Dubbelosix Posted July 22, 2020 Report Share Posted July 22, 2020 (edited) The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from: [math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J}[/math] [math]= -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math] The Bohr magneton is [math]\mu_{B} = \frac{eh}{2m}[/math] Or by manipulation [math]2m \mu_{B} = eh[/math] The spin orbit equation under the standard units are [math]\Delta H= \frac{\mu_{B}}{mc^2 (eh)} \frac{1}{R} \frac{\partial U®}{\partial R} L \cdot S[/math] By using [math]2m \mu_{B} = eh[/math] We can cancel and add one factor of the inverse mass giving [math]\Delta H= \frac{1}{2} \frac{1}{p^2}\frac{1}{R} \frac{\partial U®}{\partial R} L \cdot S[/math] The spin magnetic moment is [math]\mu_s= -g_s \mu_B \frac{S}{h}[/math] So by deduction we also have [math]\Delta H= -g_s\frac{\mu_{s}}{mc^2 e} \frac{1}{R} \frac{\partial U®}{\partial R} L[/math] Because [math]\frac{\mu_s}{\mu_{B}} = -g_s \frac{S}{h}[/math] Edited July 24, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 23, 2020 Author Report Share Posted July 23, 2020 (edited) Today we shall take one of our solutions of the Hamiltonian and apply the quantum commutator with respect to momentum. Edited July 23, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 23, 2020 Author Report Share Posted July 23, 2020 (edited) Before I do any of that it is wise to note the spin factor from Diracs equation is equal to 2 par a small discrepancy, it is very close as an approximation [math]g_s = 2[/math] If the angular momentum is replaced with J characterising a total magnetic moment we have a Lande formula. Edited July 28, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 24, 2020 Author Report Share Posted July 24, 2020 (edited) The commutation of the Hamiltonian with one of the momentum terms can be re-expressed as operators. [math]\begin{equation}[\hat H, \hat p] = [\hat T+ V, \hat p] \\=[\dfrac{\hat p^2}{2m}+ V, \hat p] \\=[\dfrac{\hat p^2}{2m}, \hat p] + [V, \hat p] \\\end{equation}[/math] Note that here, in general, the potential is a function of the radius, i.e. [math]V®[/math] Also, using the result that for any function [math]f[/math]: [math][f, \hat p]=i \hbar \dfrac{\partial f}{\partial R}[/math] From it we obtain the standard solution [math]\begin{equation}[\hat H, \hat p] = \dfrac{1}{2m}(\hat p[\hat p,\hat p]+[\hat p,\hat p]\hat p)+i \hbar \dfrac{\partial V}{\partial R}\end{equation}[/math] (Iff) [math]\dfrac{\partial V}{\partial R}\ne 0 [/math] then: [math]\begin{equation}[\hat H, \hat p] \ne 0\end{equation}[/math] The isn't hard to prove so long as the central potential has a non zero electric field. Such a field is already encoded in the spin orbit equation; [math]|E| = \frac{\partial V}{\partial R} = \frac{1}{e}\frac{\partial U®}{\partial R}[/math] Edited July 24, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 24, 2020 Author Report Share Posted July 24, 2020 (edited) OK, today I am going to do something a little more different. Instead of combining the inverse mass for a momentum like so [math]\Delta H= \frac{1}{2} \frac{1}{p^2}\frac{1}{R} \frac{\partial U®}{\partial R} L \cdot S[/math] We are going to treat it like so [math]\Delta H= \frac{1}{2mc^2} \frac{1}{m}\frac{1}{R} \frac{\partial U®}{\partial R} L \cdot S[/math] And I will explain later why as this has a much nicer theoretic interpretation Edited July 28, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 24, 2020 Author Report Share Posted July 24, 2020 (edited) OK so let's get the show on the road. The equation of interest is [math]\Delta H= \frac{1}{2mc^2} \frac{1}{m}\frac{1}{R} \frac{\partial U®}{\partial R} L \cdot S[/math] And want to concentrate on the inverse mass term which gave Bohr incredible success in calculating the various parameters of the hydrogen atom. [math]\frac{1}{m} = 4\pi^2 k_B R (\frac{e}{h})^2[/math] Not sure if other theoreticians have noticed this, but the term in the parentheses is also the Einstein slope. You can also rearrange his formula to find the Bohr orbit radius [math]R^{-1} = \frac{4\pi^2 k_B e^2 m}{h^2}[/math] Which led itself to the fine grating constant inside the atom. Plugging in his inverse mass solution into the spin orbit equation derived gives us; [math]\Delta H= \frac{2\pi^2 k_B}{mc^2}\frac{\partial U®}{\partial R} (\frac{e}{h})^2 L \cdot S[/math] We have a simplification by removing some factors from his solution as seem above. Edited July 24, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 25, 2020 Author Report Share Posted July 25, 2020 (edited) Let's finish this post, not with the word, or a bang, but a "spin" on all the physics spoke about. It will encapsulate rotation from the moment of inertia, fine grating from Bohr-Zeeman spin orbit interactions, with the Rydberg-Bohr energy transition which coincides with the rotation curve formula I proposed to answer for dark matter. In the drag reinterpretation of Newton's law, I proposed dark matter to be nothing but a drag uniformly on matter if gravity due to a binding energy with origins linked to the supermassive black hole in any typical spiral galaxy. My last post on this deducted that [math]F_{drag} \cdot \int\ dR_{disk}= 4\pi^2(\frac{m_1}{m_2}) \ddot{I}[/math] With [math]\ddot{I}[/math] as the double time derivative of rotational inertia, and is an energy binding equation. deBroglie showed that Bohr's calculation for the orbitals would satisfy a wave within a circumfrance of [math]2\pi^2 R= \lambda[/math] Then for Bohr's model I calculate [math]\frac{1}{\lambda} =\frac{1}{2\pi R} = \frac{2 \pi k_B e^2 m}{h^2}[/math] Which encodes again the Einstein-Planck slope. Distribution of one half further modifies it as simply, after rearranging [math]\frac{1}{2 m\lambda} =\frac{1}{4\pi^2 mR} = k_B(\frac{e^2}{h^2})[/math] The Boltzmann constant has dimension of Joule per Kelvin, the dimension are thus [math]k_B =\frac{J}{K} \equiv (\frac{m \ell^2}{t^2 T})[/math] By dividing through one of the lengths we should obtain the moment of inertia in a transition Rydberg formula under close inspection with consequence to the Zeeman effect of spectral lines involving the Einstein-Planck slope; [math]\frac{1}{4\pi^2 mR^2} = k_B \mathbf{R}(\frac{e^2}{n^2_f h^2} - \frac{e^2}{n^2_i h^2})[/math] Since the Rydberg constant has units if inverse length, this acting on the Boltzmann constant now has dimensions of [math]k_B \mathbf{R} \equiv (\frac{m \ell}{t^2 T})[/math] Multiplying through by t^2 and inverting the equation we retrieve the double time derivative of the moment of inertia as [math]\Delta H = 4\pi^2 \ddot{I} = \frac{T}{m \ell}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] A bit complicated tracking the units but if I trust the algebra we have reached a nice equation that theoretically unifies many different aspects of related fields in quantum mechanics. Edited August 2, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 29, 2020 Author Report Share Posted July 29, 2020 OK thought my work was over, but I have come across more interesting analysis. So will be writing this now. Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 29, 2020 Author Report Share Posted July 29, 2020 (edited) From here we notice mR^2 is the moment of inertia and using the results of the last equation we can also measure the change in the orbital radius as [math]\dot{R}= \frac{dR}{d\theta}\frac{L}{mR^2}\theta [/math] Edited August 2, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted July 30, 2020 Author Report Share Posted July 30, 2020 (edited) The quantum commentator will be our next correction factor to the inverse Bohr mass formula [math]\frac{1}{m} = 4\pi^2 k_B R (\frac{e}{h})^2[/math] To recite we showed from standard theory that [math]\begin{equation}[\hat H, \hat p] = \dfrac{1}{2m}(\hat p[\hat p,\hat p]+[\hat p,\hat p]\hat p)+i \hbar \dfrac{\partial V}{\partial R}\end{equation}[/math] [math]\frac{1}{m} = 4\pi^2 k_B R (\frac{e}{h})^2[/math] We wish again to encode the expression. To match the numerical factors we require a small modification (division of a one half) [math]\frac{1}{2m} = 2\pi^2 k_B R (\frac{e}{h})^2[/math] The correction for the Bohr atom encoded now to the quantum format becomes [math][\hat H, \hat p] = 2\pi^2 k_B R (\frac{e}{h})^2(\hat p[\hat p,\hat p]+[\hat p,\hat p]\hat p)+i \hbar \dfrac{\partial V}{\partial R}[/math] Edited July 30, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 2, 2020 Author Report Share Posted August 2, 2020 (edited) Since I briefly discussed my formula to discuss rotation curve phenomenae as the explanation behind dark matter,we may as well cover the derivation. It should be noted that the references at the end should be explored first to understand some dynamics and essentials of the investigation of the following work. When reading Newtons Principia, I was surprised to learn that he had objectively denied gravity as a drag phenomenon and in recent light of evidence discovered in which a star was found to drag space and time around with it, I felt the need to revisit his laws and implement the drag coefficient for the new physics. It has been an essential argument of mine for a while now that the supermassive black holes at the centers of most typical spiral galaxies will play the role of a spacetime polarization in which dark matter effects results from the same binding energy they harbour with trillions of solar masses.To start off, Newtons law is: [math]F = \frac{Gm^2}{R^2} = \frac{mv^2}{R}[/math] He went off to find a unification to prove that Keplers law for planetary motion around the sun was justified. To do this he took the rotation velocity [math]v_{rot} = \frac{2 \pi R}{t}[/math] And he plugged it in giving [math]G(\frac{m}{R})^2 = \frac{m(2 \pi R/t)^2}{R}[/math] By rearranging he found that Keplers third law was obtained as [math]\frac{R^3}{t^2} = \frac{Gm_{sun}}{4 \pi^2} = \frac{\mu_{sun}}{4 \pi^2}[/math] It is interesting that he wholeheartedly objected to the drag interpretation of gravity since all the planets part from one, rotates in the direction of the spin of the sun. That rule even extended to spiral galaxies in which there is no typical spiral galaxy that rotated in a direction opposite to the black hole they harbour. This was no coincidence in my eyes and strongly suggested to me that the thing was call dark matter was in fact produced from a more local effect inside of the galaxies themselves as opposed to some fundamental field rendering it superfluous and excessive. I will continue from the drag formula I created to predict a heuristic approach to explaining the drag dynamics of the supermassive black hole, from it the binding energy is found as [math]E_{binding} = F_{drag} \cdot \int dR = \frac{1}{2}f(\rho + 3P)\ V_{smbh}[/math] The drag in this case changes linearly to the volume of the supermassive black hole and I also invited to relativistic correction of 3P where f is the gravitational drag coefficient. We can rewrite the last equation as [math]E_{binding} = m_{drag}g \cdot \int dR[/math] In which we define [math]g \equiv \frac{m_1}{m_2}\frac{v^2}{R} = \frac{Gm}{R^2}[/math] Where Gm is the gravitational parameter denoted as [math]\mu[/math] and has dimensions of [math]Rv^2 = Gm[/math]. where g is the usual gravitational acceleration, by solving for the drag coefficient I get [math]f \equiv 2g(\frac{R}{v^2}\frac{V_2}{V_1}) = \frac{4 \pi^2 R^2}{v2t^2} = \frac{4 \pi^2 R}{t^2}\frac{R}{v^2} = \frac{4 \pi^2 R}{t^2}g^{-1}[/math] Distributing the factor of acceleration, we get [math]f(\frac{Gm}{R^2}) = \frac{4 \pi^2R}{t^2}[/math] and by distributing the radius squared we obtain Keplers third law under the drag coefficient interpretation of gravity: [math]f \cdot Gm = \frac{4 \pi^2R^3}{t^2}[/math] From here I wanted to get a little bit more creative, and found that by plugging in the rotational velocity into the gravitational acceleration defined near the beginning of the work I could obtain: [math]g \equiv (\frac{m_1}{m_2})\frac{4 \pi^2 R^2}{t^2 R}[/math] Multiplying through by a mass term yields Newtons force equation with a twist, mind the pun, because we obtain a quantity known as the rotational inertia [math]mR^2[/math] [math]F_{drag} \equiv mg = (\frac{m_1}{m_2})\frac{4 \pi^2 mR^2}{t^2 R}[/math] and can be rearranged for the rotational energy associated to the moment of inertia with a double time derivative [math]F_{drag} \cdot \int dR = 4 \pi^2 (\frac{m_1}{m_2}) \ddot{I_{rot}}[/math] This subtle unification of gravity drag into Newtons formula will provide a toy model, or starting point towards a revised Newtonian law that satisfies the new experimental evidence.As I stated before, the idea of a gravity drag I came to learn was even taken seriously by NASA, as you will find in the following linkTerminal Velocity (gravity and drag) (https://www.grc.nasa.gov/WWW/K-12/rocket/termvr.html) Since the construction of my own interpretation of gravity drag, cosmologists have since detected stars with much more less than the geavitational mass of the supermassive black holes found to seemingly drag spacetime around it! The link has almost all the essential ingredients within my heuristic approach, but contains an essential feature if the model is to be taken seriously, and that is, the notion of weight. Defining weight is critical under a drag theory and yet weight is remains ill-defined in general relativity but not ill-defined under Newtonian mechanics. If I was to modify the rough equations set in the link to satisfy my own approach, it not only requires the weight but also the relativistic correction and would feature as: [math]F_{drag} = \frac{1}{2} f (\rho + 3P) \frac{v^2}{R} \int dV - W[/math] In the work I provided, we remove area definitions for the volumetric interpretation, something equally important within the idea of the Ricci flow discussed in my posts about five months back. The work in NASA's page usea the frontal area, a definition found within fluid dynamics respectively from the same type of model satisfying the Fresnel drag. Further more, the fluid equations describing this I suspect to be equally important for cosmological purposes since the most important of cosmological equations are derived from using the ideal laws of fluid motion. Tomorrow, if I find time, I will define the weight under an inspection of the classical Newtonian physics and then try interpret it all under general relativity. It is not a lack of general relativity being incapable of describing weight, it is just that gravity was never a fundamental force as Newton had erroneously believed and this somewhat influenced Einstein directly in the lack of definition within general relativity. Weight is defined using the 'the most hated integral' notation (a personal beast against me since I know infinities do not exist in nature) from the following equation by Newton: To add before we cover the equation, there is a large assumption of Newton that gravity is an instantaneous force which can be easily challenged at a later date... but for now, the weight is defined through the potential energy equation ~ [math]PE = \int_{r=R}^{R = \infty} G(\frac{m}{R})^2\ dr[/math] [math]= Gm^2 \int_{R}^{\infty}\ \frac{dr}{r^2} = G\frac{m^2}{R} = G(\frac{m}{R})^2 R = mgr = W \cdot R[/math] The drag formula I derive is [math]F_{drag} = \frac{1}{2}f \rho \frac{v^2}{R} \int dV - mg[/math] One major consequence discovered a while back by myself, was that Einsteins weak equivalence explaining that gravitational mass being equal to the inertia of the system would further be extended to mean it was also related to the drag exerted on the mass. In simple terms the inertia is a resistance of a piece of matter to be moved from rest into motion. This resistance is very small and is actually due to the gravitational field locking a piece of matter in its place of rest, the extended principle follows: [math]M_{gravitational}\ g = M_{inertial}\ g\ (is\ itself\ equal\ to)\ M_{drag}\ g[/math] This extension explains what inertia is, and was something Einstein could not fully explain outside of saying it had something to do with the energy content of the system. Here by explaining it as a phenomenon of drag, sufficiently throws away a purely local effect for the influence of gravity on the surrounding system acting upon it.Moving on, the idea now is to transpose the drag formula [math]F_{drag} = \frac{1}{2}f \rho \frac{v^2}{R} \int dV - mg[/math] into the language of general relativity. To do this isn't as difficult as one might make it out to be. First of all, we must realize that the force as defined by Newton, is actually a pseudo force as defined through general relativity. Let's explain this with some detail...The Christoffel symbol can be 'loosely' thought of as being analogous to a force in Newtons equations (where mass has been set to 1 to denote that it is a constant in this formulation): [math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math] Newtonian formulation of this acceleration is [math]F = -\frac{\partial \phi}{\partial x}[/math] However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that requires quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists). It's actually a crucial component of many theories, most notably string theory.Gravity is a pseudo force and can be understood in the following (neat) and (concise and short) way: [math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math] where [math]\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial x^{\lambda}}[/math] Or more compactly as [math]\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}[/math] which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces. Now, in general relativity, there are a number of ways you can write the force, there is such a thing as the three force but its character leaves out essential dynamics which explains why curvature should even exist. In general, we wish to speak of the four force: [math]\mathbf{F} = \frac{d \mathbf{p}}{dt}[/math] The bolded characters just means we deal with the four force and four momentum. But when transposing an equation into general relativity, we like to use the correct notation such as replacing any acceleration terms (ie. g) for the Christoffel symbol [math]\Gamma[/math]. It is not always defined like this, some times authors make the Christoffel symbol dimensionless... but in our case we will define it as the acceleration as it too defines the gravitational pseudo field. So let us have this post for now sink in... before we start modifying the drag equation.We defined the acceleration from general relativity under the Christoffel symbol as [math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math] we define the metric here as a velocity term, meaning we have to redefine or correct the dimensions found in the derivative on the RHS. One way would be to put in a factor of [math]c [/math] but alternatively, we may just define [math]g_{00}[/math] as having units of velocity, and of course, the rate of change of velocity to that of time, is acceleration, so we rewrite [math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial t}[/math] Plugging this into the last term, which clearly contains the acceleration from the definition of weight in classical physics means that the half is not attached to one term as found in the NASA page, but is in fact intuitively attached to both terms, [math]E_{drag} = F_{drag} \cdot \int dR[/math] [math]= \frac{1}{2} (f\ \rho v^2 \int dV - m \frac{\partial g_{00}}{\partial t} \cdot R)[/math] (note, the time time components areb usually reserved for the stress energy, so we explicitely solved for this) Also in general relativity, the density multiplied by the factor of v^2 yields the stress energy tensor of the form [math]T_{00}[/math]. We may also plug this in and define the drag energy: [math]E_{drag} = F_{drag} \cdot \int dR[/math] [math]= \frac{1}{2}(f\ T_{00} \int dV - m\ \frac{\partial g_{00}}{\partial t} \cdot R)[/math] Now, there are other ways we can approach this, for instance, we can just plug in the Christoffel symbol, but if we do this, we can define the weight say, under the four force definition, allthewhile keeping the stress energy tensor in the first term on the RHS, but in this instance, we wane away from using the time time components for a better array for the matrices which define such four force. While I said the four force was [math]\mathbf{F} = \frac{d \mathbf{p}}{dt}[/math] we would get a slightly different version of what might be written in your standard textbook for the four force, only because we have defined the units of the Christoffel symbol with that of acceleration and massively simplifies the standard notation giving:[math]F_{\mu \nu} = g_{\mu \nu} \frac{dp}{dt} + m \Gamma_{\mu \nu}[/math] now, when the weight is redefined with the Christoffel symbol, we find it produces an extra term: [math]E_{\mu \nu} = F_{\mu \nu} \cdot \int dR[/math] [math] = \frac{1}{2}(f\ T_{\mu \nu} \int dV - g_{\mu \nu} \frac{dp}{dt} + m \Gamma_{\mu \nu} \cdot R)[/math] These are just two preliminary ways we could have achieved a modified equation for the drag under general relativity. The last question is ‘’have I gone about this the right way? ‘’ So with a quick read on stackechange, it has not made this all the more easier to grasp, but it seems I may have been on the right track, since they do have to use the four acceleration definition. Their final result however defines the force getting back the Newtonian limit as''[math] F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}[/math] As [math]c\rightarrow \infty[/math] we recover the Newtonian definition, but nobody bothers phrasing it in these terms.''Mind you, they did start with the Schwarschild metric, which led to this final result to recover Newtons equations, the denominator has featured in many papers I have read and can have many uses. What is pertinent here though is that they recognise that the definitions must require the four acceleration, or in my approach, more directly the four force. It would be interesting to gather their definition into the drag formula since the denominator is for redshift analysis. At least in below, Lubos Motl does mention that weight is not obsolete, but as I have stated before, it is very ill-defined in Einstein's general relativity to the point that such questions on its nature are hardly ever spoke aboutref: What is the weight equation through general relativity? (https://physics.stackexchange.com/questions/47379/what-is-the-weight-equation-through-general-relativity/47388#47388) On a different site, I find an answer more similar to my approach, it is said very simply that: ''In GR, the "weight" of an object is the magnitude of the object's 4-acceleration multiplied by its rest mass. The magnitude of the 4-acceleration of an object stationary on the Earth's surface is g; the magnitude of the 4-acceleration of an object stationary at an altitude above the Earth equal to the Earth's radius is g/4. Source https://www.physicsforums.com/threads/how-does-weight-work-in-general-relativity.262703/' (https://www.physicsforums.com/threads/how-does-weight-work-in-general-relativity.262703/%27) So in this sense, we did indeed take the objects four acceleration multiplied by its mass term, but the result was that I defined the acceleration with a Christoffel symbol as having those exact dimensions. After reading this link, I feel a bit more confident in this heuristic approach. **References**Fizeau experiment - Wikipedia (https://en.wikipedia.org/wiki/Fizeau_experiment) Reynolds number - Wikipedia (https://en.wikipedia.org/wiki/Reynolds_number) Drag coefficient - Wikipedia (https://en.wikipedia.org/wiki/Drag_coefficient) Conclusions Experiment and observation together fit in well to explain dark matter as an intrinsic drag effect. Aside from other strong arguments set in previous posts, I tend to find this as the "origin of inertia" insomuch that it explains why objects tend to resist motion when from a state of rest due to local gravitational fields. The reason why particles remain in motion probably does have something to do with their energy content as Einstein eluded to. Edited August 3, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 2, 2020 Author Report Share Posted August 2, 2020 (edited) The discussion of temperature of an atom transition is an interesting one because we are often told temperature only defines a collection of systems suitable for macroscopic systems alone. This can be challenged by restating that temperature is only ill-defined for fundamental particles. So long as a particle is made of other consistuents can we really entertain the motion of a temperature. The transition equation we argued from dimensional analysis showed that [math]\Delta H = 4\pi^2 \ddot{I} = \frac{T}{m \ell}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] The equation was constructed following [math]\frac{1}{2 m\lambda} =\frac{1}{4\pi^2 mR} = k_B(\frac{e^2}{h^2})[/math] The Boltzmann constant has dimension of Joule per Kelvin, the dimension are thus [math]k_B =\frac{J}{K} \equiv (\frac{m \ell^2}{t^2 T})[/math] By dividing through one of the lengths we should obtain the moment of inertia in a transition Rydberg formula under close inspection with consequence to the Zeeman effect of spectral lines involving the Einstein-Planck slope; [math]\frac{1}{4\pi^2 mR^2} = k_B \mathbf{R}(\frac{e^2}{n^2_f h^2} - \frac{e^2}{n^2_i h^2})[/math] Since the Rydberg constant has units of inverse length, this acting on the Boltzmann constant now has dimensions of [math]k_B \mathbf{R} \equiv (\frac{m \ell}{t^2 T})[/math] Plugging in those dimensions before inverting we obtained the energy equation [math]\Delta H = 4\pi^2 \ddot{I} = \frac{T}{m \ell}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] The ergodic hypothesis does clearly state that the time averaging also applies to microscopic systems, not just an assembly of atoms forming a molecule. For molecular systems the average speed is often dictated as [math]v = \sqrt{\frac{8 N_A k_BT}{\pi m}}= \sqrt{\frac{8R_{gas}T}{\pi m}}[/math] Where R(gas) is the gas constant analogous to the Boltzmann constant with N(A) as Avagadros number. Temperature is also defined through the entropy as [math]dT = \frac{dE}{k_B} =\frac{dE}{S} [/math] Where the entropy S is defined in units also of the Boltzmann constant. We will go further on this soon, maybe the pattern of where I am heading will be eliminated from the previous equations for some nice speculations from again, dimensional analysis. Only ending it the now because it is getting late here. Edited August 2, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 8, 2020 Author Report Share Posted August 8, 2020 (edited) Right, so let's try and finish this. The three equations I brought our attention to was; [math]\Delta H = 4\pi^2 \ddot{I} = \frac{T}{m \ell}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] [math]v = \sqrt{\frac{8 N_A k_BT}{\pi m}}= \sqrt{\frac{8R_{gas}T}{\pi m}}[/math] [math]dT = \frac{dE}{k_B} =\frac{dE}{S} [/math] Squaring the second equation we remove the radical [math]v^2 = \frac{8 N_A k_BT}{\pi m} = \frac{8R_{gas}T}{\pi m}[/math] Rearranging we get [math]\frac{\pi v^2}{8 N_A k_B} = \frac{T}{ m}[/math] Referring now to the first equation we had [math]\Delta H = \frac{T}{m \ell}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] We can rephrase the inverse length once again with the Rydberg constant as [math]\Delta H = 4\pi^2 \ddot{I} = \frac{T}{m} \mathbf{R}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] And notice the temperature over mass on the RHS has a further different interpretation involving the average speed as [math]\Delta H = \frac{\pi v^2}{8 N_A k_B} \mathbf{R}(n^2_f\frac{h^2}{e^2} - n_i^2\frac{h^2}{e^2})[/math] I'll come back to this because as stated, the average of microatates over time indicates that temperature does not or should not just apply to an assembly of atoms, but could indeed apply to individual atoms themselves since they are composite systems. Interpretation is important, a theory is wonderful so long as it fits experiment. The real question arises, why Avagadros number, should we be concentrating on the gas constant itself? Edited August 9, 2020 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 9, 2020 Author Report Share Posted August 9, 2020 https://www.google.com/url?q=https://www.nature.com/articles/nphys3403&sa=U&ved=2ahUKEwj8p4jGk43rAhU1onEKHWArDocQFjAOegQIBxAB&usg=AOvVaw2p1Yo8JQjAAIckkKf_S9DN Quote Link to comment Share on other sites More sharing options...

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