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191) Polyneutron model of John Fisher
Ludwik Kowalski (November 25, 2004)
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043
During my visit to Minneapolis (see the unit #188) I asked Dr. Oriani about his motivation for placing CR-39 detectors into the vapor. Most cold fusion researchers think that the unexplained nuclear phenomena (inappropriately referred to as cold fusion) occur in metals. His answer surprised me. He said I was guided by a theory of John Fisher. Up to that time I thought that not a single cold fusion phenomenon was predicted theoretically, all of them were discovered experimentally first. Who is Dr. Fisher? He was trained as a mechanical engineer, and as a mathematician. His early professional life was spent in a General Electric research center at Schenectady, NY. A photo of that retired theoretical scientist is shown in the unit #188.
John conceived polyneutrons, more than ten years ago, as particles responsible for all cold fusion phenomena. Do these theoretical particles really exist? That is an open question. This situation reminded me the history of neutrinos that were also invented theoretically by Pauli. I talked with John briefly during the tenth cold fusion conference and much longer during the last conference in Marseilles. In a private conversation he told me that many cold fusion people do not take his theory seriously. Their attitude toward me is similar to the attitude of the official scientific establishment toward all of us. That was another revelation; up to that point I was under the impression that cold fusion people do not criticize each other. I thought that they probably support each other in fighting with the common enemy -- the official scientific establishment. That establishment consists of managers of government laboratories, directors of financial support agencies, editors of journals and numerous self-appointed guardians of our current paradigm.
2) Neutrium, the first element:
The first assumption of Johns theory is the existence, in our environment, of a not-yet-recognized element number zero. Atoms of that element contain neither electrons nor protons; they are aggregates of neutrons bound by strong nuclear forces. He visualizes them as electrically neutral droplets of condensed nuclear matter. They are neutral atomic nuclei of an element that I would like to be named. All known elements have names and symbols. Symbols are needed to write nuclear reactions. Symbols of elements consist of one or two letters; the first letter is always capital, as in Zn, O, or Fe.
In this essay I will refer to the element # zero as Neutrium and I will use the symbol Nt to represent it. The name neutrium was suggested to me by John Fisher. He refers to isotopes of that element, such as 4Nt, 5Nt, 6Nt, 7Nt, etc., as polyneutrons. The symbol ANt is used for a polyneutron containing A neutrons. Unlike ordinary elements, neutrium was not discovered experimentally, it was invented by John to explain anomalies of the so-called cold fusion. This reminds us of a neutrino, a particle invented in 1930s to make sense of the apparent anomaly in beta decay.
John Fisher thinks that polyneutrons can contain more than 1000 neutrons kept together by strong nuclear forces. A neutron star, on the other hand, is not a gigantic polyneutron because its neutrons are kept together by gravitational forces. John is aware that polyneutrons of size two or three can not exist. He believes, on the basis of reference (1), that the smallest possible polyneutron consists of four neutrons. In trying to explain stability of nuclear droplets made from neutrons, and knowing that two neutrons do not stick, John brings an analogy with liquid helium. Two or three atoms of helium do not make molecules but liquid helium exists. I would like to know what other people, more knowledgeable that I am, think about that analogy.
In answering one of my questions John Fisher said that polyneutrons arise in nature as decay products of exceedingly rare precoursor nuclei. This, however, does not necessarily mean that one has to cover a lot of territory to encounter one or two of them. Suppose there is only one precursor nucleus per quintillion (1018) molecules of air. That would certainly make them exceedingly rare. But, even at such level of concentration, every cubic inch of air would still contain nearly 1000 of them to produce polyneutrons when they decay. The main point here is that atoms of neutrium are usually available to produce various effects, provided favorable conditions are encountered. What is meant by favorable conditions is explained below.
3) Unusual nuclear reactions:
Unlike electrically charged projectiles, polyneutrons are not repelled by ordinary atomic nuclei; nuclear reactions induced by them are not inhibited by repulsive coulomb forces. All cold fusion abnormalities, according to Fisher, are due to nuclear reactions that involve polyneutrons.. Another alternative is to think that cold fusion phenomena are due to ordinary neutrons. John rejects this alternative because far too few neutrons are seen in cold fusion experiments. He also rejects models postulating that coulomb barriers can somehow be lowered. In other words, Fisher thinks that the term cold fusion, as defined in 1989, is totally inappropriate. Taking coulomb barrier arguments very seriously he does not believe that two deuterium nuclei, for example, can fuse at ordinary temperatures.
A nuclear reaction can be either exothermic or endothermic, depending on its Q value. This topic is reviewed in the Appendix 1. Here is an example of a nuclear reaction involving two polyneutrons:
ANt + 18O --> 16O + (A+2)Nt . . . . . (1)
The letter A is used to indicate the atomic mass number (number of neutrons) of the polyneutron on the left side of the equation. The polyneutron on the right side contains two more neutrons than the polyneutron on the left side. This particular reaction turns 18O into a common isotope of oxygen, 16O. Another possible polyneutron reaction, according to Fisher, is:
ANt + 2H + --> 3H + (A-1)Nt . . . . . . (2)
Here the mass of the polyneutron decreases by one unit while the mass of the hydrogen increases. One can imagine many nuclear reactions of that kind. But not all of them can occur spontaneously. Most are possible only with an accelerator. Whether or not a reaction can take place spontaneously, at an ordinary temperature, depends on its Q value. The well known way of calculating Q values of nuclear reactions is reviewed in the Appendix 1. Only exothermic reactions, those whose Q values are positive, can happen spontaneously. That is why the ability to calculate Q is so important. To calculate the Q value of a reaction involving isotopes of neutrium, such as:
ANt + 18O --> 16O + BNt + CNt . . . . . . (3)
one must know exact masses, or mass excesses, of all participants. The concept of the mass excess is reviewed in the Appendix 1. Note that in this reaction 18O is transmuted to 16O and a polyneutron of size A is split into two smaller polyneutrons. Naturally, the sum B+C must be equal to A+2. Mass excesses of ordinary nuclei are well known; they can be found in many nuclear physics textbooks (3). But how can the mass excess of a polyneutron, with atomic number A be determined? According to Fisher that quantity, , the mass excess Æ(ANt), is given by:
Æ(ANt) = k*A
where A is the number of neutrons in the polyneutron and k is an unknown constant. This formula, based on the liquid drop model, is only an approximation. The term associated with the surface energy is ignored; only the volume energy term is retained. To impose a limit on the numerical value of k John notes that from the beginning of his theory he had the idea that chain reactions were involved and that atoms such as 7Li, 18O, and 2H might serve as fuels. When Oriani and I later observed a shower of about 150,000 alpha particles in the oxygen/hydrogen vapor over an active electrolysis cell, I assumed that oxygen could support a chain reaction, which requires that the Q value of reaction (1) must be positive. The initial statement that the Q of the reaction (1) is positive is essential in Fishers theory. Should we say that this statement is based on experimental facts or should we say that it is a very clever arbitrary, assumption? I want to know what a philosopher of science would say about this.
If we accept that the Q of the reaction (1) is larger than zero then, as show in the Appendix 2, k must be smaller than 1.98 MeV/u. Note that the unit of Æ(ANt) is MeV while the unit of A is u (atomic mass unit). That is why, the unit of k is MeV/u.
The other limit imposed on k i based on another experimental observation. Spontaneous generation of tritium from deuterium has been reported by cold fusion researches. Assuming that this transmutation is due to the reaction (2), Fisher states that the Q of this reaction must also be positive. This, in turn, as explained in the Appendix 2, means that k must be larger than 1.82 MeV/u. These two limits narrow the range of values of k to 1.82 < k < 1.98. The middle of that narrow region is 1.90 MeV/u. On that basis John assumes that k is equal to 1.90, plus or minus about 4.5%. By knowing k he is able to calculate Q values of many reactions involving polyneutrons, as illustrated in the Appendix 2.
4) The chain reactions process:
The 18O isotope exists in air, water and mineral surrounding us; about 0.2% of all oxygen on earth is 18O. Reaction (1) is a growth reaction. Colliding with successively encountered 18O nuclei a small polyneutron can grow and grow. Reaction (3) is a splitting reaction. Here a single polyneutron disappears and two smaller polyneutrons appear. The combination of reaction (1) and reaction (3) amounts to a chain process. Under favorable conditions (to be explained later) a polyneutron grows through frequent encounters with 18O. When it become large enough (typically when A becomes several hundred) the reaction (3) occurs and the large polyneutron splits into two smaller polyneutrons.
These two also start growing through the reaction (1) till they split through the reaction (3). After that we have four polyneutrons. Subsequent repetitions of reactions (1) and (3) bring the number of neutrons to 8, 16, 32, etc. Both reactions (growing and splitting) are exothermic and nuclear energy is released in this chain process. John compared it to a chain reaction in a typical nuclear power plant. The 18O, according to Fisher, is not the only fuel able to support a chain process. In a recent message he illustrated this with an interesting example:
ANt + 7Li --> + 4He + 1H + (A+2)Nt . . . . . . (1)
where a stable isotope 7Li, interacting with ANt, brakes into 4He and 1H. This is accompanied by the growth of the polyneutron, as in reaction (1). The Q value of this exothermic reaction is 1.39 MeV. A sequence of many (1) reactions, followed by the reaction (3), is a single step of a chain process. The fuel in this energy generating process is 7Li. Is it conceivable that accumulation of helium, reported by several cold fusion researchers is due to the reaction (1)? Answering this question John wrote: Yes some helium comes from the 7Li reaction. But some comes too from the decay of poisons and some from the decay of polyneutrons. So depending on circumstances the ratio of helium to energy can vary quite a lot from experiment to experiment. The meaning of the term poison, in this context, is explained in the next section.
5) Poisoning processes:
Suppose that the above described chain process develops spontaneously somewhere in the atmosphere, or in the ocean. Then we would have a tremendous nuclear explosion on a planetary scale. Fortunately for us, this can not happen. Why not? For the same reason for which a chain reaction in a power plant can not go on indefinitely, because the chain generates reaction products that act as poisons. In the context of a power plant poisons are fission products; they trap neutrons that would otherwise be available for the chain reaction. In the so-called cold fusion context poisons are structures composed of a polyneutron and a 16O nucleus that are stuck together like tiny drops of oil and water, each retaining its identity, but held together by a reduction of surface energy over the area of contact. Poisons can form in reactions such as:
ANt + 7Li --> (A+2)Nt + 5Li . . . . . . (8)
5Li --> 4He + 1H . . . . . . (8b)
g) Nuclear transmutations are, according to Fishers theory, produced through collisions of polyneutrons with ordinary nuclei. The above sequence transmutes lithium into helium and reaction (1) transmutes 18O into 16O. Changes in isotopic compositions of numerous elements were reported by many experimentalists.
I suppose that an appropriate exothermic nuclear reaction, or a sequence of several such reactions, can be found to explain appearance of particular products. Polyneutrons are not repelled by ordinary nuclei and reactions must often be characterized by large cross sections.
h) How would an alchemist, familiar with properties of neutrium, plan to make gold, 197Au, from a common isotope of mercury, 202Hg? The first step would be to realize that the mercury isotope, 197Hg, turns into gold through the well known process of electron capture. This has nothing to do with polyneutrons. The challenging part would be to transmute 202Hg into 197Hg. That is where the use of polyneutrons would probably be suggested, according to the following reaction:
ANt + 202Hg --> (A-5)Nt + 197Hg . . . . . .(9)
Once produced, the 197Hg will certainly turn to gold. But is the reaction (9) possible? The answer depends on its Q value. According to (2) the mass excesses of the 197Hg and 202Hg are: -30.436 MeV and -27.356 MeV, respectively. Using these numbers one finds that the Q value of (9) is +12.6 MeV. The positive sign indicates that the reaction is energetically possible. A conventional nuclear scientist, who does not believe in polyneutrons, would probably suggest to produce 197Hg from 196Hg with ordinary neutrons. This way of manufacturing gold from mercury is indisputably feasible but too expensive. It is based on conventional nuclear science.
10) Post scriptum:
Fishers presentation at the conference in Marseilles (5) had no equations. Puzzled by its content I asked John for a private lecture. He agreed. That recorded lecture was attended by three other conference participants: Russ George, William Collis (who said that he learned about Fishers polyneutron model ten years ago) and by Bielobrzeckaja-Costalariza-Nikolajeva. Most of what I wrote here is my understanding of that lecture, and of the references (4,5). I am a retired teacher who likes pedagogical challenges. Learning the model and describing it was a real challenge. The content of this essay will be modified to reflect evolution of my understanding.
Energy in nuclear reactions.
All nuclear transformations, both reactions and radioactive decays, have an energy aspect. A nuclear reaction, in general, can be described symbolically as:
L + K --> R + S
where L and K are isotopes called reactants while R and S are isotopes called products. The number of participants does not have to be limited to two on each side. A transformation with only one term on the left side is called radioactive decay. Similarly, a process with two terms on the left side but only one term on the right side is called fusion. Electrons and photons are not reaction participants in this simplified description. The energy aspect of a nuclear transformation is characterized by the so-called Q value of the process. By definition, the Q value is the difference between the sum of the masses on the left side and sum of the masses on the right side, multiplied by the square of the speed of light, c.
Q = [(sum(L+K) - sum(R+S)]*c2
A nuclear process is said to be exothermic (releasing energy) when Q is positive; it is said to be endothermic (using energy) when Q is negative. Exact masses of known nuclear isotopes have been determined with great accuracy; their values can be found in most nuclear science textbooks (2, 3). Note that Q is positive when the sum of the masses on the left is larger than the sum of the masses on the right. One way to describe this, no longer popular, is to say that Q is positive when the mass is reduced in a process and Q is negative when the mass is increased in it. The term inside the square brackets, often represented by Dm, is called the reaction mass difference. Using this term the equation (10) can be written Q=Dm*c2. Note that Q is in joules when m is in kilograms and c is in meters per second. In nuclear physics, however, Q is usually expressed in MeV and Dm in so called atomic mass units, u. Using the well known conversion factors associated with these practical units one finds that the value c2 must be equal to 931.48 MeV/u. Thus Q=9.31 MeV when Dm is 0.01 u, and vice versa. Exact masses in reference (2), given in u, can also be expressed in the units of energy. The mass of 58Ni, for example, 57.935346 u, can be expressed as 53965.616 MeV.
Reference (3), on the other hand, does not give exact masses for each isotope. What it gives instead is called the mass excess. By definition, the mass excess for an isotope (not to be confused with Dm of a reaction) is simply a difference between the exact mass of that isotope, M, and the nearest integer I. It is usually expressed in the energy units, MeV. The mass excess for 58Ni, for example, is -60.224 MeV (3). That value can be used to calculate the exact mass, M, if needed. To illustrate this let me write:
(M - I)*931.48 = -60.224
where I is 58. This gives; M=57.93534 u found in (2). It turns out that Q values of nuclear transformations can be calculated from the defining equation (10) by using mass excesses instead of exact masses. To illustrate this consider the following reaction:
1n + 27Al --> 4He + 24Na
where 1n represents a neutron. The excess masses on the left, according to (3), are +8.071 MeV for 1n and -17.194 MeV
for 27Al. Likewise, excess masses on the right are: +2.425 and -8.418 MeV, respectively. This gives -9.123 MeV for the sum of the left
and -5.993 MeV for the sum on the right. The Q value of the reaction, is the difference between the sum on the left and the sum on the right; in
this case it is equal to -3.13 MeV. Negative Q indicates that the reaction is endothermic. That reaction becomes possible only if at least 3.13 MeV
of energy is delivered, for example, in the form of kinetic energy of the neutron.
Excess masses of polyneutrons etc.
The value of k:
The mass excess of a polyneutron composed of A neutrons, as explained in the main part of this essay, is Æ(ANt) = k*A. The purpose of this appendix is to show how the Q values are calculated for the reactions involving polyneutrons. Let us see how the numerical value of k 1.90 MeV/u) was established by John Fisher. If the reaction (1) is exothermic, as assumed by Fisher, then its Q must be larger than zero. This implies that the sum of the excess masses on the left side of the equation (1) must be larger than the sum of the excess masses on the right side. Or, symbolically,
[D of 18O] + [D of ANt] > [D of 16O] + [D of (A+2)Nt] . . . . . . (10)
The excess masses of 18O and 16O are known (3); they are -0.783 and -4.737 MeV, respectively. Furthermore, [D of (A+2)Nt] is only 2*k larger than [D of ANt] Therefore:
-0.783 + k*A > -4.737 + (k*A +2*k)
This reduces to the inequality k < 1.977. The upper limit of k is established by John on the basis of reaction (2). Assuming that the Q value of that reaction is positive (otherwise the reaction would not occur spontaneously) one has:
[D of 2H] + [D of ANt] > [D of 3H] + [D of (A-1)Nt] . . . . . . (11)
The excess masses of 2H and 3H are known (3); they are 13.13 and 14.95 MeV, respectively. Therefore,
13.13 + k*A > 14.95 + k*(A-1)
The k*A cancels, because it appears on both sides of the inequality, and one has; k<1.82 MeV/u. In other words, positive Q values of the reactions (1) and (2) implies that k must be confined to a narrow range of values:
1.82 < k < 1.98
The value used by John, 1.90, is in the middle of that range. Knowing k we can calculate the Q value of any nuclear reaction involving polyneutrons. This in turn, allows us to predict which reactions can occur spontaneously and which can not.
Only reactions whose Q values are positive can take place spontaneously. Reliability of such predictions, however, are only as good as the reliability of Q values. That is where a difference between ordinary nuclear reactions and reactions involving polyneutrons become obvious. Exact masses of ordinary nuclei have been measured before the theoretical model (for calculating them) was introduced in 1930s. The model has several numerical coefficients; their values were chosen to match the experimentally measured masses. Polyneutrons, on the other hand, are hypothetical particles whose exact masses are unknown outside the theoretical model. What would the range of k be if another pair of hypothetical reactions, instead of (1) and (2), were chosen? How to decide that one choice of possible defining reactions is better than another? I suppose that such questions are premature at this preliminary stage of development. It is a miracle that a pair of reactions narrowing the range of possible values of k to plus or minus 0.4% was found by John.
Fishers theory, described in the main text of this essay, introduces molecule-like structures called composites. The 16OBNt, appearing in the reactions (4) and (5), is a typical example. The excess mass of such loosely bound structure is assumed to be the sum of the excess masses of the ordinary nucleus, such as 16O, and of the excess mass of the polyneutron, such as BNt. What is the Q value of the reaction (3)? The first step, as always, is to write down the defining equation:
Q = sum (of all D on the left side) - sum (of all D on the right side)
or, more specifically,
Q = [ D of ANt + D of 18O] - [D of 16O + D of BNt + D of CNt]
Note that to balance the numbers of nucleons the (A+18), on the left, must be equal to (16+B+C) on the right. In other words, A+2 must be equal to B+C. The excess masses of 18O and 16O, according to the reference (3) are: -0.783 MeV and -4.737 MeV, respectively. This gives
Q = [k*A + (-0.783)] - [(-4.737) - k*B - k*C]
Q = 3.254 +k*(A -B-C) = 3.254 + 2*k = 7.054 MeV
The reaction is exothermic. In the same way one can show that the Q value of the reaction (4) is equal to 4.454 MeV. Answering one of my e-mail questions John wrote:
The excess masses you are using for 1Nt, 4He, and 16O are correct. They are respectively +8.071 MeV, +2.425 MeV and -4.737 MeV. The mass excess of 18O is -0.782 MeV. My best estimate of the mass excess of polyneutron with A neutrons is 1.90*A MeV. This makes the mass excess of 16O stuck to a polyneutron with 100 neutrons, 16O100Nt, equal to -4.737 + 190 -b where b is the binding energy of the oxygen to the polyneutron drop. I don't know what b is, but I assume it is small. In any event it does not matter for most reactions. Consider for example the polyneutron growth reaction (1):
100Nt + 18O --> 102Nt + 16O
wherein the polyneutron grows by two neutrons. The energy generated in this reaction is:
(190 - 0.78 -b) - (193.8 -4.74 -b) = -3.8 -0.782 +4.737 = 0.16 MeV.
so the polyneutron growth reaction is exothermic and 18O is capable of supporting a chain reaction. Note that the b's cancel out. Now consider the decay of the composite, ANt 16O, where a polyneutron and 16O are stuck together. The composite can decay by simultaneous double beta decay and alpha decay in a reaction with the overall formula:
ANt16O --> (A-4)Nt16O + 4He
Here the energy generated is:
Q = [1.90*A - 4.74 - b] - [1.90*(A - 4) - 4.74 - b] - 2.42 = 5.18 MeV.
Keep your questions coming. I will answer them all if I can, on the first try if possible.
Other chain processes?.
1) Tables of excess masses are often shown in nuclear physics and nuclear chemistry textbooks. Here are some of the excess masses from the reference (3). They can be used to calculate Q values of reactions mentioned in my essay.
Atom Æ( of atom) in MeV
1Nt (neutron) 8.071
5Li 11.680 (According to J. Fisher; 5Li is not listed in reference (3))
160 - 4.737
18O - 0.783
27Al - 17.194
58Ni - 60.224
93 Kr - 65.6
133 Cs - 88.089
142 Ba - 77.82
197Hg - 30.735
202Hg - 27.356
ANt (large A) 1.90*A
To practice with calculations of Q values the reader is invited to verify that the two reactions shown below are endothermic (negative Q).
ANt + 7Li --> 5Li + (A+2)Nt . . . . . . (12)
ANt + 7Li --> 6Li + (A+1)Nt . . . . . . (13)
Show that their Q values are -1.80 MeV and -0.57 MeV, respectively. The reaction (14), however, is exothermic; its Q value is 9.32 MeV. That indicates that polyneutrons can grow by interacting with deuterium, for example, in heavy water.
ANt + 2H --> (A+2)Nt . . . . . (14)
ANt + 2H --> BNt + CNt . . . . . .(15)
In other words, heavy hydrogen (2H), like heavy oxygen (18O), can be a fuel in the chain process. A sequence of reactions (14) followed by the reaction (15) constitutes a single step in the energy generating chain process. This is similar to the process based on the reactions (1) and (3).
2) Suppose an electrolytic cell is used to create favorable condition for the chain process based on either reactions (1) and (3). or reactions (14) and (15). Excess heat is produced coninuously because the exothermic chain reaction is going on. Fresh fuel, 18O or 2H, is supplied and poisons are removed, as envisioned by Fisher. Suddenly one gram of 238U is introduced into the electrolyte, for example, in the form the powdered uranium nitrate salt. I am speculating about production of 235U from 238U via the following reaction:
238U + 100Nt --> 235U + 103Nt
Is this possible? No, it is not possible, because Q is equal to -8.95 MeV. The reaction is endothermic; it will not occur spontaneously, fortunately for our safety.
3) Then I envision a well known precess -- spontaneous fission of 238U according to the following reaction:
238U --> 142Ba + 93Kr + 3*(1n) . . . . . . (16)
Uranium decays by emitting two fission fragmens (barium and krypton) and three free neutrons. The above fission products are known to be beta radioactive. Using the excess masses of participants one finds that Q=166 MeV. The process is strongly exothermic but the probabilty of its occurence is extremely low (due to the fission barrier of about 6 MeV). But can this process be speeded up by polyneutrons? I am imagining, for example:
100Nt + 238U --> 142Ba + 93Kr + 103Nt . . . . . . (17)
The Q value of that reacion is 190.5 MeV. Conceptually it is similar to the reaction (1) where 7Li was split into two fragments (4He and 1H) by interacting with a polyneutron. In both cases the polyneutron grows and energy is released. The released energy in the reaction (17), however, is much larger than in the reaction (1). The reaction (17) implies that the fission barrier of the 238U is somehow overcome in the presence of the 100Nt. Note than the polyneutron of mass 103, escaping from the reaction, might interact with another 238U, etc. Is it conceivable that one can have a fast chain reaction in which 238U is generating excess heat? I do not know how to answer this question.
A revised version of the polynutrons theory, as presented iby John Fisher in Siena (May 2005), is described in a paper that can be downloaded from his website:
My comments about the new version can be seen in unit #226.
Inserted on on 7/14/2009:
A recent qoute of what John Fisher wrote can be seen at the beginning of Unit 364. And here is another message that John posted on CMNS list: Responding to a question about polyneutrons, he wrote: ÒI have returned from vacation after several weeks and now can answer your question regarding the sticking force between neutrons in a polyneutron.
First note that two neutrons are almost bound into a di-neutron. They form a resonance with energy about 0.12 MeV over a potential well about 35 MeV deep. (For references see Treatise on Heavy-Ion Science, Vol. 8, Nuclei Far From Stability, Edited by D. Allan Bromley, Plenum Press, 1989, p 313 for 0.12 MeV resonance, and also Kenneth S. Krane, Introductory Nuclear Physics, John Wiley & Sons, 1988, p 82 for 35 MeV potential well.) We see that two neutrons are very strongly attracted to each other, but not quite strongly enough to be bound.
Consider now a large cluster of neutrons having density comparable to that of ordinary nuclei. Except for surface neutrons, each neutron is surrounded by 12 other neutrons with which it has nearest-neighbor interaction. This implies a 12-fold increase in the potential well that it sees and assures that it is strongly bound to the cluster. Neutral pions are the particles to which most of the attractive force can be attributed. Although the binding energy cannot be computed from first principles (it cannot be computed for the neutron-proton clusters of ordinary nuclei either) it can be approximated by a liquid drop model with a negative volumetric energy term and a positive surface energy term. The strengths of these terms can be determined from experiment (LENR experiments) just as they have been determined by ordinary experiments for ordinary nuclei.
From LENR experiments I have been able to determine an approximate value for the volumetric binding energy term, and find it to be about half of that for ordinary nuclei.Ó
1) F.M. Marques et al., Detection of neutron clusters, Physical Review C, vol
2) K.S Krane, Introductory Nuclear Physics, John Wiley & Sons, Inc., New York, 1987.
3) G. Friedlander, J.W. Kennedy, E. S. Macias and J. M. Miller, Nuclear
and Radiochemistry third edition; John Wiley & Sons, Inc., New York,
4) J.C. Fisher, Theory of low-temperature particle showers, This paper
was presented at the 10th International Conference on Cold Fusion,
August 2003. It can be downloaded from the library at <www.lenr-canr.org>.
5) J.C. Fisher, Neutron isotope reactions, This paper was presented at the
11th International Conference on Cold Fusion, November 2003. It will be
published in the proceedings of that conference.
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