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Modified Raychauduri Equation With Poincare Symmetry And Non-Conservation


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#1 Dubbelosix

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Posted 04 April 2017 - 12:12 AM

I realized I posted this in the wrong subforum, anyway, sorry. This is not a strange claims work.

 

 

 

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
 
 
We will derive a Friedmann power equation, to do so, we use an equation I call the Friedmann Langrangian which we will derive first. 
 
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}R^2[/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}R^2 = \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}R^2 = \mathcal{L}[/math]
 
 
Going back and rearranging
 
 
[math]m\dot{R}^2- \frac{8 \pi GmR^2}{3}\rho + \frac{\Lambda mc^2R^2}{3} = \mathcal{L}[/math]
 
 
Friedmann Power Equation
 
 
Differentiation of the Langrangian gives
 
[math]m\dot{R}\ddot{R}- \frac{8 \pi GmR^2}{3}\dot{\rho} + \frac{2\Lambda mc^2R\dot{R}}{3} = \mathcal{P}[/math]
 
(just a dimensional argument)
 
 
Removing the cosmological constant, no need for it now, was only in there to show you how it related to the Langrangian of cosmology,
 
 
[math]m\dot{R}\ddot{R}- \frac{8 \pi GmR^2}{3}\dot{\rho} = \mathcal{P}[/math]
 
 
Replacing [math]\dot{\rho}[/math] with the continuity equation (which is)
 
 
[math]\dot{\rho} = \frac{\dot{a}}{a}(\rho + \frac{3P}{c^2})[/math]
 
We can see we can substitute in the equation:
 
 
[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 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}{\mu_0}[/math]
 
 
If we strip the Friedmann equation down to the basic terms of six 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 + \frac{\Lambda c^2}{3}\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
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Extra
 
 
 
 
 
 
 
Notes
 
Fluid Expansion
 
 
[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.
 
Deriving spiral binding energies
 
 

The gravitational field inside a radius [math]r = r(0)[/math] is given as
 
[math]\frac{dM}{dR} = 4 \pi \rho R^2[/math]
 
and the total mass of a star is
 
[math]M_{total} = \int 4 \pi\rho R^2 dR[/math]
 
and so can be understood  in terms of energy (where [math]g_{tt}[/math] is the time-time component of the metric),
 
[math]\mathbf{M} =  4 \pi  \int \frac{\rho R^2}{g_{tt}} dR = 4 \pi \int \frac{ \rho R^2}{(1 - \frac{R}{r})} dR[/math]
 
The difference of those two mass formula is known as the gravitational binding energy:
 
[math]\Delta M = 4 \pi \int  \rho R^2(1 - \frac{1}{(1 - \frac{R}{r})}) dR[/math]
 
Distribute c^2 and divide off the volume we get:
 
[math]\bar{\rho}_{spiral} = \rho c^2  - \frac{ \rho c^2}{(1 - \frac{R-{spiral}}{r_{spiral}})}[/math]
 
I have removed the [math]4 \pi[/math] from the equation which is really there for spherical systems. 
 
On Rotation Playing Dark Energy
 
 
[math]J = \frac{Gm^2}{v}[/math]
 
angular momentum
 
also
 
[math]J = m \omega R^2[/math]
 
Centrifugal force
 
[math]F = m \omega \times (\omega \times R) = m\omega^2 R[/math]
 
So centrifugal force is also in terms of angular momentum
 
[math]\frac{J}{R}\frac{d\theta}{dt} = m \omega^2 R[/math]
 
where [math]\omega = \frac{d\theta}{dt}[/math]
 
and the relationships also satisfy
 
[math]\omega \times R = \frac{dR}{dt} = \dot{R}[/math]
 
The Friedmann equation with rotation is
 
 
It was suggested by Arun (et al.) That rotation enters the Friedmann equation like
 
 
[math](\frac{\ddot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \omega^2[/math]
 
[1]. see references
 
It's proposed the correct derivation is not only longer, but in this form, should have a sign change for the triple cross product. Really what is in implied by the centrifugal term is a triple cross product. 
 
We will write the Friedmann equation in the following way:
 
[math]\ddot{R} = \frac{8 \pi G R}{3}\rho + \omega \times (\omega \times R)[/math]
 
Even though the centrifugal force is written with cross products [math]\omega \times (\omega \times R)[/math] it is not impossible to show it in a similar form by using the triple cross product rule 
 
[math]a \times (b \times c) = b(a \cdot c) - c(a \cdot b )[/math]
 
Using [math]\omega \cdot R = 0[/math] because of orthogonality we get
 
[math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \omega^2R[/math]
 
which justifies this form of writing it as well. If the last term is the centrifugal acceleration then the acceleration in the two frame is just, while retaining the cross product definition with positive sign,
 
[math]a_i \equiv \frac{d^2R}{dt^2}_i = (\frac{d^2R}{dt^2})_r + \omega^2 \times R[/math]
 
Expanding we get
 
[math]\ddot{R} = (\frac{d}{dt} + \omega \times)(\frac{dR}{dt} + \omega \times R)[/math]
 
[math]= \frac{d^2R}{dt^2} + \omega \times \frac{dR}{dt} + \frac{d\omega}{dt} \times R + \omega \times \frac{dR}{dt}[/math]
 
[math]= \frac{d^2R}{dt^2} + \omega \times \frac{dR}{dt} + \frac{d\omega}{dt} \times R + \omega \times ([\frac{dR}{dt}] + \omega \times R)[/math]
 
[math]= \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math]
 
Which is the four-component equation of motion which describes the pseudoforces. This gives us an equation of motion as
 
[math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math] 
 
Or simply as
 
[math]\ddot{R}_i = \frac{8 \pi G R}{3}\rho + a_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)[/math]
 
In the rotating frame we have
 
[math]\ddot{R}_r = \frac{8 \pi G R}{3}\rho + a_i - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math]
 
Since inertial systems are only a local approximation, the inertial frame of reference here may been seen to go to zero leaving us the general equation of motion for a universe
 
[math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math]
 
Using 
 
[math]a_{eul} = -\frac{d\omega}{dt} \times R[/math]
 
[math]a_{cor} = -2 \omega \times \frac{dR}{dt}[/math]
 
[math]a_{cent}= -\omega \times (\omega \times R)[/math]
 
we get
 
[math]\ddot{R} = \frac{8 \pi GR}{3}\rho + a_{eul} + a_{cor} + a_{cen}[/math]
 
 
 
 
 

 

 

Freidmann Equation Predicts Pre-Big Bang 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]
 
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.
 
[math]\frac{\mu}{k_BT} = \frac{\dot{p}}{nk_BT} - \frac{\rho + p}{nk_BT} \frac{\dot{T}}{T}[/math]
 
and related, to the famous equation of state
 
[math]n Tk_B\dot{S} = \dot{\rho} - (\rho + P)\frac{\dot{n}}{n}[/math]

 

[math] Tk_B\dot{S} + \frac{\dot{\rho}}{n} = (\frac{\rho + P}{n})\frac{\dot{n}}{n}[/math]

 

Then 

 

[math]\frac{\dot{n}}{n} = \frac{\dot{T}}{T}[/math]

 

gives simply

 

[math] Tk_B\dot{S} + \frac{\dot{\rho}}{n} = (\frac{\rho + P}{n})\frac{\dot{T}}{T}[/math]

 

Referring back again we have then a power equation in this:

 

[math]\frac{\dot{R}}{R}\frac{1}{n}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{6}[(\frac{\dot{\rho}}{n}) + 3\dot{P}(\frac{1}{n})] = \frac{8 \pi G}{6}Tk_B \dot{S}[/math]

 

We can see the equivalent terms clearly.

 

Writing them we have

 

[math]\frac{\dot{R}}{R}\frac{1}{n}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{6}[(\frac{\rho}{n}) + 3P(\frac{1}{n})]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}Tk_B \dot{S}[/math]

 

again based on fluid equation given, may not be complete, 

 

Same basic results however, for entropy we get

 

[math]\frac{8 \pi G}{6}[(\frac{\rho}{nk_BT}) + 3P(\frac{1}{nk_BT})]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}[/math]

 

which is good. The entropy remains constant and this is a definition of the thermodynamic law.

 
The Gibbs-Duhem relation is
 
[math]dp = (\rho+ p)\frac{dT}{T} + nTd(\frac{\mu}{T}) = (\rho+ p)d\log_{T} + nTd(\frac{\mu}{T})[/math]
 
Where [math]\mu[/math] is a chemical potential. From this one can get the chemical relationship in the Gibbs-Duhem relation is
 
[math]\frac{\mu}{k_BT} = \frac{\dot{p}}{nk_BT} - \frac{\rho + p}{nk_BT} \frac{\dot{T}}{T}[/math]
 
and also based on the equation of state, we can see the equation may be longer:
 
and so
 
[math] \frac{8 \pi G}{6}[\frac{\rho}{nk_BT} - (\frac{\rho}{nk_BT} + 3P(\frac{1}{nk_BT}))]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}[/math]
 
Replacing
 
[math][\frac{\rho}{nk_BT} - (\frac{\rho}{nk_BT} + 3P(\frac{1}{nk_BT}))]\frac{\dot{T}}{T}[/math]
 
There is no formal difference between this and
 
[math]\frac{\dot{\mu}}{k_BT} = (\frac{p}{nk_BT} - \frac{\rho + p}{nk_BT}) \frac{\dot{T}}{T}[/math]
 
we get
 
[math]\frac{8 \pi G}{6}[\frac{\mu}{k_BT}]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}[/math]
 
We can take small look at the physics of mixed gases in this model as well:
 
[math]\mu_B = -\frac{n_A}{n_B}\mu_A[/math]
 
[math]\frac{8 \pi G}{6}[\frac{\mu}{k_BT}]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}[/math]
 

[math]= -\frac{8 \pi G}{6}[\frac{n_A}{n_B}\frac{\dot{\mu}_A}{k_BT}][/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{1}{n}(\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.

 

 

 

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, prof. 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. )

 

 

At some point, a pre-big bang phase underwent some kind of transition - the transition is something related to topology which brought about some kind of thermodynamic increase in the system - from a super cool liquid into a vapor. At some point either between these processes, or in one of them, gravitational binding has to occur to create nucleons. I am looking into Nobel Prize winner Abdus Salam's work into strong gravity which he believed may be something related to the strong force which binds microsystems together.

 

http://dx.doi.org.sc...217732393000325

 

Abdus links the interactions to a massive spin 2 graviton in his earliest works. In that, its very precise the kinds of particles he is talking about but I was interested in it because it does at least show in theory, one can construct a theory which treats gravity as being scale dependent. I doubt though gravity is a graviton phenomenon, but the graviton physics may tell us something too deep about the theory, really, gravity is a pseudoforce, which means it is exactly the same as the centrifugal force. They are not real forces and so cannot be quantized as a field theory.

 

So I don't think after looking through there is any salvageable information for the model I am working with but it is an interesting bit of history because, it was around the same time Motz suggested the gravitational binding energy for nucleons for an early universe. I do not believe the strong force was properly understood then, with the Top Quark only being just identified and Motz (I think) erroneously presuming then the Top Quark was the relic Uniton.

 

Even though I say gravity cannot be quantized, the solution to confinement I think lies in some revelation concerning curvature. As was pointed out in the paper, it was discovered by Ne'emen and Sijacki that P^-4 propagators indicates that confinement arises from the curvature of spacetime. Interestingly, Motz wanted the early pre-big bang phase to have gravity dwarf electromagnetism by 137 factors. Interestingly, in the femtometer range the strong force matches this approximation, making it 137 times as strong as electromagnetism.  It seems reasonable then, that for Motz early universe if gravity plays the role for nucleo-binding energies, would have to occur around the same range 10^-15m, This is well below the highest threshold of the tunneling barrier, which occurs on the nanometer range 10^-9.

 
 
 
 
(notes) - in this day and age, not all models invoke pointlike scales as the origin, but it seems likely this may be very close to the case at hand. 
 
 
extra note, in this case, the magnitude of gravity is about 10^40 times weaker than electromagnetism. Gravity could grow like [math]G \cdot 10^{40}[/math] whereas electromagnetism will tend to zero [math]e \rightarrow 0[/math]
 
REF:
 
 
 
 
 
 
 
(Weak Equivalence excerpt)
 
The phase space of a black hole is just [math]\hbar^3[/math]. That is, the interior of a black hole is just one quantum of phase space filled with one photon. (Sivaram and Arum). 
 
The study of extremal rotating and charged black holes obtain unique values that are expressed fundamentally as constants for the ratio's of angular momentum to entropy and charge to entropy - Those specific relations have analogues in condensed matter physics.
 
The dimensionless constants surrounding the model allows for the phase space, a specific set of coupling analogues to the electromagnetic coupling, or analogously, the strong interaction, where it has been notd the strong interaction g is10^2 times stronger than the electromagnetism that binds nuclei together. (A similar mechanism was proposed for the super cool condensed pre-big bang phase of the universe. 
 
Sivaram and Arun have noted that those relationships are further enhanced by the Von-Klitzig constant and/or the Josephson constant which are used in superconductor physics - black holes are indicated to be diamagnetic, excluding flux just like a superconductor. Truth be told we do not know what the inside of a black hole is like, we know it stores its temperature on the horizon, presumably with the rest of the black holes information. The question is, is the interior of a black hole actually the coolest systems in existence? 
 
If so, the early universe probably did have a structure similar to a black hole, I wonder if the interior of a black hole is the closest analogue of the super cool, liquid condensed pre-big bang all-matter phase? If it is true, then it has already been explained, the pre-big bang phase undergoes a transition into a vapor radiation - this would be like a black hole evaporating - and it has been shown at least from the physics of black holes, Arun extended the weak equivalence for a black hole and I strengthened his arguments when I read his work:
 
The radius of a black hole is found directly proportional to its mas (R ~ M). The density of a black hole is given by its mass divided by its volume (ρ ~ M/V) and since the volume is proportional to the radius of the black hole to the power of three (V ~ R³) then the density of a black hole is inversely proportional to its mass radius by the second power (ρ ~ M²).
 
What does all this mean?
 
It means that if a black hole has a large enough mass then it does not appear to be very dense, which is more or less the description of our own vacuum: it has a lot of matter, around 10^80 atoms in spacetime alone, but doesn't appear very dense at all!
 
But what does this have to do with vanishing curvature? To make the temperature of a black hole go down, you need to add matter to the system. Using the following approximation we have
 
m → ∞
 
Then the temperature goes to zero 
 
T → 0
 
And for a black hole with infinite mass, the curvature tends to zero as well!
 
K → 0 
 
 
(This is more or less the description of a vacuum!)
 
 
It seems that this model is entirely consistent with the pre-big bang super cool fluid. The beginning of big bang was a hot configuration, but our pre-big bang state indicates zero temperature. If the spin of a system like a black hole varies with mass Then the universe may also have similar implications - In that kind of model though, the pre-big bang phase would not have had a lot of matter. Then through some process of non-conservation and additional energy added to a universe as it expands then we can see all the same limits above for black holes could also apply to the universe depending on scale. 
 
We can surmise this may mean that our current universe is not completely flat as suggested maybe by [math]K = 0[/math] where [math]K[/math] is the Gaussian curvature.Not only this, we know there is not an infinite amount of mass in the universe, but there is certainly a lot.So these limits should be taken as just very large numbers or very small numbers that  examples physics in extreme and complimentary/relativistic cases.

 

 

The Antimatter Problem

 

The chirality (spin) of a universe would give us reason to expect matter dominance over antimatter. 


Edited by Dubbelosix, 30 June 2017 - 01:03 AM.


#2 OceanBreeze

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Posted 05 April 2017 - 01:48 AM

 

 
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. 
 
 
 

 

 

Well, it isn’t quite that simple. It is true that the further away a galaxy is, the faster it is receding. This is a simple consequence of Hubble law; that galaxies recess with a velocity proportional to their distance, and provides good evidence that the universe is expanding, but this has nothing to say about any possible acceleration of the expansion. To determine if the expansion is either speeding up or slowing down, scientists need to use a much more complicated method than simply observing redshift.

 

 

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.

 

 

No, I don’t see any such interpretational problem, except the one that you are making by assuming that redshift alone is the basis for thinking the expansion is accelerating.

As I understand it, scientists base their conclusion that the universe is expanding on much more complex calculations than the comparatively simple observation that the further away a galaxy is, the faster it is receding. The expansion theory is based determining the “true brightness” of a certain class of supernovae (standard candles) which is then used to make a determination of its present distance (there are several other factors used to determine the present distance).
 

Then, by comparing the apparent brightness (which IS a function of redshift) to the true brightness, a calculation can be made that tells us by what factor the universe has expanded since the supernova exploded.

 

The thinking is; If the expansion of the universe is accelerating then the expansion was slower in the past, and thus the time required to expand by a given factor is longer, and the distance to the supernovae at present is larger.

 

Conversely, if the expansion is decelerating, it was faster in the past and the distance is smaller.

 

The method seems somewhat convoluted and imperfect (to me at least) but nevertheless most scientists do agree on the conclusion that the expansion is accelerating.

 

I am a bit surprised that you can be so meticulous in developing your ideas*  but in all that detail you somehow managed to overlook something this obvious that possibly contradicts you.

 

*if indeed they are yours. Several others (most notably Gödel) have theorized about a rotating universe. How does your theory differ from his? But, it is an interesting idea. Have you published this idea anywhere else, or just here?



#3 OceanBreeze

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Posted 07 April 2017 - 05:35 AM

I received a question today from one of my physicist friends, they asked,

 

 

''If the universe really is rotating, why can't we see an axis''

 

 

I explained that Hoyle and Narlikar have shown that rotation exponentially decays since the birth of a universe (hence why dark flow could be the remnant of primordial rotation), but dark flow is so very slow today that a detection of an axis is extremely unlikely, compared to when the universe was much younger and spinning much faster!

 

 

The lack of observational evidence for a rotating universe, including the non-detection of an axis of rotation, may not be such a big problem for your model. There does exist the so called “axis of evil” whereby the rotation axes of most galaxies seem to line up in a preferred orientation. That could be evidential support for a rotating universe.

 

However, what may be the biggest objection to your model is that it does not accommodate the accelerated expansion of the universe, so it is not in accord with the current cosmology and observational evidence.

 

What I am wondering is; why your model does not give a result that is in accord with that obtained from the standard FRW model as in the Einstein equation with the “artificially added” cosmological constant? I guess what I am saying is, is it possible for the angular frequency of your rotating universe to exactly reproduce the arbitrary cosmological constant that makes the hypothesis of dark energy necessary in the first place?

 

In fact, what is your calculated angular frequency? It may be somewhere in your math but I missed it.



#4 OceanBreeze

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Posted 07 April 2017 - 09:50 AM

Lack of observational evidence? I beg your pardon, who told you that?

 

There is no observational evidence the universe is spinning. Experiments involving the CMB tell us that.

 

 

"The vast majority of calculations made about our universe start with this assumption: that the universe is broadly the same, whatever your position and in whichever direction you look.

If, however, the universe was stretching preferentially in one direction, or spinning about an axis in a similar way to the Earth rotating, this fundamental assumption, and all the calculations that hinge on it, would be wrong.

Now, scientists from University College London and Imperial College London have put this assumption through its most stringent test yet and found only a 1 in 121,000 chance that the universe is not the same in all directions."



#5 OceanBreeze

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Posted 07 April 2017 - 09:51 AM

[math]\frac{\Lambda c^2}{3} = \omega \rightarrow \frac{3\omega^2}{c^2} = 10^{-56}cm^{-2}[/math]

 

This was derived by Arun et al. So its not entirely impossible to reformulate the constant of integration in Einsteins equations. 

 

What would that be in rad/s? Can I assume you are using the Hubble radius, or something else?



#6 OceanBreeze

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Posted 07 April 2017 - 11:17 AM

Can I just say, Einstein's cosmological constant was not ''artificially added'' it was a natural component that arose from the equations, he omitted to smudge his equations for a static universe.

.

 

Obviously, I am not as well versed in the mathematics as you are, but my understanding is the origin of the cosmological constant was a mathematical fix that Einstein added to the solution of his field equations when his calculation indicated the universe would collapse without it. In that sense, it didn’t arise naturally from the mathematics but had to be artificially added in order to make the math match the “supposed” reality of a static universe. Of course, now we know the universe is not static but there are other reasons for keeping the cosmological constant. It may well be that it does arise naturally in present day calculations that take into account the expanding universe but I was referring to the origin.



#7 OceanBreeze

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Posted 07 April 2017 - 11:26 AM

If an experiment has not concluded dark flow in their data, they have consistently failed. Dark flow is a real phenomenon. Secondly, the fact spiral galaxies have a preferred chirality (do you understand what that means?) then this is in fact direct evidence that the galaxies where influenced in an early rotating fluid (universe). I have seen this recent data, they absolutely, no exceptions, are wrong. If the data does't even mention previous evidence, first suggests, then thought to be refuted, only to be verified again, then the report is bias if it does not even mentions these things. 

 

You know the experiments "absolutely, no exceptions, are wrong" because they do not agree with your calculations, but they do agree with mainstream cosmology.

 

That doesn't sound good.



#8 OceanBreeze

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Posted 07 April 2017 - 11:29 AM

Yes its always the hubble radius.

 

[math]\omega[/math] has units of rad/s. 

 

yes I know omega has units of rad/s. I was asking what your calculated rotation was in rad/s, you gave it in cm/s. Ok using the Hubble radius I can convert it.



#9 OceanBreeze

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Posted 07 April 2017 - 11:31 AM

Mainstream? 

 

 

I don't know any credible scientist who has challenged my claim about whether dark flow is a real phenomenon. I don't think you understand... dark flow has the data, it is mainstream cosmology, I am telling you, the report there is bias. 

 

How can any credible scientist challenge what you have not even published?



#10 OceanBreeze

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Posted 07 April 2017 - 11:34 AM

Because I have shown it to credible scientists, published or not.

 

 

I see.

 

So, is that you, Craig? :nahnahbooboo: 


Edited by OceanBreeze, 08 April 2017 - 06:00 PM.


#11 exchemist

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Posted 12 June 2017 - 08:07 AM

If I have more insights how to make my paper ''better,'' it will be updated in the OP, otherwise I want to give further posts that will share interesting but not entirely related physics to the course I have taken in my work - reserving these posts than for these physics, we begin with:

 

 

 

The extremely simple solution we arrived at for a Friedmann equation in terms of the chemical potential was

 

 

[math]- \frac{8 \pi G}{6}[\frac{n_A}{n_B}\frac{\mu_A}{k_BT}]\frac{\dot{T}}{T} = k\dot{S}[/math]

 

 

For a constant entropy. The Gibbs energy is actually related to a mixture as

 

[math]G = \sum_i X_i \mu_i[/math]

 

and in some texts 

 

[math]\bar{G} = \mu[/math]

 

and the mixing can be given as

 

[math]\Delta G = G_{2} -G_{1} = n\mathbf{R}T(x_A \ln x_A + x_B \ln x_{B})[/math]

 

Where we have used the gas constant [math]\mathbf{R}[/math] which is formally the same as a Boltzmann constant.

 

 

 

http://staff.um.edu....2/mixtures.html

Can you explain what relevance chemical potential has to cosmology, please?

 

 I would have thought that in general, the atoms and (occasional) molecules in the cosmos are too far apart to be able to exchange energy and come to thermal equilibrium. However the definitions of both Gibbs Free Energy and chemical potential depend on temperature being defined for a system. If you have systems that are not at thermal equilibrium, temperature has no meaning, because you do not have a Boltzmann distribution among energy states. 

 

It's all very baffling. 


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#12 exchemist

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Posted 12 June 2017 - 10:30 AM

There is significance for specific phase changes of an early universe. 

 

For instance, one can hypothesize a pre-big bang phase which acted like a condensed liquid. 

 

[snip]

Thanks, so you are talking about a hypothetical state of matter in the universe before time started.

 

?! 



#13 exchemist

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Posted 12 June 2017 - 10:31 AM

Here is something on phase transitions 

 

https://www.uam.es/p...er/tema_III.pdf

No need for all that, thanks. I have a degree in chemistry. :)


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#14 exchemist

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Posted 12 June 2017 - 10:32 AM

The chemical potential is an example of intensive property to systems. 

 

In fact, according to wiki

 

''The distinction between intensive and extensive properties has some theoretical uses. For example, in thermodynamics, according to the state postulate, a sufficiently simple system consisting of a single substance requires only two independent intensive variables to fully specify the system's entire state. Other intensive properties are derived from those two variables.''

Relevance? 



#15 exchemist

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Posted 12 June 2017 - 10:33 AM

In the most developed idea's yes. That was one of the possible consequences I first read about. I did not realize my Gibbs-Duhem-Friedmann equation had any significance before I read Motz et al. Their paper suggested a similar relationship to the Gibbs-Duhem thermodynamics, which was the first paper I am aware of to introduce a phase transition for a universe.

But it is obviously nonsensical to hypothesise about anything before the start of time itself. 



#16 exchemist

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Posted 12 June 2017 - 10:40 AM

To me, that is nonsensical. 

What is?



#17 exchemist

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Posted 12 June 2017 - 10:42 AM

Not if it wasn't the start of time, or space.

 

Those theories depend on a universe that nucleates into existence without a prior reason. 

Can you direct me to a big bang model in which time and space do not start at the instant of the big bang itself? I thought that feature was common to all models of it. 


Edited by exchemist, 12 June 2017 - 10:43 AM.