Re: [Vo]:LENR and Fermi Acceleration
I agree with that. Either way you have changed the measured. On Sunday, August 19, 2012, Harry Veeder wrote: > The measuring system can either transfer energy from itself to the > system being measured or do the reverse and transfer energy from the > system being measured to itself. > > harry > > On Sun, Aug 19, 2012 at 6:58 AM, ChemE Stewart wrote: > > The act of measuring requires one to impart some energy (photons or > other) > > or matter upon the particle. Upon the object being measured, the object > may > > instantly increase in mass or change velocity. Over time this energy > will > > be transferred back to its environment as it evaporates... > > > > On Saturday, August 18, 2012, Harry Veeder wrote: > >> > >> BTW, I appear to contradict myself when I said "measuring cannot > >> increase the energy of the particle" > >> vs I agree with the claim that measuring can concentrate energy in a > >> system. In the former, I mean I don't accept the idea that measuring > >> can somehow increase the energy the particle without the transfer of > >> energy from somewhere else. > >> > >> Harry > >> > >> On Sat, Aug 18, 2012 at 7:31 PM, Harry Veeder > >> wrote: > >> > Hi LP, > >> > > >> > I haven't read the paper, but I don't disagree with claim. In fact it > >> > should not be unexpected. > >> > > >> > Even in a macroscopic system a concentration energy can come about as > >> > a result of energy being transferred from the measuring system to the > >> > system being measured. Of course, such a measuring system would be > >> > considered defective because it provides a distorted picture of the > >> > energy content of system being measured. However, classical mechanics > >> > says a measuring system can be designed in theory to have an > >> > arbitrarily small distorting effect, whereas quantum mechanics says > >> > this is not possible in theory. > >> > > >> > Harry > >> > > >> > On Sat, Aug 18, 2012 at 2:44 PM, wrote: > >> >> Hello Harry, > >> >> > >> >> To be really precise, though, an energy measurement of a particle in > a > >> >> superposition of energy eigenstates might find it in one of the > states > >> >> higher than the weighted average energy of its wavefunction. So, you > >> >> might say that the measurement increased its energy, but over many > such > >> >> measurements would just produce the mean energy of the wavefunction. > >> >> > >> >> While I am not convinced they are correct, the authors of the paper I > >> >> referenced end with the conclusion - > >> >> > >> >> "From a general perspective a phenomenon like the energy > concentration > >> >> in > >> >> a composite quantum system can indeed be motivated physically. There > >> >> exist > >> >> processes, where there is a redistribution of energy among different > >> >> system degrees of freedom making possible some amounts of system > >> >> self-organization. In particular, one could examine the possibility > of > >> >> concentrating the total energy of the system into a subset of degrees > >> >> of > >> >> freedom producing a decrease of its entropy, which in order to avoid > a > >> >> violation of the second law of thermodynamics, would compel the > release > >> >> of > >> >> energy to the environment, thus keeping the free energy constant. > This > >> >> is > >> >> possible only if the system is open..." > >> >> > >> >> "Concentrating Energy by Measurement" > >> >> http://arxiv.org/abs/1012.5868 > >> >> > >> >> Interesting theory. > >> >> > >> >> -- LP > >> >> > >> >> Harry Veeder wrote: > >> >>> Actually, I tend agree with Robin that measuring cannot increase the > >> >>> energy of the particle. My question reflects my own attempt to > >> >>> understand why it is so. Now that I have thought about it, it is > >> >>> because one doesn't measure energy per se. Most measurements are > >> >>> really the result of calculations based on measurements of length > and > >> >>> time plugged into a formula
Re: [Vo]:LENR and Fermi Acceleration
The measuring system can either transfer energy from itself to the system being measured or do the reverse and transfer energy from the system being measured to itself. harry On Sun, Aug 19, 2012 at 6:58 AM, ChemE Stewart wrote: > The act of measuring requires one to impart some energy (photons or other) > or matter upon the particle. Upon the object being measured, the object may > instantly increase in mass or change velocity. Over time this energy will > be transferred back to its environment as it evaporates... > > On Saturday, August 18, 2012, Harry Veeder wrote: >> >> BTW, I appear to contradict myself when I said "measuring cannot >> increase the energy of the particle" >> vs I agree with the claim that measuring can concentrate energy in a >> system. In the former, I mean I don't accept the idea that measuring >> can somehow increase the energy the particle without the transfer of >> energy from somewhere else. >> >> Harry >> >> On Sat, Aug 18, 2012 at 7:31 PM, Harry Veeder >> wrote: >> > Hi LP, >> > >> > I haven't read the paper, but I don't disagree with claim. In fact it >> > should not be unexpected. >> > >> > Even in a macroscopic system a concentration energy can come about as >> > a result of energy being transferred from the measuring system to the >> > system being measured. Of course, such a measuring system would be >> > considered defective because it provides a distorted picture of the >> > energy content of system being measured. However, classical mechanics >> > says a measuring system can be designed in theory to have an >> > arbitrarily small distorting effect, whereas quantum mechanics says >> > this is not possible in theory. >> > >> > Harry >> > >> > On Sat, Aug 18, 2012 at 2:44 PM, wrote: >> >> Hello Harry, >> >> >> >> To be really precise, though, an energy measurement of a particle in a >> >> superposition of energy eigenstates might find it in one of the states >> >> higher than the weighted average energy of its wavefunction. So, you >> >> might say that the measurement increased its energy, but over many such >> >> measurements would just produce the mean energy of the wavefunction. >> >> >> >> While I am not convinced they are correct, the authors of the paper I >> >> referenced end with the conclusion - >> >> >> >> "From a general perspective a phenomenon like the energy concentration >> >> in >> >> a composite quantum system can indeed be motivated physically. There >> >> exist >> >> processes, where there is a redistribution of energy among different >> >> system degrees of freedom making possible some amounts of system >> >> self-organization. In particular, one could examine the possibility of >> >> concentrating the total energy of the system into a subset of degrees >> >> of >> >> freedom producing a decrease of its entropy, which in order to avoid a >> >> violation of the second law of thermodynamics, would compel the release >> >> of >> >> energy to the environment, thus keeping the free energy constant. This >> >> is >> >> possible only if the system is open..." >> >> >> >> "Concentrating Energy by Measurement" >> >> http://arxiv.org/abs/1012.5868 >> >> >> >> Interesting theory. >> >> >> >> -- LP >> >> >> >> Harry Veeder wrote: >> >>> Actually, I tend agree with Robin that measuring cannot increase the >> >>> energy of the particle. My question reflects my own attempt to >> >>> understand why it is so. Now that I have thought about it, it is >> >>> because one doesn't measure energy per se. Most measurements are >> >>> really the result of calculations based on measurements of length and >> >>> time plugged into a formula. BTW, the same is true of measurements of >> >>> momentum. The modern physicists habit of refering to energy and >> >>> momentum as "observables" is a perscription for phenomenological >> >>> confusion. The resulting measures of length and time are only >> >>> consistent with the supposed law-like properties of energy and >> >>> momemtum on a statiscal level. >> >>> >> >>> Harry >> >>> >> >>> >> >>> >> >>> On Fri, Aug 17, 2012 at 11:31 PM, wrote: >> Hello Harry, >> >> You asked -- >> "So, the measuring instrument itself will produce energy, if it is >> used >> to precisely measure the energy of a particle?" >> >> Probably not. >> But maybe there are subtleties that obey the 2nd Law of >> Thermodynamics, >> but allow for some counterintuitive effects. For example, refer to >> -- >> >> "Concentrating Energy by Measurement" >> http://arxiv.org/abs/1012.5868 >> >> -- LP >> >> Harry Veeder wrote: >> > On Fri, Aug 17, 2012 at 8:57 PM, <
Re: [Vo]:LENR and Fermi Acceleration
The act of measuring requires one to impart some energy (photons or other) or matter upon the particle. Upon the object being measured, the object may instantly increase in mass or change velocity. Over time this energy will be transferred back to its environment as it evaporates... On Saturday, August 18, 2012, Harry Veeder wrote: > BTW, I appear to contradict myself when I said "measuring cannot > increase the energy of the particle" > vs I agree with the claim that measuring can concentrate energy in a > system. In the former, I mean I don't accept the idea that measuring > can somehow increase the energy the particle without the transfer of > energy from somewhere else. > > Harry > > On Sat, Aug 18, 2012 at 7:31 PM, Harry Veeder > wrote: > > Hi LP, > > > > I haven't read the paper, but I don't disagree with claim. In fact it > > should not be unexpected. > > > > Even in a macroscopic system a concentration energy can come about as > > a result of energy being transferred from the measuring system to the > > system being measured. Of course, such a measuring system would be > > considered defective because it provides a distorted picture of the > > energy content of system being measured. However, classical mechanics > > says a measuring system can be designed in theory to have an > > arbitrarily small distorting effect, whereas quantum mechanics says > > this is not possible in theory. > > > > Harry > > > > On Sat, Aug 18, 2012 at 2:44 PM, wrote: > >> Hello Harry, > >> > >> To be really precise, though, an energy measurement of a particle in a > >> superposition of energy eigenstates might find it in one of the states > >> higher than the weighted average energy of its wavefunction. So, you > >> might say that the measurement increased its energy, but over many such > >> measurements would just produce the mean energy of the wavefunction. > >> > >> While I am not convinced they are correct, the authors of the paper I > >> referenced end with the conclusion - > >> > >> "From a general perspective a phenomenon like the energy concentration > in > >> a composite quantum system can indeed be motivated physically. There > exist > >> processes, where there is a redistribution of energy among different > >> system degrees of freedom making possible some amounts of system > >> self-organization. In particular, one could examine the possibility of > >> concentrating the total energy of the system into a subset of degrees of > >> freedom producing a decrease of its entropy, which in order to avoid a > >> violation of the second law of thermodynamics, would compel the release > of > >> energy to the environment, thus keeping the free energy constant. This > is > >> possible only if the system is open..." > >> > >> "Concentrating Energy by Measurement" > >> http://arxiv.org/abs/1012.5868 > >> > >> Interesting theory. > >> > >> -- LP > >> > >> Harry Veeder wrote: > >>> Actually, I tend agree with Robin that measuring cannot increase the > >>> energy of the particle. My question reflects my own attempt to > >>> understand why it is so. Now that I have thought about it, it is > >>> because one doesn't measure energy per se. Most measurements are > >>> really the result of calculations based on measurements of length and > >>> time plugged into a formula. BTW, the same is true of measurements of > >>> momentum. The modern physicists habit of refering to energy and > >>> momentum as "observables" is a perscription for phenomenological > >>> confusion. The resulting measures of length and time are only > >>> consistent with the supposed law-like properties of energy and > >>> momemtum on a statiscal level. > >>> > >>> Harry > >>> > >>> > >>> > >>> On Fri, Aug 17, 2012 at 11:31 PM, wrote: > Hello Harry, > > You asked -- > "So, the measuring instrument itself will produce energy, if it is > used > to precisely measure the energy of a particle?" > > Probably not. > But maybe there are subtleties that obey the 2nd Law of > Thermodynamics, > but allow for some counterintuitive effects. For example, refer to -- > > "Concentrating Energy by Measurement" > http://arxiv.org/abs/1012.5868 > > -- LP > > Harry Veeder wrote: > > On Fri, Aug 17, 2012 at 8:57 PM, <
Re: [Vo]:LENR and Fermi Acceleration
BTW, I appear to contradict myself when I said "measuring cannot increase the energy of the particle" vs I agree with the claim that measuring can concentrate energy in a system. In the former, I mean I don't accept the idea that measuring can somehow increase the energy the particle without the transfer of energy from somewhere else. Harry On Sat, Aug 18, 2012 at 7:31 PM, Harry Veeder wrote: > Hi LP, > > I haven't read the paper, but I don't disagree with claim. In fact it > should not be unexpected. > > Even in a macroscopic system a concentration energy can come about as > a result of energy being transferred from the measuring system to the > system being measured. Of course, such a measuring system would be > considered defective because it provides a distorted picture of the > energy content of system being measured. However, classical mechanics > says a measuring system can be designed in theory to have an > arbitrarily small distorting effect, whereas quantum mechanics says > this is not possible in theory. > > Harry > > On Sat, Aug 18, 2012 at 2:44 PM, wrote: >> Hello Harry, >> >> To be really precise, though, an energy measurement of a particle in a >> superposition of energy eigenstates might find it in one of the states >> higher than the weighted average energy of its wavefunction. So, you >> might say that the measurement increased its energy, but over many such >> measurements would just produce the mean energy of the wavefunction. >> >> While I am not convinced they are correct, the authors of the paper I >> referenced end with the conclusion - >> >> "From a general perspective a phenomenon like the energy concentration in >> a composite quantum system can indeed be motivated physically. There exist >> processes, where there is a redistribution of energy among different >> system degrees of freedom making possible some amounts of system >> self-organization. In particular, one could examine the possibility of >> concentrating the total energy of the system into a subset of degrees of >> freedom producing a decrease of its entropy, which in order to avoid a >> violation of the second law of thermodynamics, would compel the release of >> energy to the environment, thus keeping the free energy constant. This is >> possible only if the system is open..." >> >> "Concentrating Energy by Measurement" >> http://arxiv.org/abs/1012.5868 >> >> Interesting theory. >> >> -- LP >> >> Harry Veeder wrote: >>> Actually, I tend agree with Robin that measuring cannot increase the >>> energy of the particle. My question reflects my own attempt to >>> understand why it is so. Now that I have thought about it, it is >>> because one doesn't measure energy per se. Most measurements are >>> really the result of calculations based on measurements of length and >>> time plugged into a formula. BTW, the same is true of measurements of >>> momentum. The modern physicists habit of refering to energy and >>> momentum as "observables" is a perscription for phenomenological >>> confusion. The resulting measures of length and time are only >>> consistent with the supposed law-like properties of energy and >>> momemtum on a statiscal level. >>> >>> Harry >>> >>> >>> >>> On Fri, Aug 17, 2012 at 11:31 PM, wrote: Hello Harry, You asked -- "So, the measuring instrument itself will produce energy, if it is used to precisely measure the energy of a particle?" Probably not. But maybe there are subtleties that obey the 2nd Law of Thermodynamics, but allow for some counterintuitive effects. For example, refer to -- "Concentrating Energy by Measurement" http://arxiv.org/abs/1012.5868 -- LP Harry Veeder wrote: > On Fri, Aug 17, 2012 at 8:57 PM, wrote: >> In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 >> 13:11:31 >> -0400 (EDT): >> Hi, >> [snip] >>>Pardon for this very late postscript, time is hard to find. >>> >>>I believe you assume a wave function totally confined in all >>> 3-dimensions. >>> This is probably not what was intended. It is easy to find papers >>>describing crystal/lattice channel conduction of much higher energy >>>particles (electrons, protons, ...). These are extended states - only >>>confined in one or two dimensions. High energy particles do not >>>necessarily break the lattice structure. >>> >>>-- LP >> >> What I meant to do was calculate the momentum (assuming a kinetic >> energy >> of >> 0.782 MeV for the proton), and divide it into h-bar/2. However it >> appears I got >> something slightly wrong the first time around. The value I get now is >> 2.57 fm >> for a proton, and 0.93 fm for the deuteron. >> >> However I don't really stand behind the entire concept. I don't think >> the energy >> of particles magically increases when they are confined. I do think >> the >> measure
Re: [Vo]:LENR and Fermi Acceleration
Hi LP, I haven't read the paper, but I don't disagree with claim. In fact it should not be unexpected. Even in a macroscopic system a concentration energy can come about as a result of energy being transferred from the measuring system to the system being measured. Of course, such a measuring system would be considered defective because it provides a distorted picture of the energy content of system being measured. However, classical mechanics says a measuring system can be designed in theory to have an arbitrarily small distorting effect, whereas quantum mechanics says this is not possible in theory. Harry On Sat, Aug 18, 2012 at 2:44 PM, wrote: > Hello Harry, > > To be really precise, though, an energy measurement of a particle in a > superposition of energy eigenstates might find it in one of the states > higher than the weighted average energy of its wavefunction. So, you > might say that the measurement increased its energy, but over many such > measurements would just produce the mean energy of the wavefunction. > > While I am not convinced they are correct, the authors of the paper I > referenced end with the conclusion - > > "From a general perspective a phenomenon like the energy concentration in > a composite quantum system can indeed be motivated physically. There exist > processes, where there is a redistribution of energy among different > system degrees of freedom making possible some amounts of system > self-organization. In particular, one could examine the possibility of > concentrating the total energy of the system into a subset of degrees of > freedom producing a decrease of its entropy, which in order to avoid a > violation of the second law of thermodynamics, would compel the release of > energy to the environment, thus keeping the free energy constant. This is > possible only if the system is open..." > > "Concentrating Energy by Measurement" > http://arxiv.org/abs/1012.5868 > > Interesting theory. > > -- LP > > Harry Veeder wrote: >> Actually, I tend agree with Robin that measuring cannot increase the >> energy of the particle. My question reflects my own attempt to >> understand why it is so. Now that I have thought about it, it is >> because one doesn't measure energy per se. Most measurements are >> really the result of calculations based on measurements of length and >> time plugged into a formula. BTW, the same is true of measurements of >> momentum. The modern physicists habit of refering to energy and >> momentum as "observables" is a perscription for phenomenological >> confusion. The resulting measures of length and time are only >> consistent with the supposed law-like properties of energy and >> momemtum on a statiscal level. >> >> Harry >> >> >> >> On Fri, Aug 17, 2012 at 11:31 PM, wrote: >>> Hello Harry, >>> >>> You asked -- >>> "So, the measuring instrument itself will produce energy, if it is used >>> to precisely measure the energy of a particle?" >>> >>> Probably not. >>> But maybe there are subtleties that obey the 2nd Law of Thermodynamics, >>> but allow for some counterintuitive effects. For example, refer to -- >>> >>> "Concentrating Energy by Measurement" >>> http://arxiv.org/abs/1012.5868 >>> >>> -- LP >>> >>> Harry Veeder wrote: On Fri, Aug 17, 2012 at 8:57 PM, wrote: > In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 > 13:11:31 > -0400 (EDT): > Hi, > [snip] >>Pardon for this very late postscript, time is hard to find. >> >>I believe you assume a wave function totally confined in all >> 3-dimensions. >> This is probably not what was intended. It is easy to find papers >>describing crystal/lattice channel conduction of much higher energy >>particles (electrons, protons, ...). These are extended states - only >>confined in one or two dimensions. High energy particles do not >>necessarily break the lattice structure. >> >>-- LP > > What I meant to do was calculate the momentum (assuming a kinetic > energy > of > 0.782 MeV for the proton), and divide it into h-bar/2. However it > appears I got > something slightly wrong the first time around. The value I get now is > 2.57 fm > for a proton, and 0.93 fm for the deuteron. > > However I don't really stand behind the entire concept. I don't think > the energy > of particles magically increases when they are confined. I do think > the > measurement uncertainty increases, but that's not the same thing as > their actual > energy. Instead, I see it as a limitation on our ability to measure, > not > a > change in the actual properties of the particle itself. > IOW the restriction applies to us, not to the particles. > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > So, the measuring instrument itself will produce energy, if it is used to precisely measure the energy
Re: [Vo]:LENR and Fermi Acceleration
Hello Harry, To be really precise, though, an energy measurement of a particle in a superposition of energy eigenstates might find it in one of the states higher than the weighted average energy of its wavefunction. So, you might say that the measurement increased its energy, but over many such measurements would just produce the mean energy of the wavefunction. While I am not convinced they are correct, the authors of the paper I referenced end with the conclusion - "From a general perspective a phenomenon like the energy concentration in a composite quantum system can indeed be motivated physically. There exist processes, where there is a redistribution of energy among different system degrees of freedom making possible some amounts of system self-organization. In particular, one could examine the possibility of concentrating the total energy of the system into a subset of degrees of freedom producing a decrease of its entropy, which in order to avoid a violation of the second law of thermodynamics, would compel the release of energy to the environment, thus keeping the free energy constant. This is possible only if the system is open..." "Concentrating Energy by Measurement" http://arxiv.org/abs/1012.5868 Interesting theory. -- LP Harry Veeder wrote: > Actually, I tend agree with Robin that measuring cannot increase the > energy of the particle. My question reflects my own attempt to > understand why it is so. Now that I have thought about it, it is > because one doesn't measure energy per se. Most measurements are > really the result of calculations based on measurements of length and > time plugged into a formula. BTW, the same is true of measurements of > momentum. The modern physicists habit of refering to energy and > momentum as "observables" is a perscription for phenomenological > confusion. The resulting measures of length and time are only > consistent with the supposed law-like properties of energy and > momemtum on a statiscal level. > > Harry > > > > On Fri, Aug 17, 2012 at 11:31 PM, wrote: >> Hello Harry, >> >> You asked -- >> "So, the measuring instrument itself will produce energy, if it is used >> to precisely measure the energy of a particle?" >> >> Probably not. >> But maybe there are subtleties that obey the 2nd Law of Thermodynamics, >> but allow for some counterintuitive effects. For example, refer to -- >> >> "Concentrating Energy by Measurement" >> http://arxiv.org/abs/1012.5868 >> >> -- LP >> >> Harry Veeder wrote: >>> On Fri, Aug 17, 2012 at 8:57 PM, wrote: In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 13:11:31 -0400 (EDT): Hi, [snip] >Pardon for this very late postscript, time is hard to find. > >I believe you assume a wave function totally confined in all > 3-dimensions. > This is probably not what was intended. It is easy to find papers >describing crystal/lattice channel conduction of much higher energy >particles (electrons, protons, ...). These are extended states - only >confined in one or two dimensions. High energy particles do not >necessarily break the lattice structure. > >-- LP What I meant to do was calculate the momentum (assuming a kinetic energy of 0.782 MeV for the proton), and divide it into h-bar/2. However it appears I got something slightly wrong the first time around. The value I get now is 2.57 fm for a proton, and 0.93 fm for the deuteron. However I don't really stand behind the entire concept. I don't think the energy of particles magically increases when they are confined. I do think the measurement uncertainty increases, but that's not the same thing as their actual energy. Instead, I see it as a limitation on our ability to measure, not a change in the actual properties of the particle itself. IOW the restriction applies to us, not to the particles. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html >>> >>> So, the measuring instrument itself will produce energy, if it is used >>> to precisely measure the energy of a particle? >>> >>> >>> Harry >>> >>> >>> >> >> > > >
Re: [Vo]:LENR and Fermi Acceleration
Actually, I tend agree with Robin that measuring cannot increase the energy of the particle. My question reflects my own attempt to understand why it is so. Now that I have thought about it, it is because one doesn't measure energy per se. Most measurements are really the result of calculations based on measurements of length and time plugged into a formula. BTW, the same is true of measurements of momentum. The modern physicists habit of refering to energy and momentum as "observables" is a perscription for phenomenological confusion. The resulting measures of length and time are only consistent with the supposed law-like properties of energy and momemtum on a statiscal level. Harry On Fri, Aug 17, 2012 at 11:31 PM, wrote: > Hello Harry, > > You asked -- > "So, the measuring instrument itself will produce energy, if it is used > to precisely measure the energy of a particle?" > > Probably not. > But maybe there are subtleties that obey the 2nd Law of Thermodynamics, > but allow for some counterintuitive effects. For example, refer to -- > > "Concentrating Energy by Measurement" > http://arxiv.org/abs/1012.5868 > > -- LP > > Harry Veeder wrote: >> On Fri, Aug 17, 2012 at 8:57 PM, wrote: >>> In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 >>> 13:11:31 >>> -0400 (EDT): >>> Hi, >>> [snip] Pardon for this very late postscript, time is hard to find. I believe you assume a wave function totally confined in all 3-dimensions. This is probably not what was intended. It is easy to find papers describing crystal/lattice channel conduction of much higher energy particles (electrons, protons, ...). These are extended states - only confined in one or two dimensions. High energy particles do not necessarily break the lattice structure. -- LP >>> >>> What I meant to do was calculate the momentum (assuming a kinetic energy >>> of >>> 0.782 MeV for the proton), and divide it into h-bar/2. However it >>> appears I got >>> something slightly wrong the first time around. The value I get now is >>> 2.57 fm >>> for a proton, and 0.93 fm for the deuteron. >>> >>> However I don't really stand behind the entire concept. I don't think >>> the energy >>> of particles magically increases when they are confined. I do think the >>> measurement uncertainty increases, but that's not the same thing as >>> their actual >>> energy. Instead, I see it as a limitation on our ability to measure, not >>> a >>> change in the actual properties of the particle itself. >>> IOW the restriction applies to us, not to the particles. >>> Regards, >>> >>> Robin van Spaandonk >>> >>> http://rvanspaa.freehostia.com/project.html >>> >> >> So, the measuring instrument itself will produce energy, if it is used >> to precisely measure the energy of a particle? >> >> >> Harry >> >> >> > >
Re: [Vo]:LENR and Fermi Acceleration
Hello Harry, You asked -- "So, the measuring instrument itself will produce energy, if it is used to precisely measure the energy of a particle?" Probably not. But maybe there are subtleties that obey the 2nd Law of Thermodynamics, but allow for some counterintuitive effects. For example, refer to -- "Concentrating Energy by Measurement" http://arxiv.org/abs/1012.5868 -- LP Harry Veeder wrote: > On Fri, Aug 17, 2012 at 8:57 PM, wrote: >> In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 >> 13:11:31 >> -0400 (EDT): >> Hi, >> [snip] >>>Pardon for this very late postscript, time is hard to find. >>> >>>I believe you assume a wave function totally confined in all >>> 3-dimensions. >>> This is probably not what was intended. It is easy to find papers >>>describing crystal/lattice channel conduction of much higher energy >>>particles (electrons, protons, ...). These are extended states - only >>>confined in one or two dimensions. High energy particles do not >>>necessarily break the lattice structure. >>> >>>-- LP >> >> What I meant to do was calculate the momentum (assuming a kinetic energy >> of >> 0.782 MeV for the proton), and divide it into h-bar/2. However it >> appears I got >> something slightly wrong the first time around. The value I get now is >> 2.57 fm >> for a proton, and 0.93 fm for the deuteron. >> >> However I don't really stand behind the entire concept. I don't think >> the energy >> of particles magically increases when they are confined. I do think the >> measurement uncertainty increases, but that's not the same thing as >> their actual >> energy. Instead, I see it as a limitation on our ability to measure, not >> a >> change in the actual properties of the particle itself. >> IOW the restriction applies to us, not to the particles. >> Regards, >> >> Robin van Spaandonk >> >> http://rvanspaa.freehostia.com/project.html >> > > So, the measuring instrument itself will produce energy, if it is used > to precisely measure the energy of a particle? > > > Harry > > >
Re: [Vo]:LENR and Fermi Acceleration
On Fri, Aug 17, 2012 at 8:57 PM, wrote: > In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 13:11:31 > -0400 (EDT): > Hi, > [snip] >>Pardon for this very late postscript, time is hard to find. >> >>I believe you assume a wave function totally confined in all 3-dimensions. >> This is probably not what was intended. It is easy to find papers >>describing crystal/lattice channel conduction of much higher energy >>particles (electrons, protons, ...). These are extended states - only >>confined in one or two dimensions. High energy particles do not >>necessarily break the lattice structure. >> >>-- LP > > What I meant to do was calculate the momentum (assuming a kinetic energy of > 0.782 MeV for the proton), and divide it into h-bar/2. However it appears I > got > something slightly wrong the first time around. The value I get now is 2.57 fm > for a proton, and 0.93 fm for the deuteron. > > However I don't really stand behind the entire concept. I don't think the > energy > of particles magically increases when they are confined. I do think the > measurement uncertainty increases, but that's not the same thing as their > actual > energy. Instead, I see it as a limitation on our ability to measure, not a > change in the actual properties of the particle itself. > IOW the restriction applies to us, not to the particles. > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > So, the measuring instrument itself will produce energy, if it is used to precisely measure the energy of a particle? Harry
Re: [Vo]:LENR and Fermi Acceleration
In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 13:11:31 -0400 (EDT): Hi, [snip] >Pardon for this very late postscript, time is hard to find. > >I believe you assume a wave function totally confined in all 3-dimensions. > This is probably not what was intended. It is easy to find papers >describing crystal/lattice channel conduction of much higher energy >particles (electrons, protons, ...). These are extended states - only >confined in one or two dimensions. High energy particles do not >necessarily break the lattice structure. > >-- LP What I meant to do was calculate the momentum (assuming a kinetic energy of 0.782 MeV for the proton), and divide it into h-bar/2. However it appears I got something slightly wrong the first time around. The value I get now is 2.57 fm for a proton, and 0.93 fm for the deuteron. However I don't really stand behind the entire concept. I don't think the energy of particles magically increases when they are confined. I do think the measurement uncertainty increases, but that's not the same thing as their actual energy. Instead, I see it as a limitation on our ability to measure, not a change in the actual properties of the particle itself. IOW the restriction applies to us, not to the particles. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:LENR and Fermi Acceleration
Pardon for this very late postscript, time is hard to find. I believe you assume a wave function totally confined in all 3-dimensions. This is probably not what was intended. It is easy to find papers describing crystal/lattice channel conduction of much higher energy particles (electrons, protons, ...). These are extended states - only confined in one or two dimensions. High energy particles do not necessarily break the lattice structure. -- LP mixent wrote: > In reply to pagnu...@htdconnect.com's message of Wed, 15 Aug 2012 > 14:54:29 > -0400 (EDT): > Hi, > [snip] >>"Brillouin's lattice stimulation reverses the natural decay of neutrons >> to >>protons and Beta particles, catalyzing this endothermic step. >> Constraining >>a proton spatially in a lattice causes the lattice energy to be highly >>uncertain. With the Hamiltonian of the system reaching 782KeV for a >> proton >>or 3MeV for a deuteron the system may be capable of capturing an >> electron, >>forming an ultra-cold neutron or di-neutron system." > > If I understand this correctly, it would require an uncertainty in > position of > less than 2.7 fm (comparable in magnitude to the size of a nucleus) for a > proton, and < 1.3 fm for a deuteron. Note that the latter is less than the > size > of the deuteron itself. > > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > > >
Re: [Vo]:LENR and Fermi Acceleration
Good questions, Robin I wish I remembered solid state physics better, but I am not sure that sure that your estimates are correct in a crystal lattice where the energy eigenfunctions are nonlocal and span the entire crystal, but acquire more nodes when they gain energy. If my memory is correct, higher energy wave functions will have sharper, more localized nodes, but will be larger than particle radii. -- LP mixent wrote: > In reply to pagnu...@htdconnect.com's message of Wed, 15 Aug 2012 > 14:54:29 > -0400 (EDT): > Hi, > [snip] >>"Brillouin's lattice stimulation reverses the natural decay of neutrons >> to >>protons and Beta particles, catalyzing this endothermic step. >> Constraining >>a proton spatially in a lattice causes the lattice energy to be highly >>uncertain. With the Hamiltonian of the system reaching 782KeV for a >> proton >>or 3MeV for a deuteron the system may be capable of capturing an >> electron, >>forming an ultra-cold neutron or di-neutron system." > > If I understand this correctly, it would require an uncertainty in > position of > less than 2.7 fm (comparable in magnitude to the size of a nucleus) for a > proton, and < 1.3 fm for a deuteron. Note that the latter is less than the > size > of the deuteron itself. > > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > > >
Re: [Vo]:LENR and Fermi Acceleration
In reply to pagnu...@htdconnect.com's message of Wed, 15 Aug 2012 14:54:29 -0400 (EDT): Hi, [snip] >"Brillouin's lattice stimulation reverses the natural decay of neutrons to >protons and Beta particles, catalyzing this endothermic step. Constraining >a proton spatially in a lattice causes the lattice energy to be highly >uncertain. With the Hamiltonian of the system reaching 782KeV for a proton >or 3MeV for a deuteron the system may be capable of capturing an electron, >forming an ultra-cold neutron or di-neutron system." If I understand this correctly, it would require an uncertainty in position of less than 2.7 fm (comparable in magnitude to the size of a nucleus) for a proton, and < 1.3 fm for a deuteron. Note that the latter is less than the size of the deuteron itself. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:LENR and Fermi Acceleration
Here's an older article I found: http://www.lenr-canr.org/acrobat/GodesRquantumfus.pdf Jeff On Wed, Aug 15, 2012 at 9:09 PM, MarkI-ZeroPoint wrote: > Lou suggests: > " If so, the effectiveness of the stimulus could be quite sensitive to > waveform shape and frequency." > > Absolutely it would... It wouldn't surprise me for Celani's and Rossi's > cells, that increasing the resistance heater temp will increase the rate of > reaction, but at some point it will peak and then begin to dampen the > reaction. Same goes for Godes' except that in his case it's the Q-wave > frequency. > > Now, one might say that increasing the power to a resistance heater, and > thus increasing the heat output, is not a 'frequency', so how is that > similar to Brillouin's Q-waves? Simple, the increasing heat causes lattice > vibrations to increase, and the frequency of those lattice vibrations will > come into resonance with atomic elements, causing an energy coupling that > normally does not occur. Rare though... frequencies have to be just right. > > -Mark > > > > >
RE: [Vo]:LENR and Fermi Acceleration
Lou suggests: " If so, the effectiveness of the stimulus could be quite sensitive to waveform shape and frequency." Absolutely it would... It wouldn't surprise me for Celani's and Rossi's cells, that increasing the resistance heater temp will increase the rate of reaction, but at some point it will peak and then begin to dampen the reaction. Same goes for Godes' except that in his case it's the Q-wave frequency. Now, one might say that increasing the power to a resistance heater, and thus increasing the heat output, is not a 'frequency', so how is that similar to Brillouin's Q-waves? Simple, the increasing heat causes lattice vibrations to increase, and the frequency of those lattice vibrations will come into resonance with atomic elements, causing an energy coupling that normally does not occur. Rare though... frequencies have to be just right. -Mark
Re: [Vo]:LENR and Fermi Acceleration
Once the energy spread of the active particles reaches the threshold needed for electron-capture, presumably the probability of their reaching much higher energies is minimal, i.e., they will be converted into neutrons before that. So the resulting neutron is born "cold" with little extra energy. -- LP Chemical Engineer wrote: > I also am not sure why a "hot" neutron might not be created as opposed to > a > "Cold Neutron" from this. If there were some type of collapsed matter > triggering the event I can understand loss of momentum to all radiation > escaping due to the extra quantum gravitational pull to be overcome. > > > On Wed, Aug 15, 2012 at 2:54 PM, wrote: > >> Brillouin's ICCF-17 paper [1] states: >> >> "Brillouin's lattice stimulation reverses the natural decay of neutrons >> to >> protons and Beta particles, catalyzing this endothermic step. >> Constraining >> a proton spatially in a lattice causes the lattice energy to be highly >> uncertain. With the Hamiltonian of the system reaching 782KeV for a >> proton >> or 3MeV for a deuteron the system may be capable of capturing an >> electron, >> forming an ultra-cold neutron or di-neutron system." >> >> "Using Q pulses tuned to the resonance of palladium and nickel hydrides >> in pressurized vessels" they induce electron capture with protons and >> deuterons. Their patent [2] allows for both electomagnetic or sonic >> pulses. >> >> From the excerpt of their paper - >> "Constraining a proton spatially in a lattice causes the lattice energy >> to >> be highly uncertain. With the Hamiltonian of the system reaching 782KeV >> for >> a proton or 3MeV for a deuteron the system may be capable of capturing >> an >> electron" >> - it seems like they may be employing "Fermi acceleration" [3] in which >> a >> particle trapped in a time-varying potential acquires large energy. >> Spatially localizing a quantum particle requires addition of energy. >> >> Possibly, as soon as the electron wave function acquires 782 keV >> components, proton electron-capture occurs resulting in a cold neutron. >> >> Fermi acceleration has been proposed before [4]. Perhaps the Energetics >> Technology results shown on CBS "60 Minutes" involving ultrasonic >> stimulation may also involve Fermi acceleration. >> >> If so, the effectiveness of the stimulus could be quite sensitive to >> waveform shape and frequency. >> >> I would be interested in any feedback. >> >> -- Lou Pagnucco >> >> >> [1] Controlled Electron Capture and the Path Toward Commercialization >> >> http://newenergytimes.com/v2/conferences/2012/ICCF17/ICCF-17-Godes-Controlled-Electron-Capture-Paper.pdf >> >> [2] U.S. Patent Application - Pub. No. US 2011/1022984 A1 - May 26, 2011 >> ENERGY GENERATION APPARATUS AND METHOD >> http://on-the-rag.com/wp-content/uploads/2012/04/US20110122984.pdf >> >> ITEM [0019]: In some embodiments of the present invention, the >> reaction may be initiated using current as the phonon initiator >> mechanism. In other embodiments of the present invention, acoustic >> energy such as sonic or ultrasonic energy can be used..." >> >> ITEM [0005]: ...a small amount of phonon energy initiates a nuclear >> reaction. Unfortunately, the first reaction creates additional phonons >> that cause a chain reaction that leads to the destruction of the >> lattice" >> >> [3] Fermi Acceleration >> >> http://statphys.skku.ac.kr/~bjkim/Teaching/ComPhys10/Labs/Fermi/Fermi.pdf >> Exponential energy growth in a Fermi accelerator >> http://www.wisdom.weizmann.ac.il/~vered/publistorder/R34_PRE10.pdf >> >> [4] Ferroelectrics for Cold Fusion >> EPRI Proceedings: 4th Int'l Conf. on Cold Fusion, Vol. 4, p.30-1 >> http://lenr-canr.org/acrobat/EPRIproceedingc.pdf >> Catalytically Induced D-D Fusion in Ferroelectrics >> >> http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97331997000400014 >> >> >> >
Re: [Vo]:LENR and Fermi Acceleration
CE, Localizing the wave function of a proton (or an electron), i.e., making it more narrow --- for instance, +--+ | | __ | | / \ | | /\ | | going from ___/ \___ to ___| |___ changes the energy spectrum of the particle wave function so that it possesses more high energy components, so that it is more likely to climb potential barriers. The first wave function has a more "certain" energy spectrum, concentrated at lower energies. The second wave function has a less certain, wider energy spectrum - with spread into the higher energies. I could be wrong, but I do not think that the paper is referring to energy uncertainty of the entire lattice. -- LP Chemical Engineer wrote: > I agree that you do have "particles" constrained thermodynamically and > spatially within the void(s) & cracks of the lattice. Why a proton causes > the lattice energy to be "uncertain" escapes me but might be true. I can > understand how some collapsed matter would keep things "uncertain" since > it is always struggling to achieve spatial and thermodynamic equilibrium > with its environment and can instantly become unstable if any matter or > energy is transferred to/from it. > > Overall I think it is a good theory worth investigating. > > > > On Wed, Aug 15, 2012 at 2:54 PM, wrote: > >> Brillouin's ICCF-17 paper [1] states: >> >> "Brillouin's lattice stimulation reverses the natural decay of neutrons >> to >> protons and Beta particles, catalyzing this endothermic step. >> Constraining >> a proton spatially in a lattice causes the lattice energy to be highly >> uncertain. With the Hamiltonian of the system reaching 782KeV for a >> proton >> or 3MeV for a deuteron the system may be capable of capturing an >> electron, >> forming an ultra-cold neutron or di-neutron system." >> >> "Using Q pulses tuned to the resonance of palladium and nickel hydrides >> in pressurized vessels" they induce electron capture with protons and >> deuterons. Their patent [2] allows for both electomagnetic or sonic >> pulses. >> >> From the excerpt of their paper - >> "Constraining a proton spatially in a lattice causes the lattice energy >> to >> be highly uncertain. With the Hamiltonian of the system reaching 782KeV >> for >> a proton or 3MeV for a deuteron the system may be capable of capturing >> an >> electron" >> - it seems like they may be employing "Fermi acceleration" [3] in which >> a >> particle trapped in a time-varying potential acquires large energy. >> Spatially localizing a quantum particle requires addition of energy. >> >> Possibly, as soon as the electron wave function acquires 782 keV >> components, proton electron-capture occurs resulting in a cold neutron. >> >> Fermi acceleration has been proposed before [4]. Perhaps the Energetics >> Technology results shown on CBS "60 Minutes" involving ultrasonic >> stimulation may also involve Fermi acceleration. >> >> If so, the effectiveness of the stimulus could be quite sensitive to >> waveform shape and frequency. >> >> I would be interested in any feedback. >> >> -- Lou Pagnucco >> >> >> [1] Controlled Electron Capture and the Path Toward Commercialization >> >> http://newenergytimes.com/v2/conferences/2012/ICCF17/ICCF-17-Godes-Controlled-Electron-Capture-Paper.pdf >> >> [2] U.S. Patent Application - Pub. No. US 2011/1022984 A1 - May 26, 2011 >> ENERGY GENERATION APPARATUS AND METHOD >> http://on-the-rag.com/wp-content/uploads/2012/04/US20110122984.pdf >> >> ITEM [0019]: In some embodiments of the present invention, the >> reaction may be initiated using current as the phonon initiator >> mechanism. In other embodiments of the present invention, acoustic >> energy such as sonic or ultrasonic energy can be used..." >> >> ITEM [0005]: ...a small amount of phonon energy initiates a nuclear >> reaction. Unfortunately, the first reaction creates additional phonons >> that cause a chain reaction that leads to the destruction of the >> lattice" >> >> [3] Fermi Acceleration >> >> http://statphys.skku.ac.kr/~bjkim/Teaching/ComPhys10/Labs/Fermi/Fermi.pdf >> Exponential energy growth in a Fermi accelerator >> http://www.wisdom.weizmann.ac.il/~vered/publistorder/R34_PRE10.pdf >> >> [4] Ferroelectrics for Cold Fusion >> EPRI Proceedings: 4th Int'l Conf. on Cold Fusion, Vol. 4, p.30-1 >> http://lenr-canr.org/acrobat/EPRIproceedingc.pdf >> Catalytically Induced D-D Fusion in Ferroelectrics >> >> http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97331997000400014 >> >> >> >
Re: [Vo]:LENR and Fermi Acceleration
I also am not sure why a "hot" neutron might not be created as opposed to a "Cold Neutron" from this. If there were some type of collapsed matter triggering the event I can understand loss of momentum to all radiation escaping due to the extra quantum gravitational pull to be overcome. On Wed, Aug 15, 2012 at 2:54 PM, wrote: > Brillouin's ICCF-17 paper [1] states: > > "Brillouin's lattice stimulation reverses the natural decay of neutrons to > protons and Beta particles, catalyzing this endothermic step. Constraining > a proton spatially in a lattice causes the lattice energy to be highly > uncertain. With the Hamiltonian of the system reaching 782KeV for a proton > or 3MeV for a deuteron the system may be capable of capturing an electron, > forming an ultra-cold neutron or di-neutron system." > > "Using Q pulses tuned to the resonance of palladium and nickel hydrides > in pressurized vessels" they induce electron capture with protons and > deuterons. Their patent [2] allows for both electomagnetic or sonic > pulses. > > From the excerpt of their paper - > "Constraining a proton spatially in a lattice causes the lattice energy to > be highly uncertain. With the Hamiltonian of the system reaching 782KeV for > a proton or 3MeV for a deuteron the system may be capable of capturing an > electron" > - it seems like they may be employing "Fermi acceleration" [3] in which a > particle trapped in a time-varying potential acquires large energy. > Spatially localizing a quantum particle requires addition of energy. > > Possibly, as soon as the electron wave function acquires 782 keV > components, proton electron-capture occurs resulting in a cold neutron. > > Fermi acceleration has been proposed before [4]. Perhaps the Energetics > Technology results shown on CBS "60 Minutes" involving ultrasonic > stimulation may also involve Fermi acceleration. > > If so, the effectiveness of the stimulus could be quite sensitive to > waveform shape and frequency. > > I would be interested in any feedback. > > -- Lou Pagnucco > > > [1] Controlled Electron Capture and the Path Toward Commercialization > > http://newenergytimes.com/v2/conferences/2012/ICCF17/ICCF-17-Godes-Controlled-Electron-Capture-Paper.pdf > > [2] U.S. Patent Application - Pub. No. US 2011/1022984 A1 - May 26, 2011 > ENERGY GENERATION APPARATUS AND METHOD > http://on-the-rag.com/wp-content/uploads/2012/04/US20110122984.pdf > > ITEM [0019]: In some embodiments of the present invention, the > reaction may be initiated using current as the phonon initiator > mechanism. In other embodiments of the present invention, acoustic > energy such as sonic or ultrasonic energy can be used..." > > ITEM [0005]: ...a small amount of phonon energy initiates a nuclear > reaction. Unfortunately, the first reaction creates additional phonons > that cause a chain reaction that leads to the destruction of the lattice" > > [3] Fermi Acceleration > > http://statphys.skku.ac.kr/~bjkim/Teaching/ComPhys10/Labs/Fermi/Fermi.pdf > Exponential energy growth in a Fermi accelerator > http://www.wisdom.weizmann.ac.il/~vered/publistorder/R34_PRE10.pdf > > [4] Ferroelectrics for Cold Fusion > EPRI Proceedings: 4th Int'l Conf. on Cold Fusion, Vol. 4, p.30-1 > http://lenr-canr.org/acrobat/EPRIproceedingc.pdf > Catalytically Induced D-D Fusion in Ferroelectrics > > http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97331997000400014 > > >
Re: [Vo]:LENR and Fermi Acceleration
I agree that you do have "particles" constrained thermodynamically and spatially within the void(s) & cracks of the lattice. Why a proton causes the lattice energy to be "uncertain" escapes me but might be true. I can understand how some collapsed matter would keep things "uncertain" since it is always struggling to achieve spatial and thermodynamic equilibrium with its environment and can instantly become unstable if any matter or energy is transferred to/from it. Overall I think it is a good theory worth investigating. On Wed, Aug 15, 2012 at 2:54 PM, wrote: > Brillouin's ICCF-17 paper [1] states: > > "Brillouin's lattice stimulation reverses the natural decay of neutrons to > protons and Beta particles, catalyzing this endothermic step. Constraining > a proton spatially in a lattice causes the lattice energy to be highly > uncertain. With the Hamiltonian of the system reaching 782KeV for a proton > or 3MeV for a deuteron the system may be capable of capturing an electron, > forming an ultra-cold neutron or di-neutron system." > > "Using Q pulses tuned to the resonance of palladium and nickel hydrides > in pressurized vessels" they induce electron capture with protons and > deuterons. Their patent [2] allows for both electomagnetic or sonic > pulses. > > From the excerpt of their paper - > "Constraining a proton spatially in a lattice causes the lattice energy to > be highly uncertain. With the Hamiltonian of the system reaching 782KeV for > a proton or 3MeV for a deuteron the system may be capable of capturing an > electron" > - it seems like they may be employing "Fermi acceleration" [3] in which a > particle trapped in a time-varying potential acquires large energy. > Spatially localizing a quantum particle requires addition of energy. > > Possibly, as soon as the electron wave function acquires 782 keV > components, proton electron-capture occurs resulting in a cold neutron. > > Fermi acceleration has been proposed before [4]. Perhaps the Energetics > Technology results shown on CBS "60 Minutes" involving ultrasonic > stimulation may also involve Fermi acceleration. > > If so, the effectiveness of the stimulus could be quite sensitive to > waveform shape and frequency. > > I would be interested in any feedback. > > -- Lou Pagnucco > > > [1] Controlled Electron Capture and the Path Toward Commercialization > > http://newenergytimes.com/v2/conferences/2012/ICCF17/ICCF-17-Godes-Controlled-Electron-Capture-Paper.pdf > > [2] U.S. Patent Application - Pub. No. US 2011/1022984 A1 - May 26, 2011 > ENERGY GENERATION APPARATUS AND METHOD > http://on-the-rag.com/wp-content/uploads/2012/04/US20110122984.pdf > > ITEM [0019]: In some embodiments of the present invention, the > reaction may be initiated using current as the phonon initiator > mechanism. In other embodiments of the present invention, acoustic > energy such as sonic or ultrasonic energy can be used..." > > ITEM [0005]: ...a small amount of phonon energy initiates a nuclear > reaction. Unfortunately, the first reaction creates additional phonons > that cause a chain reaction that leads to the destruction of the lattice" > > [3] Fermi Acceleration > > http://statphys.skku.ac.kr/~bjkim/Teaching/ComPhys10/Labs/Fermi/Fermi.pdf > Exponential energy growth in a Fermi accelerator > http://www.wisdom.weizmann.ac.il/~vered/publistorder/R34_PRE10.pdf > > [4] Ferroelectrics for Cold Fusion > EPRI Proceedings: 4th Int'l Conf. on Cold Fusion, Vol. 4, p.30-1 > http://lenr-canr.org/acrobat/EPRIproceedingc.pdf > Catalytically Induced D-D Fusion in Ferroelectrics > > http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97331997000400014 > > >
