Dubbelosix Posted August 3, 2019 Author Report Share Posted August 3, 2019 (edited) Off the rails? I am theoretical scientist, we don't tend to go off on tangents unless there are creative ways to argue it. For instance, I stated: ''We find that in a simplified expression, the Doppler effect is [math]\frac{v^2}{c^2} = 1 - \frac{1}{\gamma^2}[/math] When solving for [math]\gamma[/math] we get a very nice set of solutions: [math]\gamma = i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}[/math] and [math]\gamma = -i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}[/math] Where we notice that for this to be true, [math]v \ne c[/math]. You could go further and state that [math]v \ne - c[/math] but this is where pure algebra reaches a limit because there is no such thing as a negative speed of light.'' The steps I had in mind was to square the function [math]\gamma^2 = i^2\frac{c^2 \cdot [v^2 - c^2]}{v^4 - c^4}[/math] (The sign here comes intuitively from the imaginary form [math]i^2 = -1[/math] which is simply [math]\gamma^2 = i^2\frac{c^2 \cdot [v^2 - c^2]}{v^4 - c^4}= i^2\frac{c^2}{v^2 - c^2}[/math] The addition of a mass term would imply [math]i^2\frac{mc^2}{v^2 - c^2}[/math] The only time I have considered tachyons seriously, was from work suggesting the neutrino could move faster than light, however, this was before the LHC blunder - I got enthusiastic in those days and gave up my thoughts on tachyons for a while. Anyone who follows my work will know that I consider there to be a pre-big bang phase - I likened it to an all-dominated matter close at zero point temperatures. The allowance of this condensate in the form of tachyons could answer how light reached all the universe without inflation. Edited August 3, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

ralfcis Posted August 3, 2019 Report Share Posted August 3, 2019 (edited) Well now that you put it that way have you finally proven to yourself that i^{2 }= -1? Good first step. I thought you were self-taught. Did you confer yourself a degree in theoretical physics because I haven't seen one bit of evidence that you're legit in any way. This thread should be moved into the rubber room section. Edited August 3, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 3, 2019 Author Report Share Posted August 3, 2019 Well now that you put it that way have you finally proven to yourself that i^{2 }= -1? Good first step. Aside from the last drivel which I have no intentions of replying to, what is remarkable that we use imaginary numbers in the sign convention, in which you feel the need to be rude about what has been suggested? Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 12, 2019 Author Report Share Posted August 12, 2019 Let's not forget the bivector theory of gravity which naturally induced rotation (ie. non-zero torsion field) https://www.quora.com/q/htorqrygqxdlrpkp Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 12, 2019 Author Report Share Posted August 12, 2019 Let's talk about a purely mathematical set up for the unification of the charges in curved spacetime. All derivatives on the left hand side of the energy operator should yield the connections of a gravitational field: The energy operator is well known and is itself the wave equation: [math]i \hbar\ \frac{\partial \psi}{\partial t} = \hat{E}\psi[/math] The right hand side should become an operator that probably does not commute, since the order of space and time derivatives matter - in the case we will look at, involves the connection due to the Ricci curvature which is the ''simplest'' curvature invariant for the Riemannian manifold: [math]i \mathbf{J}\ \frac{\partial \psi}{\partial t} = \nabla \mathbf{R}\ \frac{Gm^2}{c}\frac{\partial \psi}{\partial t}[/math] In which for this case, [math]\mathbf{J}[/math] is always quantized and so we can treat this as the spin density and [math]\mathbf{R}[/math] is the usual curvature which has units of [math]\frac{1}{length^2}[/math]. Therefore this quantity is an energy: [math]\frac{Gm^2}{c}\frac{\partial \psi}{\partial t}[/math] and so the connection acting on the Ricci scalar curvature yields a density scalar. Let's refresh ourselves on the Einstein field equations, it takes in the index free notation: [math]\mathbf{G} = \mathbf{R} - \frac{1}{2}\mathbf{R} = \frac{8 \pi G}{c^4}\ \mathbf{T}[/math] With [math]\mathbf{G}[/math] being the Einstein scalar, also with units of inverse length squared and [math]\mathbf{T}[/math] being a stress energy scalar. For a simple case of a torsion free case (which is incapable of being seen in context of geometric algebra), the [math]\nabla \mathbf{R}[/math] is the derivative as found in the first reference and will abide to the Bianchi identities. The quantization of the energy operator as related to curvature dynamics is still an open question, even in this pure mathematical text, there are many avenues as you will probably know that have attempted to find the quantization as related to the connection of the gravitational field. The approach above is just food for thought. REFERENCES: https://books.google.co.uk/books?id=PZcSDAAAQBAJ&pg=PA34&lpg=PA34&dq=%E2%88%87R+-+connection+of+curvature&source=bl&ots=PcWjnEK_H8&sig=ACfU3U1zoHfpv_9qd1iI26PhQ1jUpsDL7w&hl=en&sa=X&ved=2ahUKEwiVvKP1xv3jAhX0QUEAHS1bCtUQ6AEwEHoECAoQAQ#v=onepage&q=%E2%88%87R%20-%20connection%20of%20curvature&f=false https://empg.maths.ed.ac.uk/Activities/Spin/Lecture6.pdf Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 12, 2019 Author Report Share Posted August 12, 2019 (edited) A curvature tensor would simply look like: [math]i \mathbf{J}_{\mu \nu}\ \frac{\partial \psi}{\partial t} = \nabla \mathbf{R}_{\mu \nu}\ \frac{Gm^2}{c}\frac{\partial \psi}{\partial t}[/math] Because the gravitational field is a pseudo field, we are still going to have problems justifying how the complexification of the field can be applied to a gravitational understanding since the quantization of gravity under the Wheeler de Witt equation is inherently real. So something has to give on this if there is a quantization bridge between the two invariants of the charge descriptions. Edited August 12, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 14, 2019 Author Report Share Posted August 14, 2019 (edited) Let's go back to this equation: [math]i \mathbf{J}\ \frac{\partial \psi}{\partial t} = \nabla \mathbf{R}\ \frac{Gm^2}{c}\frac{\partial \psi}{\partial t}[/math] As I believe it is possible to reconfigure the equation in terms of the contribution of curvature to the fluctuations of spacetime. It is possible to expand the Langrangian of the zero point modes on the background spacetime curvatuture in a power series [math]\mathcal{L} = \hbar c\ \mathbf{R} \int\ k\ dk... +\ \hbar c\ \mathbf{R}^2 \int \frac{dk}{k^{n-1}} + C[/math] Where [math]C[/math] is a renormalizing constant which is set to zero for flat space. It had been believed at one point that the forth power over the momenta of the particles would be zero [math]\hbar c\ \int\ k\ dk^3 = 0[/math] But interesting things happen in the curvature of spacetime, such a condition doesn't need to hold. This langrangian density, in which [math]\mathbf{R}[/math] plays the role of the curvature was first recognized by Sakharov and how the curvature can allow a fluctuation to become longer lived - perhaps even the fluctuation itself can owe its existence to the gravitational correction term. Notice in Sakharov's approach, we find the usual charge identity for the electromagnetic part [math]\hbar c[/math]. I still find the imaginary term a complication because as noted before, the gravitational field is not really a field and complex numbers are usually applied to quantum fields - so for the sake of it, we may even want to investigate a purely real form of the case: [math]\mathbf{J}\ \frac{\partial \psi}{\partial t} = \nabla \mathbf{R}\ \frac{Gm^2}{c}\frac{\partial \psi}{\partial t}[/math] Taking the derivative of the spin density allows us to consider the square of the curvature scalar and then following with it, the definition of the gravitational correction in form of the geometric power series - this also includes a small manipulation of distributing the speed of light to find the charge definitions: [math]\mathcal{L}\psi = \nabla \mathbf{J}c\ \psi\ ... +\ Gm^2\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] This is just a purely mathematical avenue, we don't know how gravity will translate into a quantum framework - certainly it cannot be treated in the normal framework of fundamental fields but the approach we have above issues a certain importance behind the gravitational interpretation of the charge, namely the term [math]Gm^2[/math]. Edited August 14, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 14, 2019 Author Report Share Posted August 14, 2019 (edited) In natural units,the gravitational fine structure constant is equal to the square of the mass of a particle [math]\alpha_G = m^2[/math] The hope or immediate realization of this are attempts to find quantization of mass depending on factors of [math]n\hbar[/math]. The [math]Gm^2[/math] can be thought of as the gravitational charge of the system analogous to the electric charge [math]e^2[/math] (as stated before - but not necessarily equal in magnitude, they are only equal in the dimensional sense). Some mass formula have been suggested in literature (see What is special about the Planck mass? Sivaram). Before we look at the more advanced mass formula that will be suggested here, let’s take a look at what some have called ‘’the mysterious’’ Weinberg formula: [math]m = (\frac{H_0\hbar^2}{Gc})^{\frac{1}{3}}[/math] In which [math]H_0[/math] is the Hubble parameter. This will correspond to either one particle mass, or Weinberg had in mind a spectral property leading to different masses. In any case, this equation cannot predict all particle masses with this description alone. Alternatively, the Weinberg formula has been written as [math]m^3 = \frac{H_0 \hbar^2}{Gc}[/math] In this formulation, it has been suggested by Arun and Sivaram that it is ''unclear'' whether now he associates the equation to three particle masses [1] or as an attempt to find a fundamental unit of mass. Either way, we can see his ultimate aim would have been to describe fundamental particle masses from cosmological features, like the Hubble constant. Schwinger, Motz (et al). have demonstrated what it means to quantize a charge: Charge and mass are in fact so similar, you can indeed describe both in similar physics. Nature seems to posit some kind of universality with the presence of an electric charge being synonymous with mass. The only exception it appears in nature is a neutrino, which is expected to have no charge and a very small mass, but if it has a very small mass (you can also argue) it may possess a vanishing, but non-zero charge with a proportionality of [math]e^2 \propto Gm^2[/math]. To try and articulate how small this charge would be, it would have a mass and charge 1 millionth of that of an electron. Either way, the ultimate idea is that large numbers can determine the small in dynamic ways. We can in fact restate Weinbergs formula for one that satisfies the gravitational interpretation of the charge. I provide this as a simple manipulation of his equation: [math]Gm^2 = \frac{H_0 \hbar^2}{mc} = \frac{H_0 \hbar^2}{p}[/math] In which we recognize a momentum term in the denominator. Let’s have a quick look at a more advanced mass formula candidate [math]m = nk[\frac{mc}{e}\frac{1}{2 \sqrt{T}}]^n M_e = nk [\frac{\sqrt{G}m}{2 e}]^n M_e[/math] Immediately we can notice the use of the gravitational charge in the last term [math]sqrt{G}m[/math] - the only difference is that it has focused on the Planck mass definition of the charge. The Planck mass should not necessarily be considered fundamental, it seems like too much a basic unit of matter for any particles we have observed in the standard model. Though the middle term is good for string dynamics and superstring tension, the last term appears to be made of more fundamental assumptions which included the gravitational charge of the system. The adjustable parameters is what allows us to predict particle masses, for example n = 2, k = 3 gives the muon mass, n = 2, k = 4 gives the pion mass, n = 3, k = 4 the \Delta resonance, n = 3, k = 6 the D meson n = 3, k = 10 the \psi, n = 4, k = 4 the upsilon particle. According to Sivaram, several more particle masses can be obtained.Anyway, the main point may have became a bit clearer: The more advanced suggested mass formula does indeed predict a wide range of particle masses, but more importantly, the Weinberg formula can be expressible also in a dimensionless form: [math]\frac{m}{m_e} = nk [\frac{\sqrt{G}m}{2 e}]^n = nk [\sqrt{\frac{H_0 \hbar^2}{mc}}\frac{1}{2 e}]^n = nk [\sqrt{\frac{H_0}{mc}}\frac{\hbar}{2 e}]^n[/math] So in this sense, we get to keep Weinberg’s attractive idea about determining fundamental parameters from cosmic parameters - dimensionless terms are well-known in physics to be the only true physical parameters of a theory. A good example would be Bernoulli's fluid equation for an example. While this is all nice, it seems from Regge trajectories, the construction of certain particles strangely coincides with multiples of the mass squared. If we go back to the formula: [math]m = nk [\frac{\sqrt{G}m}{2 e}]^n M_e[/math] Then obtaining the mass squared is a simple procedure: [math]m^2 = nk [\frac{Gm^2}{4 e^2}]^n M^2_e[/math] As shown in the previous post, the part of the Langrangian which deviates from flat space is in units of [math]G = 1[/math] is: [math]\mathcal{L}\psi = m^2\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] And so perhaps an appropriate correction to formulate an equation to predict particle masses as a contribution also from curved space may involve the replacement of the appropriate mass squared term: [math]\mathcal{L}\psi = n\mathbf{k} [\frac{Q^2_g}{4 e^2}]^n M^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n M^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] As stated in one previous post on another thread, the role of [math]\frac{Gm^2}{e^2}[/math] has played a large role in the history of physics, including applications to astrophysics - in order to know these types of relationships, I reference wiki: ''(4.5) in Barrow and Tipler (1986) tacitly defines [math]\frac{\alpha}{\alpha_G}[/math] as [math]\frac{e^2}{Gm_pm_e}[/math] -even though they do not name the [math]\frac{\alpha}{\alpha_G}[/math] in this manner, , it nevertheless plays a role in their broad-ranging discussion of astrophysics, cosmology, quantum physics, and the anthropic principle; In our case we would be defining the inverse of the formula which described this as [math]\frac{\alpha_G}{\alpha}[/math]. In what way the wave function plays a role in the Langrangian density is uncertain to me if it is even required at all. References: https://pdfs.semanticscholar.org/bfed/15fdd2752e7b26891311822f5e0b4995919d.pdf https://arxiv.org/pdf/physics/0408056.pdf https://arxiv.org/ftp/arxiv/papers/0707/0707.0058.pdf C. Sivaram, Astrophys. Sp. Sci. 207, 317 (1993); 219, 135 (1994) Edited August 14, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 14, 2019 Author Report Share Posted August 14, 2019 (edited) A further additional part wiki talks about, is definining the value of [math]\alpha_G[/math] ~ ''[math]\alpha_G[/math] has a simple physical interpretation: it is the square of the electron mass, measured in units of Planck mass. By virtue of this, [math]\alpha_G[/math] is connected to the Higgs mechanism, which determines the rest masses of the elementary particles. [math]\alpha_G[/math] can only be measured with relatively low precision, and is seldom mentioned in the physics literature.Because [math]\alpha_G = \frac{Gm^2}{\hbar c} = (t_P \omega_C)^2[/math] where [math]t_P[/math] is the Planck time, and [math]\omega_C[/math], the Compton angular frequency of the electron.'' Edited August 14, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 14, 2019 Author Report Share Posted August 14, 2019 (edited) Using these values, we can express the fine structure simply then from a previous formula: [math]Gm^2 = \frac{H_0 \hbar^2}{mc} = \frac{H_0 \hbar^2}{p}[/math] As [math]\alpha_G = \frac{Gm^2}{\hbar c} = \frac{H_0 \hbar}{mc^2} = \frac{H_0 \hbar}{pc}[/math] Edited August 14, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 16, 2019 Author Report Share Posted August 16, 2019 (edited) Let's continue on this ''pure math'' adventure.... well sort of pure math adventure, there are some educated idea's upon new formula's we should probably recognize as being potentially important for future calculations The correction of curvature to fluctuations is unified here under the Weinberg mass formula with [math]G[/math] restored:: [math]\mathcal{L}\psi = n\mathbf{k} [\frac{Q^2_g}{4 e^2}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] Weinberg's formula was: [math]m^3 = \frac{H_0 \hbar^2}{Gc}[/math] And for a matter particle, travelling freely no less, is given by the kinetic formula: [math]E_k = \frac{mv^2}{2} = \frac{p^2}{2m} = \hbar^2\ \frac{k^2}{2m}[/math] [math]Gm^3c = H_0 \hbar^2[/math] [math]\hbar^2 = \frac{Gm^2}{H_0}\ mc = \frac{2mE_k}{k^2}[/math] The most important term to remember for the following, is [math]\frac{H_0 \hbar}{pc}[/math]... why? Well [math]\frac{\hbar}{pc}[/math] indicates a relationship with the deBroglie wave-matter formula. It is accordingly related also to the uncertainties: [math]\Delta \ell \approx \frac{\hbar}{\Delta p}[/math] A matter wave is [math]\lambda = \frac{\hbar}{p} = \frac{Gm^2}{H_0 \hbar} = \frac{2mE_k}{k^2 \hbar}[/math] In which a relativistic version can also be extrapolated in literature of the form, where the numerator[math]c \rightarrow v)[/math]: [math]\frac{\hbar}{pc} = \frac{\hbar v}{Ec}(\sqrt{1 - \frac{v^2}{c^2}})[/math] Just because of the presence of the Doppler term, we can derive a simpler crunch: [math]\frac{v(c^2 - v^2)}{c^4}[/math] Or just [math]\frac{\hbar}{pc} = \frac{\hbar}{E}\frac{v(c^2 - v^2)}{c^4}[/math] I suppose it is fair game to expand this new expression by applying lastly the fraction rule, so that we would have: [math]\frac{\Delta \lambda}{c} = \frac{\hbar}{\Delta pc} = \frac{\hbar}{\Delta E} (\frac{vc^2}{c^4} - \frac{v^3}{c^4})[/math] Curious formula... Sometimes math is beautiful, but to completely associate it to physics is never easy. Distribution of [math]c[/math] and allowing us to plug in the gravitational interpretation due to gauge theory we get: [math]\Delta \lambda = \frac{\hbar}{\Delta p} = \frac{Gm^2}{\Delta E} (\frac{vc^2}{c^4} - \frac{v^3}{c^4})[/math] http://electron6.phys.utk.edu/phys250/modules/module%202/matter_waves.htm Edited August 16, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 16, 2019 Author Report Share Posted August 16, 2019 (edited) I'll continue tomorrow. [math]\mathcal{L}\psi = n\mathbf{k} [\frac{Q^2_g}{4 e^2}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] Weinberg's formula was: [math]m^3 = \frac{H_0 \hbar^2}{Gc}[/math] Rearranging we get the mass squared formula simply as: [math]Gm^2 = \frac{H_0 \hbar^2}{mc}[/math] In spirit that as the Hubble constant was small during the ground state epoch, it directly implies that: * As [math]H_0 \approx 0[/math] *Then the measured value of [math]G[/math] would be strong due to small bound universe with large gravitationalc curvature contributions. And for a matter particle, travelling freely no less, is given by the kinetic formula: [math]E_k = \frac{mv^2}{2} = \frac{p^2}{2m} = \hbar^2\ \frac{k^2}{2m}[/math] Hitting [math]Gm^2 = \frac{H_0 \hbar^2}{mc}[/math] With [math]E_k = \hbar^2\ \frac{k^2}{2m}[/math] We can be sneaky and even plug this into the opening mass-prediction formula: [math]\mathcal{L}\psi = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] Giving a formula representing kinetic energy rather than the langrangian density: [math]\frac{Gm^2k^2}{2m} = \frac{H_0}{mc}\frac{\hbar^2 k^2}{2m} [/math] As: [math]\mathcal{L}\psi = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n Gm^2_e\ \mathbf{R}^2\ \int \frac{dk}{k^{n-1}}\ \psi[/math] One way to do it is by cancelling out the common-mass factor in the denominator: [math]\frac{Gm^2k^2}{2} = \frac{H_0}{c}\frac{\hbar^2 k^2}{2} [/math] Knowing that the wave number as units of [math]inverse\ length[/math] this can remove the squared from of the Ricci curvature (while dropping the wave function as it has little use in this dimensional crunching: [math]\mathcal{L} = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n \int\ Gm^2_e\ \mathbf{R}^n\ \int \frac{dk}{k^{n-1}}\ k\ dk\ = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n \frac{H_0}{mc}\frac{\hbar^2 k^2}{2}\ \mathbf{R}^n\ \int \frac{dk}{k^{n-1}}\ [/math] We ((might)) be able to go a little simpler using the connection alone to theorize: [math]\mathcal{L} = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n \int\ \mathbf{Q}_{gravity}\ \Gamma\ \frac{dk}{k^{n-1}} = n\mathbf{k} [\frac{\alpha_G}{4 \alpha}]^n\ \Gamma Q_{gravity}\ \int\ \frac{dk}{k^{n-1}}\ dk^3 [/math] With [math]\mathbf{Q}_{gravity}[/math] as the inertial charge, or charge density. Edited August 17, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

Dubbelosix Posted August 17, 2019 Author Report Share Posted August 17, 2019 (edited) For the next ''lookback'' on previous investigations, we shall now go right back to the beginning, going back to the friedmann equation (my specialist subject) and will look at some of the more recent idea's I have had... perhaps I will start this bit later on as have a few things to do. The main equation we will work from is: [math](\frac{\dot{R}}{R}) = \frac{8 \pi G}{3} \rho + \mathbf{k}\ \frac{c^2}{a^2}[/math] With [math]mathbf{k}[/math] being the curvature constant. Be back later. We will disembowel this last equation right down to roots of its mechanics. Edited August 17, 2019 by Dubbelosix Quote Link to comment Share on other sites More sharing options...

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