How is lepton number conserved in this process?
To my knowledge, GUTCP is silent on the issue of conservation of lepton number. It only discusses the conservation of angular momentum.
optiongeek
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optiongeek: Sorry I very well understand Mills theory, but the hydrino part still has some holes!
If an electron travels below Bohr radius, then you need, as Mills states, an additional charge to compensate the centrifugal forces . This is only possible with a combined Lorenz force approach.
Thus please point me to the text, where the charge-equivalent forces/fields are generated/explained.
An other point you should explain (or ask the maestro himself): What is an energy hole photon?
I thought I had provided a reference to the explanation for the energy hole concept - please review the text between equation 5.26 and 5.27 in Vol. 1 of GUTCP (available here: http://brilliantlightpower.com…P-2016-Ed-Volume1-Web.pdf). I don't know what you mean by "combined Lorenz force approach". However, the discussion in that part of the text is pretty clear. In particular, equation 5.27 explicitly shows the solution to the boundary value problem after the energy hole has been absorbed. Note that n = 1/p is what leads to the increase in effective charge felt by the electron. -
How do two photons (each spin=1, total lepton number=0) collide to form an electron (spin=1/2, lepton number=1)?
In GUTCP, every particle has h_bar of angular momentum. The electron's spin is multi-polar, with h_bar/2 in the z-axis, and h_bar/4 along the x & y-axes. The same goes for the positron. Angular momentum is conserved during pair production. The multi-polar configuration of angular momentum results in re-interpretation of the Stern-Gerlach finding.
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Rather than try to answer your questions directly, I think the best approach is for me to post an excellent description of the pair production phenomena prepared by Jeff Driscoll. He's taken Mills' formulation and broken it down into manageable chunks. I was mystified by Mills' approach for years until I read Jeff's take on it - then it all became clear. And I mean that in a most profound way. Hope you are able to appreciate this as much as I did. . .
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Special relativity agrees with you on this detail, of course. Above you wrote, "When the radius is exactly alpha * Bohr radius, the charge is travelling at the speed of light. This is the instance of energy to mass conversion." Is that the charge of the electron that is travelling at the speed of light?
The charge velocity approaches light speed as the radius approaches alpha * Bohr radius. And in the limit, the conversion from matter to energy is accomplished.One thing I'd like to point out that I suspect you've never considered. The mechanism I describe above of a spherical resonator cavity in free space is the first concrete definition of alpha ever devised. Alpha turns out to be a blueprint for energy to matter conversion. If you want to "freeze" a photon with the proper amount of energy (511k eV) into an electron, you start with a resonator cavity with a characteristic frequency that matches that photon's frequency. Why do you want to match the characteristic frequency of the photon? Because it will cause the photon to experience infinite impedance and therefore be unable to maintain its propagation through space. Why 511k eV? Because that's the rest mass of the electron. How do you design such a resonator cavity? It has to be a sphere in free space with a radius equal to alpha * Bohr radius. When you work out the inductance and capacitance of such a resonator cavity, the characteristic frequency (when adjusted for Lorentz invariance), is exactly the frequency of the 511k eV photon. This is the answer to Feynman's conjecture - "Every good physicist should put 1/137 on his wall and worry about it". Here is a succinct version:
Alpha, the fine structure constant, is the relativistically corrected ratio to the Bohr radius of the spherical resonator cavity in free space such that the characteristic frequency matches the photon with energy equal to the electron rest mass.
If you have a better definition for alpha I'm all ears.
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Do the two leptons travel at the speed of light in Mills's telling?
No, charge/matter can never travel at light speed.
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You are describing the conversion of a massive lepton (an electron) into something, presumably not a lepton (a photon?), traveling at the speed of light. Either there is no conservation of lepton number, or a massive particle is traveling at the speed of light. In either case, we are no longer talking about classical physics.
The instance of mass/energy conversion (two-way process) is described by Mills in Chap. 28 of his book. He uses only classical laws of Newton, Maxwell, Planck, De Broglie and Lorentz. Two super-posed photons of opposite spin and energy of 511k eV strike a third body. A spherical resonator cavity of radius alpha * Bohr radius in free space (call the Transition State Orbitsphere) is created. As a resonator cavity, it has a characteristic frequency such that the two photons will experience infinite impedance and therefore "freeze" from E&M fields traveling a light speed into charge. Two particles are formed, a positively charged positron and a negatively charged electron, also as two orbitspheres. There's a lot more, including how this process results in the creation of gravity as space-time contracts. But, yes, it's all from classical physics. -
optiongeek: To my understanding of physics, when energy is released then no photon remains.., exept there was one already there.., which is excluded by the non radiation condition...
The "ground" state hydrogen atom has 511k eV stored as potential and kinetic energy in the orbiting electron so there is a large reservoir of energy that can be released. At the ground state, the electron is in force balance and the effective central charge is 1. If a photon is absorbed, the effective central charge felt by the electron will be fractional (1/n) and it will go to a higher orbit. Conversely, if an energy hole/trapped photon is absorbed, the effective central charge felt by the electron will be integral (n) and it will go to a lower orbit.The next step you must explain: If this photon generates a force, then tell us in which direction its E-field vector points! There is always a physical cause for a force - not just an energy equivalent!
Sorry, don't understand this part. A photon/trapped photon is a standing wave in the electron's orbitsphere. What force is it creating? Maybe you could read Mills' equation for the photon? Chap. 4 in GUTCP. -
optiongeek: Yes indeed you can combine the electron with a charged exiton. This may work for one step. But further exitons would have to go into different orbits I guess...
No idea what an exiton is. A trapped photon is simply the standing wave that remains when m * 27.2eV is transferred to the catalyst. And being a photon, therefore it has the property of superposition with other photons. No need for a separate orbit.(edit) Oh, and the energy from the electric field works out perfectly using nothing but classical physics and Lorentz invariance. In fact, something pretty amazing happens when you get to the 1/137 level. The energy from electron's electric field, magnetic field, and gravitational field all converge to exactly 511k eV. This is the "unified" part of GUTCP. When the radius is exactly alpha * Bohr radius, the charge is travelling at the speed of light. This is the instance of energy to mass conversion. All from classical physics.
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The Hydrino inverse Rydberg logic only works with an increased central charge. An increased central charge has so far never been seen...
I think you are mis-interpreting Mills here. Mills doesn't say that the actual nuclear charge increases, only that the "effective" charge as experienced by the electron increases. The increase comes from the electric field of the "trapped photon", which arises from the absorption of an energy hole from the catalyst.Look at the text between equation 5.26 and 5.27 in GUTCP Vol. 1 (I'm unable to copy and paste it here). It describes the "Energy Hole/Trapped Photon" concept. Think of a trapped photon/energy hole as the mirror image of a regular photon. It creates a stable standing wave in a hydrino with an electric field that increases the electric charge felt by the electron. Higher effective electric charge means that the electron orbit must shrink to maintain force balance.
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I wanted to understand the photon acording to Mills. A trapped photon could be written as
j_0(sqrt(x*x + y*y + z*z)w/c)*exp(i w t), radious less then the first zero of the spherical bessel function
You should engage with Prof. Huub Bakker of Massey University. He's done the most work on understanding Mills' photon equation and is, I believe, writing a textbook on the topic.
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Both are not correct: A typical setup of Mills is 10V peak power, with average to 5V because also Mill's needs the 5V to split the H-H bonds...
I went back and re-read the most recent independent validation reports from May 2016 and it seems I stand corrected. They describe voltages in the 5 to 10V range. I do remember Mills discussing lower voltage values on the Society for Classical Physics board but they were from at least a year ago and I'm unable to locate those discussions now. Thanks for the correction! -
I often hear of very high voltage arc discharges in connection with Mills. But let's go with your assertion of 1 V. Personally, I think it could potentially be the hydrogen that induces beta decay in already beta-unstable isotopes, such as 40K, by bringing a bound electron in close proximity to the nuclear volume, with the applied voltage possibly playing a secondary role. But this is just one idea; there could be other explanations. Do you agree that, if beta radiation could somehow be induced in Mills's setup, it would quite sufficiently explain the continuum spectrum?
Eric, very much appreciate your response. Please be assured that Mills has been very consistent over time in describing his experimental setup as low voltage with high current. A typical setup is 1v @ 10kA.
I must admit that I don't know that much about how various isotopes of H could react so I have no reason to doubt your conjecture. However I should point out that Mills has quite recently asserted that his experimental results are indifferent to the mix of H isotopes in his fuel source. I don't know that he's spent much time looking at this but he has been pretty clear that he's not loading his apparatus with a specific isotope. So for beta-unstable H to be the cause I guess we would have to assume that either he is somehow inadvertently using such an isotope. I would also point out that four other validators (research physicists from Bucknell, UNC-Ashville, and elsewhere) would also have to make the same mistake.
Can I suggest that you ask Dr. Mills about beta decay? Perhaps he has already considered it?
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Quote from optiongeek: “I've worked through and replicated the equations myself and it's absolutely the case the GUTCP is an exact match to physical reality.”
Can you elaborate on this? Do you have a link to your calculations?
Thanks for your interest. I've attached an Excel spreadsheet of my own creation that replicates the GUTCP calculations of ionization energies for 1 through 20-electron atoms. The derivation spans hundreds of pages in Mills GUTCP, but I've called out each equation by reference in the Excel cell so it should be easy to verify how I get the same values as Mills. Mills' book is available as a free download. http://brilliantlightpower.com…P-2016-Ed-Volume1-Web.pdf(edit) I should add that this is just one example of calculations using GUTCP that I have performed but have no counterpart in QED. Other examples include:
- Lepton mass ratios
- Closed-form equation for electron anomalous magnetic moment (QED version requires heavy numeric processing)
- Certain molecular properties (i.e. bond angle for H2O)
And I've only scratched the surface here. GUTCP calculates nearly every physical value. These are just the ones I've replicated on my own to date.
Fast electrons, e.g., from beta decay, will result in EUV continuum radiation with an endpoint (cutoff).
Regarding beta decay - how do you get beta decay from approximately 1v? Mills is getting EUV continuum radiation from extremely low voltage, and only when predicted catalysts are present i.e. Hydrogen and *not* Helium. -
Mills theory rejects quantum electrodynamics(QED), but with QED, the questions that QED was formulated to answer go unresolved. I need to have an answer to those questions. QED must be there in my universe.
Axil, please describe the phenomena that are solved by QED and not by GUTCP. In my experience, GUTCP not only provides answers that are just as accurate if not more so, but do so with more elegance.
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Thanks for succinctly outlining your concerns regarding the GUTCP model. I think having a thoughtful critique of the model is always a good thing.
I agree in part. I certainly would like to see a more definitive proof for the double-slit one way or the other. However, the element that has always made it difficult for me to reject GUTCP has been the extraordinary match of GUTCP's calculation of the 1-20 electron atom ionization levels with experiment. I've worked through and replicated the equations myself and it's absolutely the case the GUTCP is an exact match to physical reality. I also find it persuasive that Mills and others have seen EUV continuum radiation with cutoffs predicted by the equation 91.2nm / (n - 1)^2. I don't know how you could obtain that type of spectrum with hydrino states that weren't exactly as Mills describes - i.e. with fractional orbit radii.
Do you propose an alternate explanation?
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Quote from optiongeek: “Axil,
Please educate yourself on the reaction. There has never been a single incident in which BrLP has found evidence of a nuclear reaction in 25 years of experiments. ”
There has never been a <span style="font-size: 18pt"><span style="color: #990033">report</span></span> of a single incident in…
You are going on at length about hypothetical evidence you apparently hope to will into existence. At the same time, you pointedly ignore copious evidence to the contrary. Chief among this evidence is the presence of continuum radiation with cutoffs predicted by the formula 91.2nm / (n - 1)^2. I think you'd be more credible if you addressed this.
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Axil,
Please educate yourself on the reaction. There has never been a single incident in which BrLP has found evidence of a nuclear reaction in 25 years of experiments. At least half a dozen different experimental techniques all show the same thing - the reaction is due to hydrino formation, which is intermediate between chemical and nuclear in terms of energy release.
Also - FWIW I believe that the road-show will be invitation-only. And invitations are generally issued to qualified representatives of potential partners/suppliers/customers, etc. In terms of a demo, I would highly doubt there will be a working unit in the room. To date, the units have typically been run in "glove-boxes" due to the need to vent excess heat. Unlikely that the heat exchange assembly will be functional by early December. Maybe a live, remote shot like at the Oct. 26th demo.
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Has anyone worked through Appendix I to the point they feel comfortable with the derivations? I'm mostly OK with it (except for the discussion of the H() and G() functions). However, the conclusion uses some terms that aren't well explained. While I *think* I understand these, can anyone take a crack at providing a more intuitive justification behind the highlighted equations? Exactly what is represented by the cross-product s_n * v_n? Is omega_n the angular frequency of the emitted photon? I think s_n is the spatial frequency expressed in rad/m, and v_n is a velocity in m/sec of the current density. And radiation requires that the cross product of the two at some point on the orbitsphere is equal to the photon's wavelength. Am I understanding this correctly?
