# Paper Planned To Be Published: After Proof Reading

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### #1 Dubbelosix

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Posted 20 June 2019 - 10:17 AM

Quantization of the Black Hole Particle

To derive a quantization of the black hole, in terms of discrete quantum mechanical processes, I defined the Rydberg constant in terms of the gravitational coupling constant we get:

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

Even though the Rydberg constant was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr). In which case we may hypothesize energy levels:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Plugging in the last expressions we get an energy equation:

$\Delta E_G = \frac{n\hbar c}{\Delta \lambda} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})$

The relativistic gamma appears from the definition of the deBroglie wavelength. The ultimate goal of creating this equation, was to explore an idea I can trace back to Lloyd Motz suggesting a stable black hole. Black holes, even the microscopic kind, cannot be mathematically ruled out from physics (but later I will show in a logical sense why the would be unstable) - and if the work of Hawking is to be taken seriously, including analysis provided by Motz, then a black hole system really is deduced from a discrete set of quantum processes - those discrete processes always lead to an increase in the entropy of a system like a black hole.

The earliest model of a black hole stated that the entropy of a black hole was exactly zero - Hawking challenged this and stated if it has an entropy, then a black hole must possess a temperature and later we came to understand it as a thermal property of the black body radiation of the black hole. It may still be true, at least for the ground state black hole, that entropy may be essentially constant.

So how can we have an increasing entropy but a system with an infinite lifetime?

Something needs to give. Entropy is a measure of disorder, and if a system is infinitely stable, then the disorder remains a constant. It could be that the black hole particle at near zero temperatures exhibits a behaviour like a zeno effect. The zeno effect is when an atom is incapable of giving off energy suspended in an otherwise, infinite animation and it will remain in a ground state because its wave function is incapable of evolving from the ground state.

It is therefore similar to how an atom, ready to give up energy can be suspended infinitely in time - and the reason why it cannot give up radiation is essentially the same for the micro black hole, since the zeno effect is all about keeping an atom in the ground state. To put in a summary for clarity I have came to some conclusions: The main one being that the entropy of a stable black hole would immediately render the black hole particle as a system at absolute zero! The reason why is because if a black hole particle did not have an entropy, by first principles, it consequently cannot give up any radiation (consequence of the third law of thermodynamics) and as far as we know, systems can never actually reach absolute zero temperatures, there is reasonable argument here that the black hole particle must evaporate. In fact, I wondered immediately if there are applications of this with the micro black hole and maybe in regards to its energy levels. We start with the equation

$S = k_B \frac{Gm^2}{\hbar c}$

Where the entropy has taken on the dimensions of the Boltzmann Constant. If you plug in the mass for a black hole:

$m = \frac{c^2R}{2G}$

You will retrieve the Bekenstein entropy. It is often said that entropy has dimensions of the Boltzmann constant but it's entirely plausible, even with better arguments for, a dimensionless case of entropy in the form

$\mathbf{S} = \frac{S}{k} = \frac{Gm^2}{\hbar c}$

In which case, it is possible to identify the entropy in a dimensionless form

$\mathbf{S} = \frac{\hbar c}{k_BT}\frac{p}{4 \pi \hbar} = \frac{\hbar c}{k_BT}\frac{\sqrt{2mE}}{4 \pi \hbar} = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

In which the wave number is identified with the deBroglie wavelength

$k \equiv \frac{p}{\hbar} = \frac{\sqrt{2mE}}{\hbar}$

Because of the equivalence of the wave number the definition of the deBroglie wavelength. The SI value for the Boltzmann constant is given in units of the Hartree energy divided by the Kelvin

$\frac{E}{K} = k_B = 3.1668114(29)\times 10^{−6}$

In which the Hartree energy is simply

$E = 2\mathbf{R}\hbar c$

and so we can write it with temperature T which is also measured in Kelvin,

$S = k_B = \frac{E}{T} = \mathbf{R}\frac{\hbar c}{T}$

This can be written in such a way to introduce those quantized atomic levels ~ by making use of the formula

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

giving

$k_B = \frac{1}{T} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1T} - \frac{Gm^2}{n^2_2T})$

In which $\mathbf{R}$ is the usual Rydberg constant. The Boltzmann constant can also be expressed entirely in terms of Planck units ~

$S = \frac{m_PL^2_P}{k_BT_P t^2_P}$

$k_B = \frac{\sqrt{\frac{\hbar c}{G}} \frac{\hbar G}{c^3}}{\sqrt{\frac{\hbar c^5}{Gk^2_B}} \frac{\hbar G}{c^5}}$

The Transition of Conducting Black Holes

Another theory in literature involves a theoretical suggestion which asserts the surface of a black hole [may] be considered as a conducting surface; the argument goes like this:

‘’ Introducing an effective thickness \Delta r of the membrane, and assuming the same resistivity $R_H$ within $\Delta r$, one can relate total resistance of the black hole with the conductivity of the membrane using a model of two concentric spheres with radii $a_1 = r_s$, $a_2 = r_s + \Delta r$ where the spherical shell is filled by homogeneous conducting matter is given by

$R_H = \frac{1}{4 \pi \sigma}(\frac{1}{r_1} - \frac{1}{r_2})$

Where $\sigma$ is the conductivity. ‘’

Early on it struck me how similar this equation is, compared to our transition equation using the Rydberg constant:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Fortunately for both equations, the inverse length features and for conductivity, (which is the inverse of electrical resistivity), the resistance of a system increases with the length of the system. This must mean that in each transition from one energy to another, the conductance must weaken, as shown:

$R_H = \frac{1}{\Delta( \sigma \lambda)} = \mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

From the last equation, you can calculate conductance G as

$G = \rho (\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2}) = \frac{\mathbf{E}}{J}(\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2})$

Where \rho is the resistance and is related to conductance by $\frac{1}{\sigma}$ and $\mathbf{E}$ is the electric field where $J$ is the current. Also it is good to note, the shape operator for a sphere is mathematically described under the Weingarten equations. However, literature hints that the equation does not depend on the geometry of the system, since the resistivity and conductivity are proportionality constants; in such a case it is said it depends only on the material the system is made of but not the geometry, which robs any importance with viewing it under a shape operator. The black hole area is defined by

$A_{BH} = \frac{G^2m^2}{c^4}$

And the angular momentum per unit area is

$\mathbf{J} = \frac{J}{A_{BH}} = \frac{c^3}{G}$

Which enters the resistivity equation like;

$R_H = \frac{c^4}{G^2m^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \frac{\mathbf{J} c}{Gm^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2})= \alpha_G\mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

We have retrieved the gravitational charge (squared) in the denominator. The ratio

$\alpha_G = \frac{Jc}{Gm^2} \approx 1$

can be interpreted as a fine structure constant, as following the same idea but reached a different derivation of the same argument from Arun and Sivaram from their paper ‘’some unique constants associated to extremal black holes.’’

The equation can further be distributed with a factor of \hbar c, similar to the Rydberg formula to find the energy. we get

$\hbar c\ R_H = \frac{c^4}{G}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \alpha_G(\frac{E_1}{\sigma n^2_1} - \frac{E_2}{\sigma n^2_2})$

With this time, a classical gravitational upper limit of the force $\frac{c^4}{G}$. According to Arun and Sivaram for a black hole, the angular momentum and the square of the charge is related to the speed of light via ~

$\frac{J}{Q^2} = \frac{1}{c}$

It has crossed my mind whether it would be related more fundamentally to the gravitational aether which is defined in this work by the variable parameters $\sqrt{\epsilon_G \mu_G} = \frac{1}{c}$. Either way it was noted by the astrophysicists that it is remarkably similar to the Von Klitzig constant $\frac{\hbar}{e^2}$ or the Josephson constant in superconductors $\frac{e^2}{\hbar}$ - resistivity decreases in the case of superconductors, while it increases for metals as they increase temperatures.

We deducted not long ago that for a black hole, the conductance decreases (increase in resistivity) for a black hole as it gets smaller, this will concur nicely with the later investigation, because the equations so far tell us that in the ground state, conductance must be zero (to ensure shielding from environment) and will complement the idea of a ‘perfect insulator’ as the smallest possible radiator. While it remains curious science may predict a perfect insulator, the physics systematically argues against it due to the third law.

Differences Between The Holeum Model and Mine

Holeums were proposed by L. K. Chavda and Abhijit Chavda in 2002 and is suggested to be a form of cold ‘’dark’’ matter. The system itself is not a black hole and they consequently radiate gravitational waves as opposed to the usual electromagnetic Hawking radiation - later we will see why I do not believe a Planck particle (black hole particle) is stable at all, concerning contradictions that arise from the laws of thermodynamics, never mind two bound stable black holes.

The suggested binding energy of a bound black hole particle-pair is:

$E_n = - \frac{mc^2 \alpha_G^2}{4n^2}$

I came across the Holeum model after I constructed my approach to the quantization transition formula. I began from arguments concerning the Rydberg constant and asserted this would be a true analogous model to a ground state hydrogen atom in which the radiation inside the black hole attains its lowest quantum bound state.

Interestingly, it is claimed their model too is analogous to the ground state hydrogen atom, but notice, the models are totally different fundamentally. My approach is direct towards the treatment of a single black hole particle, not a bound pair. Notice also they used the gravitational fine structure, which was of similar contention to my own derivation that came from a modification of the Rydberg constant

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

or as the gravitational Rydberg-Hartree energy ~

$\hbar c\ \mathbf{R} = \frac{\hbar c\alpha_G^2}{4 \pi \lambda_0} = \frac{Gm^2}{4 \pi \lambda_0} = \frac{Gm^3c}{4 \pi \hbar} = Gm^2\frac{p}{4 \pi \hbar}$

• this is logical for a true analogue model since the constant is fundamental to the Bohr model derivation of the hydrogen atom it is a statement and Bohr maintained the derivation itself, is fundamental to many systems.

If for instance, we can prove later it doesn’t make sense to talk about one stable black hole particle, it would certainly not make any sense to think any stable bound black holes exist in nature.

An Argument For Temperature Variations With Black Hole Acceleration

A postulate in this work is that black holes when accelerated will obey their own radiation power law similar to an accelerated charge following the Larmor radiation law. It’s not an ad hoc hypothesis, since a micro black hole with a charge would have to obey the weak equivalence principle and give up radiation in a gravitational field in the form of the usual Hawking process.

In fact, early theories attempted to explain the electron as a black hole, and they must have crossed the question of what Larmor radiation would be: And if an electron had been a black hole, the radiation from the accelerated charged black hole would have had to expressed itself through the usual Hawking emission process.

A charge in a gravitational field should give off an electromagnetic radiation. This can be formulated in the following way:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{\sqrt{g_{tt}}(r)}{\sqrt{g_{tt}}(s)})}$

The metric $g_{tt}$ is made up of terms of $1 + \frac{\phi}{c^2}$. Higher terms then of the field are calculated like so:

$1 + \phi_1 + \phi_2 - \frac{1}{2}\phi^2_1 - \phi_1\phi_2... higher\ terms$

To first order it is just the difference formula

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

However, employing a connection made by Motz, there is a correction to the denominator of the first equation,

which he notes should be squared because there is one such term to account for in the metric ds. In our case it is simply the removal of the square root signs. A second difference between the approach I took and the one they have taken is that their denominator consists of only one such term of the metric $(1 - \frac{R}{r})$ whereas mine considered the ratio of the two signals in a relativistic way that satisfies the red shift.

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

His interpretation of the squared denominator ensures that the radiation increases as something falls towards a singularity. Of course, modern approaches are to involve situations which don't involve singularities at all, as we mentioned before.

The Motz-Kraft model is for the calculation of the luminosity from charges accelerating in gravitational fields. Quasars are essentially black holes with a dense electrically charged material around it and they produce a large luminosity for these reasons. In our case, we are saying however, the weak equivalence applies to all kinds of systems with electric charge: Including the black hole itself. In such a case, the only way a black hole could radiate is through Hawking radiation suggesting an analogue case of the Larmor radiation ~ and did in fact satisfy such a case with an uncanny analogue to the classical electron radiation par a factor of $\frac{2}{3}$ which is only an adjustable parameter - the radiation (I derived) given up by a black hole is:

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

I obtained this nice simple solution from the known inequality satisfying a black hole with mass, charge and angular momentum (in natural units):

$Q^2 + (\frac{J}{m})^2 \leq m^2$

Notice this last term in the inequality is formally identical to the classical equation describing the radiation for an electron, with the exceptions that neither the mass or radius are associated to the electron but instead the black hole:

$P = \frac{2}{3}\frac{a^2m_er_e}{c}$

Since this too describes the power emitted it would not be wrong to set it in the same physics, in a gravitational field - doing so is childsplay, I get while adopting the coefficient for a true analogue theory:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3}\frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2}$

It is also true that the luminosity is a simple calculation which measures the power emitted by an astrophysical object, but we were looking for a more general case above for any charged object. The relationships for luminosity are briefly covered ~

$L = \frac{f}{A(1 + z)^2} = \frac{\sigma T^4}{4 \pi R^2(1 + z)^2}$

You can calculate the two different luminosities as

$\frac{L_1}{L_2} = \frac{(\frac{\sigma T^4_1}{4 \pi R^2_1})}{(\frac{\sigma T^4_2}{4 \pi R^2_2})}= \frac{T^4_1 R^2_1}{T^4_2R^2_2}$

So an alternative formula to calculate power is suggested as:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3} \frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2} = \frac{2}{3} \frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}(\frac{dp}{dt})^2$

The last term arises from:

$P = \frac{2}{3}\frac{Q^2}{m^2c^3} (\frac{dp}{dt})^2$

And can be found by replacing the acceleration in our equation for $(\frac{1}{m}\frac{dp}{dt})^2$ - this is Lorentz invariant but it isn't generally relativistic.

The true generalized formula is

$P = -\frac{2}{3} \frac{Q^2}{m^2c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Since there are less dynamics in the luminosity relationship, the deduction from our equation (which involves the black hole mass, electric charge and angular momentum) is that a charged black hole will radiate more energy than one at rest: This partially makes sense of my stipulation that a moving black hole will appear hotter than one at rest, though the presence of red shift would mean it is also direction dependent. In units of $G = c = 1$ our equation to first order becomes:

$P = \frac{2}{3}\frac{a^2}{(1 - \Delta \phi)^2}(Q + \frac{J}{m})^2 \leq \frac{2}{3} mr_g\frac{a^2}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2} = \frac{2}{3} \frac{r_g}{m}\frac{1}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2}(\frac{dp}{dt})^2$

with

$\Delta \phi = \phi_1 - \phi_2$

and

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

Let’s just say a bit about the standard equations you might encounter which also attempt to calculate the power emitted by black holes. We know now that accelerated matter aroud the black hole contributes significantly to the light that comes from superluminous objects and so, this has offered a novel retake of the theory that Motz suggested first and later collaborated with Kraft.

ref: https://quantizedblackhole.quora.com/

The phase velocity of a particle or wave in a medium is

$v = \frac{c}{n}$

For light, in literature, the refractive index $n$ decreases with an increasing wavelength,

$1 < n(\lambda_{red}) < n(\lambda_{yellow}) < n(\lambda_{blue})$

The alternative statement for this is

$\frac{dn}{d\lambda} < 0$

In such a case, the dispersion is said to be normal.

https://en.wikipedia...ersion_(optics)

This however, takes us to the Fresnel drag coefficient, that is, the index of refraction is

$w_{+} = \frac{c}{n} + v$

and in the other direction

$w_{-} = \frac{c}{n} - v$

From it, Frizeau found that

$w_{+} = \frac{c}{n} + v(1 - \frac{1}{n^2})$

If we replace the concept this was used in first context of light traveling against the flow of a fluid, with gravity, then the speed of the light will be seen travelling slower with the flow of the fluid. This brings us back to the well-known concepts now, that light is not truly a constant when inside a gravitational field.

But to do this, we require dragging coefficient $f$ which involves an experiment by Arago in which that an aether drags light propagating through it with only a fraction of the mediums speed. Gravitational aether theories are not very well-explored, but if we accept that light is not a constant in general truth when gravitational fields are involved, then the dragging coefficient should be no different to the one proposed

$f = 1 - \frac{1}{n^2}$

Then in 1895, Lorentz predicted the existence of an extra term, which brings us back to the dispersion formula, which is why we mentioned it in the beginning of this post:

$w_{+} = \frac{c}{n} + v(1 - \frac{1}{n^2} - \frac{\lambda}{n} \frac{dn}{d\lambda})$

In particular, this term

$\frac{dn}{d\lambda}$

in which we had noted the relationship

$\frac{dn}{d\lambda} < 0$

Now, let's say a little about the gravitational aether, something which Einstein supported strongly, but did not appear to realize that the constancy of the speed of light in the medium of space is [not] generally constant like he believed his theory predicted.

A Gravitational Aether

Excellent arguments exist now for the existence of the gravitational permittivity and permeability with the discovery of gravitational waves. The constancy of the speed of light only holds in a vacuum - but the density of gravitation varies between celestial objects and therefore the speed of light does technically vary.

In fact, authors Masanori Sato and Hiroki Sato in their paper ‘’Gravitational wave derived from fluid mechanics applied on the permittivity and the permeability of free space’’ suggests that gravitational waves are simply fluctuations of the medium, which appears as the product of the permittivity of free space and the permeability of free space. That is, the gravitational wave is an acoustic wave in the medium - the proposal shows how the phase velocity of the fluctuation relates to the speed of light

$c = \frac{1}{\sqrt{\epsilon_G \mu_G}}$

The model has some interesting consequences, first being that permittivity and permeability are allowed to vary. A second is that the speed of light is variable in gravitational fields. Another interesting property is that while both Newtonian mechanics and Einstein’s relativity theories predict the confinement of light by gravity, neither theory defines the escape velocity or the Schwarzschild radius; in fact, the actual speed of light can only approach zero but never reach it - so in effect light is allowed to escape from a black hole.

Let’s be clear about something - I do not believe that the thickness of space (the medium) is an aether made from any particle. In fact there cannot be any motion associated to this aether because it would violate the first principles of relativity. In fact you can argue as I have already done, that any true quantization of gravity would be at odds concerning how we actually think about the roles of pseudo forces.

Many experiments have been performed to measure the value of the Newtonian $G$ but has come up with varied results and up until this year another measurement has cast a shadow over settling why we keep measuring different values for the constant - it may not be the case the constant does change, but what it may have to do with is the gravitational acceleration which is proportional to the constant. Since in this aether theory I have chosen, both permittivity and permeability will depend on $G$ ~

$\frac{1}{\epsilon_G} = 4 \pi G$

$\frac{1}{\mu_G} = \frac{c^2}{4 \pi G}$

This leaves open a question of whether the deviations in the value of G has something to do with variations spacetime permittivity and permeability. This particular theory of the aether, as a dynamical ''thickness'' of space due to varying gravitational density, the refractive index for radiation is proportional to $\sqrt{\epsilon_G \mu_G}$ (the gravitational permittivity and permeaility) and is represented as:

$n = \sqrt{\frac{\epsilon_G \mu_G}{\epsilon_0 \mu_0}}$

A high refractive index for the equation

$\frac{1}{\sqrt{\epsilon_G \mu_G}}$

causes a low speed of light (such as found round strong gravitational fields of black holes). It has been argued in literature that the refractive index is more intuitive than curvature; this suggestion is probably quite true, since curvature is the presence of a dynamic metric but we know not what causes this ''dynamic feature'' other through the presence of matter - which is well-known to tell spacetime how to curve, but still doesn't explain why the dynamic phenomenon exists. In a sense, the gravitational explanation for a refractive index supposes a type of mechanical explanation to curvature.

The formula I created under the same analysis Bohr used to model a Hydrogen atom to describe discrete transitions of a black hole losing a mass was:

$S = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

The Fresnel dragging coefficient is:

$f = 1 - \frac{1}{n^2}$

We can rearrange this in a very simple way: first we add the $\frac{1}{n^2}$

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

You could simplify further, for instance, this would give

$f + \frac{1}{n^2} = 1$

and you could even find the levels in a transition process:

$\frac{1}{n^2} = 1 - f$

And while this is interesting we explore not only this avenue, but we will first concentrate on the first equation that we featured:

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

Can you see where this equation relates to the investigation of the transition equation? First I want to solve for the quantization levels:

$\frac{1}{n^2} + \frac{1}{n^2} = 1 - f - \frac{1}{n^2}$

So how do you solve that? We will simplify the terms to make it easier and use the form

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

where the second term on the left hand side is

$f + a$

and the right hand side simplified as

$1 - b$

So that we have:

$f + a = 1 - b$

add the notation of b on both sides

$f + a + b = 1 - b + b$

simplify

$f + a + b = 1$

subtracting $f + a$ from both sides gives

$f + a + b - (f + a) = 1 - (f + a)$

which simplifies to

$b = 1 - f - a$

Plugging back in the definitions we get

$\frac{1}{n^2} + \frac{1}{n^2}= 1 - f - \frac{1}{n^2}$

Now, going back to the transition equation we will notice that it involves a negative sign on the left hand side, this is a simple procedure of distributing a negative sign:

$\frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

$- ( \frac{1}{n^2} + \frac{1}{n^2}) = - 1 - f - \frac{1}{n^2}$

giving us

$\frac{1}{n^2_1} - \frac{1}{n^2_2}= 1 + f - \frac{1}{n^2}$

The notation of $n$ is the index of refraction but in some loose way, I expect it can play the same role as the principal quantum number. There are loads of ways to continue from here, for instance, we might consider the addition of $\frac{1}{n^2}$ as a correction to the transition equation which may be similar in nature to a dispersion relation predicted by Lorentz:

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} = 1 + f$

You could from here be solving for the dragging coefficient,

$f = \frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} - 1$

But the direction I wish to take is simply one in which we retain the terms in the equation and rewrite the third term on the left in terms of the dispersion, again, first predicted by Lorentz, but formulated here in a completely new way::

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

There will be more to say on this matter as I take the equation into the transition formula I created for black holes.

Reference:

https://en.wikipedia...zeau_experiment

Again, to understand how the development of the transition equation involves the dragging coefficient, we will recognize that index of refraction must be playing a similar role to the principal quantum umber under the dimensionless entropy formula for the transition of any universal fundamental system applied to quantum or semi-classical systems:

$S = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

$\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

In which the Rydberg formula describes such transition levels

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

So by distribution we get:

$\mathbf{R}(\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}) = \frac{1 + f}{\Delta \lambda}$

Let's flip the equation for clarity

$\frac{1 + f}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Distribution of the gravitational charge $e^2 = n\hbar c = Gm^2$ will give:

$n\hbar c\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2} + e^2\frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Now we divide through by the thermodynamic energy which gives back an equation similar to the original dimensionless entropy

$\frac{n\hbar c}{k_BT}\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1 k_BT} - \frac{Gm^2}{n^2_2k_BT} + \frac{e^2}{k_BT} \cdot \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

The dimensionless drag coefficient is often described under the terms:

$f = \frac{2F}{\rho v^2 A}$

Where the large $F$ is the dragging force, $\rho$ is our case would be the density of the gravitational field and $v^2$ is the speed of the object relative to the fluid.

Before we proceed, as I always like to do, is define the formula's for clarity for further investigation:

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

$f = \frac{2F}{\rho v^2 A}$

Plugging it in simply gives us the formula, as we flip the equation

$1 + \frac{2F}{\rho v^2 A} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

This means we also have an alternative formula to investigate at a later date:

$1 + \frac{A_b}{A_f} \frac{B}{Re^2_L} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation itself is simply

$F = \frac{1}{2} \rho v^2 f A$

And has been interpreted as a statement that the drag force on [any] object is proportional to the density of the fluid medium and porportional to the relative flow speed.

Where $Re^2_L$ is the Reynolds number related to the fluid path length. The Reynolds number is defined specifically as:

$R_e = \frac{\rho v L}{\mu} = \frac{v L}{\nu}$

We have to define the notation here, where $v$, $L$ is a linear dimension which in the case of gravity, we would expect it to be gravielectromagnetic path which is by definition the linearized gravity.$\mu$ is the dynamic viscosity, in which interpreted within gravitational physics, defines the ''thickness of the gravitational field,'' or ''stickiness of the gravitational aether. Finally $\nu$ is the kinematic viscosity of the medium itself - in a simple way of visualizing viscosity, the gravitational field is a fluid and in previous investigations for a primordial spin of the universe, is much like how you have a dense object submerged in water - what happens when you include spin in the fluid, it drags the objects inside of the medium.

References:

https://en.wikipedia...Reynolds_number

https://en.wikipedia...rag_coefficient

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} = 1 + f$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + f = 1$

Plug in:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$f = \frac{2F}{\rho v^2 A}$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2F}{\rho v^2 A} = 1$

Plug in the definition of the standard drag,

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A = 1$

Or as

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A$

Simplify by removing constants:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A$

Since generic dimensions of the density is simply a mass divided by the volume, we can obtain, after distributing the velocity squared term to obtain the energy density

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{1}{A} \cdot \rho v^2 f A$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We may even write our constructed equation using the same front and back area terms:

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \cdot \rho v^2 f$

Since the density is related to the stress energy tensor through $T^{00} = \rho v^2$ we can re-write it as:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We can define the stress energy, as either gravitationally related, or by electromagnetism. In some sense, gravity and electromagnetism may in certain theories be related, maybe even to the point they are the same thing… but these are ideas to take up at a later date. Firstly, the density of electromagnetic field is given through the stress energy as:

$T^{00} = \frac{1}{c^2}(\frac{1}{2}\epsilon_0 \mathbf{E}^2 + \frac{1}{2 \mu_0}\mathbf{B}^2)$

Let’s take a few steps back, we had:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \frac{A_b}{A_f} \frac{B}{Re^2_L}\ T^{00}$

This gives us the squared quantity:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

And now we simplify and then solve for the stress energy:

$mv^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

$mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - T^{00}$

And let’s use a sign change to finish off for now.

$- mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = T^{00}$

And so we don’t end up with any idea’s of a negative mass we should entertain:

$\rho v^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2}) = T^{00}$

As noticed in a conversation online with a chemist under the moniker ''exchemist'' he had noticed an implication:

''The first line implies $f = \frac{1}{n^2} -1$

The second line says $\frac{3}{n^2} + f = 1$, so substituting for $f$, we have \frac{3}{n^2} + \frac{1}{n^2} -1 = 1[/math]

So rearranging, we have $\frac{4}{n^2}= 2$; so $\frac{2}{n^2} = 1$, hence $n^2 = 2$ and so $n = \sqrt{2}$.

Now we can solve for $f$, using the first line: $f = \frac{1}{2} - 1 = -\frac{1}{2}$.'

or $n = \sqrt{2}$ and $f = -\frac{1}{2}$'

Moving on, the way forward intended was to find three parts, loosely related to the dimensions of space itself - but by doing so, I wish to state that time is an operator and also an observable - it has always been my contention that in SR time is dependent on an observer, in GR time is an observable or it would not manifest curvature in space - and in total contradiction, in quantum mechanics, it says it is neither an operator or an observable which may present itself why there is so much confusion in uniting the theory of relativity and quantum mechanics.

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A = -f$

In  this simple case, we ignore the fancy things we can do, we just accept a simple case of removing the dimensions from the equation:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = -f$

Another simple thing to do is just take ($-1$) from both sides and we get

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} - 1 = f + 1$

If we take the first three terms of the spatial dimensions, than its last part has to be associated to the timelike dimension:

$\frac{x}{n^2} + \frac{y}{n^2} + \frac{z}{n^2} - ct = f + 1$

Notice the I have derived this in which the principal quantum numbers appear to be attached specifically to the spatial dimensions - and time has been placed in not too dissimilar to a fudge factor. Since space and time are not independent of each other, it will be under the normal sum of a three dimensional Euclidean (or Pythagorean relationship) with the negative sign indicated in the metric application on the timelike dimension. Most of us know this relationship and it is:

$s^2 = x^2 + y^2 + z^2 – (ct)^2$

If we take the derivation seriously, then the drag coefficient is dependent also on the sum of those components with the negative time component, in which the metric is simply:

$s^2 = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - c^2t^2 = \frac{2F}{\rho v^2 A} + 1$

So let's see where I was heading with this:

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + v^2t^2 - v^2t^2$

The negative of the drag is the thrust - though in this set up, I am entertaining that the last two terms associated to the space and time follow the anticommutation laws thus

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + [x,ct]^2$

But we can retrieve the ordinary drag by the distribution of the sign again

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

The last term here is known as the spacetime uncertainty principle. As always, I like to teach a little bit on the physics behind this, which will use the standard deviation operators, to understand these kinds of ''strong'' or ''stronger'' uncertainties, we can start off by defining the standard deviation for two commutators.

The standard deviation is:

$\sigma^{2}_{x,t} =\ <x^2t^2> - <x^2t^2>$

$\sigma_{x} = \sqrt{<x^2> - <x^2>}$

$\sigma_{t} = \sqrt{<t^2> - <t^2>}$

As usual the standard deviations involve the expectation value, in which the operators define the commutators:

$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

This gives us the quadradic Robertson uncertainty relation:

$\sigma_A \sigma_B \geq |\frac{1}{2i}[\hat{A},\hat{B}]| = \frac{1}{2}|[\hat{A},\hat{B}]|$

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

under standard deviation the new representation gives:

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - \sigma^{2}_{x,t} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - <x^2t^2> - <x^2t^2>$

Under quantum mechanics, time is not considered an operator, yet clearly, a spacetime commutator must require that time is manifested an observable in four dimensional space as the presence of curvature. Since we are aware that quantum mechanics treats the space dimension as an observable, it does itself have the non-trivial space operators outside of the trivial unitary operator:

$\hat{\ell} = \hat{e}_x x + \hat{e}_y y + \hat{e}_z z$

$<\frac{2F}{\rho v^2 A}>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,p} = \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

So in this case today, we have formulated the space derivatives and the time commutator as observables, something which is unavoidable in relativity when curvature is present, because again, it has to be manifested an observable if we detect a test particle moving in curved space.

Taking the concepts further:

Rearranging the standard deviation, in which we treated space and time as operators,

$<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t}$

$= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The Pressure is obtained from further rearranging:

$P = \frac{F}{A_b} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v\ = \frac{A^{-1}_{b}}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$= \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

Which removes the squared form of the metric and gives us an interpretation of the pressure related to the stress energy tensor.

Just for dimensional purposes, it could also be a related charge under the stress energy tensor:

$\frac{\hbar c}{\lambda} = \frac{1}{2n^2 \hat{x}}(x + y + z - \sigma_{x,t})\ T_{00}$

These are all just different idea's we can investigated, but it would indicate that the principal number as a discrete variable could have applications to various charges, just as it had during my investigation into the Weinberg formula for mass charges.

Rearranging the standard deviation, in which we treated space and time as operators,

$<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t}$

$= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The Pressure is obtained from further rearranging:

$P = \frac{F}{A_b} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v\ = \frac{A^{-1}_{b}}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$= \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

Which removes the dimensions from the metric. Let's move on:

An elementary form of the continuity equation for mass in hydrodynamics is:

$\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2$

Such that a mass flow rate is given as

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

with $\dot{V}$ is the volume flow rate and $\mathbf{j}$ is the mass flux.

This equation though is only true for a surface distribution of the flow. Only dimensionless equations are physically important in physics, which is why Bernoulli's equation is important for fluid mechanics -

$\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + \frac{P}{\rho} = C$

And we will work on this later. But first of all, we can also identify for the work, a dynamic pressure $q = \frac{1}{2} \rho v^2$. The force equation derived will present this quantity:

$F = \frac{A_b B}{Re^2_L} \cdot q\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

This means it related the fluid pressure weighted by the back area with the dynamic pressure as proportional to the Bejan number and inversely related to the Reynolds number as related to the stress energy:

$P = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot q\ = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

While there is so much to explore, the last obvious relationship that popped up to me was the mass flow identification which relates to the back area:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

In which we refresh our memory on the following relationships -

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

If we just concentrate on the second equality of the equation, you will notice my derivations lead to:

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b$

The standard definition of the mass flow follows the limit

$\dot{m} = \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t} = \frac{dm}{dt}$

$F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}$

The total pressure is defined as

$P + q = P_0$

In such a case, $P$ is defined as the static pressure. Recall also that $\mathbf{j}$ is the flux of $q$ and is the dynamic component of the pressure. Retrieving the dynamic component, by subtracting the static component from both sides gives:

$q = P_0 - P$

This term is pretty much identical to the equation which describes an expanding or contracting sphere under the mathematical relationships describing sonoluminscence, which in a previous blog post was derived as:

$\frac{1}{R^2}\frac{P - (P_0 - P(t))}{\rho}(1 - (\frac{T_0}{T})^3) = \frac{\ddot{R}}{R} + \frac{3}{2}(\frac{\dot{R}}{R})^2 + \frac{4 \nu \dot{R}}{V} + \frac{2S}{\rho V} + \frac{e^2}{6 \pi mV} \frac{\dddot{R}}{R} + \frac{1}{V}(\frac{e}{m})\frac{\partial \dot{U}}{\partial R}$

with

$P_{\infty} = P_0 + P(t)$

Just using the solution for the dynamic pressure it explains the force also defined under:

$F = \frac{A_bB}{Re^2_L} \cdot (P_0 - P)$

With the dynamic pressure retrieved as:

$q = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot (P_0 - P)$

The force of drag was associated to the back area, and from previous relationships we introduced from mainstream physics, we ended up being able to describe it through a series of equivalences, which ended up with a metric term (in operator notation) and a stress energy tensor.

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$\int F \cdot dx = \int F \cdot cdt = \int \frac{1}{2} \frac{B}{R^2_e} \cdot dmc^2$

and so the mass flow ~ but in such a case, we must take the speed of light as a velocity term on the force where now we set $c=1$ again

$F \cdot v = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \frac{dm}{dt} = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \dot{m}$

So a few different ways to look at the correcting factor appear. If we recall what the dragging coefficient is first of all

$f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}$

What kind of systems does the drag coefficient apply to? It applies to all kinds of sizes, but in theory it could be applied to an electron since it has been found, not to be a pointlike particle as we have been told in literature for many years, but has instead been carefully measured to be rather a spherical distribution of charge. For spherical systems, the drag coefficient is 0.47.- since we could in theory say that most particles are approximately spherical it wouldn't be too difficult to see that this number would only differ slightly depending on the structure of certain particles - for instance, a proton may appear spherical, but it's distribution inside is not entirely spherical which may mean it variates slightly in principle for different particle systems.

Retrieving the ratio of back to front area's for particular terms in the master equation, also called the wet and back area's, we can get the pressure flow:

$P = \frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot \rho v = \int\ \frac{1}{2A_f}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \mathbf{j} \cdot \frac{A_b}{A_f} = \frac{1}{2n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The first equality then has a strict relationship to the pressure flow related to the drag coefficient

$P = \frac{1}{2}\frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot q = f \cdot q$

where again, we defined the dynamic/total pressure as a simple rearranging of the mainstream formula:

$q = (P_0 - P)$

But that was just the first term, we will notice the drag coefficient is proportional to other terms from the master equation:

$P = f \cdot q = \frac{1}{2}f \cdot \rho v = \frac{1}{2}f \cdot \mathbf{j} = \frac{1}{2n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00}$

Or as twice the pressure flow related to simplified terms on the rest of the equalities:

$2P = f \cdot \rho v = f \cdot \mathbf{j} = \frac{1}{n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00} = \frac{\hat{s}^2}{n^2A_f}\ T_{00}$

These are nice simple relationships, but we can study these drag relationships now under the stress energy tensor more clearly for the future.

An elementary form of the continuity equation for mass in hydrodynamics is:

$\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2$

Such that a mass flow rate is given as

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

Using those relationships, we were able to create a drag force:

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V}$

$= \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m}$

And mass flow applies to the drag force

$F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}$

The continuity equation has many applications - for this case, let's say we rewrite it for the pressure: For instance, from it we can divide through by the energy density to get,

$\frac{P_1}{\rho_1} \cdot A_1 v_1 = \frac{P_2}{\rho_2} \cdot A_2 v_2$

This too is a continuity equation related to the Bernoulli principle, which is stated concisely in a dimensionless form as

$\frac{v^2}{2c^2} + \frac{P_1}{\rho_1} = \frac{v^2_2}{2c^2} + \frac{P_1}{\rho_1}$

To solve the velocities we obtain:

$v_1 = A_2 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}$

$v_2 = A_1 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}$

We can also entertain

$2 \dot{p} = v^2(A^2_1 \rho - A^2_2 \rho) = \rho v^2(A^2_1 - A^2_2)$

Before we go into an application involving the Friedmann equation, which is derived directly from fluid mechanics, we require to first investigate 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, often seen as a correction term to Einstein's formula - we will get to this soon because it will give us a reason to why a universe expands rather than contacts in a mechanical way.

Ok so let;s get into the cosmological fluid equation, the Friedmann equation. The most simplest form of the Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{4 \pi G}{3c^2}\rho$

One correction I made I applied in the past, inspired through the work of Motz and Kraft was that the equation need not be taken as a conservation equation - this is after all at odds with relativity in which energy is ill-defined under the Noether theorem, the distribution of a time variable indicates we get

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{3c^2}\dot{\rho}$

Motz' quite rightly stated that the equation need not be conserved, just as this previous equation indicates, as he said it was an ''unfounded assumption by Friedmann to conserve the energy in such a way) - and general relativity hints at this since not only is energy ill-defined, it is ill-defined because time itself is ill-defined (as it states there is no cosmological constant) under the Wheeler de Witt equation, in which diffeomorphism invariance is not a true time evolution of the theory. Further more, the Wheeler de Witt equation is also often taken as a statement of a conserved universe in which no changes happen inside of it, the consequence of it having no time derivative. But logic should be indicating to us that  many phase transitions have occurred and so the general statement should be taken with a pinch of salt.

Using the non-conserved form of the Friedmann equation, we can introduce the so-called continuity equation, but this is a misnomer if the derivative on the density term implies a change inside of the universe in a non-conserved way. The misnamed continuity equation takes the form:

$\dot{\rho}= -3H(\rho + P)$

where we take all terms here as the energy density and pressure. $H$ is the usual Hubble constant, but we should read this that density is not fixed, but dynamic component. From previous posts I have shown that the total pressure would imply a correction to this relativity term:

$\dot{\rho}= -3H[\rho + (P_0 - P)] = -3H[\rho + q]$

You can plug this into this into the non-conserved form of the Friedmann equation as

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{c^2}H[\rho + (P_0 - P)]$

$= \frac{4 \pi G}{c^2}H[\rho - (P_0 - P)] = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + \dot{q}]$

$\kappa = \frac{4 \pi G}{3}$

Also an important quantity, the density parameter is equivalent to

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

Later we will come back to this as it will play a role in the Bernoulli's principle. We must also consider for this following work the following work the presence of a second correction term for the total pressure of the fluid, as we had defined earlier as

$q = (P_0 - P)$

Which is a correction term, for the ''continuity equation'' a misnomer if a third derivative actually implies a diabatic, non-conserved solution for the Friedmann equation:

$\dot{\rho}= -3H(\rho + (P_0 - P)) = 3H((\rho - (P_0 - P)) = - 3H((\rho - (P_0 - P)) = 3(\dot{\rho} - \dot{q})$

The last set of terms in this equation has also featured in investigations into practically similar expanding and contracting models for sonoluminscence models - this is no accident of course since both those solutions are also derived from fluid mechanics.Plugging the main corrections into the non-conserved form of the Friedmann equation yields:

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = \frac{4 \pi G H}{c^2}(\rho - (P_0 - P))$

$= \frac{4 \pi G }{c^2}(\dot{\rho} - \dot{q})$

I have made previous corrections terms, such as introduction a zero point energy term for a pre-big bang state that existded at very near absolute zero to attempt to formulate a semi-classical model. The Friedmann equation, even in principle, cannot predict why spacetime should be flat - let me quickly explain. The density parameter, $\Omega$, is defined as the ratio of the actual (or observed) density $\rho$ to the critical density$\rho_c$ of the Friedmann universe. The relation between the actual density and the critical density determines the overall geometry of the universe; when they are equal, the geometry of the universe is flat (Euclidean). Appealing to authority, Susskind himself and Lisi Garret have expressed their opinions, that the universe is not flat - a way to articulate why this is , is because the universe would have to be infinitely large and this isn't really how we think about the universe in a scientific way, because it would take the universe an infinite steps to do so - but steps involve numbers and infinities are not numbers by definition. At any moment in time, no matter what time, the universe could be measured being finite in value always.

In a similar notion, the only way a black hole can become flat, would that the radius

$R \rightarrow 0$

Meaning the curvature of an expanding sphere must disappear

$K \rightarrow 0$

But the universe can only approach these values, but never reach it perfectly.

As wiki states itself: ''In earlier models, which did not include a https://en.wikipedia...ogical_constant term, critical density was initially defined as the watershed point between an expanding and a contracting Universe. To date, the critical density is estimated to be approximately five atoms (of https://en.wikipedia.../wiki/Monatomic https://en.wikipedia.../wiki/Hydrogen) per cubic metre, whereas the average density of https://en.wikipedia...Baryonic_matter in the Universe is believed to be 0.2–0.25 atoms per cubic meter.''

I have become a hard disbeliever against dark matter, too much mounting evidence against it over the years - I could go into loads of reasons why, but I will spare the reader. I also believe that dark energy has been misunderstood, but we will get to that later. The entire basis of the density parameter to the actual observed ratio of matter in the universe, is too many magnitudes off, which should have been taken seriously that the Friedmann equation may itself be flawed fundamentally.

Next we will get into Bernoulli's principle and how the work above will come together to explain a source of the cosmological constant and how it applies to dark energy due to a weakening of gravity in the universe - the cosmological constant has often been believed that it is not really a constant, but there are some intriguing reasons why we may have to consider new alternatives to explain this in a more rationally-mechanical way due to pressure, perhaps thermodynamic in nature.This time around i plan to write the correct formula, in whcih we retain that dark energy remains constant, but the acceleration entertained by a weakening of gravity over galactic scales, only giving the impression dark energy is soley due to it - basically, the weakening of gravity only appears to make the constant change, when it is not. The gravitational binding of the universe would arise in my picture as an effect of gravitation become weaker as it gets larger and so only appears to accelerates.

There is a density parameter which can be used in Bernoulli's principle to apply for cosmological applications. But before
we jump into that, the best way to see how Bernoulli's principle can explain the cosmological constant is literally by
writing the constant as a constant energy density in the style of conservative forces:

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

Now, let's not move to quickly, I want to now define the so-called ''density parameter'' for the Friedmann equation ~

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

And the relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^2}(\rho + P)$

So we have a number of things we can do here; first, let's plug in the density parameter and then divide through by the energy density:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\frac{P}{\rho}$

So with this equation and the other what do we do now?

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

We now plug in the co-called ''critical density'' for both formula's - but ''critical density'' is a misnomer in the absence of dark matter and the proper interpretation of dark energy:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^2}\Omega = \frac{8 \pi G}{3c^2}\frac{8 \pi G \rho}{3H^2}$

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\Omega)E = (\frac{v}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{8 \pi G \rho}{3H^2})E = \mathbf{C}_{vac}$

I will be introducing once again the Leonard extension of the Einstein equivalence to explain why a universe expands rather than contracts - ok, so we spoke about $C$ being a constant, today we know it is related to energy and the main main reason why this Freidmann expansion melding into Bernoulli's principle was not a ''an assumption'' as was challenged by someone who followed the work, because he seemed to be unaware that we work with fluid dynamics when working with an expanding universe, which relies on different pressures through various complicated parameters. In other words, the two have stronger relationships to each other than most would like to agree on, or have been unaware of.

To present the laws spoken in previous posts, a general application from the last equation would be to distribute an energy, such that the constant on the right hand side, is a constant of energy. We will leave out the right hand side until I can try and simplify it further, the real equation we should concentrate will be of the form:

$(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

where now C is no longer ''just a constant'' but is define under the units of energy defined through the vacuum contribution.

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.

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^2}\Omega = \frac{8 \pi G}{3c^2}\frac{8 \pi G \rho}{3H^2}$

It's also interesting we can find the classical upper inverse limit of the gravitational force (but with an additional gravitational constant term), which implies an inverse of the stress energy tensor on the left hand side:

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

$T_{00}^{-1}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

from all this we find a different form of the Friedmann equation:

$\frac{3H^2}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}8 \pi G \rho$

Let's go through this thoroughly

The relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}(\rho + P)$

Let's divide through by the relative density and pressure

$\frac{1}{\rho + P}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}$

But we need not take such a relativistic case, because this will drop out naturally when we take the non-conserved form of the Friedmann equation. We shall get to that soon.

I also entertained

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega$

Where we remind ourselves that the misnamed ''critical density'' is:

$\Omega = \frac{\rho}{P} = \frac{8 \pi G}{3H^2} \rho$

To retrieve the upper limit of the classical gravitational force.

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

The continuity equation is

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

and the respective energy density is

$\dot{\rho}c^2 = -H(\rho c^2 + P)$

So, we've been through this a few times, the non-conserved form of the Friedmann equation requires one extra derivative which removes one such factor of the Hubble ''constant'' (or in the last term as explained not long ago, it gives us the continuity which is strictly defined under the density and pressure parameters) - we also distribute the factor of three from the denominator:

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \dot{\rho}$

This gives by plugging in the continuity:

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{2c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \frac{\dot{R}}{R}(\rho + P)$

Which neatly gives a new interpretation/form of the expansion of a universe, in diabatic from. Distributing through an inverse factor of $8 \pi G$ to neaten it up without over-simplifying by combining components:

$\frac{3}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{1}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{ \rho}{H} = \frac{8 \pi G}{3c^4}\frac{1}{H^2} \frac{\dot{R}}{R}(\rho + P)$

And this yields

$\frac{3H^2}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4} \frac{\dot{R}}{R}(\rho + P)$

Let's talk about the Rohrlich correction now.

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 $\Delta p$ we then have

$\Delta p =\frac{v}{2c^2}E$

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

$2\Delta p=\frac{v}{c^2}E$

$2\Delta p v =\frac{v^2}{c^2}E$

In which $\frac{v^2}{c^2}$ is the Doppler shift, more appropriately given in the squared form, rather than the form Rohrlich had given, because it allows us to write the metric in a relativistic way.

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

$p′ = mv−2\Delta p = (m − \frac{E}{c^2})v$

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. To understand this pressure difference, we now include Leonards correction to the mass energy equivalence in terms of the first term in the paranthesis: To do that, we remind ourselves first of the formula we want

$\Delta p v =\frac{v^2}{2c^2}E$

and then we plug in:

$\Delta p v + (\frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

The physics of the first term, will not only explain why something expands rather than contracts, but it will also provide under its own definition, a redshift term - meaning things are moving outwards, not inwards, (ie. the universe is expanding, not contracting due to pressure differences) and because we have used a conservative form, it means it does not rely on the gravitational field soley.

Edited by Dubbelosix, 20 June 2019 - 06:11 PM.

### #2 VictorMedvil

VictorMedvil

The Human Shadow

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Posted 20 June 2019 - 04:12 PM

Quantization of the Black Hole Particle

To derive a quantization of the black hole, in terms of discrete quantum mechanical processes, I defined the Rydberg constant in terms of the gravitational coupling constant we get:

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

Even though the Rydberg constant was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr). In which case we may hypothesize energy levels:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Plugging in the last expressions we get an energy equation:

$\Delta E_G = \frac{n\hbar c}{\Delta \lambda} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})$

The relativistic gamma appears from the definition of the deBroglie wavelength. The ultimate goal of creating this equation, was to explore an idea I can trace back to Lloyd Motz suggesting a stable black hole. Black holes, even the microscopic kind, cannot be mathematically ruled out from physics (but later I will show in a logical sense why the would be unstable) - and if the work of Hawking is to be taken seriously, including analysis provided by Motz, then a black hole system really is deduced from a discrete set of quantum processes - those discrete processes always lead to an increase in the entropy of a system like a black hole.

The earliest model of a black hole stated that the entropy of a black hole was exactly zero - Hawking challenged this and stated if it has an entropy, then a black hole must possess a temperature and later we came to understand it as a thermal property of the black body radiation of the black hole. It may still be true, at least for the ground state black hole, that entropy may be essentially constant.

So how can we have an increasing entropy but a system with an infinite lifetime?

Something needs to give. Entropy is a measure of disorder, and if a system is infinitely stable, then the disorder remains a constant. It could be that the black hole particle at near zero temperatures exhibits a behaviour like a zeno effect. The zeno effect is when an atom is incapable of giving off energy suspended in an otherwise, infinite animation and it will remain in a ground state because its wave function is incapable of evolving from the ground state.

It is therefore similar to how an atom, ready to give up energy can be suspended infinitely in time - and the reason why it cannot give up radiation is essentially the same for the micro black hole, since the zeno effect is all about keeping an atom in the ground state. To put in a summary for clarity I have came to some conclusions: The main one being that the entropy of a stable black hole would immediately render the black hole particle as a system at absolute zero! The reason why is because if a black hole particle did not have an entropy, by first principles, it consequently cannot give up any radiation (consequence of the third law of thermodynamics) and as far as we know, systems can never actually reach absolute zero temperatures, there is reasonable argument here that the black hole particle must evaporate. In fact, I wondered immediately if there are applications of this with the micro black hole and maybe in regards to its energy levels. We start with the equation

$S = k_B \frac{Gm^2}{\hbar c}$

Where the entropy has taken on the dimensions of the Boltzmann Constant. If you plug in the mass for a black hole:

$m = \frac{c^2R}{2G}$

You will retrieve the Bekenstein entropy. It is often said that entropy has dimensions of the Boltzmann constant but it's entirely plausible, even with better arguments for, a dimensionless case of entropy in the form

$\mathbf{S} = \frac{S}{k} = \frac{Gm^2}{\hbar c}$

In which case, it is possible to identify the entropy in a dimensionless form

$\mathbf{S} = \frac{\hbar c}{k_BT}\frac{p}{4 \pi \hbar} = \frac{\hbar c}{k_BT}\frac{\sqrt{2mE}}{4 \pi \hbar} = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

In which the wave number is identified with the deBroglie wavelength

$k \equiv \frac{p}{\hbar} = \frac{\sqrt{2mE}}{\hbar}$

Because of the equivalence of the wave number the definition of the deBroglie wavelength. The SI value for the Boltzmann constant is given in units of the Hartree energy divided by the Kelvin

$\frac{E}{K} = k_B = 3.1668114(29)\times 10^{−6}$

In which the Hartree energy is simply

$E = 2\mathbf{R}\hbar c$

and so we can write it with temperature T which is also measured in Kelvin,

$S = k_B = \frac{E}{T} = \mathbf{R}\frac{\hbar c}{T}$

This can be written in such a way to introduce those quantized atomic levels ~ by making use of the formula

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

giving

$k_B = \frac{1}{T} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1T} - \frac{Gm^2}{n^2_2T})$

In which $\mathbf{R}$ is the usual Rydberg constant. The Boltzmann constant can also be expressed entirely in terms of Planck units ~

$S = \frac{m_PL^2_P}{k_BT_P t^2_P}$

$k_B = \frac{\sqrt{\frac{\hbar c}{G}} \frac{\hbar G}{c^3}}{\sqrt{\frac{\hbar c^5}{Gk^2_B}} \frac{\hbar G}{c^5}}$

The Transition of Conducting Black Holes

Another theory in literature involves a theoretical suggestion which asserts the surface of a black hole [may] be considered as a conducting surface; the argument goes like this:

‘’ Introducing an effective thickness \Delta r of the membrane, and assuming the same resistivity $R_H$ within $\Delta r$, one can relate total resistance of the black hole with the conductivity of the membrane using a model of two concentric spheres with radii $a_1 = r_s$, $a_2 = r_s + \Delta r$ where the spherical shell is filled by homogeneous conducting matter is given by

$R_H = \frac{1}{4 \pi \sigma}(\frac{1}{r_1} - \frac{1}{r_2})$

Where $\sigma$ is the conductivity. ‘’

Early on it struck me how similar this equation is, compared to our transition equation using the Rydberg constant:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Fortunately for both equations, the inverse length features and for conductivity, (which is the inverse of electrical resistivity), the resistance of a system increases with the length of the system. This must mean that in each transition from one energy to another, the conductance must weaken, as shown:

$R_H = \frac{1}{\Delta( \sigma \lambda)} = \mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

From the last equation, you can calculate conductance G as

$G = \rho (\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2}) = \frac{\mathbf{E}}{J}(\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2})$

Where \rho is the resistance and is related to conductance by $\frac{1}{\sigma}$ and $\mathbf{E}$ is the electric field where $J$ is the current. Also it is good to note, the shape operator for a sphere is mathematically described under the Weingarten equations. However, literature hints that the equation does not depend on the geometry of the system, since the resistivity and conductivity are proportionality constants; in such a case it is said it depends only on the material the system is made of but not the geometry, which robs any importance with viewing it under a shape operator. The black hole area is defined by

$A_{BH} = \frac{G^2m^2}{c^4}$

And the angular momentum per unit area is

$\mathbf{J} = \frac{J}{A_{BH}} = \frac{c^3}{G}$

Which enters the resistivity equation like;

$R_H = \frac{c^4}{G^2m^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \frac{\mathbf{J} c}{Gm^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2})= \alpha_G\mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

We have retrieved the gravitational charge (squared) in the denominator. The ratio

$\alpha_G = \frac{Jc}{Gm^2} \approx 1$

can be interpreted as a fine structure constant, as following the same idea but reached a different derivation of the same argument from Arun and Sivaram from their paper ‘’some unique constants associated to extremal black holes.’’

The equation can further be distributed with a factor of \hbar c, similar to the Rydberg formula to find the energy. we get

$\hbar c\ R_H = \frac{c^4}{G}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \alpha_G(\frac{E_1}{\sigma n^2_1} - \frac{E_2}{\sigma n^2_2})$

With this time, a classical gravitational upper limit of the force $\frac{c^4}{G}$. According to Arun and Sivaram for a black hole, the angular momentum and the square of the charge is related to the speed of light via ~

$\frac{J}{Q^2} = \frac{1}{c}$

It has crossed my mind whether it would be related more fundamentally to the gravitational aether which is defined in this work by the variable parameters $\sqrt{\epsilon_G \mu_G} = \frac{1}{c}$. Either way it was noted by the astrophysicists that it is remarkably similar to the Von Klitzig constant $\frac{\hbar}{e^2}$ or the Josephson constant in superconductors $\frac{e^2}{\hbar}$ - resistivity decreases in the case of superconductors, while it increases for metals as they increase temperatures.

We deducted not long ago that for a black hole, the conductance decreases (increase in resistivity) for a black hole as it gets smaller, this will concur nicely with the later investigation, because the equations so far tell us that in the ground state, conductance must be zero (to ensure shielding from environment) and will complement the idea of a ‘perfect insulator’ as the smallest possible radiator. While it remains curious science may predict a perfect insulator, the physics systematically argues against it due to the third law.

Differences Between The Holeum Model and Mine

Holeums were proposed by L. K. Chavda and Abhijit Chavda in 2002 and is suggested to be a form of cold ‘’dark’’ matter. The system itself is not a black hole and they consequently radiate gravitational waves as opposed to the usual electromagnetic Hawking radiation - later we will see why I do not believe a Planck particle (black hole particle) is stable at all, concerning contradictions that arise from the laws of thermodynamics, never mind two bound stable black holes.

The suggested binding energy of a bound black hole particle-pair is:

$E_n = - \frac{mc^2 \alpha_G^2}{4n^2}$

I came across the Holeum model after I constructed my approach to the quantization transition formula. I began from arguments concerning the Rydberg constant and asserted this would be a true analogous model to a ground state hydrogen atom in which the radiation inside the black hole attains its lowest quantum bound state.

Interestingly, it is claimed their model too is analogous to the ground state hydrogen atom, but notice, the models are totally different fundamentally. My approach is direct towards the treatment of a single black hole particle, not a bound pair. Notice also they used the gravitational fine structure, which was of similar contention to my own derivation that came from a modification of the Rydberg constant

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

or as the gravitational Rydberg-Hartree energy ~

$\hbar c\ \mathbf{R} = \frac{\hbar c\alpha_G^2}{4 \pi \lambda_0} = \frac{Gm^2}{4 \pi \lambda_0} = \frac{Gm^3c}{4 \pi \hbar} = Gm^2\frac{p}{4 \pi \hbar}$

• this is logical for a true analogue model since the constant is fundamental to the Bohr model derivation of the hydrogen atom it is a statement and Bohr maintained the derivation itself, is fundamental to many systems.

If for instance, we can prove later it doesn’t make sense to talk about one stable black hole particle, it would certainly not make any sense to think any stable bound black holes exist in nature.

An Argument For Temperature Variations With Black Hole Acceleration

A postulate in this work is that black holes when accelerated will obey their own radiation power law similar to an accelerated charge following the Larmor radiation law. It’s not an ad hoc hypothesis, since a micro black hole with a charge would have to obey the weak equivalence principle and give up radiation in a gravitational field in the form of the usual Hawking process.

In fact, early theories attempted to explain the electron as a black hole, and they must have crossed the question of what Larmor radiation would be: And if an electron had been a black hole, the radiation from the accelerated charged black hole would have had to expressed itself through the usual Hawking emission process.

A charge in a gravitational field should give off an electromagnetic radiation. This can be formulated in the following way:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{\sqrt{g_{tt}}(r)}{\sqrt{g_{tt}}(s)})}$

The metric $g_{tt}$ is made up of terms of $1 + \frac{\phi}{c^2}$. Higher terms then of the field are calculated like so:

$1 + \phi_1 + \phi_2 - \frac{1}{2}\phi^2_1 - \phi_1\phi_2... higher\ terms$

To first order it is just the difference formula

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

However, employing a connection made by Motz, there is a correction to the denominator of the first equation,

which he notes should be squared because there is one such term to account for in the metric ds. In our case it is simply the removal of the square root signs. A second difference between the approach I took and the one they have taken is that their denominator consists of only one such term of the metric $(1 - \frac{R}{r})$ whereas mine considered the ratio of the two signals in a relativistic way that satisfies the red shift.

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

His interpretation of the squared denominator ensures that the radiation increases as something falls towards a singularity. Of course, modern approaches are to involve situations which don't involve singularities at all, as we mentioned before.

The Motz-Kraft model is for the calculation of the luminosity from charges accelerating in gravitational fields. Quasars are essentially black holes with a dense electrically charged material around it and they produce a large luminosity for these reasons. In our case, we are saying however, the weak equivalence applies to all kinds of systems with electric charge: Including the black hole itself. In such a case, the only way a black hole could radiate is through Hawking radiation suggesting an analogue case of the Larmor radiation ~ and did in fact satisfy such a case with an uncanny analogue to the classical electron radiation par a factor of $\frac{2}{3}$ which is only an adjustable parameter - the radiation (I derived) given up by a black hole is:

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

I obtained this nice simple solution from the known inequality satisfying a black hole with mass, charge and angular momentum (in natural units):

$Q^2 + (\frac{J}{m})^2 \leq m^2$

Notice this last term in the inequality is formally identical to the classical equation describing the radiation for an electron, with the exceptions that neither the mass or radius are associated to the electron but instead the black hole:

$P = \frac{2}{3}\frac{a^2m_er_e}{c}$

Since this too describes the power emitted it would not be wrong to set it in the same physics, in a gravitational field - doing so is childsplay, I get while adopting the coefficient for a true analogue theory:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3}\frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2}$

It is also true that the luminosity is a simple calculation which measures the power emitted by an astrophysical object, but we were looking for a more general case above for any charged object. The relationships for luminosity are briefly covered ~

$L = \frac{f}{A(1 + z)^2} = \frac{\sigma T^4}{4 \pi R^2(1 + z)^2}$

You can calculate the two different luminosities as

$\frac{L_1}{L_2} = \frac{(\frac{\sigma T^4_1}{4 \pi R^2_1})}{(\frac{\sigma T^4_2}{4 \pi R^2_2})}= \frac{T^4_1 R^2_1}{T^4_2R^2_2}$

So an alternative formula to calculate power is suggested as:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3} \frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2} = \frac{2}{3} \frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}(\frac{dp}{dt})^2$

The last term arises from:

$P = \frac{2}{3}\frac{Q^2}{m^2c^3} (\frac{dp}{dt})^2$

And can be found by replacing the acceleration in our equation for $(\frac{1}{m}\frac{dp}{dt})^2$ - this is Lorentz invariant but it isn't generally relativistic.

The true generalized formula is

$P = -\frac{2}{3} \frac{Q^2}{m^2c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Since there are less dynamics in the luminosity relationship, the deduction from our equation (which involves the black hole mass, electric charge and angular momentum) is that a charged black hole will radiate more energy than one at rest: This partially makes sense of my stipulation that a moving black hole will appear hotter than one at rest, though the presence of red shift would mean it is also direction dependent. In units of $G = c = 1$ our equation to first order becomes:

$P = \frac{2}{3}\frac{a^2}{(1 - \Delta \phi)^2}(Q + \frac{J}{m})^2 \leq \frac{2}{3} mr_g\frac{a^2}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2} = \frac{2}{3} \frac{r_g}{m}\frac{1}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2}(\frac{dp}{dt})^2$

with

$\Delta \phi = \phi_1 - \phi_2$

and

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

Let’s just say a bit about the standard equations you might encounter which also attempt to calculate the power emitted by black holes. We know now that accelerated matter aroud the black hole contributes significantly to the light that comes from superluminous objects and so, this has offered a novel retake of the theory that Motz suggested first and later collaborated with Kraft.

ref: https://quantizedblackhole.quora.com/

The phase velocity of a particle or wave in a medium is

$v = \frac{c}{n}$

For light, in literature, the refractive index $n$ decreases with an increasing wavelength,

$1 < n(\lambda_{red}) < n(\lambda_{yellow}) < n(\lambda_{blue})$

The alternative statement for this is

$\frac{dn}{d\lambda} < 0$

In such a case, the dispersion is said to be normal.

https://en.wikipedia...ersion_(optics)

This however, takes us to the Fresnel drag coefficient, that is, the index of refraction is

$w_{+} = \frac{c}{n} + v$

and in the other direction

$w_{-} = \frac{c}{n} - v$

From it, Frizeau found that

$w_{+} = \frac{c}{n} + v(1 - \frac{1}{n^2})$

If we replace the concept this was used in first context of light traveling against the flow of a fluid, with gravity, then the speed of the light will be seen travelling slower with the flow of the fluid. This brings us back to the well-known concepts now, that light is not truly a constant when inside a gravitational field.

But to do this, we require dragging coefficient $f$ which involves an experiment by Arago in which that an aether drags light propagating through it with only a fraction of the mediums speed. Gravitational aether theories are not very well-explored, but if we accept that light is not a constant in general truth when gravitational fields are involved, then the dragging coefficient should be no different to the one proposed

$f = 1 - \frac{1}{n^2}$

Then in 1895, Lorentz predicted the existence of an extra term, which brings us back to the dispersion formula, which is why we mentioned it in the beginning of this post:

$w_{+} = \frac{c}{n} + v(1 - \frac{1}{n^2} - \frac{\lambda}{n} \frac{dn}{d\lambda})$

In particular, this term

$\frac{dn}{d\lambda}$

in which we had noted the relationship

$\frac{dn}{d\lambda} < 0$

Now, let's say a little about the gravitational aether, something which Einstein supported strongly, but did not appear to realize that the constancy of the speed of light in the medium of space is [not] generally constant like he believed his theory predicted.

A Gravitational Aether

Excellent arguments exist now for the existence of the gravitational permittivity and permeability with the discovery of gravitational waves. The constancy of the speed of light only holds in a vacuum - but the density of gravitation varies between celestial objects and therefore the speed of light does technically vary.

In fact, authors Masanori Sato and Hiroki Sato in their paper ‘’Gravitational wave derived from fluid mechanics applied on the permittivity and the permeability of free space’’ suggests that gravitational waves are simply fluctuations of the medium, which appears as the product of the permittivity of free space and the permeability of free space. That is, the gravitational wave is an acoustic wave in the medium - the proposal shows how the phase velocity of the fluctuation relates to the speed of light

$c = \frac{1}{\sqrt{\epsilon_G \mu_G}}$

The model has some interesting consequences, first being that permittivity and permeability are allowed to vary. A second is that the speed of light is variable in gravitational fields. Another interesting property is that while both Newtonian mechanics and Einstein’s relativity theories predict the confinement of light by gravity, neither theory defines the escape velocity or the Schwarzschild radius; in fact, the actual speed of light can only approach zero but never reach it - so in effect light is allowed to escape from a black hole.

Let’s be clear about something - I do not believe that the thickness of space (the medium) is an aether made from any particle. In fact there cannot be any motion associated to this aether because it would violate the first principles of relativity. In fact you can argue as I have already done, that any true quantization of gravity would be at odds concerning how we actually think about the roles of pseudo forces.

Many experiments have been performed to measure the value of the Newtonian $G$ but has come up with varied results and up until this year another measurement has cast a shadow over settling why we keep measuring different values for the constant - it may not be the case the constant does change, but what it may have to do with is the gravitational acceleration which is proportional to the constant. Since in this aether theory I have chosen, both permittivity and permeability will depend on $G$ ~

$\frac{1}{\epsilon_G} = 4 \pi G$

$\frac{1}{\mu_G} = \frac{c^2}{4 \pi G}$

This leaves open a question of whether the deviations in the value of G has something to do with variations spacetime permittivity and permeability. This particular theory of the aether, as a dynamical ''thickness'' of space due to varying gravitational density, the refractive index for radiation is proportional to $\sqrt{\epsilon_G \mu_G}$ (the gravitational permittivity and permeaility) and is represented as:

$n = \sqrt{\frac{\epsilon_G \mu_G}{\epsilon_0 \mu_0}}$

A high refractive index for the equation

$\frac{1}{\sqrt{\epsilon_G \mu_G}}$

causes a low speed of light (such as found round strong gravitational fields of black holes). It has been argued in literature that the refractive index is more intuitive than curvature; this suggestion is probably quite true, since curvature is the presence of a dynamic metric but we know not what causes this ''dynamic feature'' other through the presence of matter - which is well-known to tell spacetime how to curve, but still doesn't explain why the dynamic phenomenon exists. In a sense, the gravitational explanation for a refractive index supposes a type of mechanical explanation to curvature.

The formula I created under the same analysis Bohr used to model a Hydrogen atom to describe discrete transitions of a black hole losing a mass was:

$S = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

The Fresnel dragging coefficient is:

$f = 1 - \frac{1}{n^2}$

We can rearrange this in a very simple way: first we add the $\frac{1}{n^2}$

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

You could simplify further, for instance, this would give

$f + \frac{1}{n^2} = 1$

and you could even find the levels in a transition process:

$\frac{1}{n^2} = 1 - f$

And while this is interesting we explore not only this avenue, but we will first concentrate on the first equation that we featured:

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

Can you see where this equation relates to the investigation of the transition equation? First I want to solve for the quantization levels:

$\frac{1}{n^2} + \frac{1}{n^2} = 1 - f - \frac{1}{n^2}$

So how do you solve that? We will simplify the terms to make it easier and use the form

$f + \frac{1}{n^2} = 1 - \frac{1}{n^2} + \frac{1}{n^2}$

where the second term on the left hand side is

$f + a$

and the right hand side simplified as

$1 - b$

So that we have:

$f + a = 1 - b$

add the notation of b on both sides

$f + a + b = 1 - b + b$

simplify

$f + a + b = 1$

subtracting $f + a$ from both sides gives

$f + a + b - (f + a) = 1 - (f + a)$

which simplifies to

$b = 1 - f - a$

Plugging back in the definitions we get

$\frac{1}{n^2} + \frac{1}{n^2}= 1 - f - \frac{1}{n^2}$

Now, going back to the transition equation we will notice that it involves a negative sign on the left hand side, this is a simple procedure of distributing a negative sign:

$\frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

$- ( \frac{1}{n^2} + \frac{1}{n^2}) = - 1 - f - \frac{1}{n^2}$

giving us

$\frac{1}{n^2_1} - \frac{1}{n^2_2}= 1 + f - \frac{1}{n^2}$

The notation of $n$ is the index of refraction but in some loose way, I expect it can play the same role as the principal quantum number. There are loads of ways to continue from here, for instance, we might consider the addition of $\frac{1}{n^2}$ as a correction to the transition equation which may be similar in nature to a dispersion relation predicted by Lorentz:

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} = 1 + f$

You could from here be solving for the dragging coefficient,

$f = \frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} - 1$

But the direction I wish to take is simply one in which we retain the terms in the equation and rewrite the third term on the left in terms of the dispersion, again, first predicted by Lorentz, but formulated here in a completely new way::

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

There will be more to say on this matter as I take the equation into the transition formula I created for black holes.

Reference:

https://en.wikipedia...zeau_experiment

Again, to understand how the development of the transition equation involves the dragging coefficient, we will recognize that index of refraction must be playing a similar role to the principal quantum umber under the dimensionless entropy formula for the transition of any universal fundamental system applied to quantum or semi-classical systems:

$S = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

$\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

In which the Rydberg formula describes such transition levels

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

So by distribution we get:

$\mathbf{R}(\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}) = \frac{1 + f}{\Delta \lambda}$

Let's flip the equation for clarity

$\frac{1 + f}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Distribution of the gravitational charge $e^2 = n\hbar c = Gm^2$ will give:

$n\hbar c\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2} + e^2\frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Now we divide through by the thermodynamic energy which gives back an equation similar to the original dimensionless entropy

$\frac{n\hbar c}{k_BT}\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1 k_BT} - \frac{Gm^2}{n^2_2k_BT} + \frac{e^2}{k_BT} \cdot \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

The dimensionless drag coefficient is often described under the terms:

$f = \frac{2F}{\rho v^2 A}$

Where the large $F$ is the dragging force, $\rho$ is our case would be the density of the gravitational field and $v^2$ is the speed of the object relative to the fluid.

Before we proceed, as I always like to do, is define the formula's for clarity for further investigation:

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

$f = \frac{2F}{\rho v^2 A}$

Plugging it in simply gives us the formula, as we flip the equation

$1 + \frac{2F}{\rho v^2 A} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

This means we also have an alternative formula to investigate at a later date:

$1 + \frac{A_b}{A_f} \frac{B}{Re^2_L} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation itself is simply

$F = \frac{1}{2} \rho v^2 f A$

And has been interpreted as a statement that the drag force on [any] object is proportional to the density of the fluid medium and porportional to the relative flow speed.

Where $Re^2_L$ is the Reynolds number related to the fluid path length. The Reynolds number is defined specifically as:

$R_e = \frac{\rho v L}{\mu} = \frac{v L}{\nu}$

We have to define the notation here, where $v$, $L$ is a linear dimension which in the case of gravity, we would expect it to be gravielectromagnetic path which is by definition the linearized gravity.$\mu$ is the dynamic viscosity, in which interpreted within gravitational physics, defines the ''thickness of the gravitational field,'' or ''stickiness of the gravitational aether. Finally $\nu$ is the kinematic viscosity of the medium itself - in a simple way of visualizing viscosity, the gravitational field is a fluid and in previous investigations for a primordial spin of the universe, is much like how you have a dense object submerged in water - what happens when you include spin in the fluid, it drags the objects inside of the medium.

References:

https://en.wikipedia...Reynolds_number

https://en.wikipedia...rag_coefficient

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} = 1 + f$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + f = 1$

Plug in:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$f = \frac{2F}{\rho v^2 A}$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2F}{\rho v^2 A} = 1$

Plug in the definition of the standard drag,

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A = 1$

Or as

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A$

Simplify by removing constants:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A$

Since generic dimensions of the density is simply a mass divided by the volume, we can obtain, after distributing the velocity squared term to obtain the energy density

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{1}{A} \cdot \rho v^2 f A$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We may even write our constructed equation using the same front and back area terms:

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \cdot \rho v^2 f$

Since the density is related to the stress energy tensor through $T^{00} = \rho v^2$ we can re-write it as:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We can define the stress energy, as either gravitationally related, or by electromagnetism. In some sense, gravity and electromagnetism may in certain theories be related, maybe even to the point they are the same thing… but these are ideas to take up at a later date. Firstly, the density of electromagnetic field is given through the stress energy as:

$T^{00} = \frac{1}{c^2}(\frac{1}{2}\epsilon_0 \mathbf{E}^2 + \frac{1}{2 \mu_0}\mathbf{B}^2)$

Let’s take a few steps back, we had:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \frac{A_b}{A_f} \frac{B}{Re^2_L}\ T^{00}$

This gives us the squared quantity:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

And now we simplify and then solve for the stress energy:

$mv^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

$mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - T^{00}$

And let’s use a sign change to finish off for now.

$- mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = T^{00}$

And so we don’t end up with any idea’s of a negative mass we should entertain:

$\rho v^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2}) = T^{00}$

As noticed in a conversation online with a chemist under the moniker ''exchemist'' he had noticed an implication:

''The first line implies $f = \frac{1}{n^2} -1$

The second line says $\frac{3}{n^2} + f = 1$, so substituting for $f$, we have \frac{3}{n^2} + \frac{1}{n^2} -1 = 1[/math]

So rearranging, we have $\frac{4}{n^2}= 2$; so $\frac{2}{n^2} = 1$, hence $n^2 = 2$ and so $n = \sqrt{2}$.

Now we can solve for $f$, using the first line: $f = \frac{1}{2} - 1 = -\frac{1}{2}$.'

or $n = \sqrt{2} and [math]f = -\frac{1}{2}$'

Moving on, the way forward intended was to find three parts, loosely related to the dimensions of space itself - but by doing so, I wish to state that time is an operator and also an observable - it has always been my contention that in SR time is dependent on an observer, in GR time is an observable or it would not manifest curvature in space - and in total contradiction, in quantum mechanics, it says it is neither an operator or an observable which may present itself why there is so much confusion in uniting the theory of relativity and quantum mechanics.

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A = -f$

In  this simple case, we ignore the fancy things we can do, we just accept a simple case of removing the dimensions from the equation:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = -f$

Another simple thing to do is just take ($-1$) from both sides and we get

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} - 1 = f + 1$

If we take the first three terms of the spatial dimensions, than its last part has to be associated to the timelike dimension:

$\frac{x}{n^2} + \frac{y}{n^2} + \frac{z}{n^2} - ct = f + 1$

Notice the I have derived this in which the principal quantum numbers appear to be attached specifically to the spatial dimensions - and time has been placed in not too dissimilar to a fudge factor. Since space and time are not independent of each other, it will be under the normal sum of a three dimensional Euclidean (or Pythagorean relationship) with the negative sign indicated in the metric application on the timelike dimension. Most of us know this relationship and it is:

$s^2 = x^2 + y^2 + z^2 – (ct)^2$

If we take the derivation seriously, then the drag coefficient is dependent also on the sum of those components with the negative time component, in which the metric is simply:

$s^2 = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - c^2t^2 = \frac{2F}{\rho v^2 A} + 1$

So let's see where I was heading with this:

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + v^2t^2 - v^2t^2$

The negative of the drag is the thrust - though in this set up, I am entertaining that the last two terms associated to the space and time follow the anticommutation laws thus

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + [x,ct]^2$

But we can retrieve the ordinary drag by the distribution of the sign again

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

The last term here is known as the spacetime uncertainty principle. As always, I like to teach a little bit on the physics behind this, which will use the standard deviation operators, to understand these kinds of ''strong'' or ''stronger'' uncertainties, we can start off by defining the standard deviation for two commutators.

The standard deviation is:

$\sigma^{2}_{x,t} =\ <x^2t^2> - <x^2t^2>$

$\sigma_{x} = \sqrt{<x^2> - <x^2>}$

$\sigma_{t} = \sqrt{<t^2> - <t^2>}$

As usual the standard deviations involve the expectation value, in which the operators define the commutators:

$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

This gives us the quadradic Robertson uncertainty relation:

$\sigma_A \sigma_B \geq |\frac{1}{2i}[\hat{A},\hat{B}]| = \frac{1}{2}|[\hat{A},\hat{B}]|$

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

under standard deviation the new representation gives:

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - \sigma^{2}_{x,t} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - <x^2t^2> - <x^2t^2>$

Under quantum mechanics, time is not considered an operator, yet clearly, a spacetime commutator must require that time is manifested an observable in four dimensional space as the presence of curvature. Since we are aware that quantum mechanics treats the space dimension as an observable, it does itself have the non-trivial space operators outside of the trivial unitary operator:

$\hat{\ell} = \hat{e}_x x + \hat{e}_y y + \hat{e}_z z$

$<\frac{2F}{\rho v^2 A}>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,p} = \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

So in this case today, we have formulated the space derivatives and the time commutator as observables, something which is unavoidable in relativity when curvature is present, because again, it has to be manifested an observable if we detect a test particle moving in curved space.

Taking the concepts further:

Rearranging the standard deviation, in which we treated space and time as operators,

$<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t}$

$= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The Pressure is obtained from further rearranging:

$P = \frac{F}{A_b} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v\ = \frac{A^{-1}_{b}}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$= \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

Which removes the squared form of the metric and gives us an interpretation of the pressure related to the stress energy tensor.

Just for dimensional purposes, it could also be a related charge under the stress energy tensor:

$\frac{\hbar c}{\lambda} = \frac{1}{2n^2 \hat{x}}(x + y + z - \sigma_{x,t})\ T_{00}$

These are all just different idea's we can investigated, but it would indicate that the principal number as a discrete variable could have applications to various charges, just as it had during my investigation into the Weinberg formula for mass charges.

Rearranging the standard deviation, in which we treated space and time as operators,

$<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t}$

$= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The Pressure is obtained from further rearranging:

$P = \frac{F}{A_b} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v\ = \frac{A^{-1}_{b}}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$= \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

Which removes the dimensions from the metric. Let's move on:

An elementary form of the continuity equation for mass in hydrodynamics is:

$\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2$

Such that a mass flow rate is given as

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

with $\dot{V}$ is the volume flow rate and $\mathbf{j}$ is the mass flux.

This equation though is only true for a surface distribution of the flow. Only dimensionless equations are physically important in physics, which is why Bernoulli's equation is important for fluid mechanics -

$\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + \frac{P}{\rho} = C$

And we will work on this later. But first of all, we can also identify for the work, a dynamic pressure $q = \frac{1}{2} \rho v^2$. The force equation derived will present this quantity:

$F = \frac{A_b B}{Re^2_L} \cdot q\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

This means it related the fluid pressure weighted by the back area with the dynamic pressure as proportional to the Bejan number and inversely related to the Reynolds number as related to the stress energy:

$P = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot q\ = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}$

While there is so much to explore, the last obvious relationship that popped up to me was the mass flow identification which relates to the back area:

$F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

In which we refresh our memory on the following relationships -

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

If we just concentrate on the second equality of the equation, you will notice my derivations lead to:

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b$

The standard definition of the mass flow follows the limit

$\dot{m} = \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t} = \frac{dm}{dt}$

$F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}$

The total pressure is defined as

$P + q = P_0$

In such a case, $P$ is defined as the static pressure. Recall also that $\mathbf{j}$ is the flux of $q$ and is the dynamic component of the pressure. Retrieving the dynamic component, by subtracting the static component from both sides gives:

$q = P_0 - P$

This term is pretty much identical to the equation which describes an expanding or contracting sphere under the mathematical relationships describing sonoluminscence, which in a previous blog post was derived as:

$\frac{1}{R^2}\frac{P - (P_0 - P(t))}{\rho}(1 - (\frac{T_0}{T})^3) = \frac{\ddot{R}}{R} + \frac{3}{2}(\frac{\dot{R}}{R})^2 + \frac{4 \nu \dot{R}}{V} + \frac{2S}{\rho V} + \frac{e^2}{6 \pi mV} \frac{\dddot{R}}{R} + \frac{1}{V}(\frac{e}{m})\frac{\partial \dot{U}}{\partial R}$

with

$P_{\infty} = P_0 + P(t)$

Just using the solution for the dynamic pressure it explains the force also defined under:

$F = \frac{A_bB}{Re^2_L} \cdot (P_0 - P)$

With the dynamic pressure retrieved as:

$q = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot (P_0 - P)$

The force of drag was associated to the back area, and from previous relationships we introduced from mainstream physics, we ended up being able to describe it through a series of equivalences, which ended up with a metric term (in operator notation) and a stress energy tensor.

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}$

$\int F \cdot dx = \int F \cdot cdt = \int \frac{1}{2} \frac{B}{R^2_e} \cdot dmc^2$

and so the mass flow ~ but in such a case, we must take the speed of light as a velocity term on the force where now we set $c=1$ again

$F \cdot v = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \frac{dm}{dt} = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \dot{m}$

So a few different ways to look at the correcting factor appear. If we recall what the dragging coefficient is first of all

$f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}$

What kind of systems does the drag coefficient apply to? It applies to all kinds of sizes, but in theory it could be applied to an electron since it has been found, not to be a pointlike particle as we have been told in literature for many years, but has instead been carefully measured to be rather a spherical distribution of charge. For spherical systems, the drag coefficient is 0.47.- since we could in theory say that most particles are approximately spherical it wouldn't be too difficult to see that this number would only differ slightly depending on the structure of certain particles - for instance, a proton may appear spherical, but it's distribution inside is not entirely spherical which may mean it variates slightly in principle for different particle systems.

Retrieving the ratio of back to front area's for particular terms in the master equation, also called the wet and back area's, we can get the pressure flow:

$P = \frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot \rho v = \int\ \frac{1}{2A_f}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \mathbf{j} \cdot \frac{A_b}{A_f} = \frac{1}{2n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00}$

The first equality then has a strict relationship to the pressure flow related to the drag coefficient

$P = \frac{1}{2}\frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot q = f \cdot q$

where again, we defined the dynamic/total pressure as a simple rearranging of the mainstream formula:

$q = (P_0 - P)$

But that was just the first term, we will notice the drag coefficient is proportional to other terms from the master equation:

$P = f \cdot q = \frac{1}{2}f \cdot \rho v = \frac{1}{2}f \cdot \mathbf{j} = \frac{1}{2n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00}$

Or as twice the pressure flow related to simplified terms on the rest of the equalities:

$2P = f \cdot \rho v = f \cdot \mathbf{j} = \frac{1}{n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00} = \frac{\hat{s}^2}{n^2A_f}\ T_{00}$

These are nice simple relationships, but we can study these drag relationships now under the stress energy tensor more clearly for the future.

An elementary form of the continuity equation for mass in hydrodynamics is:

$\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2$

Such that a mass flow rate is given as

$\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A$

Using those relationships, we were able to create a drag force:

$F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V}$

$= \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m}$

And mass flow applies to the drag force

$F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}$

The continuity equation has many applications - for this case, let's say we rewrite it for the pressure: For instance, from it we can divide through by the energy density to get,

$\frac{P_1}{\rho_1} \cdot A_1 v_1 = \frac{P_2}{\rho_2} \cdot A_2 v_2$

This too is a continuity equation related to the Bernoulli principle, which is stated concisely in a dimensionless form as

$\frac{v^2}{2c^2} + \frac{P_1}{\rho_1} = \frac{v^2_2}{2c^2} + \frac{P_1}{\rho_1}$

To solve the velocities we obtain:

$v_1 = A_2 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}$

$v_2 = A_1 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}$

We can also entertain

$2 \dot{p} = v^2(A^2_1 \rho - A^2_2 \rho) = \rho v^2(A^2_1 - A^2_2)$

Before we go into an application involving the Friedmann equation, which is derived directly from fluid mechanics, we require to first investigate 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, often seen as a correction term to Einstein's formula - we will get to this soon because it will give us a reason to why a universe expands rather than contacts in a mechanical way.

Ok so let;s get into the cosmological fluid equation, the Friedmann equation. The most simplest form of the Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{4 \pi G}{3c^2}\rho$

One correction I made I applied in the past, inspired through the work of Motz and Kraft was that the equation need not be taken as a conservation equation - this is after all at odds with relativity in which energy is ill-defined under the Noether theorem, the distribution of a time variable indicates we get

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{3c^2}\dot{\rho}$

Motz' quite rightly stated that the equation need not be conserved, just as this previous equation indicates, as he said it was an ''unfounded assumption by Friedmann to conserve the energy in such a way) - and general relativity hints at this since not only is energy ill-defined, it is ill-defined because time itself is ill-defined (as it states there is no cosmological constant) under the Wheeler de Witt equation, in which diffeomorphism invariance is not a true time evolution of the theory. Further more, the Wheeler de Witt equation is also often taken as a statement of a conserved universe in which no changes happen inside of it, the consequence of it having no time derivative. But logic should be indicating to us that  many phase transitions have occurred and so the general statement should be taken with a pinch of salt.

Using the non-conserved form of the Friedmann equation, we can introduce the so-called continuity equation, but this is a misnomer if the derivative on the density term implies a change inside of the universe in a non-conserved way. The misnamed continuity equation takes the form:

$\dot{\rho}= -3H(\rho + P)$

where we take all terms here as the energy density and pressure. $H$ is the usual Hubble constant, but we should read this that density is not fixed, but dynamic component. From previous posts I have shown that the total pressure would imply a correction to this relativity term:

$\dot{\rho}= -3H[\rho + (P_0 - P)] = -3H[\rho + q]$

You can plug this into this into the non-conserved form of the Friedmann equation as

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{c^2}H[\rho + (P_0 - P)]$

$= \frac{4 \pi G}{c^2}H[\rho - (P_0 - P)] = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + \dot{q}]$

$\kappa = \frac{4 \pi G}{3}$

Also an important quantity, the density parameter is equivalent to

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

Later we will come back to this as it will play a role in the Bernoulli's principle. We must also consider for this following work the following work the presence of a second correction term for the total pressure of the fluid, as we had defined earlier as

$q = (P_0 - P)$

Which is a correction term, for the ''continuity equation'' a misnomer if a third derivative actually implies a diabatic, non-conserved solution for the Friedmann equation:

$\dot{\rho}= -3H(\rho + (P_0 - P)) = 3H((\rho - (P_0 - P)) = - 3H((\rho - (P_0 - P)) = 3(\dot{\rho} - \dot{q})$

The last set of terms in this equation has also featured in investigations into practically similar expanding and contracting models for sonoluminscence models - this is no accident of course since both those solutions are also derived from fluid mechanics.Plugging the main corrections into the non-conserved form of the Friedmann equation yields:

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = \frac{4 \pi G H}{c^2}(\rho - (P_0 - P))$

$= \frac{4 \pi G }{c^2}(\dot{\rho} - \dot{q})$

I have made previous corrections terms, such as introduction a zero point energy term for a pre-big bang state that existded at very near absolute zero to attempt to formulate a semi-classical model. The Friedmann equation, even in principle, cannot predict why spacetime should be flat - let me quickly explain. The density parameter, $\Omega$, is defined as the ratio of the actual (or observed) density $\rho$ to the critical density$\rho_c$ of the Friedmann universe. The relation between the actual density and the critical density determines the overall geometry of the universe; when they are equal, the geometry of the universe is flat (Euclidean). Appealing to authority, Susskind himself and Lisi Garret have expressed their opinions, that the universe is not flat - a way to articulate why this is , is because the universe would have to be infinitely large and this isn't really how we think about the universe in a scientific way, because it would take the universe an infinite steps to do so - but steps involve numbers and infinities are not numbers by definition. At any moment in time, no matter what time, the universe could be measured being finite in value always.

In a similar notion, the only way a black hole can become flat, would that the radius

$R \rightarrow 0$

Meaning the curvature of an expanding sphere must disappear

$K \rightarrow 0$

But the universe can only approach these values, but never reach it perfectly.

As wiki states itself: ''In earlier models, which did not include a https://en.wikipedia.org/wiki/Cosmological_constant term, critical density was initially defined as the watershed point between an expanding and a contracting Universe. To date, the critical density is estimated to be approximately five atoms (of https://en.wikipedia.org/wiki/Monatomic https://en.wikipedia.org/wiki/Hydrogen) per cubic metre, whereas the average density of https://en.wikipedia.org/wiki/Baryons#Baryonic_matter in the Universe is believed to be 0.2–0.25 atoms per cubic meter.''

I have become a hard disbeliever against dark matter, too much mounting evidence against it over the years - I could go into loads of reasons why, but I will spare the reader. I also believe that dark energy has been misunderstood, but we will get to that later. The entire basis of the density parameter to the actual observed ratio of matter in the universe, is too many magnitudes off, which should have been taken seriously that the Friedmann equation may itself be flawed fundamentally.

Next we will get into Bernoulli's principle and how the work above will come together to explain a source of the cosmological constant and how it applies to dark energy due to a weakening of gravity in the universe - the cosmological constant has often been believed that it is not really a constant, but there are some intriguing reasons why we may have to consider new alternatives to explain this in a more rationally-mechanical way due to pressure, perhaps thermodynamic in nature.This time around i plan to write the correct formula, in whcih we retain that dark energy remains constant, but the acceleration entertained by a weakening of gravity over galactic scales, only giving the impression dark energy is soley due to it - basically, the weakening of gravity only appears to make the constant change, when it is not. The gravitational binding of the universe would arise in my picture as an effect of gravitation become weaker as it gets larger and so only appears to accelerates.

There is a density parameter which can be used in Bernoulli's principle to apply for cosmological applications. But before
we jump into that, the best way to see how Bernoulli's principle can explain the cosmological constant is literally by
writing the constant as a constant energy density in the style of conservative forces:

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

Now, let's not move to quickly, I want to now define the so-called ''density parameter'' for the Friedmann equation ~

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

And the relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^2}(\rho + P)$

So we have a number of things we can do here; first, let's plug in the density parameter and then divide through by the energy density:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\frac{P}{\rho}$

So with this equation and the other what do we do now?

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

We now plug in the co-called ''critical density'' for both formula's - but ''critical density'' is a misnomer in the absence of dark matter and the proper interpretation of dark energy:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^2}\Omega = \frac{8 \pi G}{3c^2}\frac{8 \pi G \rho}{3H^2}$

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\Omega)E = (\frac{v}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{8 \pi G \rho}{3H^2})E = \mathbf{C}_{vac}$

I will be introducing once again the Leonard extension of the Einstein equivalence to explain why a universe expands rather than contracts - ok, so we spoke about $C$ being a constant, today we know it is related to energy and the main main reason why this Freidmann expansion melding into Bernoulli's principle was not a ''an assumption'' as was challenged by someone who followed the work, because he seemed to be unaware that we work with fluid dynamics when working with an expanding universe, which relies on different pressures through various complicated parameters. In other words, the two have stronger relationships to each other than most would like to agree on, or have been unaware of.

To present the laws spoken in previous posts, a general application from the last equation would be to distribute an energy, such that the constant on the right hand side, is a constant of energy. We will leave out the right hand side until I can try and simplify it further, the real equation we should concentrate will be of the form:

$(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

where now C is no longer ''just a constant'' but is define under the units of energy defined through the vacuum contribution.

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.

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^2}\Omega = \frac{8 \pi G}{3c^2}\frac{8 \pi G \rho}{3H^2}$

It's also interesting we can find the classical upper inverse limit of the gravitational force (but with an additional gravitational constant term), which implies an inverse of the stress energy tensor on the left hand side:

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

$T_{00}^{-1}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

from all this we find a different form of the Friedmann equation:

$\frac{3H^2}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}8 \pi G \rho$

Let's go through this thoroughly

The relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}(\rho + P)$

Let's divide through by the relative density and pressure

$\frac{1}{\rho + P}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}$

But we need not take such a relativistic case, because this will drop out naturally when we take the non-conserved form of the Friedmann equation. We shall get to that soon.

I also entertained

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega$

Where we remind ourselves that the misnamed ''critical density'' is:

$\Omega = \frac{\rho}{P} = \frac{8 \pi G}{3H^2} \rho$

To retrieve the upper limit of the classical gravitational force.

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

The continuity equation is

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

and the respective energy density is

$\dot{\rho}c^2 = -H(\rho c^2 + P)$

So, we've been through this a few times, the non-conserved form of the Friedmann equation requires one extra derivative which removes one such factor of the Hubble ''constant'' (or in the last term as explained not long ago, it gives us the continuity which is strictly defined under the density and pressure parameters) - we also distribute the factor of three from the denominator:

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \dot{\rho}$

This gives by plugging in the continuity:

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{2c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \frac{\dot{R}}{R}(\rho + P)$

Which neatly gives a new interpretation/form of the expansion of a universe, in diabatic from. Distributing through an inverse factor of $8 \pi G$ to neaten it up without over-simplifying by combining components:

$\frac{3}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{1}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{ \rho}{H} = \frac{8 \pi G}{3c^4}\frac{1}{H^2} \frac{\dot{R}}{R}(\rho + P)$

And this yields

$\frac{3H^2}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^4} \frac{\dot{R}}{R}(\rho + P)$

Let's talk about the Rohrlich correction now.

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 $\Delta p$ we then have

$\Delta p =\frac{v}{2c^2}E$

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

$2\Delta p=\frac{v}{c^2}E$

$2\Delta p v =\frac{v^2}{c^2}E$

In which $\frac{v^2}{c^2}$ is the Doppler shift, more appropriately given in the squared form, rather than the form Leonard had given, because it allows us to write the metric in a relativistic way.

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

$p′ = mv−2\Delta p = (m − \frac{E}{c^2})v$

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. To understand this pressure difference, we now include Leonards correction to the mass energy equivalence in terms of the first term in the paranthesis: To do that, we remind ourselves first of the formula we want

$\Delta p v =\frac{v^2}{2c^2}E$

and then we plug in:

$\Delta p v + (\frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

The physics of the first term, will not only explain why something expands rather than contracts, but it will also provide under its own definition, a redshift term - meaning things are moving outwards, not inwards, (ie. the universe is expanding, not contracting due to pressure differences) and because we have used a conservative form, it means it does not rely on the gravitational field soley.

Looks good though you may want to change your references from wikipedia, in the publishing community it is frowned upon to use wikipedia as a creditable source as anyone can edit it. Use a math book or something. Here is a Link to physics journal A: Mathematical and theoretical https://iopscience.i...urnal/1751-8121 , if you wish to publish it via IOP science in a journal.

Edited by VictorMedvil, 20 June 2019 - 04:18 PM.

### #3 Dubbelosix

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Posted 20 June 2019 - 04:24 PM

True, but you can't change fundamental theorems, willy-nilly, which is why I will let it pass - but if not accepted, obviously I would change it... I expect a lot of proof reading anyway, like constants and things like that.

### #4 VictorMedvil

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Posted 20 June 2019 - 04:28 PM

True, but you can't change fundamental theorems, willy-nilly, which is why I will let it pass - but if not accepted, obviously I would change it... I expect a lot of proof reading anyway, like constants and things like that.

They will check all of that dubbel and make sure it all matches you better have boned up on your math before publishing and checked it yourself several times, make sure there are zero errors or it will be rejected.

Edited by VictorMedvil, 20 June 2019 - 04:29 PM.

### #5 Dubbelosix

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Posted 20 June 2019 - 04:58 PM

They will check all of that dubbel and make sure it all matches you better have boned up on your math before publishing and checked it yourself several times, make sure there are zero errors or it will be rejected.

I expect it to be rejected at least once if lucky, twice if lucky enough. Even Hawking and Penroses singularity theorems where rejected the first time around.

### #6 GAHD

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Posted 20 June 2019 - 10:19 PM

Don't take this the wrong way, but I'm leaning towards the Alternative subforum as the right place for this to have been posted.

### #7 Dubbelosix

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Posted 20 June 2019 - 10:48 PM

Don't take this the wrong way, but I'm leaning towards the Alternative subforum as the right place for this to have been posted.

Without alternatives, science would not progress, so I would never take it the wrong way. I will profess though, it surely could not be based on quality. I have spent a long time on this.