Lasted edited by Andrew Munsey, updated on June 15, 2016 at 1:38 am.
To: Dr. Marco Nardecchia and Dr. Sophie Renner
Experts on physics beyond the Standard Model
University of Cambridge
Dears Dr. Marco Nardecchia and Dr. Sophie Renner
In the interview by Marianne Freiberger entitled LHC glimpses hint of new physics, Dr Nardecchia says about some anomalies unable to be explained by the Standard Model:
"Usually the explanation is that the strong interaction effects aren't properly under control [in the calculations]. But now we have two anomalies, one of them being very clean. This could really mean that there is some new physics here."
Aong the last decade several experiments are debunking the Standard Model, and often we read interviews where some academic physicists recognize that new puzzles are requiring the development of a New Physics.
Sometimes the need of a New Physics is not even mentioned. For instance, in the paper The quantum vacuum as the origin of the speed of light the authors propose that the space is filled by particles and antiparticles:
“We show that the vacuum permeability and permittivity may originate from the magnetization and the polarization of continuously appearing and disappearing fermion pairs.”
Well, a space filled by particles and antiparticles is actually a New Physics, in spite of the authors did not mention it.
However, the need of a New Physics is present along the development of the Standard Model since its birth 80 hears ago, because several unsolved puzzles were pointing out that some fundamental laws are missing in the Standard Model.
For instance, consider the deuteron , formed by proton+neutron.
The proton’s charge distribution is spherical, having electric quadrupole moment zero, Q(p)=0 (confirmed by experiments). The neutron has no charge, and therefore it has electric quadrupole moment is also zero, Q(n)=0 (confirmed by experiments).
Therefore it’s IMPOSSIBLE (by considering the laws of the Standard Nuclear Physics ) for the deuteron to have electric quadrupole moment different of zero. From the laws of the Standard Model the deuteron MUST have zero electric quadrupole moment, Q(D)=0.
But in 1939 an experiment detected that deuteron has Q(D)= 2.73×10?27cm2, and therefore at that time the nuclear theorists would have to conclude the following: the way they have adopted for the development of the Standard Nuclear Physics is wrong, and a New Physics is need. And of course at that time some voices started to claim that a New Physics is need.
However, instead of admiting that it is impossible to explain the electric quadrupole moment of the deuteron by considering the model of neutron considered in the Standard Model, the nuclear theorists tried to solve the puzzle by introducing new puzzles.
Let us analyze how the puzzle was solved, by considering that the deuteron in the ground state is a mixture of the states 3S1 and 3D1:
“Recently1 the magnetic moments of the proton, neutron and deuteron have been measured with great precision. It is found that the moment of the deuteron differs from the algebraical sum of the moments of the proton and neutron by an amount much greater than the experimental error. The discovery of the electric quadrupole moment of the deuteron2 in 1939 already indicated that the ground-state of the deuteron is not of pure 3S character but contains an admixture of 3D. The observed magnetic moment can be explained if the percentage admixture is 4 per cent”.
Nuclear Forces and the Magnetic Moment of the Deuteron
The physical meaning of the states 3S1 and 3D1 are shown in the Figure 1:
According to the Standard Model, the deuteron at the ground state is a mixture of the states 3S1 and3D1, as follows:
· 96% of the time the deuteron has no orbital momentum, l=0 ,
· and along 4% of the time it has rotation with momentum l=2
First of all, we have the following fundamental question:
· why a hell the deuteron exits the most stable state 3S1 (with no rotation, l=0), and it starts to rotate, with angular momentum l=2 ??
After all, there is need energy so that to occur the changing from 3S1 to 3D1, and obviously the question is: where the deuteron gests such energy from?
But the solution adopted in the Standard Model is also incompatible with the experiments, as shown in the Figure 2.
Suppose that three experiments A, B, and C, have been made as follows: 100 measurements have been made in each experiment, with the total of 300 measurements:
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Figure 2 shows statistical results similar to what we have to expect from the Standard Model, by considering that the deuteron stays 96% of the time at the 3S1 state and 4% of the time at the state 3D1:
In the collection of the experiments A, 94 measurements have to obtain Q(3S1)=0, and 6 measurements have to obtain Q(3D1) = 68,2x10-27cm2.
In the collection of the experiments B, 96 measurements have to obtain Q(3S1)=0, and 4 measurements have to obtain Q(3D1) = 68,2x10-27cm2.
· In the collection of the experiments C, 98 measurements have to obtain Q(3S1)=0, and 2 measurements have to obtain Q(3D1) = 68,2x10-27cm2.
So, in the total, the experiments would have to get:
· 288 measurements with the value Q(3S1) = 0
· and 12 measurements with the value Q(3D1) = 68,2×10?27cm2.
Besides, if the deuteron existed in the 3D1 state, when it decays to the state 3S1 it had to emit energy, since it lost the energy of rotation from l=2 to l=0. But emission of photons by the deuteron has never been detected.
Other reason why 3D1 state is impossible to occur is explained ahead.
Let’s apply the equation for the angular orbital moment as function of the magnetic dipole moment for the case of the 3D1 state of deuteron, as shown in the figure of the link ahead:
The sum of the nuclear magnetic moments of proton+neutron is:
+2,7896 – 1,9103 = +0,8793
The nuclear magnetic moment for the deuteron is +0,857 , measured by experiments.
The difference is +0,8783 – 0,857 = 0,022
But the deuteron stays only 4% of the time at the state 3D1, and therefore in such state it has to produce a negative magnetic moment equal to:
mi = – 0,022 /0,04 mN = - 0,55 miN.
So, from mi = (e/2m).L we have for the deuteron:
mi = [e/2(mP + mN)].L , where:
mP and mN are the masses of proton and neutron
L= (mP + mN).V.R
0,55mN = [1,6x10-19/2.( mP + mN) ].( mP + mN).V.R
Being 0,87fm the radius of the proton, the radius of the orbit of the proton’s center (and the center of the neutron) up to the center of the deuteron is:
R= 1,95 – 0,87 = 1,08fm
0,55miN = 1,6x10-19.V.1,08x10-15/2
But miN = 5,05x10-27 J/T
0,55 x 5,05x10-27 = 0,864 x 10-34 .V
V= 3,215x107 m/s
Let us calculate the work of the required force for moving the center of the proton and the center of the neutron along a distance 1,0fm from the center of the deuteron (to move away each one of them).
As the distance between the center of proton to the center of deuteron is 1,08fm, when the center of the proton moves away 1fm from the center of the deuteron (and the center of the neutron also moves away 1fm), then the distance between the two centers of the proton and neutron will be 1,08x2fm + 2fm = 4,16fm, and with such distance between them the strong nuclear force between the proton and neutron will not actuate.
The changing of the centripetal force on the proton (and on the neutron) moving from a radius R= 1,08fm up to a radius R= 2,08fm is:
Fc = mV2 . int[ 1/R ](1,08 to 2,08)
Fc = 1,67x10-27 x (32,15x106)2 x [ln2,08 – ln1,08]/10-15
Fc = 1726 x (0,732 – 0,077) = 1131,3 N
The total displacement “d” traveled by the center of the proton and by the center of the neutron under the action of the force Fc is:
The work is:
L = Fc.d = 1131,3 x 2x10-15 = 2263 x 10-15 J = 22,6x10-13 J
The deuteron’s binding energy is 2,2MeV = 3,5x10-13 J , and so it is 6,5 times weaker than the energy produced by the centripetal force Fc to put the proton and neutron with their centers separated by a distance 4,16fm within the deuteron.
Thereby the deuteron at the state 3D1 will not withstands the action of the centripetal forces on the proton and neutron, and the deuteron will break.
This conclusion is reinforced by the fact that deuteron’s 3D1 state was never confirmed by experiments. If the deuteron existed in the 3D1 state, obviously it would be of extreme interest for the nuclear physicists to confirm experimentally its existence, because such confirmation would confirm the theory according to which the deuteron has non null electric quadrupole moment as consequence that it is a mixture of the states 3S1 and 3D1.
If the deuteron existed in the 3D1 state, the experiments would measure the following nuclear properties:
· Magnétic moment
· Electric quadrupole moment
In resume, it’s IMPOSSIBLE to explain the electric quadrupole moment and the magnetic moment for the deuteron, by considering the model of neutron adopted in the Standard Nuclear Physics. Because a deuteron formed with the theoretical neutron model proposed in Standard Model cannot have Q= 2,7×10?31 m2 and m = +0,857miN.
Actually a deuteron formed by the neutron considered in the Standard Model must have:
· Q= 0
· mi = +0,8793miN
The values Q= 2.7×10?27 cm2 and mi =+0,857miN for the deuteron measured by experiments can be obtained theoreticall only by considering a new model of neutron, as proposed in my paper ANOMALOUS MASS OF THE NEUTRON, published in the book Quantum Ring Theory, and also published as a paper in the Andrea Rossi’s blog Journal of Nuclear Physics:
In the paper it is calculated the electric quadrupole moment, getting the value Q= + 2.7×10–31m2 , and the magnetic moment getting mi =+0,857miN, without the need of adopting unacceptable nonsenses (similar to the nonsenses adopted in the theory of the deuteron based on the neutron model of the Standard Nuclear Physics).
The nuclear theorists cannot avoid the use of nonsenses in some theories based on the nuclear models of the Standard Model, because as the Standard Model was developed from some mistaken assumptions, it is impossible to avoid puzzles impossible to be solved. And so, any time when a new experiment debunks the Standard Model, the nuclear theorists try desperately to save the theory, by proposing nonsenses. For instance, in 2009 the Physical Review Letters has published the paper “Atomic nucleus of beryllium is three times as large as normal due to halo”.
For the first time, scientists had measured the size of a one-neutron halo with lasers, and the measurement proved that nucleons are not bound within the nuclei by the strong force, because in the 4Be11 the halo-neutron is 7fm far away from the rest of the cluster, and since the strong force actuates in a maximum distance shorter than 3fm, it is obvious that the neutron is not bound via the strong force in the Be11.
Trying to save the Standard Nuclear Theory, the nuclear theorist Dr. Wilfried Nörtershäuse has proposed the following desperate solution:
“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'. “
But beyond the fact that Nörtershäuse’s theory is very strange, because he is proposing a sort of neutron which behaves like a rubber band used by dressmakers, his theory is also unacceptable, because:
1) Suppose that Nörtershäuser’s theory was viable and wise, and the neutron indeed could have the strange property of behaving like a rubber band. However his theory cannot explain other experimental fact: the 4Be11 decay produces the stable isotope 5B11, and there is no way to explain it by considering the Nörtershäuser’s hypothesis.
Indeed, Nörtershäuser’s hypothesis is also unacceptable because of the feature of the decay of the nucleus 4Be11, as explained ahead:
2) He could argue that the halo-neutron is weakly linked to the cluster, and it exits the nucleus after the 13,81 seconds just because of the weak link. However this is no true, because in 97% of decays the 4Be11 transmutes to 5B11, and therefore the neutron does not exits the nucleus. In 4Be11 the neutron decays in a proton and electron, and the proton goes back to the cluster. If the strong nuclear force was responsible for the cohesion of nuclei as the nuclear theorists suppose, the proton could never go back to the cluster, because in a distance of 7fm it cannot interact with the cluster via strong force, and the classical Coulomb repulsion between the cluster and the proton would be so strong that the proton would be expelled from the 4Be11, and 5B11 could not be formed in 97% of the 4Be11 decay.
3) Therefore, even if the bizarre Nörtershäuser’s solution was viable for the explanation of the halo neutron in a distance of 7fm from the rest of the nucleus, however the 5B11 would never be formed from the decay of the 4Be11, according to his solution.
4) And in his paper Nörtershäuser did not propose any explanation for the formation of the isotope 5B11 from the decay of the 4Be11. He only tried to explain how a neutron could be kept in a distance of 7fm.
So, Nörtershäuser solution is unacceptable, and therefore it is impossible to explain the 7fm distance of the neutron in the Be11 by considering the current nuclear models based on the Standard Nuclear Physics. The distance of 7fm detected in the experiment suggests that nucleons are not bound in the nuclei via the strong nuclear force, as predicted in Quantum Ring Theory.
So, dears Dr. Marco Nardecchia and Dr. Sophie Renner,
concerning the words "But now we have two anomalies, one of them being very clean. This could really mean that there is some new physics here", I hope you may realize that many new experiments made in the last 10 years are showing the need of a New Physics in a way so much strinking, and the most wise decision of the community of physicists must be face the unavoidable:
there is need to review the Standard Model by starting from the beginning,
as for instance from the model of the neutron