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
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^−27cm^2,
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
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:
“Recently 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 deuteron 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
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
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:
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^-27cm^2.
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^-27cm^2.
· In the collection of the experiments C, 98 measurements
have to obtain Q(3S1)=0, and 2 measurements have to obtain Q(3D1) =
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^−27cm^2.
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
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
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= 3,215x10^7 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
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 = mV^2 . int[ 1/R ](1,08 to 2,08)
Fc = 1,67x10^-27 x (32,15x10^6)^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 m^2 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 cm^2 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^–31m^2 , 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
there is need to review the Standard Model by starting from the beginning,
as for instance from the model of the neutron