Pre Big Bang State

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#1 Dubbelosix

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Posted 03 January 2019 - 08:43 AM

An article came to my attention, it speaks of a big bang as simply a phase:

https://www.space.co...rrwL34rBhbfbdYE

This has parallel to a theory I suggested, except it involves no big bounce.

#2 Flummoxed

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Posted 03 January 2019 - 11:51 AM

An article came to my attention, it speaks of a big bang as simply a phase:

https://www.space.co...rrwL34rBhbfbdYE

This has parallel to a theory I suggested, except it involves no big bounce.

I tend to favour Verlindes and others work, which are significantly more developed than yours, suggesting gravity is fundamentally an emergent property of cosmic entropy. This emergent property of entropy of gravity, if true, would also indicate that gravitational inertia, dark matter and dark energy are simply manifestations of cosmic entropy. Which tend to make your argument pointless.

You mention you have an antimatter problem in your quora link, what is it?

#3 Dubbelosix

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Posted 03 January 2019 - 04:05 PM

It only tends to make my argument pointless, if you are personally bias for another theory: Keep in mind though this is just a subjective belief of your own.

I don't like Verlinde's theory simply because it is based on holographic physics; and I am not one to think we live in a holographic universe. But each to their own.

#4 Dubbelosix

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Posted 03 January 2019 - 04:07 PM

The antimatter problem is well-known, it asks why there is an excess of matter over antimatter. In the rotating model of the universe, the antimatter problem goes away because the universe would have a primordial preferred direction of spin (Lorentz violating theory) and so explains why it prefers one chirality over the other.

#5 Flummoxed

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Posted 03 January 2019 - 04:21 PM

The antimatter problem is well-known, it asks why there is an excess of matter over antimatter. In the rotating model of the universe, the antimatter problem goes away because the universe would have a primordial preferred direction of spin (Lorentz violating theory) and so explains why it prefers one chirality over the other.

You misunderstand entropic gravity.

The antimatter problem is just another outdated theory, but each to there own

#6 Dubbelosix

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Posted 03 January 2019 - 04:48 PM

The antimatter problem [is not] a theory but a matter of experimental and observational fact. It is well-known that there is an excess of matter over antimatter in the universe...

... look, I can handle the other two idiots because they pretend to know more physics than they really know, and I thought I was getting along with you, but you increasingly come across as just as disingenuous about your own knowledge,.

#7 Dubbelosix

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Posted 03 January 2019 - 04:50 PM

I have never in my life, had a debate on this issue, with actual scientists, for anyone of them to state this was an ''outdated theory.'' I haven't heard so much rubbish.

#8 Dubbelosix

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Posted 03 January 2019 - 04:51 PM

#9 Flummoxed

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Posted 04 January 2019 - 09:20 AM

"

Neither the standard model of particle physics, nor the theory of general relativity provides a known explanation for why this should be so, and it is a natural assumption that the universe be neutral with all conserved charges.[3] The Big Bangshould have produced equal amounts of matter and antimatter. Since this does not seem to have been the case, it is likely some physical laws must have acted differently or did not exist for matter and antimatter. Several competing hypotheses exist to explain the imbalance of matter and antimatter that resulted in baryogenesis. However, there is as of yet no consensus theory to explain the phenomenon. As remarked in a 2012 research paper, "The origin of matter remains one of the great mysteries in physics"

"

This statement alone in your wiki link suggests something is wrong with current theories, the mystery points out no one knows yet.

I have never in my life, had a debate on this issue, with actual scientists, for anyone of them to state this was an ''outdated theory.'' I haven't heard so much rubbish.

Sorry I dont follow religious belief in theories that dont work

The antimatter problem [is not] a theory but a matter of experimental and observational fact. It is well-known that there is an excess of matter over antimatter in the universe...

I thought I was getting along with you, but you increasingly come across as just as disingenuous about your own knowledge,.

The antimatter problem suggests the theories are wrong and outdated.

Disingenuous, interesting observation

#10 Dubbelosix

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Posted 04 January 2019 - 01:19 PM

"

Neither the standard model of particle physics, nor the theory of general relativity provides a known explanation for why this should be so, and it is a natural assumption that the universe be neutral with all conserved charges.[3] The Big Bangshould have produced equal amounts of matter and antimatter. Since this does not seem to have been the case, it is likely some physical laws must have acted differently or did not exist for matter and antimatter. Several competing hypotheses exist to explain the imbalance of matter and antimatter that resulted in baryogenesis. However, there is as of yet no consensus theory to explain the phenomenon. As remarked in a 2012 research paper, "The origin of matter remains one of the great mysteries in physics"

"

''This statement alone in your wiki link suggests something is wrong with current theories, the mystery points out no one knows yet. ''

So what is your point? This statement is such a red-herring, yesterday you were claiming that [the antimatter problem was an outdated theory] which, according to your attempt of a backtrack, had nothing specifically to do with the statement from wiki saying maybe something is wrong with current theory.

#11 Flummoxed

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Posted 05 January 2019 - 07:30 AM

''This statement alone in your wiki link suggests something is wrong with current theories, the mystery points out no one knows yet. ''

I will give you 100% for that observation

I have not backtracked on anything, all I have done is ask questions.

As has been pointed out, if observations do not match theories, then something is seriously wrong with the theories.

#12 Dubbelosix

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Posted 05 January 2019 - 11:44 AM

It wasn't my observation, if you cannot recall, you said that. That is why it is in quotation.

#13 Dubbelosix

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Posted 14 January 2019 - 05:03 AM

I noticed you said if things do not match observation they are wrong, and I totally agree! Maybe you could do me a favor and read through my article again ---- we do not deviate from a model which had a hot history, the question is where the hot history arose from, since a hot big bang with low entropy violates the third law of thermodynamics. To make sense out of this problem, I argue we MUST consider a cold pre-big bang state which went under some phase transition from a dense, all-matter (cold) region into a radiation vapour phase (of radiation). The fact is, this model still fits all observational evidence, and arguably, makes more sense about big bang not being the true origin phase.

#14 Dubbelosix

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Posted 14 January 2019 - 06:24 AM

I came back to this because I wanted to draw up some more math, here specifically the update later on blog.

$(\frac{\ddot{R}}{R} + \frac{kc^2}{a})\Theta = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma)$

Recall:

The fluid expansion $\Theta = \frac{\dot{R}}{R}$ for a homogeneous sphere. The particle production rate is (in a short-non-conserved and irreversible phase transition):

$\Gamma = \frac{\dot{n}}{n}$

Also recall that there are a series of equivalences (that gauge the rate of change in a number of factors within standard physics):

$\Theta^{\mu;}_{\mu;} = \frac{\dot{n}}{n} = \frac{\dot{a}}{a} = \frac{\dot{T}}{T} = \frac{\dot{R}}{R}$

$\mu; = n; = R; = a; = T;$

There are unique ways we can view the Friedmann equation that had not properly registered with me until recently. That is, the third derivative does not [need] to be defined by a the rate of radius change, but instead measured by derivatives in temperature anistropies - when fashioning this, we will change the Friedmann equation into a more recognizable form, but is totally equivalent in a dimensional sense, but the interpretation may be different, (something we will come back to later) - and so the equation we form is:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]\frac{\dot{R}}{R}$

The motivation behind this is totally identical to the situation I wrote for anistropies in the Friedmann equation except [it] featured on the right hand side, quick memory boost of what that equation looked like, this time with a zero point energy term, in the form of the Sakharov gravitational correction for longer lived virtual particles leading to irreversible pressure dynamics in the long-run:

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{6}(\rho + \frac{3P}{c^2} + \hbar c \int k^3 dk)\frac{\dot{T}}{T}$

In the case of phase transitions, a change in a feature of a physical system, topological or not, often involves the absorption or emission of energy from the system, resulting in a transition of that system to another state. We where clear about what a third derivative in time ultimately means, and it expresses the non-conserved form of a Friedmann equation.

In statistical mechanics a scale invariance is a feature of phase transitions. The key observation is that near a phase transition or critical point, fluctuations occur at all length scales, and thus one should look for an explicitly scale-invariant theory to describe the phenomena (the only feature in the model which remains invariant regardless of scale is the zero point field and fluctuations! Scale invariance is a feature of objects or laws that do not change if scales of length, energy, or other variables - this brings into question also how to further interpret the pre big bang phase transition, in context of scale invariance. Daunting? Not too much, remember the theory I present explains the phase transition from a liquid to radiation vapour, then back into the current condensed and dominated region but diluted over great distances (after all, all visible matter covers only about 1% of all spacetime) $\rightarrow$ we take the following from wiki:

A phenomenon known as universality is seen in a large variety of physical systems. It expresses the idea that different microscopic physics can give rise to the same scaling behaviour at a phase transition. A canonical example of universality involves the following two systems:

As stated, the third derivative implies non-conservation, the derivatives do not tell us when the non-conservation happened, but I suspect it happened during the collapse of the first state of the universe - a degenerate condensed matter region of cold thermodynamics and low entropy. Let's quickly go back to the equation that had the third derivatives on the anistropy terms:

$\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]\frac{\dot{R}}{R}$

While fluctuations occur at all lengths, temperature is by contrast, scale-invariant: We measure the change in the radius of a universe on the left hand side this time. The key feature to notice here by writing it in this way, we no longer care about third derivatives in the radius, the universe could be any scale and undergo through any phase transition and would still alter the cosmological radius in exactly the same way. The pre-big bang state, says the universe would have been small, not necessarily under Planck scales. But large enough that it is not on the scale of atoms. And since you cannot confine a particle to scales smaller than the wave length, this should by principle [always] forbid these types of singularities we often think about from arising. The third derivative over the temperature then, may be describing the three era's that our model leads already hints at:

1. A cold dominated region undergoing a phase $\frac{\dot{T}_1}{T}$

2. The cold region degenerated into a radiation vapor phase indicating higher powers $\frac{\dot{T_2}}{T}$

3. Then a third derivative, perhaps describing the re-cooling of a universe back into a matter-dominated phase, the current cosmological era $\frac{\dot{T}_3}{T}$

But these are only idea's to chew on for now.

Edited by Dubbelosix, 14 January 2019 - 10:22 AM.

#15 Dubbelosix

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Posted 14 January 2019 - 11:45 AM

Now let us go to a simplified version of the Friedmann equation, but retaining the new definition using the temperature terms for anistropies:

$\frac{\dot{T}}{T} (\frac{\ddot{T}}{T} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\dot{\rho}$

It is deceiving in the math of the continuity of $\dot{\rho}$ since it defined as the density plus another term, a correction term if you like, in the form of a pressure term, which is why more than often you encounter it in a similar form to:

$\frac{\dot{T}}{T} (\frac{\ddot{T}}{T} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\rho + 3P)\frac{\dot{R}}{R}$

But what do I mean that it is deceiving concerning fluid continuity? Well a time derivative on the density does NOT explicitly tell us anything about a non-conservation phase, in fact, derivatives on the density or pressure will yield an over-all conservation:

$\frac{\partial \rho}{\partial t} + \nabla \cdot (PV) = 0$

The fluid continuity however, is more appropriately written, with simplification further under natural units of $8 \pi G = c = 1$:

$\dot{\rho} = \frac{\dot{T}}{T}(\rho + 3P)= \frac{\dot{R}}{R}(\rho + 3P)= (\rho + 3P)\frac{\dot{n}}{n}$

In terms of first the temperature anistropies, then the radius, but the final term is the proposed breaker, since it measures a non-conserved form of particle production and in such a case, it is unavoidable to see this in terms of a non-conserved form drifting from continuity simply because it implies a non-zero, non-conserved quantity.

But we know for sure from the left hand side we must be dealing with non-conservation in this respect, again, because of the third derivative in time. This would obviously imply ~

$\dot{\rho} = \frac{\dot{T}}{T}(\rho + 3P)= \frac{\dot{R}}{R}(\rho + 3P)= \frac{\dot{n}}{n}(\rho + 3P) \ne 0$

I decided to take a look at third derivatives of temperature to accidently come across  Playing around with the opening equation, we will add some extra physics in to help us dimensionally-navigate our way around the next investigation, so we write:

$\frac{\dot{R}}{R} (\frac{k_B\ddot{T}}{k_BT} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\dot{\rho}$

So this is ok too, we simply expressed the temperature in terms of its equivalent energy using the Boltzmann constant which is, also for a quick fact, the same dimensions as a gas constant. We have also, the rate of change of the radius as a coefficient this time on the left hand side. Again, because of the third derivative, we must assume the derivative on the right hand side is involved with a non-conserved form of the density. When we plug in, the discontinuous equation we simply have the correction pressure term:

$\frac{\dot{R}}{R} (\frac{k_B\ddot{T}}{k_BT} + \frac{kc^2}{a}) = (\rho + 3P)\frac{\dot{n}}{n}$

This... unique representation of the Friedmann equatioon allows us to involve factors of radius, temperature and particle density creation. It doesn't matter if we write it out this way to propose a theoretical route, again, all that really matters is that third derivative is present - in fact, you can shuffle those terms as you wish... Here is a good example with us taking the particle production term on the left hand side, the physics, though, can lead to different forms of a Friedmann equation. If we take

$n = \frac{N}{V}$

as the number density then we can rewrite our equation, in natural units of $8 \pi G = c = 1$

$\frac{\dot{n}}{n} (\frac{k_B\ddot{T}}{k_BT} + \frac{kc^2}{a}) = (\rho + 3P)\frac{\dot{R}}{R}$

For atoms, there is a very good approximation for the mass and the particle number,

$n = \frac{PN_A}{k_BT} = \frac{PN_A}{\mathbf{R}T}$

as mentioned before, the Boltzmann constant $k_B$ is equal in regards to the gas constant $\mathbf{R}$ and Avagadro's constant is $N_A$. Distribution of the temperature is

$n k_B T = PN_A = \frac{PN_A}{\mathbf{R}T}k_B T$

and conventionally related as:

$\mathbf{R} = k_B N$

This should imply a correction to the equation in presence of a dimensionless particle number attached as a coefficient to the temperature;

$n k_B T = PN_A = \frac{PN_A}{\mathbf{R}T}N k_B T$

Integration of the volume element will easily yeild:

$\frac{N}{2}mv^2 = \int\ \frac{3}{2}N k_B T dV = \int\ PN_A\ dV = \int \frac{PN_A}{\mathbf{R}T}N k_B T\ dV$

or even as

$\frac{1}{2}mv^2 = \int\ \frac{3}{2} k_B T dV = \int\ P\ (\frac{N_A}{N})\ dV = \int \frac{PN_A}{\mathbf{R}T} k_B T\ dV$

and if

$\frac{k_BT}{\mathbf{R}T} \approx 1$ then the last term can also be written as

$\int \frac{PN_A}{\mathbf{R}T} Nk_B T\ dV = mN_AN$

and further it is possible to construct:

$\frac{k_B}{\mathbf{R}}\int\ PN_A\ (\frac{\dot{T}}{T})\ dV = mN_A\frac{\dot{N}}{N}$

Edited by Dubbelosix, 15 January 2019 - 03:42 AM.

#16 Dubbelosix

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Posted 15 January 2019 - 04:46 AM

Thermodynamic energy can be represented as continuity equation:

$n Tk_B\dot{S} = \dot{\rho} - (\rho + P)\frac{\dot{T}}{T} =\dot{\rho} - (\rho + P)\frac{\dot{R}}{R} = \dot{\rho} - (\rho + P)\frac{\dot{n}}{n}$

By introducing the heat per unit particle

$d\bar{q} = \frac{dQ}{dN}$

the above component reduces to a Gibbs expression from the Gibbs equation.

$\frac{\mu}{T} = \frac{\dot{p}}{nT} - \frac{\rho + p}{nT} \frac{\dot{T}}{T}$

You can form a similar equation from the thermodynamic law:

$dE = dQ - p dV$

$TdS = dq = d(\frac{\rho}{n}) + pd(\frac{1}{n})$

$Tk_B\dot{S} + \frac{\dot{\rho}}{n} = (\frac{\rho + P}{n})\frac{\dot{T}}{T}$

We can compare these results with the ordinary Gibbs-Duhem equation

$dp = (\rho+ p)\frac{dT}{T} + nTd(\frac{\mu}{T}) = (\rho+ p)d\log_{T} + nTd(\frac{\mu}{T})$

A relationship (to construct a Raychauduri equation) is of the form:

$\dot{H} + H^2 \equiv \frac{\ddot{R}}{R} = \frac{\ddot{T}}{T} = \frac{\ddot{n}}{n}$

The point behind these identities is that it allows us to write the Friedmann equation in terms of the chemical potential and other ways. For instance, the Friedmann equation a post back had been considered:

$\frac{\dot{T}}{T} (\frac{\ddot{T}}{T} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\dot{\rho}$

The left hand side can be constructed then like:

$\frac{\dot{T}}{T} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\dot{\rho}$

The right hand side can then be constructed as:

$\frac{\dot{T}}{T} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\rho} - (\rho + P)\frac{\dot{R}}{R}) = \frac{8 \pi G}{3}(\dot{\rho} - (\rho + P)\frac{\dot{n}}{n})$

If we think about it in terms of the particle creation gauge $\frac{\dot{n}}{n}$ then it should not be difficult to see the Friedmann equation when it substitutes the heat per particle becomes a Gibbs-like equation:

$\frac{\dot{T}}{T} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\rho} - (\frac{\rho}{n} + P(\frac{1}{n}))\dot{n}) = \kappa T\dot{S}$

Here we have given entropy dimensions of the Boltzmann constant, and as we can see, the Friedmann equation is now concerned with the thermodynamics of the expanding universe. There is still a particle creation going on, and usually we reserve that for an irreversible pressure term related to that creation. In fact one such term in the equation looked so familiar to the entropy production which can reserve parts of an equation for irreversible thermodynamic processes. For instance, such an entropy looks like:

$\frac{dS}{dt} = \sum_k \frac{\dot{Q}_k}{T_K} + \sum_k \dot{S}_k + \sum_k \dot{S}_{ik}$

and it is the subscript notation of $\dot{S}_{ik}$ which represents entropy production, not only due to internal processes, but it refers to an entropy with irreversible processes.

So, not only can we write the Friedmann equation as

$\frac{\dot{T}}{T} (\dot{H} + H^2 + \frac{kc^2}{a}) = -\sum_k\ \frac{8 \pi G}{3}(\dot{\rho}_k - (\frac{\rho_k}{n} + P_{ik}(\frac{1}{n}))\dot{n})$

But it also should be possible to write it as:

$\frac{\dot{T}}{T} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( Q_k + T S_k + T S_{ik})\dot{n}$

$= \frac{8 \pi G}{3}\ \sum_k\ ( \mathbf{Q}_k + T \mathbf{S}_k + T \mathbf{S}_{ik})\dot{N}$

or equivalently:

$\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( Q_k + T S_k + T S_{ik})\dot{n}$

Edited by Dubbelosix, 15 January 2019 - 08:48 AM.

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