Dubbelosix Posted April 4, 2017 Report Share Posted April 4, 2017 (edited) It feels out of place to post my work here, it says in the forum they are generally interesting ideas but without any support from scientific evidence. My work is not strange at all and I hope you will come to like it. Abstract With inflation theories leading naturally to a multiverse, a new explanation perhaps is warranted to explain the rapid inflation phase. In this work, I offer a solution in which rotation plays the role of dark energy in the universe. A fast enough spin is expected to result in a centrifugal force inside the universe capable of pushing all objects away from each other. I will offer more insights into why rotation might be expected in a universe and possible idea's in order to falsify the theory. Friedmann Langrangian First off, we derive a Friedmann power equation. Without the curvature component just for now, the Friedmann equation is [math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda mc^2}{3}[/math] Rearrange [math]\dot{R}^2 = \frac{8 \pi GR^2}{3}\rho + \frac{\Lambda mc^2}{3}[/math] Putting all terms on the left hand side, after distributing a mass term, we get a Langrangian, [math]m\dot{R}^2 - \frac{8 \pi GmR^2}{3}\rho + \frac{\Lambda mc^2}{3} = \mathcal{L}[/math] How did I know this was a Langrangian? Simply because there is already a standard Langrangian out there and it has identical terms [math]\mathcal{L}(R,\dot{R}) = T - U = \frac{1}{2}m\dot{R}^2 + \frac{GM^2}{R} + \frac{mc^2}{6} \Lambda R^2[/math] To make the numerical part all one needs to do is distribute a factor of [math]1/2[/math] on the LHS [math]\frac{1}{2}m\dot{R}^2 - \frac{8 \pi GmR^2}{6}\rho + \frac{\Lambda mc^2}{6} = \mathcal{L}[/math] Going back and rearranging [math]\frac{1}{2}m (\frac{\dot{R}}{R})^2- \frac{8 \pi Gm}{6}\rho + \frac{\Lambda mc^2}{6} = \mathcal{L}[/math] Friedmann Power Equation Differentiation of the Langrangian gives [math]\frac{1}{2}m \frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi Gm}{6}\dot{\rho} + \frac{\Lambda mc^2}{6}\frac{\dot{R}}{R} = \mathcal{P}[/math] Removing the cosmological constant, no need for it now, was only in there to show you how it related to the Langrangian of cosmology (and now using scale factor) [math]\frac{1}{2}m \frac{\dot{a}}{a}\frac{\ddot{a}}{a} - \frac{8 \pi Gm}{6}\dot{\rho} = \mathcal{P}[/math] (note, we could have carried the derivative like we had in the other terms in the following way:) [math]\frac{1}{2}m \frac{\dot{a}}{a}\frac{\ddot{a}}{a} - \frac{8 \pi Gm}{6}\rho\frac{\dot{a}}{a}} = \mathcal{P}[/math] But it doesn't have the pressure term, which is why we have done it this way, to replace for the fluid equation). Replacing [math]\dot{\rho}[/math] with the continuity equation (which is) [math]\dot{\rho} = \frac{\dot{a}}{a}(\rho + \frac{3P}{c^2})[/math] [math]\frac{1}{2}m \frac{\dot{a}}{a}\frac{\ddot{a}}{a} - \frac{8 \pi Gm}{6}\frac{\dot{a}}{a}(\rho + \frac{3P}{c^2}) = \mathcal{P}[/math] By rearranging you can see how (before the fluid equation was plugged in) it is exact to the form we will come to use. (replacing scale factor for R just for now) [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{3}\dot{\rho} = \mathcal{P}[/math] and clearly substitution leads to the form [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{3}\frac{\dot{R}}{R}(\rho + \frac{3P}{c^2})[/math] The full derivation of the modified Raychauduri equation goes as follows: A power equation derived (shown previously) from the Friedmann euation was given as: [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\dot{\rho} = \mathcal{P}[/math] The fluid equation of cosmology is [math]\dot{\rho} = \frac{\dot{R}}{R}(\rho + 3P)[/math] We obtain after substitution [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\frac{\dot{R}}{R}(\rho + \frac{3P}{c^2}) = \mathcal{P}[/math] The modified law for irreversible particle production we have [math]dE = dQ - PdV + (\frac{\rho + P}{n}) dN[/math] and [math]\dot{E} = \dot{Q} - P\dot{V} + (\frac{\rho + P}{n}) \dot{N}[/math] We'll go back to curvature now, but we'll make an argument for the form: [math]\frac{kc^2}{a^2} \frac{\dot{a}}{a}[/math] The curvature can be given as [math]kc^2 = - \frac{2U}{mx^2}[/math] Divide through by [math]a^2[/math] [math]\frac{kc^2}{a^2} = - \frac{2U}{mR^2}[/math] where [math]R = ax[/math] and [math]\frac{kc^2}{a^2}\frac{\dot{R}}{R} = - \frac{2\dot{U}®}{mR^2}[/math] Remember this coefficent [math]\frac{\dot{R}}{R}[/math] as it is important to this work. There is a direct equivalence between [math]\frac{\dot{R}}{R} = \frac{\dot{a}}{a}[/math]. Plugging in the curvature to where it should be, after some more rearranging, we can end up with an equation (which is my preferred form, now with only scale factors) [math]\frac{\dot{a}}{a}(\frac{\ddot{a}}{a} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2}}{n})n \frac{\dot{a}}{a}[/math] Deriving all this properly, we find there is a fluid expansion coefficient on each term, giving meaning to how [math]\rho, P[/math] and [math]\Gamma[/math] are related, where [math]\Gamma[/math] is the particle production rate and is encoded in the fluid expansion [math]\Theta[/math], therefore, arguably, a more correct version of the Raychaudhuri equation) involves the fluid expansion distributed on all the terms, which in natural units [math]8 \pi G = c = 1[/math] the final equation takes the form by making use of [math]\dot{H} + H^2 = \frac{\ddot{a}}{a}[/math] [math]\dot{H}\Theta + H^2\Theta + \frac{k}{a^2}\Theta = (\frac{\frac{\rho}{3} + \frac{P}{c^2}}{n})n \Theta[/math] And in flat space approximation its just (without irreversible particle production now) [math]\dot{H}\Theta + H^2\Theta = (\frac{\rho}{3} + P) \Theta[/math] I conclude from this derivation that theta is the fluid expansion and without it, we'd just have the Raychaudhuri equation - we didn't plug it in ad hoc though, we showed that these are real artifacts that remained from the derivation we used. So one could argue Raychaudhuri's equation was incomplete and this is the correct version above. Unless you imply that energy remains constant in this equation, then the fluid expansion also naturally implies that conservation is violated since the fluid expansion acting on [math]\frac{\ddot{a}}{a}}[/math] just gives an equivalent form the non-conserved Friedmann equation [math]\frac{\dot{a}}{a}\frac{\ddot{a}}{a}[/math]. Not only does it link all the vital subjects of the universe together. there is reason to think then the fluid expansion being shared on these components are doing something special in terms of non-conservation. Excerpt ~ This has been added after I made this page, rotation plays an important feature in my early universe. Rotation is capable of stealing bulk energy from the vacuum because rotation lowers the Friedmann energy levels. This can be understood if the system is closed and cannot steal rotational energy from anything outside of it. Since rotation requires energy, it is likely vacuum bulk energy went towards the primordial rotation and also offers solutions to the cosmological vacuum energy problem, dark flow and even acceleration since rotation would produce an internal centrifugal force field. If you derive the Friedmann equation with rotation and derive the master equation in the same way, I get: [math]\frac{\dot{a}}{a}(\frac{\ddot{a}}{a} + \frac{kc^2}{a^2}) = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2}}{n})n\frac{\dot{a}}{a} + \omega^2\frac{\dot{a}}{a}[/math] surprise surprise, fluid expansion coefficient attached to rotation as well! [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2}}{n})n\Theta + \omega^2\Theta[/math] If rotation really should enter the picture, this would produce a torsion field which enters the equation with a negative sign. Rotation would not be unusual if we considered gravity as part of the full Poincare group of spatial symmetries which involves the spin and torsion (Venzo de Sabbata) also see Sivaram. For instance, Torsion enters the Poisson equation [math]\nabla^2 \phi = 4 \pi G(\rho - \mathbf{k}\sigma^2)[/math] Where [math]\mathbf{k}[/math] is [math]\frac{G}{c^4}[/math] and [math]\sigma[/math] is the spin density which can be calculated the following way: [math]\sigma = \frac{J}{V} = \frac{m \omega R^2}{L^3} = \frac{m vR}{L^3}[/math] so [math]\mathbf{k}\sigma^2 = \frac{Gm^2v^2R^2}{c^4L^6} = \frac{Gm^2}{c^2L^4}[/math] I give the following then with rotation and torsion [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2} - \mathbf{k} \sigma^2}{n})n\Theta + \omega^2\Theta[/math] Rearrange and plug the cosmological constant [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2} - \mathbf{k} \sigma^2}{n})n\Theta + \frac{\Lambda c^2}{3}\Theta + \omega^2\Theta[/math] We can maybe see some relationship between large spin and a cancellation of cosmic energy in the cosmological constant. A large enough spin will give a large torsion field that can cancel out the large primordial energies of the universe towards the unified field scale, perhaps leaving a small positive residual energy, just like observation contends. For theories that leave out singularitries, this can be avoided a number of ways, but the current popular way (especially within bounce theories) is to invoke short scale quantum effects that act like a repulsive gravity. It is likely though, this kind of correction won't be needed, as it is known that Torsion offers non-singular solutions to the universe as well. Because rotation would result in a ''dark flow'' in our universe, we need more study on this phenomenon to perhaps falsify or maybe add evidence to the conjecture. The Expansion of the Universe If the universe had a fast spin, with a large torsion field that has dropped in strength since BB (due to an exponential decay of the rotational property of the universe) then why is the universe accelerating today, where the rotation acts like dark energy? Surely, if rotation has decayed since BB, then the universe shouldn't actually be accelerating! It may come about that we have the expansion picture of our universe wrong. Just using a bit of logic, something doesn't seem right, let me explain. We infer on the acceleration of the universe by noticing that the further galaxies are in fact receding at an ever-increasing rate. However here comes the interpretational problem, scientists have taken this to mean that the universe must be accelerating, but we are often told in cosmology, the further you actually look into space, the further in the past cone of the universe you are actually measuring! See the possible problem here? We are inferring on a state about the universe today, from systems that actually exist in the past state of the universe - that wouldn't actually mean the universe is accelerating today, on the contrary, it would suggest the universe has in fact slowed down! Therefore, I don't see this as a problem, I think mainstream has applied the wrong interpretation to the evidence. On one hand, we are told past states of our universe can be found the further we look, yet on the same hand, we are expected to take these furthest receding systems as an evidence the universe is expanding today. There is no reason, if these systems truly exist in our past, should we infer on them to have a meaning about the current state of the universe. Bulk Energy Transfer Particle creation in curved spaces, is not the only method in which we get a non-conservation of the universe. Rotation is expected to lower the Friedmann energy levels in a universe. We saw an example of the algebra of the Friedmann equation with torsion and spin where the torsion entered the equation with a negative sign (which was capable of leaving a residual small value of the cosmological constant). Bulk Energy is the energy contained within the universes vacuum, it is possible there is a direct Bulk Energy exchange with the rotational property of the universe (on the horizon). Filchenkov has shown in his own work how the rotation alters the primordial energies of the early Friedmann universe (see the links below). It is easy to understand why rotation would steal energy from the vacuum like this, if the universe truly was closed - particles can exchange rotational energy, but if nothing exists outside of the universe (closed universe) then there has to be energy from somewhere to compensate for the rotation of the universe. This can explain why there are large energies arising in the cosmological constant which aren't accounted for by observation. Pro's and Con's Pro's 1. Rotation explains dark energy as an internal centrifugal force field. 2. Rotation has been shown by Hoyle and Narlikar to exponentially decay. It is suggested in this work that dark flow is the residual left over of primordial rotation. 3. Rotation naturally involves torsion and torsion is capable of answering what dark matter essentially is as a phenomenon. 4. Torsion has been known to avoid singular regions of spacetime, something I welcome because singularities are not physical. 5. Rotation explains the spiral structure of galaxies and can explain why most galaxies prefer a specific handedness (or chirality). Galaxies tend to align themselves with the primordial rotation of the universe. 6. Because the vacuum energy has to go towards the rotational property of a closed universe, it can explain why there are discrepancies in the calculation of vacuum energy by a factor of 10^122 magnitudes too small. 7. Equally, rotation then also explains why there is such a small cosmological constant. It also explains it algebraically since torsion enters the equations with a negative sign (large torsion cancels a large cosmological constant) leaving behind a small, positive value like observed. Con's 1. Rotation lowers primordial Friedmann energy levels below the tunneling probability barrier. So nucleation of a universe is in question. Zero Point Modes in Space One of the studies of my work has been to find new models to describe various density parameters (part of the effective density) that would hold significance in the early universe and maybe later. One of those investigation left me wondering about the role of the virtual particle in spacetime. I came to the conclusion that [math]\rho[/math] in the master equation must refer to the on-shell mass components of a universe: [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2} - \mathbf{k} \sigma^2}{n})n\Theta + \frac{\Lambda c^2}{3}\Theta + \omega^2\Theta[/math] Note, [math]P = \rho c^2[/math] [math]\frac{P}{c^2} = \rho[/math] (On the dimensional relationships. Pressure term on its own is an energy density) ie. [math]P = \frac{F}{A} = \frac{E}{V}[/math] So how do we even begin to describe the fluctuations of spacetime? I show it is possible under a formulation by Sakharov who attempted to calculate the zero point modes over curved spacetime. We don't need to formulate his full expression for curvature because thankfully, a curvature term is already present in the Friedmann equation [math]\frac{kc^2}{a^2}[/math] And it seems that it is usual convention just to leave this term out when dealing with flat space. The zero point modes are given as [math]\mathcal{L}_{density} = \hbar c \int\ k^3\ dk[/math] where [math]k[/math] here is the wavenumber, so not to be confused with [math]kc^2[/math] the curvature term or even [math]-\mathbf{k}\sigma^2[/math] which may be related to the string tension of unified field theories. Deriving an appropriate representation of the modes in the context of the Freidmann equation is surprisingly not too difficult. The Friedmann equation can be written as [math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{2U}{mR^2}[/math] since the energy of the zero point modes is [math]\hbar c R \int\ k\ dk[/math] This differs slightly from the energy density [math]\hbar c \int\ k^3\ dk^2[/math] Plugging in [math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{2\hbar c}{mR^2} R \int\ k\ dk[/math] [math](\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \frac{2\hbar c}{mR} \int\ k\ dk[/math] and again, rearranging we find [math]m\dot{R}^2 = \frac{8 \pi GmR^2}{3}\rho + 2\hbar c R \int\ k\ dk[/math] Which recovers the Sakharov zero point field term. You can construct the power equation from this, but what we want for the master equation is a zero point term written in terms of mass density (for the effective density) and then a fluid expansion coefficient will give it its non-conserved formulation, which is already attached to the effective density part. If the energy density of the modes is [math]\rho_{off} = \hbar c \int\ k^3 dk[/math] Then this can be plugged in, where now we redefine [math]\mathbf{k} = \frac{G}{c^2}[/math] and [math]\rho[/math] (the on-shell effective density component) is an energy density description as well, we have [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3c^2}(\frac{\rho_{on} + \rho_{off} + 3P - \mathbf{k} \sigma^2}{n})n\Theta + \frac{\Lambda c^2}{3}\Theta + \omega^2\Theta[/math] Friedmann's Makeup Kenath Arun has given a simple description of gravitational binding energy, we extend it for a more appropriate relativistic form in which you calculate the difference of gravitational fields of spiral galaxies to obtain the gravitational binding energy density: [math]\rho_{binding} = \rho_{spiral}c^2(1 - \frac{1}{(1 - \frac{2Gm^{2}_{spiral}}{c^2R_{spiral}})})[/math] Sivaram also suggested a direct calculation of binding energies between particles in the universe, [math]\frac{\ddot{a}}{a} = \frac{4 \pi G m^4c^3}{3\hbar}(1 - \frac{Gm^2}{\hbar c})[/math] Which is also pleasant... In an older study, to get the non-conservation, I took the derivative of this equation, we now know you don't need to do that with a fluid expansion coefficient. We will implement both into separate formulations of the final equation. Electromagnetic density consists of two terms [math]\rho_{EM} = \frac{1}{2} \epsilon_0 \mathbf{E}^2 + \frac{1}{2} \frac{\mathbf{B}^2}{\nu_0}[/math] [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2} - \mathbf{k} \sigma^2}{n})n\Theta + \omega^2\Theta[/math] If we strip the Friedmann equation down to the basic terms of effective density parameters, including cosmological constant and rotation, we have: [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3c^2}(\frac{\rho_{on} + \rho_{off} + \rho_{pressure} + \rho_G + \rho_{EM} - \rho_{\sigma}}{n})n\Theta + \omega^2\Theta[/math] Seeing the universe in light of these reasonable energy density parameters, just shows us how complicated the universe may actually be, the effective density part now consists of density due to on-shell particles, density due to off-shell particles, density due to pressure, density due to gravitational binding, density due to primordial electromagnetic fields and the density due to torsion. I haven't spoke about primordial magnetic fields or electric fields in the universe - but its a very important question for the unification theories, in which some investigations have shown that gravity and electric fields may have a complimentary existence, ie. charge vanishes from the early universe as the strength of gravity increases! If we put the main terms back in, I will take the irreversible particle out of the parenthesis, just remember what it is supposed to be doing, keeping it in there makes it look terribly messy. [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3c^2}(\rho_{on} + \hbar c \int k^3\ dk + 3P + \rho_{spiral}c^2(1 - \frac{1}{(1 - \frac{2Gm^{2}_{spiral}}{c^2R_{spiral}})}) +\frac{1}{2} \epsilon_0 \mathbf{E}^2 + \frac{1}{2} \frac{\mathbf{B}^2}{\mu_0} - \mathbf{k} \sigma^2)\frac{n\Theta}{n} + \frac{\Lambda c^2}{3}\Theta + \omega^2\Theta[/math] ref http://cbc.arizona.edu/~salzmanr/480b/statt02/statt02.html https://arxiv.org/pdf/1512.03100.pdf https://iktp.tu-dresden.de/bndschool/notes_tytgat_cosmology.pdf https://arxiv.org/pdf/1512.03100.pdf http://cds.cern.ch/record/405307/files/9910483.pdf https://books.google.co.uk/books?id=afLLCQAAQBAJ&pg=PA209&lpg=PA209&dq=modified+Raychaudhuri+equation&source=bl&ots=VxH-Jd_Llm&sig=6Ew5GZhEV3Uk8Hia9hwGi6AP4-w&hl=en&sa=X&ved=0ahUKEwi75NHf1-3SAhUE7RQKHTY-Bso4ChDoAQgmMAI#v=onepage&q=modified%20Raychaudhuri%20equation&f=false https://arxiv.org/pdf/gr-qc/0511123.pdf https://benthamopen.com/contents/pdf/TOAAJ/TOAAJ-5-7.pdf http://sci-hub.bz/10.1007/BF02187817 http://www.ast.cam.ac.uk/sites/default/files/Lec05_PC_7Nov.pdf https://arxiv.org/pdf/1006.5191.pdf http://sci-hub.ac/10.1007/s10509-011-0770-2 Extra https://books.google.co.uk/books?id=Y6LvPkD9UyoC&pg=PA126&lpg=PA126&dq=spin+density,+torsion&source=bl&ots=KF77H_-zor&sig=flQHP5Cr1w5X11WbfFAeTp-jXGY&hl=en&sa=X&ved=0ahUKEwiz3Zb5pK3RAhUrL8AKHeadD5sQ6AEIPjAE#v=onepage&q=spin%20density%2C%20torsion&f=false https://arxiv.org/pdf/gr-qc/0009025v1.pdf https://books.google.co.uk/books?id=_WFpDQAAQBAJ&pg=PA284&lpg=PA284&dq=M.Fil+Chenkov+Quantum+Tunneling+of+a+Rotating+Universe&source=bl&ots=qsvr7wgaPn&sig=lDUvxzMnTfX1nfHvk-v0blm5Pmo&hl=en&sa=X&redir_esc=y#v=onepage&q=M.Fil%20Chenkov%20Quantum%20Tunneling%20of%20a%20Rotating%20Universe&f=false https://books.google.co.uk/books?id=X-nICgAAQBAJ&pg=PA135&lpg=PA135&dq=M.Fil+Chenkov+Quantum+Tunneling+of+a+Rotating+Universe&source=bl&ots=JTdEk58o9e&sig=y0W5aaeRXfsHJa0LnehnnSbmxJ8&hl=en&sa=X&redir_esc=y#v=onepage&q=M.Fil%20Chenkov%20Quantum%20Tunneling%20of%20a%20Rotating%20Universe&f=false https://books.google.co.uk/books?id=YavUCgAAQBAJ&pg=PA285&lpg=PA285&dq=tunneling+universe,+rotation,+gamow+factor&source=bl&ots=4zthfMSYYZ&sig=niopLu8GcSC3KsPirHWPYWZznog&hl=en&sa=X&ved=0ahUKEwiviveIu-bOAhXrJcAKHUBFDX44ChDoAQgqMAI#v=onepage&q=tunneling%20universe%2C%20rotation%2C%20gamow%20factor&f=false https://books.google.co.uk/books?id=_WFpDQAAQBAJ&pg=PA284&lpg=PA284&dq=M.+Fil%E2%80%99chenkov+rotation+friedmann+equation&source=bl&ots=qswq4pb7Qn&sig=NEAt1sHpxhMlCQgjYYjVvhLjvXM&hl=en&sa=X&ved=0ahUKEwjz2MWR3-3SAhViD8AKHVg3AuUQ6AEIHjAA#v=onepage&q=M.%20Fil%E2%80%99chenkov%20rotation%20friedmann%20equation&f=false Notes [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}(\frac{\rho + 3P}{n})n\frac{\dot{R}}{R} = \mathcal{P}[/math] By introducing the heat per unit particle [math]d\bar{q} = \frac{dQ}{dN}[/math] the above component reduces to a Gibbs expression from the Gibbs equation. The process of particle creation in the sense above can be thought of being induced gravitationally (gravitational particle production) and in fact, the particle number coupled to the gravitational field looks like; [math]N^{\mu}_{:\mu} \equiv n \sigma \Gamma \equiv s \Gamma \ne 0[/math] The [math]\Gamma[/math] symbol here is actually known as the particle production rate and for a non-conserved universe, [math]n\Gamma[/math] has a non-zero value. For the adiabatic universe, particle production rate is simply [math]\dot{n} + \Theta n = 0[/math] In our case we have [math]\dot{n} + \Theta n = n\Gamma[/math] and Theta represents [math]\Theta^{\mu}_{:\mu} = 3H = 3 \frac{\dot{a}}{a}[/math] Mathematical Dynamics of Torsion and Relation to the Spin Torsion doesn't just appear because of rotation, mathematically speaking, they are in fact related. They are not seperate terms as they may appear to be in the master equation [math]\dot{H}\Theta + H^2\Theta + \frac{kc^2}{a^2}\Theta = \frac{8 \pi G}{3}(\frac{\rho + \frac{3P}{c^2} - \mathbf{k} \sigma^2}{n})n\Theta + \frac{\Lambda c^2}{3}\Theta + \omega^2\Theta[/math] where the torsion is encapsulated by the term [math]- \mathbf{k} \sigma^2[/math] and the angular frequency is [math]\omega^2[/math] - though in their relationship, it isn't an angular frequency but instead a rotational relationship related to the angular action of the system. For instance, Sabbata shows that torsion is in fact related to spin (we see this relation by noticing [math]\sigma^2[/math] is the spin density. Torsion [math]Q[/math] and spin [math]S[/math] can be related as [math]Q = \frac{4 \pi G}{c^2}S[/math] Notice then, the spin [math]S[/math] is not the same as the angular frequency [math]\omega[/math] as such, which measures the time it takes for one revolution. Instead, [math]S[/math] has more similarity to an action term [math]\hbar[/math]. So an interaction energy can also be given [math]Q \cdot S = \frac{4 \pi G}{c^2}S^2[/math] or in context of mass [math]\frac{1}{c^2} Q \cdot S = \frac{4 \pi G}{c^4}S^2[/math] We can see, this is just a form of [math]-\mathbf{k}\sigma^2[/math] - in fact, to get that form, you just reduce the units down to a mass and distribute the density [math]\frac{1}{c^2} Q \cdot \frac{S}{V} = \frac{4 \pi G}{c^4} \sigma^2[/math] again, with [math]\sigma[/math] playing the role of spin density with units [math]\frac{mvR}{L^3}[/math]. This part just helps to disseminate the spin-torsion relationships for clarity of understanding. Pre-Big Bang Liquid Condensed All-Matter Phase A previous equation in my past work involved a power equation represented in Freidmann's cosmology [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\dot{\rho} = \mathcal{P}[/math] The fluid equation of cosmology is [math]\dot{\rho} = \frac{\dot{R}}{R}(\rho + 3P)[/math] We obtain after substitution [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}\frac{\dot{R}}{R}(\rho + \frac{3P}{c^2}) = \mathcal{P}[/math] The modified law for irreversible particle production we have [math]dE = dQ - PdV + (\frac{\rho + P}{n}) dN[/math] and [math]\dot{E} = \dot{Q} - P\dot{V} + (\frac{\rho + P}{n}) \dot{N}[/math] The modified power equation is therefore [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}(\frac{\rho + 3P}{n})\dot{N} = \mathcal{P}[/math] If we just take a look at the differential form of the modified first law of thermodynamics we have [math]dE = dQ - pdV + (\frac{\rho + p}{n})dN[/math] Solving for the key expression (the last expression on the RHS involving the effective densities) I get; [math](\frac{\rho + p}{n})dN = dE - dQ + pdV[/math] We know this is a key feature in one of our terms in our power equation: [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}(\frac{\rho + 3P}{n})\dot{N} = \mathcal{P}[/math] The mathematical theory does allow for a stranger equation consisting of many more terms: [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}(\dot{E} - \dot{Q} + p\dot{V}) = \mathcal{P}[/math] Which might be interesting to look at another time - what is interesting though is that if you introduce the heat per unit particle, the thermodynamic expression should dissolve into something related to the Gibbs equation. The heat per unit particle is given as [math]d\bar{q} = \frac{dQ}{N}[/math] So that the first law reduces to [math]d(\frac{\rho}{n}) = d\bar{q} - qd(\frac{1}{n})[/math] Which is the Gibbs equation. (see references). So it seems possible at first glance, that a fusion of the Gibbs physics can happen in the Friedmann cosmology, especially in the context of irreversible processes. Going back to the thermodynamic law, we have [math]dE = dQ - p dV[/math] Just like what we saw before. Now we must introduce the energy density as [math]\rho = \frac{E}{V}[/math] We can write the thermodynamic law as a Gibbs equation and that is given as [math]TdS = dq = d(\frac{\rho}{n}) + pd(\frac{1}{n})[/math] And of course, the time derivative is [math]T\dot{s} = \dot{q} = (\frac{\dot{\rho}}{n}) + \dot{p}(\frac{1}{n})[/math] Entropy is naturally dimensionless, but in some cases, entropy can be defined by the number of microscopic thermodynamioc systems multiplied by the Boltzmann constant, [math]S = k_B\ \ln \Omega[/math] And the relationship [math]dS = \frac{dQ}{T}[/math] is important as we recognize [math]TdS = dQ[/math] Which is the heat energy. The key fact is that entropy has units of the Boltzmann constant, but it is naturally dimensionless It implements like in our approach, as: [math]T k_B\dot{s} = \dot{q} = (\frac{\dot{\rho}}{n}) + \dot{p}(\frac{1}{n})[/math] Just always remember, the Boltzmann constant is always attached to the temperature to convert the units properly into energy. Implementing that I get an equation which looks like: [math]m \dot{R}\ddot{R} - \frac{8 \pi Gm R^2}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})] = \mathcal{P}[/math] it is possible to retrieve the more identifiable form of a Friedmann equation, just divide through by mass, rearrange the radius: [math]\frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})] = \frac{2\dot{E}}{mR^2(t)}[/math] Where [math]R(t) = a(t)r[/math] [math]\frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})] = \frac{2\dot{E}}{ma(t)r^2}[/math] Note* we have a factor of 6 because of some differentiation many works back, but I have shown it is probably truer as [math]8 \pi G/3[/math]. The final part can be recognized as the curvature term. [math]\frac{\dot{R}}{R}\frac{\ddot{R}}{R} - \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})] = \frac{kc^2}{a}\frac{\dot{R}}{R}[/math] we'll show this: (EDIT) - There may be a negative sign attached to it which is related to the potential. [math]k = - \frac{2E R^2}{c^2r^2}[/math] [math]kc^2 = -\frac{2E}{mr^2}[/math] divide off by the scale factor [math]\frac{kc^2}{a} = - \frac{2E}{ma^2(t)r^2} = - \frac{2E}{mR^2}[/math] Which is of course, the curvature. If the negative sign exists, then the equation rearranges like [math]\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})][/math] It was explained by Motz in his paper (see references), that a universe will collapse to a point (also see notes); the acceleration is always negative [math]\ddot{R} < 0[/math] and will collapse so long as... [math]\rho + 3c^{-2}(P + \frac{u}{3}) > 0[/math] ... in the Friedmann universe. This is necessarily so because the pressure is positive [math]P > 0[/math]. There is some relic notation in here, we will change this for modern representations. Motz goes on to explain how a differentiation of the Friedmann equation will lead to a non-conservation, an important departure he says from a model in which it has taken the unwarranted assumption that the over all rest mass should remain the same. He said the basis of that assumption is that thermonuclear transformation of mass into energy in the stellar interiors has reduced the mass of the universe only negligably and increased the entropy only slightly. [a source needed for this] Motz et al. further explains that this assumption is not valid if the initial state of the universe was cold. I argue in this work, Motz may be correct in the assumption, perhaps his model needs a tweak. He does explain that a radiationless state dominated by what he called ''Unitons'' could explain such a state, where the gravitational attractions between any two such particles exceeded their mutual electrostatic attraction. Indeed, with some consideration of this mechanism, it seems perfectly plausible for some relic non-electromagnetically-interacting system of particles could explain the cold baby universe. Their mutual electrostatic interactions would be dwarfed by gravity 137 times - he expects them to have coalesced very rapidly into gravitationally-bound triplets that he would come to identify as nucleons. This is where Motz-Kraft theory may suffer problematics, since none of the quark particles today contain the kind of gravitational mass required for a relic Planck particle. There is an alternative to their Uniton model. Franck Wilczek has shown (initially with some skepticism) that there may be more to the story in the unification picture. Gravity and electromagnetism may in fact hold a complimentary existence (see references). The unification picture requires (at least logically) that gravity increases as you approach the initial singularity (if indeed there is one at all). What he has discovered, at least mathematically is that as gravity increases, electromagnetism will tend to zero! This means, the early universe is already free of electromagnetic interactions in Wilczek's universe, based on the most modern approach. This means, any relic Planck particles that did exist as a degenerate gas, will not have interacted electromagnetically. These particles would have a mutual electrostatic interactions dwarfed by many magnitudes of gravity - from this point, if the radius [math]R[/math] of this initial state was [math]>0[/math] and its temperature equal to zero, then the energy released during this gravitational collapse would have produced the big bang and inflated the universe from its initial compact state. Motz did emphasize that this remarkable transition of the universe from a static state to an expanding state required a vast change of phase from all-matter to a phase consisting of a gas radiation phase. Motz and Kraft go on to derive a formula which boils down to a pressure term related essentially to the Gibbs equation! This interested me because I too saw a relationship between Friedmann cosmology and the Gibbs equation. Noticing that for diabatic systems the first law of thermodynamics changes into a Gibbs equation and so altered the effective density parameter. The equation I arrived at with Friedmann and Gibbs physics was: [math]\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})][/math] They notice their solution is simple yet a profound result, essentially the Gibbs equation for an isothermal reversible change of a phase of liquid to vapor for infinitessimal differential volume of change gives a singularity free universe. The model I found striking for the counter-intuitive notion that we may have to start thinking about the baby universe as a cold dominated matter phase in the absence of electromagnetic interactions. It was only by a realization I made linking possible importance between Wilczek's work and this older work that I had read many years ago. If Wilczek is indeed correct (like many independent scientists are trying to prove), then Motz and Krafts early universe no longer looks strange. It would actually make sense. Based on all the current models of significance in the theoretical world, I have came to some final conclusions about the universe at large. 1). To follow the intrinsic rules of spatial symmetries, the universe is part of the full Poincare Group leading to a primordial universal rotation. 2) The rotation naturally leads to an intrinsic torsion field. 3) It also leads to an intrinsic centrifugal force field which will replace inflation theories which are being heavily attacked right now in the academic community. 4) A dynamical infusion of Friedmann cosmology with re-definitions of the thermodynamic law as expressed as a Gibbs equation may lead to an interpretation of the universe which had a pre-big bang all-matter liquid phase. 5) The pre-big bang phase was a gas of degenerate Planck particles that underwent a phase transition from a liquid state to a vapor. The phase transition from all-matter liquid to radiation vapor states generated the expansion pushing the universe out of the dense Planck era. The pre-big bang phase would be gravity dominated only. As a universe expands, gravity weakens and electromagnetism dominates. Based on all the current models of significance in the theoretical world, I have came to some final conclusions about the universe at large. 1). To follow the intrinsic rules of spatial symmetries, the universe is part of the full Poincare Group leading to a primordial universal rotation. 2) The rotation naturally leads to an intrinsic torsion field. 3) It also leads to an intrinsic centrifugal force field which will replace inflation theories which are being heavily attacked right now in the academic community. 4) A dynamical infusion of Friedmann cosmology with re-definitions of the thermodynamic law as expressed as a Gibbs equation may lead to an interpretation of the universe which had a pre-big bang all-matter liquid phase. 5) The pre-big bang phase was a gas of degenerate Planck particles that underwent a phase transition from a liquid state to a vapor. The phase transition from all-matter liquid to radiation vapor states generated the expansion pushing the universe out of the dense Planck era. The pre-big bang phase would be gravity dominated only. As a universe expands, gravity weakens and electromagnetism dominates. There has been a lot more study, equations and information of other types inbetween these five premises which create the foundation of my entire theory. Inflation needs to be scrapped, there are other solutions. It just so happens, when you extend the Poincare group to the Friedmann cosmology, it naturally creates the repulsion required to push the universe out the Planck era. Or at least, contributed to pushing it out which would be more accurate to say. The pre-big bang phase was a super cool static region that had to undergo some transition in the absence of electromagnetism: the phase transition occurs because of a change in the metric itself, not to do with the actual dynamics of any systems inside of it (which was an important realization made by my friend, Matti Pitkanen, creator of the 40-year old extraordinary theory of Topological Geometrogenesis (or TGD). He also linked important to the Hagedorn temperature, which is something I am yet to investigate. ) My time writing about the Friedmann equation and my idea's about the origin of the universe is coming to an end here. I am getting tired now of venturing into this. I just wanted to plant a seed out there that could help a future scientist or someone aspiring to be similar to look at it and consider it and maybe take it forward to the next levels of hypothesis. I'll be concentrating now on other things, but I will be occasionally coming back to answer and reply to certain things. Hope you enjoyed my contributions! ref. https://arxiv.org/pdf/1512.03100.pdf http://sci-hub.bz/10.1111/j.1749-6632.1992.tb17071.x https://www.newscientist.com/article/mg20827853.600-why-the-early-universe-was-free-of-charge/ Edited June 4, 2017 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
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