[General] Compton and de Broglie wavelengththe "error"

[Richard Gauthier]

Hi Andre,
    Thanks for this helpful summary. I want to confirm that when you write your formula: 
E2 = p2 c2 + (mc2 )2 where m=ymoc2
that E = y mo c^2   and p=y mo v   in the usual way. If not, what  are your equations  for E and p of a particle with speed v?
       best,
             Richard

> On Dec 12, 2017, at 8:15 PM, André Michaud <srp2 at srpinc.org> wrote:
> 
> Hi Richard,
> 
> Ok, I will try to explain as clearly as possible, but for my explanation to make any sense, you will need to get hold of Abraham's analysis of Kaufmann's data and verify for yourself the calculations on top of also analyzing the Marmet derivation and the analysis I carried out afterwards (see below for links to the papers).
> 
> There is not a single equation in all of my derivations that I have not calculated values with to confirm. I even often put the figures directly in my papers where I thought readers should do these calculations themselves to make certain, so they would not have to laboriously go look for examples on their own.
> 
> I used two calculators, the simple scientific Casio fx-991 that has about 40 standard constants preprogrammed, that makes most calculations easy:
> 
> https://www.amazon.com/Casio-fx-991MS-Scientific-Calculator-Display/dp/B0000VILI2 <https://www.amazon.com/Casio-fx-991MS-Scientific-Calculator-Display/dp/B0000VILI2>
> and a Ti-89 Titanium for the more complex cases:
> 
> https://www.amazon.com/Texas-Instruments-TI-89-Titanium-Calculator/dp/B000KUIKGQ/ref=sr_1_3?s=electronics&ie=UTF8&qid=1513133901&sr=1-3&keywords=ti-89+titanium+graphing+calculator <https://www.amazon.com/Texas-Instruments-TI-89-Titanium-Calculator/dp/B000KUIKGQ/ref=sr_1_3?s=electronics&ie=UTF8&qid=1513133901&sr=1-3&keywords=ti-89+titanium+graphing+calculator>
> Kaufmann was an experimentalist. He did not interpret his data. He simply collected it, gave his opinion and published his findings.
> 
> Max Abraham analyzed it and found that the longitudinal non-linear energy growth curve obeyed the gamma factor established by Voigt, and that the transverse non-linear mass increment growth curve obeyed the gamma factor growth curve applied to half the rest mass of the electron.
> 
> Poincare studied the results and confirmed.
> 
> I studied the results and agree to Max Abraham conclusions.
> 
> What is more, Marmet`s derivation demonstrates that the magnetic field of an accelerating electron increases with velocity according to the transverse energy/mass growth curve established by Max Abraham.
> 
> That`s all.
> 
> Certainly you can see that the (mo c2 )2 element of SR equation  E2 = p2 c2 + (mo c2 )2 is invariant, which goes counter the Kaufmann data regarding the transverse mass increase.
> 
> And that the (mc2 )2 element of electromagnetic equation  E2 = p2 c2 + (mc2 )2 where m=ymoc2 varies with velocity. This element is meant to correspond to the transversally measurable mass of the Kaufmann experiment and of Marmet's derivation.
> 
> If you correlate the increase in mass of element (mc2 )2 element of the electromagnetic equation with Marmet's derivation that confirms that the magnetic field of the electron in motion increases with the velocity, you will be able to verify that the increase in magnetic mass corresponds to this factor and to the figures collected by Kaufmann.
> 
> From Marmet's conclusion, it can be seen that the rest magnetic mass of the electron corresponds to exactly half the invariant rest mass of the electron, ad that this part of the mass is the one that seems to grow transversally according to the gamma factor. Actually, finer analysis reveals that this transverse mass increment belongs to the carrying energy of the electron. This mass increament is measurable longitudinally as well as transversally, just like the invariant rest mass of the electron.
> 
> I proposed the electromagnetic energy-momentum version simply so that people could see the difference with respect to the SR version, but in reality,  I think that this SR energy-momentum version (and even the electromagnetic version) are in my view just unrequired complicated hair-splitting.
> 
> I admit that I am now getting a little mixed up into all these half of this, half of that. But the real derivations with correct forms from the Marmet derivation are in these 2 papers:
> 
> http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/2257 <http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/2257>
> and http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197 <x-msg://82/and%20http:/www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197>
> And the Marmet paper is here:
> 
> http://www.newtonphysics.on.ca/magnetic/index.html <http://www.newtonphysics.on.ca/magnetic/index.html>
> If you can make abstraction of this antiquated notion of kinetic energy converting into so called "potential energy" that was axiomatically drilled in to all of us, for the duration of your analysis, this would help a lot in understanding these derivations. If you try to think of energy as always being "active kinetic energy", this would help a lot also.
> 
> Best Regards
> 
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/
> http://www.srpinc.org/
> 
> 
> On Tue, 12 Dec 2017 18:48:22 -0800, Richard Gauthier wrote:
> 
> Hello André,
> You wrote:"The SR energy-momentum equation as formulated is simply wrong, and demonstrably so with any electron deflection experiment".
> Could you pleaseexplain in a concise way why you say that any electron deflection experiment would demonstrate that thestandard relativistic energy-momentum equation E^2 = p^2 c^2 + (mo c^2 )^2 wrong, where E= gamma mo c^2 and p=gamma mo v? This would be really helpful. Thanks.
> Richard
>  
>  
>> On Dec 11, 2017, at 4:19 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
>>  
> 
> Hi Richard,
> 
> Then, I'm afraid that we will have to agree that we disagree on this issue and on the issue of the existence of potential energy, because this really has nothing to do with the trispatial geometry. It has to do with the adiabatic nature of the energy induced in elementary charged partcles by the Coulomb force.
> 
> The corrected equation does agree with the Coulomb equation and the adiabatic nature of the energy it induces in charged electromagnetic particles as a function of distance and with the confirming Kaufmann experiment, so there is no physical way that the energy adiabatically induced in a moving electron could be equal to only its momentum related translational half. The SR energy-momentum equation as formulated is simply wrong, and demonstrably so with any electron deflection experiment.
> 
> The total measurable kinetic energy induced is made of the momentum half (p=gamma mov) plus the electromagnetic half (gamma mo - mo), both of which are measurable longitudinally as confirmed by the Kaufmann experiment and the Coulomb equation, while only the electromagnetic half (gamma mo - mo) can be measured transversally, as also experimentally proven by Kaufmann. Moreover, this electromagnetic structure of the carrying energy of the electron is confirmed by the derivation of the gamma factor from strict electromagnetic considerations in direct line from the Biot-Savart equation via Marmet`s derivation.
> 
> The trispatial geometry only brings an a posteriori mechanical explanation to this internal structure of elementary charged particles.
> 
> The Kaufmann data is on record and can be verified by anybody.
> 
> I simply agree with Abraham and Poincare on this issue after analyzing the Kaufmann data myself and analysingthe also confirming Marmet derivation.
> 
> I also thank you for your patience.
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Mon, 11 Dec 2017 15:26:35 -0800, Richard Gauthier wrote:
> 
> I’m resending this message because I just received a “bounce” message from the “Nature of Light and Particles” server. I’ve resubscribed as requested. I don’t know if this last message I sent to the group went through.
> Richard
>  
> Hello André,
> When in the equation below equation (5.65) in your Appendix B, you separated theγfrom the mo v as you mentioned, this has no effect on the equality of the equation. And the p above the mo v is still equal to p=γmo v despite your separating out the mo v fromγin the expression below it. So for the six equations starting from equation (B3) and including the one above equation (B5), the equations are all mathematically unchanged in their content, and are all still consistent with the conventional relativistic energy-momentum equation E^2 = p^2 c^2 + (mo c^2)^2 . And the p in the equation above equation (B5) is STILL equal to p=γmo v as it was in equation (B3) and as it was defined at the top of your Appendix B. So shifting theγaway from the mo v term has really accomplished nothing mathematically (or physically).
>  
> Then from the equation before equation (B5) to equation (B5) in moving theγ^2 p^2 c^2 term from the left side of the equation to the right side of the equation to get the p^2 c^2 term (withoutγ^2)on the right, you have dropped theγ^2 WITHOUT ANY JUSTIFICATION, since p in this equation (before dropping theγ^2) is still p=γmo v and not p(Newton) = mo v as you have claimed (incorrectly).
>  
> Then in the line after Equation (B5) you incorporated theγ^2 in the γ^2 E^2 term to become E^2 on the left-hand side of the equation. I renamedγE as E(André) =γE to avoid the confusion of keeping the same letter E for two different terms E=γmoc^2andE(André)=γE=γ^2moc^2.
>  
> So your equation (5.70 (your relativistic energy-momentum equation) has two mistakes in it, if you want to keep the meaning in equation (5.70) of E asγmo c^2 and p=γmo v, the same as you defined them at the beginning of the appendix. The corrected equation (5.70), keeping the meaning of E and p as the same as at the beginning, is:
>  
> γ^2 E^2 =γ^2 p^2 c^2 + (mc^2) ^2 (5.70 corrected) where m is defined as m=γmo
>  
> where if you divide each term byγ^2 you get
>  
> E^2 = p^2 c^2 + (mo c^2) ^2 which is STILL the conventional relativistic energy-momentum equation.
>  
> OR if you want to keep the larger energy quantity E(André)=γE (this seems to be your intention), then equation (5.70 becomes)
>  
> E(André)^2 =γ^2 p^2 c^2 + (mc^2)^2 (5.70 corrected) where m is defined as m=γmo .
>  
> This corrected equation is also mathematically equivalent to the conventional relativistic energy-momentum equation.
>  
>  
> You said that your relativistic energy-momentum equation (5.70) has nothing to do with your tri-space theory. If so, you will see objectively that what I have said above about your derivation of the relativistic energy-momentum equation (which is the purpose of your Appendix B according to its title) is correct. Your shifting ofγ’s in your Appendix B derivation can be seen as related to deriving your trii-space hypothesis and its novel kinetic energy relationships. Otherwise there is no need to have a new relativistic energy-momentum equation when the present one is fully-confirmed experimentally. I don’t expect you to immediately accept what I’ve said above, but I’m also patient. ; )
>  
> You have said that the conventional relativistic energy-momentum equation does not allow for the experimentally-confirmed increase of inertial mass of an electron with velocity as m=γmo. But M=γmo IS (though it is out of fashion to say so) the transverse or relativistic inertial mass of the electron (or other object) with longitudinal speed v (and also as used in relativistic kinematics calculations in large synchrotron accelerators every day to keep the accelerated electrons moving in circular orbits). This inertial mass M=γmo, which increases with velocity v, is just not the invariant rest mass mo, and now just called m (which is defined as being independent of the speed of the object), and is generally combined in the momentum formula p=γmov. (I derive this inertial mass result M=γmo for my internally-helically-moving electron model inhttps://www.academia.edu/25599166/Origin_of_the_Electrons_Inertia_and_Relativistic_Energy_Momentum_Equation_in_the_Spin_Charged_Photon_Electron_Model <https://www.academia.edu/25599166/Origin_of_the_Electrons_Inertia_and_Relativistic_Energy_Momentum_Equation_in_the_Spin_Charged_Photon_Electron_Model>starting on page 5 . And I obtain this result without modifying the conventional relativistic energy-momentum equation. In fact I show that the conventional relativistic energy-momentum equation can be interpreted as the Pythagorean vector relations among the electron’s transverse internal momentum component moc, its longitudinal external momentum component p=γmovand its total vector momentum P=γmoc.
>  
> Again, thanks for your patience.
> Richard
>  
>> On Dec 10, 2017, at 3:51 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard,
> 
> My comments in red below
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Sun, 10 Dec 2017 14:53:20 -0800, Richard Gauthier wrote:
> 
> Hello André,
> Thank you for yourquick reply. One quick question. You wrote:
>> Not exactly. I say that the first two gammas are incorporated into the new energy E and into the (mov definition of the p term) so that the equation above becomes my reformed relativistic energy-momentum equation.
>> 
> 
> But in the first line of equations in your Appendix B you define relativistic momentum as : p=γmov (5.65) and you usethis equation to obtain yourequation (B2). 
> 
> Yes. but in the line immediately following line (5.65), note that I mathematically separate the gamma factor from the ratio p/mov, so the mathematical group "mov" is henceforth separated from the gamma factor, and is by very definition the Newtonian mov of the Newtonian p(Newton)=mov definition.
> 
> 
> Then you combinedequation (B2)with the relativistic energy equation (B1) to get the equation above (B5) ,
> 
> You move too quickly to (B5) Richard. If you look at the intervening lines you will see that I actually get equation (B4) from combining (B2) with (B1).
> 
> It is here that the disconnect between the SR version and the (Andre) version occurs, because if you look at how (B4) is arrived at, you will also understand how the ratio gamma/gamma is arrived at, and how tempting it is to simplify it to 1/1 before proceeding.
> 
> If this simplification is carried on, and if it was assumed (as you do) that p is relativistic by default, then you end up with the SR version E2 = p2 c2 + (mo c2)2, that cannot account for the relativistic/electromagnetic mass increment.
> 
> But from a strict mathematical viewpoint, if you simplify the gamma/gamma ratio to 1, you actually remove it entirely from the equation since only Newtonian occurrences of mo remain in equation (B4) --->>>> including in the inner definition of p, since from the second line on, the Newtonian mov group had been mathematically separated from the gamma factor.
> 
> This is why the gamma/gamma ratio must not be simplified to 1/1 but rather squared in the following line (to match the exponent level of the left side of the equation), so each gamma occurrence can be reunited with its Newtonian mo companion for the mass to become relativistic again: m=gamma mo.
> 
> I hope this clarifies the sequence.
> 
> Do not worry about me losing patience. This won't happen.
> 
> Best Regards
> 
> Andre
> 
> 
> 
> which is consistent with the conventionalrelativistic energy momentum equation E^2 = p^2 c^2 + (mo c^2)^2 as long as we agree to define m as m=gamma mo. But thenyou unaccountablydropped the originalγ from your earlier expression (5.65) for relativistic momentum p =γmo v , and now say (above in red) you are incorporating the newγ into "the mo v definition of the p term”, so that your new equation for p as p=γ mo vis the SAME as your original equation for p=γ mo v, even though you have incorporated anadditionalγ into it. So until this point (unaccountably dropping the originalγ from p=γ mo v (5.65) in your derivation, before incorporating the newγ)is cleared up, it’s difficult for me to continue to follow your line of reasoning.
> 
> Thanks for your patience.
> Richard
>  
>> On Dec 10, 2017, at 12:56 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard,
> 
> Glad that you did some calculations, so we now each have personally verified common figures to relate to.
> 
> As before, my comments inline in red.
> 
> Best Regards
> 
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Sat, 9 Dec 2017 21:51:20 -0800, Richard Gauthier wrote:
> 
> HelloAndré,
> 
> I’ve accepted your calculator challenge, though I don’t have a hand-held calculator handy and am using my computer’s version of a hand-held calculator athttps://www.google.com/search?q=calculator&cad=h <https://www.google.com/search?q=calculator&cad=h> .
> 
> 
> First, let us observe that your energy value4.359743805E-18 Joulesis the value of the Coulomb potential of an electron in the ground state of a Bohr hydrogen atom, at a distance from the proton of one Bohr radius
> 
> ao =5.29177 x 10^-11 m.
> 
> 
> Taking the electric constant k=8.987551787369 x 10^9 and the electric charge e = 1.6021765 x 10^-19 C we get
> 
>  
> Potential energy PE = -k e^2/ao = - (8.987551787369 x 10^9) (1.6021765 x 10^-19)^2 /(5.29177 x 10^-11)
> 
> = - 4.35974 x 10^-18 Joules (to six significant figures) which (except for the minus sign of the negative potential energy) matches your above energy value4.359743805E-18 Joulesthat you found using your hand calculator.
> 
> No. Ok. It is here that we have a major disconnect.
> 
> NOT POTENTIAL ENERGY, REAL KINETIC ENERGY induced by the Coulomb force:
> 
> E=e2/(4 pi eps_0 ao) Ref: equations (4) and and (10) in this paper:
> 
> https://www.omicsonline.org/open-access/on-adiabatic-processes-at-the-elementary-particle-level-2090-0902-1000177.pdf <https://www.omicsonline.org/open-access/on-adiabatic-processes-at-the-elementary-particle-level-2090-0902-1000177.pdf>
> Half of this 4.35974 x 10^-18 Joules is Momentum related kinetic energy (see equations (6) and (7)) and the other half transversally oscillates electromagnetically and is the velocity relatred relativistic mass increment (or in the trispatial model, the electromagnetic mass increment - See equation (11)).
> 
> 
> 
> Butwhat does the above calculation have to do with your reformed relativistic energy momentum equation? Your equation
> 
> Because the Momentum related kinetic energy is the traditional momentum energy (see equation (7)) and the second half is the added mass contributed by the Coulomb force at this distance from the proton (see equation (11)) that amounts to m = gamma mo.
> 
> gamma^2 E^2 - gamma^2 p^2 = (mc^2)^2 ,where E= gamma mo c^2 , p= gamma mo v and m= gamma mo(your definition of m)
> 
> is the equation above your equation B5 in your Appendix B, and is consistent with the standard relativistic energy-momentumequation.You then say that the first 2 gammas are incorporated into new energy E and momentum p terms sothat the equation above becomes your reformed relativistic energy-momentum equation
> 
> Not exactly. I say that the first two gammas are incorporated into the new energy E and into the (mov definition of the p term) so that the equation above becomes my reformed relativistic energy-momentum equation.
> 
> E^2 = (pc)^2 + (m c^2)^2 Equation (5.70) where m=gamma mo (your definition of m)
> 
> Yes, even into the (gamma mov subcomponents of p) which becomes mv within the definition of p.
> 
> But to avoid confusion between your new gamma-incorporated E and gamma-incorporated p and the originalrelativistic terms E and p, let’s call your new gamma-incorporated E and p terms E(André) = gamma E and p(André) = gamma p.
> 
> Almost. Ok for the gamma-incorporated E to become E(Andre), but the gamma-incorporated p becomes p(Andre)=(gamma mov).
> 
> Your reformed relativistic energy-momentum equation, thus relabeled, becomes
> 
> E(André)^2 = p(André)^2 c^2 + (mc^2)^2 where E(André)= gamma E = gamma x gamma mo c^2 = gamma^2 mo c^2 , and p(André) = gamma p = gamma x gamma mo v = gamma^2 mo v , and m = gamma mo (in your definition of m .)
> 
> No. Where p(André) = gamma p(Newton) = gamma mo v
> 
> Now, using your reformed relativistic-energy momentum equation (5.70) you (in your tri-space model) calculate a new kinetic energy KE(André) (using the relabeled reformed relativistic energy-momentum equation) to be
> 
> KE (André) = E(André) – mo c^2
> 
> = gamma^2 mo c^2 –mo c^2 = (gamma^2 -1 ) mo c^2
> 
> No. See above. gamma is not squared.
> 
> while traditionally, relativistic kinetic energy is calculated with the conventional relativistic energy-momentum equation as
> 
> KE(conventional) = E – mo c^2
> 
> = gamma mo c^2 – mo c^2 = (gamma -1 )mo c^2 .
> 
> It still is for the momentum related energy.
> 
> I find it difficult to continue commenting further, because the equation is now wrong all along.
> 
> Until we come to common terms with the fact that the Coulomb force induces "real" kinetic energy and not POTENTIAL energy, we will not be able to come to a common understanding.
> 
> The problem is that with the new paradigm, the old concept of momentum involving conservation of energy via mutual conversion of momentum energy converting to potential energy makes no sense at all, since the kinetic energy turns out to be an adiabatically inducedreally existing substance, that remains induced even when the velocity of a charged particle is impeded from being expressed.
> 
> So I think that we should discuss this particular issue first.
> 
> Best Regards
> 
> Andre
> 
> In the limit of very small velocities compared to c, such as in the Bohr atom in your calculations, the above expressions for KE(André) and KE(conventional) become
> 
> KE (André) = (1/(1-v^2/c^2) - 1) mo c^2
> 
> => (1 + v^2/c^2 -1) mo c^2 for v<<c
> 
> = (v^2/c^2) mo c^2
> 
> = mo v^2 i.e twice the traditional non-relativistic value of kinetic energy. And
> 
>  
> KE(conventional) = E – mo c^2 = (gamma -1 ) mo c^2
> 
> = ( {1/sqrt(1 – v^2/c^2)} – 1) mo c^2
> 
> => (1 + ½ v^2/c^2 - 1) mo c^2 for v<<c
> 
> = ½ (v^2/c^2) mo c^2
> 
> = ½ mo v^2 i.e the traditional non-relativistic value of kinetic energy.
> 
>  
> The result is that in the limit of v<<c, KE(André) = 2 x KE (conventional).
> 
> In the Bohr atom at any energy level it is found by a standard total energy calculation that
> 
> Etotal = PE(coulomb) + KE (conventional) = -ke^2/r + ½ mo v^2 where r is the distance between the proton and the electron and v is the speed of the electron in its orbit (the small speed of the proton is neglected here.)
> 
> For the first Bohr orbit where r = ao we get
> 
> Etotal = PE(coulomb) + KE (conventional)
> 
> -13.6 eV = -27.2 eV + 13.6 eV
> 
> where -13.6 eV is the electron’s total energy (equals minus the electron’s ionization energy from the ground state), -27.2 eV is the electron’s potential energy and +13.6 eV is the circling electron’s kinetic energy.
> 
> The magnitude 27.2eV of the potential energy above is the value (in Joules) that André calculates with his hand calculator:
> 
> E = 27.2 eV x 1.6021765 x 10^-19 J/eV
> 
> = 4.36 x10^-18 Joules
> 
> which to 3 sig figs (because of 3 sig figs in 27.2 eV) corresponds to André’s calculated value4.359743805E-18 Joules.The value of the kinetic energy KE of the electron in the Bohr atom’s ground state is 13.6 eV.
> 
> So if André uses his reformed relativistic energy-momentum equation above, he will calculate the kinetic energy KE(André) to be twice the value of the ground state kinetic energy of the electron in the Bohr atom i.e. 2 x 13.6 eV = 27.2 eV = 4.359743805E-18 Joules(his calculated value), which, except for the minus sign in the actual potential energy PE = -27.2 eV , is the same magnitude as the electron’s Coulomb electric potential energy.
> 
> André wrote previously:
> 
> Equation (5.70) from which you subtract mo will give you the same energy as the Coulomb equation for the electron on the rest orbital of the hydrogen atom, while the standard SR equation minus mo will give you only its momentum energy.
> 
> The proof to me that equation (5.70) is right, is that it gets the same energy as the Coulomb equation (4.359743805E-18 j) in excess of the energy of the electron rest mass for the hydrogen rest orbital, which is the actual total energy of the system, as confirmable with the Coulomb equation.
> 
> The fact that the SR version does not, tells me that there is a clear disconnect between SR and Maxwell, because the Coulomb equation emerges from Maxwell's first equation (Gauss equation for the electric field), and that the SR version does not give me the total energy of the system on my hand held calculator that the Coulomb equation gives.
> 
> I have no other argument in support.
> 
> Please do the calculation.
> 
>  
> What I have shown above is that the extra gamma in André’s gamma-incorporated equation for energy E(André) = gamma^2 mo c^2 is what causes André’s predicted electron non-relativistic kinetic energy mo v^2, or 27.2 eV = 4.359743805E-18 Joules, to be twice the value of the electron’s kinetic energy 13.6 eV in the ground state of the Bohr atom.
> 
> Of course, this does not in any way prove that André’s equation (5.70) is a correct relativistic energy-momentum equation just because, when given the kinetic energy of an electron in the ground state of a Bohr hydrogen atom, it predicts a kinetic energy that is twice this kinetic energy, whose magnitude happens to equal the magnitude of the electron’s potential energy in the ground state in the Bohr hydrogen atom. This potential energy PE of an electron in a Bohr atom is always the negative of twice the kinetic energy of the electron in the Bohr atom, due to the Coulomb attractive force in the Bohr atom. André’s reformed relativistic energy-momentum equation always predicts twice the nonrelativistic kinetic energy given by the conventional non-relativistic kinetic energy formula ½ mo v^2.
> 
> I hope this is helpful.
> 
> Richard
> 
>> On Dec 7, 2017, at 5:57 AM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard,
> 
> I do not "insist" that E2= p2c2+mo2c4 is Newtonian. I "observe" by calculation that it is Newtonian in the sense that it doesn't account for the relativistic mass increase.
> 
> Even if you are convinced that I am mistaken, my Casio hand held calc tells me I am not mistaken.
> 
> This has nothing to do with the 3-spaces hypothesis.
> 
> We could argue like this in circle for years Richard.
> 
> This is very simple. The momentum energy that propels an electron is p= gamma mo v2, and the relativistic mass being propelled is gamma mo. It is as simple as that, as demonstrated by Kaufmann, gamma mo having been measured transversally.
> 
> Please take a hand calculator and "observe" the difference.
> 
> Equation (5.70) from which you subtract mo will give you the same energy as the Coulomb equation for the electron on the rest orbital of the hydrogen atom, while the standard SR equation minus mo will give you only its momentum energy.
> 
> The proof to me that equation (5.70) is right, is that it gets the same energy as the Coulomb equation (4.359743805E-18 j) in excess of the energy of the electron rest mass for the hydrogen rest orbital, which is the actual total energy of the system, as confirmable with the Coulomb equation.
> 
> The fact that the SR version does not, tells me that there is a clear disconnect between SR and Maxwell, because the Coulomb equation emerges from Maxwell's first equation (Gauss equation for the electric field), and that the SR version does not give me the total energy of the system on my hand held calculator that the Coulomb equation gives.
> 
> I have no other argument in support.
> 
> Please do the calculation.
> 
> Best Regards
> 
> André
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Wed, 6 Dec 2017 16:44:30 -0800, Richard Gauthier wrote:
> 
> Hello André,
> 
> I don’t know why you insist thatE2= p2c2+mo2c4 is Newtonian when it is constructed from the relativistic relations E =γmoc2 and p = γmov and the rest massmo. I think that it is you who are mistaken here, and not Wikipedia as well as and many standard textbooks that now often setmo=min the first equation above as in present high-energy physics usage. Seehttps://en.wikipedia.org/wiki/Energy–momentum_relation <https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation>, which states: "Inphysics <https://en.wikipedia.org/wiki/Physics>, the energy-momentum relation, or relativistic dispersion relation, is therelativistic <https://en.wikipedia.org/wiki/Theory_of_relativity>equation <https://en.wikipedia.org/wiki/Equation>relating any object'srest (intrinsic) mass <https://en.wikipedia.org/wiki/Invariant_mass>, totalenergy <https://en.wikipedia.org/wiki/Energy>, andmomentum <https://en.wikipedia.org/wiki/Momentum>."If in this equationγis set equal to 1 (as you mention), so that v=0, the equation becomesmo2c4= 0 +mo2c4, which is pretty useless and is NOT a Newtonian equation. Using the above relativistic energy-momentum equation for very small (non-zero) velocities compared to c, we get KE = E -moc 2 => 1/2mov2 which IS useful as non-relativistic kinetic energy and is the standard KE formula for small velocities compared with c.
> 
> But for the sake of argument let's use YOUR DEFINITION m= γ mo(I know that many others use this equation also when talking about relativistic mass or inertial mass.) With this definition, the relativistic energy E and momentum p terms in the above relativistic energy-momentum equation become E=mc2 and p=mv, whilemo= m/γ. The standard relativistic energy-momentum equation above becomes
> 
> E2= p2c2+(m/γ)2c4
> 
> = p2c2+ m2c4/γ2
> 
> Multiply both sides of the equation by γ2 and we get:
> 
> γ2E2=γ2p2c2+m2c4 (I)
> 
> This corresponds to the equation above (B5) in your Appendix B:
> 
> γ2E2-γ2p2c2=m2c4 where m = γ mo .
> 
> But when you bring the second term -γ2p2c2above over to the right side of the equation, you intentionally drop theγ2,saying (in your explanation to me) thatγis incorporated into the momentum term p=mov . But in the equation (I) above , p is already p=mv =γmov so you can’t incorporate thisγinto the momentum equation p=mv =γmov a SECOND TIME— you will get p=γm v =γ2mov, which has no obvious relativistic momentum meaning. So although your derivation at the top of Appendix B starts out with relativistic expressions E=γmoc2, and p= γ mov, these RELATIVISTIC starting expressions for your derivation were apparently forgotten during the derivation at equation (B5), where you incorporatedγinto the NON-RELATIVISTIC expressions E= moc2 and p= mov at this point in your derivation. Then you arrive at a NEW relativistic energy-momentum equation:
> 
> E2=p2c2+m2c4 (5.70) or E2=p2c2+γ2mo2c4 since you definem=γmo .
> 
> So clearly your equation (5.70), with m=γmo ,is not the standard relativistic energy-momentum equation that you claim to be deriving in Appendix B. No one will object if you DEFINE a new E by this new, non-standard equation (5.70) where you define m=γmo, but this new E equation is not correctly derived from the standard relativistic energy-momentum equation above. Your new E is no longer given by E=mc2 =γmoc2, whichever you like since you define m=γmo. Rather, your new E is bigger than the standard E=γmoc2, maybe in just the way you need it to be bigger than the standard E for your tri-space hypothesis, which carries extra kinetic energy in its electric and magnetic spaces.
> 
> 
> I’ve tried to be as clear as I can in pointing out what I see as serious errors in your derivation of the relativistic energy-momentum equation. If I am mistaken here I will be pleased to hear how.
> 
> best regards,
> 
> Richard
> 
>  
>  
>  
>  
>  
>> On Dec 4, 2017, at 8:36 AM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard
> 
> A small complementary comment added in blue further down.
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Mon, 04 Dec 2017 10:48:51 -0500, André Michaud wrote:
> 
> 
> Hi Richard
> 
> My comments inline in red for consistency.
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Sun, 3 Dec 2017 22:00:45 -0800, Richard Gauthier wrote:
> 
> Hello André,
> In yourarrival at equation (B5) during your derivation of E 2= p 2c 2 +m 2c4in your book’s Appendix B, you made a simple algebra mistake by apparently accidentally dropping the factor ofγ2from theγ2p2c2term in the line aboveequation (B5):
> γ 2E 2 -γ2p 2c2 = (mc 2)2 where m=γmo
> 
> Not apparent nor accidental. I did not drop it at all. I incorporated it to p.
> 
> I thought this inclusion was obvious, so I did not number this line, since m=γmoby definition. In this line, I treated only the right side of the equation to replace ymo by its relativistic equivalent m.
> 
> If you look at second line before this line, you will see that the right side of the equation is (γ2mo 2c4) which is the same as (γmo c2) 2 and since ymo=m, then (γmo c2) 2 is the same as (mc2) 2 which gives you the form you find in the equation before equation (B5), note the comment to the right of the equation: ("where m=γmo").
> 
> In other words, it is where m was established as being the same as ymo.
> 
> 
> when you moved that term from the left side to the right side of the equation to obtain your equation (B5):
>  
> γ 2E 2 =p 2c 2 + (mc 2) 2 where p=γmov (B5).
>  
> Without this algebra mistake your equation (B5) would read
>  
> γ 2E 2 =γ 2p 2c 2 + (mc 2)2 where p=γmov (B5)
> No. Same explanation as above.
> 
> In this line, I transfer (γ 2p 2c 2) from the left side of the equation to the right side (changing its sign), and incorporate the gamma factor into the (mo v) definition of (p): since p=mov, then (γ 2(mov) 2c 2) is the same as ((γmov) 2c 2) which is the same as ((mv) 2c 2) and finally the same as (p2c 2) which gives the form you find in equation (B5)
> 
> Sincep=γmovis included in your original statement at the top of this appendix, stating the same equation in (B5) has no effect on your derivation and doesn’t correct the algebra mistake mentioned above.
> The y factor was separated from all occurrences of mo from the get go at the start of the derivation. Line (B5) is the first time it is combined to mo to establish relativistic form m.
> 
> This is where m is mathematically established as being the same as ymo even in p.
> 
> Then you deletedγ2from the termγ2E2inyour equation (B5) (now you wouldhave to delete both of theγ2's in the corrected equation (B5) ), stating thatγis dimensionless (though it is certainly not equal to one unless v=0 which you imply that it is not because the equation derived from this equation (B5), you claim, is the relativistic energy-momentum equation that fits yourtri-space theory.
> No. I do not delete it. I "incorporate it" into the value of E. In reality you have y2 (E(before))2 = (E(after))2. Which means that the value of E(after) is the same as the product of yE. Since y is dimensionless, this multiplication does not change the fact that the value remains in joules.
> 
>  
> You also state that E 2 = p2c2 +mo2c4is the non-relativistic energy-momentum equation or the Newtonian equation (which it is not, it is the correct relativistic energy-momentum equation.)
> It is non-relativistic and moreover it provides only the momentum energy, not the total energy of the system. It is equivalent to K= ½ mv2.
> 
> Note here that to obtain the energy in excess of the rest mass, this rest mass must be subtracted from the total energy obtained:
> 
> E=sqrt(p2c2+ (moc2)2) - moc2 with the SR equation
> 
> And
> 
> E=sqrt(p2c2+ (mc2)2) - moc2 with equation (5.70)
> 
> 
> If you calculate the total energy of the system with this equation, you get only half the total energy that can be calculated with the Coulomb equation at the corresponding axial distance between the electron and the proton in the hydrogen atom for example: (2.179871902E-18 j)
> 
> But if you do the same calculation with equation (5.70) you obtain the same energy as with the Coulomb equation (4.359743805E-18 j) which is proof enough to me that equation (5.70) is correct and that the SR version where the y/y term of equation (B4) is simplified to 1 is wrong.
> 
> So when you deletedγ2 fromthe term γ2E2your equation (B5) you should also have deletedγ2 from the term γ2p2c2in the corrected equation (B5) , andyou should not forget to delete the impliedγ2 from the third term (mc2)2 in (B5) which equals (γmoc2) 2 = γ2(moc2)2 in your deriviation, since for you, m =γmo
>  
> Deletingγ2from each of the three terms in your corrected equation (B5) leaves the following equation:
>  
> E2 =p2c2 + (mo c2)2
> which is what you call the non-relativistic equation or the Newtonian equation.
> 
> Actually the above equation, as mentioned above, IS the correct relativistic energy-momentum equation, here written withmoinstead of the usual m which equalsmo (and not gammamo).
>  
> What you call therelativistic energy-momentum equation:
>  
> E2 =p2c2 + (γmoc2)2 or E2 =p2c2 + (mc2)2 where you say m = gammamo is not a correct equation at all, since the right side of your “equation”is larger than the left side of the equation due to the presence ofγ in the right side of your“equation”. E and p are defined (by you) in the standardway as E=γmoc 2 and p=γmov .
>  
> You conclude your appendix:
>  
> "At face value, fusing the last occurrence of the squared γ factor with the energy (γ2E2) may seem to be problematic, but considering that this factor is a dimensionless quantity, it can be multiplied with the energy component without any adverse effect for the integrity of the equation."
> 
>  
> Yes,yourcalculations above, and your statement above, are unfortunately quite problematic, and lead to an incorrect relativisticenergy-momentum equation which is not even an equation, due to the two sides being unequal.
> 
> I hope you will take my remarks above in the helpful spirit in whichthey are intended, and make any necessary corrections accordingly.
> 
> Dear Richard, I do take your remarks in the helpful spirit that you intend, but there is one thing that supersedes all arguments. It is mathematical proof by calculation:
> 
> If you can mathematically prove that SR equation (E2 =p2c2 + (mo c2)2) gives the correct energy value (4.359743805E-18 j) that the Coulomb force induces at mean rest orbital distance from the proton (5.291772083E-11 m) in a hydrogen atom, then I will concede.
> 
> In the mean time, I observe by actual calculation that relativistic equation (E2=p2c2+ (mc2)2) gives the right value (4.359743805E-18 j)
> 
> Best Regards
> 
> André
> 
>  
> On Nov 30, 2017, at 11:10 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard,
> 
> There is no error in defining terms, wich is a standard mathematical practice from which logical derivations can be made.
> 
> Modern usage or not, (m = gamma mo) simply is the formal mathematical definition of m as representing the physical mass of a really existing electron (that can never be devoid of energy in excess of its rest mass in physical reality), while mo is the NIST recognized fundamental constant corresponding to the theoretical mass of the electron at rest (with no energy added), a state in which the electron cannot be in in physical reality.
> 
> If " modern high energy physics usage" as you say, has been neglectful in keeping the definition clear, then " modern high energy physics usage" should review its neglectful usage.
> 
> I do not understand where you saw that gamma could always be equal to 1. It is by structure a variable dimensionless value that depends only on the value of the variable embedded ratio.
> 
> Taking a pocket scientific calculator with real figures to equations and actually doing the calculations as I suggested to confirm that the results obtained correspond to physical reality is the only way to clear such issues. This is the mandatory condition for understanding the validity of any equation.
> 
> Failing this, no constructive discussion can be had. Only those doing this with the equations of the string of equations where Newton's kinetic energy equation is progressively upgraded to full relativistic status, for example, will be able verify their validity, and how they logically derive from the Biot-Savart equation via Marmet`s derivation, and how the standard relativistic equations can be derived from them.
> 
> Just looking at the equations will not do it and can only regretfully end in all too common futile discussions, which is not something that I engage in.
> 
> Since you are convinced that (E = gamma E) and that m is equals (gamma mo) by formal mathematical definition are incorrect, then regretfully, no amount of discussion will convince you that the derivation is correct.
> 
> Best Regards
> ---
> 
> 
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Thu, 30 Nov 2017 21:41:23 -0800, Richard Gauthier wrote:
> 
> HelloAndré (and all)
>  
> I am by no means an expert in relativistic kinematics, but I’ve tried to understand the relativistic energy-momentum equation clearly, and also the formulas E=mc^2 and E=gamma mc^2 (which one is correct generally? The second one! The first is only correct for a resting object. You see the problem?) There is one pitfall in writing equations of basic relativity which some people still make, and that is to write m= gamma mo . This is a pitfall because in modern high energy physics usage, m has replaced mo and means exactly the same as mo, i.e. the “rest mass” or “invariant mass” of a particle, which doesn’t change with a particle's velocity. The expression m=gamma mo is therefore incorrect since m=mo. In your appendix you use the relation m= gamma mo between equations (B4) and (B5) and this creates other problems in your derivation. For example, you write that the equation E^2 = p^2 c^2 + mo^2 c^4 is the non-relativistic version of the relativistic equation E^2 = p2 c^2 + m^2 c^4 where actually these two equations are BOTH the relativistic energy-momentum equation, written using mo in one case and m in the other. They are the same because m= mo. Then you write
> And finally, since γ is dimensionless, then E=γE and we obtain equation
> 
> (5.70): ...
>  
> Gamma is dimensionless but that doesn’t mean gamma always equals 1, so clearly "E= gamma E" is incorrect.
> It’s hard to get into discussing physical theories when there may be basic terminology errors that need clarification and/or correction.
> Richard
>  
>>  
> 
>> On Nov 29, 2017, at 5:52 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard,
> 
> Forgotten added embedded inline comment in blue below, regarding where where the extra 1/2 mv^2 of kinetic energy of the particle go when the particle slows down to v=0 and gives up its standard kinetic energy 1/2 mv^2 to its environment.
> 
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Wed, 29 Nov 2017 18:40:12 -0500, André Michaud wrote:
> 
> 
> Hi Richard (and all)
> 
> I'll comment in red.
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Wed, 29 Nov 2017 11:53:35 -0800, Richard Gauthier wrote:
> 
> Hello André (and all),
>  
> What you seem to be saying is that the transverse velocity component of each of the two charges in yourtri-space composite photon model changes from zero to c and back to zero (twice) in one transverse cycle from your electric space to your magnetic space, while the photon as a whole moves forward with a longitudinal velocity of c in normal space. This tells me that the minimum total vector velocity (the transverse velocity component combined vectorially with the longitudinal velocity) of the chargeduring a transverse cycle is c (when the transverse component of the charge's velocity is zero), while the maximum total vector velocity of the charge (combining its its maximum transverse velocity component c with the constant longitudinal velocity component c) is c sqrt(2). I don’t see how compartmentalizing your composite photon’s total motion into three mutually orthogonal spaces (where the component of the charge’s speed is less than or equal to c in each of these three component spaces) gets you“off the hook” from this resultant superluminal speed of up to c sqrt(2) of the two charges in your composite photon model as a whole.
> You are thinking about this as if this was occurring in 4D normal space. In the trispatial geometry, the minor unit vectors in each space operate independently of those in in the other two. It is the Major unit vectors that represent the separate spaces orientations that can be mutually correlated since it is the 3 spaces themselves that are perpendicular to each other.
> 
> Remember that these 3 spaces are only 3D-exploded occurrences of the triple orthogonal vector relation of Maxwell's electromagnetic wave fields configuration. What happens in electrostatic space is now coupled to Maxwell's electric field's major unit vector, what happens in magnetostatic space is coupled to Maxwell's magnetic field's major unit vector, the cross product of which is then the normal space major unit vector coupled in the phase velocity of the particle moving at c in normal space, propelled by half the photon's energy (velocity related momentum energy), which remains unidirectional within normal space (propelling the particle). This remains the same as in original Maxwell.
> 
> I can't get past the words of your kinetic energy formula explanation which leads to a particle kineticenergy that is twice the standard, experimentally accepted (and experimentally verified over many decades ofhigh-energy physics research) relativistic kinetic energy formula ofKE=moc2(γ-1). Your derived formulaKE=2moc2(γ-1) gives, in the limit of very small velocities v << c, the formula KE= 2 x 1/2 mv^2 = mv^2 while the standard relativistic KE formula in the limit of very small velocitiesv << cgives KE = 1/2 mv^2 , the standard non-relativistic kinetic energy formula which is equal to the work W= Fxd done on a particle to give it that kinetic energy 1/2 mv^2. Where did your extra 1/2 mv^2 of energy come from in your formula (apparently representing a serious violation of conservation of energy)? And also, where does your extra 1/2 mv^2 of kinetic energy of the particle go when the particle slows down to v=0 and gives up its standard kinetic energy 1/2 mv^2 to its environment?
> Well, seems to me that it always was double. You put down the equation yourself:
> 
> K=1/2 mv2 so doesn't this mean that 2K=mv2?
> 
> Let's now look at energy calculated from momentum, we have p=mv, so v=p/m
> 
> If we replace v by p/m in K=1/2 mv2 we end up with K=m (p/m)2/2 and finally
> 
> K=p2/2m and then 2K=p2/m
> 
> Seems to me that this is consistent with my derivations, no?
> 
> There always was twice the amount of kinetic energy energy involved in calculating mass velocities from Newton. The pecularity is that it was coherently divided by 2 to isolate the momentum component, since he did not know what to do with the other half (mass increase with velocity was an unknown concept at the time).
> 
> The second K is the part that converts to the transversely oscillating half, which is the relativistic mass increment that was measured by Kaufmann.
> 
> Newton`s equations were ok in this regard. The only missing parameter was the gamma factor, that he couldn't know about and that allows calculating the energy according to the proper electromagnetism compliant growth curve.
> 
> I invite you to try any relativistic velocities calculations with my equations. You will see that they all pan out. The proof is in the pudding as the saying goes.
> 
> Looks like it is the SR version that is erroneous: KE=moc2(γ-1), because it calculates only the momentum amount, not the other half that goes into velocity related mass increment. It is not even in sync with Newton at low velocities. Seems to me that it should be corrected to KE=2moc2(γ-1) to even conform to Newton at low velocities.
> 
> regarding where where the extra 1/2 mv^2 of kinetic energy of the particle go when the particle slows down to v=0 and gives up its standard kinetic energy 1/2 mv^2 to its environment.
> 
> Remember that the only way an electron can be naturally stopped in physical reality is for it to stabilize in resonance state in an atomic orbital.
> 
> when this occurs, a photon carries away its momentum translational complement of carrying energy, but since it is subject to the Coulomb force between it and the nucleus, it is synchronously induced with a replacement equal amount of kinetic energy that now applies pressure towards the nucleus. The other half that makes up the relativistic mass increment remains unaffected.
> 
> Ref: https://www.omicsonline.org/open-access/on-adiabatic-processes-at-the-elementary-particle-level-2090-0902-1000177.pdf <https://www.omicsonline.org/open-access/on-adiabatic-processes-at-the-elementary-particle-level-2090-0902-1000177.pdf>
> Yes, I think that kinetic energy is made of something physical (in its broadest sense.) But the word “substance” — from Latin “substantia”= “being, essence” -- like the word “matter” from Latin “mater”= “mother"), is a metaphysical term that cannot be clearly defined by physics experiments. Even the word “physical” is from Latin “physica” = “things related to nature” and is a metaphysical term.
> 
> From my perspective, if it can be scattered against, it is physically there. So it has to be made of "something that physically exists to be scattered against". Confirmed out of any doubt in my view in countless scattering experiments. From my analysis, this substance is physically existing kinetic energy.
> Are you familiar with the demonstration that the well-known relativistic energy-momentum equation E^2 = p^2 c^2 + m^2 c^4 for a particle of mass m is mathematically equivalent to (E/c)^2 = p^2 + (mc)^2 where mc can be interpreted as the transverse circling internal linear momentum of a resting particle of mass m, p is the particle's longitudinal linear momentum p=gamma mv and E/c is the Pythagorean total linear momentum P=E/c = gamma mc of the particle?
> Regarding the energy-momentum equation E^2 = p^2 c^2 + m^2 c^4, I invite you to do a google search about it. You will discover that everywhere it is erroneously written E^2 = p^2 c^2 + mo^2 c^4.
> 
> Note the (mo).
> 
> You will also find it in immediate access on wikipedia, which is bound to misinform all those who trust the site.
> 
> This is an erroneous representation because the gamma factor is simplified out of this form of the equation, and reduces it to Newton`s non-relativistic form. I append the complete and proper derivation that was published as Appendix B in my recently published monograph.
> 
> 
> These relations can be seen in Figure 1 of my spin-1/2 charged half-photon electron model in my article athttps://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength <https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength>. What is called a charged photon in this figure and article I am now calling a charged half-photon. Nothing is superluminal in this figure.
> I had a look. Looks quite interesting as all your other papers. I downloaded a copy for reading at leasure.
> 
> Best Regards
> 
> André
> 
>  
> 
>  
>>  
>> On Nov 28, 2017, at 9:20 AM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>  
> 
> Hi Richard and all,
> 
> I understand that your composite photon model involves no extra energy associated with the internal superluminal motion of the charged half-photons.
> 
> But the thing is (not meant here as a negative point) that by structure, in your model, the half photons longitudinally progress on spiral paths at some distance from the axis of motion of the linearly moving photon, they have to move "by structure" at a velocity slightly higher than c otherwise, they could not keep up with the axial linear motion that is defined as being c. Just geometric logic irrespective of any math considerations.
> 
> I'll try to explain more clearly about what I understand of kinetic energy, but to remain with the photon structure, by contrast to your "longitudinally" spiralling half photons (moving charges), in the trispatial model, the two charges are oscillating "transversally" with respect to the direction of motion, which is in sync with the established understanding that electromagnetic energy oscillates transversally.
> 
> As the related "kinetic energy" of which the half photons are made cyclically transfers from twin-particle state (charges state) to single magnetic particle state in the trispatial model, it accelerates from zero transverse velocity at maximum transverse extent to transverse velocity c at mid transfer to zero transverse velocity again when completely transferred to magnetostatic space. Then in reverse motion, from zero transverse velocity at maximum magnetic spherical extent to transverse velocity c at mid transfer to twin electric charges state to zero transverse velocity at maximum half photon state. Ready for the next transverse reciprocating cycle again. All the while, the complete photon translationally progresses at c, propelled by the other half of the photon's total quantum of energy, that remains permanently unidirectional while the transversally oriented half oscillates.
> 
>  
>> This is why c is never exceeded in the trispatial model.
>> 
>> Now to the nature of kinetic energy.
>> 
>> You mention that when I call kinetic energy a “substance”, this confuses you.
>> 
>> I know. I observed that it confuses everybody.
>> 
>> I have a question that may help you seeing what I mean. In you own photon model, what do you think your half photons are made of really? Charges, would you say! What are charges then? We don't really know.
>> 
>> But whatever what we name "charges" may be, don't you think that a physically existing "substance" has to be involved, that they must be quantities of "something" that physically exists?
>> 
>> Now what could this "something" be?
>> 
>> Remember that we talked about the 13.6 eV electromagnetic photon that is evacuated when an electron is captured by an ionized hydrogen atom (a proton)?
>> 
>> You certainly observed already that this amount of energy corresponds exactly to the kinetic energy that can be calculated due to the acceleration of the electron until it reaches the velocity related to the Bohr radius, which is 2187691.253 m/s classical (2187647.561 m/s relativistic).
>> 
>> Since the electron is now "immobilized", we know that this energy is indeed the energy corresponding to the maximum momentum that the electron had before "hitting the wall" so to speak. Now instead of converting to "potential energy" as classical physics assume, "we observe" that it escapes as an electromagnetic photon.
>> 
>> This means that unidirectional momentum related kinetic energy that propelled the electron is of the same nature as electromagnetic energy, and that in fact the complete complement of any electromagnetic quantum is actual kinetic energy "that-did-not-convert-to-potential-energy", half of which transfers to transverse orientation to henceforth electromagnetically oscillate, in transverse reciprocating motion (in the trispatial geometry), and spiralling about the axis of motion in your model.
>> 
>> This is why I concluded that "kinetic" energy is a "physically existing substance".
>> 
>> When 1.022 MeV photon sare made to convert to massive electro-positron pairs, this means that even the mass of electrons and positrons can also only be the very same substance also. What else could it be since the mother photon was entirely made of only this kinetic energy substance. Whatever it may turn out to really be in reality. Whatever it really is, it has to be a really and physically existing "substance", that we need to study. That's my conclusion.
>> 
>> Now to the extra fact or 2 that you mention,
>> 
>> You mention "KE=moc2(γ-1) wheremois usually just written m"
>> 
>> To remain strictly consistent with mathematical representation, m is the relativistic mass of the electron that was measured by transverse interaction by Kaufmann 100 years ago, which is made of mo + [KE=moc2(γ-1)]/c2. Here [KE=moc2(γ-1)]/c2 is the relativistic mass "increment".
>> 
>> Now KE=moc2(γ-1) also happens to also be equal to the unidirectional momentum related kinetic energy amount that propels the electron at velocity v (embedded in the gamma factor).
>> 
>> Consequently, the complete complement of kinetic energy that an electron possesses at any velocity in excess of its own mo energy value is KE=moc2(γ-1) + [KE=moc2(γ-1)]/c2 both of which physically exist (my conclusion), and can be summarily calculated with:
>> 
>> KE=2moc2(γ-1)
>> 
>> This stems from correlating the Kaufmann electron deflection experiment with Marmet's discovery that the magnetic field of a moving electron increases in sync with its velocity. Explained at the beginning of the "From Classical to Relativistic Mechanics via Maxwell" paper, but as analyzed completely in this other paper:
>> 
>> http://www.gsjournal.net/Science-Journals/Essays/View/2257 <http://www.gsjournal.net/Science-Journals/Essays/View/2257>
>> 
>> You ask: "Are you claiming that a moving massive particle has twice as much kinetic energy as is predicted by standard relativity kinematics?"
>> 
>> My answer is YES.
>> 
>> Grounded on the Kaufmann experiment PLUS Marmet's converging derivation, that is one half remaining unidirectional (momentum kinetic energy) plus the other equal amount transferring to transverse orientation to henceforth oscillate electromagnetically in reciprocal swing between twin component electric and single component magnetic states, displaying omnidirectional inertia (mass) like the invariant rest mass mo of the electron.
>> 
>> Hope this helps understanding how I see "kinetic energy".
>> 
>> Best Regards
>> 
>> ---
>> André Michaud
>> GSJournal admin
>> http://www.gsjournal.net/ <http://www.gsjournal.net/>
>> http://www.srpinc.org/ <http://www.srpinc.org/>
>> 
>> On Mon, 27 Nov 2017 22:15:30 -0800, Richard Gauthier wrote:
>> 
>> HelloAndré (and all),
>> 
>> Thank you.
>>  
>> I want to clarify that in my composite photon model, there is no extra energy associated with the internal superluminal motion of the spin-1/2 charged half-photons composing the composite photon model, whose superluminal energy quanta both move helically at c sqrt(2). The energy of each half-photon is given by E=hf = h/lambda where f is the frequency of the half-photon (here E is not the energy of the composite photon which is 2E) and lambda is the wavelength of the half photon. Remember that the half-photon makes two helical turns per half-photon wavelength, so its apparent frequency 2f (due to its double-looping, like the zitterbewegung frequency) is twice its energy-related frequency f, and its apparent wavelength lambda/2 (due to its double-looping per wavelength lambda) is half of its energy-related wavelength lambda.
>> I’ve been working on understanding your tri-space approach to the photon and electron. I am trying to understand your approach to kinetic energy of a massive particle. In one place you call kinetic energy a “substance”, which confused me. What does this mean? Is this substance different from energy?
>> 
>> Also, in standard special relativity kinematics, the formula for the kinetic energy of a massive particle is
>> 
>> KE=moc2(γ-1) wheremois usually just written m
>> 
>> while in your article "From “Classical to Relativistic Mechanics via Maxwell”athttp://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197 <http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197>
>> you write: "So from equation (42) we can now directly calculate the associated kinetic energy even if we know only the relativistic velocity of a particle
>> 
>> K=2moc2(gamma-1) (43)
>> 
>> You write the same equation in a slightly different form in your article “On De Broglie’s Double-particle Photon Hypothesis”
>> 
>> at: https://www.omicsonline.org/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.php?aid=70373 <https://www.omicsonline.org/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.php?aid=70373>:
>> 
>> “ From equation (39) can be derived the following equation that allows calculating the kinetic energy that must be communicated to an electron for it to move at relativistic velocity v, when only this velocity is known:
>> 
>> x = 2a(γ-1) (40)
>> 
>> Where “x” is the added kinetic energy, “a” is the energy making up the rest mass of the electron and γ is the Lorentz gamma factor. Any relativistic velocity plugged into the gamma factor will allow obtaining the amount of kinetic energy required for the particle to move at this velocity.”
>> 
>> So my question is, where does this extra factor of 2 come from in thetheoretical derivation of your formula for the relativistic kinetic energy of a massive particle as a function of gamma, compared to the standard relativistic formula for kinetic energy, which is well-established experimentally?Are you claiming that a moving massive particle has twice as much kinetic energy as is predicted by standard relativity kinematics?
>> 
>> Richard
>> 
>>> On Nov 27, 2017, at 4:51 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>>  
> Hi Richard,
> 
> I agree with you here that this is an incremental progress over the previous view, Moreover it is now in sync with de Broglie's original twin halph-photons hypothesis.
> 
> As I already mentioned, possibly the best that can be had in 4D geometry.
> 
> Note that In the trispatial geometry, all of this is accomplished without the energy involved ever exceeding the speed of light.
> 
> Best Regards
> ---
> André Michaud
> GSJournal admin
> http://www.gsjournal.net/ <http://www.gsjournal.net/>
> http://www.srpinc.org/ <http://www.srpinc.org/>
> 
> On Mon, 27 Nov 2017 16:15:08 -0800, Richard Gauthier wrote:
> 
> HI Ray (and all),
> You are right that an electron logically can’t be composed of a photon which is composed of an electron-positron pair (or other dipole particle structure) where each half of the dipole is composed of a photon which is composed of… ad infinitum. I think that an electron that is composed from its beginning as a spin-1/2 charged half-photon (not a photon) solves this logical dilemma. I think most of us started out thinking that an electron, if it is composed of a light-speed object, must be composed of a uncharged photon of spin 1 which somehow curls up to become a charged spin-1/2 electron with rest mass m. But I now think that this approach has become a dead-end. Rather, a pre-electron (a spin-1/2 charged half-photon) produced with pre-positron in e-p pair production from a sufficiently energetic spin-1 photon (having net charge zero) starts out already electrically charged and having spin 1/2, so it doesn’t have to change from a spin-1 uncharged particle into a a spin-1/2 charged particle by curling up and losing half its spin while gaining its charge and mass due to its being curled up in a double-loop. Rather the spin-1/2 charged photon retains the spin-1/2 and electric charge e and inertial mass (and its rest mass m) that it started with coming (with a positron) from a photon in e-p pair production. This light-speed (in its longitudinal direction) charged spin-1/2 particle composing an electron is not curled up anyway when the electron it composes is moving very highly relativistically at v < c. Then the light-speed particle composing the relativistic electron retains its charge e, its spin-1/2 and its mass m while moving at light-speed c (longitudinally) along its helical trajectory with forward helical angle theta given by cos (theta) = v/c and with an electron velocity (along its helical axis) of v < c. The spin-1/2 charged photon maintains its spin-1/2 at highly relativistic velocities because it retains the same internal superluminal speed c sqrt (2) (and transverse momentum component value p= h/lambda and its helical radius value R= lambda/4pi) that it had while composing the composite photon from which it emerged in e-p pair production. Only now lambda is much shorter when the electron is very highly relativistic. So at highly relativist velocities the charged photon’s spin remains Sz = R x p = lambda/4pi x h/lambda = h/4pi = hbar/2 = spin-1/2 . To me this understanding represents real, if incremental, progress in understanding photons and electrons (and particles in general.)
> Richard
>> On Nov 27, 2017, at 12:54 PM, Ray Fleming <rayrfleming at gmail.com <mailto:rayrfleming at gmail.com>> wrote:
>> Richard,
>>  
>> As I mentioned. The De Broglie quote in your recent paper expresses that the half-wavelength photons are Dirac Fermions. So a photon can be treated a a series of pairs of Dirac Fermions.
>>  
>> So I see these models of an electron as a photon are saying that the electron is made of a pair of Dirac Fermions while not showing why the electron is matter with negative charge rather than the opposite. Then there is the circularity issue an electron is made of a photon which is an electron-positron pair, which are both photons, which are both electron-positron pairs,...
>>  
>> Of course Dirac Fermions will behave like Dirac Fermions, but unless we break the circular logic we cannot achieve anything definitive. For my part I see The Dirac Fermion, the electron as what is fundamental. That said, I do not have a model for it. I do appreciate all the fine work on the problem.
>>  
>> One thing a lot of papers have in common is talking about the point like nature of an electron, while several including my own derivation of electron mass show that it relates to a size around the Compton wavelength. If we are to understand the true structure of an electron we need to perform scattering experiments with electrons and photons rather than protons, and at lower energies that allow us to see what is going on an the Compton scale. As I say in a couple of my books, proton scattering to find the size of an electron is simply an attempt to find the diameter of the center of a hole.
>>  
>> Ray
>> On Mon, Nov 27, 2017 at 2:25 PM, Richard Gauthier <richgauthier at gmail.com <mailto:richgauthier at gmail.com>> wrote:
>> Hello John, Martin, Vivian, Chip,André, Grahame, Albrecht, Rayand all,
>> 
>> Three of our members, that I know of, have derived the de Broglie wavelength in different ways from our double-looping-photon-like-object electron models having spin-1/2: John (and Martin), Vivian and myself. I don’t know if Grahame, André, Chip or Albrecht have derived the de Broglie wavelength from their electron models (and if so, where), but I would like to know.
>> 
>> The three de Broglie wavelength derivations from the above electron models are at:
>>  
>>  
>> 1. Is the electron a photon with toroidal topology?”, J.G. Williamson and M.B.van der Mark, http://www.cybsoc.org/electron.pdf <http://www.cybsoc.org/electron.pdf>, section 6, pp15-16.
>> 2 “A Proposal on the Structure and Properties of an Electron”, VNE Robinson,https://www.academia.edu/10819172/A_Proposal_on_the_Structure_and_Properties_of_an_Electron <https://www.academia.edu/10819172/A_Proposal_on_the_Structure_and_Properties_of_an_Electron> , section 9, p.13
>> 
>> 3. “Electrons are spin 1/2 charged photons generating the de Broglie wavelength”, Richard Gauthier,https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength <https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength>, section 11, pp 9-11. What I call “charged photons” in my article I am now calling “charged half-photons”, but this does not affect the derivation.
>> 
>> Since we are focusing on the validity of the de Broglie wavelength relation in this email thread, I would like to know if anyone, besides myself, sees any serious error in any of the three de Broglie wavelength derivations above. If there is an error in the derivation in my electron model, I would certainly like to know what it is, and I think that the others feel the same about theirs. Thanks!
>> 
>>  
>> Richard
>> 
>>> On Nov 24, 2017, at 3:41 PM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>> Hello Chip,
>>> 
>>> You touch an important point by highlighting that at α*cvelocity, would-be dilation or contraction tiny at these velocities, since it lies in the very low relativistic velocity range.
>>> 
>>> A note however regarding motion at such velocity on a "trajectory" about the nucleus of the hydrogen atom, Heisenberg concluded that the electron did not have to translate at any velocity to remain captive on the ground state orbital. From my analysis from the trispatial perspective, I tend to agree with him. Considering how both electrons have to remain by structure midway between the two protons in a hydrogen molecule for their covalent bounding to be logically explainable, this seems to be factual from my perspective.
>>> 
>>> Best Regards
>>> ---
>>> André Michaud
>>> GSJournal admin
>>> http://www.gsjournal.net/ <http://www.gsjournal.net/>
>>> http://www.srpinc.org/ <http://www.srpinc.org/>
>>> 
>>> On Fri, 24 Nov 2017 17:13:54 -0600, "Chip Akins"wrote:
>>> Hi Albrecht and Andre
>>>  
>>> First, Albrecht, I agree that the de Broglie wave, as envisioned by de Broglie, leaves much unexplained, and may well be simply wrong, even though it sort of fits partial mathematical descriptions which seem to be shedding some light on the atomic orbitals in a narrow set of circumstances. This failure of the de Broglie hypothesis to work in all circumstances is a part of the reason I started looking into this subject more. The de Broglie wave also seems to fit double slit experiments, but does not really offer a foundation of physical cause. It just seems to work that way without really disclosing the physical reasons for the de Broglie wave itself. So I think we should look for a better foundation, a causal and concrete explanation, instead of building elaborate theoretical structure on speculation which still remains unexplained. So, yes, there are occasions where the speculation of de Broglie can be applied, and we get the right numbers, but that does not mean the theory is correct and we should stop looking for the actual answers.
>>>  
>>> Andre is onto something when he looks for a relationship between the fields of the proton and the fields of the electron to sort out these issues of the quantization of orbitals. However I will have to do some more math to see if the magnetic field relationships can actually be the answer. At this point, prior to doing the requisite math, I think there is also the possibility that certain dynamics of the electric fields created by the proton and electron will explain the quantization of orbitals. But this is premature speculation. And is sort of a moot point because the dynamics of electric fields are the cause of magnetic fields.
>>>  
>>> There exists a beat frequency which is ¼ the de Broglie wavelength, and this beat frequency is a natural condition of the electron in the significantly sub-light speed orbital (example: a mean circular path at α*cvelocity), so it requires no speculation about dilation or contraction (which are tiny at these velocities).
>>>  
>>> So I think you are both quite correct to look into these issues. We have a lot to gain by reexamining our theoretical basis.
>>>  
>>> Chip
>>>  
>>> From:General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org <mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org>]On Behalf OfAlbrecht Giese
>>> Sent:Friday, November 24, 2017 4:25 PM
>>> To:general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org>
>>> Subject:Re: [General] Compton and de Broglie wavelengththe "error"
>>>  
>>> Hi André, Chip, and all,
>>> if we discuss de Broglie's concept of a particle wave, we should in my view refer to his original work and not to others who have used the results (well understood or misunderstood) in other applications.
>>> So, de Broglie in original:
>>> It is of course correct that de Broglie did not just “assume” his wave but he has deduced it from considerations about relativity. But his deduction is based on a severe error as I have explained in detail earlier. So, let’s do it again.
>>> De Broglie has seen a logical conflict between the Einstein- Planck relation (1) E=h*frequency and (2) relativistic dilation; because according to (1) the frequency has to increase at motion and according to (2) dilation will cause the frequency to decrease. But his concern is an error as this conflict does not exist. Because we have to look at an interaction of particles, which is the relevant situation. Any interaction sees frequencies which are increased by the Doppler effect. And the Doppler effect gives an over-compensation of the normal relativistic slow down so that both frequencies above will fit on their own. The same result is achieved if the temporal Lorentz transformation is properly applied. - For de Broglie's new wave no justification exists at all.
>>> The comment of two of you that a single electron does not produce an interference pattern is of course correct. One electron only produces one dot on the screen. But if we assume that a bunch of electron flies to the multi-slit with same speed then the argument works. There will be an interference pattern behind the multi-slit. But if we transform the experiment into the frame of the electrons then the momentum of the electrons is zero, and so the wavelength is infinite, and seen from that frame no interference pattern can occur. But it does occur, also visible for a co-moving observer, and that shows that de Broglie's idea is erroneous. - I have shown in calculations (but not in this place) why under certain circumstances the impression occurs that de Broglie is correct. But in general it is wrong. De Broglie's approach violates Galileo's relativity as well as Lorentzian relativity.
>>> You have mentioned the good results of the use of the de Broglie wave to determine the quantization of atomic orbits. It is true that it works, but it has a similar problem like for the scattering of electrons. Assume a hydrogen atom moving into axial direction with a similar speed as the speed of the electrons in the orbits. Then the resulting momentum of the orbiting electrons increases by about 40% seen from the frame at rest. So the de Broglie wavelength has to decrease by this factor and the energy of these states has to change accordingly. But in practice there will be a much smaller energy change. So also in this case de Broglie fails at a more thorough look.
>>> In the mails there have been some considerations about what de Broglie did "have in mind". But what he had in mind he has written in his PhD thesis. Anything about the energy states of atoms came later and by others (like Schrödinger and Bohr).
>>> Now I will be wondering about objecting arguments.
>>> Albrecht
>>>  
>>> 
>>> I thank you for your answers and arguments. I will now answer to it, of course. Which means to repeat my arguments of the last three weeks here where I have given argument which seem to have been overlooked.
>>>  
>>> Am 24.11.2017 um 01:20 schrieb Richard Gauthier:
>>> Hi John,André, Chip and all,
>>> Deriving the de Broglie wavelength of an electron model without superluminal motion is easy (in hindsight, since de Broglie did it using special relativity.) But try getting, without superluminal motion, the spin-1 of a non-pointlike photon model (for a photon-in-a-box or otherwise) AND the spin-1/2 of a highly relativistic non-pointlike electron model. In either case there will be some longitudinal momentum Plong, at light speed for a photon model and at very near light speed for a highly relativistic electron model, as well as some significant locally transverse linear momentum Ptrans (even if the net transverse linear momentum of the photon model is zero as in the double-helix photon model) that generates spin Sz = R x Ptrans = 1 hbar for a photon model or 1/2 hbar for a highly relativistic electron model . A longitudinal light-speed or near-light-speed linear momentum vector plus a significant local transverse linear momentum vector gives a diagonal local linear momentum vector with a corresponding diagonal velocity vector whose magnitude is greater than c. Putting a photon model’s or electron model's transverse oscillatory motion, that generates its spin, into two different transverse dimensional spaces is ingenious, but if the photon is to move along longitudinally as a whole and not leave the two transverse dimensional spaces behind, I think there will still be some diagonal superluminal motion. I would be happy to see a proved counterexample.
>>> Richard
>>>  
>>> On Nov 23, 2017, at 12:19 PM, John Williamson <John.Williamson at glasgow.ac.uk <mailto:John.Williamson at glasgow.ac.uk>> wrote:
>>>  
>>> Hi Richard and everyone,
>>> 
>>> You do not need to add anything. "Superluminal" is not needed. If you consider light-in-a-box (including light in a box of its own making) the de Broglie wavelength follows from the beat frequencies of the proper relativistic transformations of the light going with the motion and that going against. Remeber, one needs to consider BOTH the Doppler shift AND the SR transformations. Then everything works. Martin is writing a definitive paper on this.
>>>  
>>> Regards, John.
>>> From:General [general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org <mailto:general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org>] on behalf of Richard Gauthier [richgauthier at gmail.com <mailto:richgauthier at gmail.com>]
>>> Sent:Thursday, November 23, 2017 6:36 PM
>>> To:srp2 at srpinc.org <mailto:srp2 at srpinc.org>; Nature of Light and Particles - General Discussion
>>> Subject:Re: [General] Compton and de Broglie wavelengththe "error"
>>> 
>>> Hello André, Chip, John and all,
>>>  
>>> I also think that there is “an additional factor” that settles an electron into an atomic resonant state. In my view the electron is composed of this additional factor, a charged superluminal energy quantum that circulates and generates quantum waves having the de Broglie wavelength. These quantum waves self-resonate in regions around an atomic nucleus. When an available resonant region around an atomic nucleus is found, the superluminal energy quantum settles into this region and continues to emit quantum waves that for some period of time maintain it in this resonance state in the atom. The electron is more likely to be detected wherever the amplitude of this resonant state (the electron’s eigenfunction for this state) is larger.
>>> 
>>> This idea is not fully developed but is hinted at in “Transluminal Energy Quantum Model of a Spin-½ Charged Photon Composing an Electron”,“Electrons Are Spin-½Charged Photons Generating the de Broglie Wavelength”,“The Charged-Photon Model of the Electron Fits the Schrödinger Equation”and “The Charged-Photon Model of the Electron, the de Broglie Wavelength, and a New Interpretation of Quantum Mechanics" athttps://richardgauthier.academia.edu/research#papers <https://richardgauthier.academia.edu/research%23papers>. What I called a charged photon in theses articles I am now calling a charged half-photon.
>>> 
>>> Richard
>>>  
>>>  
>>> On Nov 23, 2017, at 8:52 AM, André Michaud <srp2 at srpinc.org <mailto:srp2 at srpinc.org>> wrote:
>>>  
>>> Hi Chip, and all
>>> 
>>> You write: "I prefer the second option, there is some additional factor interacting with the electron, to cause these quantized orbitals, and understand from Andre’s writings that he feels the same way."
>>> 
>>> You are exactly right about what I think. I came to the same conclusion as yourself (the second option) way back in fact when I finally lighted up to the fact that the wave function originally was related to electrons orbitals by Schrödinger because he was inspired in this direction by a conclusion of de Broglie that electrons had to be captive in some form of resonance state about nuclei.
>>> 
>>> I think that this was sort of lost sight of in the community due to the acrimonious debate that raged on afterwards between the proponents of the Copenhagen school and the determinists, which indeed was fundamentally whether the first or second option applied in physical reality.
>>> 
>>> After I came to the second option conclusion, I started to look around for descriptions of this resonance state that could be related to the wave function but found nothing, as if the only option that had been explored was the first one, with which the Heisenberg solution was in harmony and also later Feynman's path integral.
>>> 
>>> To me, the idea of "resonance" always made me think of a vibrating guitar string, whose shape and extent of the volume visited by the transversally oscillating string can be described by the wave function.
>>> 
>>> I suspected that this might have been what de Broglie had in mind also, and became convinced that the electron could remain localized while being captive within the theoretical volume defined by the wave function, on an axial resonance trajectory (sort of stochastic maybe to some extent) that may be describable mathematically and that could be due to electric versus magnetic interaction between the electron and the nuclei.
>>> 
>>> I see that you lean in a similar direction Chip. I have explored the possible electric vs magnetic potential explanation to a large extent, but I am at a loss as to how to exactly mathematize the localized resonance trajectory proper within the volume definable by the wave function. You seem to be better equipped mathematically than me to address such an issue, with your¼ de Broglie wavelengthexploration.
>>> 
>>> For a general overview of how the trispatial geometry allows defining this type of electromagnetic electron equilibrium states involving both electric and magnetic aspects of energy, here is my final paper on the whole concept:
>>> 
>>> https://www.omicsonline.org/open-access/gravitation-quantum-mechanics-and-the-least-action-electromagneticequilibrium-states-2329-6542-1000152.pdf <https://www.omicsonline.org/open-access/gravitation-quantum-mechanics-and-the-least-action-electromagneticequilibrium-states-2329-6542-1000152.pdf>
>>> Even though it involves an entirely new paradigm that may feel very unfamiliar at first, I hope it nevertheless makes some sense to you.
>>> 
>>> Best Regards
>>> 
>>> ---
>>> André Michaud
>>> GSJournal admin
>>> http://www.gsjournal.net/ <http://www.gsjournal.net/>
>>> http://www.srpinc.org/ <http://www.srpinc.org/>
>>> 
>>> On Thu, 23 Nov 2017 05:16:52 -0600, "Chip Akins"wrote:
>>> 
>>> Hi All
>>> But in all this, regarding de Broglie’s wavelength and the electron orbitals, there is still something missing.
>>> Either we have to assume that the electron occupies the entire circumference of the orbital simultaneously by its wavefunction, or there is some additional factor interacting with the electron, to cause these quantized orbitals.
>>> I prefer the second option, there is some additional factor interacting with the electron, to cause these quantized orbitals, and understand from Andre’s writings that he feels the same way.
>>> In the hydrogen atom there is a simple, naturally occurring cause, for a “matter wave” which is exactly ¼ the de Broglie wavelength. This “matter wave” is a beat frequency created by the perceived frequency difference with motion, of the outer radius and inner radius of the electron as it circulates about the proton. I found this to be interesting, and wanted to share this observation.
>>> Chip
>>> From:General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org <mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org>]On Behalf OfAndré Michaud
>>> Sent:Wednesday, November 22, 2017 10:52 PM
>>> To:general at lists.natureoflightandparticles.org <mailto:general at lists.natureoflightandparticles.org>
>>> Subject:Re: [General] Compton and de Broglie wavelengththe "error"
>>> Hello John,
>>> You are absolutely right.
>>> 
>>> In fact de Broglie derived this relation with respect to the values of the Bohr ground state orbit energy parameters.
>>> 
>>> Heisenberg did the same, except that he formulated the relation so that it could account for a precision drift of the chosen velocity on either side of the selected velocity value about the ground orbit of the Bohr atom.
>>> 
>>> In 1923, he himself expressed his uncertainty principle as delta_x delta_p equal-or-larger-than h, which is the same as delta_x approx_equal to h / (m delta_v_x), which is fundamentally de Broglie's single valued h/mv for the Bohr ground state orbit.
>>> 
>>> This is at the origin of Heisenberg's statistical solution.
>>> 
>>> Best Regards ---
>>> André Michaud
>>> GSJournal admin
>>> http://www.gsjournal.net/ <http://www.gsjournal.net/>
>>> http://www.srpinc.org/ <http://www.srpinc.org/>
>>> 
>>> On Thu, 23 Nov 2017 03:17:31 +0000, John Williamson wrote:
>>> 
>>> Dear Albrecht,
>>> 
>>> Your error is more fundamental than you know. See below in green.
>>> From:General [general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org <mailto:general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org>] on behalf of Viv Robinson [viv at universephysics.com <mailto:viv at universephysics.com>]
>>> Sent:Wednesday, November 22, 2017 10:49 PM
>>> To:Albrecht Giese; Nature of Light and Particles - General Discussion
>>> Subject:Re: [General] Compton and de Broglie wavelengththe "error"
>>> Dear Albrecht,
>>> IMHO you have a fundamental flaw in your first paragraph below. A single electron cannot generate an interference pattern, any more than can a single photon. An observer moving with a single electron will, if the screen is angled towards him, see only a single spot where the electron impinged upon that screen. That is all. If he repeats that observation say 10,000 times he will still only see on spot each time the electron impinges upon the screen. If the spots are recorded, each time he travels with another electron he will see an interference image slowly appear because it is dependent upon the frame of reference of the slit and screen. The motion of the observer does not interfere with that pattern.
>>> Sincerely
>>> Vivian Robinson
>>> On 23 November 2017 at 8:24:21 AM, Albrecht Giese (phys at a-giese.de <mailto:phys at a-giese.de>) wrote:
>>> Dear André,
>>> the "error" which I see for de Broglie is his assumed relation lambda = h / momentum .
>>> Your error, and this is an error not an "error" is that you assume that de Broglie "assumed lambda = h / momentum. Louis de Broglie did not assume lambda = h / momentum - he derived it. From relativity. Please do not assume what you think other people assume. Remember, de Broglie was very smart, and this relation had to come from somewhere, no? It would be instructive for you to understand the how and why he did this before making uninformed comments on it.
>>> This relation fails at any linear transformation. Take as an example the scattering of electrons at a multi-slit. If you look at it from the rest frame of the multi-slit then de Broglie's wavelength describes correctly the generated interference pattern. However, if this situation is observed by someone moving at the side of the electron the result is completely wrong. Assume as an extreme situation that the observer moves together with the electron. Then in the frame of the observer the electron has the momentum = 0 and so the wavelength is infinite. This means: no interference! But the pattern does of course not disappear and will be visible to the observer. This shows that de Broglie does not even fulfil Galileo's physical rule of relativity believed and proven since 600 years.
>>> Regarding the particle mass: My equation is simple: m = h(bar) / (c*R) , where R is the radius of the particle. And R can be easily determined by use of the known magnetic momentum of the particle.
>>> The mag. momentum of a circling elementary charge is classically: mm = (1/2)*c*e0*R
>>> The mag. moment of particles is known. So, R can be determined. This R inserted into the equation above yields the particle mass with an accuracy of about 10-3. - This is now based only on the strong force. If the result is corrected by the influence of the electrical charge, this yields the Landé factor in case of the electron. This applied yields the mass with an accuracy of 2*10-6.
>>> References for this are:www.ag-physics.org/rmass <http://www.ag-physics.org/rmass>andwww.ag-physics.org/electron <http://www.ag-physics.org/electron>.
>>> Hope this explains it. Otherwise please ask.
>>> 
>>> Albrecht
>>> Am 18.11.2017 um 22:54 <http://airmail.calendar/2017-11-18%2022:54:00%20AEST>schrieb André Michaud:
>>> Dear Albrecht,
>>> 
>>> I must say that I don't see as "errors" conclusions that were drawn before more precise knowledge was discovered. For example, I don't think that Newton made an "error" by not immediately concluding to the possibility the fixed velocity of light. He simply did not know about it because this had not yet been discovered.
>>> 
>>> The same for de Broglie in my opinion, he worked with the knowledge available a the time.
>>> 
>>> As i understand it, what we call the de Broglie wave is simply a representation of the sum of the energies of the rest mass of the electron plus the translational energy related to its momentum. How can this be wrong at the general level, unless I misunderstand the whole concept?
>>> 
>>> As for Hönl and the mass of the electron, I was meaning this rhetorically. I simply mean that any solution that exactly provides the exact mass of the electron as experimentally measured by numerous means can only be a proper description, so your description has to be correct. The exact mass of the electron has been experimentally confirmed for over 1 century. I do not know where to look to examine your solution. Can you provide a link?
>>> 
>>> ---
>>> André Michaud
>>> GSJournal admin
>>> http://www.gsjournal.net/ <http://www.gsjournal.net/>
>>> http://www.srpinc.org/ <http://www.srpinc.org/>
>>> 
>>> On Sat, 18 Nov 2017 21:56:34 +0100 <http://airmail.calendar/2017-11-19%2006:56:34%20AEST>, Albrecht Giese wrote:
>>> Dear André,
>>> there is no doubt that de Broglie has made great contributions to the development of physics. So, if there is an anniversary in honour of him and even the Nobel price, then as many as possible of his achievements are of course presented.
>>> My concern, however, refers to a specific result of his early activities. The assumed necessity to introduce the "harmony of waves" and to deduce the "de Broglie" wavelength are based on a logical error and on a misunderstanding of SR.
>>> It is a quite funny situation that in spite of this error his result seems usable to explain certain physical processes. It is one goal of my physical activities to understand this. In one fundamental case I have found an explanation. That is the scattering of electrons at a double / multiple slit. If such experiment is viewed from a specific inertial frame (the one normally used), de Brolgie's calculation conforms to the measurement. However in any other frame it fails. - I can explain why the de Broglie wave seems to work even though it is erroneous. (Not here but I can give you a reference if you want it.)
>>> Regarding Hönl I do not understand what you say. Hönl did NOT get a correct mass by assuming only the electrical force in the electron. He was wrong by a factor of about 300 as I wrote earlier. But the calculation which I did is correct with high precision and the formula does not have any free parameters, only the standard ones. I do not know any other model which has this. Do you? Then please give me a reference.
>>> Best regards
>>> Albrecht
>>> _______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List atsrp2 at srpinc.org <mailto:srp2 at srpinc.org>
>>> 
>>> Click here to unsubscribe <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/srp2%40srpinc.org?unsub=1&unsubconfirm=1>
>>> _______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List atrichgauthier at gmail.com <mailto:richgauthier at gmail.com>
>>> <a href="http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1 <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1>">
>>> Click here to unsubscribe
>>> </a>
>>>  
>>> _______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List atrichgauthier at gmail.com <mailto:richgauthier at gmail.com>
>>> <a href="http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1 <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1>">
>>> Click here to unsubscribe
>>> </a>
>>>  
>>>  
>>> _______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List at phys at a-giese.de <mailto:phys at a-giese.de>
>>> <a href="http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/phys%40a-giese.de?unsub=1&unsubconfirm=1" <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/phys%40a-giese.de?unsub=1&unsubconfirm=1>>
>>> Click here to unsubscribe
>>> </a>
>>>  
>>>  
>>>  
>>> Diese E-Mail wurde von Avast Antivirus-Software auf Viren geprüft.
>>> www.avast.com <https://www.avast.com/antivirus>
>>>  
>>> _______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List atsrp2 at srpinc.org <mailto:srp2 at srpinc.org>
>>> 
>>> Click here to unsubscribe <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/srp2%40srpinc.org?unsub=1&unsubconfirm=1>_______________________________________________
>>> If you no longer wish to receive communication from the Nature of Light and Particles General Discussion List atrichgauthier at gmail.com <mailto:richgauthier at gmail.com>
>>> <a href="http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1 <http://lists.natureoflightandparticles.org/options.cgi/general-natureoflightandparticles.org/richgauthier%40gmail.com?unsub=1&unsubconfirm=1>">
>>> Click here to unsubscribe
>>> </a>

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.natureoflightandparticles.org/pipermail/general-natureoflightandparticles.org/attachments/20171212/7073b19a/attachment.html>