Jump to content


Photo
- - - - -

Pre-Big Bang Phase And The Implications For Physics


  • Please log in to reply
46 replies to this topic

#1 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 04:43 AM

Since the creation of my idea concerning the non-conserved phase of the early universe, scientists are starting to realize, only just in recent articles that there is now a growing mathematical evidence that the universe most likely went through a phase (ie. The universe did not happen spontaneously, it appears from a series of complicated phase transitions that (may be) cyclic in nature).

 

Certainly, my own model could satisfy that, which is why I never disliked Penroses idea's, I just disagreed with his approach. I do find it interesting though, that before this became fashionable, I was writing on these types of phase transitions so that we could

 

1. Explain why the notion of energy cannot be created nor destroyed as only a classical concept.

 

2. The universe is not just a classical system, obeying the characteristic laws we see on an everyday basis. It has a more fundamental connection to quantum mechanics, which makes the classical laws look as though they are approximations to a much deeper understanding of non-conservation, leading to a diabatic universe.

 

In a conversation on facebook, a poster wrote a comment concerning a quantum experiment which (as most headlines of scipop articles tend to do), exaggerated what the scientists found. I refer, to the Wheeler Delayed Choice and including that which is an age-old question for modern physics, ''does reality exist when not looking at it?''

 

First of all, the conversation on the creation of energy as a quintessential element to a unification theory -  the conversation went like this, word by word:

 

Gareth Meredith: Its a very good question, not enough people understand it. In fact, the conservation of energy is a classical concept, quantum field theory violates the energy conservation all the time, just for very short periods. This is how energy came into existence, it had to happen as a short non-conserved phase of an early universe - the real question was if anything existed before this non-conserved phase? To have such a concept, at one point in the universes history, it can easily be argued that energy is created and destroyed all the time, in fact, in quantum field theory is all about non-conservation, it just happens for a very short period of time.

Joni Joni Energy is neither created nor destroyed?

Gareth Meredith Again, this is not true. That statement above is a classical concept.

Joni Joni You know I am a student and learned that energy neither created nor destroyed

Gareth Meredith: I didn't know you are a student - but what you have done is learn a little bit about why classical physics is an approximation, whilst the universe itself is a quantum system. We are quantum systems, and quantum mechanics does allow the creation of energy and can destroy it just as simply. We use what is called creation and annihilation operators, which is the ''quantum language'' of something far more significant than the classical ham that people tend to parrot.

 

Then another poster talking about the delayed choice experiment, noting that the scipop article seems to be saying we are shaping the universe by looking into the distant past:

Graham Croucher: Gareth Meredith Yes, I have sympathy with that view. The apparent subjectivity we impose on something simply by looking at it, may be due to some fault in our reasoning.


Gareth Meredith: Well, it goes back to archaic belief systems like, the sun and all heavenly bodies rotated round the Earth, and for some reason, scientists have still not quite got past this whole ''we are important so that everything can exist.'' Everything exists because we are important, now that's a different but much more interesting concept.
But it is still probably a very essential question, but we are not special in the sense of observers. Particles where interacting long before any conscious entity existed to notice such things. Unless you believe in a god. The retrocausal effect of creating something now by observing the past is related to the creator of that theory, the Wheeler delayed experiment.

 

And I continued:

 

Gareth Meredith: But whether it means there is no objective reality, that's just a load of rubbish, for if there was no objective reality, then there would be nothing to observe.

 

 

I am personally starting to get more faith in science, but it first has to admit it's mistakes, including people who think they know a lot but actually do not and in a way these people are holding physics back by keeping in line with absolutist ''facts.'' But my faith has been restored and now with the growing evidence, I intend finally to get my pre-big bang, diabatic phase of the universe as a correction principle to Penrose's cyclic universe theory. I noticed a while back there was one inconsistency with his proposed model and that was, that his cyclic model is not entirely symmetric, the distant past was much more dense than the final radiation phase. I haven't seen this discrepancy mentioned before, but I find it curious that he has not even attempted to explain it... maybe he was aware of it, but doesn't quite have an answer yet?

 

Either way, the scientific community appears to be, in the words of Frankunfurter, ''wising up.''


 


Edited by Dubbelosix, 29 April 2019 - 05:04 AM.


#2 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 04:55 AM

The Phase Transition Equation (as I call it) is not however a unification equation, it's a phase transition formula for how a universe can change from a condensed form, possibly an all-matter or all-energy form pre existing the big bang as we know it...

 

https://prebigbangst...-Big-Bang-Phase

 

The Unification Equation, involves not unification per se, but will pave a way towards the unification process: The instructions from the modified Friedmann equation I developed applied non-classical laws to the Friedmann equation

 

 

The energy of each oscillator needs to be attributed to the Planck law as (using dimensionless form of entropy this time):

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( [n_k + n S_k + n S_{ik}]\ \frac{\hbar \omega}{\frac{\hbar \omega}{k_BT} - 1})\frac{\dot{T}}{T}[/math]

 

Absorbing the derivative into the entropy (entropy production) and allowing the correction of zero point energy we get:

 

[math]\frac{\dot{T}}{T}(\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( [\dot{n}_k + n_k \dot{S}_k + n_k \dot{S}_{ik}]\ \frac{\hbar \omega}{\frac{\hbar \omega}{k_BT} - 1} + \dot{n}\frac{\hbar \omega}{2})[/math]

 

This shows that temperature cannot go to zero. And so from this the calculated thermal application (not too disimilar to entropic gravity theories);

 

[math]\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{3072 G}{60}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ PV = \frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U}V = \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\dot{N}k_BT [/math]

 

ref. https://prebigbangst...ropy-Production

 



#3 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 06:31 AM

Anyway, next time I will return to this topic, we shall investigate what the last equation may be telling us. I haven't quite grasped the true way forward from the entropic equation, but I have some idea's in store.



#4 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 08:59 AM

Ok... so we are almost there... only dimensionless constants are physical in nature. First we take the differential logarithm. You do that dividing through by the volume:

 

[math]\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ \frac{dV}{V} = \frac{3072 G}{60}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ P = \frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U} = \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\dot{N}\frac{k_BT}{V} [/math]

 

and so this becomes

 

[math]\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ d\log_V = \frac{3072 G}{60}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ P = \frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U} = \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\dot{\rho}_{thermal} [/math]

 

In which the last identity has to be translated as a thermal energy density [math]\rho_{thermal}[/math]. Since the pressure is also a type of energy density, we can simply the entire equation in a dimensionless format, but we will invite a new notation which will weight out the density terms - the notation we will use will be [math]\rho_{\alpha}[/math]:

 

[math]\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V = \frac{3072 G}{60}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \frac{P}{\rho_{\alpha}} = \frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \frac{\mathbf{U}}{\rho_{\alpha}} = \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\frac{\dot{\rho}_{thermal}}{\rho_{\alpha}} [/math]

 

Again, only dimensionless equations hold any physical meaning in physics, this has simplified the equation then, in physical terms. Have we learned anything new? Maybe a little, but the last part is interpretation, which is the hardest part of understanding the physics.

 

 

EDIT: Of course, its not truly dimensionless, but pretty close. Making it dimensionless would retrieve the original Friemdann equations. The remaining dimension is a factor of inverse time.


Edited by Dubbelosix, 29 April 2019 - 09:02 AM.


#5 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 09:05 AM

But this is what I have found interesting from the approach, the extra time derivative implies non-conservation, which means in a simpler form, a dimensionless case would be:

 

 

[math]\int\ \frac{T}{T_0}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\frac{\rho_{thermal}}{\rho_{\alpha}} [/math]

 

 

Edit: Can also be written as

 

 

 

[math]\int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{T_0}{T})\frac{\rho_{thermal}}{\rho_{\alpha}} [/math]


Edited by Dubbelosix, 29 April 2019 - 09:13 AM.


#6 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 10:08 AM

A compactly written form of the arguments can be found at my blog:

 

 

https://prebigbangst...-Big-Bang-Phase



#7 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 10:15 AM

I felt like I have not demonstrated in a written sense, the significance of the dimensionless roles according to fluid dynamics and so, came across this nice article on the importance of those roles:

 

https://en.wikipedia...fluid_mechanics

 

The universe is modeled according to fluid dynamics, much like how I recognized the equation describing the sonoluminscence was a phenomenon of the same dynamic features. I still believe we can learn from both of those approaches.


Edited by Dubbelosix, 29 April 2019 - 10:16 AM.


#8 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 29 April 2019 - 11:38 PM

I just realized how badly written this was yesterday, I didn't seem to be articulating it properly. It's not that the equation has been reduced to a dimensionless form, as there still exists an issue though of the gravitational constant, which means the equation isn’t entirely dimensionless - but there are components here made in a dimensionless fashion in such a way that it will appease fluid dynamics - for instance, check out Bernoulli's principle.



#9 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 30 April 2019 - 10:49 AM

But this is what I have found interesting from the approach, the extra time derivative implies non-conservation, which means in a simpler form, a dimensionless case would be:

 

 

[math]\int\ \frac{T}{T_0}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\frac{\rho_{thermal}}{\rho_{\alpha}} [/math]

 

 

Edit: Can also be written as

 

 

 

[math]\int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{T_0}{T})\frac{\rho_{thermal}}{\rho_{\alpha}} [/math]

 

 

 

 

Now, we apply the physics of Bernoulli's equation - which may tell us something about the nature of the universe and perhaps its own expansion.

 

In short, the equation of Bernoulli implies that a decrease in the speed of a fluid occurs in symbiosis with a decrease of pressure (and/or) a decrease in the potential contained in the fluids potential energy. Most of the hard work has been done... Let's rewrite the equation in the style of Bernoulli's equation... doing so however, we find corrections to the last equation - but before we encounter these corrections, we will make it that this will be a special case in which the equation is not limited by the gravitational force; in doing this, we can completely remove the dimensions of the gravitational constant from the equation making it truly dimensionless - but there is additional important physics, for instance, we call such an approach one which satisfies the conservative force.

 

What is a conservative force?

 

Basically, the conservative force implies that the total work done in a moving system between say, two points or locations in its history is independent of the path! This means the cyclic nature of a universe, if it has one, implies the total net work done by a conservative force is exactly zero. This is actually very close to an investigation I made into gravielectromagnetism, which explored how Sciama defined the gravielectromagnetic field. But we will come to that soon... First we rewrite the equation in the style of Bernoulli while keeping in mind, the coefficient on the pressure over density term has a thermal coefficient, meaning it has a thermal interpretation

 

 

[math]\int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{T_0}{T})\frac{P}{\rho} [/math]

 

We can take [math]P[/math] and [math]\rho[/math] as both the density and pressure components of the universe, making it independent of the gravitational force is relatively easy. But first we recognize a correction to the equation in which the velocity of the entire system is described through the application on the right hand side as

 

[math]\int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + (\frac{T_0}{T})\frac{P}{\rho}) [/math]

 

To make the equation independent of [math]G[/math] we can do something clever, but it takes us right back to the gravielectromagnetic interpretation I studied provided from the work of Sciama - he worked in the cgs unit base for the gravitational constant, in other words,

 

[math][G] = \frac{\mathrm{cm}^3}{\mathrm{g}\ \mathrm{s}}[/math]

 

[math][c] = \frac{\mathrm{cm}}{\mathrm{s}} [/math]

 

[math][M] = \mathrm{g} [/math]

 

[math][r] = \mathrm{cm} [/math]

 

[math][\phi] = \left[ -\frac{M}{r} \right] = \frac{\mathrm{g}}{\mathrm{cm}} [/math]

 

[math]\left[\frac{\phi}{c^2}\right] = \frac{\mathrm{g}\ \mathrm{s}^2}{\mathrm{cm}^3} =\left[ \frac{1}{G} \right][/math]

 

The ratio then [math]\frac{\Phi}{c^2}[/math] has the same units as [math]\frac{1}{G}[/math] - in other words, using these units for the gravitational constant, we remove it from the equation from the simple procedure of

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + (\frac{T_0}{T})\frac{P}{\rho}) [/math]

 

When the field is not limited to a gravitational interpretation we require a further corrective factor

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) [/math]

 

In which [math]\Psi[/math] designated the potential field of the universe. This means the entire equation equates to a constant, not dissimilar at all to Bernoulli's equation

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

This model, finally is becoming a unified equation for the universe.... but there is one final thing to do... and that will be the last highlight of the thread, an implication no less, of a driving force which we may consider as an interpretation of the cosmological constant. To do so, we require the work of Fritz Rohrlich who demonstrated that the force on a moving system (as in opposition to one at rest) is important when considering the mass-energy equivalence.

 

The model he proposed considered in a frame that moves with velocity to the left, the driving force moving to the left is redshifted, while the driving force moving to the right is blueshifted. The blue light carries more momentum than the red light, so that the momentum of the light in the moving frame is not balanced - in other words, a cosmological constant is interpreted here-on-in as a non-balanced force since the energy is carrying some net momentum to the right.

 

The object has not changed its velocity before or after the emission, however, in this frame it has lost some right-momentum to the energy driving it in a particular direction. The only way it could have lost momentum is by losing mass - this may be also a statement of non-conservation and is not only quintessentially tied to Poincaré's radiation paradox, it also solves it.

 

So the right-moving energy carries extra momentum [math]\Delta p[/math] we then have

 

[math]\Delta p =\frac{v}{2c^2}E[/math]

 

The left-moving energy will carry a little less momentum, by the same quantity [math]\Delta p[/math] such that the total right-momentum in the energy is twice the value of [math]\Delta p[/math]. This is the right-momentum energy lost from the system (universe)

 

[math]2\Delta p=\frac{v}{c^2}E[/math]

The momentum of the universe moving in the directional frame after the emission is reduced by the amount of 

 

[math]p′ = mv−2\Delta p = (m − \frac{E}{c^2})v[/math]

 

So the change in the universes mass is equal to the total energy lost divided by the speed of light squared - the big implication here is that any emission of energy can be carried in a two-step process in which energy used by the universe is converted to mass, while the emission of an energy is accompanied by a loss of the mass in the universe. The same physics can be theoretically understood technically through the appearance of the same quantity found on the right hand side of finally.... the unified equation

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

Which concludes the investigation in such a way, I feel confident this will be telling us the nature behind the impetus, or driving force we call the cosmological constant.


Edited by Dubbelosix, 30 April 2019 - 02:29 PM.


#10 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 30 April 2019 - 02:42 PM


 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho_{\alpha}}\ d\log_V =  \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

There is of course, a further interpretation that requires an attention - the numerical ratio of [math]\frac{3072}{60}[/math], what is important about it? In general relativity we have coefficients say, of [math]8 \pi G[/math] but such a modification was introduced to simplify several approaches in physics. The fact I ended up with the number [math]3072[/math] is certainly intriguing, since it has a very special importance involving prime factors. The number [math]3072[/math] is known as a ''composite number'' which consists of two prime factors multiplied together. Not only this but it also has a total of [math]22[/math] divisors.

 

It doesn't tell us very much, but it is curious for this reason. The divisors are

 

[math]1,2,3,4,6,8,12,16,24,32,48,64,96,128,192,256,284,512,768,1024,1536[/math]

 

whilst the factorization is

 

[math]2^{10} \times 3[/math]


Edited by Dubbelosix, 01 May 2019 - 10:41 AM.


#11 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 30 April 2019 - 02:48 PM

There is of course, a further interpretation that requires an attention - the numerical ratio of [math]\frac{3072}{60}[/math], what is important about it? In general relativity we have coefficients say, of [math]8 \pi G[/math] but such a modification was introduced to simplify several approaches in physics. The fact I ended up with the number [math]3072[/math] is certainly intriguing, since it has a very special importance involving prime numbers. The number [math]3072[/math] is known as a ''composite number'' which consists of two prime numbers multiplied together. Not only this but it also has a total of [math]22[/math] divisors.

 

It doesn't tell us very much, but it is curious for this reason. The divisors are

 

[math]1,2,3,4,6,8,12,16,24,32,48,64,96,128,192,256,284,512,768,1024,1536[/math]

 

whilst the factorization is

 

[math]2^{10} \times 3[/math]

 

Further, the number [math]3072[/math] is not a triangular number, while the number (denominator) [math]60[/math] is.



#12 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 30 April 2019 - 04:17 PM

After more analysis, this is becoming even more interesting, consider that we rewrite the ratio using the factorization

 

[math]\frac{2^{10} \times 3}{60}[/math]

 

can be simplified as

 

[math]2^{10} \cdot 0.05[/math]

 

This is because of the result of a simplification

 

[math]\frac{3}{60} = 0.05[/math]

 

Why is it interesting?

 

The p-value is defined as a probability, under what is known as the null hypothesis

 

:(meaning at times denoted as opposed to denoting the alternative hypothesis) on a population variate, for the variate to be observed as a value equal to or more extreme than the value observed. I now quote wiki:

 

 

''When the p-value is calculated correctly, this test guarantees that the type I error rate is at most α. For typical analysis, using the standard α = 0.05 cutoff, the null hypothesis is rejected when p < .05 and not rejected when p > .05. The p-value does not, in itself, support reasoning about the probabilities of hypotheses but is only a tool for deciding whether to reject the null hypothesis.''

 

A null hypothesis already exists from the model, we have talked about it in the sense of conserved fields. Further, wiki has defined information on this subject as

 

''When the null hypothesis is true, if it takes the form 

 

[math]H_0; \theta = \theta_0[/math]

 

, and the underlying random variable is continuous, then the probability distribution of the p-value is uniform on the interval [0,1]. By contrast, if the alternative hypothesis is true, the distribution is dependent on sample size and the true value of the parameter being studied.[2][21]''

 

Now we must ask, what kind of continuous nature are we talking about? The language of mathematics can be very vague and I am yet to understand the full complexities of these statements, but it would certainly ring a bell in the case of [a] spacetime found as a consequence implied by relativity. Uniformity, also has implications to cosmology since the universe largely appears uniform in all directions we see. Are we getting to a truth, or is this a red-herring? I am not entirely sure, but further investigation is needed.


Edited by Dubbelosix, 30 April 2019 - 04:40 PM.


#13 Flummoxed

Flummoxed

    Explaining

  • Members
  • 607 posts

Posted 01 May 2019 - 03:23 AM

Would it be possible to include a link indicating what each variables are and the units are using. Other wise too much is left to the imagination.

For instance i assume your are differentiating pressure over volume, maybe it is another variable time perhaps I am not sure, I assume you are trying to link this into the HUP via one of your links but this is absolutely not clear how. 

 

At the moment all I am reading is Smoke and mirrors, as others might be as well.



#14 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 01 May 2019 - 04:57 AM

You can't accuse something of being smoke and mirrors just because you are having difficulty following it... however, I will very much so write out all the terms and their meanings, later.



#15 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 01 May 2019 - 07:25 AM

[math]R[/math] - radius

[math]H[/math] is the hubble constant

[math]G[/math] the gravitational constant

[math]n[/math] particle number density

[math]N[/math] is the particle number

[math]S[/math] is the entropy

[math]\mathbf{S}[/math] entropy density

[math]T[/math] is the temperature

[math]k[/math] is the curvature constant

[math]k_B[/math] the Boltzmann constant

[math]\mathbf{U}[/math] notation for the energy density

[math]\rho[/math] also alternative notation for energy density

[math]P[/math] is the pressure

[math]N[/math] particle number

[math]c[/math] the speed of light

[math]v[/math] the velocity

[math]\zeta[/math] is the Zeta Riemann function

[math]V[/math] is the volume

[math]d\log_V[/math] differential logarithm of the volume

[math]\phi[/math] - is the gravitational potential (under the units of a cgs system for the gravitational constant) when weighted by a factor of [math]c^2[/math])

and finally,

[math]\hbar[/math] is Planck's constant.

If you need further assistance in understanding anything, do not hesitate to ask.


Edited by Dubbelosix, 01 May 2019 - 07:29 AM.


#16 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 01 May 2019 - 07:39 AM

It should also be noted, the ratio

 

[math]\frac{\zeta(4)}{\zeta(3)}[/math] is part of a very special definition - the Zeta Riemann function is deeply connected to the prime number system and a powerful tool for definite integrals.

 

For further understanding, or reference to what has been stated, you can read up on this:

 

http://mathworld.wol...taFunction.html


Edited by Dubbelosix, 01 May 2019 - 10:45 AM.


#17 Dubbelosix

Dubbelosix

    Creating

  • Members
  • 2443 posts

Posted 01 May 2019 - 08:53 AM

Anyway... moving on, we have to try and simplify some of these expressions (number coefficients) down, in order to record what we find, and later see if we can find something more important.

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho}\ d\log_V = \frac{3072}{60} \frac{\zeta(4)}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

With

 

[math]\zeta(4) = \frac{\pi^4}{90}[/math]

 

with it weighted by

 

[math]\zeta(3) = 1.2020569032[/math]

 

[math]\frac{\phi}{c^2} \cdot \int\ (\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\frac{1}{\rho}\ d\log_V = \frac{3072}{60 \cdot 90} \frac{\pi^4}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

[math]= \frac{3072}{5400} \frac{\pi^4}{\zeta(3)}(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho}) = \mathbf{C}[/math]

 

and

 

[math]3072 \cdot \pi^4 = 299240.728[/math]

 

final simplification further yields

 

[math]\frac{299240.728}{1.2020569032 \cdot 5400}[/math]

 

and we get

 

[math]\frac{299240.728}{6491.10728} = 46.1001051[/math]

 

Of course, this doesn't tell me a lot at face value, but its important we keep track of these factors, because there are a few different ways we can try and manipulate these constants, maybe something interesting can be found behind them.


Edited by Dubbelosix, 01 May 2019 - 10:51 AM.