Dubbelosix 152 Posted July 17 Report Share Posted July 17 (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 Later, the planetary model was superseded by wave mechanics, the idea was simple enough 4.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 Sunday at 09:52 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted July 17 Author Report Share Posted July 17 I have a second result, which was just as important but will write this up tomorrow. The point is that my ad hoc derivation before, used a Coulmb constant, when really its a new constant in its own right when derived more carefully. Quote Link to post Share on other sites

Dubbelosix 152 Posted July 17 Author Report Share Posted July 17 (edited) How I came to realize how the mass charge and the electric charge required that G and k(B ) are analogous and that their product must yield a new unseen constant was pretty straightforward. The new equation e = 4 π²/ Φ ⋅ (r³/t²) Is now the equation replacing the previous result which was e ~ 4 π²/k(B ) ⋅ (r³/t²) I noticed the force if written as F = G(Mm/r²) = m(v²/r) = k(B ) (Ze/ r²) Replacing, v = 2 π r/t in the central expression and multiplying through by r² yields Gm² = 4 π² m(r³/t²) = k(B ) e² This was how i came to realise that from inspecting the first and last expression that it had to satisfy a charge as √G m = √k(B ) e And from the second last equation there, while e ~ 4 π²/k(B ) ⋅ (r³/t²) May not have been entirely right, you could rearrange the results to find e² = 4 π² m/k(B ) ⋅ (r³/t²) As being dimensionally true from rearranging Gm² = 4 π² m(r³/t²) = k(B ) e² Its the charge squared which plays an interesting role in the next work. Ill let you know when its written out. Edited Sunday at 09:54 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted Saturday at 08:17 PM Author Report Share Posted Saturday at 08:17 PM (edited) Right lets finish this off with a result arguably more interesting than the last. Yes its true we can modify the spin orbit equation for the deviation of how the orbit deviates from a circle, called the eccentricity. Now... it will not play a part as such, i just want to bring our attention to the spin orbit equation itself and the Bohr inverse mass, can you recall what I got when plugging Bohrs result into it? The spin orbit equation is B = 1/2emc² ⋅ 1/r ⋅ ∂U(r)/∂r ⋅ J And one thing I did was plug in Bohrs inverse mass, 1/m = 4 π² k(B ) e² r/ h² And this produced B = 4 π² k(B ) e² r/ h² ⋅ 1/2ec² ⋅ 1/r ⋅ ∂U(r)/∂r ⋅ J or more compactly B = 4 π² k(B ) e / 2h² c² ⋅ ∂U(r)/∂r ⋅ J The factor of 2 in the denominator might be an adjustable parameter ie. It may not even be there. I then multiplyy a factor of e on both sides giving eB = 4 π² k(B ) e² / 2h² c² ⋅ ∂U(r)/∂r ⋅ J Then we find the Lorentz force as e(v x B)= 4 π² k(B ) e² / 2h² c² ⋅ ∂U(r)/∂r ⋅ Jv Jv has dimensions of mvr and (hc)² dimensions of m² v² r² ⋅ v², but we also know from a prevoous eq that the charge squared is e² = 4 π² m/k(B ) ⋅ (r³/t²) Ill show tomorrow what we find when we crunch all of this together. Edited Sunday at 09:58 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 09:39 AM Author Report Share Posted Sunday at 09:39 AM (edited) Right, lets finish this. We continue by plugging in e² = 4 π² m/k(B ) ⋅ (r³/t²) into e(v x B)= 4 π² k(B ) e² / 2h² c² ⋅ ∂U(r)/∂r ⋅ Jv And this now gives e(v x B) = 4 π² m / k(B ) ⋅ (r³/t²) ⋅ 4 π² k(B ) / 2h² c² ⋅ ∂U(r)/∂r ⋅ Jv And we now simplify, k(B ) falls out and we now attach m to Jv for clarity. We now get e(v x B) = 4 π² ⋅ (r³/t²) ⋅ 4 π² / 2h² c² ⋅ ∂U(r)/∂r ⋅ Jmv lets not be quick to multiply our constants yet, so now we want to simplify Jmv with h² c² , I'll write the dimensions out, we get. Remember what we said about the dimensions of Jv, and those contained in (hc)², as physicists preparing to act like a mathematician, we do many things to our equations in hope of finding something relevant. Some techniques, more complicated than others, cancelling terms based on dimensional analysis, is one of the easier ones (so long as you know what you are doing). We absord the mass term into the Jv part, both give Jmv = m²v² r And h² c² = m²v²r² v² cancelling them out gives us 4 π² Jmv / 2h² c² = 4 π²/rv² (we exchanye c as a velocity variable) Keplers orbital expression was found by Newton as (r³/t²) = Gm/4 π² And the gravitational parameter is Gm = rc² So, by conclusion, 4 π²/rc² is the exact inverse of Gm/4 π² or an equality inverse to (r³/t²). But it has a factor of two in the denominator from our derivation 4 π²/2rv² But this might be an adjustable parameter, meaning it might fall out of the theory. It might not, but we keep this in mind. What we discover is an equation now of the form e(v x B) = 4 π² ⋅ 4 π²/2rv² (r³/t²) ⋅ ∂U(r)/∂r if the inverse of 4 π²/rv² is Gm/4 π² Then it is the inverse of Keplers (r³/t²) meaning strictly that 4 π²/2rv² (r³/t²) Is a dimensionless parameter now. Only dimensionless objects in physics are important. We go as far to name it with new notation, a Kepler-Newton number denoted as N and we might now theorise an equation as e(v x B) = 4 π² ⋅ N ⋅ ∂U(r)/∂r N = 4 π²/2rv² (r³/t²) Edited Sunday at 10:02 PM by Dubbelosix Quote Link to post Share on other sites

JeffreysTubes8 1 Posted Sunday at 07:21 PM Report Share Posted Sunday at 07:21 PM 9 hours ago, Dubbelosix said: as physicists preparing to act like a mathematician, Which you are neither. Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 08:18 PM Author Report Share Posted Sunday at 08:18 PM (edited) 58 minutes ago, JeffreysTubes8 said: Which you are neither. Im a scientist all right. I can understand your discontempt, perhaps a dash of jealousy... when you cant do equations, incapable of using scientific doctrine, you resort to the most dumbfounded literacy, to those who do not know science might sound smart, while those who do read, learn and become aquainted with it, can only laugh at you. In a way, i do feel sorry for you. All the time though you waste away writing sheer nonesense, that sorryness i feel deminishes as a troll will be a troll to himself first. The more time you write about nonesense like tachyon fields, dark matter travelling faster than light, the more I come to realize, ill never look at you the same way again, because lets face it, you have no intentions learning science... and preaching the craziest of theories hoping for some attention... well, its these people i feel sorry for. You might dupe some hill billy with no education, but you wont dupe me or others. Edited Sunday at 08:20 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 08:21 PM Author Report Share Posted Sunday at 08:21 PM (edited) If you were capable of writing coherent literature on science, that would be acceptable. Forget the equations part. That alone would suffice to be taken at least half way seriously. Edited Sunday at 08:28 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 08:26 PM Author Report Share Posted Sunday at 08:26 PM Last bit of advice, go spend a few years learning some real science. I only attacked your threads because it was of the highest form of pseudoscience. Ive been reading textbooks since the agd of 16. Im now 36, so youve got a long way to go. And it wasnt just me, others responded in similar ways. Very few read your posts. This is why. Youre a charlatan. Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 09:32 PM Author Report Share Posted Sunday at 09:32 PM (edited) Anyway, ignoring the last posts, unless you want to know why I responded like I did to him, heres a nice research gate of academics explaining various reasons why dimensionless physical mathematical objects are important. Its a thing I used recently also in the modified spin orbit, where the eccentricity, also dimensionless, was implemented to explain how the orbit around the nuclei of atoms deviating from a near perfect circilar motion. Eccentricity may also play a role in thd equations above, but all we did for now at least, was crunch down some equations using some college minor algebra. My next step will be to work on a more complete model that will use the results above, and hopefully try and explain why some orbits give up radiation by being (in the real sense accelerated) rather than a planetary or satellite model which follows free fall which is absent of Larmor - cyclotron radiation principles. Hopefully in the next few weeks and make a writeup for some physicist friends to chew over. https://www.researchgate.net/post/What-is-the-purpose-of-dimensionless-equations Edited Sunday at 09:34 PM by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted Sunday at 11:23 PM Author Report Share Posted Sunday at 11:23 PM Heres a new line of enquiry, before bedtime. Lets put allay some mysticism behind why pi even appears in the equation F = e(v x B) = 4 π² ⋅ N ⋅ ∂U(r)/∂r N = 4 π²/2rv² (r³/t²) Why is pi squared? First, it wasnt originally, it just happened that we had to square it because v²/r described the circular inward motion of mv²/r in the force equation. This was well known to Newton. The velocity is actually the circumfrence swept by a period of time, v = 2 π r/t so 2 π r Is actually the circumfrence of a circle. But when we talk about area, a mathematician will always say, A = π r ² No coefficient of 2, but the radius in squared. A physicist would agree with this, but might start being all fancy and write an area related to gravity and fundamental constants like A = r ∫ dr = 2Gℏ/c³ Nothing wrong with it, its actually very interesting, as the quantum version picks up that factor of 2, along with three fundamental aspects of nature. Lets remind ourselves of some things about the model we are envisioning in the OP. The particle (electron) guided by a wave presence, some of the waves are in phase, others not, regardless, the electron is following a Kepler planetary orbit around some fixed axis and as a result, will sweep an area. A physicist knows exactly what this means, any force per area, will exert a pressure. Pressure is a similar animal to stress energy, im the sense it os encoded within the effective parameters describing the matrix T_ij in Einsteins equations, the stress tensor by another name. I will write out the covarianf equations later for curved paths in magnetic fields another day, but for sake of keeping this all simple, the magnetic cross product Lorentz force we looked at came in the form of F = e(v x B) = 4 π² ⋅ N ⋅ ∂U(r)/∂r If we divide through an area term of either A = π r ² Or A = π r ²/4 we should get vack a pressure, often denoted as P. We now write P = 4 π² / π r ² ⋅ N ⋅ ∂U(r)/∂r simple arithmatic tells us we should get back two possible new equations that only differ by a numerical constant P = 4π/r ² ⋅ N ⋅ ∂U(r)/∂r P = 8π/r ² ⋅ N ⋅ ∂U(r)/∂r But hasnt it messed the rquation up? Afterall, did we not get the factor of 4 π² because we where talking about circular motion? And why is pi squared now just pi? More to the point, i suppose, does it mean anything? Welcome to the world of interpretation. Its one thing deriving an equation, its another correctly assigning an interpretation to it. Now that we speak of pressure, one part of the intetpretation may be bodly connected to two gravitational equations in physics, one being the Einstein equation(s) G(Tensor) = 8πG/c⁴ T_ij where we have only a factor of π and its not squared in any way. When you see the stress tensor, think of pressure and density 3P + ρ both have nearly the same units, pressure P just has a correction of the speed of light squared in the denominator. Their addition like this is called a relatovistic correction, youll find it in the Friedmann equation for example, which is a direct solution of his his pseudofield equations. The factor of 3 attached to pressure I think arises from the spatial dimensions of space. Another equation which has one factor of pi which springs to mind, is the Poisson equation and looks like ∇g = -4πρ the del, or nabla operator, can be applied to 1, 2 or 3 dimensions, when higher we use the Box operator, or deAlembertian. Youllbe taught often that the gravitational field is irrotational, but im quite sure physics will have this wrong somewhere if gravity should compliment the full Poincare group of space symmetries, that which is rotational and allows torsion. Tjis means the cross product on the connection of the field cannot be zero in such a case. The dot product on torsion however should be zero, since that would be the analogue of a gravitational monopole, as would the dot product in a magnetic and electric charge would be its ectromagnetic monopole, and ecperimentation so far says neither exists. So its not too weird afterall that pi shows up in a non squared form. One last thing, the new dimensionless coefficient I named as the Kepler-Newton number had so.e additional constants and factors of pi thrown in there N = 4 π²/2rv² (r³/t²) Its plausible to state that the two in the denominator, can play some role. Further the pressure might be something obscure, like a Poincare stress, one that might even be gravitational in nature. The pressure equations where, with the dimensionless coefficient written out in full form would be P = 4π/r ² ⋅ 4 π²/2rv² (r³/t²) ⋅ ∂U(r)/∂r P = 8π/r ² ⋅ 4 π²/2rv² (r³/t²) ⋅ ∂U(r)/∂r Careful mot to remove the 4 π² in the numerator, attached to the gravitational parameter, at least the remainjng factor of two in the denominator simplifies the frontline factors in the following ways: P = 2π/r ² ⋅ 4 π²/rv² (r³/t²) ⋅ ∂U(r)/∂r P = 4π/r ² ⋅ 4 π²/rv² (r³/t²) ⋅ ∂U(r)/∂r And maybe its important that it does so in such a way. Quote Link to post Share on other sites

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