of the relative motion between the nucleons that collide.

Probably strong force is only a kind of dynamic gravity: the relative motion of the nucleons in collision change the magnitude of the gravitational interaction between them.

That`s why two neutrons do not bound so that to form a dineutron: because there is not enough relative motion between them in the dineutron, so that there is not strong force interaction between them.

Therefore, so that to develop the Nuclear Physics in the correct way, there is need before everything to discover the fundamental laws that rule the interaction of nucleons within the nuclei.

This is just one of the proposals of Quantum Ring Theory. In the nuclear model of QRT the nucleons are not bound within the nuclei via strong force.

But how does to explain why the protons are not expelled from the nuclei via the Coulomb repulsions?

In QRT the proton (and the electron) have two concentric fields. The internal field is named Principal Field Sn(p), and the outer field is named Secondary Field Sn(e) .

The Coulomb repulsions are caused by the secondary field Sn(e).

But in order to pack together two nucleons so that to form a nucleus, there is need to perforate the secondary field. And when the nucleons are inside the secondary field, they do not experience anymore the Coulomb repulsions.

And how does occur the stability of nuclei?

The stability of light nuclei occurs thanks to the equilibrium of two forces: the electric force on protons yield by the central 2He4 (such force is induced by the rotation of the protons about the central 2He4), and the other is the centripetal force on the protons.

From such nuclear model we can explain the stability and instability impossible to be explained via current nuclear models. For instance, it explains why 4Be8 is no stable, although from the current nuclear models (

which work via strong force) 4Be8 would have to be stable.The stability of light nuclei (some of them impossible to be explained via strong force) is explained in my paper

Stability of Light Nuclei, published in the Rossi`s JNP:http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

So, we live today the

Voigt Era of Nuclear Physics.The new nuclear model proposed in my Quantum Ring Theory represents for Nuclear Physics the same what Bohr hydrogen model of atom had represented for the development of Atomic Physics.

How many years the nuclear physicists will waste before to realize it, only the time will show us.

In particular, my theory predicts that a free proton has a radius very larger than a proton confined within the nuclei. In my theory it is predicted that proton`s radius within the nuclei is R= 0,26fm.

Up to now, the radius of proton had been measured via proton-electron scattering. It has obtained a radius in order of 0,8fm.

A new experiment, to be made between 2015 and 2016, will measure the proton`s radius via proton-muon scattering. As the muon is more massive than the electron, I hope that in the proton-muon scattering their collision will reproduce a similar interaction as occurred within the nuclei, and so in this new experiment the proton`s radius to be obtained will be a little larger than 0,26fm (but very shorter than 0,8fm).

Also, the experiments can show if the shrinkage depends on the velocity of collision proton-muon: faster relative motion proton-muon yielding stronger shrinkage.

If the experiments to be made in 2015-2016 confirm the proton radius predicted in my theory, perhaps the nuclear physicists will decide to consider seriously the new nuclear model proposed in my Quantum Ring Theory.

Regards

wlad

Dear JR

I have another question to be answered by you.

Except 4Be8, all the nuclei with Z=N=parr are stable.

Look:

2He4

6C12

8O16

1020Ne

12Mg24

14Si28

etc

etc

etc

All they are stable, but 4Be8 is not.

There is not any satisfactory explanation yet.

In the paper linked bellow they try to explain the instability of 8Be by that same unacceptable method used by Heisenberg so that to explain why two neutrons do not are bound by the strong force, so that to form the dineutron, that is, via mathematics:

http://adsabs.harvard.edu/abs/1971NuPhA.168..438B

Of course the mathematics cannot create a force so that to avoid 4 protons and 4 neutrons to be bound strongly via strong nuclear force.

This is typical. When the theorists do not succeed to explain something about the aggregation of nucleons within the nuclei,

by considering the fundamental principles of current Nuclear Theory, they use to try to explain by the use of the mathematics,by rejecting the principles of Nuclear Physicsin which they believe.So, dear JR,

I would like you explain us why 4Be8 is no stable,

BY CONSIDERING THE PRINCIPLES OF CURRENT NUCLEAR PHYSICS.Regards

wlad

Wladimir wrote:

”

The mystery of 7Be quadrupole moment

… bla, bla, bla …

END OF THE MYSTERY

”

I am rather convinced that it has not been measured. However, it is important to remember that the table you refer to is just a list of references with associated experimental data which may or may not be complete. Had there been any measurements that for some reason deliberately had not been included in the table (perhaps because they indicated a value of zero) you should anyway be able to point to the original publication. Did you find any such data that could support your claims? Otherwise your claims are just bla, bla, bla.

Wladimir wrote:

”

I only see that both you share the opinion that the neutron has a fantastic property of expanding itself like a rubber.

However the scattering experiments, or any other experiment, did never show that the neutron has such a fantastic property of expansion.

Obviously you can explain everything if you adopt crazy fantastic explanations.

But instead of to propose that neutron has a crazy fantastic elasticity, which allow to it to behave like a rubber, you can explain it by another most reliable explanation, as follows:

you can claim that the halo neutron is kept thanks to the help of the Santa Claus.

”

It seems to me that you still believe neutrons are small hard spheres orbiting the center of gravity. Do you understand the concept of wave functions used in the page you used as a reference?

Wladimir wrote:

”

However, the value quoted by you is Q(b) = +0,11 .

While concerning 12C, 36Ar, 60Zn my theory predicts µ= 0 , as the updated table 2006 indeed quotes for 36Ar.

Excited 12C also has non-null Q(b)= +0,06, but it has i=2 and µ= 0.

”

No. You are either looking in the 2005 table or in the wrong row. In the 2011 version on

http://www-nds.iaea.org/nsdd/indc-nds-0594.pdf

page 24 you will find two rows for the 2+ state of 36Ar. The first row gives the value of µ=+1.0(4), the second Q=0.11(6). This violates your prediction regarding the magnetic moment and seems to disprove your theory as such.

Hope this helps,

Erik

Joe wrote in July 17th, 2013 at 6:16 PM

Why could not the 2 excess neutrons of a nucleus have their magnetic moments point in OPPOSITE directions, thereby giving a total magnetic moment of zero (presence of symmetry) to the nucleus?

COMMENT

They can.

For instance it occurs in non-excited 20Ca42, 20Ca44, 92U238, etc.

regards

wlad

Wladimir,

Your argument from almost the very beginning is “they

musthave measured it by now”. It wasn’t convincing then, and it isn’t convincing now. It’s also silly to claim that they couldn’t put a result consistent with zero in the table. In fact the table does quote individual measurements that are consistent with zero, and in at least one case it quotes an upper limit from a measurements that did not see a value.If you’re going to read “as yet unknown” as “measured but it’s zero and we don’t believe it can be zero” then there’s no point trying to have a meaningful discussion with you because you’re back to just making up measurements based on your desire that they be there and give what you want. If you think there’s been a measurement, find a reference. If not, stop coming up with nonsensical justifications for your nonsensical claim that a blank space means that the value is measured to be exactly zero. Or are you going to claim that the dozens and dozens of excited states in heavier nuclei that don’t list a Q(b) value have Q=0?

p.s. the new tables also don’t leave blank spaces for everything that’s zero. Please don’t make us correct every single mistake you made based on the 2005 tables one more time with the newer tables.

Erik wrote in July 17th, 2013 at 5:45 PM

————————————————-

I found an updated table (2011) by Stone on the home page of IAEA. One recent (2006) item of interest is the first excited state of 36Ar (2+) which is claimed to have a magnetic moment of 1 nm. You will find the new table using the link below:

http://www-nds.iaea.org/nsdd/indc-nds-0594.pdf

————————————————–

COMMENT

Thank you, Erik

However, the value quoted by you is Q(b) = +0,11 .

While concerning 12C, 36Ar, 60Zn my theory predicts µ= 0 ,

as the updated table 2006 indeed quotes for 36Ar.Excited 12C also has non-null Q(b)= +0,06, but it has i=2 and µ= 0.

regards

wlad

The mystery of 7Be quadrupole momentDears JR, Erik, and Joe

I found an article in which the authors had calculated the Q(b) of 7Be:

http://cds.cern.ch/record/344733/files/9802003.pdf

It seems that Q(b) of 7Be defies the theories of nuclear physics, because:

1- From current nuclear models 7Be cannot have Q(b)=0 , or at least very close to zero.

2- In spite of the theorists calculate the Q(b), and they find Q(b) different of zero, however the experiments do not succeed to find any value different from zero.

In that paper quoted by JR,

Large-basis shell-model calculations for p-shell nuclei, the authors had calculated it to be:Q(b)= -4.631 fm^2

http://cds.cern.ch/record/351635/files/9804014.pdf

In this present paper quoted now by me, they calculated it to be:

Q(b) -> between -6fm^2 and -7fm^2

In the paper the authors say:

——————————————

In Ref. [9] we have shown that the zero-energy cross section of the 7Be(p; γ)8B reaction scales linearly with the,

unfortunately yet unknown, quadrupole moment of 7Be.——————————————

In the end of the paper (last paragraph before the Summary) they say:

——————————————-

We note again that

a measurement of the 7Be quadrupole moment would place some additional constraints on the consistency of our calculations.For the complete 4He+3He+p model calculation the simultaneous reproduction of the indicators predict Q7 to be in the range −(5:5−6:0) e fm2. However, this value is smaller than the one (Q7= −6:9 e fm2[9]) obtained if we chose the cluster size parameters such to reproduce the quadrupole moment of the analog nucleus 7Li. Does this already point to the necessity of a further enlargement of the model space beyond the 4He + 3He + p three-cluster model which would then also efect our results obtained for 7Be, e.g.,

change the 7Be quadrupole moment?——————————————-

Then, the mystery:1-

Of course the measuement of 7Be Q(b) is very important for the advance of the theoretical calculations.

2-

The paper quoted here had been published in 1998. And JR told us that the paper quoted by him is about 15 years old.

3-

So,

along 15 years the experimental physicists did not succeed to measure the Q(b) of 7Be ???????4-

I dont think so, since such measurement is very important, as the authors of the paper had pointed out, when they wrote:

———————————————

We note again that a measurement of the 7Be quadrupole moment would place some additional constraints on the consistency of our calculations.

———————————————

5-

It is hard to believe that the experimental physicists did not succeed to measure it.Of course they had tried several times,but they did find NOTHING, because the Q(b) of 7Be isvery close to zero, as pointed out in my paper published here in Rossi`s JNP:It it is stated in the Figure 47 , page 48, of my paper

Stability of Light Nuclei:The experiments do not detect the electric quadrupole moment of 4Be7 because it is very small, near to zero, because…http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

6-

Of course the nuclear tables cannot publish the value zero for the Q(b) of 7Be, because the tables never quote the value zero, since there is no way to assert that the value is really zero, because it can be very near to zero, but not zero, due to uncertainty of measurement.

7-

However, no matter if Q(b) of 7Be is zero or near to zero, the fact is the following:

————————————————-

There is no way to calculate Q(b) of 7Be near to zero, by considering the current nuclear models, because:A) it has A= 7 (odd),

and therefore it cannot have a spherical distribution of chargeB) it has non-null spin and non-null magnetic moment, and so it cannot behave like 8O16, which is non-spherical (the experiments do not detect Q(b) of 8O16 because it has i=0 and µ= 0 ).

————————————————-

Therefore it seems the mistery is solved.

The theorists say:

—————————————————

…

unfortunately yet unknown, quadrupole moment of 7Be—————————————————

because they expect a value very different from zero.

However, as the experiments do not succeed to measure a Q(b) far away of zero (as they expect), then then they claim that it is

unfortunately yet unknown.END OF THE MYSTERYregards

wlad

Wladimir,

Why could not the 2 excess neutrons of a nucleus have their magnetic moments point in OPPOSITE directions, thereby giving a total magnetic moment of zero (presence of symmetry) to the nucleus?

All the best,

Joe

Erik wrote in July 17th, 2013 at 10:26 AM

——————————————-

If you had read and understood the last paragraph in your link it had been obvious that Dr. Nörtershäuser, JR and I share the same opinion in this case:

Thus, it is highly likely that the neutron can expand into classically forbidden distances, thereby inducing the expansive ‘heiligenschein’.——————————————–

COMMENT

I only see that both you share the opinion that the neutron has a fantastic property of expanding itself like a rubber.

However the scattering experiments, or any other experiment, did never show that the neutron has such a fantastic property of expansion.

Obviously you can explain everything if you adopt crazy fantastic explanations.

But instead of to propose that neutron has a crazy fantastic elasticity, which allow to it to behave like a rubber, you can explain it by another most reliable explanation, as follows:

you can claim that the halo neutron is kept thanks to the help of the Santa Claus.regards

wlad

Wladimir wrote:

”

Dear JR,

from the sequence of multiples 6C12 in the structure of light nuclei, I had showed that:

excited 6C12 , 18Ar36, 30Zn60 have to have µ= 0.

Do you know any article where their magnetic moments are calculated?

I was looking for excited 6C12 in the google, but I did find nothing.

I am convinced that current nuclear models cannot calculate the magnetic moments of excited 6C12, 18Ar36, and 30Zn60.

I would like you prove me wrong.

regards

wlad

”

I found an updated table (2011) by Stone on the home page of IAEA. One recent (2006) item of interest is the first excited state of 36Ar (2+) which is claimed to have a magnetic moment of 1 nm. You will find the new table using the link below:

http://www-nds.iaea.org/nsdd/indc-nds-0594.pdf

Reference is given in the right column as always.

Hope this helps,

Erik

Dear Wladimir,

You say: “

from the sequence of multiples 6C12 in the structure of light nuclei, I had showed that: excited 6C12 , 18Ar36, 30Zn60 have to have µ= 0.”You have not showen that, you have argued that your approach (QRT, I presume) predicts this. So I’m not terribly interested in whether conventional theory makes the same prediction or not. In any case, my goal is not to try and disprove QRT, as I do not wish to invest the time that would be required to understand it first. My point was only that your claims that conventional nuclear theory has a myriad of problems was not, in fact, based on a realistic evaluation of the data or the theory.

Yona:

You can email to

info@leonardocorp1996.com

Warm Regards,

A.R.

JR wrote in July 17th, 2013 at 8:43 AM

——————————

Dear Wladimir,

At best, you’ve claimed to show examples where current techniques to perform the calculations are not good enough (and probably not expected to be good enough), but given your poor success rate with claims we could directly evaluate, I’m not convinced you’ve shown anything at all of meaning.

——————————–

Dear JR,

from the sequence of multiples 6C12 in the structure of light nuclei, I had showed that:

excited 6C12 , 18Ar36, 30Zn60 have to have µ= 0.

Do you know any article where their magnetic moments are calculated?

I was looking for excited 6C12 in the google, but I did find nothing.

I am convinced that current nuclear models cannot calculate the magnetic moments of excited 6C12, 18Ar36, and 30Zn60.

I would like you prove me wrong.

regards

wlad

JR wrote in July 17th, 2013 at 8:43 AM

Dear Wladimir,

Where to begin? Erik saved some time by pointing out that the two papers have different units, so your imagined factor of 100 errors are wrong. Your newest factor of 100 error comes from randomly multiplying the numbers in the table by Stone by 100 but then quoting the original units.

COMMENT

Dear Jr,

sorry, I made a mistake. Erik is wright.

The article uses the same unity for the magnetic moments, but it uses a different unity for Q(b). That`s why my mistake.

I did the same in my paper ELECTRIC QUADRUPOLE moment in my book QRT (page 141), where I calculate the Q(b) of 8O18.

I made the calculations in fm^2, and in the end of the article I d hatransfered them to barns, so that to compare with the values of Stone table.

regards

wlad

Dear Joe,

I did not quote the 26Fe in the sequence of 6C12 multiples posted yesterday.

Now I am quoting it.

——————————————–

6C12 – one incomplete hexagonal floor

Ex …….. i …….. µ

4438 …… 2 …….. ????

The structure of 6C12 is perflectly symmetric

——————————————–

——————————————–

12Mg24 – one complete hexagonal floor + 6C12

Ex …….. i …….. µ

1369 …… 2 …….. +1,02

The structure of 12Mg24 is NO symmetric. That` why it has µ= +1,02

——————————————–

——————————————–

14Si28 – two complete hexagonal floors

Ex …….. i …….. µ

1779 …… 2 …….. +1,1

It has not the 6C12 in its structure.

That`s why it has µ= +1,1

——————————————–

——————————————–

18Ar36 – two complete hexagonal floors + 6C12

Ex …….. i …….. µ

1970 …… 2 …….. ????

Note that 18AR36 is similar to 6C12, because:

* there are two complete hexagonal floors in 18Ar36

* In 18AR36 the 6C12 takes place between the two hexagonal floors.

So, the structure of 18Ar36 is perflectly symmetric

That`s why excited 18Ar36 has µ= 0 ,as also happens with 6C12.

——————————————–

——————————————–

20Ca40 – three complete hexagonal floors

Ex …….. i …….. µ

3737 …… 3 …….. +1,6

It has not the 6C12 in its structure.

That`s why it has µ= +1,6

——————————————–

——————————————–

24Cr50 – three complete hexagonal floor + 6C12 + 2n

Ex …….. i …….. µ

783 ……. 2 …….. +1,24

The structure of 24Cr50 is NO symmetric, because of the two neutrons.

That` why it has µ= +1,24

——————————————–

——————————————–

26Fe52 – four complete hexagonal floors

Not quoted in Stone table

However, the excited 26Fe52 cannot have µ= 0, because it has not the 6C12 in its structure (in the middle of the four hexagonal floors.

——————————————–

——————————————–

26Fe54 – four complete hexagonal floors + 2 neutrons

Ex …….. i …….. µ

1408 …… 2 …….. +2,1

It has not the 6C12 in its structure, and it is not perfectly symmetric.

That`s why it has µ= +2,1

——————————————–

——————————————–

30Zn60 – four complete hexagonal floors + 6C12

Unfortunatelly it is not quoted in Stone table.

It has PERFECT SYMMETRY + 6C12

By considering the sequence of multiples 6C12 with perfect symmetry,

the excited 30Zn60 must have 1=2 and µ= 0, as the excited 6C12 and 18Ar36.——————————————–

——————————————–

30Zn62 – four complete hexagonal floors + 6C12 + 2n

Ex …….. i …….. µ

954 …… 2 …….. +0,7

* In 30Zn62 the 6C12 takes place in the middle of the four hexagonal floors.

However the two neutrons broke the perfect symmetry

That`s why excited 30Zn62 has µ= +0,7

——————————————–

Regards

wlad

Dear Andrea Rossi

Is it possible to call you? i want to ask you some question.

Warm Regards

Wlad wrote:

http://www.uni-mainz.de/eng/13031.php

Dr. Nörtershäuser does not share your opinion.

But of course he is wrong. He does not understand Physics.

Sure that all the physicits know nothing about Physics in that university.

Go to teach Physics to them.

If you had read and understood the last paragraph in your link it had been obvious that Dr. Nörtershäuser, JR and I share the same opinion in this case:

The measurements revealed that the average distance between the halo neutrons and the dense core of the nucleus is 7 femtometers. Thus, the halo neutron is about three times as far from the dense core as is the outermost proton, since the core itself has a radius of only 2.5 femtometers. “This is an impressive direct demonstration of the halo character of this isotope. It is interesting that the halo neutron is thus much farther from the other nucleons than would be permissible according to the effective range of strong nuclear forces in the classical model,” explains Nörtershäuser. The strong interaction that holds atoms together can only extend to a distance of between 2 to 3 femtometers. The riddle as to how the halo neutron can exist at such a great distance from the core nucleus can only be resolved by means of the principles of quantum mechanics: In this model, the neutron must be characterized in terms of a so-called wave function. Because of the low binding energy, the wave function only falls off very slowly with increasing distance from the core. Thus, it is highly likely that the neutron can expand into classically forbidden distances, thereby inducing the expansive ‘heiligenschein’.

Wladimir wrote:

“and I have the unpleasure to announce that your argument imply that ALL THE OTHER values of the paper are wrong, since all they are near to those exhibited in Stone`s table.

Look:

3Li8

Stone : Q(b) = +2,208 barn

paper : Q(b) = +3,11 fm^2 = +0,0311 barn

3Li9

Stone : Q(b) = -2,085 barn

paper : Q(b) = -2,74 fm^2 = -0,0274 barn

5B10

Stone : Q(b) = +5,643 barn

paper : Q(b) = +8,472 fm^2 = +8,472 barn

.. bla, bla, bla …

”

I do not know which version of the table you use. I use the version from Atomic Data and Nuclear Data Tables 90 (2005) pp 75–176. In this version values of about 0.03, 0.03 and 0.08 b are given for 8Li, 9Li and 10B respectively, which are in good agreement with the values used in the paper.

You have done it again! Please go on humiliating yourself!

Joe wrote in July 17th, 2013 at 3:27 AM

Wladimir,

How does QRT explain the lack of symmetry (non-zero magnetic moment) in a nucleus that contains 2 or 3 COMPLETE hexagonal floors?

COMMENT

Joe,

all nuclei with complete hexagonal floors having pair number of neutrons have null magnetic moment

When they are excited, one nucleon changes its spin

In the Fig. 3.1, page 146, of my book Quantum Ring Theory it is proposed that when 8O16 is excited with Ex= 6130, one nucleon of the central 2He4 changes its spin, and also a deuteron of the hexagonal floor changes its spin, so that the excited nucleus has spin i=3.

regards

wlad

Dear Wladimir,

Where to begin? Erik saved some time by pointing out that the two papers have different units, so your imagined factor of 100 errors are wrong. Your newest factor of 100 error comes from randomly multiplying the numbers in the table by Stone by 100 but then quoting the original units.

From here, let’s go back to your oldest comments/mistakes first, and go forward. I pointed to 3 papers that calculated the magnetic moment of 7Li in conventional frameworks, which is what you requested. You point out that the 7Be value is way off. So first of all, just because one calculation is off doesn’t mean that underlying picture is off; some nuclei can be calculated more reliably, others less reliably. Since your really mean 11Be, not 7Be, it’s not so surprising that this result would be less reliable (and I’ll note that the calculation is 15 years old, things do get better with time). Actually, I think you have a fundamental misunderstanding of the issues involved in these calculations, but I’ll make that a separate post.

You also appear to claim that because the ‘GMC and GFMC’ calculations involve using computers, they must simply be an attempt to fit the theory to the data. That’s just nonsense.

11Be –> 11B decay. You’re still wrong, for all the same reasons. You haven’t added anything new to the argument that I can see, so the old responses still apply. Now it is true that the coulomb force falls off as 1/r, while the strong force (in this context) falls off as exp(-r). So yes, there is a point at which the coulomb force is stronger, and that may well include 7fm. However, these are not balls on strings or sticks. The halo neutron does not sit 7fm apart from the other nucleons at all time. It has a wavefunction, with an average (or probably root-mean-square) distance of 7fm from the core, but it does spend time close to the core as well, and the binding effect from it’s overall distribution is much stronger than that coming from the coulomb potential if it were to turn into a proton.

Long description of what your model predicts for states where we don’t actually know the magnetic moment: irrelevant. We’re talking about whether conventional models can describe nuclei, not your model. And even if we were evaluating your model (and I have no such desire), we’d want to compare to known quantities, not made up ones.

You then take some more random quotes about individual calculations and their assumptions/limitations to paint all of conventional nuclear physics as arbitrary and wrong. That too, is incorrect. I’ll discuss the broader point in another message. I’ll simply state for now that differentiating between “the shell model” and “the core model” have more to do with how one evaluates the nuclear structure given a specific interaction, they aren’t different interactions (and, in fact, there is no such thing as “the core model”, but we could put a real approach in it’s place and the same would hold true). In addition, numerical convergence for a monte carlo calculation is related to numerical issues, and doesn’t tell you anything about the underlying model being good or bad, just that that approach is not well suited to that observable.

Finally, the fact that people are still making measurements and trying to do better calculations does mean that we do not yet have a complete set of tools to answer all questions about nuclear structure. There are many thing that we can not yet predict, and it may be that there are some that we will not be able to predict with our current understanding of nuclear physics. But the fact that some things can’t yet be calculated does not imply that the model is entirely wrong. And despite all your bluster, you have yet to present an example where the results are at odds with the conventional picture. At best, you’ve claimed to show examples where current techniques to perform the calculations are not good enough (and probably not expected to be good enough), but given your poor success rate with claims we could directly evaluate, I’m not convinced you’ve shown anything at all of meaning.

Erik wrote in July 17th, 2013 at 7:15 AM

————————————————

4- There is nothing strange about a weakly bound neutron. Even using the simple shell model you can understand why the neutron does not fall back into the nucleus and the proton is not falling out. You only need basic understanding of how it works.

————————————————

COMMENT

Dear Erik

I suggest you to go teaching Physics for the physicists in the Institute of Nuclear Chemistry of the Johannes Gutenberg University Mainz.

Look what said Dr. Nörtershäuser:

——————————————–

This is an impressive direct demonstration of the halo character of this isotope. It is interesting that the halo neutron is thus much farther from the other nucleons than would be permissible according to the effective range of strong nuclear forces in the classical model.

By studying neutron halos,

scientists hope to gain further understandingof the forces within the atomic nucleus that bind atoms together,taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics.———————————————

http://www.uni-mainz.de/eng/13031.php

Dr. Nörtershäuser does not share your opinion.

But of course he is wrong. He does not understand Physics.

Sure that all the physicits know nothing about Physics in that university.

Go to teach Physics to them.

regards

wlad

ERRATA:

in my last post the correct is

5B10

Stone : Q(b) =

+5,643 barn/b>paper : Q(b) = +8,472 fm^2 =

+0,08472 barnErik wrote in July 17th, 2013 at 4:51 AM

==================================================Wlad wrote:

”

Dears JR and Erik

There is an error in that paper which link had been posted by JR:

Large-basis shell-model calculations for p-shell nuclei

http://cds.cern.ch/record/351635/files/9804014.pdf

Look at the table in the page 16 :

4Be7 : Values of Q(b):

CALCULATION = +3,245

EXPERIMENT = +5,29(4)

However, according to the Jones nuclear table the correct value is:

EXPERIMENT: +0,0529(4)

… bla, bla, bla …

I have the pleasure to announce that the unit used in the paper is fm^2 while the unit used in the table is barn. Believe it or not: 100fm^2=1b, so the experimental values are identical. There might be other errors in the paper, so please go on humiliating yourself.================================================

COMMENT

and I have the unpleasure to announce that your argument imply that

ALL THE OTHER values of the paper are wrong, since all they are near to those exhibited in Stone`s table.Look:

3Li8

Stone : Q(b) =

+2,208 barnpaper : Q(b) = +3,11 fm^2 =

+0,0311 barn3Li9

Stone : Q(b) =

-2,085 barnpaper : Q(b) = -2,74 fm^2 =

-0,0274 barn5B10

Stone : Q(b) =

+5,643 barnpaper : Q(b) = +8,472 fm^2 =

+8,472 barnetc

etc

etc

Dear Erik

thank you.

Because thanks to your comment now we realize that ALL THE VALUES of the paper ARE WRONG.

Only that value of 4Be7, pointed by you, is correct:————————————————

4Be7

Stone: Q(b)=

+0,0529 barnpaper : Q(b)= +3,245 fm^2 =

+0,03245 barn————————————————

Thank you, Erik

you really has helped us to understand that the paper is a shitregards

wlad

Again I happen to interfere in a discussion betwen Wladimir and JR. I am perfectly aware that JR may answer himself, so I need to apologize for my behaviour.

Wlad wrote:

”

1- There is not Coulomb repulsion on the halo neutron in the 4Be11

2- If your argument should be correct, the halo neutron would not exist in the 4Be11, because the neutron would go back to the nucleus 4Be11.

3- As you claim that a halo proton (under Coulomb repulsion) in the new-born 5B11 goes back to the nucleus 5B11, then why a halo neutron in 4Be11 does not succeed to go back to the nucleus 4Be11 ???? (since there is not any Coulomb repulsion on it).

4- From your argument the 4Be11 cannot have halo neutron

”

1- You can put it that way.

2- Using the shell model (did you google yet?) two neutrons are in the 1s1/2 state, four in the 1p3/2 and the last one (although it may be critisized you may refer to this as the “halo”) in the 1p1/2 state. Due to the Pauli principle (please use google if you are not familiar with this one) the halo neutron can not move into already occupied states. So the neutron will be stuck in the halo state.

3- When you replace one of the neutrons with a proton there will be available 1p3/2 states because only the 1s1/2 and two 1p3/2 states are used by the four protons in 11Be. So the proton may go into more “thightly bound” states than the original halo neutron.

4- There is nothing strange about a weakly bound neutron. Even using the simple shell model you can understand why the neutron does not fall back into the nucleus and the proton is not falling out. You only need basic understanding of how it works.

Arthur B:

As I said, the E-Cat technology is undergoing extremely rigorous tests.

Warm Regards,

A.R.

Wlad wrote:

”

Dears JR , Erik , and Joe

The missing of µ for the excited 6C12 in nuclear tables is not due to what Erik had supposed.

He said:

” I think they have simply not been measured yet for various reasons “

Actually there is a very interesting sequence of values of µ which depends on the multiples of 6C12 in the formation of nuclei.

… bla, bla, bla …

”

Still you have not presented any reference or experimental data that suggests that the magnetic moment of the first excited state in 12C would equal zero. That your model says so does not convince anyone and does not prove anything.

You also refer to the excited states as the excited state. Are you aware that each nucleus may have several excited states with may have the same spin and parity? How is the spin of these states (and the ground state) derived in your model?

Wlad wrote:

”

Dears JR and Erik

There is an error in that paper which link had been posted by JR:

Large-basis shell-model calculations for p-shell nuclei

http://cds.cern.ch/record/351635/files/9804014.pdf

Look at the table in the page 16 :

4Be7 : Values of Q(b):

CALCULATION = +3,245

EXPERIMENT = +5,29(4)

However, according to the Jones nuclear table the correct value is:

EXPERIMENT: +0,0529(4)

… bla, bla, bla …

”

I have the pleasure to announce that the unit used in the paper is fm^2 while the unit used in the table is barn. Believe it or not: 100fm^2=1b, so the experimental values are identical. There might be other errors in the paper, so please go on humiliating yourself.

Wladimir,

How does QRT explain the lack of symmetry (non-zero magnetic moment) in a nucleus that contains 2 or 3 COMPLETE hexagonal floors?

All the best,

Joe

Dear Dr. Rossi,

Thank you very much for taking the time to respond to my posting. The fact that you are all working very, very hard is really good news. And I fully understand why it is not yet time to talk. However, I do look forward to further developments sooner than later.

Kind Regards

Arthur B

Svein Utne:

It is more easy for a camel to pass through a needle eye than for an E-Cat be certified to be used on a balloon ( so far and for the next 20 years).

Warm Regards,

A.R.

Arthur B:

We are working very, very hard. This is not time for talks.

Warm Regards,

A.R.

Hi Dr. Rossi,

I interpret your silence as “No news is good news”.

I guess that you now have your first E-Cat model available for market and it is now in or close to whole sale production. I’m thinking we will all know soon than later.

Dear Andrea,

It would be nice to use the Hot E-cat for heating hot air balloons. Also it would make some eye opener to make a circumnavigation with only some E-Cats instead of tons of propane. Maybe Richard Branson would like to make a new try?

Regards

Svein Utne

JR wrote in July 15th, 2013 at 11:57 AM

——————————————–

Dear Wladimir:

Your request: “please explain to orsobubu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.”

My reply:

http://arxiv.org/pdf/1201.0724.pdf

———————————————

Dear JR,

I have read the paper, and the conclusion is the following:

the nuclear theorist did not discover yet how the nucleons interact within the nuclei.So, sometimes they use a shell-model, and sometimes they use core-model.

Look what the authors say in the SUMMARY AND OUTLOOK:

————————————————-

“For certain observables that are more sensitive to long-range correlations (the RMS radius, electric quadrupole moments, and reduced quadrupole transition probabilities)we were unable to obtain full convergence, though they also compare favorably with alternate methods and interactions.”—————————————————

Now look at what is written in the article

Magnetic moments of short-lived excited nuclear states: measurements and challengeshttp://users.uoa.gr/~tmertzi/ClassFiles/Refs/koller01.pdf

—————————————————

Summary and future

The measurements of the magnetic moments of short-lived nuclear excited states has provided over the years a unique perspective into the microscopic underpinnings of their structure.

In many cases, these measurements have made it possible to determine fairly accurately whether the nuclei were behaving more as a collection of single-particles or more in a collective mode.Accurate measurements over the periodic table, in single nuclei as well as within isotopic chains,

have placed constraints on nuclear modelsand on theinterplay between different descriptions, as well as on thechoices of nucleon-nucleon interactions.In particular, the versatility, sensitivity and time structure of the transient ﬁeld technique has generated a whole gamut of experiments which focused on the evolution of nuclear properties as a function of energy, angular momentum and isotopic spin for both discrete nuclear states and for states in the continuum.

The ability to extend these experiments in the future, with the development of improved technology and the advent of copious beams of radioactive nuclei, coupled to theoretical progress in large scale

shell-model calculations, Monte Carlo techniques, andwill greatly expand the reach ofalgebraic methods

our understanding of nuclear structure.The combination of the measurements of magnetic moments of short-lived excited states via the transient ﬁeld technique together with the measurements of magnetic dipole and electric quadrupole moments of long-lived nuclei via a plethora of other techniques,

will playan important and fundamental role in future studies of nuclear structure.

—————————————————-

Therefore, dear JR,

there is not yet a consistent knowledge about the structure of nuclei in current Nuclear Physics.

The nuclear theorists are making experiments like a blind man touching a closed box, shaking it, smelling it, so that to discover what exists within the box. Perhaps there are apples within the box, which do not glue one each other, or perhaps there are small spherical iron magnetized balls which interact together forming some cluster.

It is hard to believe that it is possible to get reliable understanding on the true structure of nuclei from such lack of knowledge about how the nucleons are forming the nuclear structures.

regard

wlad

Dears JR , Erik , and JoeThe missing of µ for the excited 6C12 in nuclear tables is not due to what Erik had supposed.

He said:

” I think they have simply not been measured yet for various reasons “Actually there is a very

interesting sequenceof values of µ which depends on themultiplesof 6C12 in the formation of nuclei.According to my new nuclear model, there is a central 2He4, and hexagonal floors of deuterons are distributed about the 2He4.

The reason why the excited 6C12 has µ = 0 is shown in the sequence of figures between Fig. 26 and Fig. 29 (page 37) of the paperStability of Light Nuclei:http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

Look at the interesting sequence (see it in the Stone`s nuclear table):

——————————————–

6C12 – one incomplete hexagonal floor

Ex …….. i …….. µ

4438 …… 2 ……..

?The structure of 6C12 is

perflectly symmetric——————————————–

——————————————–

12Mg24 – one complete hexagonal floor + 6C12

Ex …….. i …….. µ

1369 …… 2 …….. +1,02

The structure of 12Mg24 is

NO symmetric. That` why it has µ= +1,02——————————————–

——————————————–

14Si28 – two complete hexagonal floors

Ex …….. i …….. µ

1779 …… 2 …….. +1,1

It has not the 6C12 in its structure.

That`s why it has µ= +1,1

——————————————–

——————————————–

18Ar36 – two complete hexagonal floors + 6C12

Ex …….. i …….. µ

1970 …… 2 ……..

?Note that 18AR36 is similar to 6C12, because:

* there are two complete hexagonal floors in 18Ar36

* In 18AR36 the 6C12 takes place between the two hexagonal floors.

So, the structure of 18Ar36 is

perflectly symmetricThat`s why excited 18Ar36 has µ= 0 ,as also happens with 6C12.——————————————–

——————————————–

20Ca40 – three complete hexagonal floors

Ex …….. i …….. µ

3737 …… 3 …….. +1,6

The structure of 20Ca40 is

NO symmetric. That` why it has µ= +1,6——————————————–

——————————————–

24Cr50 – three complete hexagonal floor + 6C12 + 2n

Ex …….. i …….. µ

783 ……. 2 …….. +1,24

The structure of 24Cr50 is

NO symmetric, because of the two neutrons.That` why it has µ= +1,24

——————————————–

CONCLUSION:

The excited 6C12 and 18Ar36 have both them µ= 0, and this is the reason why they are not quoted in the nuclar tables.Dear Joe,do you remember that you had asked to me about the laws that rule the excitation of the light nuclei ?

So, now we have a good track for understanding the law that rules the changing of the spin of the deuterons in the flux n(o):

it depends on the symmetry of the multiples of 6C12.Regards

wlad

ERRATA:In my last post the correct is:

Look what is written in the Figure

37, page 48, of my paper Stability of Light Nuclei, published in the Rossi`s blog:Dears JR and Erik

There is an error in that paper which link had been posted by JR:Large-basis shell-model calculations for p-shell nucleihttp://cds.cern.ch/record/351635/files/9804014.pdf

Look at the table in the page 16 :4Be7 :Values of Q(b):CALCULATION = +3,245

EXPERIMENT = +5,29(4)However, according to the Jones nuclear table the correct value is:

EXPERIMENT: +0,0529(4)Now consider the nucleus

4Be7:1- It has A=7 (odd), and therefore it cannot have Q(b) near to zero, according to nuclear models.

2- But 4Be9 has A=9, and therefore its Q(b) must larger than that of 4Be7, because it has one excess neutron.

3- Q(b) of 4Be7 is not quoted in the Jones nuclear table, and you both have claimed the reason is not because it has Q(b)=0, but actually because by an unknown reason it was not measured in the experiments.

4- But 4Be9 has Q(b) = +0,0529 , which is very small (near to zero).

5-

Therefore 4Be7 needs to have Q(b) very close to zero, which descredit the nuclear models.Pay attention that in the paper published in the link above the authors had calculated

Q(b)= +3,245, which is the value expected from the current nuclear models, and not a value Q(b)= +0,0529 , near to zero.Look what is written in the Figure 47 , page 48, of my paper

Stability of Light Nuclei, published in the Rossi`s blog:The experiments do not detect the electric quadrupole moment of 4Be7 because it is very small, near to zero, because…http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

So, dears JR and Erik,

please explain to orsobubu what quantum mechanics,, nuclear theory, etc…. actually say about these observations.

regards

wlad

Dear Wladimir, Thanks for your iformation regards aether and the faster than light experiments. Very interesting and I don’t have a problem. Regards Eric Ashworth.