Re: Cathode (Cometary) Commentary
Robin van Spaandonk wrote: In reply to Edmund Storms's message of Sun, 20 Nov 2005 11:34:48 -0700: Hi, [snip] Essentially correct, but be careful not to confuse Hy (neutral) with what I have been designating hyh (Hydrinohydride) which carries a negative charge (or Hy- if you prefer that notation). So, we have three possible combinations: 1: A proton with one electron in the normal Bohr quantum state (i.e. the usual hydrogen atom) Obviously correct. 2. A proton with one or more electrons in Mills quantum states, which would be a very stable negative Hy ion, which I would designate Hy-, Hy--, etc, depending on the number of electrons. Hy- is the only possible negative ion AFAIK. This is because the first electron (the hydrinos own electron) essentially neutralizes the protons charge, while the second electron that turns Hy into Hy- neutralizes the magnetic field of the first electron. It's as though you joined two bar magnets together with their opposing poles together, then put keepers on them. Most of the magnetic flux flows through the keepers directly to the other magnet completing the circle. In short there is very little noticeable external flux available to form additional bonds. This is also the true reason why S shells of all atoms are limited to two electrons. In short even though the hydrino is shrunken, it is still a spherical shell, and hence essentially an S shell. More electrons are not possible, because both magnetic and electrostatic attraction forces have been used up, i.e. there are no more protons in the nucleus to generate electrostatic force which could hold on to additional electrons. 3. A proton with one or more electrons in Mills quantum states and zero or one electron in the usual Bohr quantum state, which would be a stable molecule. In this case the other ion would have a positive charge equal to the total number of electrons associated with the proton. If the other ion is a proton, I would designate this as HyH, which would be neutral. I believe what you are talking about here is an ionic bond between Hy- and a proton (though the wording is a little confusing). This may not exist as such, but rather as a dihydrino molecule. (Essentially a shrunken hydrogen molecule). Here is a list of possible combinations, where Hy is a proton having one electron in a Mills quantum level and H is a proton having an electron in a Bohr level. Which of these do you call a dihydrino molecule? 1. Hy + H, where one Bohr electron exchanges between Bohr levels in the Hy atom and the H atom. 2. Hy + Hy, where the Mills electrons exchange between the two atoms. No Bohr electrons are present. In my description, Hy is a proton with one Mills electron and Hy- is a proton with two Mills electrons. If the latter made a bond, it would be 3. Hy- + H+, where no Bohr electrons would be present. I suggest that #2 is not possible and #3 would produce a much weaker bond than #1. Regards, Ed A corresponding compound with a Hy-- would be HyH2, or HyNa2 or HyCa, if other elements are used. Does this describe your understanding? As mentioned above, I don't think there is any Hy ion more negative than Hy-. [snip] Never heard of it. Reference? Oriani, R.A. Anomalous Heavy Atomic Masses Produced by Electrolysis. in The Seventh International Conference on Cold Fusion. 1998. Vancouver, Canada: ENECO, Inc., Salt Lake City, UT. Fisher, J.C., Polyneutrons as agents for cold nuclear reactions. Fusion Technol., 1992. 22: p. 511. Thanks. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Mon, 21 Nov 2005 10:18:05 -0700: Hi, [snip] In this case the other ion would have a positive charge equal to the total number of electrons associated with the proton. If the other ion is a proton, I would designate this as HyH, which would be neutral. I believe what you are talking about here is an ionic bond between Hy- and a proton (though the wording is a little confusing). This may not exist as such, but rather as a dihydrino molecule. (Essentially a shrunken hydrogen molecule). Here is a list of possible combinations, where Hy is a proton having one electron in a Mills quantum level and H is a proton having an electron in a Bohr level. Which of these do you call a dihydrino molecule? 1. Hy + H, where one Bohr electron exchanges between Bohr levels in the Hy atom and the H atom. If there are Bohr levels in the Hy atom, then this may be possible. (I see no reason why there wouldn't be, however I would expect them to be modified). I'm pretty much in the dark here. Mills' past spectrum experiments may help resolve the issue. He also has calculations for the levels I think, but not sure off hand where to find them (most of his book is calculations, so it's a bit like looking for a needle in a haystack). 2. Hy + Hy, where the Mills electrons exchange between the two atoms. No Bohr electrons are present. If both Hy are the same size, then this is the dihydrino molecule. It may only exist when they *are* the same size (my guess). In my description, Hy is a proton with one Mills electron and Hy- is a proton with two Mills electrons. If the latter made a bond, it would be 3. Hy- + H+, where no Bohr electrons would be present. Apparently you meant If the latter made a bond ...with an H+. When a bare proton approaches Hy-, I think the two Mills electrons would be equally shared between both protons, resulting in a dihydrino molecule. I asked Mills about the reaction:- Hy + p - Hy2+ (positive dihydrino molecular ion) and he thought that would happen, so I assume that Hy- + p - Hy2 (dihydrino molecule) also works. (p = H+ = bare proton). Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Robin van Spaandonk wrote: In reply to Edmund Storms's message of Sat, 19 Nov 2005 15:19:06 -0700: Hi, [snip] Why? In a perfect ionic compound, solidity results from the binding energy of positive and negative ions. IOW the attractive force between ions of opposite charge pulls the ensemble together. There is no real need for electrons to be interchanged at a local level as would be the case in a covalent bond. Granted, with normal substances there is more often a polar bond than a pure ionic bond. In short, the hyh bond with a positive ion would be the most extreme ionic bond imaginable. You may calculate the degree of electron sharing if you wish, but given an ionization potential of around 70 eV for hyh[n=1/16], I think you will find that it is so negligible as to be immeasurable. A normal ionic compound results from electrons being moved from one atom to the other. For example, in making NaCl, the electron moves from the Na atom to reside for most of the time at the Cl atom. This is different from the situation with Hy, which I'm trying to understand, so be patient. When Hy is involved, the situation involves a preionized atom, so to speak, which as a negative charge that can not be removed by chemical interaction. Consequently for it to form a bond, the other atom must also be preionized to form a positive ion. Essentially correct, but be careful not to confuse Hy (neutral) with what I have been designating hyh (Hydrinohydride) which carries a negative charge (or Hy- if you prefer that notation). So, we have three possible combinations: 1: A proton with one electron in the normal Bohr quantum state (i.e. the usual hydrogen atom) 2. A proton with one or more electrons in Mills quantum states, which would be a very stable negative Hy ion, which I would designate Hy-, Hy--, etc, depending on the number of electrons. 3. A proton with one or more electrons in Mills quantum states and zero or one electron in the usual Bohr quantum state, which would be a stable molecule. In this case the other ion would have a positive charge equal to the total number of electrons associated with the proton. If the other ion is a proton, I would designate this as HyH, which would be neutral. A corresponding compound with a Hy-- would be HyH2, or HyNa2 or HyCa, if other elements are used. Does this describe your understanding? The presence of electrons in the Hy states would, I expect, alter the energy level of the Bohr quantum states. As a result, a range of properties would be expected depending on the energy level of electrons in the other atom, how many Hy electrons were present, and what quantum levels they occupied. snip That's the general concept, though it would essentially be a negatively charged neutron, effectively reducing the atomic number by 1. This is because it would orbit the nucleus inside the K shell, so from the point of view of the electrons, the nuclear charge would be reduced by 1. (Actually I'm guessing here. The size of the hydrino is still up in the air somewhat as far as I'm concerned). Besides it will depend on which Hy- combines with which positive ion. On the other hand, I would expect such a structure to be so close to being neutral that interaction with the electron quantum states would not be possible. This seems to be an idea worth exploring. Would this explain the Fisher-Oriani super heavy carbon? Never heard of it. Reference? Oriani, R.A. Anomalous Heavy Atomic Masses Produced by Electrolysis. in The Seventh International Conference on Cold Fusion. 1998. Vancouver, Canada: ENECO, Inc., Salt Lake City, UT. Fisher, J.C., Polyneutrons as agents for cold nuclear reactions. Fusion Technol., 1992. 22: p. 511. Regards, Ed snip
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Sun, 20 Nov 2005 11:34:48 -0700: Hi, [snip] Essentially correct, but be careful not to confuse Hy (neutral) with what I have been designating hyh (Hydrinohydride) which carries a negative charge (or Hy- if you prefer that notation). So, we have three possible combinations: 1: A proton with one electron in the normal Bohr quantum state (i.e. the usual hydrogen atom) Obviously correct. 2. A proton with one or more electrons in Mills quantum states, which would be a very stable negative Hy ion, which I would designate Hy-, Hy--, etc, depending on the number of electrons. Hy- is the only possible negative ion AFAIK. This is because the first electron (the hydrinos own electron) essentially neutralizes the protons charge, while the second electron that turns Hy into Hy- neutralizes the magnetic field of the first electron. It's as though you joined two bar magnets together with their opposing poles together, then put keepers on them. Most of the magnetic flux flows through the keepers directly to the other magnet completing the circle. In short there is very little noticeable external flux available to form additional bonds. This is also the true reason why S shells of all atoms are limited to two electrons. In short even though the hydrino is shrunken, it is still a spherical shell, and hence essentially an S shell. More electrons are not possible, because both magnetic and electrostatic attraction forces have been used up, i.e. there are no more protons in the nucleus to generate electrostatic force which could hold on to additional electrons. 3. A proton with one or more electrons in Mills quantum states and zero or one electron in the usual Bohr quantum state, which would be a stable molecule. In this case the other ion would have a positive charge equal to the total number of electrons associated with the proton. If the other ion is a proton, I would designate this as HyH, which would be neutral. I believe what you are talking about here is an ionic bond between Hy- and a proton (though the wording is a little confusing). This may not exist as such, but rather as a dihydrino molecule. (Essentially a shrunken hydrogen molecule). A corresponding compound with a Hy-- would be HyH2, or HyNa2 or HyCa, if other elements are used. Does this describe your understanding? As mentioned above, I don't think there is any Hy ion more negative than Hy-. [snip] Never heard of it. Reference? Oriani, R.A. Anomalous Heavy Atomic Masses Produced by Electrolysis. in The Seventh International Conference on Cold Fusion. 1998. Vancouver, Canada: ENECO, Inc., Salt Lake City, UT. Fisher, J.C., Polyneutrons as agents for cold nuclear reactions. Fusion Technol., 1992. 22: p. 511. Thanks. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Sun, 20 Nov 2005 11:34:48 -0700: Hi, [snip] Never heard of it. Reference? Oriani, R.A. Anomalous Heavy Atomic Masses Produced by Electrolysis. in The Seventh International Conference on Cold Fusion. 1998. Vancouver, Canada: ENECO, Inc., Salt Lake City, UT. Having just read this, I am somewhat bemused. The notion of carbon with hundreds of neutrons is a bit rich even for me. Note that the Pd labeled 0-113 was washed in HCl after use. I have previously suggested that Cl- is a complex former for some metals, which is why it forms part of aqua regia. The other component of aqua regia is nitric acid, which due to the presence of the N5+ ion functions as an oxidizer. Combine an oxidizer with a complex former and noble metals (e.g. Pd) relinquish their electrons and form a complex. In this case, all the ingredients are present for this process. The Cl- was present due to the HCl wash and the oxidizer was present in the form of pure oxygen, in which the metal was baked for hours. In short, I wouldn't be in the least surprised to see a Pd complex forming comprising Pd with multiple Cl- ions attached. My guess is that the mass works out just about in the neighborhood of those measured. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Jones Beene's message of Wed, 16 Nov 2005 11:37:01 -0800: Hi, [snip] BTW for those (from Oz ;-) who are sure to correct some of my past posted details (and I appreciate that), and since my original rough calculation for expected di-hydrino density was too hasty, here is something based on what we know about H2 - with the assumption that liquid Hy2 (Hy2 being the di-hydrino) is very similar to liquid H2 except in melting point. I expected the melting point for N=1/2 to be over 1000 K and the others to be solid, but I am certain that there will be a wide divergence of opinion as to those numbers and as to the phase change points. Indeed. Since the phase change point depends on van der Waals forces which are higher order derivatives of the electrostatic and magnetic forces, I'm not even going to hazard a guess. (My mental models don't handle higher orders very well ;) [snip] Since the Oort cloud will contain few cations, This isn't necessarily so. For a small hyh, just about anything else may look like a cation. (If it's got a nucleus, I'll take it! ;) [snip] Based then on extending the H2 model, liquid n=1/3 hydrinos should weigh in at 1.89 gm/cc and n=1/4 would be 4.48 and n=1/7 which Mills claims to have samples of (but in the form of ionic bound chemical hydrides) would be 24 gm/cc - about the density of uranium. Note however this important detail - that even though the density of shrunken di-hydrino molecules would be every high, the same does not apply to hydrino hydrides (ionic bonding), since the (more shrunken variety) hydrino would probably be actually located most of the time *inside* the orbital cloud (not unusual) and likely around the k-shell of the resultant hydride (unusual), which would be only slightly larger, and perhaps even less dense than before. Indeed, or as I mentioned in my other email, the resultant atom may just end up looking like an isotope of a different element. Nevertheless, I cannot imagine this material, even in the first shrinkage state, staying gaseous at STP. And I still think many hydrinos should be found in ocean minerals, particularly the alkalis, and that could be the Mills' catalyst connection. Why sodium would not be a catalyst carrier (i.e and active Mills catalyst) is not certain. Perhaps many alkalis are in that category because the nucleus is expressing a lesser positive charge (near-field than expected). But what eliminates sodium? I hate to disappoint you, but perhaps it's because Mills is right about what forms a good catalyst? (Personally, I happen to think he's been a bit too liberal in what he allows, but I think he is closer than you are). Could it be that hydrino catalysts are only catalytic for one reason - that is because they have a natural affinity for hydrino-hydriding (with the natural population of solar-derived hydrinos) and therefore already contain primordial amounts of hydrinos? IOW they are not the catalyst themselves, they just carry the primordial hydrino, which is the true catalyst. I don't think this is the primary explanation. I have my own theory, with which I will regale you another time. ;) However I think you might be correct in as much as any atom harboring hyh, might potentially yield it up. However this is only going to happen under circumstances where ionizing radiation is present. Given that the n=1/2 would float on water, density-wise (although likely completely soluble) they could be anywhere, if they are indeed of primordial (and ongoing origin) in PPM or PPB ratios - and this is especially true of potassium salts - and in other Mills 'catalysts' but why not sodium... hmm. Actually this may disprove my notion of deeply buried hyh. Since the only natural sodium isotope is Na23, if you bury hyh in it you would end up with what should behave as stable Ne24. But there is no such animal. Therefore we can fairly safely assume that either the whole concept is wrong, or the chemical changes are more drastic than I thought, and the resultant atom doesn't behave enough like Ne to pass for a noble gas. Perhaps hyh just heads straight for the nucleus, gets stripped of its second electron and then wanders off again. There is something wrong with this scenario energy wise, but I can't put my finger on it yet. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Robin van Spaandonk wrote: In reply to Edmund Storms's message of Fri, 18 Nov 2005 14:49:56 -0700: Hi, [snip] Yea, I changed my mind based on the way you described how the Hy is thought to behave. Note that most of the behavioral aspects are my interpretation, not necessarily Mills' opinion. That's ok, we might even arrive at a better understanding than he has. Just to be clear, both electrons are in fractional quantum states according to Mills. (Otherwise the binding energy of the second electron wouldn't increase with shrinkage level). Yes, that is what I initially assumed. However, for a compound to form, the normal quantum levels must be involved. Why? In a perfect ionic compound, solidity results from the binding energy of positive and negative ions. IOW the attractive force between ions of opposite charge pulls the ensemble together. There is no real need for electrons to be interchanged at a local level as would be the case in a covalent bond. Granted, with normal substances there is more often a polar bond than a pure ionic bond. In short, the hyh bond with a positive ion would be the most extreme ionic bond imaginable. You may calculate the degree of electron sharing if you wish, but given an ionization potential of around 70 eV for hyh[n=1/16], I think you will find that it is so negligible as to be immeasurable. A normal ionic compound results from electrons being moved from one atom to the other. For example, in making NaCl, the electron moves from the Na atom to reside for most of the time at the Cl atom. This is different from the situation with Hy, which I'm trying to understand, so be patient. When Hy is involved, the situation involves a preionized atom, so to speak, which as a negative charge that can not be removed by chemical interaction. Consequently for it to form a bond, the other atom must also be preionized to form a positive ion. Of course, this is easily done. You would propose that if Hy were bubbled through a solution of Na+ Cl- in H2O, a compound should form having the formula NaHy. In a similar fashion, Hy bubbled through an acid should result in HHy. In addition to ionic bonding, both compounds have the potential for some covalent bonding as electrons from the normal atom briefly occupy normal energy levels in the Hy structure. I suggest this addition of pure ionic and covalent components makes the bond exceptionally strong, not just the ionic part. I suggest the ionic part can not be any stronger than the stability of the positive ion respect to regaining its electron from other sources in the environment. Does this fit with your understanding? The Hy levels might be filled by one or more electrons, but these only give the assembly a negative charge, rather like a really big electron. ...or a very small negative ion. Bonds are formed by electrons interacting between similar quantum levels. Bonds are formed by two forces. Electrostatic, and magnetic. Pure covalent bonds are pure magnetic bonds. Pure Ionic bonds are pure electrostatic bonds. Polar bonds are a mixture of the two. In the case of hyh, because the second electron is so tightly bound, the bond is the purest electrostatic bond of all compounds. IMO. No, I don't agree. The positive ion can get an electron from many sources other than the Hy. Consequently, the bond is no more stable than any other pure ionic bond. Thermodynamic stability is based on the components of the compound being returned to their elemental state. Returning NaHy to its elements would be equivalent to adding an electron to the Na+ and doing nothing with the Hy, because its state can not be changed. Hy is essentially a pure element when viewed from a chemical viewpoint. This is something the Hy electrons can not do. However these Hy electrons would modify the energetics of normal quantum levels and cause such compounds to have unusual properties without the Hy electrons being directly involved. Yes, this is also possible, particularly if the hyh can replace a deeper electron from the normal shell. Such an atom may simply appear to the outside world as an the original atom with a proton converted to a neutron, and a neutron added. E.g. if one started out with K39 and hyh replaced an inner electron, then the result may look like Ar40, both chemically and for SIMS. If D were used iso H, then it would look like Ar41. Because of this possibility of fooling SIMS, it's imperative that NAA also be used to identify new atoms in CF experiments. (Tightly bound hydrinos can't fool NAA). This is an interesting possibility. The question is, can a highly reduced Hy actually act like a neutron that is stuck to an atom outside of the nucleus? On the other hand, I would expect such a structure to be so close to being neutral that interaction with the electron quantum states would not be possible. This seems to be an idea worth exploring. Would this explain the
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Sat, 19 Nov 2005 15:19:06 -0700: Hi, [snip] Why? In a perfect ionic compound, solidity results from the binding energy of positive and negative ions. IOW the attractive force between ions of opposite charge pulls the ensemble together. There is no real need for electrons to be interchanged at a local level as would be the case in a covalent bond. Granted, with normal substances there is more often a polar bond than a pure ionic bond. In short, the hyh bond with a positive ion would be the most extreme ionic bond imaginable. You may calculate the degree of electron sharing if you wish, but given an ionization potential of around 70 eV for hyh[n=1/16], I think you will find that it is so negligible as to be immeasurable. A normal ionic compound results from electrons being moved from one atom to the other. For example, in making NaCl, the electron moves from the Na atom to reside for most of the time at the Cl atom. This is different from the situation with Hy, which I'm trying to understand, so be patient. When Hy is involved, the situation involves a preionized atom, so to speak, which as a negative charge that can not be removed by chemical interaction. Consequently for it to form a bond, the other atom must also be preionized to form a positive ion. Essentially correct, but be careful not to confuse Hy (neutral) with what I have been designating hyh (Hydrinohydride) which carries a negative charge (or Hy- if you prefer that notation). Of course, this is easily done. You would propose that if Hy were bubbled through a solution of Na+ Cl- in H2O, a compound should form having the formula NaHy. In a similar fashion, Hy bubbled through an acid should result in HHy. This would then be the neutral dihydrino molecule. Some time back I asked Mills directly whether he thought that Hy + proton - Hy2+ and he said that he thought it would. In addition to ionic bonding, both compounds have the potential for some covalent bonding as electrons from the normal atom briefly occupy normal energy levels in the Hy structure. I suggest this addition of pure ionic and covalent components makes the bond exceptionally strong, not just the ionic part. Only experiment will tell. I suggest the ionic part can not be any stronger than the stability of the positive ion respect to regaining its electron from other sources in the environment. Does this fit with your understanding? I find myself forced to agree, if the Hyh is not buried within the shell structure of the positive ion. Furthermore, the statement you make must also be true of all ordinary ionic compounds. So, while reducing the positive ion with a free electron determines the upper limit of the bond strength[1], the Hyh compounds should nevertheless be stronger than other ionic compounds, because of the small size of the Hyh. [snip] Bonds are formed by two forces. Electrostatic, and magnetic. Pure covalent bonds are pure magnetic bonds. Pure Ionic bonds are pure electrostatic bonds. Polar bonds are a mixture of the two. In the case of hyh, because the second electron is so tightly bound, the bond is the purest electrostatic bond of all compounds. IMO. No, I don't agree. The positive ion can get an electron from many sources other than the Hy. Consequently, the bond is no more stable than any other pure ionic bond. But *no* ionic bond is stronger than the upper limit implied by neutralizing the positive ion. In short, sorted in order of increasing bond strength, we have: 1) Normal ionic bond. 2) Hyh ionic bond. 3) Neutralization energy of positive ion by free electron(s). [snip] This is an interesting possibility. The question is, can a highly reduced Hy actually act like a neutron that is stuck to an atom outside of the nucleus? That's the general concept, though it would essentially be a negatively charged neutron, effectively reducing the atomic number by 1. This is because it would orbit the nucleus inside the K shell, so from the point of view of the electrons, the nuclear charge would be reduced by 1. (Actually I'm guessing here. The size of the hydrino is still up in the air somewhat as far as I'm concerned). Besides it will depend on which Hy- combines with which positive ion. On the other hand, I would expect such a structure to be so close to being neutral that interaction with the electron quantum states would not be possible. This seems to be an idea worth exploring. Would this explain the Fisher-Oriani super heavy carbon? Never heard of it. Reference? [snip] I doubt it, because, while the hyh may not be able to leak away, the electron it replaces can leak away. This would still leave a neutral charge over all. Yes, over all. But immediately at the surface of the metal, the positive changes left behind would generate a voltage gradient. Would this gradient be large enough to do something unusual? I suspect that here you may be confusing Hy
Re: Cathode (Cometary) Commentary
In reply to Robin van Spaandonk's message of Sun, 20 Nov 2005 12:57:17 +1100: Hi, [snip] I wrote: Furthermore, the statement you make must also be true of all ordinary ionic compounds. So, while reducing the positive ion with a free electron determines the upper limit of the bond strength[1], the Hyh compounds should nevertheless be stronger ..then promptly forgot all about the [1]. [1] Of course, by analogy, the same goes for removing an electron from the negative ion. The actual upper limit on the bond strength is then the minimum of the two possibilities. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode( Cometary) Commentary
Robin, The thoughts remind me of the functions of a water softener. Resin beads attract the hardness in water and uses salt to release the attracted solids for flushing. There is a remarkable similarity in your description of the actions you observe and our workin liquid vortex studies. Richard
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Fri, 18 Nov 2005 08:50:40 -0700: Hi, [snip] Thanks Robin, the situation is getting clearer. However, I still have some questions. In summary, the model you are describing assumes one electron is in a fractional quantum state (Hy) and the additional electron is in a normal quantum level. Here you contradict yourself. See your own words here below[1]. Just to be clear, both electrons are in fractional quantum states according to Mills. (Otherwise the binding energy of the second electron wouldn't increase with shrinkage level). (If I'm not mistaken the energy levels of the two electrons are very different, but the physical location is almost identical according to Mills - i.e. there is very little difference in radius - which come to think of it, doesn't make a lot of sense to me). Presumably, the normal quantum level has been modified by the presence of a charge between it and the positive nucleus, if we assume a classical structure. As a result, the normal electron is less tightly held depending on the level of the Hy electron, as you note. This would mean the the chemical bond between Hy- and M+ would be less energetic than if normal hydrogen were involved. Apparently, when the lowest Hy levels are occupied, the atom becomes essentially inert because the proton charge is almost totally neutralized by an electron that can not be lost or modified. Since the second electron has a binding energy curve (for want of a better term), it would be helpful here if you elucidate your remarks with level numbers. (lowest and highest are terms I try to avoid, because they depend on one's point of view. I.e. is 1 the highest or lowest level?) As for heat after death, I suggest a simpler explanation is possible. While electrolysis is ongoing, deuterium is available to the active surface from the current. However, when current is stopped, deuterium is available from the interior during the deloading process. Therefore, life after death would be most apparent when a massive Pd sample is used. Also possible. I didn't say the explanation I gave was the only one, I just said that it was neat. Reality will be determined by experiment. [snip] Robin van Spaandonk wrote: In reply to Edmund Storms's message of Thu, 17 Nov 2005 14:03:14 -0700: [snip] [1] OK, you propose that two or more electrons occupy fractional quantum levels at the same proton. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Robin van Spaandonk wrote: In reply to Edmund Storms's message of Fri, 18 Nov 2005 08:50:40 -0700: Hi, [snip] Thanks Robin, the situation is getting clearer. However, I still have some questions. In summary, the model you are describing assumes one electron is in a fractional quantum state (Hy) and the additional electron is in a normal quantum level. Here you contradict yourself. See your own words here below[1]. Yea, I changed my mind based on the way you described how the Hy is thought to behave. Just to be clear, both electrons are in fractional quantum states according to Mills. (Otherwise the binding energy of the second electron wouldn't increase with shrinkage level). Yes, that is what I initially assumed. However, for a compound to form, the normal quantum levels must be involved. The Hy levels might be filled by one or more electrons, but these only give the assembly a negative charge, rather like a really big electron. Bonds are formed by electrons interacting between similar quantum levels. This is something the Hy electrons can not do. However these Hy electrons would modify the energetics of normal quantum levels and cause such compounds to have unusual properties without the Hy electrons being directly involved. Because the charge is stable, the Hy should act like a really heavy electron when focused by electric and magnetic fields. In fact, if they were caused to bombard a metal plate, they could be used to build up very high static potentials. Unlike electron, they could not leak away by conduction. This might produce some unusual effects. (If I'm not mistaken the energy levels of the two electrons are very different, but the physical location is almost identical according to Mills - i.e. there is very little difference in radius - which come to think of it, doesn't make a lot of sense to me). Presumably, the normal quantum level has been modified by the presence of a charge between it and the positive nucleus, if we assume a classical structure. As a result, the normal electron is less tightly held depending on the level of the Hy electron, as you note. This would mean the the chemical bond between Hy- and M+ would be less energetic than if normal hydrogen were involved. Apparently, when the lowest Hy levels are occupied, the atom becomes essentially inert because the proton charge is almost totally neutralized by an electron that can not be lost or modified. Since the second electron has a binding energy curve (for want of a better term), it would be helpful here if you elucidate your remarks with level numbers. (lowest and highest are terms I try to avoid, because they depend on one's point of view. I.e. is 1 the highest or lowest level?) I'm assuming Hy1 is a level closest to the Bohr zero level and Hy22 is a level that releases the most energy when it is occupied, in which the electron occupies an orbit close to the proton. regards, Ed As for heat after death, I suggest a simpler explanation is possible. While electrolysis is ongoing, deuterium is available to the active surface from the current. However, when current is stopped, deuterium is available from the interior during the deloading process. Therefore, life after death would be most apparent when a massive Pd sample is used. Also possible. I didn't say the explanation I gave was the only one, I just said that it was neat. Reality will be determined by experiment. [snip] Robin van Spaandonk wrote: In reply to Edmund Storms's message of Thu, 17 Nov 2005 14:03:14 -0700: [snip] [1] OK, you propose that two or more electrons occupy fractional quantum levels at the same proton. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Fri, 18 Nov 2005 14:49:56 -0700: Hi, [snip] Yea, I changed my mind based on the way you described how the Hy is thought to behave. Note that most of the behavioral aspects are my interpretation, not necessarily Mills' opinion. Just to be clear, both electrons are in fractional quantum states according to Mills. (Otherwise the binding energy of the second electron wouldn't increase with shrinkage level). Yes, that is what I initially assumed. However, for a compound to form, the normal quantum levels must be involved. Why? In a perfect ionic compound, solidity results from the binding energy of positive and negative ions. IOW the attractive force between ions of opposite charge pulls the ensemble together. There is no real need for electrons to be interchanged at a local level as would be the case in a covalent bond. Granted, with normal substances there is more often a polar bond than a pure ionic bond. In short, the hyh bond with a positive ion would be the most extreme ionic bond imaginable. You may calculate the degree of electron sharing if you wish, but given an ionization potential of around 70 eV for hyh[n=1/16], I think you will find that it is so negligible as to be immeasurable. The Hy levels might be filled by one or more electrons, but these only give the assembly a negative charge, rather like a really big electron. ...or a very small negative ion. Bonds are formed by electrons interacting between similar quantum levels. Bonds are formed by two forces. Electrostatic, and magnetic. Pure covalent bonds are pure magnetic bonds. Pure Ionic bonds are pure electrostatic bonds. Polar bonds are a mixture of the two. In the case of hyh, because the second electron is so tightly bound, the bond is the purest electrostatic bond of all compounds. IMO. This is something the Hy electrons can not do. However these Hy electrons would modify the energetics of normal quantum levels and cause such compounds to have unusual properties without the Hy electrons being directly involved. Yes, this is also possible, particularly if the hyh can replace a deeper electron from the normal shell. Such an atom may simply appear to the outside world as an the original atom with a proton converted to a neutron, and a neutron added. E.g. if one started out with K39 and hyh replaced an inner electron, then the result may look like Ar40, both chemically and for SIMS. If D were used iso H, then it would look like Ar41. Because of this possibility of fooling SIMS, it's imperative that NAA also be used to identify new atoms in CF experiments. (Tightly bound hydrinos can't fool NAA). Because the charge is stable, the Hy should act like a really heavy electron when focused by electric and magnetic fields. In fact, if they were caused to bombard a metal plate, they could be used to build up very high static potentials. Unlike electron, they could not leak away by conduction. This might produce some unusual effects. I doubt it, because, while the hyh may not be able to leak away, the electron it replaces can leak away. This would still leave a neutral charge over all. [snip] Since the second electron has a binding energy curve (for want of a better term), it would be helpful here if you elucidate your remarks with level numbers. (lowest and highest are terms I try to avoid, because they depend on one's point of view. I.e. is 1 the highest or lowest level?) I'm assuming Hy1 is a level closest to the Bohr zero level and Hy22 is a level that releases the most energy when it is occupied, in which the electron occupies an orbit close to the proton. g . Yes, but it wasn't the definition of 1 or 22 that was ambiguous, but rather the definition of high and low. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Well Robin, as you eventually concluded, rapid collapse of a local superconducting site would not supply the necessary energy to make a neutron because local conservation of energy would still have to occur. Energy might be concentrated in a few individual electrons, but the total number of electrons having this energy would be small, hence the total available energy would also be small. Frankly, I do not see how sufficient energy can be concentrated in a region of space occupied by a proton to allow it to be converted into a neutron. As far as I know, no demonstrated example of such conversion has been published. Do you know of any demonstrated example of a proton being converted to a neutron using an electron of any size? Regards, Ed Regards, Ed Robin van Spaandonk wrote: In reply to Edmund Storms's message of Wed, 16 Nov 2005 13:12:18 -0700: Hi, [snip] I imagine that the unique structure is not only based on the arrangement of atoms, but also on the arrangement of electrons in uncharacteristic ways. I believe the prediction that the structure is a room temperature superconductor is correct. Such a structure would potentially be able to support a coherent electron wave as well as dissipate the released energy within this electron structure. [snip] That's the opening I have been waiting for. ;) Several years ago (late 90's?) I posted a message to this forum suggesting that superconducting segments that suddenly cease to superconduct would accelerate charged particles to high energy, and that such microscopic particle accelerators might be involved in CF. I would now like to embroider on this a little. First, when a small section of cathode becomes superconducting, essentially all local current will flow through it rather than through the surrounding metal. That in turn results in a strong local magnetic field around the superconducting segment. If the mag. field strength gets too high, the material suddenly ceases to be a superconductor, and the field collapses (very rapidly). Due to V = -L dI/dt, the voltage accelerating free charged particles in the segment can be locally very high. Now we switch our point of view to the heavy electrons in the Widom/Larsen paper, and discover that we now potentially have another source of heavy electrons. Once the mag. field has collapsed, if the arrangement of atoms is unchanged, they are free to once again become superconducting, and the whole cycle can begin again. Theoretically, this process could repeat at quite high frequencies, provided that the power supply continues to feed it[1]. We do however have a problem. In order to form a neutron, the electron needs to have about 7-800 keV of kinetic energy. The neutron in turn will on average yield about 5 MeV of energy, by attaching itself to another nucleus, so at least 1 in 6 of the electrons that are energetic enough has to form a neutron, just to break even[2]. To make matters worse, probably nowhere near enough electrons are going to have that much energy anyway. All of which may go some way toward explaining why CF is frequently only OU by a few percent. [1] This may be enhanced by placing a fast responding capacitor directly across the electrodes at the cell (with short leads). [2] This depends somewhat on your definition of breaking even. Since every neutron formed results in excess energy, and electrons that don't form neutrons simply convert their energy into heat that is counted in the output anyway, this process is always at least somewhat OU. How much so, only experiment will tell. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Ed and Robin, RVS: Don't you believe in Hy-hydrides, or that they may bind to positive ions? ES: I have a hard time, Robin, understanding how a chemical bond can form with a Hy. The electrons are in energy states that are far removed from the states in normal atoms and the states are not compatible in a quantum sense. Normal chemistry recognizes three types of bonds, which are ionic, covalent and metallic. Each requires the electron be removed from the atom for some length of time. This time is longest for ionic bonds and shortest for a metallic bond. How long can the electron associated with a Hy be located elsewhere? How does this loss happen without the energy lost during formation of a Hy being returned to the electron? GH: Is the Hy-hydride bond a stronger version of the weak bond (.6 ev) that is found in negative hydrogen ions (H-) ? I don't know how Ed would classify this bond but it seems to me that getting the first electron closer to the proton could increase this bond strength. Robin, can you clarify how Mills explains the bond and even calculates the bond strength for many 1/n Hy-hydrides? George Holz Varitronics Systems
Re: Cathode (Cometary) Commentary
Edmund Storms writes, I have a hard time, Robin, understanding how a chemical bond can form with a Hy. The electrons are in energy states that are far removed from the states in normal atoms and the states are not compatible in a quantum sense. With the huge exception of another hydrino of the same redundant ground state. In that situation a chemical bond is highly favored, no? Jones
Re: Cathode (Cometary) Commentary
Vorts, Sure hope everyone is following this thread close.. there is meat on them bones,, reminds me of the Beta atmosphere thread.. hopefully, there is a way to blend the two. Jump in there Grimer. Richard - Original Message - From: Jones Beene [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Thursday, November 17, 2005 10:56 AM Subject: Re: Cathode (Cometary) Commentary Edmund Storms writes, I have a hard time, Robin, understanding how a chemical bond can form with a Hy. The electrons are in energy states that are far removed from the states in normal atoms and the states are not compatible in a quantum sense. With the huge exception of another hydrino of the same redundant ground state. In that situation a chemical bond is highly favored, no? Jones
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Thu, 17 Nov 2005 09:31:01 -0700: Hi, [snip] Well Robin, as you eventually concluded, rapid collapse of a local superconducting site would not supply the necessary energy to make a neutron because local conservation of energy would still have to occur. Energy might be concentrated in a few individual electrons, but the total number of electrons having this energy would be small, hence the total available energy would also be small. Frankly, I do not see how sufficient energy can be concentrated in a region of space occupied by a proton to allow it to be converted into a neutron. As far as I know, no demonstrated example of such conversion has been published. Do you know of any demonstrated example of a proton being converted to a neutron using an electron of any size? [snip] All electron capture reactions are examples of this, including the rare case of electron capture by two protons in the Sun, leading to D formation. However to be fair, these all result in more kinetic energy being released than consumed, while the reaction above consumes more kinetic energy than it releases. To find an example of this, I expect one would have to look at particle beam experiments. Mind you, the number of conversions is also limited by the fact that it's a weak force mediated interaction, so it's going to be much more rare than I previously thought. You win, I think I'll drop this. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Thu, 17 Nov 2005 08:52:53 -0700: Hi, [snip] Don't you believe in Hy-hydrides, or that they may bind to positive ions? I have a hard time, Robin, understanding how a chemical bond can form with a Hy. The electrons are in energy states that are far removed from the states in normal atoms and the states are not compatible in a quantum sense. Normal chemistry recognizes three types of bonds, which are ionic, covalent and metallic. Each requires the electron be removed from the atom for some length of time. This time is longest for ionic bonds and shortest for a metallic bond. How long can the electron associated with a Hy be located elsewhere? 0. How does this loss happen without the energy lost during formation of a Hy being returned to the electron? The loss doesn't happen at all. On the contrary it's the other ion that suffers the loss, the hydrino suffers a gain. The hydrinohydride is the negative ion. It can form ionic bonds with positive ions of other atoms. When forming a coating on a metal, think of it as a substitute for O--, and the layer formed as analogous to an oxide layer. [snip] Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to George Holz's message of Thu, 17 Nov 2005 11:33:33 -0500: Hi, [snip] GH: Is the Hy-hydride bond a stronger version of the weak bond (.6 ev) that is found in negative hydrogen ions (H-) ? Yes. I don't know how Ed would classify this bond but it seems to me that getting the first electron closer to the proton could increase this bond strength. Robin, can you clarify how Mills explains the bond and even calculates the bond strength for many 1/n Hy-hydrides? The bond strength between the hydrino and the second electron is based on the magnetic field attraction between the two electrons. Mills does the calculation in his book, and even provides a table of binding energies for different hydrino sizes. However this is based on radius, and given that I'm still trying to figure out which of us is right about the radius, I'm not sure yet whether or not I need to recalculate the bond strengths. (I believe he only uses magnetic field energy because the hydrino is essentially a neutral particle, hence the second electron experiences no electrical field, only the magnetic field of the first electron - that's his reasoning AFAIK). Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
Robin van Spaandonk wrote: In reply to Edmund Storms's message of Thu, 17 Nov 2005 08:52:53 -0700: Hi, [snip] Don't you believe in Hy-hydrides, or that they may bind to positive ions? I have a hard time, Robin, understanding how a chemical bond can form with a Hy. The electrons are in energy states that are far removed from the states in normal atoms and the states are not compatible in a quantum sense. Normal chemistry recognizes three types of bonds, which are ionic, covalent and metallic. Each requires the electron be removed from the atom for some length of time. This time is longest for ionic bonds and shortest for a metallic bond. How long can the electron associated with a Hy be located elsewhere? 0. How does this loss happen without the energy lost during formation of a Hy being returned to the electron? The loss doesn't happen at all. On the contrary it's the other ion that suffers the loss, the hydrino suffers a gain. The hydrinohydride is the negative ion. It can form ionic bonds with positive ions of other atoms. When forming a coating on a metal, think of it as a substitute for O--, and the layer formed as analogous to an oxide layer. OK, you propose that two or more electrons occupy fractional quantum levels at the same proton. Presumably the same kind of limitations exist here as in normal quantum levels, i.e. no two electrons can occupy the same quantum level. These electrons can not leave the proton to interact with other atoms but their charge is felt by positive ions with which an ionic compound is formed. Is this what you have in mind? I expect you would also conclude that such compounds are very good insulators and would interact chemically only with material in which ions are present. In other words, no metallic or covalent interaction would be possible. Regards, Ed Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Thu, 17 Nov 2005 14:03:14 -0700: Hi, [snip] The loss doesn't happen at all. On the contrary it's the other ion that suffers the loss, the hydrino suffers a gain. The hydrinohydride is the negative ion. It can form ionic bonds with positive ions of other atoms. When forming a coating on a metal, think of it as a substitute for O--, and the layer formed as analogous to an oxide layer. OK, you propose that two or more electrons occupy fractional quantum levels at the same proton. Presumably the same kind of limitations exist here as in normal quantum levels, i.e. no two electrons can occupy the same quantum level. These electrons can not leave the proton to interact with other atoms but their charge is felt by positive ions with which an ionic compound is formed. Is this what you have in mind? With reservation[1], yes. That's Mills' definition of hydrinohydride AFAIK. (Poor name IMO). I expect you would also conclude that such compounds are very good insulators and would interact chemically only with material in which ions are present. In other words, no metallic or covalent interaction would be possible. Indeed, I would make that assumption, with the same reservation mentioned above, i.e. [1] it is true for Hydrinohydride that has shrunk past about level 3-4. However for the *first couple of levels* the second electron is only bound very weakly, so *this* hyh (my abbrev.) is actually a strong reductor, akin to a metal atom. (According to Mills this is paradoxically also true for levels beyond about 22, and after 24 the second electron doesn't bind at all). The strongest bond is for level 16 if I remember correctly. BTW, note that if hit by an energetic particle, the hyh can be knocked out of its lattice, and deprived of its second electron (this may require a second collision, depending on the energy of the colliding particle), so that it once again becomes a plain hydrino. Once this has happened, the hydrino can undergo further shrinkage reactions. Also of interest is that hyh because it is very small, sits much closer to its positive ion in its lattice. If this ion is itself small e.g. B+++ then the chances of a fusion reaction are enhanced enormously (many orders of magnitude), particularly as the two are continually in close proximity, so that the confinement time in the Lawson criterion is effectively infinite. Of course the greater the shrinkage of the hyh, the shorter will be the half-life of such reactions. IOW looking at the Lawson criterion, the density is huge, and the confinement time is unlimited, but the temperature is low. Another thing to consider is that the hyh may actually sit *inside* the electron shells of the other positive ion, effectively displacing an existing electron. If this is possible, because of its huge mass, it should orbit the positive ion effectively at the radius of the hyh. (Analogous to muonic molecules, but where the negative muon is replaced by hyh). Actually, it probably couldn't get that close, because by then the attractive force exerted on the second electron by the nucleus of the positive ion would remove it from the hydrino. This is unfortunately an aspect that Mills appears unwilling to entertain. It would however IMO provide a very neat explanation for heat after death. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode (Cometary) Commentary
In reply to George Holz's message of Thu, 17 Nov 2005 16:29:11 -0500: Hi, [snip] Of course, the spherical symmetry and neutrality make the concept and calculation obvious. I was hindered here by still basing my thinking on my point electron/ZPE based theory of Hydrinos. So the bond energy would be dependent on the distance between a Hydrino orbitsphere and a normal orbitsphere with magnetic dipoles aligned? Something like that. You have basically reached my limits now, so you will need to look it up in Mills' book if you want to go any deeper. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
Re: Cathode ( Cometary) Commentary
Robin wrote.. IOW looking at the Lawson criterion, the density is huge, and theconfinement time is unlimited, but the temperature is low.Another thing to consider is that the hyh may actually sit*inside* the electron shells of the other positive ion,effectively displacing an existing electron. If this is possible,because of its huge mass, it should orbit the positive ioneffectively at the radius of the hyh. (Analogous to "muonicmolecules", but where the negative muon is replaced by hyh).Actually, it probably couldn't get that close, because by then theattractive force exerted on the second electron by the nucleus ofthe positive ion would remove it from the hydrino. This is unfortunately an aspect that Mills appears unwilling toentertain. It would however IMO provide a very neat explanationfor "heat after death". Interesting explanation of temperature. Some our work in liquid vortex studies reveal adifferential of temperatures that give me pause. Would like for Fred and Jones to enter their comments. Richard
Re: Cathode (Cometary) Commentary
Ed Storms Apparently, a very unusual structure is required that is not present in ordinary matter. The various theories have been so unsuccessful in guiding research because they are based on the properties of normal material. Everyone can probably agree on that part, and in fact, catalysts usually depend on structure at some level of observation. If however LENR relates to more than 'just' geometric structure, such as the structure of a metal which is hydrided with a very tightly bound hydride which modifies the external geometry - that would be interesting. I wish my electron microscope was not on the fritz today ;-) BTW for those (from Oz ;-) who are sure to correct some of my past posted details (and I appreciate that), and since my original rough calculation for expected di-hydrino density was too hasty, here is something based on what we know about H2 - with the assumption that liquid Hy2 (Hy2 being the di-hydrino) is very similar to liquid H2 except in melting point. I expected the melting point for N=1/2 to be over 1000 K and the others to be solid, but I am certain that there will be a wide divergence of opinion as to those numbers and as to the phase change points. The liquid density of H2 is .07 gm/cc. Unfortunately, no solid density is available, as it would demand extreme pressure. Since the Oort cloud will contain few cations, one can expect almost all solar hydrinos deposited there to be paired together molecularly, and in a similar fashion to H2. except with the smaller orbital and higher resultant density. Based then on extending the H2 model, liquid n=1/3 hydrinos should weigh in at 1.89 gm/cc and n=1/4 would be 4.48 and n=1/7 which Mills claims to have samples of (but in the form of ionic bound chemical hydrides) would be 24 gm/cc - about the density of uranium. Note however this important detail - that even though the density of shrunken di-hydrino molecules would be every high, the same does not apply to hydrino hydrides (ionic bonding), since the (more shrunken variety) hydrino would probably be actually located most of the time *inside* the orbital cloud (not unusual) and likely around the k-shell of the resultant hydride (unusual), which would be only slightly larger, and perhaps even less dense than before. Nevertheless, I cannot imagine this material, even in the first shrinkage state, staying gaseous at STP. And I still think many hydrinos should be found in ocean minerals, particularly the alkalis, and that could be the Mills' catalyst connection. Why sodium would not be a catalyst carrier (i.e and active Mills catalyst) is not certain. Perhaps many alkalis are in that category because the nucleus is expressing a lesser positive charge (near-field than expected). But what eliminates sodium? Could it be that hydrino catalysts are only catalytic for one reason - that is because they have a natural affinity for hydrino-hydriding (with the natural population of solar-derived hydrinos) and therefore already contain primordial amounts of hydrinos? IOW they are not the catalyst themselves, they just carry the primordial hydrino, which is the true catalyst. Given that the n=1/2 would float on water, density-wise (although likely completely soluble) they could be anywhere, if they are indeed of primordial (and ongoing origin) in PPM or PPB ratios - and this is especially true of potassium salts - and in other Mills 'catalysts' but why not sodium... hmm. Jones
Re: Cathode (Cometary) Commentary
Jones Beene wrote: Ed Storms Apparently, a very unusual structure is required that is not present in ordinary matter. The various theories have been so unsuccessful in guiding research because they are based on the properties of normal material. Everyone can probably agree on that part, and in fact, catalysts usually depend on structure at some level of observation. If however LENR relates to more than 'just' geometric structure, such as the structure of a metal which is hydrided with a very tightly bound hydride which modifies the external geometry - that would be interesting. I wish my electron microscope was not on the fritz today ;-) I imagine that the unique structure is not only based on the arrangement of atoms, but also on the arrangement of electrons in uncharacteristic ways. I believe the prediction that the structure is a room temperature superconductor is correct. Such a structure would potentially be able to support a coherent electron wave as well as dissipate the released energy within this electron structure. BTW for those (from Oz ;-) who are sure to correct some of my past posted details (and I appreciate that), and since my original rough calculation for expected di-hydrino density was too hasty, here is something based on what we know about H2 - with the assumption that liquid Hy2 (Hy2 being the di-hydrino) is very similar to liquid H2 except in melting point. I expected the melting point for N=1/2 to be over 1000 K and the others to be solid, but I am certain that there will be a wide divergence of opinion as to those numbers and as to the phase change points. I take the opposite view, that the melting point would be far lower than H2. In fact, I would predict that Hy would never be a liquid, at least on this world. Formation of a condensed phase requires some attraction between the atoms. Helium has the lowest melting point because the atoms have the least attraction. What would be the basis of attraction between Hy atoms? The electron can not leave the nucleus for even a brief time, thus no covalent attraction is possible. Also, how is Hy pairing possible? This requires electron exchange, which would appear to be impossible in these atoms. I also suggest that any compound formation would require the Hy atom to be bounded in a cage formation where it would be trapped in a structure that is more physical than chemical. I'm open to suggestions as to why I'm full of s---. Regards, Ed The liquid density of H2 is .07 gm/cc. Unfortunately, no solid density is available, as it would demand extreme pressure. Since the Oort cloud will contain few cations, one can expect almost all solar hydrinos deposited there to be paired together molecularly, and in a similar fashion to H2. except with the smaller orbital and higher resultant density. Based then on extending the H2 model, liquid n=1/3 hydrinos should weigh in at 1.89 gm/cc and n=1/4 would be 4.48 and n=1/7 which Mills claims to have samples of (but in the form of ionic bound chemical hydrides) would be 24 gm/cc - about the density of uranium. Note however this important detail - that even though the density of shrunken di-hydrino molecules would be every high, the same does not apply to hydrino hydrides (ionic bonding), since the (more shrunken variety) hydrino would probably be actually located most of the time *inside* the orbital cloud (not unusual) and likely around the k-shell of the resultant hydride (unusual), which would be only slightly larger, and perhaps even less dense than before. Nevertheless, I cannot imagine this material, even in the first shrinkage state, staying gaseous at STP. And I still think many hydrinos should be found in ocean minerals, particularly the alkalis, and that could be the Mills' catalyst connection. Why sodium would not be a catalyst carrier (i.e and active Mills catalyst) is not certain. Perhaps many alkalis are in that category because the nucleus is expressing a lesser positive charge (near-field than expected). But what eliminates sodium? Could it be that hydrino catalysts are only catalytic for one reason - that is because they have a natural affinity for hydrino-hydriding (with the natural population of solar-derived hydrinos) and therefore already contain primordial amounts of hydrinos? IOW they are not the catalyst themselves, they just carry the primordial hydrino, which is the true catalyst. Given that the n=1/2 would float on water, density-wise (although likely completely soluble) they could be anywhere, if they are indeed of primordial (and ongoing origin) in PPM or PPB ratios - and this is especially true of potassium salts - and in other Mills 'catalysts' but why not sodium... hmm. Jones
Re: Cathode (Cometary) Commentary
Ed The electron can not leave the nucleus for even a brief time, thus no covalent attraction is possible. ...methinks you are just trying to get rid of these critters (but by gravity instead of the normal levity ;-) Covalent bonding is indeed seemingly impossible with any other element than another hydrino, one can suspect that much - BUT there is such a strong preference for the interlocking wave functions of two bound electrons that I think a paring of hydrinos is almost a foregone conclusion in QM terms. The classic case of covalent bonding, where the hydrogen molecule forms by the overlap of the wavefunctions of the electrons of the respective hydrogen atoms in an interaction which is characterized as an exchange interaction, is surprisingly strong. Why would this change with a smaller radius? In fact the interchange bonding should make this molecule almost like a helium nulceus, only with stronger bonding. Once the Hy2 forms it is going to take a very high energy photon to ionize it. That would definitely bolster your view that the liquid state is unlikely. Perhaps you are correct on that point about no liquid phase, and that does seems to be the majority opinion - but I think the pairing to di-hydrino is inevitable. And the potential density-equalent is there, even in a gas, and it might have low comparative mobility - if QM wavefunctions are applicable at that geometry, and since the Hy pair will be comparatively small, dense and slow - plus according to Mills, will have a positive near field ! then they would tend to lodge in the orbitals of whatever cation is available, no? But this is NOT Mills' version of events exactly. Come to think of it - maybe Mills got that part slightly wrong and what we have is always the di-hydride instead of the hydride? Perhaps that is part of the reason that sodium isn't a catalyst (its wavefunction interaction does not fit with the Hy2 ?). Hmm. Jones BTW - since several Hy2 molecules, if they do express a positive near field, should be able to be bound with a single electron that would seem to open up the prospect of charged dense strucutre - like the buckyball or icosohedron... yet the charge is hidden. Is one of these the nucleating agent for an EVOs ?
Re: Cathode (Cometary) Commentary
ERRATA: should be helium atom not nulceus... ;-) not to mention, what IS a nulceus anyway? And does anyone know why spell checkers will often miss simple errors? Perhaps that is also an interlocking wave function but with the computer owner?
Re: Cathode (Cometary) Commentary
In reply to Edmund Storms's message of Wed, 16 Nov 2005 13:12:18 -0700: Hi, [snip] to be impossible in these atoms. I also suggest that any compound formation would require the Hy atom to be bounded in a cage formation where it would be trapped in a structure that is more physical than chemical. [snip] Don't you believe in Hy-hydrides, or that they may bind to positive ions? Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
