Re: [Vo]:Use magnetic fld to enhance effective mass of e-
In reply to Mark Iverson-ZeroPoint's message of Fri, 30 Dec 2011 16:11:45 -0800: Hi Mark, [snip] Robin: Thanks for the comments, and I see your chicken-n-egg argument... As I prefaced my comment about Horace's calcs, I'm not sure if this is relevant either... Please note that in many cases I am just doing a brain-dump in the hopes of triggering some creative thinking. :-) OTOH, I'm not so sure I agree that, as you say, ... in an ordinary magnet many (most?) of the atomic [magnetic] fields are aligned... Magnetic materials are composed of 'magnetic domains'; regions where the magnetic moments are more or less aligned in the same direction. However, adjacent domains are randomly oriented, diminishing the effect for the bulk material and, thus, the *external* magnetic field is *much less* than what one would find in an individual domain. I am curious... if one were to look at the individual atoms (10^6 to 10^9) in one of these 'magnetic domains', what percentage of the magnetic moments are parallel??? Google magnetic saturation. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
In reply to Horace Heffner's message of Fri, 30 Dec 2011 15:32:09 -0900: Hi Horace, [snip] In the process of changing their spin axes the particles can (in a non QM interpretation) precess, due to torque on the spin axis. When this happens the particles can radiate, and flip their spins into alignment. If they can radiate, wouldn't that imply that the ground state electrons of atoms in an externally applied magnetic field can radiate as they approach the nucleus? In a magnetic field the spins of (quantum) particles tend to be aligned either with the magnetic field, or opposed to it. If opposed, a particle will tend to eventually flip into a matching spin. Two particles can experience an attracting force if their poles are aligned N-S N-S However, if they are in orbitals, they align with opposed spin, like so: N S | | S N which is still an attracting mode. If they aligned in the opposing directions they would repel. This is partially the basis of the Pauli Exclusion principle. The spin axes of electrons tend to align with an orbital axis, not perpendicular to it. A pair of electrons sharing other quantum states in an atom will have one spin up and the other down, i.e opposed spins. They will have a magnetic attraction force, a negative potential, but one which at atomic size distances is nominal. At nuclear distances magnetic forces become very large. This tendency of particle spins to align in a magnetically attracting way, creating a potential energy, is called spin coupling. [snip] BTW, as Mark pointed out, the chicken-'n-egg reference was in relation to combining your theory with someone else's, not a criticism of your theory. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
In reply to Mark Iverson-ZeroPoint's message of Thu, 29 Dec 2011 13:48:59 -0800: Hi, [snip] Primarily for the theorists in the Collective. This from the Ni-H yahoo group... -Mark I try to explain it: All you have to do is, to put the electron from the H-atom nearer to the nucleus and Fusion will happen. From the K-electron capture from Be-7 I know, that a faktor 4 is enough. So, how can this be done? Idea comes from Muon, where it is proved, so just enhance the effective mass of the surrounding electron. Vektorpotential A = 1/2 B * r (B orthogonal A, B=const, r is distance) For Fusion, A = sqr(5.405961)*mc/e=0.004 Tesla*meter (to enlarge elektron energy about 782.333keV from proton to Neutron) You can't enlarge the energy with a static magnetic field. Even if all the potential energy existing between electron and proton were converted to kinetic energy, you would still be 782 keV short. Furthermore, r if I'm not mistaken needs to be 1/4 the size of an atom, in order to get shrinkage by a factor or 4, not several cm (or I have completely misunderstood how this is supposed to work). Such a small radius would require a magnetic field vastly stronger than anything humanity has so far managed to create. (225 million tesla for a 782 keV electron). Actually this takes no account of the fact that the electron would be relativistic at that energy, but it gives a rough idea of what would be needed. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Use magnetic fld to enhance effective mass of e-
Robin: If one looks at it macroscopically, then your criticism is understandable, however, one must keep in mind the environment of the H or D loaded lattice at the dimensions of a few atoms. When you get ALL magnetic domains aligned in a small region (a few 10s, 100s of atoms), magnetic fields can become quite large... I'm not sure if this is relevant either, but here is what Horace calculated in his model: If you look at the spreadsheet I provided in 2007, you will see the magnetic field of the electron on the deuteron in the D+e deflated state is given as 4.0210e+14 Tesla. That's about 6 orders of magnitude greater than your 225e6. -Mark -Original Message- From: mix...@bigpond.com [mailto:mix...@bigpond.com] Sent: Thursday, December 29, 2011 11:48 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Use magnetic fld to enhance effective mass of e- In reply to Mark Iverson-ZeroPoint's message of Thu, 29 Dec 2011 13:48:59 -0800: Hi, [snip] Primarily for the theorists in the Collective. This from the Ni-H yahoo group... -Mark I try to explain it: All you have to do is, to put the electron from the H-atom nearer to the nucleus and Fusion will happen. From the K-electron capture from Be-7 I know, that a faktor 4 is enough. So, how can this be done? Idea comes from Muon, where it is proved, so just enhance the effective mass of the surrounding electron. Vektorpotential A = 1/2 B * r (B orthogonal A, B=const, r is distance) For Fusion, A = sqr(5.405961)*mc/e=0.004 Tesla*meter (to enlarge elektron energy about 782.333keV from proton to Neutron) You can't enlarge the energy with a static magnetic field. Even if all the potential energy existing between electron and proton were converted to kinetic energy, you would still be 782 keV short. Furthermore, r if I'm not mistaken needs to be 1/4 the size of an atom, in order to get shrinkage by a factor or 4, not several cm (or I have completely misunderstood how this is supposed to work). Such a small radius would require a magnetic field vastly stronger than anything humanity has so far managed to create. (225 million tesla for a 782 keV electron). Actually this takes no account of the fact that the electron would be relativistic at that energy, but it gives a rough idea of what would be needed. Regards, Robin van Spaandonk
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
In reply to Mark Iverson-ZeroPoint's message of Fri, 30 Dec 2011 11:08:00 -0800: Hi Mark, [snip] Horace's calculation has nothing to do with alignment of magnetic fields in clusters, which can't produce such huge fields anyway. (Consider that in an ordinary magnet many (most?) of the atomic fields are aligned, and the total field is pitiful by comparison to what would be needed.) Robin: If one looks at it macroscopically, then your criticism is understandable, however, one must keep in mind the environment of the H or D loaded lattice at the dimensions of a few atoms. When you get ALL magnetic domains aligned in a small region (a few 10s, 100s of atoms), magnetic fields can become quite large... I'm not sure if this is relevant either, but here is what Horace calculated in his model: If you look at the spreadsheet I provided in 2007, you will see the magnetic field of the electron on the deuteron in the D+e deflated state is given as 4.0210e+14 Tesla. That's about 6 orders of magnitude greater than your 225e6. I haven't checked Horace's calculation, but let's take it at face value. 1) That doesn't necessarily mean that such an orbital is possible. 2) It is a far cry from the intent of the original author that you quoted, who proposed applying an external magnetic field. This is becoming a form of circular reasoning: If we had a strong field we could force the electron into a tight orbital that would then produce a strong magnetic field. Perhaps the Lenz effect means that what one is actually calculating may be the degree to which the electron fights the field, i.e. the field strength one would need to enforce to ensure that the electron remained in the orbital? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Use magnetic fld to enhance effective mass of e-
Robin: Thanks for the comments, and I see your chicken-n-egg argument... As I prefaced my comment about Horace's calcs, I'm not sure if this is relevant either... Please note that in many cases I am just doing a brain-dump in the hopes of triggering some creative thinking. :-) OTOH, I'm not so sure I agree that, as you say, ... in an ordinary magnet many (most?) of the atomic [magnetic] fields are aligned... Magnetic materials are composed of 'magnetic domains'; regions where the magnetic moments are more or less aligned in the same direction. However, adjacent domains are randomly oriented, diminishing the effect for the bulk material and, thus, the *external* magnetic field is *much less* than what one would find in an individual domain. I am curious... if one were to look at the individual atoms (10^6 to 10^9) in one of these 'magnetic domains', what percentage of the magnetic moments are parallel??? -Mark -Original Message- From: mix...@bigpond.com [mailto:mix...@bigpond.com] Sent: Friday, December 30, 2011 1:13 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Use magnetic fld to enhance effective mass of e- In reply to Mark Iverson-ZeroPoint's message of Fri, 30 Dec 2011 11:08:00 -0800: Hi Mark, [snip] Horace's calculation has nothing to do with alignment of magnetic fields in clusters, which can't produce such huge fields anyway. (Consider that in an ordinary magnet many (most?) of the atomic fields are aligned, and the total field is pitiful by comparison to what would be needed.) Robin: If one looks at it macroscopically, then your criticism is understandable, however, one must keep in mind the environment of the H or D loaded lattice at the dimensions of a few atoms. When you get ALL magnetic domains aligned in a small region (a few 10s, 100s of atoms), magnetic fields can become quite large... I'm not sure if this is relevant either, but here is what Horace calculated in his model: If you look at the spreadsheet I provided in 2007, you will see the magnetic field of the electron on the deuteron in the D+e deflated state is given as 4.0210e+14 Tesla. That's about 6 orders of magnitude greater than your 225e6. I haven't checked Horace's calculation, but let's take it at face value. 1) That doesn't necessarily mean that such an orbital is possible. 2) It is a far cry from the intent of the original author that you quoted, who proposed applying an external magnetic field. This is becoming a form of circular reasoning: If we had a strong field we could force the electron into a tight orbital that would then produce a strong magnetic field. Perhaps the Lenz effect means that what one is actually calculating may be the degree to which the electron fights the field, i.e. the field strength one would need to enforce to ensure that the electron remained in the orbital? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
There is no Chicken and egg problem with my theory for the following reasons: 1. The electron is periodically close to the nucleus. When the electron is close to the nucleus, at the range indicated, the magnetic fields are super intense, and spin coupling occurs. 2. An external magnetic field, while it can not add to an electron's momentum, it can greatly increase the probability of spin alignment of deflated state hydrogen and a magnetic nucleus, thus increasing the probability of tunneling of the deflated hydrogen into the nucleus. 3. In the media suggested, the nanoparticles are smaller than typical magnetic domains in iron. 4. The presence of very fine magnetic gradients, as from nano- particles separated by dielectrics, further increases the magnetic potential and adds to the energy gained by the tunneling To understand why magnetic binding occurs just get a couple small strong magnets a throw them up into the air together and at each other. If they get close enough they will always attract and smash together. When two charged particles with spin approach, they act like a couple magnets spinning about their magnetic axis. If not already aligned, their poles will experience a force which brings unlike poles in orientation with each other. In the process of changing their spin axes the particles can (in a non QM interpretation) precess, due to torque on the spin axis. When this happens the particles can radiate, and flip their spins into alignment. In a magnetic field the spins of (quantum) particles tend to be aligned either with the magnetic field, or opposed to it. If opposed, a particle will tend to eventually flip into a matching spin. Two particles can experience an attracting force if their poles are aligned N-S N-S However, if they are in orbitals, they align with opposed spin, like so: N S | | S N which is still an attracting mode. If they aligned in the opposing directions they would repel. This is partially the basis of the Pauli Exclusion principle. The spin axes of electrons tend to align with an orbital axis, not perpendicular to it. A pair of electrons sharing other quantum states in an atom will have one spin up and the other down, i.e opposed spins. They will have a magnetic attraction force, a negative potential, but one which at atomic size distances is nominal. At nuclear distances magnetic forces become very large. This tendency of particle spins to align in a magnetically attracting way, creating a potential energy, is called spin coupling. On Dec 30, 2011, at 3:11 PM, Mark Iverson-ZeroPoint wrote: Robin: Thanks for the comments, and I see your chicken-n-egg argument... As I prefaced my comment about Horace's calcs, I'm not sure if this is relevant either... Please note that in many cases I am just doing a brain-dump in the hopes of triggering some creative thinking. :-) OTOH, I'm not so sure I agree that, as you say, ... in an ordinary magnet many (most?) of the atomic [magnetic] fields are aligned... Magnetic materials are composed of 'magnetic domains'; regions where the magnetic moments are more or less aligned in the same direction. However, adjacent domains are randomly oriented, diminishing the effect for the bulk material and, thus, the *external* magnetic field is *much less* than what one would find in an individual domain. I am curious... if one were to look at the individual atoms (10^6 to 10^9) in one of these 'magnetic domains', what percentage of the magnetic moments are parallel??? -Mark -Original Message- From: mix...@bigpond.com [mailto:mix...@bigpond.com] Sent: Friday, December 30, 2011 1:13 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Use magnetic fld to enhance effective mass of e- In reply to Mark Iverson-ZeroPoint's message of Fri, 30 Dec 2011 11:08:00 -0800: Hi Mark, [snip] Horace's calculation has nothing to do with alignment of magnetic fields in clusters, which can't produce such huge fields anyway. (Consider that in an ordinary magnet many (most?) of the atomic fields are aligned, and the total field is pitiful by comparison to what would be needed.) Robin: If one looks at it macroscopically, then your criticism is understandable, however, one must keep in mind the environment of the H or D loaded lattice at the dimensions of a few atoms. When you get ALL magnetic domains aligned in a small region (a few 10s, 100s of atoms), magnetic fields can become quite large... I'm not sure if this is relevant either, but here is what Horace calculated in his model: If you look at the spreadsheet I provided in 2007, you will see the magnetic field of the electron on the deuteron in the D+e deflated state is given as 4.0210e+14 Tesla. That's about 6 orders of magnitude greater than your 225e6. I haven't checked Horace's calculation, but let's take it at face value. 1) That doesn't necessarily
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
I wrote: However, if they are in orbitals, they align with opposed spin, like so: N S | | S N which is still an attracting mode. I should note that should say opposed poles, not opposed spin. A nucleus with negative mu has spin reversed with respect to the poles. I explained this on page 14 of: http://www.mtaonline.net/~hheffner/NiProtonRiddle.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Use magnetic fld to enhance effective mass of e-
Horace: The reference to the chicken-n-egg was not with your theory... sorry for the misunderstanding. -Original Message- From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Friday, December 30, 2011 4:32 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Use magnetic fld to enhance effective mass of e- There is no Chicken and egg problem with my theory for the following reasons: 1. The electron is periodically close to the nucleus. When the electron is close to the nucleus, at the range indicated, the magnetic fields are super intense, and spin coupling occurs. 2. An external magnetic field, while it can not add to an electron's momentum, it can greatly increase the probability of spin alignment of deflated state hydrogen and a magnetic nucleus, thus increasing the probability of tunneling of the deflated hydrogen into the nucleus. 3. In the media suggested, the nanoparticles are smaller than typical magnetic domains in iron. 4. The presence of very fine magnetic gradients, as from nano- particles separated by dielectrics, further increases the magnetic potential and adds to the energy gained by the tunneling To understand why magnetic binding occurs just get a couple small strong magnets a throw them up into the air together and at each other. If they get close enough they will always attract and smash together. When two charged particles with spin approach, they act like a couple magnets spinning about their magnetic axis. If not already aligned, their poles will experience a force which brings unlike poles in orientation with each other. In the process of changing their spin axes the particles can (in a non QM interpretation) precess, due to torque on the spin axis. When this happens the particles can radiate, and flip their spins into alignment. In a magnetic field the spins of (quantum) particles tend to be aligned either with the magnetic field, or opposed to it. If opposed, a particle will tend to eventually flip into a matching spin. Two particles can experience an attracting force if their poles are aligned N-S N-S However, if they are in orbitals, they align with opposed spin, like so: N S | | S N which is still an attracting mode. If they aligned in the opposing directions they would repel. This is partially the basis of the Pauli Exclusion principle. The spin axes of electrons tend to align with an orbital axis, not perpendicular to it. A pair of electrons sharing other quantum states in an atom will have one spin up and the other down, i.e opposed spins. They will have a magnetic attraction force, a negative potential, but one which at atomic size distances is nominal. At nuclear distances magnetic forces become very large. This tendency of particle spins to align in a magnetically attracting way, creating a potential energy, is called spin coupling. On Dec 30, 2011, at 3:11 PM, Mark Iverson-ZeroPoint wrote: Robin: Thanks for the comments, and I see your chicken-n-egg argument... As I prefaced my comment about Horace's calcs, I'm not sure if this is relevant either... Please note that in many cases I am just doing a brain-dump in the hopes of triggering some creative thinking. :-) OTOH, I'm not so sure I agree that, as you say, ... in an ordinary magnet many (most?) of the atomic [magnetic] fields are aligned... Magnetic materials are composed of 'magnetic domains'; regions where the magnetic moments are more or less aligned in the same direction. However, adjacent domains are randomly oriented, diminishing the effect for the bulk material and, thus, the *external* magnetic field is *much less* than what one would find in an individual domain. I am curious... if one were to look at the individual atoms (10^6 to 10^9) in one of these 'magnetic domains', what percentage of the magnetic moments are parallel??? -Mark -Original Message- From: mix...@bigpond.com [mailto:mix...@bigpond.com] Sent: Friday, December 30, 2011 1:13 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Use magnetic fld to enhance effective mass of e- In reply to Mark Iverson-ZeroPoint's message of Fri, 30 Dec 2011 11:08:00 -0800: Hi Mark, [snip] Horace's calculation has nothing to do with alignment of magnetic fields in clusters, which can't produce such huge fields anyway. (Consider that in an ordinary magnet many (most?) of the atomic fields are aligned, and the total field is pitiful by comparison to what would be needed.) Robin: If one looks at it macroscopically, then your criticism is understandable, however, one must keep in mind the environment of the H or D loaded lattice at the dimensions of a few atoms. When you get ALL magnetic domains aligned in a small region (a few 10s, 100s of atoms), magnetic fields can become quite large... I'm not sure if this is relevant either, but here is what Horace calculated in his model: If you look at the spreadsheet I provided in 2007, you will see
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
On Dec 30, 2011, at 9:58 PM, Mark Iverson-ZeroPoint wrote: Horace: The reference to the chicken-n-egg was not with your theory... sorry for the misunderstanding. My mistake. Sorry. Any excuse to post on my theory. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
On Thu, Dec 29, 2011 at 4:48 PM, Mark Iverson-ZeroPoint zeropo...@charter.net wrote: Primarily for the theorists in the Collective… This from the Ni-H yahoo group... Add a smitch of neodymium and a dash of boron and you probably have a WMD! :-) HNY! T
Re: [Vo]:Use magnetic fld to enhance effective mass of e-
On Dec 29, 2011, at 12:48 PM, Mark Iverson-ZeroPoint wrote: Primarily for the theorists in the Collective… This from the Ni-H yahoo group... -Mark I try to explain it: All you have to do is, to put the electron from the H-atom nearer to the nucleus and Fusion will happen. From the K-electron capture from Be-7 I know, that a faktor 4 is enough. So, how can this be done? Idea comes from Muon, where it is proved, so just enhance the effective mass of the surrounding electron. Vektorpotential A = 1/2 B * r (B orthogonal A, B=const, r is distance) For Fusion, A = sqr(5.405961)*mc/e=0.004 Tesla*meter (to enlarge elektron energy about 782.333keV from proton to Neutron) and from this B=0.008 Tesla (r=1m). For a 5 cm chamber diameter, it is B = 0.16 Tesla (if I am right :-)). Iron is also important, because it has a high Curie temperature. For Nickel is T Curie 360 Celsius, for Iron T Curie 768 Celsius. So, the iron in the chamber enlarges the magnetic field from outside by about a factor 1000. Dietmar Something being overlooked here is that a single iron atom in a nano- cluster of about 100 Ni atoms can magnetize the entire cluster, without an external field. Add a bit of iron (and in some cases copper) to the Ni, heat treat with hydrogen, and you have mu metal. This can increase the permeability by a factor of 40! See my comments on this at: http://www.mail-archive.com/vortex-l@eskimo.com/msg59662.html http://www.mail-archive.com/vortex-l@eskimo.com/msg44662.html There are many forms of mu metal. I noted a specific mu metal composition as an example: 80% Ni, 14% Fe, 5% Mo, 0.5% Mn, plus trace S, Si, C, P. This is a very good protium cold fusion lattice prospect. Curie temp about 454°C. The saturation induction is surprisingly low though, at 7500 gauss. Permeability is 325,000! Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
