Dubbelosix Posted July 17, 2021 Report Share Posted July 17, 2021 (edited) I wrote up a piece recently and realized my previous model though insightful, required a new constant. I could never have formally realized how close my previous result was to this modified approach and I feel confident it tells us something very deep about the stable orbits, a planetary model akin to Keplers model, motivated to explain why electrons in the stable orbits explain why they do not radiate due to them being in a state of free fall. Before we explore the revised model, ill quickly rehash why I found this approach important again. I did, last year, come to derive a set of equation based from intuition, I later was informed that a similar model was pursued outside of my own model in the past, but I believe regardless, the reasoning I set my model on the exact derivations leading to those assumptions are in themselves unique. Unique as it is, it is like a starting off point to a more simpler model than the one that preceded mine. First of all, why did I pursue a free falling model of an electron the in the stable orbits of an atom? 1. Electrons cannot be at rest in a stable equilibrium 2. Electrons cannot be in motion pursuing ellipitical orbits These two basic premises of quantum theory are at odds with each other. One thing was certain, if electrons had motion then Bohr was right: 3. Only non-stable orbits with motion can allow atoms to radiate. So with the help of quantum theory, Bohr created a model of the atom where electrons followed ''special orbits'' and these ''special orbits'' where the ones where electrons had been able to whizz around without giving off the usual radiation we would expect from a moving charge. Later, the planetary model was superseded by wave mechanics, the idea was simple enough 4. That electrons did not move around the nucleus, instead they existed as a wave spread out statistically in space. But with this, there was a catch. DeBroglie, the true inventor of the wave particle duality model, for all states of matter, never said that his wave mechanics specifically said this. From experiments, like the photoelectric effect and gamma scattering, we knew the particle had to exist both sometimes as a particle, other times a wave. The inseparability of the wave from the particle, lead to his famous wave hypothesis, stating that the particle was accompanied by wave. The wave itself was unobservable however, only today using very special techniques, computers and special equipment have been able to indirectly see the wave nature inside of particles. It still doesn't tell us at what time the electron would act as a particle, unless directly observed, by which time the wave would collapse and all that would remain, would be the particle. Yeah, quantum theory was weird. In spirit of deBroglie, I'd like to carry on his strong assertion that particles where guided by waves, so that we can in some way rationalise the weird nature of quantum mechanics into a regime that is more acceptable for a willing and rational mind. Certainly, why cannot a particle be guided by its wave? Matter was guided by curvature in space, and it was this correspondence of the two ideas where I linked perhaps a unity between the strange wave mechanics of deBroglie to that of GR. One stated that matter told space how to curve and space told matter how to move, whereas particles told waves how to spread, and the waves told particles how to move. Maybe wave mechanics and curvature where closely related. This was my first motivation. We'll learn as my theory progresses, that the orbits described by the moving particle also contain their own curvatures, their own geometries. By inviting the weak equivalence principle, we would further learn how to allow a particle to move in an orbit without giving off any radiation. My derivation was at best, rudimentary. To do it, I just needed to know some basic laws, like Newtons laws, Keplers laws and some electrostatic equations of motion. In the more complicated arguments I explain that this acceleration disappears in the ground state, again due to it being in a state of free fall... Or does it? You might come to learn that accelerated charged objects experience only radiation, and this is true. Sometimes you might hear, ''the acceleration from a free falling body,'' and so both those statements might seem a bit at odds. What it means is that the body has to be freely falling in a gravitational field, not affected by an external source that adds to that acceleration in some way, otherwise it would obey the relativistic Larmor equation for circular orbits, or those followimg ellipses. There is still existing in this framework, the notion of wave mechanics in the theory from a de Broglie guiding wave model and would quicly make it semi classical , but I'm interested only in the case where the wave pilots the electron as curvature and waves may be saying the same thing since again. Newton would have found when he was deriving Keplers results, a moment of clarity from his own laws, it was Kepler who actually guessed his work! Kepler not once ever derived his equations that described motion of planetary bodies, it was all trial and error. And some excellent guesswork. We also note that Gm = r(s)v^2 as the gravitational parameter. This is important to Keplers work and Newton found it, but it features strongly in a second investigation I made which will be for a following day. I really only came to derive my own conclusions after being inspired by a book by Rogers where he stated that maybe particles obeyed Kepler's laws inside the atom, so I rewrote the gravitational theory in the language of electrostatics, so I was very surprised afterwards to learn that my idea of a free falling model of an electron had been speculated more thoroughly than my rudimentary model gave. I had no insight of a free falling model before this as I drew on a theoretical model from weak equivalence to explain why the ground state electron did not radiate. Though I claim my model is quite a bit more simpler, I think it gives a clearer insight how I fell upon these ideas which appeared to exist in literature outside of my independent model; Michał Gryziński - Wikipedia In his work, and no I haven't read the actual paper, but there is a Langrangian derived. It appears as a Kinetic term and a potential term and the very last term describing the spin and orbit of the equation, the spin orbit equation is something I am pretty well read up on. I haven't gone as far to describe any of the results I came to in terms of a Langrangian. In regards To Michals approach, I had already been postulating on how different orbits could be more accurately described and more recently suggested a correction term to the spin orbit equation as the eccentricity of the electrons orbit, a dimensionless parameter, which if an electron was moving around the nucleus, in a real way, eccentricity would become part and parcel of the dynamics Lets do the math now, because its a bit complicated how I came to it. I had two results, both as important as the other. Let me remind the reader, I came to explain that the charge relationship following a Bohr orbit, using the results of Kepler later confirmed from Newtons classical equations was Ze ~ 4 π² m /k(B)(R³/t²) Where (R³/t²) is the Kepler law for various orbits and notice and keep in mind, k(B) the Coulomb or Boltzmann constant. The latter will come to have a makeover soon. I decided, if we talk about the Hydrogen atom, it must be set equal to one. As a stringent result, a common thing to do is if the mass refers to an electron mass, this too can be set to 1. The resulting equation, which I wrote as an equality, I now write as an approximation e ~ 4 π²/k(B) ⋅ (R³/t²) because my new investigation imposes something important. For any model recognising the importance of the apparent symmetries of gravitation and the electrostatic laws, we are accostomed to notice the objects of similar nature as F = G(Mm/r²) And F = k(B) (Ze/ r²) Authors being bold this is not a coincidence, said for this reaon, that k(B) and G, are strict analogues of each other which obey inverse square laws. In this sense, Newtons masses become the analogues of charge. In the past,in models I wrote about, Id often say for that reason, Gm² was a gravitational charge with units of charge squared, but current research adds a tweak, one might argue its a matter of convention, or the units you chose to work, where now I modify it and convinced myself, the real model must satify for each unit of mass charge, we must now state its equal to √G m = √k(B) e This is to satisfy a model where F = G(Mm/r²) = k(B) (Ze/ r²) To get Keplers guess, from hard derivation from Newtons law, we note its important to notice that v²/r describes inward acceleration from a curved path in orbit, and that each revolution is a velocity satisfying v = 2 π r/t Now we write G(Mm/r²) = m(v²/r) = k(B) (Ze/ r²) Plugging in the revolution you will get G(Mm/r²) = m(2 π r/t)²/r = k(B) (Ze/ r²) Collecting the radius terms, we get GMm = 4 π² m(r³/t²) = k(B) Ze Notice in the central expression, we got the Kepler law of orbital motion. Lets rewrite this a new way. Say we break up all mass charges, the Coulomb charges and make them equal the result derived using Kepler orbital motion; √G M ⋅ √G m = √k(B) Z ⋅ √k(B) e = m (v²/r) ⋅ r² = 4 π² m(r³/t²) We notice we can cancel the small mass on the first and last expressions of the equation and gives √G M ⋅ √G = 1/m ⋅ (√k(B) Z ⋅ √k(B) e) = (v²/r) ⋅ r² = 4 π² (r³/t²) And is the same as saying GM = 1/m ⋅ (k(B) Ze) = (v²/r) ⋅ r² = 4 π² (r³/t²) And if the remaining mass charge M is always the analogue of the charge and a square root of the Coulomb constant we can get √G ⋅ √k(B) ⋅ e = 4 pi² (r³/t²) How, see 1). At the end By absorbing √G and √k(B) we rewrite a new constant, its a mixture of both gravitational and electrical constants, and we must respect its a new object we will call Φ. Rearranging we finally get e = 4 π²/ Φ ⋅ (r³/t²) From inspection of this with my original result which ignored this more careful derivation, its still surprising how close it was to e ~ 4 π²/k(B) ⋅ (r³/t²) Where we had an equation, set the electron mass which is small m=1 and the nuclear charge Z=1 for the most basic model available, the hydrogen atom. 1). Lets be more clear why this is. We can accept that √G M ⋅ √G/r² = 1/mr² ⋅ (√k(B) Z ⋅ √k(B) e) = (v²/r) = 4 π² (r²/t²r) This is an acceleration equation. mr² in the denominator of the second expression is further the rotational inertia. Alternatively, we can write (√k(B) Z ⋅ √k(B) e)/r² = m (v²/r) = 4 π² m (r²/t²r) Multipling through by √G we get √G (√k(B) Z ⋅ √k(B) e)/r² = √Gm (v²/r) = √Gm (4 π² r²/t²r) Since √k(B) Z = √Gm we can divide by through √k(B) Z and allow √GM/√k(B) Z ~ 1 giving us √G (√k(B) e)/r² = (v²/r) = 4 π² r²/t²r Now multiplying through by r² gives us, dropping the central expression √G √k(B) e = 4 π² (r³/t²) Divind through by √G √k(B) = Φ gives us our result we already confirmed e = 4 π²/Φ (r³/t²) This second way to derive itmay be simpler to understand. Edited July 18, 2021 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
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