Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 6:04 PM, mix...@bigpond.com wrote: In reply to Harry Veeder's message of Sat, 23 Mar 2013 04:04:55 -0400: Hi, [snip] The weak force doesn't actually present a barrier. It presents a chance that something will occur. Electrons and protons don't normally combine into neutrons because their combined mass is inadequate. It's 782 keV short of the mass of a free neutron. However, they could combine to form a reduced mass neutron as part of the nucleus of a heavier atom. This does in fact happen with some isotopes. It's called electron capture (EC). In this case, even though the mass of the proton is also reduced, the net result (an isotope of the previous element in the periodic table), is sufficiently more stable than the initial isotope to more than make up for the difference. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html ok, so they don't normally combine because the rules of QM forbid it or at least say it is highly improbable below some energy/temperature level. It's actually classically forbidden. 1 baseball + a second baseball does not make 3 baseballs. I don't understand your analogy. Aren't we talking about 1e combining with 1p to make 1n? So QM provides the barrier, but I presume it becomes more likely if you substantially increase the kinetic energy of the two particles. The necessary kinetic energy is converted into the extra mass required to form the neutron and this is, roughly speaking, what the W-L theory proposes. Only very roughly. They don't actually explain where the extra energy comes from. Note that *extra* is about 1 1/2 times the mass of an electron. However, it would be interesting to speculate if the kind of low energy electron capture you describe could happen more frequently. In other words, If reduced mass neutrons could be made easily (from a free electron and free proton/deuteron) would this one miracle be able to explain observations without the additional miracles required by the W-L theory. Note that they can only be made *in* another nucleus, or at the very least, very, very close to it, such that the ensuing neutron(s) are immediately captured. I am just speculating on the possibility that ionized hydrogen could be made into a reduced mass neutron since the reduced mass neutron won't produce radioactive isotopes needing gamma shielding. Ionized hydrogen, i.e. a proton, would become a reduced mass neutron by colliding with a free electron. As the electron approaches the proton it would be on a trajectory where the combination is a inevitable outcome. This scenario is based on on a analogy from celestial mechanics where different approach trajectories can result in different outcomes: collision, stable orbit, escape. In most environments, free electrons and protons form atoms, i.e. systems with stable orbits. However, the environment of some lattices would tend to channel protons and electrons on to paths such that they are bound to collide. This is included in Horace's theory, see http://www.mtaonline.net/~hheffner/DeflationFusion.pdf and also a possibility with Hydrino fusion. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Yes many CF theorys involve neutrons or an entity that appears neutrally charged from the point of view of another nucleus. Harry
Re: [Vo]:Graneau Questions
In reply to Harry Veeder's message of Sun, 24 Mar 2013 20:36:40 -0400: Hi, [snip] It's actually classically forbidden. 1 baseball + a second baseball does not make 3 baseballs. I don't understand your analogy. Aren't we talking about 1e combining with 1p to make 1n? Yes, but that's the problem. 1p + 1e is not enough to make 1n. In fact you would need the mass of about an extra 1 1/2 electrons to make an n. e.g. 1p + 2 1/2 e ~= 1n, but then you have too much negative charge. In order to make 1n from 1p and only 1e, at least one of the two needs to be moving very fast before the collision, so that the kinetic energy can be converted to the needed extra mass, according to m = E/c^2. [snip] Note that they can only be made *in* another nucleus, or at the very least, very, very close to it, such that the ensuing neutron(s) are immediately captured. I am just speculating on the possibility that ionized hydrogen could be made into a reduced mass neutron since the reduced mass neutron won't produce radioactive isotopes needing gamma shielding. Ionized hydrogen, i.e. a proton, would become a reduced mass neutron by colliding with a free electron. As the electron approaches the proton it would be on a trajectory where the combination is a inevitable outcome. This scenario is based on on a analogy from celestial mechanics where different approach trajectories can result in different outcomes: collision, stable orbit, escape. In most environments, free electrons and protons form atoms, i.e. systems with stable orbits. However, the environment of some lattices would tend to channel protons and electrons on to paths such that they are bound to collide. They are also oppositely charged, so they naturally attract one another. That improves the chances of a collision a lot. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sun, Mar 24, 2013 at 10:50 PM, mix...@bigpond.com wrote: In reply to Harry Veeder's message of Sun, 24 Mar 2013 20:36:40 -0400: Hi, [snip] It's actually classically forbidden. 1 baseball + a second baseball does not make 3 baseballs. I don't understand your analogy. Aren't we talking about 1e combining with 1p to make 1n? Yes, but that's the problem. 1p + 1e is not enough to make 1n. In fact you would need the mass of about an extra 1 1/2 electrons to make an n. e.g. 1p + 2 1/2 e ~= 1n, but then you have too much negative charge. In order to make 1n from 1p and only 1e, at least one of the two needs to be moving very fast before the collision, so that the kinetic energy can be converted to the needed extra mass, according to m = E/c^2. Ok, I will give up on formation of reduce mass neutrons outside a nucleus. Instead I will focus on how an electron and proton could collide to form neutral-like entity. [snip] Note that they can only be made *in* another nucleus, or at the very least, very, very close to it, such that the ensuing neutron(s) are immediately captured. I am just speculating on the possibility that ionized hydrogen could be made into a reduced mass neutron since the reduced mass neutron won't produce radioactive isotopes needing gamma shielding. Ionized hydrogen, i.e. a proton, would become a reduced mass neutron by colliding with a free electron. As the electron approaches the proton it would be on a trajectory where the combination is a inevitable outcome. This scenario is based on on a analogy from celestial mechanics where different approach trajectories can result in different outcomes: collision, stable orbit, escape. In most environments, free electrons and protons form atoms, i.e. systems with stable orbits. However, the environment of some lattices would tend to channel protons and electrons on to paths such that they are bound to collide. They are also oppositely charged, so they naturally attract one another. That improves the chances of a collision a lot. Yes, like asteriods and planets. harry
Re: [Vo]:Graneau Questions
On Wed, Mar 20, 2013 at 4:14 PM, mix...@bigpond.com wrote: In reply to Harry Veeder's message of Mon, 18 Mar 2013 21:40:15 -0400: Hi, [snip] On Mon, Mar 18, 2013 at 5:10 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sun, 17 Mar 2013 22:56:07 -0700: Hi, [snip] On Sun, Mar 17, 2013 at 2:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. I take it that when physicists refer to a potential barrier, they mean specifically an electrostatic potential barrier, and not simply an energy threshold that must be overcome? Eric A barrier usually implies an impediment that gets in the way of a reaction that would otherwise release energy. However the formation of a neutron from a proton and an electron does not release energy, it consumes it. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Some sort of barrier is necessary to stymie the mutual attraction of electrons and protons otherwise matter would quickly consist of nothing but neutrons. Isn't this barrier supplied by the weak force? The weak force doesn't actually present a barrier. It presents a chance that something will occur. Electrons and protons don't normally combine into neutrons because their combined mass is inadequate. It's 782 keV short of the mass of a free neutron. However, they could combine to form a reduced mass neutron as part of the nucleus of a heavier atom. This does in fact happen with some isotopes. It's called electron capture (EC). In this case, even though the mass of the proton is also reduced, the net result (an isotope of the previous element in the periodic table), is sufficiently more stable than the initial isotope to more than make up for the difference. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html ok, so they don't normally combine because the rules of QM forbid it or at least say it is highly improbable below some energy/temperature level. So QM provides the barrier, but I presume it becomes more likely if you substantially increase the kinetic energy of the two particles. The necessary kinetic energy is converted into the extra mass required to form the neutron and this is, roughly speaking, what the W-L theory proposes. However, it would be interesting to speculate if the kind of low energy electron capture you describe could happen more frequently. In other words, If reduced mass neutrons could be made easily (from a free electron and free proton/deuteron) would this one miracle be able to explain observations without the additional miracles required by the W-L theory. Harry
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Wed, 20 Mar 2013 19:41:08 -0400: Hi, [snip] On Wed, Mar 20, 2013 at 7:38 PM, Terry Blanton hohlr...@gmail.com wrote: On Wed, Mar 20, 2013 at 7:37 PM, Terry Blanton hohlr...@gmail.com wrote: On Wed, Mar 20, 2013 at 4:03 PM, mix...@bigpond.com wrote: I would much appreciate a reference to the measurement of the *magnetic moment* of *free* electrons. Isn't that how a cathode ray tube works? Have you ever put a permanent magnet in front of one? Without a magnetic moment, how do you scan the tube? This is due to the fact that an electron undergoing translational motion creates a magnetic field. It isn't an indication that the electron is rotating on it's own axis, and thus has an intrinsic magnetic field. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to Harry Veeder's message of Sat, 23 Mar 2013 04:04:55 -0400: Hi, [snip] The weak force doesn't actually present a barrier. It presents a chance that something will occur. Electrons and protons don't normally combine into neutrons because their combined mass is inadequate. It's 782 keV short of the mass of a free neutron. However, they could combine to form a reduced mass neutron as part of the nucleus of a heavier atom. This does in fact happen with some isotopes. It's called electron capture (EC). In this case, even though the mass of the proton is also reduced, the net result (an isotope of the previous element in the periodic table), is sufficiently more stable than the initial isotope to more than make up for the difference. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html ok, so they don't normally combine because the rules of QM forbid it or at least say it is highly improbable below some energy/temperature level. It's actually classically forbidden. 1 baseball + a second baseball does not make 3 baseballs. So QM provides the barrier, but I presume it becomes more likely if you substantially increase the kinetic energy of the two particles. The necessary kinetic energy is converted into the extra mass required to form the neutron and this is, roughly speaking, what the W-L theory proposes. Only very roughly. They don't actually explain where the extra energy comes from. Note that *extra* is about 1 1/2 times the mass of an electron. However, it would be interesting to speculate if the kind of low energy electron capture you describe could happen more frequently. In other words, If reduced mass neutrons could be made easily (from a free electron and free proton/deuteron) would this one miracle be able to explain observations without the additional miracles required by the W-L theory. Note that they can only be made *in* another nucleus, or at the very least, very, very close to it, such that the ensuing neutron(s) are immediately captured. This is included in Horace's theory, see http://www.mtaonline.net/~hheffner/DeflationFusion.pdf and also a possibility with Hydrino fusion. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
I had the opportunity to speak to Peter about this, and I was led to believe that they were indeed mystified by what they had found, but also felt that they needed some kind of hypothesis in order to get the paper published. I know from bitter experience that it is very difficult to get a paper published which simply contains mysterious results. Maybe the reviewers were hoodwinked, or maybe they realised that they made no sense but recognised the problem and allowed it through. I chased this up, right back to the early experiments in 1907, and I am convinced that there is something odd/interesting going on. Nigel The 1907 paper notes that they came in one morning to find the janitor half dead after having touched some of the high voltage batteries, making them realise that they needed to take more care. On 15/03/2013 22:18, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:05:34 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 4:49 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html OK, so were the peer reviewers for J. Plasma Physics off their rockers? I suspect they were as mystified by the results as everyone else and, unable to come up with an explanation of their own, simply let it stand. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 5:52 PM, mix...@bigpond.com wrote: This is due to the fact that an electron undergoing translational motion creates a magnetic field. It isn't an indication that the electron is rotating on it's own axis, and thus has an intrinsic magnetic field. Okay, but the free electron still has spin and a magnetic moment. http://www.chemistry.mcmaster.ca/esam/Chapter_4/section_2.html http://prd.aps.org/abstract/PRD/v3/i8/p1728_1 There are many experiments which have demonstrated this.
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Sat, 23 Mar 2013 18:33:44 -0400: Hi, [snip] On Sat, Mar 23, 2013 at 5:52 PM, mix...@bigpond.com wrote: This is due to the fact that an electron undergoing translational motion creates a magnetic field. It isn't an indication that the electron is rotating on it's own axis, and thus has an intrinsic magnetic field. Okay, but the free electron still has spin and a magnetic moment. http://www.chemistry.mcmaster.ca/esam/Chapter_4/section_2.html An electron in an s orbital is not *free*. http://prd.aps.org/abstract/PRD/v3/i8/p1728_1 There are many experiments which have demonstrated this. This is behind a pay-wall, but I get the impression that it's a theoretical paper, not an experimental one. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 7:45 PM, mix...@bigpond.com wrote: This is behind a pay-wall, but I get the impression that it's a theoretical paper, not an experimental one. Well, if you are right, the ionization energy to free an electron must include the negative spin momentum energy. Otherwise it's magic!
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Sat, 23 Mar 2013 19:51:16 -0400: Hi, [snip] On Sat, Mar 23, 2013 at 7:45 PM, mix...@bigpond.com wrote: This is behind a pay-wall, but I get the impression that it's a theoretical paper, not an experimental one. Well, if you are right, the ionization energy to free an electron must include the negative spin momentum energy. Otherwise it's magic! What exactly is negative spin momentum energy? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 8:04 PM, mix...@bigpond.com wrote: What exactly is negative spin momentum energy? You say the electron has spin when in orbit; but, when free, has no spin momentum. If so, the energy to totally ionize an electron, free it from the nucleus, must also eliminate the spin momentum energy, ie stop the electron spin, right?
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Sat, 23 Mar 2013 20:07:07 -0400: Hi, [snip] On Sat, Mar 23, 2013 at 8:04 PM, mix...@bigpond.com wrote: What exactly is negative spin momentum energy? You say the electron has spin when in orbit; but, when free, has no spin momentum. If so, the energy to totally ionize an electron, free it from the nucleus, must also eliminate the spin momentum energy, ie stop the electron spin, right? I don't think it has any spin (about it's own axis) when in orbit either. However I do think that when bound, it has two sorts of angular momentum, one of which is usually referred to as spin, but misunderstood as angular momentum about it's axis, which it isn't. Consider the following:- An electron in a circular path has angular momentum the vector of which is directed along the axis of the path. This is what's commonly referred to as spin (s) (IMO). Now take that same circle and stretch it into an ellipse. For the moment let the nucleus reside at one of the foci of the ellipse. Now let the entire ellipse swing around the focus like a hoola hoop. We have a second form of angular momentum (l). Note that the electron itself is still following the original trajectory around the perimeter of the ellipse as well, so it now has two forms of angular momentum. Remove it from the atom altogether, and it has neither. Note also, that without the elongation of the ellipse, there is no second form possible, hence an electron in an s orbital has only spin. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 5:27 PM, mix...@bigpond.com wrote: Now let the entire ellipse swing around the focus like a hoola hoop. We have a second form of angular momentum (l). Note that the electron itself is still following the original trajectory around the perimeter of the ellipse as well, so it now has two forms of angular momentum. Remove it from the atom altogether, and it has neither. Interesting idea regarding an angular momentum from the precession of an ellipsoid probability distribution around the nucleus. According to Terry's link [1], the putative electron spin will always either be aligned with or against a magnetic field. What's to make this kind of precession not be a spherical one, e.g., such that the movement of the ellipsoid over time rather than being planar instead cancels out any magnetic moment? [1] http://www.chemistry.mcmaster.ca/esam/Chapter_4/section_2.html
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 8:27 PM, mix...@bigpond.com wrote: Consider the following:- I'll have to cogitate on that.
Re: [Vo]:Graneau Questions
On Sat, Mar 23, 2013 at 5:40 PM, Eric Walker eric.wal...@gmail.com wrote: What's to make this kind of precession not be a spherical one, e.g., such that the movement of the ellipsoid over time rather than being planar instead cancels out any magnetic moment? To attempt an answer to my own question, the precession might act like a weathervane, where the wind (the magnetic field) begins to blow through it -- the weathervane swivels and aligns itself with the oncoming wind. Similarly, perhaps the precession would align with the magnetic field. Eric
Re: [Vo]:Graneau Questions
In reply to Eric Walker's message of Sat, 23 Mar 2013 17:40:02 -0700: Hi, On Sat, Mar 23, 2013 at 5:27 PM, mix...@bigpond.com wrote: Now let the entire ellipse swing around the focus like a hoola hoop. We have a second form of angular momentum (l). Note that the electron itself is still following the original trajectory around the perimeter of the ellipse as well, so it now has two forms of angular momentum. Remove it from the atom altogether, and it has neither. Interesting idea regarding an angular momentum from the precession of an ellipsoid probability distribution around the nucleus. According to Terry's link [1], the putative electron spin will always either be aligned with or against a magnetic field. What's to make this kind of precession not be a spherical one, e.g., such that the movement of the ellipsoid over time rather than being planar instead cancels out any magnetic moment? I have just presented a simple model that sort of makes sense of experimental data, without any miracles being required, and effectively does away with the unimaginable concept of a point particle with intrinsic angular momentum that QM seems to call for. I'm not sure that it will stand up to thorough scrutiny. However take a look at Mills' derivation of the spin of a spherical shell electron. It may at least partially answer your question. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to Eric Walker's message of Sat, 23 Mar 2013 18:01:48 -0700: Hi, [snip] On Sat, Mar 23, 2013 at 5:40 PM, Eric Walker eric.wal...@gmail.com wrote: What's to make this kind of precession not be a spherical one, e.g., such that the movement of the ellipsoid over time rather than being planar instead cancels out any magnetic moment? To attempt an answer to my own question, the precession might act like a weathervane, where the wind (the magnetic field) begins to blow through it -- the weathervane swivels and aligns itself with the oncoming wind. Similarly, perhaps the precession would align with the magnetic field. Indeed. Note that the entire ellipse also carries the charge of the electron, and hence two separate magnetic fields are created. This is why j is the vector sum of l s. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Mon, 18 Mar 2013 19:49:19 -0400: Hi, [snip] On Mon, Mar 18, 2013 at 5:37 PM, mix...@bigpond.com wrote: I suspect that *free* electrons (don't) have any spin momentum. And no magnetic moment? Come now! I would much appreciate a reference to the measurement of the *magnetic moment* of *free* electrons. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to Harry Veeder's message of Mon, 18 Mar 2013 21:40:15 -0400: Hi, [snip] On Mon, Mar 18, 2013 at 5:10 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sun, 17 Mar 2013 22:56:07 -0700: Hi, [snip] On Sun, Mar 17, 2013 at 2:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. I take it that when physicists refer to a potential barrier, they mean specifically an electrostatic potential barrier, and not simply an energy threshold that must be overcome? Eric A barrier usually implies an impediment that gets in the way of a reaction that would otherwise release energy. However the formation of a neutron from a proton and an electron does not release energy, it consumes it. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Some sort of barrier is necessary to stymie the mutual attraction of electrons and protons otherwise matter would quickly consist of nothing but neutrons. Isn't this barrier supplied by the weak force? The weak force doesn't actually present a barrier. It presents a chance that something will occur. Electrons and protons don't normally combine into neutrons because their combined mass is inadequate. It's 782 keV short of the mass of a free neutron. However, they could combine to form a reduced mass neutron as part of the nucleus of a heavier atom. This does in fact happen with some isotopes. It's called electron capture (EC). In this case, even though the mass of the proton is also reduced, the net result (an isotope of the previous element in the periodic table), is sufficiently more stable than the initial isotope to more than make up for the difference. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Wed, Mar 20, 2013 at 4:03 PM, mix...@bigpond.com wrote: I would much appreciate a reference to the measurement of the *magnetic moment* of *free* electrons. Isn't that how a cathode ray tube works?
Re: [Vo]:Graneau Questions
On Wed, Mar 20, 2013 at 7:37 PM, Terry Blanton hohlr...@gmail.com wrote: On Wed, Mar 20, 2013 at 4:03 PM, mix...@bigpond.com wrote: I would much appreciate a reference to the measurement of the *magnetic moment* of *free* electrons. Isn't that how a cathode ray tube works? Have you ever put a permanent magnet in front of one?
Re: [Vo]:Graneau Questions
On Wed, Mar 20, 2013 at 7:38 PM, Terry Blanton hohlr...@gmail.com wrote: On Wed, Mar 20, 2013 at 7:37 PM, Terry Blanton hohlr...@gmail.com wrote: On Wed, Mar 20, 2013 at 4:03 PM, mix...@bigpond.com wrote: I would much appreciate a reference to the measurement of the *magnetic moment* of *free* electrons. Isn't that how a cathode ray tube works? Have you ever put a permanent magnet in front of one? Without a magnetic moment, how do you scan the tube?
Re: [Vo]:Graneau Questions
In reply to Eric Walker's message of Sun, 17 Mar 2013 22:56:07 -0700: Hi, [snip] On Sun, Mar 17, 2013 at 2:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. I take it that when physicists refer to a potential barrier, they mean specifically an electrostatic potential barrier, and not simply an energy threshold that must be overcome? Eric A barrier usually implies an impediment that gets in the way of a reaction that would otherwise release energy. However the formation of a neutron from a proton and an electron does not release energy, it consumes it. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to pagnu...@htdconnect.com's message of Sun, 17 Mar 2013 20:43:13 -0400 (EDT): Hi Lou, [snip] Robin, I believe that the 782 keV represents a steep electroweak barrier that repels electrons from protons at a very close range where it overwhelms the coulomb attractive force. Since the proton is a quark bag, the equations governing the complete interaction become quite challenging - too much for me to tackle. I don't think the electrons are actually repelled. They either react or they don't (pure chance). If they don't then their kinetic energy just carries them away again. Much as a comet passing the Sun. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to Terry Blanton's message of Sun, 17 Mar 2013 18:22:01 -0400: Hi, [snip] On Sun, Mar 17, 2013 at 5:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. What is missing is sufficient mass to form a neutron. This can however be overcome if the mass difference is supplied in the form kinetic energy. How does the required energy compare to the spin momentum of the electron? I suspect that *free* electrons have any spin momentum. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Mon, Mar 18, 2013 at 5:37 PM, mix...@bigpond.com wrote: I suspect that *free* electrons (don't) have any spin momentum. And no magnetic moment? Come now!
Re: [Vo]:Graneau Questions
On Mon, Mar 18, 2013 at 5:10 PM, mix...@bigpond.com wrote: In reply to Eric Walker's message of Sun, 17 Mar 2013 22:56:07 -0700: Hi, [snip] On Sun, Mar 17, 2013 at 2:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. I take it that when physicists refer to a potential barrier, they mean specifically an electrostatic potential barrier, and not simply an energy threshold that must be overcome? Eric A barrier usually implies an impediment that gets in the way of a reaction that would otherwise release energy. However the formation of a neutron from a proton and an electron does not release energy, it consumes it. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Some sort of barrier is necessary to stymie the mutual attraction of electrons and protons otherwise matter would quickly consist of nothing but neutrons. Isn't this barrier supplied by the weak force? Harry
Re: [Vo]:Graneau Questions
Robin, It's been a long time since I looked at it, but a bare, high kinetic energy e-p collision (not just a coulombic deflection) can emit an unpredictable variety of subatomic particle sprays which must, of course, satisfy all conservation laws. An e-p collision involving collective electric or magnetic forces may be quite different. If we just use Newtonian physics and view an electron as a classical charged particle traveling in a ballistic current, or in an arc, its kinetic energy (KE) may be well below 782 keV, but the magnetic field it couples to can possess enormous momentum, allowing it to surmount potential barriers greater than KE. A mechanical analogy: A basketball must surmount the gravitational barrier of a 1m high ramp. If its initial kinetic energy is equivalent to the barrier's potential energy g X 1[m](g = gravit const), it rolls up and over. However, a suffiently strong constant wind can push it over the ramp, even if the ball never reaches a speed of g X 1[m]. I believe that a classical electron encountering a barrier in an intense current slows and receives continual coulombic and magnetic momentum kicks similar to having a strong wind at its back. My guess (and it is only that) is that bare e-p collisions cannot produce the thunderstorm results. Cheers, Lou Pagnucco Mixent wrote: In reply to pagnu...@htdconnect.com's message of Sat, 16 Mar 2013 19:09:49 -0400 (EDT): Hi, [snip] I could be mistaken, but I think that e-p free space bare collisions over 782 keV will result in all kinds of subatomic particle shards and debris, but seldom in a single neutron. What sort of shards do you expect from a proton and an electron? Bremsstrahlung probably, but I can't imagine what shards there might be. However I agree with you that neutron formation would probably be rare. The question is, would it be common enough to explain the results? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
One of the interesting characteristics about the P - E collision is that there is no standard coulomb repulsion between these two. I assume that the only reason that they do not join together is because of some form of quantum mechanical process. It is interesting that the electron seeks close companionship with the proton, but not too close. Two protons are repelled as expected by the coulomb barrier. Dave -Original Message- From: pagnucco pagnu...@htdconnect.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Mar 17, 2013 12:53 pm Subject: Re: [Vo]:Graneau Questions Robin, It's been a long time since I looked at it, but a bare, high kinetic energy e-p collision (not just a coulombic deflection) can emit an unpredictable variety of subatomic particle sprays which must, of course, satisfy all conservation laws. An e-p collision involving collective electric or magnetic forces may be quite different. If we just use Newtonian physics and view an electron as a classical charged particle traveling in a ballistic current, or in an arc, its kinetic energy (KE) may be well below 782 keV, but the magnetic field it couples to can possess enormous momentum, allowing it to surmount potential barriers greater than KE. A mechanical analogy: A basketball must surmount the gravitational barrier of a 1m high ramp. If its initial kinetic energy is equivalent to the barrier's potential energy g X 1[m](g = gravit const), it rolls up and over. However, a suffiently strong constant wind can push it over the ramp, even if the ball never reaches a speed of g X 1[m]. I believe that a classical electron encountering a barrier in an intense current slows and receives continual coulombic and magnetic momentum kicks similar to having a strong wind at its back. My guess (and it is only that) is that bare e-p collisions cannot produce the thunderstorm results. Cheers, Lou Pagnucco Mixent wrote: In reply to pagnu...@htdconnect.com's message of Sat, 16 Mar 2013 19:09:49 -0400 (EDT): Hi, [snip] I could be mistaken, but I think that e-p free space bare collisions over 782 keV will result in all kinds of subatomic particle shards and debris, but seldom in a single neutron. What sort of shards do you expect from a proton and an electron? Bremsstrahlung probably, but I can't imagine what shards there might be. However I agree with you that neutron formation would probably be rare. The question is, would it be common enough to explain the results? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sun, Mar 17, 2013 at 10:03 AM, David Roberson dlrober...@aol.com wrote: I assume that the only reason that they do not join together is because of some form of quantum mechanical process. It is interesting that the electron seeks close companionship with the proton, but not too close. I would imagine the potential barrier imposed by the need to generate a huge W- boson virtual particle in connection with the weak interaction is responsible for some of this. Eric
Re: [Vo]:Graneau Questions
In reply to pagnu...@htdconnect.com's message of Sun, 17 Mar 2013 12:53:23 -0400 (EDT): Hi, [snip] Robin, It's been a long time since I looked at it, but a bare, high kinetic energy e-p collision (not just a coulombic deflection) can emit an unpredictable variety of subatomic particle sprays which must, of course, satisfy all conservation laws. That depends on how high the kinetic energy is, and that gets complicated. I know of no particles other than the electron/positron, the neutrino/anti-neutrino and photons(?) that have a mass less than 782 keV. From what I have gathered on the Internet, the total voltage available to a lightning bolt probably lies in the multi megavolt range, somewhere between 10-100 MV. However the actual instantaneous energy acquired by an individual particle will depend on the free path it has, so will mostly be a lot less than if it were able to accelerate over the full length of the bolt. An e-p collision involving collective electric or magnetic forces may be quite different. If we just use Newtonian physics and view an electron as a classical charged particle traveling in a ballistic current, or in an arc, its kinetic energy (KE) may be well below 782 keV, but the magnetic field it couples to can possess enormous momentum, allowing it to surmount potential barriers greater than KE. A mechanical analogy: A basketball must surmount the gravitational barrier of a 1m high ramp. If its initial kinetic energy is equivalent to the barrier's potential energy g X 1[m](g = gravit const), it rolls up and over. However, a suffiently strong constant wind can push it over the ramp, even if the ball never reaches a speed of g X 1[m]. I believe that a classical electron encountering a barrier in an intense current slows and receives continual coulombic and magnetic momentum kicks similar to having a strong wind at its back. I am happy to accept alternative methods by which energy may be acquired. :) My guess (and it is only that) is that bare e-p collisions cannot produce the thunderstorm results. I think some neutrons will be created this way, but I have no idea how many. BTW another possibility is that some particles will acquire enough energy (by whatever means), to split a deuteron resulting in a free proton and neutron. I also have no idea how many this method would produce, however perhaps combined with direct neutron formation it might explain the results. D-D fusion is of course also a possibility, but I think this would produce very few due to the scarcity of D and thus the very low chances of two of them getting together. Unless of course there is some chemical/physical heavy water enrichment process occurring in some raindrops. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
In reply to pagnu...@htdconnect.com's message of Sun, 17 Mar 2013 12:53:23 -0400 (EDT): Hi, [snip] but the magnetic field it couples to can possess enormous momentum, allowing it to surmount potential barriers greater than KE. BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. What is missing is sufficient mass to form a neutron. This can however be overcome if the mass difference is supplied in the form kinetic energy. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Sun, Mar 17, 2013 at 5:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. What is missing is sufficient mass to form a neutron. This can however be overcome if the mass difference is supplied in the form kinetic energy. How does the required energy compare to the spin momentum of the electron?
Re: [Vo]:Graneau Questions
Robin, I believe that the 782 keV represents a steep electroweak barrier that repels electrons from protons at a very close range where it overwhelms the coulomb attractive force. Since the proton is a quark bag, the equations governing the complete interaction become quite challenging - too much for me to tackle. Yes - the barrier can be overcome by a purely kinetic collision, but I wager that bare collisions do not yield as many e+p--n reactions as are (claimed to be) observed in high current/voltage dense matter systems. I am not sure. These reactions may not occur at all. Hopefully, experiments will give us a definitive answer soon. -- Lou Pagnucco mixent wrote: In reply to pagnu...@htdconnect.com's message of Sun, 17 Mar 2013 12:53:23 -0400 (EDT): Hi, [snip] but the magnetic field it couples to can possess enormous momentum, allowing it to surmount potential barriers greater than KE. BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. What is missing is sufficient mass to form a neutron. This can however be overcome if the mass difference is supplied in the form kinetic energy. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
Interesting thought. Are you suggesting the energy could be supplied by a reduction in collective electron spin? - i.e., by raising collective e-spin entropy? Terry Blanton wrote: On Sun, Mar 17, 2013 at 5:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. What is missing is sufficient mass to form a neutron. This can however be overcome if the mass difference is supplied in the form kinetic energy. How does the required energy compare to the spin momentum of the electron?
Re: [Vo]:Graneau Questions
On Sun, Mar 17, 2013 at 8:47 PM, pagnu...@htdconnect.com wrote: Interesting thought. Are you suggesting the energy could be supplied by a reduction in collective electron spin? - i.e., by raising collective e-spin entropy? Not really my idea; but one that changed Don Hotson's life from being a physicist to a land surveyor because he asked inconvenient questions regarding the spin energy of the virtual electron/positron pair.
Re: [Vo]:Graneau Questions
On Sun, Mar 17, 2013 at 2:50 PM, mix...@bigpond.com wrote: BTW there is no potential barrier here. The proton and the electron carry opposite charges, so they are attracted to one another, rather than repelled. I take it that when physicists refer to a potential barrier, they mean specifically an electrostatic potential barrier, and not simply an energy threshold that must be overcome? Eric
Re: [Vo]:Graneau Questions
I think lightning is the discharge from the buildup of charge in the atmosphere created from the surface LENR of orbital quantum micro black holes of entropy and the cooling condensing, rain and snow is triggered as they extract entropy from the surrounding gaseous atmosphere along cold fronts and such. It is just vacuum energy and we live in a very non smooth spacetime, even on Earth. Stewart Darkmattersalot.com On Saturday, March 16, 2013, wrote: Yes. There are a number of papers proposing a counter-intuitive environment-to-system heat energy concentration based on non-thermal entropy exchanges (e.g. from spin baths) and/or taylored quantum measurement wavefunction collapses. Also, the anomalous effects surrounding lightning may be relevant: Lightning strikes produce free neutrons, and we’re not sure how - Low energy neutrons not due to cosmic rays or any other previously known source. http://arstechnica.com/science/2012/03/nuclear-lightening/ Observation of thundercloud-related gamma rays and neutrons in Tibet http://arxiv.org/pdf/1204.2578.pdf -- Lou Pagnucco Jones Beene wrote: Well, it would not be nonsense if there was gain from the zero point field. That kind of gain is expected to carry ambient heat with it - with the side effect of cooling the surroundings. From: James Bowery Well, of course he would retract the nonsense about ambient energy. [...]
Re: [Vo]:Graneau Questions
In reply to pagnu...@htdconnect.com's message of Sat, 16 Mar 2013 01:09:56 -0400 (EDT): Hi, [snip] Lightning strikes produce free neutrons, and were not sure how - Low energy neutrons not due to cosmic rays or any other previously known source. http://arstechnica.com/science/2012/03/nuclear-lightening/ Lightning is a high voltage particle accelerator, so how about p + e- (782 keV) - n directly? (There are plenty of protons available from the water in raindrops.) Following on from there, suppose that the neutron then fuses with another nucleus releasing about 8 MeV in energy. Result 0.8 MeV in 8 MeV out. Amplification factor ~10. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
Robin, Possibly. If the p + e- -- n reaction actually occurs (as per W-L), though, my guess is that electrons borrow just enough energy from their neighbors to climb the 782 keV electroweak barrier - just like the atom (or electrons) on the tip of an arrow borrows energy from the other atoms in the arrow to penetrate a target. The point atom would just bounce off the target if it only possessed its own kinetic energy and was uncoupled from the arrow body. I could be mistaken, but I think that e-p free space bare collisions over 782 keV will result in all kinds of subatomic particle shards and debris, but seldom in a single neutron. Cheers, Lou Pagnucco mixent wrote: Lightning strikes produce free neutrons, and were not sure how - Low energy neutrons not due to cosmic rays or any other previously known source. http://arstechnica.com/science/2012/03/nuclear-lightening/ Lightning is a high voltage particle accelerator, so how about p + e- (782 keV) - n directly? (There are plenty of protons available from the water in raindrops.) Following on from there, suppose that the neutron then fuses with another nucleus releasing about 8 MeV in energy. Result 0.8 MeV in 8 MeV out. Amplification factor ~10. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
Maybe. Who can do the math on quantum black holes, though? Daunting. ChemE Stewart wrote: I think lightning is the discharge from the buildup of charge in the atmosphere created from the surface LENR of orbital quantum micro black holes of entropy and the cooling condensing, rain and snow is triggered as they extract entropy from the surrounding gaseous atmosphere along cold fronts and such. It is just vacuum energy and we live in a very non smooth spacetime, even on Earth. Stewart Darkmattersalot.com On Saturday, March 16, 2013, wrote: Yes. There are a number of papers proposing a counter-intuitive environment-to-system heat energy concentration based on non-thermal entropy exchanges (e.g. from spin baths) and/or taylored quantum measurement wavefunction collapses. Also, the anomalous effects surrounding lightning may be relevant: Lightning strikes produce free neutrons, and were not sure how - Low energy neutrons not due to cosmic rays or any other previously known source. http://arstechnica.com/science/2012/03/nuclear-lightening/ Observation of thundercloud-related gamma rays and neutrons in Tibet http://arxiv.org/pdf/1204.2578.pdf -- Lou Pagnucco Jones Beene wrote: Well, it would not be nonsense if there was gain from the zero point field. That kind of gain is expected to carry ambient heat with it - with the side effect of cooling the surroundings. From: James Bowery Well, of course he would retract the nonsense about ambient energy. [...]
Re: [Vo]:Graneau Questions
In reply to pagnu...@htdconnect.com's message of Sat, 16 Mar 2013 19:09:49 -0400 (EDT): Hi, [snip] I could be mistaken, but I think that e-p free space bare collisions over 782 keV will result in all kinds of subatomic particle shards and debris, but seldom in a single neutron. What sort of shards do you expect from a proton and an electron? Bremsstrahlung probably, but I can't imagine what shards there might be. However I agree with you that neutron formation would probably be rare. The question is, would it be common enough to explain the results? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
An interesting paper - a printable copy is at URL: Arc-liberated chemical energy exceeds electrical input energy http://ose.accomazzi.net/files/15115795-Graneau-Paper-on-Water-Explosions.pdf There are also seven papers that cite this one, availble thru google. -- Lou Pagnucco James Bowery wrote: Quoting http://www.scribd.com/doc/15115795/Graneau-Paper-on-Water-Explosions The internal-energy diï¬erence between the cold fog expelled from the accelerator must be made up by atmospheric heat â that is, essentially by solar energy. No other energy source appears to be available for replacing the extracted kinetic energy. This 'cold fog' expansion with high kinetic energy drawing from ambient heat would seem to violate Carnot's Law. Has anyone found a peer-reviewed replication of Graneau's phenomenon? PS: Is Graneau's peer-reveiewed paper, inked above, what set McKubre off on his quest for the Papp phenomenon via Bob Rohner because he had at least plausible evidence of 10 to 1 return on energy from a similar phenomenon?
Re: [Vo]:Graneau Questions
Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The difference in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? Obviously it made sense to the peer reviewers. On Fri, Mar 15, 2013 at 3:21 PM, James Bowery jabow...@gmail.com wrote: Quoting http://www.scribd.com/doc/15115795/Graneau-Paper-on-Water-Explosions The internal-energy difference between the cold fog expelled from the accelerator must be made up by atmospheric heat – that is, essentially by solar energy. No other energy source appears to be available for replacing the extracted kinetic energy. This 'cold fog' expansion with high kinetic energy drawing from ambient heat would seem to violate Carnot's Law. Has anyone found a peer-reviewed replication of Graneau's phenomenon? PS: Is Graneau's peer-reveiewed paper, inked above, what set McKubre off on his quest for the Papp phenomenon via Bob Rohner because he had at least plausible evidence of 10 to 1 return on energy from a similar phenomenon?
Re: [Vo]:Graneau Questions
In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Fri, Mar 15, 2013 at 4:49 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html OK, so were the peer reviewers for J. Plasma Physics off their rockers?
Re: [Vo]:Graneau Questions
In reply to James Bowery's message of Fri, 15 Mar 2013 17:05:34 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 4:49 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html OK, so were the peer reviewers for J. Plasma Physics off their rockers? I suspect they were as mystified by the results as everyone else and, unable to come up with an explanation of their own, simply let it stand. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Fri, Mar 15, 2013 at 5:18 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:05:34 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 4:49 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html OK, so were the peer reviewers for J. Plasma Physics off their rockers? I suspect they were as mystified by the results as everyone else and, unable to come up with an explanation of their own, simply let it stand. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html That's very problematic given Figure 6. They're showing 10 gain in that figure from E7 to E12. How can plasma physicists who are staking their careers on billions of dollars of investment to get to near break be so obsessively sadistic toward claims like PF which were far more modest, and let this slide?
RE: [Vo]:Graneau Questions
If my memory serves me correctly, it was Dr. Peter Graneau who wrote: Ampere-Neumann Electrodynamics of Metals which was advocating longitudinal forces. In the latest issue of JSE is this interesting article: The Journal of Scientific Exploration has just published its latest issue at http://journals.sfu.ca/jse/index.php/jse. We invite you to review the Table of Contents here and then visit our website to review articles and items of interest. Research Articles Longitudinal Electromagnetic Waves? The Monstein-Wesley Experiment Reconstructed by Edward J. Butterworth, Charles B. Allison, Daniel Cavazos, Frank M. Mullen -Mark Iverson From: James Bowery [mailto:jabow...@gmail.com] Sent: Friday, March 15, 2013 3:25 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Graneau Questions On Fri, Mar 15, 2013 at 5:18 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:05:34 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 4:49 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 16:43:42 -0500: Hi, [snip] Quoting from the conclusion of the article they reiterate the explanation of the source of energy: The di?erence in the latent heat between fog and bulk water is eventually restored by heat in the atmosphere, which allows the fog to condense and return to earth. Does this make any sense to anyone? No. (I also think that their explanation is highly likely to be not even wrong. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html OK, so were the peer reviewers for J. Plasma Physics off their rockers? I suspect they were as mystified by the results as everyone else and, unable to come up with an explanation of their own, simply let it stand. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html That's very problematic given Figure 6. They're showing 10 gain in that figure from E7 to E12. How can plasma physicists who are staking their careers on billions of dollars of investment to get to near break be so obsessively sadistic toward claims like PF which were far more modest, and let this slide?
Re: [Vo]:Graneau Questions
In reply to James Bowery's message of Fri, 15 Mar 2013 17:24:47 -0500: Hi, [snip] That's very problematic given Figure 6. They're showing 10 gain in that figure from E7 to E12. How can plasma physicists who are staking their careers on billions of dollars of investment to get to near break be so obsessively sadistic toward claims like PF which were far more modest, and let this slide? ...the cynic in me would say because this is obvious BS, and thus harmless, while PF stood a chance of being a real threat. ;) The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. There may well be excess energy liberated during water arcs, but the source is almost certainly not as claimed by the Graneaus. Some form of Hydrino /or nuclear reaction is a far better candidate. Note that Mills claims that individual H2O molecules (not liquid water where the intermolecular Hydrogen bonds are still intact), is a catalyst. In an electrical arc in water, one might reasonably expect both atomic H and individual H2O molecules to be present. Various forms of Oxygen which may also act as Mills catalysts are also likely to be present. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Graneau Questions
On Fri, Mar 15, 2013 at 8:04 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:24:47 -0500: Hi, [snip] That's very problematic given Figure 6. They're showing 10 gain in that figure from E7 to E12. How can plasma physicists who are staking their careers on billions of dollars of investment to get to near break be so obsessively sadistic toward claims like PF which were far more modest, and let this slide? ...the cynic in me would say because this is obvious BS, and thus harmless, while PF stood a chance of being a real threat. ;) The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. There may well be excess energy liberated during water arcs, but the source is almost certainly not as claimed by the Graneaus. Some form of Hydrino /or nuclear reaction is a far better candidate. Note that Mills claims that individual H2O molecules (not liquid water where the intermolecular Hydrogen bonds are still intact), is a catalyst. In an electrical arc in water, one might reasonably expect both atomic H and individual H2O molecules to be present. Various forms of Oxygen which may also act as Mills catalysts are also likely to be present. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Its one thing to promote an obvious BS theory, as you describe. Its quite another to promote obvious BS scientific data. My problem with Figure 6 is of the latter, not the former type of BS. The Enlightenment equivalent of Satan Worship is publishing false experimental data since a main, if not THE main, point of the Enlightenment was Experiment over Argument. So one can understand the plasma physicists going on a witch hunt if they genuinely believed PF to be publishing false experimental data. But if that's the case, how much worse is a factor of 10 energy gain and yet nary a peep out of the high priests.
Re: [Vo]:Graneau Questions
On Fri, Mar 15, 2013 at 6:04 PM, mix...@bigpond.com wrote: The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. This seems like a big mistake, then. I wonder if an errata have been published. Eric
Re: [Vo]:Graneau Questions
In reply to James Bowery's message of Fri, 15 Mar 2013 20:16:02 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 8:04 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:24:47 -0500: Hi, [snip] That's very problematic given Figure 6. They're showing 10 gain in that figure from E7 to E12. How can plasma physicists who are staking their careers on billions of dollars of investment to get to near break be so obsessively sadistic toward claims like PF which were far more modest, and let this slide? ...the cynic in me would say because this is obvious BS, and thus harmless, while PF stood a chance of being a real threat. ;) The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. There may well be excess energy liberated during water arcs, but the source is almost certainly not as claimed by the Graneaus. Some form of Hydrino /or nuclear reaction is a far better candidate. Note that Mills claims that individual H2O molecules (not liquid water where the intermolecular Hydrogen bonds are still intact), is a catalyst. In an electrical arc in water, one might reasonably expect both atomic H and individual H2O molecules to be present. Various forms of Oxygen which may also act as Mills catalysts are also likely to be present. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Its one thing to promote an obvious BS theory, as you describe. Its quite another to promote obvious BS scientific data. My problem with Figure 6 is of the latter, not the former type of BS. The Enlightenment equivalent of Satan Worship is publishing false experimental data since a main, if not THE main, point of the Enlightenment was Experiment over Argument. So one can understand the plasma physicists going on a witch hunt if they genuinely believed PF to be publishing false experimental data. But if that's the case, how much worse is a factor of 10 energy gain and yet nary a peep out of the high priests. ...but the data is quite possibly valid, though one may perhaps argue about the various E values assigned. If no one did, I would suspect that it's because they thought it either wasn't worth the effort, or they were secretly hoping that the paper would be taken seriously and would distract others long enough for them to get ahead. But then I can't read their minds (unless they are in range ;^), so who knows what really went through their heads. You might ask them, but I think referees are anonymous? BTW a factor of 10 is in the ballpark for Hydrino reactions. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Graneau Questions
It should be noted that George Hathaway was a co-author on several of the Graneau papers. He retracted some of conclusions: http://www.mail-archive.com/vortex-l@eskimo.com/msg26685.html “I published a rebuttal of the Graneau excess-energy claims a letter to the editor of Infinite Energy Magazine V12 #71 2007 (pg 4). In it, I claim that the conclusions which I published together with the Graneaus in Jnl. of Plamsa Physics were not logically able to be derived from the experiments we performed together. In other words, while there may be some gain mechanism in water subject to electric arc discharges, it has not been proven by experiment.” From: Eric Walker The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. This seems like a big mistake, then. I wonder if an errata have been published. Eric
Re: [Vo]:Graneau Questions
Well, of course he would retract the nonsense about ambient energy. HOWEVER Did Hathaway retract the experimental data presented? If not, then the comparison of E7 to E12 still stands as true with very little in the way of inference. On Fri, Mar 15, 2013 at 9:07 PM, Jones Beene jone...@pacbell.net wrote: It should be noted that George Hathaway was a co-author on several of the Graneau papers. He retracted some of conclusions: ** ** http://www.mail-archive.com/vortex-l@eskimo.com/msg26685.html ** ** “I published a rebuttal of the Graneau excess-energy claims a letter to the editor of Infinite Energy Magazine V12 #71 2007 (pg 4). In it, I claim that the conclusions which I published together with the Graneaus in Jnl. of Plamsa Physics were not logically able to be derived from the experiments we performed together. In other words, while there may be some gain mechanism in water subject to electric arc discharges, it has not been proven by experiment.” ** ** ** ** ** ** ** ** *From:* Eric Walker ** ** The sentence:- The loss of intermolecular bond energy in the conversion from liquid to fog must be the source of the explosion energy. ... is the problem. First, they have the sign of intermolecular bond energy wrong. When water *forms* Hydrogen bonds, energy is *released*, ergo, to *break* them *requires* energy, it doesn't magically produce more. The whole solar energy nonsense follows on from this first mistake. ** ** This seems like a big mistake, then. I wonder if an errata have been published. ** ** Eric
RE: [Vo]:Graneau Questions
Well, it would not be nonsense if there was gain from the zero point field. That kind of gain is expected to carry ambient heat with it - with the side effect of cooling the surroundings. From: James Bowery Well, of course he would retract the nonsense about ambient energy.
Re: [Vo]:Graneau Questions
Touché On Fri, Mar 15, 2013 at 9:44 PM, Jones Beene jone...@pacbell.net wrote: Well, it would not be nonsense if there was gain from the zero point field. ** ** That kind of gain is expected to carry ambient heat with it – with the side effect of cooling the surroundings. ** ** ** ** *From:* James Bowery ** ** Well, of course he would retract the nonsense about ambient energy. ** ** ** **
RE: [Vo]:Graneau Questions
Yes. There are a number of papers proposing a counter-intuitive environment-to-system heat energy concentration based on non-thermal entropy exchanges (e.g. from spin baths) and/or taylored quantum measurement wavefunction collapses. Also, the anomalous effects surrounding lightning may be relevant: Lightning strikes produce free neutrons, and were not sure how - Low energy neutrons not due to cosmic rays or any other previously known source. http://arstechnica.com/science/2012/03/nuclear-lightening/ Observation of thundercloud-related gamma rays and neutrons in Tibet http://arxiv.org/pdf/1204.2578.pdf -- Lou Pagnucco Jones Beene wrote: Well, it would not be nonsense if there was gain from the zero point field. That kind of gain is expected to carry ambient heat with it - with the side effect of cooling the surroundings. From: James Bowery Well, of course he would retract the nonsense about ambient energy. [...]
