.
by
by
Sundar Narayan
Lambton College – Sarnia, Ontario, Canada
Abstract
This paper derives a formula for the lifetime of an unbound or free neutron and shows that neutron lifetime can be related to Newton’s gravitational constant, G, providing a much-needed theoretical formula for G, enabling G to be computed with greater accuracy than today’s experiments allow.
Another equally accurate formula for G is derived based on the properties of the virtual electrons that very briefly exist in a quantum vacuum.
Also, Newton’s law of gravity and Coulomb’s electrostatic law are derived from the same equation, providing a simple proof of the well-known connection between these two laws.
Dear Andrea,
I was wondering, why does the plaque text contain periods where there should be commas?
Sincerely,
The Punctuation Police
Greg Leonard:
I agree.
About the photos: you will find them here asap.
Warm Regards
A.R.
Dear AR
A very worthy tribute.
I look forward to seeing it when the pictures of the new plant are published
Dear Joseph Fine,
thanks for reading the “article” and for the comment about the already used name.
I think will change from neuter (-um) to masculine (-us) (always in the second declension): Hydronius.
The shortening will remain Hyd, which suggests how the particles so far managed to “hide” from the researcher’s’ eyes. And Hyd could also be the Hyde version of Dr Hydrogen when using the electron potion.
Regards
Andrea Calaon
To all our Readers:
One year ago Prof Sven Kullander passed away. We applied to the 1 MW plant installed in the factory of the Customer of IH the following plate, in memory of this giant of the scientific world:
” To Prof. Sven Kullander.
The first industrial plant working with the new LENR technology is dedicated to Prof. Sven Kullander, professor emeritus in high energy physics in Uppsala University and member of the Swedish Royal Academy of Science.
His scientific spirit and great skill, human and cultural standards and neverending enthusiasm have been of monumental value for the completion of this plant .
Andrea Rossi and the whole Team that manufactured this plant “
Dear Andrea Calaon,
Thank you for your kind answer. In the paper you say that Hyd is “picometrically neutral”. If this is taken to mean that its effective size with respect to reactions with metal nuclei is 1 pm, then Hyd’s mean free path in nickel is only 10 microns. If the Hyd (assuming it exists) mostly reacts inelastically with a nucleus it encounters (that is, if it usually causes a nuclear reaction of some sort instead of being scattered elastically), then Hyds should not be able to leak outside the device although there should be some transmutations taking place in the innermost few tens of microns layer of the Al2O3 reactor wall unless Al and O would be for some reason “inert” nuclei with respect to such process.
best regards, pekka
TO ALL OUR READERS:
From yesterday through today we suffered an attack that has put off line our blog. The IT Guy of our Team has restored the connection few minutes ago. Thanks to him ( Daniele Princiotto, an Italian informatic engineer) for his outstanding job.
Probably some comment has been lost in the meantime in the spam, therefore the authors of spammed comments are kindly invited to send them again.
Warm Regards,
A.R.
Andrea Calaon,
The term Hydronium ( “Hyd” ) is already used in Chemistry as the H3O+ cation.
http://en.wikipedia.org/wiki/Hydronium ( But ‘Hyd’ is new. )
Thanks for your interesting article.
Joseph Fine
Dear Pekka Janhunen,
thank you for the interest in my theory and for the comment.
The equation of Schroedinger (what I am saying is valid for the Pauli and Dirac equations as well) allows to calculate the energy levels of electron orbitals because it describes a system where the electron remains “separate” from the nucleus. The electron Zitterbewegung is free from “intruders” and the electron orbitals are the stationary solutions that match the ZB intrinsic rotation with the orbital motion. The formation of Hyd is a very special event, in fact Cold Fusion is still debated today …
It is possible that in other, although rare, occasions humans or the universe produced Hyd, however Hyd are picometrically neutral and do not stably add up to nuclei as neutrons do. They are difficult to detect and probably have never been produced massively so far. If I remember well Randell Mills even says that they are responsible for the Diffuse interstellar bands and dark matter. I do not agree, but I am not surprised by the fact that Hyd have so far passed undetected.
The formation of Hyd must be exothermic because there must be a force that overcomes what I called the “orbital repulsion”, the tendency of electrons not to be confined at distances of less than its ZB size (383 [fm]). And the magnetic attraction force (possibly with an associated potential) accelerates the charges towards the new configuration liberating some energy.
The Hyd are not bound to the electronic structure of the metal, neither with the conduction nor with the valence electrons. Therefore none of the physical or chemical properties of the metal should change at the formation of Hyd, apart from the “disappearance” of hydrogen nuclei and electrons. Once the Hyd are formed they should behave similarly to the neutrons, and be able to cross matter easily, apart from nuclear encounters with reactions on the way. And even these reactions are not very evident because they do not generate rains of gammas or other energetic particles.
I suspect the formation of one Hydronium liberates as much as 1.745 [MeV], while the formation of a Deuteronium 1.445 [MeV].
Best Regards
Andrea
Dear Andrea Calaon,
Your paper is interesting. But of course, a lot is unclear. Below are some quick thoughts.
The idea of a compact Hyd is good, but I’m sceptical that a Hyd could leave the metal and exist as a free particle because quantum states of single hydrogen atoms are well known and do not allow such solution and because if such compact Hyds would exist, we probably would have seen them already in nature and in astrophysics. That said, it might be possible that some form of “Hyd” can form and exist inside the many-electron environment of the metal where a larger number of electrons might spend part of their time near the hydrogen nucleus to provide charge neutralisation and posssibly electric current to make magnetic effects.
If such “Hyd” exist, its formation could be more or less exothermic (i.e. we don’t know). If it’s strongly exothermic, then the Hyd should have a hard time leaving the metal because that would require the same energy to be put back from some external source to turn the “Hyd” back to normal hydrogen atom. Existence of such exothermic “Hyds” inside the metal might even increase the melting point of the nickel particle. But if the formation of Hyd would be only weakly exothermic, then these effects would be correspondingly mild.
I might have more comments later concerning the paper.
regards, pekka
Dear Dr. Rossi , this link talks about the E cat and even in India. Congratulations for your work that will change the world .
Next month’s edition of Current Science magazine of the Indian Academy of Sciences will feature low energy nuclear reactions on its cover, a sign of grudging acceptance by the scientific community.
http://www.thehindubusinessline.com/news/science/do-not-forget-the-other-nuclear/article6818560.ece
Andrea Calaon:
Specialists are working on the electric power production.
Warm Regards,
A.R.
Wladimir,
If your nuclear model can explain the Rossi Effect, does this mean that E-Cats can be used to remediate nuclear waste?
All the best,
Joe
JR,
Thanks for completing the scenario. I had trouble locating a grams-to-mole conversion factor for hydrogen.
All the best,
Joe
Dear All,
It has been a long time since my last post on my LENR “theory”. In the meantime I have changed it significantly.
Now the beta-decaying H4 is no more needed, while three neutral pseudo-particles appeared.
There is never a Coulomb barrier to be overcome and the fractionation happens through the emission of photons during the acceleration of the involved particles.
The reactions happen in two stages. In the First Stage the neutral pseudo-particles form in the very special conditions Edmund Storms calls Nuclear Active Environment. In the Second Stage the neutral pseudo-particles, which can almost freely cross solid matter, react with the nuclei they encounter.
The Second Stage reactions take place at practically no excess kinetic energy, so that only the most stable and least energetic nuclei can form.
I analysed the results of the Hot-Cat test through the theory, as well as the experiments of Iwamura and Mizuno.
Here is the link if you would like to have a look:
lenr-calaon-explanation.weebly.com/
I hope some reader will be so kind to put her/his comment/critic/suggestion in the blog section.
Thank you Andrea for publishing this shameless advertising.
Regards
Andrea Calaon
Dear Andrea Rossi,
Sorry to insist, but I am convinced that you do not need to fiddle much with gas heated cats, which anyway emit some CO2 … You have reactors with surface temperature in excess of 1,300 [C], and electric-to-heat COP above 14 (min. 3.5/0.25 = 14), so a system with a heat-electric efficiency of 50% would already give 50-100/14=42 units of electric power for each 100 units of heat produced. Not bad for a start!
The single supercritical CO2 turbine has an efficiency around 50%, while with heat recuperators and the rest of the plant, considering water separation and compressors, surpasses 60%.
If IH manages to enter the consortium of NET Power, LLC, CB&I, Exelon Corporation, and 8 Rivers Capital, for the building of a 50 [MWt] plant in Texas:
http://news.toshiba.com/press-release/corporate/toshiba-supplies-first-kind-supercritical-co2-turbine-new-thermal-power-gene
it will be able to show electricity production at 25 [MW] electric scale.
No emissions whatsoever. New technology, almost no fuel cost. In Texas with the world best energy companies. The whole world would buy it.
My guess is that the plant will have 60% efficiency, so that, with Cold Fusion modules with a COP of 15 you will have 60-100/15 = 53% heat to electricity conversion. You do not need gas, it is a complication useful only for staying in an old technology market. You will succeed with the gas heated cat, but I am not sure the best move is to concentrate efforts on gas.
I imagine that from low temperature plants for heating and the domestic units IH will have the revenues to stay in the cutting edge technology like supercritical CO2 for electricity production at >50% overall efficiency.
Best regards
Andrea Calaon
Joe wrote in January 22nd, 2015 at 5:50 PM
Wladimir,
2.——————————————————-
You then stated that this flux was directed toward Earth alone:
“As a flux with 2s10^6 neutrons/cm².sec is emitted toward the earth direction”
———————————————————-
Joe,
I did not say that the flux 2×10^6 neutrons/cm².sec hits the earth.
I did mean to say the following:
1- There is emission of two types moving in contrary direction:
1.a) one flux goes moving toward a radial direction leaving out the sun
1.b) and the other flux goes moving toward a radial direction going to the center of the sun
2- The flux 1.a is 2×10^6 neutrons/cm².se, and 2×10^-2 neutrons/cm².sec hits the earth
Actually the quantity of neutrons emitted is actually very bigger than (2*10^-2 neutrons/cm^2*sec) * (area of sphere in cm^2/sphere) pointed by you, because there is emission of neutrons in NO radial direction. They do not hit the earth, however most of them hit the center of the sun.
regards
wlad
Joe,
Note that a quick glance at the paper Wladimir linked to suggests that they found NO neutrons coming from the sun, with enough data taken to set an upper limit of possible neutron flux at the 4×10-2 level.
But even taking that upper limit (and the other assumptions) and assuming that every single neutron is captured, it’s again a completely negligible effect. Your estimate was 10^41 individual neutrons vs. 10^33 grams of hydrogen. One gram is about 10^24 protons, making 10^58 protons total and giving one neutron for every million-billion protons. So again, barring a maths errors, a totally negligible amount.
Wladimir,
1. The flux toward Earth is 2*10^-2 neutrons/cm^2*sec. If we assume that this flux is the same in every direction from the Sun at the same distance that Earth is from the Sun, the number of neutrons per SPHERE per second should be the following:
(2*10^-2 neutrons/cm^2*sec) * (area of sphere in cm^2/sphere)
This number is much bigger than your present number and actually goes toward strengthening your hypothesis since you will need as big of a number as possible to show that all the 1H1 in the Sun should have been converted to 1H2.
2. You then stated that this flux was directed toward Earth alone:
“As a flux with 2s10^6 neutrons/cm².sec is emitted toward the earth direction”
This is illogical.
3. How do you know that the emission of neutrons is uniform throughout the surface of the Sun? Other matter seems to be mostly ejected by way of solar winds that usually are directed toward nearby bodies (planets, comets, etc). Therefore, the proposition that neutrons may be entering the Sun seems dubious.
4. If neutrons were entering the Sun with the same flux as emitted, the total number entering would be 10^41 in the span of 1 billion years. (Use the formula from (1) above.) These would have to combine with the 10^33 grams of 1H1 in the Sun to produce 1H2.
5. 1H2 is very sensitive to gamma photons. Even if neutrons travel further than photons, neutrons would undergo elastic scattering with the 1H1 in the Sun. This would slow them down and they would finally be absorbed by the 1H1 to form 1H2. But these might release gamma photons in the vicinity of other newly formed 1H2 in the Sun and have those 1H2 dissociate as a result. There would be no need to have photons travel through dense matter (which they can not do anyway) to destroy 1H2; this would occur locally instead.
6. There are two possibilities for the dominance of 1H1 in the Sun:
i) Not enough free neutrons to create 1H2 and subsequently 2He4 (see (4) above).
ii) Enough free neutrons to create 1H2 but quickly dissociating due to resulting gamma photons (see (5) above). A small portion of 1H2 would happen to combine quickly enough to form 2He4 and avoid the destructive consequence of being hit by gamma photons.
All the best,
Joe
Sterling Allan:
Thanks for the updating,
Warm Regards,
A.R.
I finally got another digest compiled.
LENR-to-Market Digest — January 22, 2015 – Highlights include: McKubre reports on a variation of Rossi’s 3rd party test in Lugano by Russian senior scientist, Alexander Parkhomov; info on pre-ordering E-Cats; 1 MW plant test updates; preparing for mass production; Brillouin’s travels and progress; MFMP “dog bone core test” progress. (PESN; January 22, 2015)
Joe wrote in January 22nd, 2015 at 2:26 AM
Wladimir,
1. —————————————
You write,
“Therefore the total flux of neutrons emitted by the sun, and leaving out the sun is:
2×10^-2 x 10^8 = 2×10^6 neutrons/cm².sec”
followed by
“As a flux with 2s10^6 neutrons/cm².sec is emitted toward the earth direction”.
Why do you say that the Sun’s TOTAL flux of neutrons is directed toward the Earth ONLY?
————————————————————-
I did not say that
the flux measured by experiments, hiting the earth, is 2×10^-2
The flux of neutrons emitted by the sun is spread by a surface A
The surface of the earth is S.
Then the total flux emitted by the sun is 2×10^-2 x A/S
2. ————————————————-
You write,
“then also a flux with the same intensity is emitted by the sun going toward the center of the sun”
How do you know this?
—————————————————–
because the sun will not say: “I prefer emit neutrons along one direction only”
3. ————————————————–
You write,
“and they [neutrons] hit the hydrogen of the sun, and so all the hydrogen of the sun would have to be converted to deuterium, along billion years”.
i) Have you calculated the ratio of neutrons to 1H1 in the Sun?
————————————————————
There is no need. We can use the calculation made by Dr. JR, based on the Avogadro number
ii) ————————————————
1H2 is very sensitive to photons. How does it survive in the Sun’s environment?
—————————————————-
photons do not cross matter. Have you ever seen a light crossing matter?
But neutrons can do it.
iii) —————————————————–
1H2 has a strong tendency to couple up and become 2He4. Since 2He4 is second only to 1H1 in abundance in the Sun, what makes you think that 1H2 exists at all in the Sun?
———————————————————–
Ok, then let us change the point, as follows:
all the hydrogen of the stars would have to be converted to 2He4, and so the hydrogen could not exist in the universe.
4. —————————————————
You write,
“As the earth and the planets were formed by matter coming from the sun, the water in the earth would have to be formed by D2O”
i) Standard theory describes planets as being formed simultaneously with their star in a disk of dust, and not by matter coming from the star.
———————————————————-
then ask to the authors of the standar theory to explain how heavy elements like uranium were formed from the dust
ii)—————————————————
Standard theory says water was bought to Earth by meteorites.
——————————————————-
And who did put water in the meteorites?
The god Neptune or Poseidon?
.
regards
wlad
Wladimir,
1. You write,
“Therefore the total flux of neutrons emitted by the sun, and leaving out the sun is:
2×10^-2 x 10^8 = 2×10^6 neutrons/cm².sec”
followed by
“As a flux with 2s10^6 neutrons/cm².sec is emitted toward the earth direction”.
Why do you say that the Sun’s TOTAL flux of neutrons is directed toward the Earth ONLY?
2. You write,
“then also a flux with the same intensity is emitted by the sun going toward the center of the sun”
How do you know this?
3. You write,
“and they [neutrons] hit the hydrogen of the sun, and so all the hydrogen of the sun would have to be converted to deuterium, along billion years”.
i) Have you calculated the ratio of neutrons to 1H1 in the Sun?
ii) 1H2 is very sensitive to photons. How does it survive in the Sun’s environment?
iii) 1H2 has a strong tendency to couple up and become 2He4. Since 2He4 is second only to 1H1 in abundance in the Sun, what makes you think that 1H2 exists at all in the Sun?
4. You write,
“As the earth and the planets were formed by matter coming from the sun, the water in the earth would have to be formed by D2O”
i) Standard theory describes planets as being formed simultaneously with their star in a disk of dust, and not by matter coming from the star.
ii) Standard theory says water was bought to Earth by meteorites.
All the best,
Joe
JR wrote in January 15th, 2015 at 8:12 AM
Joe,
1) ————————————————-
You’re right to ask about the details of the neutrons, as the details of where they are formed and how likely they are to form deuterons is important. But even if we ignore that and assume that all 5×10^10 neutrons/second are captured, that’s a tiny number in this context.
At about 3×10^7 seconds per year, you have roughly 10^18 captured per year, and 10^27 in a billion years. With two hydrogen per water molecule and 6×10^23 molecules per mole, you need 10^24 to convert one mole of water (which is about 18 grams). So this gives 1000 moles, about 18kg, of water being converted to D20 over a billion years. Not even a drop in the bucket.
—————————————————————
Dear JR
I made a mistake here:
Instead of 13.000km = 13.000.000m = 13×10^7cm
the correct is:
13.000km = 13.000.000m = 13×10^8cm
Then the total of deuterium formed is actually 1800kg over a billion hear.
But you are right.
Not even a drop in the bucket
However,
it only means that the background of neutrons produced in the earth is not enough to convert all the hydrogen existing the earth to deuterium
The sun is formed basically by hydrogen:
http://zebu.uoregon.edu/~soper/Sun/fusionsteps.html
But energetic neutrons are produced in the solar atmosphere by solar cosmic rays, and the emission of solar neutrons at the earth is 2×10^-2neutrons/cm².sec
http://adsabs.harvard.edu/full/1969SoPh….6..339F
The distance sun-earth is 150×10^6 km
So the area of the sphere which receives the flux of neutrons is:
A = 3,14 x (2 x 1,5×10^8)² = 3×10^17km²
The area of the disk due to the earth diameter d= 13×10^3 km is:
S = 3,14 x (13×10^3)² = 5×10~8 km²
The ratio between the areas A and S is:
A/S = 3×10^17/5×10^8 = 10^8
Therefore the total flux of neutrons emitted by the sun, and leaving out the sun is:
2×10^-2 x 10^8 = 2×10^6 neutrons/cm².sec
Of course neutrons with low energy are also formed, by as they have low velocity, and they decay in 15 minutes, they do not arrive to the earth.
As a flus with 2s10^6 neutrons/cm².sec is emitted toward the earth direction, then also a flux with the same intensity is emitted by the sun going toward the center of the sun.
Therefore a very high flux of neutrons is emitted toward the center of the sun, and they hit the hydrogen of the sun, and so all the hydrogen of the sun would have to be converted to deuterium, along billion years.
As the earth and the planets were formed by matter comming from the sun, the water in the earth would have to be formed by D2O
regards
wlad
Orsobubu:
Thank you, interesting,
Permanent Regards,
A.R.
An (old) nice article, never linked here before:
http://climate.nasa.gov/news/864/
BroKeeper:
We are working on both the issues you cited.
Real data will be supplied at the end of the R&D and test period.
So far I can’t say anything consistent.
Warm Regards,
A.R.
Dear Dr. Rossi,
Could you give us any insight whether the team has made any further strides increasing self-sustain-mode time? If so, could you predict a commercial industrial and domestic E-Cat surpassing a COP of 10?
Also, could you predict maintenance frequency near a year versus the previous six month periods?
With much respect,
BroKeeper
Georgehants:
Thank you, but what is important now is to work for the present and the future. History is a consequence that usually is written by the winners, in many cases lack of respect for deads ( as said Sitting Bull). I am not very much interested to it. What counts is to make working plants, the rest is not my problem.
Warm Regards,
A.R.
Gian Luca:
We are studying the issue. Probably we will make agreements with well consolidated existing organizations.
Warm Regards,
A.R.
Dear A.R.
you have already thought about how to manage service to customers for optimal distribution of domestic ECAT? This important part of management will be a source of exceptional work. Will be managed by your dealer?
greetings
Dear Mr. Rossi, now that Cold Fusion is moving very fast,who do you think will go down in history for being the first to use a Cold Fusion device, to supply clean water for those millions suffering and dying in this World.
Best wishes.
Frank Acland:
Yes, the pre-orders will be handled as follows, if and when the domestic units will be put on the market: all the persons that have sent a pre-order will receive an offer: if they will confirm the order along the offer they will receive the E-Cat. Priority will not be a big issue, because when we will decide to put the domestic E-Cat in the market it will be produced in big numbers, so that it is likely that the delivery term will not be a big issue.
Warm Regards,
A.R.
Dear Andrea,
It’s good news to hear that you are focusing on the domestic E-Cats — I think many people are excited about the prospect of this product.
In 2011 you allowed people to pre-order these units, and many signed up. Will these people have priority to purchase the first units that are available?
Many thanks,
Frank Acland
Andreas Moraitis:
Correct. Sorry for my typo !
Thank you for the additional two links.
Warm Regards,
A.R.
Dear Andrea Rossi,
This might be the site that you meant: http://www.nndc.bnl.gov/
Below are two links to other useful websites. It needs only a few clicks to obtain the isotopic data. The first site displays the data for all isotopes of an element on a single page. The second site provides additional information, for example on possible decay chains:
http://education.jlab.org/itselemental/
http://periodictable.com/
Best regards,
Andreas Moraitis
Curiosone:
To find all the information about isotopes I use
http://www.nndc.bnl.gov/
It is based upon the Segre chart. Very useful and easy to consult.
Warm Regards,
A.R.
Enrico Billi:
What do I expect?
1- successful (I hope) completion of the R&D and test of the 1 MW plant in operation in the factory of the Customer of IH
2- completion of the R&D for the domestic unit
3- lavolale, lavolale, lavolale, lavolale, lavolale, lavolale…
Warm Regards,
A.R.
Curiosone:
As soon as possible we will publish photographies of the plant in the factory of the Customer, provided we will get the necessary authorization.
Warm Regards,
A.R.
JC Renoir:
As I said in past, the fact that these giants have approached LENR field without bias is an important achievement of our work, in the interest of all.
Warm Regards,
A.R.
D. Travchenko:
Thank you for the information, obviously I am honoured of this.
Warm Regards,
A.R.
D. Travchenko:
No, I do not know, but I am glad of that. I repeat, though, that, since I cannot know the particulars, I cannot comment these tests.
Warm Regards,
A.R.
Dear Andrea,
wish you a great 2015. What news should we expect for the new year?
Best regards and lavoLaLe lavoLaLe
Enrico Billi
blog: billienrico.wordpress.com
Dr Rossi:
Probably you know that many labs in Russia are replicating your effect
DT
Dr Rossi:
The MIT Group is talking also of your experiments.
Warm Regards,
DT
Dear Andrea Rossi:
Lockeed Martin, NASA, Areva, MIT, Bill Gates, Shell…what do you think about the fact that you have raised interest in LENR in such companies?
JCRenoir
When will we be able to see photographies of the 1 MW plant in the factory of the customer ?
W.G.
Can you explain where we can find all the characteristics of the existing atomic isotopes ?
W.G.
Dave Lafleur:
Thank you,
Warm Regards,
A.R.
Thank you for your reply. My dad used to burn old railroad ties in Chicago for heat.
Good luck to all you innovators.