# Aether Structure for unification between gravity and electromagnetism

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by
retired, author of the Quantum Ring Theory
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In the book Quantum Ring Theory I had proposed a double-field model for elementary particles (composed by two concentric fields), therefore a field model fundamentally different of the mono-field model considered in the Quantum Electrodynamics  (QED).
The inner field, named principal field Sp, gyrates and induces the outer field, named secondary field Sn.  In the book, published in 2006, it was considered that the outer field Sn gyrates.
In this model, the outer field Sn is responsible for the electric charge of the particles as the electron, the proton, etc.
Later in 2010 I changed the  double-field model, by considering that the outer field Sn does not gyrates.  However, in 2014, after a long discussion with the reader Mr.Joe in the Comments of the Journal of Nuclear Physics, he drew our attention to two key points:
1. An outer field Sn induced by the rotation of an inner field Sp must have rotation.
2. A mono-field model violates the monopolar nature of the electric charge in the even-even nuclei with Z=N, because they have null magnetic moment, but as all the nuclei have rotation then the even-even nuclei with Z=N would have to have non-null magnetic moment (because the rotation of the positive charge of the proton would have to induce a magnetic moment). Therefore QED violates the monopolar nature of the electric charge in the case of the even-even nuclei with Z=N.
3. A double-field model in which the outer field Sn gyrates would have to induce a magnetic field in the case of even-even nuclei with Z=N, if we consider the field Sn in the classical sense of Euclidian space.  But the space considered in Quantum Ring Theory is not Euclidian, in order that the rotation of the field Sn never induces magnetic fields, and this is the reason why the even-even nuclei with Z=N have null magnetic moment.
Here we will analyse these questions in details.
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### 538 comments to Aether Structure for unification between gravity and electromagnetism

• BroKeeper

Thank you for clarifying the infinite conundrum within theoretical
physics equations versus more common mathematical equations (I think).
Again on behalf of us normal’s, Thank you!
BK

• Joe

1. How did you determine the exact distribution of Q between Sp and Sn?

2. Are you saying that the central area of the proton’s body-ring might have radius r = 0 in order to account for the lack of a varying aether density in Sp?

All the best,
Joe

Joe,
even if the fields Sn and Sp´have charges with the same value, X=-Y= +1,6×10^-19C, the charges of Sn and Sp will not cancel each other.
Because the charge of Sp is concentrated in a region with radius in the order of 1fm, while the charge of Sn is distributed along a radius in the order of 10.000fm.
As the electric charge decreases with the square of the distance, the charge of the field Sp will be very very weak in a distance of the order of 10.000fm.
So, the electric charge of the proton is q = _1,6×10^-19C.

The same happens within the nuclei.

Two charges with the same value and contrary signals can cancel one each other only when they have the same distribution.
For instance, the field Sn of the neutron is formed by the overlap between the fields Sn of the electron and the field Sn of the proton.
As the two fields Sn of the proton and electron have the same size (R= 10^-11m), then the charge of the neutron is null.

regards

• Andrea Rossi

Robert Curto:
Thank you for the interesting information.
Warm Regards
A.R.

• Robert Curto

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Ft Lauderdale, FL
USA

• Andrea Rossi

BroKeeper:
You are infinitely welcome!
Warm Regards,
A.R.

Andrea Rossi wrote in February 24th, 2015 at 12:10 AM

BroKeeper:
In Physics the situation is dramatically changed with the Relativity and the Quantum Theory: both made of infinity a nonsense when you are working with equations to resolve problems of Physics.
————————————————————————–
Dear BroKeeper ,
as Relativity and Quantum Theory are two incomplete theories, since they do not consider a structure for the space (aether), the infinity sometimes appears in the equations.
In order to solve the puzzle, the theorists introduced a mathematical artifice known as “renormalization”

The mathematical artifice actually makes no sense, because the quantum theorists use to make subtraction between two infinite quantities..
For instance:
5 infinite – 2 infinite = 3 infinite

Such arithmetic obviously makes no sense, because infinite is infinite. And so 5 infinite is not greater than 3 infinite

regards

• BroKeeper

An infinite thank yous to you. 🙂
BK

Joe wrote in February 24th, 2015 at 2:13 AM

1. —————————————————-

Q = (q of Sp) + (q of Sn)

Q would have to be measured as +8 for 4Be, for example. But this is obviously not the case in reality.
——————————————————————

Joe,
you did not understand.

Sn has charge X = +1,9×10^-19 C
Sp has charge Y = -0,3×10^-19 C

The charge of a proton is Q = X + Y = +1,6×10^-19 C

4Be has 4 protons, and therefore its charge will be +4×1,6×1,6×10^-19 C
Therefore 4Be has charge +4

2. ———————————————————-
Did you not originally have Sp and Sn as having opposite electric charges? If so, the Q of every nucleus would be equal to zero, which is obviously not the case in reality.
————————————————————-

explained in 1

3. ————————————————————
Why does only Sn have a varying aether density (in order to always cancel magnetic moment) but not Sp? In your F(M) paradigm, Sp has an aether just like Sn. (Of course, if Sp had a varying aether density too, it too would have a canceled magnetic moment. That would mean a total magnetic moment of zero for every nucleus, which is obviously not the case in reality.)
—————————————————————

Joe,
probably there is a central very small core around the body of the proton (with radius in order of 5fm, where the density of the aether is practically near to constant.
This can be explained as consequence of the fact that around the body of the proton there is a very high dense field of permeabilitons (see Fig. 2.2 in the present paper).

Consider a central region C in the center of the proton’s body-ring.
The density of the aether at the right of the region C and at the left of the point C cannot decrease if the region C has radius zero, because if the radius is zero there is a singularity in the center of the region C.

regards

• Joe

1. In your new scenario of

Q = (q of Sp) + (q of Sn)

Q would have to be measured as +8 for 4Be, for example. But this is obviously not the case in reality.

2. Did you not originally have Sp and Sn as having opposite electric charges? If so, the Q of every nucleus would be equal to zero, which is obviously not the case in reality.

3. Why does only Sn have a varying aether density (in order to always cancel magnetic moment) but not Sp? In your F(M) paradigm, Sp has an aether just like Sn. (Of course, if Sp had a varying aether density too, it too would have a canceled magnetic moment. That would mean a total magnetic moment of zero for every nucleus, which is obviously not the case in reality.)

All the best,
Joe

• Andrea Rossi

BroKeeper:
The concept of infinity could need infinite time to talk about it.
You can find infinite papers talking of it. We must make a distinction related to the field of application of the concept of infinity: it has completely different meaning in Philosophy, in Religion, in Mathematics and in Physics.
Infinity is an important religious and phylosophical concept, and under a phylosophical and/or religious point of view infinity can help to formulate hypotesis. In Mathematics, obviously, infinity exists and is at the base of mathematical pillars: it is not a case that one of the inventors of the infinitesimal calculus was a philosopher. In Physics the situation is dramatically changed with the Relativity and the Quantum Theory: both made of infinity a nonsense when you are working with equations to resolve problems of Physics.

• Robert Curto

Thanks Steven, for the detailed explanation, I really appreciate it.
Robert

Andrea Rossi wrote in February 23rd, 2015 at 1:31 PM

When in SSM the E-Cat, obviously, in terms of Physics has a consumption that makes a nonsense the infinite. During the SSM the COP is quite high. To put a zero below the line of fraction is a nonsense.
————————————————————-

Dear Andrea
actually the definition of COP in terms of Physics is wrong (or insatisfactory, if you prefer this word), because the COP is defined regarding a close system, where the energy stored in the space is not considered.

In order to get the correct COP for the eCat working in the self sustained mode there is need a new definition of the COP, by considering the energy stored in the space.
The zero below the line of fraction becomes non-zero

regards

Joe wrote in February 23rd, 2015 at 2:17 PM

Sn can not act as a Faraday shield to Sp since electricitons are not the equivalent of electrons. With the Faraday shield, an electric field induces a flow of electrons, preventing the electric field from extending further beyond the shield. But an electric field could not induce a flow of electricitons since an electric field would actually be composed of electricitons. Therefore, the electric field of Sp could not be prevented from extending beyond Sn.
———————————————————-

Joe,
suppose the following:

1- The outer field Sn of the proton has charge X
2- The inner field Sp´of the proton has charge Y
3- The total charge of the proton is q = X+Y = 1,6×10^-19 Coulomb

A nucleus with Z quantity of protons has:

a- the outer total field Sn with charge Z.X
b- the inner total field of protons with charge Z.Y
c- and therefore the total charge of the nucleus is q = Z.(X+Y) = Z.(1,6X10^-19)Coulomb

Therefore,
what the experiments measure is actually the the total electric charge composed by Sn and Sp.

When the proton is moving about the central 2He4 of a nucleus, the magnetic moment induced by the proton depends on the direction of the flux n(o) regarding the z-axis:
http://peswiki.com/index.php/Image:Two_protons_symmetric_captured_by_central_2He4.png
The flux n(o) crossing the proton in the side Douglas is UP, while in the side Ana the flux n(o) crossing the proton is DOWN.

.

In the case of the electron, its charge is q = X+Y = 1,6×10^-19 Coulomb, where X is the charge of the outer field Sn, and Y is the charge of the inner field Sp.

regards

• BroKeeper

Dear Andrea,

All my life the concept of ‘infinity’ has been part of everyday mathematical certainty. Now reading your comment “In Physics when you have an equation that contains the “infinite” means you are in error”. I am now confronted with a paradigm mind-busting concept infinity does not exist. I stumbled across on the same day an article titled “Infinity Is a Beautiful Concept – And It’s Ruining Physics” by physicist Max Tegmark: http://blogs.discovermagazine.com/crux/2015/02/20/infinity-ruining-physics/. It is a philosophical argument viewing infinity as the “greatest crisis facing modern physics” and appears to mirror your recent comment. Having a Doctor’s Degree in Physics Philosophy yourself, could you shed more light on this controversial principle? Is the infinite time/space concept a convenient way for our limited physical minds to accept this universe? Thanks.
With much respect, BroKeeper

• Steven N. Karels

Robert,

a. How much does 8000 gallons of diesel cost? Assume the price is around \$3.00USD per gallon. \$24,000USD
b. Fuel cost per hour: \$24,000USD / 8 hours = \$3,000USD per hour.
c. Type of fuel eCat needs for heating: See Andrea Rossi – natural gas or electricity or maybe something else.
d. How much energy is needed (excluding heat losses): heat of fusion for water plus heat needed to raise the thawed water to “warm water”. Convert the 180 tons to metric and apply heat of fusion and heat to raise the temperature. This will be how much energy per hour is needed. Then convert it to Watts. I estimated around 3.3 MW of heating was required.
e. Next assume a design COP: e.g., 3, 6 or some other number. Divide the power level obtained in d. by the COP. This tells you how much input energy is needed (e.g., natural gas or electricity).
f. Regardless of COP, some additional electrical energy will be required for control, display and running pumps, etc.
g. Cost to operate: Add the fuel replacement costs (eCat) plus the input energy costs and compare it with the costs in b.
h. Operating time: depends on the city location and the amount of snow. eCat works best when operated continuously (days, weeks or months at a time). But the city will want the snow removed quickly so the operating time may be very limited. A detailed cost analysis is needed before applying this technology.
i. External energy type: It is probably more convenient to use diesel or other portable fuel rather than electricity. Where do you plug in for 3.3 MW/COP of electricity? Control electrical power could be provided by a portable generator.

• Joe

Sn can not act as a Faraday shield to Sp since electricitons are not the equivalent of electrons. With the Faraday shield, an electric field induces a flow of electrons, preventing the electric field from extending further beyond the shield. But an electric field could not induce a flow of electricitons since an electric field would actually be composed of electricitons. Therefore, the electric field of Sp could not be prevented from extending beyond Sn.

All the best,
Joe

• Andrea Rossi

You consider the space filled by aether, not me, therefore our conclusions follow a different logic.
When in SSM the E-Cat, obviously, in terms of Physics has a consumption that makes a nonsense the infinite. During the SSM the COP is quite high. To put a zero below the line of fraction is a nonsense.
Warm Regards,
A.R.

• Robert Curto

Steven, first I want to thank you for your response.
Snow Dragon is the name of the company I am talking about.
You can go on their website, They are very good at responding to emails.
We have exchanged 17 emails, Jennifer has responded to all of my emails in about one hour.

The snow melt machine that can melt 180 tones per hour ( 600 to 1,800 cubic yards ) per hour has a 3,000 gallon Tank for diesel.
It lasts 8 to 9 hours !
It does not require very hot water as I said, it talks about warm water.

If you have time to help me with my limited knowledge, I would appreciate it.
I understand you need electric power for the pumps etc.
I THINK you said you need diesel or national gas to run the E-Cat ?
This is what I THINK I know about one E-Cat.
It’s fuel is a small amount of low cost nickel, hydrogen gas, and a secret catalysis. The fuel is replaced after 6 months.
I understand it will take hundreds of E-Cats to warm the water to melt the snow.
But I am trying to compare the cost of the fuel.
How much does 3,000 gallons of diesel cost…..every 8 hours ?
Robert Curto

Andrea Rossi wrote in February 23rd, 2015 at 5:53 AM

Steven N. Karels:
In Physics when you have an equation that contains the “infinite” means you are in error. I agree with you and Robert that this field of application could be particularly fit.
—————————————————–

The equations get infinite, of course, when we consider the space as empty.
Because as an empty space cannot store energy, the equation gives infinite.

But when we consider the space filled by aether, we get a value different of infinite, since energy is being supplied by the aether.

But I am curious, dear Andrea:
when the Ecat is working in the self-sustained model, what is the value of the COP?

regards

• Andrea Rossi

Steven N. Karels:
In Physics when you have an equation that contains the “infinite” means you are in error. I agree with you and Robert that this field of application could be particularly fit.
Warm Regards,
A.R.

• Steven N. Karels

Robert,

That is not correct. I was suggesting that, perhaps, Andrea Rossi might find a way for the eCat reaction to only require an energy source during initial start-up and then either none or minimal additional thermal energy during continuous operation. This is effectively the same as having the so called COP become very high (or infinite). There would still be a need for some electrical input for pumps, controls and indications, etc.

• Robert Curto

Steven, am I correct ?
The E-Cat does not need natural gas or diesel to produce heat.
It produces heat with it’s own fuel.
Robert Curto

Joe wrote in February 21st, 2015 at 2:09 PM

1. You state,
“The charge of the field Sp of each proton is not detectable.”

But this can not be true in your F(M) model which includes electricitons (and magnetons) within Sp. Since magnetic moment due to the magnetons of Sp is detectable, electric charge must logically also be detectable due to lack of symmetry in the distribution of electricitons within the Sp of nuclei that are not even-even Z=N.
—————————————————————-

NO, Joe,
there is difference between the way of the electricitons and magnetons be captured.

Look at the fields of the proton:
http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png

In the magnetic fields M(+) and M(-), the magnetons are bound thanks to the their interaction with the field of permeabilitons involving the body-ring of the proton (see Fig. 2.2 in the present paper).
So, the magnetic field is able to spread in the space, without being blocked by the outer electric field Sn of the proton, formed by electricitons e(+) captured by the flux n(o).

But as the outer electric field Sn is formed by electricitons e(+) captured by, the electric charge of the inner electric field are blocked.
The phenomenon happens similar to the Faraday cage:
“A Faraday cage or Faraday shield is an enclosure formed by conductive material or by a mesh of such material. Such an enclosure blocks external static and non-static electric fields by channeling electricity through the mesh.
Faraday cages cannot block static or slowly varying magnetic fields”

The Faraday cage cannot block magnetic fields because the magnetons of the magnetism are spread in the space thanks to the permeability given by the permeabilitons of the aether.

Therefore,
the electricity and the magnetism spread in the space via two different mechanisms.

.

When a nucleus is aligned by an external magnetic field (produced in experiments), the outer electric field Sn of the nucleus takes the non-spherical shape shown in the figure:
http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

When a nucleus is not aligned by an external magnetic field, due to the chaotic rotation of the nucleus the outer electric field Sn takes the shape of a spherical field, as shown ahead for the 2He4 nucleus:
http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE2.png
All the inner electric fields Sp of the protons wihin a nucleus are blocked by that outer spherical electric field Sn involving the nucleus.

In the case of the even-even nuclei with Z=N, as they have null magnetic moment, they cannot be aligned by an external field, and therefore their outer field Sn is spherical and involves completely the nucleus, and it works similar to a Faraday cage, blocking all the inner electric charges Sp of the protons.

regards

• Andrea Rossi

Steven N. Karels:
Could be an interesting field of application.
Warm Regards,
A.R.

Joe wrote in February 21st, 2015 at 2:29 PM

2. A neutron does not necessarily need to be bound by the gravitational flux n(o). It can also be bound by a spin-interaction as is the usual case in the F(M) paradigm.
————————————————–

Joe,
in the 3Li7 of the figure:
http://peswiki.com/index.php/Image:FIGURE-8-substitute3Li7-28Ni.png
the neutron is bound by spin-interaction because it is the unique possibility, since as the neutron has no electric charge, it cannot have interaction by the magnetic F(M) force.

But in the case of a 3Li7 having possibility to be bound by strong force and by spin-interaction force, the neutron must be bound by the stronger of the two forces, which is the strong force.

regards

• Steven N. Karels

Robert and Andrea Rossi,

My rough calculations indicated that 3 or 4 MW eCat units could melt the 180 Tons per hour of snow to water. So the sizing is not too distant from your current machine. Probably should use natural gas or diesel as the heat source (assuming the eCat is not self-sustaining). Probably a very good candidate as the application would be a slow start-up and long duration – ideal for eCat technology. Trucks would bring the snow to a location near a natural river and the melted water returned to the environment. Please consider it.

• Robert Curto

Steven, February 22 at 4:23A
I agree with you 100%
The snow melting machines today run on Diesel fuel, or if it is available on natural gas.
The last thing you want to do is add more emissions in the middle of a city,
24 hours a day.
They heat water very hot, that is used to melt the snow. They have to keep the water very hot.
Some of the large machines can melt 180 tons per hour.
Plus i believe the fuel cost of an E-Cat will be a lot less expensive.
Robert Curto
Ft. Lauderdale, FL
USA

• Andrea Rossi

Steven N. Karels:
Well, this is an application that could fit the E-Cat, as every situation in which you need heat. Nevertheless, plowing apart, snow is what gives to New Hampshire part of its beauty. I used to make jogging on the iced surface of the Massabesic Lake in this season: so fascinating…
Warm Regards,
A.R.

• Steven N. Karels

Dear Andrea Rossi,

A self-sustaining eCat heating unit to melt snow would be very useful here in the Northeast of the US. We have had so much snow that our snowblowers can barely move the snow to the high snow mounds. Perhaps another future consumer product?

• Andrea Rossi

AlainCo:
Thermionic generation is very interesting, but we did not find yet an industrialized product ready to work with acceptable efficiency.
Warm Regards
A.R.

• Joe

2. A neutron does not necessarily need to be bound by the gravitational flux n(o). It can also be bound by a spin-interaction as is the usual case in the F(M) paradigm.

All the best,
Joe

• Joe

1. You state,
“The charge of the field Sp of each proton is not detectable.”

But this can not be true in your F(M) model which includes electricitons (and magnetons) within Sp. Since magnetic moment due to the magnetons of Sp is detectable, electric charge must logically also be detectable due to lack of symmetry in the distribution of electricitons within the Sp of nuclei that are not even-even Z=N.

All the best,
Joe

• Dear Dot. Rossi,

Did you study, at least on paper, the idea to use Thermionic converters. It seems Soviet used rugged thermionic converters (like in TOPAZ).
The temperature of Lugano’s E-cat are inside the usual working temperature (1500-2000K), but like for steam it is probably not so simple …
Are there reasons to reject that idea?

Joe wrote in February 20th, 2015 at 6:10 PM

1. ——————————————————-
Do not worry about specific calculations. It is important that the theoretical foundation is logical.
———————————————————-

Joe,
by considering gravity (strong force) interaction between nucleons introduces some phylosophical inconsistences.

For instance, the 3Li7 has only one neutron, and it is weakly bound.
If you consider the binding energy due to strong force, the neutron and the deuteron of the 3Li7 would be bound to the central 2He4 with the same force.
But the neutron is weakly bound.
See the structure of the 3Li7:
http://peswiki.com/index.php/Image:FIGURE-8-substitute3Li7-28Ni.png

If the neutron were bound by strong force, the 3Li7 could not have the structure shown in the figure.

4. ———————————————————
A possibility might exist where the gravitational fluxes n(o) are actually color charges. In standard theory, there are three colors and three anti-colors.
————————————————————

Joe,
in the case there is need to consider the color charges for explaining how quarks are bound, then there is need to introduce three additional particles to the structure of the aether, beyond those proposed in my present paper.
Beyond the particles e(+), m(+), p(+), P(+), g(+), G(+), and their antiparticles, there is need to introduce more the particles R(+), Y(+), B(+), and their antiparticles (red, yellow, blue).

The structure will be:

e(+), m(+), p(+), P(+), g(+), G(+), R(+), Y(+), B(+)
e(-), m(-), p(-), P(-), g(-), G(-), R(-), Y(-), B(-)

regards

Joe wrote in February 21st, 2015 at 2:27 AM

1) ———————————————————-
You state,
“[…] only the outer secondary field is detected in macroscopic experiments.”

If that were true, why would we be detecting magnetic moment emanating from Sp and only from Sp? The problem here is that you have organized your model in order to account for a null magnetic moment in the case of even-even Z=N nuclei, ignoring that in non-null cases, an electric charge would then also necessarily be detectable even within your paradigm of fluxes-influencing-charge-at-the-level-of-Sp.
—————————————————-

No, Joe,
I am not ignoring them.
We are not speaking about charges, we are speaking about magnetic moments.

There are two sort of magnetic moments generated by each proton within the field Sp:

1- The magnetic moment induced by the rotation of the charge of the proton about the z-axis of the nucleus.
In the case of even-even nuclei with Z=N the total magnetic moment is zero, because two symmetric protons cancel each other their magneti fields (but not in the case of other nuclei with Z different of N, and they contribute for the total magnetic moment of the nucleus).

2-The magnetic moment induced by the spin of each proton (equal to +2,793). It is the magnetic moment which contributes for the total magnetic moment of the nucleus.

2) ——————————————————–
And since that electric charge at the level of Sp would be detectable, it would also necessarily be added to the electric charge found in Sn to give a final electric charge that guaranteed is wrong by empirical standards. In short, in order to save magnetic moments in your F(M) paradigm, you have unknowingly sacrificed electric charge.
————————————————————–

The charge of the field Sp of each proton is not detectable.
The outer secondary electric field Sn is induced by the rotation of the flux n(o) of the principal field Sp of each proton.

regards

• Joe

You state,
“[…] only the outer secondary field is detected in macroscopic experiments.”

If that were true, why would we be detecting magnetic moment emanating from Sp and only from Sp? The problem here is that you have organized your model in order to account for a null magnetic moment in the case of even-even Z=N nuclei, ignoring that in non-null cases, an electric charge would then also necessarily be detectable even within your paradigm of fluxes-influencing-charge-at-the-level-of-Sp. And since that electric charge at the level of Sp would be detectable, it would also necessarily be added to the electric charge found in Sn to give a final electric charge that guaranteed is wrong by empirical standards. In short, in order to save magnetic moments in your F(M) paradigm, you have unknowingly sacrificed electric charge.

All the best,
Joe

Joe,
I drew a picture, in order you may understand easily my explanation in the previous comment:

http://peswiki.com/index.php/Image:Two_protons_symmetric_captured_by_central_2He4-1.png

regards

Joe wrote in February 20th, 2015 at 6:10 PM

3. Magnetic moments that are induced by the rotation of electric charge are independent of the gravitational fluxes n(o). We see this clearly at the macroscopic level where the direction of rotation is the only factor deciding the orientation of magnetic moments. Gravitational fluxes n(o) do not exist at this level.
———————————————————–

Joe,
the situation in the macroscopic leve is different. The charge is monopolar in the macroscopic level.

In the microscopic level, there is need to consider two fields produced by each proton in that nucleus 4Be6 shown in the figure:
http://peswiki.com/index.php/Image:Two_protons_symmetric_captured_by_central_2He4.png

The two fields are:

1- The outer total secondary electric field of the nucleus 4Be6:
It is formed by the overlap of the 4 secondary fields of the 4 protons. It is responsible for the Coulomb electric charge of the 4Be6.
The direction of the flux n(o) of the protons has NOT influence in this total field of the 4Be6, and this is the reason why the electric charge of the 4Be6 is monopolar.

2- The inner principal electric field of each proton:
The signal of the magnetic moment induced by the rotation of this principal field depends on the direction of the flux n(o).
Such property of the inner principal field cannot be detected, because only the outer secondary field is detected in macroscopic experiments.

regards

• Joe

1. Do not worry about specific calculations. It is important that the theoretical foundation is logical.

2. There is no reason why any neutron must necessarily be retained by gravitational fluxes n(o). You have already given examples of deuterons having spin-interaction with more than one neutron. I even asked you what the upper limit is for the ratio of neutrons-to-deuterons, and you replied that you had not considered it.

3. Magnetic moments that are induced by the rotation of electric charge are independent of the gravitational fluxes n(o). We see this clearly at the macroscopic level where the direction of rotation is the only factor deciding the orientation of magnetic moments. Gravitational fluxes n(o) do not exist at this level.

4. A possibility might exist where the gravitational fluxes n(o) are actually color charges. In standard theory, there are three colors and three anti-colors. These might correspond to three Douglas fluxes and three Ana fluxes. They would be responsible for holding the quarks together. And a spillover of these strong fluxes would be responsible for holding nucleons together; these would be the residual fluxes that you are always discussing. The orbiting nucleons would actually be continually relayed from one color/anti-color flux pair to another. The present calculations would remain unaltered.

All the best,
Joe

• Andrea Rossi

Andrea Calaon:
As I already said, I cannot give further information regarding this issue. We will talk about electric power when we will deem our R&D to be mature for the market.
Warm Regards
A.R.

• Andrea Calaon

Dear Andrea,
I have a few questions about the electrical energy production:
– did you use a Turbine, a Sterling Engine or a Steam Engine?
– has the thermal energy been produced solely from the LENR source (E-Cat/H-Cat), or the LENR was added to a chemical source?
– Have you used a series of E-Cat and H-Cat to heat the fluid (if any)?
Warm/Hot regards
Andrea Calaon

• Andrea Rossi

John:
I confirm that in our work of R&D we have also made experiments related to the production of electric power. I cannot give information of our R&D work, until we procuce something really working in a satisfactory mode.
Warm Regards,
A.R.

Monopolar nature of the electric charge

Dear Joe,
there is a serious problem with my present paper Aether Structure for unification between gravity and electromagnetism.

Because in the paper Stability of Light Nuclei I had considered the interaction between the nucleons via magnetic force. I made several calculations in that paper, the stability of the nuclei was explained very well, the calculations of magnetic moments were agree to the experimental data, that model was able to explain the halo neutron of the Be11, etc.

By considering the nucleons bound via gravity (strong force) there is no way to explain the properties of the light nuclei (for instance, there is no way to explain why 3Li8 is no stable, since the eighth neutron would be bound via strong force.

So:
1- There is need to keep the hypothesis that protons are bound within the nuclei via magnetic force, as considered in the paper Stability of Light Nuclei

2- The reason why the even-even nuclei with Z=N have null magnetic moment must be explained from another mechanism of that proposed in the present paper Aether Structure for unification between gravity and electromagnetism.

I was thinking about the question, and I think the puzzle must be solved as I explain ahead.

Look at the figure showing two protons captured by a central 2He4. The proton in the side Douglas has spin-up, and the proton in the side Ana has spin-down:

http://peswiki.com/index.php/Image:Two_protons_symmetric_captured_by_central_2He4.png

But note that the flux n(o) crossing the proton in the side Douglas is UP, while in the side Ana the flux n(o) crossing the proton is DOWN.

Then we have to conclude that a magnetic moment induced by an elementary electric charge (as the charge of the proton) depends on the direction of the flux n(o) crossing the particle, regarding the axis of rotation of the particle.

Therefore, the two electric charges of the two protons shown in the figure induce two magnetic moments with contrary signals, and therefore they cancel one each other.

.

Joe,
no matter what is the solution for solving the puzzle by considering a nuclear model working with the particles of the aether, however there is an important point to be considered:

IT IS IMPOSSIBLE TO SOLVE THE PUZZLE BY SUPPOSING THE HYPOTHESIS OF THE EMPTY SPACE, AS IS CONSIDERED FOR THE WHOLE NUCLEAR MODELS BASED ON THE STANDARD MODEL.

The puzzle can be solved only by considering a structure for the space.

regards

• John

Dear Mr. Rossi,

Is the now confirmed production of electricity–however preliminary–something that was achieved very recently, say in the past few months? I believe this is the first time that you have confirmed electricity production from the Hot Cat!

Best Regards,

John

• Andrea Rossi

Italo R.:
Probably
Warm Regards
A.R.

• Andrea Rossi

Mark Saker:
1- yes, but still at R&D primary stage level.
2- no
Warm Regards
A.R.

• Andrea Rossi

Bernie Koppenhofer:
We re studying the theoretical issues deriving from the report, that obviously is correct.

We are making intense theoretical work on the results and we are making a reconciliation, but so far we are not ready to give further information about this issue, which is also bound to restricted data.
Warm Regards
A.R.

• Bernie Koppenhofer

Dr. Rossi: After the third party test showed amazing isotopic changes I am wondering if you have made similar isotopic tests on your research reactors or the industrial reactor. Of course I do not expect you to give us the results but just a confirmation that the isotopic changes the third party reported were not a fluke or error would help us E-Caters. Thanks again for this site and the recent pictures and answering our questions.

• Mark Saker

Dear Andrea,

Thanks for answering my question although I am still left a little unclear.

1) During your R&D can you confirm that you have produced electricity using the e-cat?

2) If so, have you recycled the electricity into controlling the e-cat creating a closed loop or is that in the future?

Many Thanks – especially for the pictures, it gives us all something to chew over 🙂