Wormhole Metric...... How is this screwed up.

[Mordred]

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]

29 minutes ago, 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.

 

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 ω_s and R_s

Edited November 22, 2017 by Vmedvil

[Mordred]

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]

5 minutes ago, 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 ?

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]

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]

4 minutes ago, 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?

Where I accounts for density, counting each object at a radius as individual objects, which have a total mass of M_b or ∑M = M₁ + M₂ + M₃ ..... M_n

Where I_s or ∑ I = M₁R₁² + M₂R₂² + M₃R₃² + ..... + M_nR_n²

Edited November 22, 2017 by Vmedvil

[Mordred]

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]

13 minutes ago, 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

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]

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]

21 minutes ago, 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.

 

 

 

And no it does not take in account anything to do with Hubble's constant, but that is handled by ω_{s ,}  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]

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]

7 minutes ago, 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.

 

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]

1 hour ago, 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. 

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]

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]

9 minutes ago, 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.

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]

1 minute ago, 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.

Math is built on concepts

[Mordred]

Every definition under physics is mathematically precise. 

[SuperPolymath]

I had hoped my concepts had been adopted into said maths before this threads conception. I was wrong

[Vmedvil]

1 minute ago, SuperPolymath said:

I had hoped my concepts had been adopted into said maths before this threads conception. I was wrong

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]

Even still if you read my thread, there's  @ least a dozen articles that show a necessity for my lines of thought here

[Mordred]

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]

39 minutes ago, 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

 

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.

∇E_b(x,y,z) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s²/2M_b) + V)²/C²))^1/2))M_bC²

Then it would read 

∇E_b(x,y,z) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s²/2M_b) + (HΔx(1/3.08567782*10¹⁹)))²/C²))^1/2))M_bC²

In any case, I will let you latex that now.

Edited November 22, 2017 by Vmedvil

[Mordred]

Start with a box volume [math]V=L^3[/math] with periodic boundary conditions. Apply the Schrodinger equation to give the energy and momentum eugenstates. where the possible momentum values are [math]\vec{p}=\frac{h}{L}(n_1\hat{x}+n_2\hat{y}+n_3\hat{z})[/math] [math](n_i=0,\pm1,\pm2...)[/math] the state density in momentum space ie number of states [math]\Delta p_x,\Delta P_y\Delta p_z[/math] is thus [math]\frac{L^3}{h^3}=\frac{V}{h^3}[/math] and the state density in phase space is [math](\vec{x},\vec{p})[/math] is [math]\frac{1}{h^3}[/math] if the particle has g internal degrees of freedom (eg spin) the density of states is [math]\frac{g}{h^3}=\frac{g}{(2\pi)^2}[/math]

[math](\hbar=\frac{h}{2\pi}=1)[/math] this is true even in the relativistic case. The particle energy if given by [math]E(\vec{p})=\sqrt{p^2+m^2}[/math] where [math]p=|\vec{p}|[/math] 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

[math]f(\vec{p}=\frac{1}{e^{(E-\mu)/T}\pm1}[/math] 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 [math]\mu[/math] 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 [math]\frac{g}{(2\pi)^3}f(\vec{p}[/math] we get the particle density in (ordinary space) by integrating over the momentum space to arrive at the following quantities.

number density [math] n=\frac{g}{(2\pi)^3}[/math]

energy density [math]\rho=\frac{g}{2\pi^3}\int E(\vec{p}f(\vec{p})d^3p[/math]

pressure [math]p=\frac{g}{(2\pi)^3}\int\frac{|\vec{p}|^2}{3E}f(\vec{p}d^3p[/math]

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]

14 minutes ago, 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

 

 

 

So, if I steal the Hubble constant and that energy density for matter density does that fix it.

∇E_b(x,y,z) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s²/2M_b) + (HΔx(1/3.08567782*10¹⁹)))²/C²))^1/2))M_bC²

Where H

 

Then ρ =(g/(2π)³)∫E(p⃗ f(p⃗ )d³p

 

Edited November 22, 2017 by Vmedvil

[Mordred]

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