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by
Valeriy Y.Tarasov
E-mail: vytarasov@yahoo.com
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Abstract
The h-space theory is a variant of unified physical theory – a theory of everything.
This theory was built de novo, as the existing physical theories are incompatible and so unsuitable for unification.
A new approach is needed, and has been developed by re-evaluating the definitions of primary physical concepts.
The starting point for the re-evaluation was the following equation – Et = mvL, where energy – E, time – t, length – L, mass – m, velocity – v.
Analysis of these physical concepts resulted in the construction of a unique equation of the primary concepts such as space, length, energy and velocity.
From this, models could be developed that explain all well-known physical phenomena.
In addition, h-space theory predicts phenomena rejected by the current mainstream theories, such as limits to gravitational and electrostatic interactions, and the possibility of cold fusion (as a consequence of the electric charge definition, a modification of Coulomb’s law and the definitions of elementary particles in h-space theory).
The final section of this article describes a number of experimental tests that could be used to verify the h-space theory.
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Read the whole article
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JonJon:
This is a tricky question, because in Physics “infinity” is a nonsense and when an equation involves infinity it is considered wrong.
To be correct we can say this: during the ssm the consumption of electric power from the grid of the E-Cat is zero. Under a more complex concept, there is a consumption of energy that resolves the infinity issue.
Warm Regards,
A.R.
Andrea,
During ssm (self sustained mode), the Ecat Cop is infinity as the input power is turned off, except for the control electronics?
Robert Curto wrote in December 8th, 2014 at 1:51 AM
Dear JR: We miss your counterpoint to what Wladimir Guglinsky writes.
We loved your corrections based on the Standard Model and many of us miss you.
I see that Dr. Andrea Rossi prefers to stay out of theoretical confrontations, even if he repeatedly said he stays inside the Standard Model to explain everything, so we need your counterpoint.
————————————————-
Robert,
perhaps they prefer stay out because even if a reasonable theory can be found based on the “Low Radiation Fusion Through Bound Neutron Tunneling” proposed by Dr. Oscar Gullstroem , however is very hard to explain with the same theory how alpha emission with 15MeV can be produced in Mosier-Boss experiment (and how can occur the nuclear fission of the Pd nucleus hit by particles with low energy).
Because there is a paradox in the “Low Radiation Fusion Through Bound Neutron Tunneling” proposed by Dr. Oscar Gullstroem, as follows:
1) He is proposing a theory for the Rossi-Effect, so that to explain how particles with low energy can surpass a high Coulomb barrier.
2) However, in Mosier-Boss experiment many particles with very high energy are emitted (as neutrons 14MeV and alphas with 15MeV)
3) So, the paradox is evident:
if the solution on the puzzle on how particles with low energy can surpass a high Coulomb barrier is by proposing that low energetic bound neutron tunneling can cross it, then how can those low energy neutrons produce emissions with high energy????
regards
wlad
Frank Acland:
We are aiming at that.
Warm Regards,
A.R.
Dear Andrea,
When you say “the ssm should result to be by far the main mode of operation of the 1MW E-Cat” does this mean that for the great majority of the time the input power is turned off?
If so, this should result in tremendous energy savings for the customer, shouldn’t it?
Many thanks,
Frank Acland
Wladimir, it seems that the fluorescent bulbs LENR history was already debunked two years ago:
http://www.e-catworld.com/2014/12/05/compact-flourescent-lightbulbs-we-are-surrounded-by-lenr-in-our-own-homes-gordon-docherty/#comment-1727599591
Buck, thanx for the marvelous article on NYtimes. It shows clearly how economic interests drive the political process in an advanced capitalistic/imperialistic economy. Most interesting of all are 1500 disillusioned readers’ comments, as they suddenly discover that democracy never existed as they imagined. The article also demonstrate how federalism is at work, absolutely not enhancing the possibilities for small communities to vote for their representatives, for citizens rights to breath in a clean environment and for endangered lesser prairie chicken to survive; instead, federalism is the best way, in a complex society, to manage the superior interests of the ruling class in spite of impossible, demagogic and idealistic central government electoral promises.
Joe
December 7th, 2014 at 8:27 PM
Wladimir,
1. ) ————————————
You write,
“The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.”
But you said that the purpose of the “hole” was to allow positively charged particles to bypass the Coulomb barrier.
Only yesterday you wrote,
“The proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barrier.”
So, can a proton not enter a nucleus through the “hole” along the z-axis?
——————————————————–
yes,
by HOT FUSION in the Sun and any engine workng via hot fusion
2. ) ——————————————
You write,
“The neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.”
i) What is the electron that is “captured”?
———————————————
the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.
ii) —————————————
What is the mechanism that sets the intrinsic spin, and therefore the intrinsic magnetic moment, of the newborn neutron AGAINST the induced magnetic moment of the rotating 2s1 electron in order that the neutron be repelled into the target nucleus?
———————————————–
http://peswiki.com/index.php/Image:Why_neutron_is_repelled_by_the_electron_orbit.png
regards
wlad
Dear JR: We miss your counterpoint to what Wladimir Guglinsky writes.
We loved your corrections based on the Standard Model and many of us miss you.
I see that Dr. Andrea Rossi prefers to stay out of theoretical confrontations, even if he repeatedly said he stays inside the Standard Model to explain everything, so we need your counterpoint.
Robert Curto
Ft. Lauderdale Florida
USA
Tom Conover:
The 1 MW E-Cat is a low temperature plant, it reaches max T= 120°C.
What I wrote regarding the high temperatures is related to the Hot Cat.
Warm Regards,
A.R.
Readers of the JONP. This is how I see the coulomb barrier enigma. The barrier is formed of 90 degree angles of various field resistance from zero to an achievable maximum. Structure is cubic, formed of four pyramids of a horizontal plane, apex inner most and two pyramids apex inner most on a vertical plane. This represents the geometric layout not the actual.
These four pyramids have to be set one above another on a vertical plane in a spiral format. Each pyramid occupies a quadrant of a quadrangle and ascends a pyramid structure from its base. Thereby divide a pyramid into four levels. At the base of the top level in one of the quadrants mark (p) for proton zone. At the base of the level beneath but advanced by 90 degrees in a quadrant marked (n) for neutron zone. At the base of the level beneath but advanced by 90 degrees in a quadrant mark (e) for the electron zone and finally at the base of the level beneath mark (a) for aether/electrosphere zone. Thereby to ascend the pyramid which is of a cubic dimension being of four horizontal pyramids you enter at the base level of the base pyramid and ascend on a spiral of four 90 degree angles to get to the next level. To reach the apex of the pyramid from its base a total of sixteen 90 degree angles have to be completed.
Every 360 degrees of ascent requires a vertical descent back to the negative zone being a south pole of the cube. Upon exiting the charge will ascend upon the exterior and enter the cube at the next level above at the base of the above pyramid and repeat the process until the apex is reached. Consequently, when the circuit is completed the charge will enter the apex of the pyramid being the absolute north gate and exit at the absolute south gate being in a central position of the quadrangle. A positive charge represents a compressed negative and a negative charge represents an expanded positive. Four nickel isotopes represent four horizontal pyramids that in combination represent a cubic dimension of nickel.
The pyramids revolve to remain neutral but the coulomb barrier is not solid with regards a 360 degree coverage. It may appear to be so due to r.p.ms. of the pyramid. Because of inertia caused by a moving economy flow system which is within every structure/charge, the charge is able to escape from its structure being in an accelerated state as an absolute with regards its charge potential. Thereby, the base of the preceeding pyramid is able to capturte, increase it in charge potential and accelerate it in velocity. Upon negotiating the macro cube of four isotopes with no further gate to cross, transmutation of the fourth isotope takes place.
Between the apex of one pyramid and the base of a preceeding one is a gate/empty space/almost empty and thereby there are minor gates within a cube and major gates between cubes. Thereby for a charge to complete six circuits of a cube being of six pyramids the seventh circuit requires a higher level of a unique vibration that requires a transmutation of one pyramid into the base of the preceeding cube. Out of one cube into a higher dimension which entails the first pyramid/apex of the lower cube becoming the last/base pyramid of the higher cube and untimately the last becomes the first. Such is evolution. If the geometric activity is as such, then to line up more so the horizontal pyramids would aid in the reaction. To introduce hydrogen into the atmosphere could in theory bring a state of chaos into the activity that would result in the production of heat which would contribute to the process by producing more overall activity. Transmutation, I believe, is the major event because without it evolution would not exist and thereby purpose would be a none factor. I put this as a consideration. Regards Eric Ashworth.
Andrea,
On November 4th, 2014 you replied to Bob that the maximum temperature an operating e-cat can produce is at peak 1,400°C, and that you had achieved that temperature in e-cat operation. Does that mean that the regular e-cat 1MW masterpiece can perhaps easily achieve temps higher than 1,000°C during normal operation?
Is the peak temperature still 1,400°C for the e-cat 1MW plant?
Thank you!
Tom Conover
Buck:
As I always said, LENR must integrate in the complex world of energy sources.
Warm Regards,
A.R.
Andrea,
about three weeks ago, you responded to a news article in the (Persian) Gulf News describing LENR phenomena as fact and LENR technology as imminent, stating that “now the competition is very serious.”
link>> http://gulfnews.com/news/world/india/indian-government-urged-to-revive-cold-fusion-1.1413814
Would you include the following article from the New York Times “Energy Firms in Secretive Alliance with Attorneys General” as another example of “now the competition is very serious”?
link>> http://www.nytimes.com/2014/12/07/us/politics/energy-firms-in-secretive-alliance-with-attorneys-general.html?hp&action=click&pgtype=Homepage&module=first-column-region&_r=0
Wladimir,
1. You write,
“The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.”
But you said that the purpose of the “hole” was to allow positively charged particles to bypass the Coulomb barrier.
Only yesterday you wrote,
“The proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barrier.”
So, can a proton not enter a nucleus through the “hole” along the z-axis?
2. You write,
“The neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.”
i) What is the electron that is “captured”?
ii) What is the mechanism that sets the intrinsic spin, and therefore the intrinsic magnetic moment, of the newborn neutron AGAINST the induced magnetic moment of the rotating 2s1 electron in order that the neutron be repelled into the target nucleus?
All the best,
Joe
Charles Jameson wrote in December 7th, 2014 at 7:06 PM
Dear Wladimir
Your focus has been on the interaction between Ni and Li but Rossi and others have said that hydrogen is a key element. If I understand correctly your theory does not require it to explain the Rossi Effect.
—————————————————
Charles,
you did understand correctly.
regards
wlad
Andrea Rossi wrote in December 7th, 2014 at 6:44 PM
JC Renoir:
I am also taking in strong account the 2 papers published in November by Dr Oscar Gullstroem on Ecatworld.
————————————————
I would suggest to Dr. Oscar Gullstroem to apply his theory on “Low Radiation Fusion Through Bound Neutron Tunneling” so that to explain how neutrons with 14MeV , protons with 10MeV, and alphas with 15MeV, can be produced in the Mosier-Boss experiment, since the energy available is only 2,2MeV (the binding energy of the deuteron used in the experiment for the cold fusion Pd-1H2):
III. CONCLUSIONS
In this communication, the use of microscopic examination, sequential etching, and LET spectrum
analysis to identify the energetic charged particles and neutrons responsible for the tracks in CR-39 detectors used in Pd/D co-deposition experiments were discussed. These analytical techniques gave complementary results .
The energetic particles were identified as 2.45 MeV neutrons, 3-10 MeV protons, 2-15 MeV alphas, and 14.1 MeV neutrons.
[…]
The only known source of these long range alphas is fission reactions. This suggests that the new elements observed on the cathodes are caused by fissioning of the Pd nucleus
.
http://newenergytimes.com/v2/conferences/2012/ICCF17/papers/Mosier-Boss-Its-Not-Low-Energy-Paper-ICCF17-pp.pdf
The question is:
how protons (or neutrons) with low energy 2,2MeV (resulted from the fusion 1H2-1H2) can be able to produce the fission of the Pd nucleus ?
I can see only one answer:
the proton produced by the 1H2-1H2 fusion is accelerated, it decays in a neutron which is also accelerated until to get 10MeV, and it hits the Pd nucleus, causing its fission (the neutron is not absorbed by the Pd nucleus, as the neutron is aborbed by the Ni nucleus in the Rossi-Effect, because the energy of the neutron in the Mosier-Boss experiment is too much high for its absorption by the Pd nucleus)
But perhaps I am wrong.
And then the question is to know how Dr Oscar Gullstroem can explain it with the theory of the Fusion Through Bound Neutron Tunneling.
regards
wlad
Dear Wladimir
Although I am a retired engineer with no experience in nuclear physics I commend you on the simplicity and clarity of your theory. Your diagrams are a big help and I think I do have a basic understanding of what you are proposing. I do have a question though. In your theory explaining the Rossi Effect what part does hydrogen play. Your focus has been on the interaction between Ni and Li but Rossi and others have said that hydrogen is a key element. If I understand correctly your theory does not require it to explain the Rossi Effect.
Thanks
Charles Jameson
Curiosone:
The ssm should result to be by far the main mode of operation of the 1MW E-Cat. This thanks to the new control system and this also is the reason of most of the difficulties we have to overcome. Honestly, our Team is making a masterpiece, among many difficulties.
Warm Regards,
A.R.
JC Renoir:
I am perusing the books of Prof. Nornam Cook “Models of the Atomic Nucleus” ( Springer, 2nd edition 2010) and “Nuclear Models” of Prof. Walter Greiner and Prof. Joachim Maruhn (Springer 1996).
I am also taking in strong account the 2 papers published respectively on October 25 and on November 18 2014 by Carl-Oscar Gullstroem on Ecatworld:
” Low radiation fusion through bound neutron tunnelling” and ” Collective neutron reduction model for neutron transfer reaction”.
Warm Regards,
A.R.
ERRATA:
In my previous comment, where it written:
In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Mg.
the correct is:
In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Hg.
Herb Gillis wrote in December 6th, 2014 at 11:44 AM
2) —————————————–
I don’t know what the emitter is. There is no obvious source of hydrogen or deuterium in a fluorescent bulb. Possibly it might be the phosphors used. These are typically rare earth metal iodides (such as those of yttrium or europium). There may be other things in the bulbs as well. Certainly there is glass, and argon.
———————————————
Herb,
there are other ways to produce cold fusion, without the need of emitters.
For instance:
a) fusion p+e -> n.
This occurs in the Conte-Pieralice experiment (a different version of the Don Borghi experiment)
In their paper published in Infinite Energy Magazine in 1999, they say:
“During the electrolytic process, a beam of electrons was forced to hit the aluminium cathode, and so induced a direct interaction of the protons and the electrons. A radioactive source of 90Sr + 90Y was utilized, having activity about 2.39 KBq (overall uncertainty 10%). The radioactive source was a disk having a diameter of 20 mm with implanted 90Sr + 90Y by electrodeposition. It was fixed external to the cell and the electron beam was collimated in order to hit directly the surface of the aluminium cathode with a geometry estimated to be 2 at a good approximation.
We should add that the aim of the experiment was to measure the possible synthesis of a proton and an electron into a neutron. It was not our purpose to effect calorimetric measurements. It was an unexpected result that we discovered that the aluminium cathode melted in the water during the experiment.
All previous calculations that we performed, accounting for the radioactive source at such low activity, excluded the possibility of the melting of the cathode. The aluminium cathode melted only when we used the radioactive source and not in the course of the same experiment but without the beta source.
We were not equiped to follow and measure the temperature rise in the cell. In a repetition of the experiment we are considering measuring parameters to investigate the melting of the cathode.”
b) Other possibility is: a coherent flux of protons (hydrogen) moving along the magnetic vector of a magnetic field target a nucleus aligned by the magnetic field. A proton can enter within a nucleus by the “hole” in the Coulomb barrier of the nucleus.
In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Mg.
regards
wlad
Andrew1:
Good luck to Vessela Nikolova’s book; you are right, the music is worth listening.
Warm Regards,
A.R.
Joe
December 7th, 2014 at 1:45 AM
Wladimir,
1. ) —————————————–
You write,
“Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.”
As I said in a previous post, this could happen, but it would be a very rare event due to the very precise direction needed for the initial velocity of the newborn proton.
———————————————–
Joe,
in the decay of the neutron the proton is emitted along the z-axis of the 3Li7, and it is easy to undestand why.
Look at for instance the structure of the 3Li7:
FIG. 1:
http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png
We have:
1- The electron gyrates about the proton in the structure of the neutron
2- When the neutron starts to decay, the radius orbit of the electron moving about the proton starts to increase.
3- Note in the Fig. 1 above that the orbit of the electron is perpendicular to the z-axis of the 3Li7.
4- So, the magnetic moment of the electron’s orbit is growing up
5- Then the proton is emitted along the z-axis, because the proton is forced by the Lorentz force to take the direction perpendicular to the electron’s orbit, which coincides with the z-axis of the 3Li7.
This is shown in the Figure 2:
FIG. 2:
http://peswiki.com/index.php/Image:Fig._2-proton_direction_forced_by_Lorentz_force.png
.
2. ) —————————————
You write,
“This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.”
I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.
————————————————–
Joe,
what you say makes no sense.
The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.
The electron has negative charge, and it has not repulsion with the Coulomb barrier, and so it can cross it in any point.
regards
wlad
Hello Dr. Rossi, I’ve just seen the book trailer of your biography on the Vessela Nikolova’s blog: http://www.ecat-thenewfire.com/blog/book-trailer-ecat-thenewfire/. The music is worthy of your discovery! Best wishes, Andrew
In the 1MW plant operating in the factory of Industrial Heat’s customer will be the self sustaining mode important, or it will not be used, as in the Lugano test of the Hot Cat?
W.G.
Which books are you helping with to study the theory behind the so called Rossi Effect?
Thank you,
JC Renoir
Subject: use of 50Sn as receptor, so that to get COP higher than 3
Dear Joe,
50Sn has ten stable isotopes, while 28Ni has only five. So, if one succeeds to get cold fusion with 50Sn, while 28Ni gets a COP of 3 the 50Sn will supply a COP very higher than 3.
Andrea Rossi already told in the JoNP that he had tested all the sort of combinations, and he concluded that Ni is the best fuel for the eCat. For instance, he said that 46Pd is not a good element. Then probably Andrea Rossi already had tested 50Sn, and concluded that it is also not a good element to react with 3Li7.
However it is easy to understand why Andrea Rossi did not succeed to get a god COP by using 46Pd and 50Sn combined with 3Li7. We can see it in the Fig. 1 ahead,
FIG. 1:
http://peswiki.com/index.php/Image:Fig._1-50Sn_as_RECEPTOR_in_eCat.png
After the decay 3Li7 -> 4Be7, the proton of the 4Be7 is pulled by the orbit of the electron 2s1, and when the proton crosses the plane of the orbit the proton by electron capture transmutes to neutron, which is now pushed by the orbit 2s1 with a repulsion force.
So, in the reactor using Ni-3Li7 as fuel, the final acceleration on the neutron is not so high, because the radius orbit of the electron 2s1 is not large.
But in the reactor using Pd-3Li7 the final acceleration on the neutron is very faster, because the radius orbit of the electron 2s1 is very larger. And so the neutron trespasses the 46Pd nucleus, and is not captured.
The situation is worst in the case of a fuel composed by Sn-3Li7, because the final velocity of the neutron will be very larger than it was occurred by using as fuel Pd-3Li7.
That’s why Andrea Rossi did not succeed to get a good COP by using 46Pd.
However, the acceleration on the proton by the orbit 2s1 must be very higher than the acceleration on the neutron, because:
A) The proton is accelerated by magnetic interaction, It has magnetic moment 2,793, and also by Coulomb attraction with the electron 2s1.
B) But the neutron is accelerated only by the magnetic interaction, and its magnetic moment is only 1,913, and it is not pushed by Coulomb repulsion
Therefore, if one discovers a suitable emitter to be used with the 50Sn, maybe he can succeed to use 50Sn as a receptor, and the COP will be very high than the COP 3 obtained by Ni-3Li7.
The emitter to be used with 50Sn must have a pair number of protons, in order to have a have a second orbit parallel to the the orbit 2s1, as shown in the Figure 2.
FIG. 2:
http://peswiki.com/index.php/Image:Fig._2-50Sn_as_RECEPTOR_in_eCat.png
Let us analyse what can be the best candidates to be the emitter:
4Be9 as emitter for 50Sn :
Abundance: 100%
With the decay of the neutron, 4Be9 transmutes to 5B9.
The half-life of 5B9 is 800×10-21 s, and it transmutes to 4Be8 by emtting a proton. 4Be8 is unstable, and decays emitting two 2He4.
The structures of 4Be9 and 4B10 are shown in the Figure 3.
FIG. 3:
http://peswiki.com/index.php/Image:Fig._3-structures_of_4Be9_and_5B9.png
After the decay of the 5B9, the proton exits the nucleus along the z-axis, and goes to target the 50Sn nucleus, accelerated by the resultant of magnetic forces due to the orbits 2s1 and 2s2.
As the neutron in the 4Be9 is strongly bound than the neutron in the 3Li7, then a higher oscillatory electromagnetic field must be applied than that used in the Rossi’s Ni-3Li7 fusion.
.
6C13 as emitter for 50Sn:
Abundance: 1,1% . There is need to enrich the fuel with C13 isotopes.
The structure of excited 6C13 is shown in the Figure 4.
FIG. 4:
http://peswiki.com/index.php/Image:Fig._4-structures_of_EXCITED_6C13.png
The structures of the 6C13 and the 7N13 are shown in the Figure 5:
FIG. 5:
http://peswiki.com/index.php/Image:Fig._5-structures_of_6C13_and_7N13.png
After the decay of excited 6C13 to 7N13 by the decay of the neutron to proton. In normal conditions, the 7N13 has half-life of 9,9 minutes, and it transmutes to 6C13 by beta decay. But before the decay the proton can be captured by the orbit of the electron 1p1, and goes to target the 50Sn nucleus.
.
8.O17 as emitter for 50Sn:
Abundance: 0,2%
.
10Ne21 as emitter for 50Sn:
Abundance: 0,27%
.
12Mg25 as emitter for 50Sn:
Abundance: 10%
Magnesium 12Mg24 has a structure similar to carbon 6C12 (24Mg = 12C + one complete hexagonal floor).
Figure 6 shows how 12Mg24 changes its structure when it captures a neutron and it is formed the 12Mg25.
FIG. 6:
http://peswiki.com/index.php/Image:Fig._6-structures_of_12Mg24_and_12Mg25.png
In 12Mg25 the neutron is bound to the deuteron D-2 via spin-interaction in a similar way as the neutron is bound to the deuteron in the 3Li7.
With the decay of the neutron, the newborn 13Al25 has half-life of 7,18s , and in normal conditions it decays to 25Mg by beta decay.
So, before 13Al25 to decay, its proton will be extracted by the resultant of the magnetic force due to the orbits 1p5 and 1p6, and the proton goes to target the 50Sn nucleus.
.
Joe,
I think would be of interest to perform experiments, so that to verify if by using 50Sn a higher COP than 3 can be obtained, if the suitable emitter is used by suitable conditions of excitation of the emitter.
What do you think about?
Regards
wlad
eernie1 wrote in December 6th, 2014 at 5:40 PM
Wlad,
That is my point. In a normal molecular configuration where the valence electron remains in its 2S1 energy state, no neutron emission is observed. In a LENR reaction, the valence electron is driven into a higher energy state where it can influence the removal of the loose neutron in the 3Li7 nucleus by its more energetic electrosphere. But since the orbit would normally be larger and the influence less strong because of the 1/r^2 rule, we must reduce the radius of the orbit to retain the higher interactive strength.
—————————————–
Dear Eernie,
I and Joe are concluding that there is not emission of neutron in the Rossi-Effect.
Our conclusion is that the neutron decays in a proton, and the proton exits the nucleus.
Our conclusio is because if the neutron would exit the 3Li7, there is no way to explain the 10MeV neutrons emitted in the Pamela Mosier-Boss experiment from the fusion 1H2 +1H2 -> 1H3.
In the Rossi-Effect, the neutron is also accelerated , but with energy lower than 10MeV, because Ni has 28 protons, while Pd has 46 protons, and so the 2s1 orbit in the Pd nucleus has larger radius than in the Ni nucleus. That’s why in Rossi-Effect the neutron is captured by the Ni, while in Mosier-Boss experiment the neutron is NOT captured by the Pd nucleus, because the neutron is moving too must faster than the maximum speed which allows its capture.
The neutron acceleration is impossible if it exits the 3Li7 in the form of a neutron, because:
1- when the neutron exits the 3Li7, it is accelerated due to magnetic attraction with the magnetic field induced by the 2s1 orbit.
2- but after crossing the plane of the 2s1 orbit, the neutron must be attracted by the orbit, and so it will be stoped before to target the nucleus Ni.
That’s why now Joe and me are considering that the neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.
So, the neutron can target the Ni nucleus only if it decays in a proton before to exit the 3Li7 in the Rossi-Effect.
regards
wlad
Wladimir,
1. You write,
“Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.”
As I said in a previous post, this could happen, but it would be a very rare event due to the very precise direction needed for the initial velocity of the newborn proton. The reason for this is due to the Lorentz force which contains velocity, v, as a factor. You must not simply look at the fact that 2s1 might induce a magnetic moment and imagine that a CHARGED particle would be automatically pulled to it. For example, if the initial velocity of the proton were parallel to the plane of rotation of 2s1, it would rotate itself in its own plane and never travel along the z-axis. Here are two figures explaining this:
https://en.wikipedia.org/wiki/File:Lorentz_force.svg
https://en.wikipedia.org/wiki/File:Cyclotron_motion.jpg
2. You write,
“This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.”
I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.
All the best,
Joe
Joe wrote in December 6th, 2014 at 4:12 PM
1) —————————————
If the newborn proton is immediately affected by 2s1, then no 4Be7 could ever be formed. The proton’s initial velocity would ensure the presence of a Lorentz force acting upon the proton, keeping the proton from ever settling in a gravitational flux n(o) and thereby preventing the formation of a 4Be7. (The role of the other electrons is not important since there would be cancellation of effect due to symmetry.)
——————————————————–
Joe,
the proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barier.
Therefore, the proton will exit the 4Be7 moving about the z-azis direction.
This also happens in the alpha decay of 92U238. The experiemnts have shown that 2He4 is emitted in a radial line, i.e. along the z-axis which passes by the center of the nucleus 92U.
The 2He4 exits the 92U by the “hole” in the Coulomb barrier because its energy is 4MeV, and the Coulomb barier has 8MeV, and so the 2He4 can exist the 92U only trhough the z-axis.
This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.
Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.
regards
wlad
Wlad,
That is my point. In a normal molecular configuration where the valence electron remains in its 2S1 energy state, no neutron emission is observed. In a LENR reaction, the valence electron is driven into a higher energy state where it can influence the removal of the loose neutron in the 3Li7 nucleus by its more energetic electrosphere. But since the orbit would normally be larger and the influence less strong because of the 1/r^2 rule, we must reduce the radius of the orbit to retain the higher interactive strength.
Regards.
eernie1 wrote in December 6th, 2014 at 11:27 AM
Wlad,
The normal valence electron of the 3Li7 is a 2s1 electron. However in its normal state within the molecules it is associated with there is no neutron emitted from the nucleus of the 3Li7. I think the bound electrons must be excited in some manner up to at least the 2p1 energy level before the force of the electrosphere can get involved in dislodging the Halo neutron of the 3Li7 nucleus. The orbit of the electron must also shrink due to an external negative field(perhaps from a negative Hydrogen ion)applied to the exterior of the atomic electron configuration.
———————————————-
not necessarily, Eernie
within the molecules, the nuclei of the atoms are not aligned towards the z-axis as happens in the cold fusion reactors.
So, in the molecules the electric field of the atoms is spherical, there is not any electron orbit perpendicular to the z-axis, and therefore the neutron and the deuterons of the 3Li7 are not submitted to an oscillation along any specific direction, as occurs in the cold fusion reactors, where they get oscillation toward the z-axis.
We have to take caution, in order do not get wrong conclusions by the comparison between the behavior of the atoms in normal conditions and the special conditions existing within a cold fusion reactor.
regards
wlad
Wladimir,
We are not dealing with natural circumstances whereby the 3Li7 may become 4Be7 after a certain amount of time. We are dealing with a technology (E-Cat) that might be artificially decaying the neutron much faster than natural and then having the newborn proton pulled toward a likewise artificially placed 2s1 electron rotation. In this scenario, no 4Be7 could ever be formed since the newborn proton would not naturally place itself in the gravitational flux n(o) as you have pictured it. The way that you imagine it, the proton would behave the same independently of the presence or absence of a rotating 2s1 electron. This is illogical. There is contradiction even between your statements:
“1- The orbit of 2s1 is not able to extract the proton from the newborn 4Be7.”
“4- […] the orbit of the 2s1 electron is pulling the proton, and so the proton exists the 4Be7.”
If the newborn proton is immediately affected by 2s1, then no 4Be7 could ever be formed. The proton’s initial velocity would ensure the presence of a Lorentz force acting upon the proton, keeping the proton from ever settling in a gravitational flux n(o) and thereby preventing the formation of a 4Be7. (The role of the other electrons is not important since there would be cancellation of effect due to symmetry.)
All the best,
Joe
Wladimir Guglinsky:
In response to your two questions:
1) Mercury transmutation in CFL bulbs is discussed at the link below, with references to original analytical paper (from 2013):
http://www.e-catworld.com/2014/12/05/compact-flourescent-lightbulbs-we-are-surrounded-by-lenr-in-our-own-homes-gordon-docherty/
2) I don’t know what the emitter is. There is no obvious source of hydrogen or deuterium in a fluorescent bulb. Possibly it might be the phosphors used. These are typically rare earth metal iodides (such as those of yttrium or europium). There may be other things in the bulbs as well. Certainly there is glass, and argon.
All this raises some further questions:
– What if one added H2 to one of these bulbs?
– What if nickel and lithium vapors were added?
– What about a combination of Ni, Li, and H2 vapors:
– with the original mercury vapor,
– or without the original mercury vapor?
Regards; HRG.
Wlad,
The normal valence electron of the 3Li7 is a 2s1 electron. However in its normal state within the molecules it is associated with there is no neutron emitted from the nucleus of the 3Li7. I think the bound electrons must be excited in some manner up to at least the 2p1 energy level before the force of the electrosphere can get involved in dislodging the Halo neutron of the 3Li7 nucleus. The orbit of the electron must also shrink due to an external negative field(perhaps from a negative Hydrogen ion)applied to the exterior of the atomic electron configuration.
Regards.
Dear Joe,
in my theory for the explanation on the Rossi-Effect the ORBITS of the electrons (in both Ni and 3Li7) have a SYMMETRICAL distribution about the z-axes of both Ni and 3Li7.
Neverthelless,
you are exhibiting many resctrictions against my theory.
I am not criticizing you. You actually are doing a good work, because we need to eliminate all the restrictions regarding the theory.
However, think about the other theories, as those proposed by Stoyan Sarg, Widom-Larsen, Peter Hagelstein, Fedir Mykhaylov, etc.
From their theories it is simply impossible to explain cold fusion.
As it is impossible to explain cold fusion from any theory where the Coulomb barrier is spherical
Even by Stoyan Sarg theory is not possible to explain cold fusion, because:
a) his Coulomb barrier is non-spherical, however there is not a “hole” in the barrier.
b) the orbits of the electrons will not have a symmetrical distribution about the z-axis.
What do you think about ?
Do you think is possible to explain cold fusion by considering a spherical Coulomb barrier?… as are trying Widom-Larsen, Peter Hagelstein, Fedir Mykhaylov, and others. …
regards
wlad
Dear Herb,
also remember that in order to have cold fusion there is to have a receptor and an emitter.
In Fleischmann-Pons cold fusion the emitter is supplied by heavy water, which has deuterons. Their experiment does not work with hydrogen
In the Rossi-Effect, the emitter is 3Li7.
What would be the emitter in the mercury bulbs?
regards
wlad
Herb Gillis wrote in December 5th, 2014 at 9:10 PM
Wladimir Guglinsky:
Can your theory explain the apparent transmutation of mercury in fluorescent bulbs? The evidence for transmutation looks pretty strong, and it seems to be a neutron transfer process- – very reminiscent of what was observed in Lugano report (with nickel and lithium).
If the two processes are indeed similar then it would seem to imply that LENR is not necessarily a solid state phenomenon. Perhaps its time to start looking at ionic liquids??
——————————————————-
Herb,
I did not know this phenomenon on the transmutation of mercury.
Is there an analysis of the elements before and after a buld beging to work ?
regards
wlad
Joe wrote in December 5th, 2014 at 8:35 PM
Wladimir,
1. ) —————————————————
You have created a further alteration to your model by the introduction of 4Be7. Why would the newly born proton settle to help create a 4Be7 when a 2s1 is acting on it and disrupting any potential creation of a 4Be7?
——————————————————–
I did not understand your question.
The proton does not help to create the 4Be7.
The 4Be7 is resulted from the decay of the neutron to proton in the 3Li7.
The neutron decays because:
a) it is weakly bound to the deuteron via spin-interaction
b) the deuteron has oscillation in the 3Li7, due to repulsion with the central 2He4
c) the oscillatory electromagnetic field applied within the eCat causes the decay of the neutron
5) So, the 4Be7 is formed due to the decay of the neutron
The situation is:
1- The orbit of 2s1 is not able to extract the proton from the newborn 4Be7.
2- Due to repulsion between the proton and the deuteron in the newborn 4Be7, the proton and the deuteron take the places shown in the figure:
FIG. 1:
http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png
3- In the structure of the 4Be7, the proton begins to oscillate aling the z-axis, due to the acton of the Coulomb attraction with the electrons 1×1 and 1s2:
FIG. 2:
http://peswiki.com/index.php/Image:Oscilation_of_the_proton_in_4Be7.png
4- Finally the amplitude of the oscillation increases to much, because the orbit of the 2s1 electron is pulling the proton, and so the proton exists the 4Be7.
2. ) ——————————————
When a neutron decays, the resultant particles separate from each other, so there must be an initial nonzero velocity for the proton. Therefore , the Lorentz force does have its role to play and must be carefully determined in order to ascertain the viability of your model.
————————————————
Yes,
but with that initial velocity the proton is not able to exit the newborn 4Be7 (otherwise, the 4Be7 would have to decay in fraction of seconds in normal conditions, however it has half-life 53,2 days).
In the 3Li7, the deuteron-neutron bound via spin-interaction are kept within the 3Li7 thanks to the following equilibrium:
1- The centrifugal force actuates on the deuteron-neutron
2- The magnetic force of the proton avoids the deuteron-neutron to be expelled from the 3Li7
3- When the neutron decays in a proton, due to repulsion the proton and the deuteron take those symmetric positions regarding to the central 2He4.
4- The centripetal force on the newborn proton than when it was acting on the deuteron-neutron. That’s why the orbit of the proton decreases, and the structure of the 4Be7 gets the equilibrium shown in the Fig. 1.
5- The newborn 4Be7 has a half-live of 53,2 days, and therefore that initial velocity of the proton due to the decay of the neutron is lost (in the form of heat).
6- But as already explained, the action of the electrons 1s1 and 1s2 together with the orbit of 2s1 apply an oscilaltory motion on the proton about the z-axis, the amplitude increases, and finally in fraction of seconds the proton exits the 4Be7.
regards
wlad
Wladimir Guglinsky:
Can your theory explain the apparent transmutation of mercury in fluorescent bulbs? The evidence for transmutation looks pretty strong, and it seems to be a neutron transfer process- – very reminiscent of what was observed in Lugano report (with nickel and lithium).
If the two processes are indeed similar then it would seem to imply that LENR is not necessarily a solid state phenomenon. Perhaps its time to start looking at ionic liquids??
Regards; HRG.
Wladimir,
1. You have created a further alteration to your model by the introduction of 4Be7. Why would the newly born proton settle to help create a 4Be7 when a 2s1 is acting on it and disrupting any potential creation of a 4Be7?
2. When a neutron decays, the resultant particles separate from each other, so there must be an initial nonzero velocity for the proton. Therefore , the Lorentz force does have its role to play and must be carefully determined in order to ascertain the viability of your model.
All the best,
Joe
Robert Curto:
Thank you for the information,
Warm Regards,
A.R.
Dr. Rossi, the second edition is out.
Google:
AN IMPOSSIBLE INVENTION SECOND EDITION
Robert Curto
Ft. Lauderdale, Florida
USA
Joe wrote in December 5th, 2014 at 1:07 PM
Wladimir,
So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?
———————————————-
Joe,
also note that the two electrons 1s1 and 1s2 of the newborn 4Be7 apply a rapid oscillation on the proton along the z-axis, because the Coulomb attraction on the proton changes the direction of its motion, as shown in the figure:
http://peswiki.com/index.php/Image:Oscilation_of_the_proton_in_4Be7.png
As the orbit of the 2s1 applies a strong magnetic force on the proton along the z-axis, the amplitude of the oscillation along the z-axis starts to increase, and so quickly the proton exits the 4Be7, going along the z-axis accelerated by the force applied by the orbit 2s1.
regards
wlad
ERRATA:
In my last post, where it is written:
But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the exits of the 4Be7 nucleus begins to be accelerated along the z-axis, by starting from an initial velocity zero.
the correct is:
But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the proton exits the 4Be7 nucleus beginning to be accelerated along the z-axis, by starting from an initial velocity zero.
Joe wrote in December 5th, 2014 at 1:07 PM
Wladimir,
The direction of initial velocity of the newly born proton would be the most important factor for accomplishing a nucleon transfer. The reason for this is shown in the Lorentz force:
F = q ( E + v x B )
As you can see, it is not just a simple matter of summing electric and magnetic forces, although it does include that. If v is not parallel to the z-axis, the proton may never strike the plane of rotation. So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?
—————————————————————-
Joe,
the initial velocity of the proton is zero, because the half-life of the newborn 4Be7 is 53,2 days.
So, after the decay of the neutron, the proton stays in the 4Be7.
Look at the structure of the 4Be7 after the decay of the neutron in the 3Li7:
http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png
You can see it also in the Figure 38 at the page 49 of the paper Stability of Light Nuclei:
http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf
The 53,2 days of the half-life of the 4Be7 occurs in normal conditions.
But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the exits of the 4Be7 nucleus begins to be accelerated along the z-axis, by starting from an initial velocity zero.
regards
wlad
Wladimir,
The direction of initial velocity of the newly born proton would be the most important factor for accomplishing a nucleon transfer. The reason for this is shown in the Lorentz force:
F = q ( E + v x B )
As you can see, it is not just a simple matter of summing electric and magnetic forces, although it does include that. If v is not parallel to the z-axis, the proton may never strike the plane of rotation. So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?
All the best,
Joe
Daniel De Caluwé wrote in December 5th, 2014 at 5:56 AM
@Wladimir,
Why does the more losely bound neutron in 3Li7 decays? Due to the excitation by the alternating magnetic field? And does this give enough energy to let it decay? How much energy actually is needed to let that neutron decay in a proton + electron, and are all necessary conditions fulfilled to let it happen?
————————————-
Daniel,
free neutrons decay in proton+electron in about 15 minutes.
Within the nuclei the neutrons are stable, because they are bound to deuterons via spin-interaction.
The spin-interaction between a deuteron and a neutron within the nuclei reinforces the binding energy between the electron and the proton within the structure of the neutron.
But a neutron weakly bound (as in the 3Li7) is in the verge of having decay (there is very little lacking for it to decay, because it is poorly connected to the deuteron).
So, with the excitation of the 3Li7 through the oscillatory electromagnetic field, two things may occur:
a) with NO excitation, the neutron has a normal oscillatory motion about the z-axis in the 3Li7 nucleus, because of magnetic interactions with the other deuterons.
b) with excitation, the spin-interaction deuteron-neutron becomes weaker, because the oscillation of the neutron about the z-axis increases its amplitude .
c) the excitation of the neutron exceeds the tolerable limit, and the neutron decays.
regards
wlad
Joe,
in the figure posted by me earlier it is shown only the Coulomb force on the proton:
http://peswiki.com/index.php/Image:For%C3%A7as_sobre_o_proton_no_3Li7.png
But there is also a atrong magnetic force applied by the orbit of the electron 2s1, pulling the proton along the z-axis.
So, the proton acquires a helical trajectory which axis is the z-axis.
The radius of the helical trajectory (about the z-axis) increases as the proton progresses.
But the proton travels the distance 10^-11m between the nucleus and the orbit 2s1 in fraction of seconds, and so the growth of the radius of the helical trajectory is despicable when the proton hits the orbit of 2s1.
Thefore the proton hits the orbit 2s1 practically in the center of the orbit
regards
wlad