Mordred Posted November 22, 2017 Posted November 22, 2017 (edited) Where would you find this operator in the hydrodynamic equations pertianing to an adiabatic fluid expansion ? (adiabatic meaning no net inflow/outflow of heat). The FRW fluid equations is derived using an adiabatic and isentopic fluid. Ok let me dig up the peer review upper boundary to a rotating universe. I had posted this to one of Dubblesix threads when he first proposd his idea here. Here is the arxiv on the upper bound. https://arxiv.org/abs/0902.4575 Models of a rotating universe have been studied widely since G{\"o}del \cite{1}, who showed an example that is consistent with General Relativity (GR). By now, the possibility of a rotating universe has been discussed comprehensively in the framework of some types of Bianchi's models, such as Type V, VII and IX \cite{2,3}, and different approaches have been proposed to constrain the rotation. Recent discoveries of some non-Gaussian properties of the Cosmic Microwave Background Anisotropies (CMBA) \cite{nG1,nG2,nG3,nG4,nG5,nG6,nG7}, such as the suppression of the quadrupole and the alignment of some multipoles draw attention to some Bianchi models with rotation \cite{bi1,bi2}. However, cosmological data, such as those of the CMBA, strongly prefer a homogeneous and isotropic model. Therefore, it is of interest to discuss the rotation of the universe as a perturbation of the Robertson-Walker metric, to constrain the rotating speed by cosmological data and to discuss whether it could be the origin of the non-Gaussian properties of the CMBA mentioned above. Here, we derive the general form of the metric (up to 2nd-order perturbations) which is compatible with the rotation perturbation in a flat Λ-CDM universe. By comparing the 2nd-order Sachs-Wolfe effect \cite{4,5,6,7,8} due to rotation with the CMBA data, we constrain the angular speed of the rotation to be less than 10−9 rad yr−1 at the last scattering surface. This provides the first constraint on the shear-free rotation of a ΛCDM universe. Edited November 22, 2017 by Mordred
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 3:42 AM, Mordred said: Where would you find this operator in the hydrodynamic equations pertianing to an adiabatic fluid expansion ? (adiabatic meaning no net inflow/outflow of heat). The FRW fluid equations is derived using an adiabatic and isentopic fluid. Ok let me dig up the peer review upper boundary to a rotating universe. I had posted this to one of Dubblesix threads when he first proposd his idea here. Here is the arxiv on the upper bound. https://arxiv.org/abs/0902.4575 Models of a rotating universe have been studied widely since G{\"o}del \cite{1}, who showed an example that is consistent with General Relativity (GR). By now, the possibility of a rotating universe has been discussed comprehensively in the framework of some types of Bianchi's models, such as Type V, VII and IX \cite{2,3}, and different approaches have been proposed to constrain the rotation. Recent discoveries of some non-Gaussian properties of the Cosmic Microwave Background Anisotropies (CMBA) \cite{nG1,nG2,nG3,nG4,nG5,nG6,nG7}, such as the suppression of the quadrupole and the alignment of some multipoles draw attention to some Bianchi models with rotation \cite{bi1,bi2}. However, cosmological data, such as those of the CMBA, strongly prefer a homogeneous and isotropic model. Therefore, it is of interest to discuss the rotation of the universe as a perturbation of the Robertson-Walker metric, to constrain the rotating speed by cosmological data and to discuss whether it could be the origin of the non-Gaussian properties of the CMBA mentioned above. Here, we derive the general form of the metric (up to 2nd-order perturbations) which is compatible with the rotation perturbation in a flat Λ-CDM universe. By comparing the 2nd-order Sachs-Wolfe effect \cite{4,5,6,7,8} due to rotation with the CMBA data, we constrain the angular speed of the rotation to be less than 10−9 rad yr−1 at the last scattering surface. This provides the first constraint on the shear-free rotation of a ΛCDM universe. Expand Well, there is a problem with that when one considers small scale things it is not Homogeneous and Isotropic, this cannot just define the big it has to define both. I defines sub structures like Quantum mechanics and QFT which makes homogeneous too simple. Where The Special Relativity part makes Isotropic too simple. You cannot have it Isotropic and Homogeneous when taken in account the things GR doesn't, GR assumes it is a perfect fluid and not tons of small interactions. Where GR is taken account by Schwarzchild Metric and Kerr Metric as ωs and Rs Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 (edited) Lol not exactly a scientific definition. How big ? how small? how do you define this boundary ? What is the point where curvature becomes measurable? ie where light paths convergence/divergence begins to be potentially measurable ? Lets use the example at 10^{43} seconds where the observable universe is contracted to roughly the radius of a grapefruit. Does curvature matter? this is a state that can be accurately described strictly by its temperature. All particles are in thermal equilibrium. Yet once electroweak symmetry breaking occurs you get rapid inflation. Edited November 22, 2017 by Mordred
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 4:15 AM, Mordred said: Lol not exactly a scientific definition. How big ? how small? how do you define this boundary ? What is the point where curvature becomes measurable? ie where light paths convergence/divergence begins to be potentially measurable ? Expand depends on the total mass of an object or set of objects, which defines everything from that, really this is coded kinda like a physics engine for a video game. I is like saying object. Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 Why would it depend on the objects mass and not its density to thermodynamic relations via the equations of state and thermal equilibrium of particle species?
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 4:21 AM, Mordred said: Why would it depend on the objects mass and not its density to thermodynamic relations via the equations of state and thermal equilibrium of particle species? Expand Where I accounts for density, counting each object at a radius as individual objects, which have a total mass of Mb or ∑M = M1 + M2 + M3 ..... Mn Where Is or ∑ I = M1R12 + M2R22 + M3R32 + ..... + MnRn2 Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 Well lets properly define this and avoid the confusion. I have a good reference for you to read http://www.wiese.itp.unibe.ch/lectures/universe.pdf :" Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis. See chapters 3 and 4 covering the Bose-Einstein and Fermi-Dirac statistics
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 4:27 AM, Mordred said: Well lets properly define this and avoid the confusion. I have a good reference for you to read http://www.wiese.itp.unibe.ch/lectures/universe.pdf :" Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis. See chapters 3 and 4 covering the Bose-Einstein and Fermi-Dirac statistics Expand Where it is like that if R goes to L, the Lagrangian forms are compatible, it is in that form being the same as radius with a different geometry operator being 1/12 or 1/2 or whatever, where 1/12 means rod or QFT geometry. Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 No that's not how thermal dynamic equilibrium works. Not in Cosmology applicaions. https://en.m.wikipedia.org/wiki/Decoupling_(cosmology) There is a very distinct relation between the particle species involved and when they decouple in relationship to rates of expansion.
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 4:51 AM, Mordred said: No that's not how thermal dynamic equilibrium works. Not in Cosmology applicaions. https://en.m.wikipedia.org/wiki/Decoupling_(cosmology) There is a very distinct relation between the particle species involved and when they decouple in relationship to rates of expansion. Expand And no it does not take in account anything to do with Hubble's constant, but that is handled by ωs , I thought that Kerr Metric took in account for that as a spinning BH where expansion is the spin of the object. Where a =ω(I/M) , all I did is take the integral of it. Where this is super nested. Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 No a spinning object by definition in anistropic and inhomogeneous you won't get a uniform temperature distribution. So how thermal equilibrium would work under the Godel type metrics gets rather complex. One of the most common problems with posters trying to model build toy universes is skipping numerous key lessons one would get via training at a university. One of the most common areas of study most don't understand is how thermodynamics relate to expansion rates and thermal decoupling.
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 5:11 AM, Mordred said: No a spinning object by definition in anistropic and inhomogeneous you won't get a uniform temperature distribution. So how thermal equilibrium would work under the Godel type metrics gets rather complex. One of the most common problems with posters trying to model build toy universes is skipping numerous key lessons one would get via training at a university. One of the most common areas of study most don't understand is how thermodynamics relate to expansion rates and thermal decoupling. Expand Well, that explains alot if I took the integral wrong there, see I acted as if k was a constant in the integral of that. Edited November 22, 2017 by Vmedvil
SuperPolymath Posted November 22, 2017 Posted November 22, 2017 (edited) On 11/22/2017 at 4:15 AM, Mordred said: Lol not exactly a scientific definition. How big ? how small? how do you define this boundary ? What is the point where curvature becomes measurable? ie where light paths convergence/divergence begins to be potentially measurable ? Lets use the example at 10^{43} seconds where the observable universe is contracted to roughly the radius of a grapefruit. Does curvature matter? this is a state that can be accurately described strictly by its temperature. All particles are in thermal equilibrium. Yet once electroweak symmetry breaking occurs you get rapid inflation. Expand There's no evidence that documents anything before CMB, such as some mythical "grapefruit sized singularity". Idk about what 006 & vmedvil have said about rotating universes, but my concept is that the CMB is all you get. & all you need. So as the observable universe ages, black holes grow. I proposed that they grow by consuming mass by merging with the microscopic black holes (beneath the Planck length of 1.6x10^-35 metres for radiation & about the Planck length for the center's center of atomic nuclei between the shifting protons & neutrons falling into the core like the bright quasar of the heart of a galaxy) inherent within all matter. This matter stripped of its mass is just Cherenkov radiation that gets jettisoned back put into space before propagating away from said macroscopic BH. When these super God black holes get get large enough, during big rip, adjacent waves of Cherenkov radiation around CDM-like collections of evaporating mother black holes pass through each other. By the two the mother black holes have shrunk into infinitesimal versions of their former selves (micro-black holes) the curtain of space-time has flowed from their collective anti de sitter space into ours, pulling all these Cherenkov radiation waves into a convergence point equal to the total area of the CMB, restarting the process & spawning the first atoms of a low entropy body out of a quark-gluon pool that looks exactly like the CMB. That's a small taste of the out-of-box ideas that I presented in support of a rotating universe. Edited November 22, 2017 by SuperPolymath
Mordred Posted November 22, 2017 Posted November 22, 2017 (edited) Out of the box ideas are absolutely useless unless you can them up with the correct mathematics. This is physics not guess work and unsupported conjectures based on personal ideas. Drawings are not compliant to the mathematical rigor required as per the link on the other forum. Edited November 22, 2017 by Mordred
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 5:18 AM, Vmedvil said: Well, that explains alot if I took the integral wrong there, see I acted as if k was a constant in the integral of that. Expand Well, No I get the samething either way yes it agrees with the Einstein tensor and Gobel's Universe model as ω2 = u Edited November 22, 2017 by Vmedvil
SuperPolymath Posted November 22, 2017 Posted November 22, 2017 On 11/22/2017 at 5:25 AM, Mordred said: Out of the box ideas are absolutely useless unless you can them up with the correct mathematics. This is physics not guess work and unsupported conjectures based on personal ideas. Expand Math is built on concepts
Mordred Posted November 22, 2017 Posted November 22, 2017 Every definition under physics is mathematically precise.
SuperPolymath Posted November 22, 2017 Posted November 22, 2017 I had hoped my concepts had been adopted into said maths before this threads conception. I was wrong
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 5:35 AM, SuperPolymath said: I had hoped my concepts had been adopted into said maths before this threads conception. I was wrong Expand We have not gotten to that part yet we are proofing this, where your model enters when ω2 = k2 + m2 , somehow this needs to take account for expansion the equation which that is a solution to. Edited November 22, 2017 by Vmedvil
SuperPolymath Posted November 22, 2017 Posted November 22, 2017 Even still if you read my thread, there's @ least a dozen articles that show a necessity for my lines of thought here
Mordred Posted November 22, 2017 Posted November 22, 2017 Ok lets do this the correct way. Lets start with particles in a box in terms of thermal equilibrium. This will take me a bit to latex in so bear with me
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 5:40 AM, Mordred said: Ok lets do this the correct way. Lets start with particles in a box in terms of thermal equilibrium. This will take me a bit to latex in so bear with me Expand Which I know the sign of expansion will be positive like a Friedmann equation to matter density, but I wanna see what you type up in latex. Could it be as simple as steal the hubble constant and put it as velocity being in units of kilometers per second per megaparsec. Like V = H * Δx(Kiloparsec) where Kiloparsec = 3.086e+19 meters. ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb) + V)2/C2))1/2))MbC2 Then it would read ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb) + (HΔx(1/3.08567782*1019)))2/C2))1/2))MbC2 In any case, I will let you latex that now. Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 (edited) Start with a box volume V=L^3 with periodic boundary conditions. Apply the Schrodinger equation to give the energy and momentum eugenstates. where the possible momentum values are \vec{p}=\frac{h}{L}(n_1\hat{x}+n_2\hat{y}+n_3\hat{z}) (n_i=0,\pm1,\pm2...) the state density in momentum space ie number of states \Delta p_x,\Delta P_y\Delta p_z is thus \frac{L^3}{h^3}=\frac{V}{h^3} and the state density in phase space is (\vec{x},\vec{p}) is \frac{1}{h^3} if the particle has g internal degrees of freedom (eg spin) the density of states is \frac{g}{h^3}=\frac{g}{(2\pi)^2} (\hbar=\frac{h}{2\pi}=1) this is true even in the relativistic case. The particle energy if given by E(\vec{p})=\sqrt{p^2+m^2} where p=|\vec{p}| don't confuse this with pressure. There is two classifications of particles Bosons and fermions. In thermodynamic equilibrium the distribution function is given by the following equation f(\vec{p}=\frac{1}{e^{(E-\mu)/T}\pm1} where the + is the bosons and - for the fermions. If f<1 f gives the probability that a state is occupied. (Paulis exclusion principle). The equilibrium distribution function has two distinct parameters T the temperature and the chemical potential \mu The temperature is related to the energy density of the system ad the chemical potential is related to the number density (n) of particles in the system. The particle density in phase space is the density of states times their occupation number \frac{g}{(2\pi)^3}f(\vec{p} we get the particle density in (ordinary space) by integrating over the momentum space to arrive at the following quantities. number density n=\frac{g}{(2\pi)^3} energy density \rho=\frac{g}{2\pi^3}\int E(\vec{p}f(\vec{p})d^3p pressure p=\frac{g}{(2\pi)^3}\int\frac{|\vec{p}|^2}{3E}f(\vec{p}d^3p now particles can be in kinetic equilibrium, thermal equilibrium and chemical equilibrium. see the uwe jen weise article I linked earlier to cover how the above applies to nucleosynthesis Edited November 22, 2017 by Mordred
Vmedvil Posted November 22, 2017 Author Posted November 22, 2017 (edited) On 11/22/2017 at 6:22 AM, Mordred said: Start with a box volume V=L3 woth periodic boundary conditions. Apply thee Schrodinger equation to give the energy and momentum eugenstates. where the possible momentum values are p⃗ =hL(n1x^+n2y^+n3z^) (ni=0,±1,±2...) the state density in momentum space ie number of states Δpx,ΔPyΔpz is thus L3h3=Vh3 and the state density in phase space is (x⃗ ,p⃗ ) is \frac{1}{h^3}[/math] if the particle has g internal degrees of freedom (eg spin) the density of states is gh3=g(2π)2 (ℏ=h2π=1) this is true even in the relativistic case. The particle energy if given by E(p⃗ )=p2+m2−−−−−−−√ where p=|p⃗ | don't confuse this with pressure. There is two classifications of particles Bosons and fermions. In thermodynamic equilibrium the distribution function is given by the following equation f(p⃗ =1e(E−μ)/T±1 where the + is the bosons and - for the fermions. If f<1 f gives the probability that a state is occupied. (Paulis exclusion principle). The equilibrium distribution function has two distinct parameters T the temperature and the chemical potential μ The temperature is related to the energy density of the system ad the chemical potential is related to the number density (n) of particles in the system. The particle density in phase space is the density of states times their occupation number g(2π)3f(p⃗ we get the particle density in (ordinary space) by integrating over the momentum space to arrive at the following quantities. number density n=g(2π)3 energy density ρ=g2π3∫E(p⃗ f(p⃗ )d3p pressure p=g(2π)3∫|p⃗ |23Ef(p⃗ d3p now particles can be in kinetic equilibrium, thermal equilibrium and chemical equilibrium. see the uwe jen weise article I linked earlier to cover how the above applies to nucleosynthesis Expand So, if I steal the Hubble constant and that energy density for matter density does that fix it. ∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb) + (HΔx(1/3.08567782*1019)))2/C2))1/2))MbC2 Where H Then ρ =(g/(2π)3)∫E(p⃗ f(p⃗ )d3p Edited November 22, 2017 by Vmedvil
Mordred Posted November 22, 2017 Posted November 22, 2017 (edited) If you apply the above to each particle species to each corresponding equation of state derived using the above and using the Bose-Einstein/Fermi_Dirac statistics and can maintain a homogenous and isotropic state under rotation as the volume increases then yes. However keep in mind the article I posted on the upper boundary limits I posted earlier which applies the above. If you go through that arxiv link you will see that it also applies these statistics under rotation inclusive to the FRW fluid equations. Edited November 22, 2017 by Mordred
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