Re: [Vo]:Holmlid, Mills & muons
Keep in mind as well that Holmlid adduces not only muons, but kaons and pions as well. Once we introduce (negative) kaons, we have the following decays to deal with: [image: Inline image 1] The neutral pion assures us that there will either be penetrating gammas or positrons, which lead to 511 keV annihilation photons, a signature that is easy to pick up and that will pass through thin shielding. The energy balance for kaons does not make sense to me; but, then again, neither does that for pions or muons. If we go along with Holmlid and allow negative kaons, we must either also allow positive and neutral kaons, or we must come up with a reason for why they don't occur. But it doesn't matter; negative kaons are no doubt not being detected in the first place. They are a merely means to an end, explaining, however tenuously, where the muons come from. Eric
Re: [Vo]:Holmlid, Mills & muons
I believe that when the muon decays, if it is a negative muon, it decays into an electron and a pair of neutrinos. If it is a positive muon, it decays into a positron and 2 neutrinos. Before it decays, if it enters the electronic structure of an atom (likely in condensed matter), then it quickly descends into the innermost orbital, giving off soft x-rays in the process. The resulting muonic atom has a greater chance of internal conversion (I think). If the muon enters a molecule, like a D2 or a D-T in particular, muon catalyzed fusion is highly likely with different products. I am still in study mode for the muon interaction with matter. On Mon, Nov 14, 2016 at 3:40 PM, Bob Cook wrote: > What is the mode of "decay" of free muons and, separately, in condensed > matter? > > They seem not to produce any high energy EM nor radioactive products. If > they did, I would assume this would have been reported unless it was > intended to remain a secret. > > I consider based on reported muon models of Hatt and Stubbs and deep > elastic electron scattering experiments with muons and protons, electons > and positrons should be observed during muon decay, if high energy gammas > do not show up. > > Regarding these ideas, I question the designation of a muon as a lepton > (primary) particle. The scattering experiments suggest a different type of > particle--more akin to a proton or a neutron. > > Bob Cook > -- > *From:* Russ George > *Sent:* Monday, November 14, 2016 9:57 AM > *To:* vortex-l@eskimo.com > *Subject:* RE: [Vo]:Holmlid, Mills & muons > > > The idea that the muons are interacting in solid matter with the electrons > not the nuclei of atoms is very compelling to me. Indeed this may well > explain two mysteries of my cold fusion muon/mischegunons, that is that > very few are escaping the experiment cells. That what I have detected is > the dwindling remains of the reaction is very compelling and as well > explains why so few cold fusion experiments have detected any such > emanations. The time dilation effect that effectively increases the > cross-section of materials just works very well indeed. > > > > This speaks to the growing revelations on silver being a valuable > constituent in a range of experiments. Silver of course has a very complete > electron cloud, as such it might well be the best material for engaging > with the muon/mischugenon nuclear ash. This would help me a lot in > understanding why it just happens that I have found silver so useful (as > has Mills) it is not the neutron cross section of silver it is the muon > cross-section! > > > > > > *From:* Bob Higgins [mailto:rj.bob.higg...@gmail.com] > *Sent:* Monday, November 14, 2016 8:38 AM > *To:* vortex-l@eskimo.com > *Subject:* Re: [Vo]:Holmlid, Mills & muons > > > > In this discussion, Jones presumes muons to be traveling at light speed: > > The muon is an unstable fermion with a lifetime of 2.2 microseconds, > which is an eternity compared to most beta decays. Ignoring time > dilation, this would mean that muons, travelling at light speed, would be > dispersing > and decaying in an imaginary sphere about 600 meters from the reactor. > > > > There are a number of things wrong with this. First, most commonly > encountered muons are cosmogenic and have 100MeV-GeV energies. At these > energies, the muon is traveling at a significant fraction of the speed of > light (but not at the speed of light) and as such experiences time dilation > in its decay. Because of time dilation, the stationary observer sees the > cosmogenic muon decay to be much longer than 2.2 microseconds. This is why > cosmogenic muons can travel 50-100 miles to the Earth's surface without > having decayed. > > What Holmlid has reported is "10MeV/u" as a measurement for his muons - > this is a measure of velocity squared. One u (atomic mass unit) is 931 > MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured > X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's > report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is > only an approximation for small velocity compared to c; as the velocity > increases special relativity must be invoked in the solution. Special > relativity would reduce the velocity from this equation as it started > approaching c, so the actual velocity will be somewhat less than 0.1c for > Holmlid's particles, and a slight time dilation would be experienced. > > So, if Holmlid's particles were muons, and if Mills was creating the same > at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 > meters. However, muons being charged, are well stopped in condensed matter > because the particle doesn't have to run into a nucleus to be scattered, > just run into the dense electronic orbitals. The more dense the condensed > matter, the greater the stopping power for the muon. > > If muons were being generated with a v^2 of 10MeV/u, I doubt an
Re: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons
Hi Fran, I am unable to imagine how something special would happen in that case. A muon in slow motion may have a greater chance of interaction if its energy is near the ionization energy of the atoms upon which it is incident - but this is only a small energy - less than 10eV. At higher energy, it is probably more likely that the muon is going to ionize the atom and then scatter at lower energy. The distances are so small in condensed matter that the scattering will happen rapidly and will reduce the muon to the sweet spot wherein it can interact with the chemical (electronic) structure of the next atom it meets. How would a brief passage though a Casimir geometry alter these behaviors? On Mon, Nov 14, 2016 at 2:12 PM, Roarty, Francis X < francis.x.roa...@lmco.com> wrote: > Bob, what if the “muon” doesn’t have to achieve light speed but rather > becomes so “suppressed” think traveling thru a tiny Casimir cavity that the > muons actual speed inside the cavity where vacuum wavelengths are dilate by > suppression appears to achieve negative light speed relative to observers > outside the cavity where vacuum wavelengths are not suppressed.. IMHO > catlitic action is a weak cousin to Casimir action and the longer > wavelengths we consider suppressed are actually still present from the > perspective of a local observer in the cavity.. the calculations of decay > and distance traveled are then complicated by their Pythagorean > relationship to the spacetime inside these cavities traveling distances we > instwead perceive as dilation… but not just the dilation from their spatial > displacement, rather the cavities push this dilation in the opposite > direction and to some extent cancel? > > Always out on a limb, > > Fran > >
[Vo]:Article: Physicists Observe Rydberg Molecule for The First Time
Physicists Observe Rydberg Molecule for The First Time http://flip.it/s9Qrqn
Re: [Vo]:Holmlid, Mills & muons
What is the mode of "decay" of free muons and, separately, in condensed matter? They seem not to produce any high energy EM nor radioactive products. If they did, I would assume this would have been reported unless it was intended to remain a secret. I consider based on reported muon models of Hatt and Stubbs and deep elastic electron scattering experiments with muons and protons, electons and positrons should be observed during muon decay, if high energy gammas do not show up. Regarding these ideas, I question the designation of a muon as a lepton (primary) particle. The scattering experiments suggest a different type of particle--more akin to a proton or a neutron. Bob Cook From: Russ George Sent: Monday, November 14, 2016 9:57 AM To: vortex-l@eskimo.com Subject: RE: [Vo]:Holmlid, Mills & muons The idea that the muons are interacting in solid matter with the electrons not the nuclei of atoms is very compelling to me. Indeed this may well explain two mysteries of my cold fusion muon/mischegunons, that is that very few are escaping the experiment cells. That what I have detected is the dwindling remains of the reaction is very compelling and as well explains why so few cold fusion experiments have detected any such emanations. The time dilation effect that effectively increases the cross-section of materials just works very well indeed. This speaks to the growing revelations on silver being a valuable constituent in a range of experiments. Silver of course has a very complete electron cloud, as such it might well be the best material for engaging with the muon/mischugenon nuclear ash. This would help me a lot in understanding why it just happens that I have found silver so useful (as has Mills) it is not the neutron cross section of silver it is the muon cross-section! From: Bob Higgins [mailto:rj.bob.higg...@gmail.com] Sent: Monday, November 14, 2016 8:38 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Holmlid, Mills & muons In this discussion, Jones presumes muons to be traveling at light speed: The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. There are a number of things wrong with this. First, most commonly encountered muons are cosmogenic and have 100MeV-GeV energies. At these energies, the muon is traveling at a significant fraction of the speed of light (but not at the speed of light) and as such experiences time dilation in its decay. Because of time dilation, the stationary observer sees the cosmogenic muon decay to be much longer than 2.2 microseconds. This is why cosmogenic muons can travel 50-100 miles to the Earth's surface without having decayed. What Holmlid has reported is "10MeV/u" as a measurement for his muons - this is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is only an approximation for small velocity compared to c; as the velocity increases special relativity must be invoked in the solution. Special relativity would reduce the velocity from this equation as it started approaching c, so the actual velocity will be somewhat less than 0.1c for Holmlid's particles, and a slight time dilation would be experienced. So, if Holmlid's particles were muons, and if Mills was creating the same at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 meters. However, muons being charged, are well stopped in condensed matter because the particle doesn't have to run into a nucleus to be scattered, just run into the dense electronic orbitals. The more dense the condensed matter, the greater the stopping power for the muon. If muons were being generated with a v^2 of 10MeV/u, I doubt any would escape Mills' reactor vessel. On Sat, Nov 12, 2016 at 9:23 AM, Jones Beene mailto:jone...@pacbell.net>> wrote: For those who suspect that the Holmlid effect and the Mills effect are related, no matter what the proponents of each may think, here is a further thought from the fringe ... about one of the possible implications. Holmlid has suggested that a very high flux of muons can be produced by a subwatt laser beam. Mills uses an electric arc and will probably offer a real demo of the Suncell® at some point. No one doubts that it works but an extended demo will be needed... therefore, even if everything seen thus far is little more than PR fluff, we could have a worrisome situation in response to a much longer demo. Since Mills is applying higher net power to reactants (even if Holmlid's laser p
Re: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons
Axil's post is one interpretation of QM, other could be that the QM fields represents real fields e.g. no particles in space. This means that you can view QM as billiard with fields in stead of balls and things get to be much less mystic. Also Mills is starting to get real evidences of over unity now. With that comes his theory that after all have guided him to success, which means that when the suncell, if it works, start to get noticed, then Mills theory might as well become the standard way of interpretting physics. His theory have non of the mysteries in QM and can be viewed as billiard with fields in stead of balls using classical thinking. I myself are pretty certain that the theory are the best way to view the world but it is difficult to come to this conclusion. His book is hard to see through. On Mon, Nov 14, 2016 at 10:40 PM, Axil Axil wrote: > We are talking Quantum Mechanics here, not billards. In QM, > superposition means that the muon can be in many places at once while > it is in the entangled state. Distance does not matter. Where the muon > ends up is based on decoherence of what has entangled the muon with > the LENR reaction. It is all random and not predictable. > > A fundamental difference between classical physics and quantum theory > is the fact that, in the quantum world, certain predictions can only > be made in terms of probabilities > > A travelling particle > > As an example, take the question whether or not a particle that starts > at the time tA at the location A will reach location B at the later > time tB. > > Classical physics can give a definite answer. Depending on the > particle's initial velocity and the forces acting on it, the answer is > either yes or no. In quantum theory, it is merely possible to give the > probability that the particle in question can be detected at location > B at time tB. > > The path integral formalism, which was invented by the US physicist > Richard Feynman, is a tool for calculating such quantum mechanical > probabilities. Feynman's recipe, applied to a particle travelling from > A to B, is the following. > > Step 1: Consider all possibilities for the particle travelling from A > to B. Not only the boring straight-line approach, but also the > possibility of the particle turning loopings and making diverse > detours. > > There exists an infinity of possibilities. The particle can visit > New York, Ulan Bator, or even the moon or the Andromeda Galaxy before > arriving at its destination. Last but not least, it does not contain > information about velocities. The first part of the particle's > trajectory may be travelled at break-neck speed and the final > millimetres at a snail's pace - or the other way around, or completely > different; another infinity of possibilities. In short, for the first > step, take into account all ways of travelling from A to B, however > outlandish they may seem. > > The second step is to associate a number with each of these > possibilities (not quite the kind of number we're used to from school, > but we will not bother with the difference here). Finally, the numbers > associated with all possibilities are added up - some parts of the sum > canceling each other, others adding up. The resulting sum tells us the > probability of detecting the particle that started out at A at the > location B at the specified time. Physicists call such a sum over all > possibilities a path integral or sum over histories. > > > > > > > > > > On Mon, Nov 14, 2016 at 4:12 PM, Roarty, Francis X > wrote: > > Bob, what if the “muon” doesn’t have to achieve light speed but rather > > becomes so “suppressed” think traveling thru a tiny Casimir cavity that > the > > muons actual speed inside the cavity where vacuum wavelengths are dilate > by > > suppression appears to achieve negative light speed relative to > observers > > outside the cavity where vacuum wavelengths are not suppressed.. IMHO > > catlitic action is a weak cousin to Casimir action and the longer > > wavelengths we consider suppressed are actually still present from the > > perspective of a local observer in the cavity.. the calculations of decay > > and distance traveled are then complicated by their Pythagorean > relationship > > to the spacetime inside these cavities traveling distances we instwead > > perceive as dilation… but not just the dilation from their spatial > > displacement, rather the cavities push this dilation in the opposite > > direction and to some extent cancel? > > > > Always out on a limb, > > > > Fran > > > > From: Bob Higgins [mailto:rj.bob.higg...@gmail.com] > > Sent: Monday, November 14, 2016 11:38 AM > > To: vortex-l@eskimo.com > > Subject: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons > > > > > > > > In this discussion, Jones presumes muons to be traveling at light speed: > > > > The muon is an unstable fermion with a lifetime of 2.2 microseconds, > which > > is an eternity compared to most beta decays. Ignoring time dilation, this > > would
Re: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons
We are talking Quantum Mechanics here, not billards. In QM, superposition means that the muon can be in many places at once while it is in the entangled state. Distance does not matter. Where the muon ends up is based on decoherence of what has entangled the muon with the LENR reaction. It is all random and not predictable. A fundamental difference between classical physics and quantum theory is the fact that, in the quantum world, certain predictions can only be made in terms of probabilities A travelling particle As an example, take the question whether or not a particle that starts at the time tA at the location A will reach location B at the later time tB. Classical physics can give a definite answer. Depending on the particle's initial velocity and the forces acting on it, the answer is either yes or no. In quantum theory, it is merely possible to give the probability that the particle in question can be detected at location B at time tB. The path integral formalism, which was invented by the US physicist Richard Feynman, is a tool for calculating such quantum mechanical probabilities. Feynman's recipe, applied to a particle travelling from A to B, is the following. Step 1: Consider all possibilities for the particle travelling from A to B. Not only the boring straight-line approach, but also the possibility of the particle turning loopings and making diverse detours. There exists an infinity of possibilities. The particle can visit New York, Ulan Bator, or even the moon or the Andromeda Galaxy before arriving at its destination. Last but not least, it does not contain information about velocities. The first part of the particle's trajectory may be travelled at break-neck speed and the final millimetres at a snail's pace - or the other way around, or completely different; another infinity of possibilities. In short, for the first step, take into account all ways of travelling from A to B, however outlandish they may seem. The second step is to associate a number with each of these possibilities (not quite the kind of number we're used to from school, but we will not bother with the difference here). Finally, the numbers associated with all possibilities are added up - some parts of the sum canceling each other, others adding up. The resulting sum tells us the probability of detecting the particle that started out at A at the location B at the specified time. Physicists call such a sum over all possibilities a path integral or sum over histories. On Mon, Nov 14, 2016 at 4:12 PM, Roarty, Francis X wrote: > Bob, what if the “muon” doesn’t have to achieve light speed but rather > becomes so “suppressed” think traveling thru a tiny Casimir cavity that the > muons actual speed inside the cavity where vacuum wavelengths are dilate by > suppression appears to achieve negative light speed relative to observers > outside the cavity where vacuum wavelengths are not suppressed.. IMHO > catlitic action is a weak cousin to Casimir action and the longer > wavelengths we consider suppressed are actually still present from the > perspective of a local observer in the cavity.. the calculations of decay > and distance traveled are then complicated by their Pythagorean relationship > to the spacetime inside these cavities traveling distances we instwead > perceive as dilation… but not just the dilation from their spatial > displacement, rather the cavities push this dilation in the opposite > direction and to some extent cancel? > > Always out on a limb, > > Fran > > From: Bob Higgins [mailto:rj.bob.higg...@gmail.com] > Sent: Monday, November 14, 2016 11:38 AM > To: vortex-l@eskimo.com > Subject: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons > > > > In this discussion, Jones presumes muons to be traveling at light speed: > > The muon is an unstable fermion with a lifetime of 2.2 microseconds, which > is an eternity compared to most beta decays. Ignoring time dilation, this > would mean that muons, travelling at light speed, would be dispersing and > decaying in an imaginary sphere about 600 meters from the reactor. > > > > There are a number of things wrong with this. First, most commonly > encountered muons are cosmogenic and have 100MeV-GeV energies. At these > energies, the muon is traveling at a significant fraction of the speed of > light (but not at the speed of light) and as such experiences time dilation > in its decay. Because of time dilation, the stationary observer sees the > cosmogenic muon decay to be much longer than 2.2 microseconds. This is why > cosmogenic muons can travel 50-100 miles to the Earth's surface without > having decayed. > > What Holmlid has reported is "10MeV/u" as a measurement for his muons - this > is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. > In Holmlid's units of measure (MeV/u), call the amount measured X, then the > velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a > measure of 10 MeV/u, one gets sqrt(10/931)*
RE: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons
Bob, what if the “muon” doesn’t have to achieve light speed but rather becomes so “suppressed” think traveling thru a tiny Casimir cavity that the muons actual speed inside the cavity where vacuum wavelengths are dilate by suppression appears to achieve negative light speed relative to observers outside the cavity where vacuum wavelengths are not suppressed.. IMHO catlitic action is a weak cousin to Casimir action and the longer wavelengths we consider suppressed are actually still present from the perspective of a local observer in the cavity.. the calculations of decay and distance traveled are then complicated by their Pythagorean relationship to the spacetime inside these cavities traveling distances we instwead perceive as dilation… but not just the dilation from their spatial displacement, rather the cavities push this dilation in the opposite direction and to some extent cancel? Always out on a limb, Fran From: Bob Higgins [mailto:rj.bob.higg...@gmail.com] Sent: Monday, November 14, 2016 11:38 AM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons In this discussion, Jones presumes muons to be traveling at light speed: The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. There are a number of things wrong with this. First, most commonly encountered muons are cosmogenic and have 100MeV-GeV energies. At these energies, the muon is traveling at a significant fraction of the speed of light (but not at the speed of light) and as such experiences time dilation in its decay. Because of time dilation, the stationary observer sees the cosmogenic muon decay to be much longer than 2.2 microseconds. This is why cosmogenic muons can travel 50-100 miles to the Earth's surface without having decayed. What Holmlid has reported is "10MeV/u" as a measurement for his muons - this is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is only an approximation for small velocity compared to c; as the velocity increases special relativity must be invoked in the solution. Special relativity would reduce the velocity from this equation as it started approaching c, so the actual velocity will be somewhat less than 0.1c for Holmlid's particles, and a slight time dilation would be experienced. So, if Holmlid's particles were muons, and if Mills was creating the same at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 meters. However, muons being charged, are well stopped in condensed matter because the particle doesn't have to run into a nucleus to be scattered, just run into the dense electronic orbitals. The more dense the condensed matter, the greater the stopping power for the muon. If muons were being generated with a v^2 of 10MeV/u, I doubt any would escape Mills' reactor vessel. On Sat, Nov 12, 2016 at 9:23 AM, Jones Beene mailto:jone...@pacbell.net>> wrote: For those who suspect that the Holmlid effect and the Mills effect are related, no matter what the proponents of each may think, here is a further thought from the fringe … about one of the possible implications. Holmlid has suggested that a very high flux of muons can be produced by a subwatt laser beam. Mills uses an electric arc and will probably offer a real demo of the Suncell® at some point. No one doubts that it works but an extended demo will be needed… therefore, even if everything seen thus far is little more than PR fluff, we could have a worrisome situation in response to a much longer demo. Since Mills is applying higher net power to reactants (even if Holmlid’s laser provides more localized power) there is a chance that some portion of the energy produced escapes the sun-cell as muons. If Holmlid gets millions of muons per watt of coherent light, what will be the corresponding rate be from an electric arc? If anything like this scenario turns out to be the accurate, then any muons produced will decay at a predictable distance away from the reactor, thus they could have been missed by BrLP in testing thus far. The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. Thus, the effect of radioactive decay could be significant at unexpected distance– and Mills may never had imagined that this is a problem. Fortunately, humans are exposed to a constant flu
[Vo]:LENR INFO tomorrow contacting ACS
http://egooutpeters.blogspot.ro/2016/11/nov-14-2016-lenr-info.html peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
RE: [Vo]:Holmlid, Mills & muons
The idea that the muons are interacting in solid matter with the electrons not the nuclei of atoms is very compelling to me. Indeed this may well explain two mysteries of my cold fusion muon/mischegunons, that is that very few are escaping the experiment cells. That what I have detected is the dwindling remains of the reaction is very compelling and as well explains why so few cold fusion experiments have detected any such emanations. The time dilation effect that effectively increases the cross-section of materials just works very well indeed. This speaks to the growing revelations on silver being a valuable constituent in a range of experiments. Silver of course has a very complete electron cloud, as such it might well be the best material for engaging with the muon/mischugenon nuclear ash. This would help me a lot in understanding why it just happens that I have found silver so useful (as has Mills) it is not the neutron cross section of silver it is the muon cross-section! From: Bob Higgins [mailto:rj.bob.higg...@gmail.com] Sent: Monday, November 14, 2016 8:38 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Holmlid, Mills & muons In this discussion, Jones presumes muons to be traveling at light speed: The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. There are a number of things wrong with this. First, most commonly encountered muons are cosmogenic and have 100MeV-GeV energies. At these energies, the muon is traveling at a significant fraction of the speed of light (but not at the speed of light) and as such experiences time dilation in its decay. Because of time dilation, the stationary observer sees the cosmogenic muon decay to be much longer than 2.2 microseconds. This is why cosmogenic muons can travel 50-100 miles to the Earth's surface without having decayed. What Holmlid has reported is "10MeV/u" as a measurement for his muons - this is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is only an approximation for small velocity compared to c; as the velocity increases special relativity must be invoked in the solution. Special relativity would reduce the velocity from this equation as it started approaching c, so the actual velocity will be somewhat less than 0.1c for Holmlid's particles, and a slight time dilation would be experienced. So, if Holmlid's particles were muons, and if Mills was creating the same at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 meters. However, muons being charged, are well stopped in condensed matter because the particle doesn't have to run into a nucleus to be scattered, just run into the dense electronic orbitals. The more dense the condensed matter, the greater the stopping power for the muon. If muons were being generated with a v^2 of 10MeV/u, I doubt any would escape Mills' reactor vessel. On Sat, Nov 12, 2016 at 9:23 AM, Jones Beene mailto:jone...@pacbell.net> > wrote: For those who suspect that the Holmlid effect and the Mills effect are related, no matter what the proponents of each may think, here is a further thought from the fringe … about one of the possible implications. Holmlid has suggested that a very high flux of muons can be produced by a subwatt laser beam. Mills uses an electric arc and will probably offer a real demo of the Suncell® at some point. No one doubts that it works but an extended demo will be needed… therefore, even if everything seen thus far is little more than PR fluff, we could have a worrisome situation in response to a much longer demo. Since Mills is applying higher net power to reactants (even if Holmlid’s laser provides more localized power) there is a chance that some portion of the energy produced escapes the sun-cell as muons. If Holmlid gets millions of muons per watt of coherent light, what will be the corresponding rate be from an electric arc? If anything like this scenario turns out to be the accurate, then any muons produced will decay at a predictable distance away from the reactor, thus they could have been missed by BrLP in testing thus far. The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. Thus, the effect of radioactive decay could be significant at unexpected distance– and Mills may never ha
RE: [Vo]:Holmlid, Mills & muons
Bob, You are conflating two or more different Holmlid papers… He is unambiguous. The 10MeV particles are clearly stated to be “mainly protons from the fusion process and deuterons ejected by proton collisions” (see the abstract from the paper). The muon observations are from other papers, not this one… Jones The 10 MeV paper is here: https://arxiv.org/abs/1302.2781 From: Bob Higgins What Holmlid has reported is "10MeV/u" as a measurement for his muons - this is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is only an approximation for small velocity compared to c; as the velocity increases special relativity must be invoked in the solution. Special relativity would reduce the velocity from this equation as it started approaching c, so the actual velocity will be somewhat less than 0.1c for Holmlid's particles, and a slight time dilation would be experienced. So, if Holmlid's particles were muons, and if Mills was creating the same at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 meters. However, muons being charged, are well stopped in condensed matter because the particle doesn't have to run into a nucleus to be scattered, just run into the dense electronic orbitals. The more dense the condensed matter, the greater the stopping power for the muon. If muons were being generated with a v^2 of 10MeV/u, I doubt any would escape Mills' reactor vessel. On Sat, Nov 12, 2016 at 9:23 AM, Jones Beene wrote: For those who suspect that the Holmlid effect and the Mills effect are related, no matter what the proponents of each may think, here is a further thought from the fringe … about one of the possible implications. Holmlid has suggested that a very high flux of muons can be produced by a subwatt laser beam. Mills uses an electric arc and will probably offer a real demo of the Suncell® at some point. No one doubts that it works but an extended demo will be needed… therefore, even if everything seen thus far is little more than PR fluff, we could have a worrisome situation in response to a much longer demo. Since Mills is applying higher net power to reactants (even if Holmlid’s laser provides more localized power) there is a chance that some portion of the energy produced escapes the sun-cell as muons. If Holmlid gets millions of muons per watt of coherent light, what will be the corresponding rate be from an electric arc? If anything like this scenario turns out to be the accurate, then any muons produced will decay at a predictable distance away from the reactor, thus they could have been missed by BrLP in testing thus far. The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. Thus, the effect of radioactive decay could be significant at unexpected distance– and Mills may never had imagined that this is a problem. Fortunately, humans are exposed to a constant flux of muons due to cosmic rays, and the flux is well-tolerated. Nevertheless, this detail is worth noting – and should Mills or his associates start to feel a bit ill from the exposure – possibly an unseasonal sun tan, then we can identify a culprit. The effects could be felt more in a remote office - than in the lab … which is curious.
RE: [Vo]:Holmlid, Mills & muons
From: Bob Higgins In this discussion, Jones presumes muons to be traveling at light speed… That is an oversight. It should read “a large fraction of light speed”… .
Re: [Vo]:Holmlid, Mills & muons
In this discussion, Jones presumes muons to be traveling at light speed: The muon is an unstable fermion with a lifetime of 2.2 microseconds, which is an eternity compared to most beta decays. Ignoring time dilation, this would mean that muons, travelling at light speed, would be dispersing and decaying in an imaginary sphere about 600 meters from the reactor. There are a number of things wrong with this. First, most commonly encountered muons are cosmogenic and have 100MeV-GeV energies. At these energies, the muon is traveling at a significant fraction of the speed of light (but not at the speed of light) and as such experiences time dilation in its decay. Because of time dilation, the stationary observer sees the cosmogenic muon decay to be much longer than 2.2 microseconds. This is why cosmogenic muons can travel 50-100 miles to the Earth's surface without having decayed. What Holmlid has reported is "10MeV/u" as a measurement for his muons - this is a measure of velocity squared. One u (atomic mass unit) is 931 MeV/c^2. In Holmlid's units of measure (MeV/u), call the amount measured X, then the velocity of the particle is sqrt(X/931)*c. For Holmlid's report of a measure of 10 MeV/u, one gets sqrt(10/931)*c = 0.104c. This is only an approximation for small velocity compared to c; as the velocity increases special relativity must be invoked in the solution. Special relativity would reduce the velocity from this equation as it started approaching c, so the actual velocity will be somewhat less than 0.1c for Holmlid's particles, and a slight time dilation would be experienced. So, if Holmlid's particles were muons, and if Mills was creating the same at a v^2 of 10MeV/u, then the range in a vacuum would be on the order of 60 meters. However, muons being charged, are well stopped in condensed matter because the particle doesn't have to run into a nucleus to be scattered, just run into the dense electronic orbitals. The more dense the condensed matter, the greater the stopping power for the muon. If muons were being generated with a v^2 of 10MeV/u, I doubt any would escape Mills' reactor vessel. On Sat, Nov 12, 2016 at 9:23 AM, Jones Beene wrote: > For those who suspect that the Holmlid effect and the Mills effect are > related, no matter what the proponents of each may think, here is a further > thought from the fringe … about one of the possible implications. Holmlid > has suggested that a very high flux of muons can be produced by a subwatt > laser beam. > > Mills uses an electric arc and will probably offer a real demo of the S > uncell® at some point. No one doubts that it works but an extended demo > will be needed… therefore, even if everything seen thus far is little > more than PR fluff, we could have a worrisome situation in response to a > much longer demo. > > Since Mills is applying higher net power to reactants (even if Holmlid’s > laser provides more localized power) there is a chance that some portion of > the energy produced escapes the sun-cell as muons. If Holmlid gets > millions of muons per watt of coherent light, what will be the > corresponding rate be from an electric arc? If anything like this scenario > turns > out to be the accurate, then any muons produced will decay at a > predictable distance away from the reactor, thus they could have been > missed by BrLP in testing thus far. > > The muon is an unstable fermion with a lifetime of 2.2 microseconds, > which is an eternity compared to most beta decays. Ignoring time > dilation, this would mean that muons, travelling at light speed, would be > dispersing and decaying in an imaginary sphere about 600 meters from the > reactor. Thus, the effect of radioactive decay could be significant at > unexpected distance– and Mills may never had imagined that this is a > problem. Fortunately, humans are exposed to a constant flux of muons due > to cosmic rays, and the flux is well-tolerated. > > Nevertheless, this detail is worth noting – and should Mills or his > associates start to feel a bit ill from the exposure – possibly an > unseasonal sun tan, then we can identify a culprit. > > The effects could be felt more in a remote office - than in the lab … > which is curious. > >
Re: [Vo]:Holmlid, Mills & muons
Hi all EDITED TO CORRECT AN ARRITHMETIC ERROR! ~600 m not 6000m By the way the density of the incidents has to be distributed across a sphere that is approximately 1,440,000 π (pi) meters squared. Then you have to plug in the distribution curve to get cubed meters for area. The numbers are very big Hence why I think the density will be very small. It is also why I think putting dense shielding round such a source may increase the reaction density in a smaller sphere making the effect more measurable but why I think putting shielding round such a source may be more dangerous than letting the such a source propagate out to a safe dispersal range. If LENR works in the way suggested it may be that rules about no lead tungsten within x meters might apply. Unless we go for Axil's 10ft dense walls option. Have to so the math. Kind Regards walker On 14 November 2016 at 13:13, Ian Walker wrote: > Hi all > > By the way the density of the incidents has to be distributed across a > sphere that is approximately 144,000,000 π (pi) meters squared. > > Then you have to plug in the distribution curve to get cubed meters for > area. > > The numbers are very big > > Hence why I think the density will be very small. > > It is also why I think putting dense shielding round such a source may > increase the reaction density in a smaller sphere making the effect more > measurable but why I think putting shielding round such a source may be > more dangerous than letting the such a source propagate out to a safe > dispersal range. If LENR works in the way suggested it may be that rules > about no lead tungsten within x meters might apply. Unless we go for Axil's > 10ft dense walls option. > > Have to so the math. > > Kind Regards walker > > On 14 November 2016 at 12:49, Ian Walker wrote: > >> Hi all >> >> With that size of sphere, 6000m radius, I am guessing, from experience >> the density of interactions will be only a little above natural background. >> You need to know the surface area of the sphere. Then the distribution >> curve for the straight line from the source; then calculate peak and the >> nominal width of the curve, probably a narrow bell curve. >> >> I did some work on ballistics, including terminal ballistics, looking at >> shrapnel density and effective radius of devices, chance of a hit at a >> certain range from the explosion. These reduce to a near statistically zero >> probability on a logarithmic curve as you progress further from the point >> source. You alter the force of the terminal explosive to produce shrapnel >> that is still travelling at killing speed at a density of one hit per >> person size. Having the shrapnel still moving at killing speed beyond this >> range, is a waste of explosive charge and increase the risk of collateral >> damage (killing those you had not intended) so you set the charge fit for >> purpose. >> >> The effect we are looking at is similar. >> >> But the key thing is that the sphere will describe a circle round the >> source, varying due to density of objects like walls in the path that is >> centred on the source. This would be the experiment to do. >> >> As I said spread across such a large sphere the density will be very low. >> >> Slowing down the particles with dense shielding materials would decrease >> the size of the sphere at that direction and increase the density of the >> radiation at the calculable distance from the source. This would give proof >> of the particle nature. >> >> Kind Regards walker >> >> On 14 November 2016 at 04:12, Axil Axil wrote: >> >>> mischugnons... >>> >>> I might know what they are. They have made themselves visible in the >>> research of Keith Fredericks that can be found here: >>> >>> http://restframe.com/ >>> >>> I have described the mischugnons as metalized hydrogen crystals and >>> how they work, how they store GeV levels of power, how they manifest a >>> monopole field, and how they catalyze the LENR reaction. Their >>> description starts with Holmlid, shows how the metallic hydrogen's >>> structure produces spin waves through hole superconductivity and >>> whispering gallery wave, how they can store massive amounts of energy, >>> and how that energy can be projected as monopole flux lines to >>> catalyzed proton and neutron weak force decay to produce mesons as >>> seen by Holmlid. >>> >>> Keith Fredericks calls the tachyons but they are just a monopole like >>> quasiparticle that Holmlid and LENR reactors can created using a >>> catalyst. >>> >>> >>> >>> >>> >>> >>> On Sun, Nov 13, 2016 at 7:13 PM, Russ George >>> wrote: >>> > In many many experiments over the years the mischugnons have made their >>> > presence irrefutably known. It is a thrilling time just now in cold >>> fusion >>> > as there are many confirmations and affirmations of the choirs >>> existence, >>> > we’ve been hearing their voices for nearly 30 years and just now the >>> > theatrical smoke is beginning to clear just enough that we can see the >>>
Re: [Vo]:Holmlid, Mills & muons
Hi all By the way the density of the incidents has to be distributed across a sphere that is approximately 144,000,000 π (pi) meters squared. Then you have to plug in the distribution curve to get cubed meters for area. The numbers are very big Hence why I think the density will be very small. It is also why I think putting dense shielding round such a source may increase the reaction density in a smaller sphere making the effect more measurable but why I think putting shielding round such a source may be more dangerous than letting the such a source propagate out to a safe dispersal range. If LENR works in the way suggested it may be that rules about no lead tungsten within x meters might apply. Unless we go for Axil's 10ft dense walls option. Have to so the math. Kind Regards walker On 14 November 2016 at 12:49, Ian Walker wrote: > Hi all > > With that size of sphere, 6000m radius, I am guessing, from experience the > density of interactions will be only a little above natural background. You > need to know the surface area of the sphere. Then the distribution curve > for the straight line from the source; then calculate peak and the nominal > width of the curve, probably a narrow bell curve. > > I did some work on ballistics, including terminal ballistics, looking at > shrapnel density and effective radius of devices, chance of a hit at a > certain range from the explosion. These reduce to a near statistically zero > probability on a logarithmic curve as you progress further from the point > source. You alter the force of the terminal explosive to produce shrapnel > that is still travelling at killing speed at a density of one hit per > person size. Having the shrapnel still moving at killing speed beyond this > range, is a waste of explosive charge and increase the risk of collateral > damage (killing those you had not intended) so you set the charge fit for > purpose. > > The effect we are looking at is similar. > > But the key thing is that the sphere will describe a circle round the > source, varying due to density of objects like walls in the path that is > centred on the source. This would be the experiment to do. > > As I said spread across such a large sphere the density will be very low. > > Slowing down the particles with dense shielding materials would decrease > the size of the sphere at that direction and increase the density of the > radiation at the calculable distance from the source. This would give proof > of the particle nature. > > Kind Regards walker > > On 14 November 2016 at 04:12, Axil Axil wrote: > >> mischugnons... >> >> I might know what they are. They have made themselves visible in the >> research of Keith Fredericks that can be found here: >> >> http://restframe.com/ >> >> I have described the mischugnons as metalized hydrogen crystals and >> how they work, how they store GeV levels of power, how they manifest a >> monopole field, and how they catalyze the LENR reaction. Their >> description starts with Holmlid, shows how the metallic hydrogen's >> structure produces spin waves through hole superconductivity and >> whispering gallery wave, how they can store massive amounts of energy, >> and how that energy can be projected as monopole flux lines to >> catalyzed proton and neutron weak force decay to produce mesons as >> seen by Holmlid. >> >> Keith Fredericks calls the tachyons but they are just a monopole like >> quasiparticle that Holmlid and LENR reactors can created using a >> catalyst. >> >> >> >> >> >> >> On Sun, Nov 13, 2016 at 7:13 PM, Russ George >> wrote: >> > In many many experiments over the years the mischugnons have made their >> > presence irrefutably known. It is a thrilling time just now in cold >> fusion >> > as there are many confirmations and affirmations of the choirs >> existence, >> > we’ve been hearing their voices for nearly 30 years and just now the >> > theatrical smoke is beginning to clear just enough that we can see the >> > outlines of the choir, it’s a big one. It’s not the single voices that >> make >> > the music of the choir so wonderful it is the combination of them all. >> > Perhaps it is a Gregorian harmony they are singing. >> > >> > >> > >> > >> > >> > From: Eric Walker [mailto:eric.wal...@gmail.com] >> > Sent: Sunday, November 13, 2016 3:44 PM >> > To: vortex-l@eskimo.com >> > Subject: Re: [Vo]:Holmlid, Mills & muons >> > >> > >> > >> > Ok. So you've survived the stinkers and the peanut gallery and the >> > charlatans, the high priests, the prelates and the faithful of >> physics. In >> > your own experiments you've seen muons or mischugenon. >> > >> > >> > >> > On Sun, Nov 13, 2016 at 5:32 PM, Russ George >> wrote: >> > >> > >> > >> > What is interesting is that the real data has always shone most brightly >> > even when the signal was incredibly poorly understood. That’s the >> benefit of >> > longevity and dedication the real shining bits tend to agglomerate into >> an >> > understandable thing. Such is the case
Re: [Vo]:Holmlid, Mills & muons
Hi all With that size of sphere, 6000m radius, I am guessing, from experience the density of interactions will be only a little above natural background. You need to know the surface area of the sphere. Then the distribution curve for the straight line from the source; then calculate peak and the nominal width of the curve, probably a narrow bell curve. I did some work on ballistics, including terminal ballistics, looking at shrapnel density and effective radius of devices, chance of a hit at a certain range from the explosion. These reduce to a near statistically zero probability on a logarithmic curve as you progress further from the point source. You alter the force of the terminal explosive to produce shrapnel that is still travelling at killing speed at a density of one hit per person size. Having the shrapnel still moving at killing speed beyond this range, is a waste of explosive charge and increase the risk of collateral damage (killing those you had not intended) so you set the charge fit for purpose. The effect we are looking at is similar. But the key thing is that the sphere will describe a circle round the source, varying due to density of objects like walls in the path that is centred on the source. This would be the experiment to do. As I said spread across such a large sphere the density will be very low. Slowing down the particles with dense shielding materials would decrease the size of the sphere at that direction and increase the density of the radiation at the calculable distance from the source. This would give proof of the particle nature. Kind Regards walker On 14 November 2016 at 04:12, Axil Axil wrote: > mischugnons... > > I might know what they are. They have made themselves visible in the > research of Keith Fredericks that can be found here: > > http://restframe.com/ > > I have described the mischugnons as metalized hydrogen crystals and > how they work, how they store GeV levels of power, how they manifest a > monopole field, and how they catalyze the LENR reaction. Their > description starts with Holmlid, shows how the metallic hydrogen's > structure produces spin waves through hole superconductivity and > whispering gallery wave, how they can store massive amounts of energy, > and how that energy can be projected as monopole flux lines to > catalyzed proton and neutron weak force decay to produce mesons as > seen by Holmlid. > > Keith Fredericks calls the tachyons but they are just a monopole like > quasiparticle that Holmlid and LENR reactors can created using a > catalyst. > > > > > > > On Sun, Nov 13, 2016 at 7:13 PM, Russ George > wrote: > > In many many experiments over the years the mischugnons have made their > > presence irrefutably known. It is a thrilling time just now in cold > fusion > > as there are many confirmations and affirmations of the choirs existence, > > we’ve been hearing their voices for nearly 30 years and just now the > > theatrical smoke is beginning to clear just enough that we can see the > > outlines of the choir, it’s a big one. It’s not the single voices that > make > > the music of the choir so wonderful it is the combination of them all. > > Perhaps it is a Gregorian harmony they are singing. > > > > > > > > > > > > From: Eric Walker [mailto:eric.wal...@gmail.com] > > Sent: Sunday, November 13, 2016 3:44 PM > > To: vortex-l@eskimo.com > > Subject: Re: [Vo]:Holmlid, Mills & muons > > > > > > > > Ok. So you've survived the stinkers and the peanut gallery and the > > charlatans, the high priests, the prelates and the faithful of physics. > In > > your own experiments you've seen muons or mischugenon. > > > > > > > > On Sun, Nov 13, 2016 at 5:32 PM, Russ George > wrote: > > > > > > > > What is interesting is that the real data has always shone most brightly > > even when the signal was incredibly poorly understood. That’s the > benefit of > > longevity and dedication the real shining bits tend to agglomerate into > an > > understandable thing. Such is the case it seems with Holmlid’s ‘muons’, > > there are too many coincidences coming together to ignore his > contributions > > to what is becoming a choir. > > > > > > > > What are those coincidences that lead one inevitably to the conclusion > that > > Holmlid is seeing muons, and that he's seeing the same thing you believe > > you've been seeing? You speak with enough confidence to lead me to > believe > > that you've read his work, are quite familiar with it and are able to > > support your position with concrete details. > > > > > > > > As for being the tutor or free simple sound-bite tour-guide sorry I have > > neither the time nor inclination to help the reluctant. There is so much > to > > do and so little time to do it. As Thomas Edison so aptly put it long > ago, > > “The thing I lose patience with most is the clock, its hands move too > fast.” > > > > > > > > Alas it's not for my edification that you should answer these questions. > > It's for your own credibility! You'v