Schmelzer Aether Reinterpretation

An interesting consequence of treating gravity as a quasifluid-filling aether, comes right back to cosmology itself, so long as the universes geometric evolution satisfies a Ricci flow that shrinks the positive curvature and as a result, expands the regions of negative curvature, leading to an all-scale homogeneity. The smoothing out of those metric irregularities may be seen as playing a role in the evolution of the universe in which we observe large scale flatness.

Of course, there appears a missing piece of the interpretation of gravity, in which it seems to have escaped scientists notice that its role in the vacuum is absolutely analogous to a drag phenomenon, not so different to how objects are dragged while submerged in a liquid and that drag, in the classical physics, is determined by the objects speed and the density of the fluid in response to the medium it travels in. Gravity is very similar in regards that differing field strengths can result in the extreme case of trapping even light, but even I am not so certain that light is trapped and the information paradox has to be avoided. The idea is that light could actually escape but it depended on a radically new, but plausible interpretation of a gravitational aether in which gravitational waves themselves are analogous to accoustic waves. We covered this earlier so no point to delve too deeply into that model again. The main point though I wanted to add, that the expansion of a universe could be explained by a Ricci flow, my mathematical prediction from relativity showed that the expansion of a universe, under a misnomer of dark energy, is really a thrust - the opposite of gravitational drag.

The mathematical form found that the stress energy we relate to a dark energy, had in fact an inverse drag coefficient, with an all-crucial negative sign indicating that there is some thrust preference over a collapsing model. I began to call the thrust, the intrinsic impetus. If the thrust was much larger than the drag in early cosmology then the universe could easily break out of the strong gravitational binding. An immediate consequence of this is whatever thrust energy is left over will now become dominant and inexorably appear as if it was changing over time... So why is this? The answer is surprisingly simple, it comes down to the fact that as galaxies recede further from each other, the weaker gravity becomes and so the thrust appears to amplify over time. In this respect the Hubble parameter would vary not according to a changing cosmological energy, but a changing gravitational field.

While the notion of the thrust does not explain why it happens, the Ricci curvature could be in fact directly the mechanism behind it. Moreover, it would play an important role in work I have read which I have come to call the Schiemzer Aether model, in which we can successfully argue not only for a mobile aether model but also one that can be measurable... If the gravitational field is the aether, then we observe the effects of the geometric evolution all the time in our universe. It is even believed by some the most distant galaxies are receding faster than light and this too would owe the phenomenon to the flow of spacetime, though I must admit I have become skeptical of the acceleration based on a number of reasons which no cosmologist to date has satered my curiosity of a possible breakdown in understanding of distant light signals, but this is not for the main debate of this paper, so we can cart blanch this one for one.

The interesting thing is the Ricci flow is responsible for the thrust, then it would have direct implication in the Schmelzer model, or at least part of it. My motivation to continue a line of investigation to a scientist posting online science forums is often the last thing I do, but I found the basic premise behind the harmonic gauge of the metric satisfying, though not agreeing on the interpretations it led to.

Geometric Evolution

The Ricci flow does what it says on the tin, it is a geometric evolution equation which describes how curvature can flow in the universe. For the universe to expand, the Ricci flow would have had to feed off positive curvature to then lead to negatively expanding space. The Ricci flow, less commonly understood, is the heat equation for a Riemannian manifold and also follows its own with diffusion laws.

The Modern Crisis

If the manifold is a sphere with positive curvature, then Ricci flow collapses the manifold to a point in some given finite time - for cosmic analogy, this would be the collapsing hypothesis. While This shows that the Ricci flow sometimes cannot be continued for all time, it is often indicated to run into mathematical singularities. In my own training, it is custom to think singularities are indication the physics has broken down so where, so I remain skeptical of the physical existence of these singular solutions. Under more technical babble, if the manifold is an Einstein manifold with negative curvature, then Ricci flow will expand it and it is here I will seek out the mechanism behind the cosmological thrust/impetus.

But there is a big crisis right now in cosmology and while both results from many different investigation s at first glance appear contradictory, a splendid and elegant set of explanations can be given to drive them away. On one hand, one group say the universe is flat, but when scientists say flat, they are usually meaning the curvature presently measurable in a directions, and since, expectedly there is very little curvature, it doesn't address the distribution of matter in which you could impose a shape in the universe. Instead of trying to compile arguments why one is favored over the other, it seems that maybe the universe is spherical and appears flat at the same time. The flatness of the universe could be part of a much larger system and is an illusion. A good analogy is how we the Earth possesses curvature, but someone sitting on the surface may think the land appears flat - Lest I unwillingly start debates on the flat-earthers, let's quickly move on.

How can a sphere be hyperdimensional but have no curvature?

No right minded scientists says there is no curvature, only that there is so little it is expected to be negligible. Some prominent scientists even suspect we may even detect such a small curve. It is true, that a very small sphere has a very large curvature compared to a very large sphere - the closest thing we have in nature to a perfect sphere, is a black hole and such a system will suffice in the next statements.

A small black hole also had a very large curvature while large black holes will tend to lose curvature as it grows. But this may be just a limit and not a true physical application in which curvature can disappear from a universe.

If the case is that we are inside a black hole, the flatness problem may also be related to these strange facts that where discovered about relative density many years ago. Simply put, a universe that does not appear dense will not appear to have much curvature present in it. The only thing the observers inside the black hole and outside of it can agree on is that the system contains a lot of mass... And our universe does indeed contain a lot of mass even though it covers roughly 1% of spacetime itself, as it harbors over a massive (10^80 x 3) of particles, give or take a few powers of ten for room for error. The number has to be multiplied by a factor of three to account for the dimensions of space (volume).

So to summarize the crisis, we may not have a crisis at all but rather a deeper puzzle related to a number of things we must take in into respect, such as a sphere being capable of approaching cosmic flatness, homogeneous distribution of galaxies in such a way that you could infer a shape. Contradictory appearing and sounding at first, the two may go hand in hand when explaining the apparent oxymoronic crossroads of experimental conclusions.

It's been known to me for a while the Friedmann equation contains some serious issues in the way of the experimental varification of actual spacetime density - as I will come to explain, and have since taken it to mean Friedmanns model is largely incompatible for the correct description of our universe because current density has been set to the critical density to ensure it produces a flat universe, but I have since came to think we had the model wrong long before the crisis of the shape of the universe became popular this year and I had even from time to time, stressed the importance of why the universe could not be completely flat by arguing it would violate temperature laws under Aruns approach. The universe for instance cannot be flat because this would imply an infinite radius,and we observe no such thing, in fact the universe appears quite safely finite at any given frame of time. So long as Aruns limits holds, and we have faith in quantum mechanics, namely zero point energy fields, the the temperature of a universe can never be zero, but can come extremely close to it! And so long as there is a nonzero temperature, this will forbid the following limits in which we expect the curvature to be precisely zero, but again, this would also mean a non-sensical infinite radius.

The Reinterpretation of the Schmelzer Aether.

You will often see the Ricci flow defined in terms of

∂_0 g^μν = -2R^μν

(a variation of the metric, here with units of inverse length is directly related to the Ricci curvature determining the Geometric evolution of a system). But this is not volumetric, in which we expect vital physics associated to the volume of a space, just as telle-motion gave rise to a volumetric understanding of curvature. Here, R is the curvature tensor, not to be confused with the radius of curvature. However, being a very nice formula for the curvature, it will not satisfy a compact manifold, in other words, this geometric evolution equation does not preserve the volume as explained. In order to rectify this so it has a broad physical application, one has to use the renormalized Ricci flow of the form (working in natural units of c=1);

∂_0 g^μν = -2R^μν + 2/n R g^μν

The units might look a bit odd... We will get into this later as the metric is specifically designed.

An interesting property that would have implication in the evolution of an expanding geometry, is that if one did not take into consideration a three dimensional structure, its very possible again to collapse the size of the system. In similar fashion, when deriving the drag equation in three dimensions for my model of gravity, did I find expanding solutions giving rise to the thrust. Obviously other important reasons exist in why we would want to preserve the notion of volume, obvious ones such as the desire to calculate energy densities in the geometric flow.

While we understand the curvature tensor and curvature scalar from tensors, the equation above is actually not a tensor equation, because the time derivative of a tensor is strictly not a tensor. Often we see the flow equation with lower index terms, but as all will learn in a first year course of general relativity, the metric lowers or raises the index, which is all we have done to obtain the last equation,

G^μν = - 2R^μν + 1/2 R g^μν = 8πG(Newton)/c^4 T^μν

This equation certainly is a tensorial set of equations. The flow then would have to be interpreted in a way to preserve some kind of tensorial structure, so that the timelike partial derivative be replaced by a timelike covariant derivative... Because the metric is given in units of inverse length, there will be commutation laws between the connections, known universally as the spacetime uncertainty relationships. Moving on, Schmelzer has defined a reasonable harmonic gauge capable of describing the aether, known from standard literature on GR to be;

ρ = g^(00) √-g

ρv^i = g^(0i) √-g

ρv^iv^j + σ^ij = g^(ij) √-g

For Schmelzer, the aether has a dynamic speed, however we have disagreed on the interpretation of the motion of the aether defined strictly as being the gravitational field, probably the only element we reasonably agree on. His choice of gauge allows us to analyse the motion of the aether under the Ricci flow. I do not assign any particle to the field and if there was to be a physical interpretation of ρv^iv^j, it would have to be related to the dynamic pressure, which is known in literature to be given as

Q = ρ_0 - ρ_s = 1/2 ρv^2

From Bernoulli's law, with ρ_0 being the total and ρ_s being the static part of the pressure.

And is in fact due to the kinetic velocity of particles in a fluid. Since spacetime is often modelled under the perfect fluid equation, (which Schelmzer appears to be aware of as conditions that set conservation laws) under adabiatic models, but the kinetic pressure still has real physical interpretation inside of spacetime and the role of thermodynamics will be equally important in its own way for expanding and collapsing models, I suspect.

The Friedmann Equation Does Not Fit Observation

The density parameter is defined as the ratio of the actual (or observed) density to the critical density 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 (exactly) flat (or Euclidean). In earlier models, which did not include a 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 monatomic hydrogen) per cubic metre, whereas the average density of ordinary matter in the Universe is believed to be 0.2–0.25 atoms per cubic metre. This should have immediately rang alarm bells because dark matter has not sufficiently been demonstrated with enough evidence to infer whether it exists or not, and the making of density equal to the critical density appears at face value, only an assumption.

Because (g) is the Riemannian metric and is often dimensionless but in this case, it is not. I have always been a bit confused by the dimensions of the Ricci flow as it is universally written as the metric is often dimensionless and the curvature R has units of inverse length squared, but here as was explained to me a while back, the metric contains the dimensions of

g = [L^(-2)] [t]

That is, the inverse of the square of length and a time coordinate.

The choices of gauge from the Schmelzer appears distinctively to depend on a gravitational aether with an assigned Doppler shift term;

v/c = g^(i0) / g^(00) = Δλ / λ

Being fundamentally constructed from the metric, the gravitational aether is dynamically responsible for a shift in the light. We remind ourselves that

γ = 1/(√1 - v^2/c^2) = dt/dT

With T being the proper time. I suspected that the proper time could be evaluated from Birkoff's theorem for the Schwarzchild metric

dT = (1 - r_s/R) dt +...

I choose this because it satisfies dT as the clock of an observer at a distance R (radius) in a spherical homogeneous system (possibly ideal for a largely homogeneous spherical universe). And γ goes under standard Lorentz convention. But I have since left this to the back of the mind. It may also be more appealing under spherical coordinates.

A number of reasons exist that could argue why these relationships exist in a plausible mathematical sense. Certainly, the relationship between the Lorentz contraction and redshift under special relativity produced the correct observed shift in the universe than what classical physics had attempted before.

Going back to Schmelzers aether, the parameters set from the harmonic condition was

ρ = g^(00) √-g

ρv^i = g^(0i) √-g

ρv^iv^j + σ^ij = g^(ij) √-g

And he defined the box operator (which is the dAlembertian, a four dimensional metric) as:

□ = ∂_μ g^μv √-g ∂_v

(and based on geometric algebra, it would not be hard to produce a bivector form of these equations which would employ the Poincare symmetries allowing it to feature in the space of rotations, but when working in four dimensions, great care has to be taken due to the cross product as I will explain at the end in the mathematical notes)

And of course one can identify the density

□φ = ρ

Which if memory recalls, Nordstrom was the first to identify this latter relationship, but really I want the reader to concentrate on the identity ∂_μ g^μv √-g ∂_v because it arises famously from the continuity, which we will cover. I pointed out first the relationship

∂_0 g^μv / ∂ √-g = - 2 √-g (∂_0 g^μv / ∂ g)

And by using the definition of the Ricci flow we extend it as

∂_0 g^μv / ∂ √-g = - 2 √-g (∂_0 g^μv / ∂ g) = 4 √-g (R^μv / ∂ g)

This features the gauge of choice for the aether interpretation in the numerator. It seemed relevant at the time to point it out. In hindsight there may be a relationship of the factor of 4 with the identity

g_μv g^μv = 4

I also explained from other mainstream identities, that the curvature with a diffusion φ defining a mass flow is also encapsulated in the following relationships;

g^μv R_μv φ = g^μv ∇_μ∇_v φ

= g^μv ∇_μ (∂_v φ) = g^μv (∂_μ∂_v + Γ^σ_μv ∂_σ) φ

= - 1/√-g ∂_μ (g^μv √-g ∂_v) φ

(see notes for a derivation of the Ricci flow)

And viola! What have we ended up with? Its just a slight variation on the four dimensional operator defined by Schmelzer and standard general relativity as;

□ = ∂_μ g^μv √-g ∂_v

It is also the representation of a gravitational-like Poisson equation, or Ricci flow with a diffusion coefficient

α(η_μν ∂^μ∂^ν ϕ)= α □ϕ = α(∂^2ϕ/∂τ^2+∇^2ϕ)

(it's also known in literature that it can be wise to distinguish the number of derivatives as □^2 since you also use a first derivative form of the operator in physics)

Which had been represented in curved space, otherwise it would have only satisfied special relativity, but when taking the case into general relativity, the four dimensional dAlembertian defined according to Schmelzer will arise from common application of the curvature tensor.

What is the physical meaning behind these identities? Let's first identify the two covariant derivatives as

[∇_μ, ∇_v] φ = 4πG/c^2 ∇_μv j = - R_μv φ

Where [j] is known as the mass current density... Let us just look at the first term, the dimensions are simple{ enough;

G · Mass/length^3

Since

φ = Gm/r

And each connection has an inverse length. It can be noticed then we retrieve Poisson equation,

∇· g = - 4πGρ

Or by a substitution into the Gauss Law.

∇ · (-∇)g = - 4πGρ

But this only holds for an irrotational case of the grav magnetic field meaning there would be no torsion correction. If the acceleration is defined by the Christoffel symbol then we have classically

∇ · Γ = - 4πGρ

The mass flow can be seen as playing the same role as the electromagnetic flux. The gravielectric field, with a magnetic potential is in standard form;

E = - ∇φ - 1/c (∂A/∂t)

And the gradient of the field;

∇ · E = ∇ · (-∇) φ - 1/c^2 (∂^2A/∂t^2)

However, the minus sign ensures that the Ricci flow is well defined for sufficiently small positive times; if the sign is changed, then the Ricci flow would usually only be defined for small negative times. (This is similar to the way in which the heat equation can be run forwards in time, but not usually backwards in time.)

Conclusions

I had to personally reject the largest part of Schmelzer's interpretation of a velocity assigned to a matter field defining the aether, though this objection could appear moot, since after all, would not the gravitational field simply reduce to the stress energy of the fields inside of spacetime? Sure, but we have to be very careful because there has been no sufficient explanation to how a classical pseudo field as the aether itself, be assigned to a particle field in a way identical to how we quantize the fields of nature? There are five classical pseudo forces known to me and it is a priori that none of them are quantized in any way. Gravity too being a psuedo force cannot be hastily interpreted into a particle model without a more sound argument supporting that case. To date I have found no such evidence but rather a massive misunderstanding that seems to have continued into modern day.

Further, while the velocity is assignable in his model, he adds that it is unobservable. While I argued Popper philosophy, he argued scientism, so we ended up not meeting agreement here either. I explained that the velocity could be assignable to the metric itself and explained Ricci flow, but it appears as if he did not find it very appealing. I did on the other hand knowing how important the velocity issue is. I saw that by making a gravitational field the aether, we were talking about it in quasifluid terms, in which the Universe follows the ideal fluid equation and gravitation acted as a drag on objects as it passed through it, following curved geodesics. Fortunately, we agree that the speed of light is spatially variable and has been an important factor of my own investigations into black holes in the past.

It is true though, in some sadistic arm-wavey sort of way, that the gravitational field would be associated to its own stress energy tensor as Schmelzer has stated... And because all forms of matter and energy contribute to gravitation, it is easy to see why you might conclude it is here you find the matter field associated to gravity - indeed, all the matter fields for that matter of fact. But I do not find this compelling enough to accept a transition of the gravitational force to be modelled as a matter field but rather more fundamentally to the geometry of space itself and possibly not isolated to a single field as I accept his premise that all fields contribute to such dynamics... Much like how Wheeler pointed out, the increase of energy in a particle is not owed to some intrinsic change in the systemy, but owes its phenomenon of increasing in size due to its geometric properties alone.

Being left with geometry alone, along with the common definition of gravity itself, it was not completely rid of mobility after all as relativity previously concluded, as we are allowed the Ricci flow - its not a matter field but it does involve the way space moves and evolves over time which can be measured and it seems to add, that space is a fundamental thing as a priori to any final theory. Matter fields contribute or sometimes even causes the flow but this is not always the case, such as when a metric can move at super relativistic speeds, inexorably dragging galaxies along with it, while locally moving below the speed of light. This may even be related if not the direct cause of an inflationary period... As Hawking stated himself, the inflation period led to different parts of the universe expanding at different rates, though probably a bit more detailed than this.

Viewing gravity as an aether was an idea that I too considered before my encounter with Schmelzer and I was initially romanced by the conditions set for the aether in question, it just ended to be interpretation we differ on - big differences in fact to how we view this aether. While it is true we cannot extrapolate all meaning of science from observation, it still stands to reason that to extrapolate any meaningful results requires observational backup, otherwise it cannot be falsified. While this does not bother the other author, it troubles me quite a bit that there should be an undetectable motion in the aether.

# On The Schmelzer Aether

Started By
Dubbelosix
, Dec 04 2019 04:39 AM

4 replies to this topic

### #1

Posted 04 December 2019 - 04:39 AM

### #2

Posted 05 December 2019 - 08:10 AM

But regardless, the discussion did lead me on to investigate these things further, one thing that stood out was that we don't even need to talk about geometry alone to describe this flow.

Some basics first, the scalar curvature is

R = g^μv ∇_μ∇_v

The curvature tensor is

R_μv = g_μv R = ∇_μ∇_v = ∇_μ (∂_v φ) = ∂_μ∂_v + Γ^σ_μv ∂_σ

Variation of the action with respect to the metric in general relativity under standard theory is written in the form, without a potential as

A / φ = - 1/√-g (∂_μ √-g ∂^μ) φ

The variational result has led to a relationship with the d'Alemertian operator,

g_μv □ = ∂_μ √-g ∂_v

With g having units of inverse length squared and one inverse of time as found in the Ricci flow allows us to write it as

g^μv □_μv = ∇^μ ∂_μ √-g ∇^v ∂_v

□_μv = [∇^μ, ∇^v] = (∂_μ + Γ_μ)√-g(∂_v + Γ_v)

= √-g (∂_μ ∂_v + ∂_μ Γ_v + Γ_μ ∂_v + Γ_μ Γ_v)

The torsion typically arises from the last terms (which are under commutation laws)

[Γ_μ, Γ_v] = Γ_μ Γ_v - Γ_v Γ_μ

The Ricci flow is

∂_0 g^μv(φ) / ∂ √-g = - 2 √-g (∂_0 g^μv(φ) / ∂ g) = 4 √-g (R^μv / ∂g)

While a geometric flow is does not preserve the volume so we write,

∂_0 g_μν(t) / ∂ √-g = - √-g (4R_μν + R g_μν) / ∂g

But here is the interesting thing that leads us to a new equation for the Ricci flow where we notice the renormalized case is identical to the Einstein equations of the form

G_μν = R_μν - D/2Rg_μν

When you plug in for the spacetime dimension D=4 you get 4R_μν - 2Rg_μν and when divided through by four we get R_μν - 1/2Rg_μν. From here we remind ourselves we can rewrite the factor as the stress energy tensor

G_μν = R_μν - 1/2 g_μν (-kT)

Allowing us to define

R_μν = k (T_μν - 1/2 g_μν T)

This means the Ricci flow has to be associated to an energy flow, we don't need to meddle about now with the volumetric case because we will find the ordinary Ricci flow wi suffice,

∂_0 g_μν(t) = - 2R_μν

And redefine under the equivalent form of the curvature tensor using the equation we stated before

R_μν = k (T_μν - 1/2 g_μν T)

And the Ricci flow becomes a stamement about the stress energy tensor.

∂_0 g_μν(t) = - k (2T_μν - g_μν T)

Some basics first, the scalar curvature is

R = g^μv ∇_μ∇_v

The curvature tensor is

R_μv = g_μv R = ∇_μ∇_v = ∇_μ (∂_v φ) = ∂_μ∂_v + Γ^σ_μv ∂_σ

Variation of the action with respect to the metric in general relativity under standard theory is written in the form, without a potential as

A / φ = - 1/√-g (∂_μ √-g ∂^μ) φ

The variational result has led to a relationship with the d'Alemertian operator,

g_μv □ = ∂_μ √-g ∂_v

With g having units of inverse length squared and one inverse of time as found in the Ricci flow allows us to write it as

g^μv □_μv = ∇^μ ∂_μ √-g ∇^v ∂_v

□_μv = [∇^μ, ∇^v] = (∂_μ + Γ_μ)√-g(∂_v + Γ_v)

= √-g (∂_μ ∂_v + ∂_μ Γ_v + Γ_μ ∂_v + Γ_μ Γ_v)

The torsion typically arises from the last terms (which are under commutation laws)

[Γ_μ, Γ_v] = Γ_μ Γ_v - Γ_v Γ_μ

The Ricci flow is

∂_0 g^μv(φ) / ∂ √-g = - 2 √-g (∂_0 g^μv(φ) / ∂ g) = 4 √-g (R^μv / ∂g)

While a geometric flow is does not preserve the volume so we write,

∂_0 g_μν(t) / ∂ √-g = - √-g (4R_μν + R g_μν) / ∂g

But here is the interesting thing that leads us to a new equation for the Ricci flow where we notice the renormalized case is identical to the Einstein equations of the form

G_μν = R_μν - D/2Rg_μν

When you plug in for the spacetime dimension D=4 you get 4R_μν - 2Rg_μν and when divided through by four we get R_μν - 1/2Rg_μν. From here we remind ourselves we can rewrite the factor as the stress energy tensor

G_μν = R_μν - 1/2 g_μν (-kT)

Allowing us to define

R_μν = k (T_μν - 1/2 g_μν T)

This means the Ricci flow has to be associated to an energy flow, we don't need to meddle about now with the volumetric case because we will find the ordinary Ricci flow wi suffice,

∂_0 g_μν(t) = - 2R_μν

And redefine under the equivalent form of the curvature tensor using the equation we stated before

R_μν = k (T_μν - 1/2 g_μν T)

And the Ricci flow becomes a stamement about the stress energy tensor.

∂_0 g_μν(t) = - k (2T_μν - g_μν T)

### #3

Posted Yesterday, 09:52 AM

ON A STRESS ENERGY FLOW SUBJECTED TO STATISTICS

Let's cover some basics first, Einstein's equations are;

G_μv = - 2R_μν + 2/n R g_μν = g_μν k T

Where

T = 1/2 ρcu

k = 8πG/c^4

And the scalar curvature is

R = g^μv ∇_μ∇_v

The curvature tensor is

R_μv = g_μv R = ∇_μ∇_v = ∇_μ (∂_v φ) = ∂_μ∂_v + Γ^σ_μv ∂_σ

Generally will not commute even in classical physics,

[∇_μ, ∇_v] = (∂_μ + Γ_μ) (∂_v + Γ_v)

With g having units of inverse length squared and one inverse of time as found in the Ricci flow allows us to write it as

g^μv □_μv = ∇^μ ∂_μ √-g ∇^v ∂_v

□_μv = [∇^μ, ∇^v] = (∂_μ + Γ_μ)√-g(∂_v + Γ_v)

= √-g (∂_μ ∂_v + ∂_μ Γ_v + Γ_μ ∂_v + Γ_μ Γ_v)

The torsion typically arises from the last terms (which are under commutation laws)

[Γ_μ, Γ_v] = Γ_μ Γ_v - Γ_v Γ_μ

The Ricci flow is

∂_0 g^μv / ∂ √-g = - 2 √-g (∂_0 g^μv / ∂ g) = 4 √-g (R^μv / ∂g)

While a geometric flow is does not preserve the volume so we write,

∂_0 g_μν(t) / ∂ √-g = - √-g (4R_μν + R g_μν) / ∂g

But here is the interesting thing that leads us to a new equation for the Ricci flow where we notice the renormalized case is identical to the Einstein equations of the form

G_μν = R_μν - D/2Rg_μν

When you plug in for the spacetime dimension D=4 you get 4R_μν - 2Rg_μν and when divided through by four we get R_μν - 1/2Rg_μν. From here we remind ourselves we can rewrite the factor as the stress energy tensor

G_μν = R_μν - 1/2 g_μν (-kT)

Allowing us to define

R_μν = k (T_μν - 1/2 g_μν T)

This means the Ricci flow has to be associated to an energy flow, we don't need to meddle about now with the volumetric case because we will find the ordinary Ricci flow will suffice,

∂_0 g_μν(t) = - 2R_μν

And redefine under the equivalent form of the curvature tensor using the equation we stated before

R_μν = k (T_μν - 1/2 g_μν T)

And the Ricci flow becomes a stamement about the flow contributed by the stress energy tensor and is attributed to the manifolds geometry evolution as,

∂_0 g_μν(t) = - k (2T_μν - g_μν T)

Since the Ricci flow is a heat flow, there may be a universal application for the formula in regards that the stress energy tensor is applied to all types of fields in nature. A heat flow under the stress energy will require a proper meld of the right physics. If we make all metric terms equal in dimensions would help keep in clearer, involving some small modifications,

∂_0 g_μν( R) = - k (2 ∂_0 T_μν - ∂_0 g T)

The right hand side can be reinterpreted into four dimensions with first derivatives in □ giving with c =1. And to get the four dimensional case as you will know simply requires replacing the complete 3-space operator ∇ for its punctual cousin □ sticking to first order equations in the process. Three dimensional case

∂_0 g_μν( R) = - 2k ∇ T_μν

= - 2k (∂_x T_μν + ∂_y T_μν + ∂_z T_μν)

And also for calculation the squared form of this

g_μν ∂^0 R^μν( R) ∂_0 R g_μν = - 2k g_μν [∇, ∇] T_μν T^μν

= - 4k^2 (∂_x T_μ + ∂_y T_μ + ∂_z T_μ)(∂_x T_v + ∂_y T_v + ∂_z T_v)

No one taught me how to do this, but it is perfectly acceptable it seems to interpret a volumetric flow in this way. And the four dimensional case is thus a modification of;

∂_0 g_μν(vR) = - 2 □ kT_μν

= - k (2 ∂_x T_μν + 2 ∂_y T_μν + 2 ∂_z T_μν - ∂_0 g_μν T)

The d'Alemertian operator is inside this last equation where g_μν(R, t) just means a metric with Riemannian curvature dimension (typically) of inverse length squared so g(t) would be concerned with time alone ang g(x... N) same idea here but is related to the dimensions position length derivative alone. Here we would relate any metric without a dependent term as dimensionless for clarity.

We are allowed to implicate the d'Alemertian because of the presence of the time derivative, a feature which del operator lacks.

∂_x + ∂_y + ∂_z - ∂_0

Which is a low limit relativistic case and for this we had to choose then first derivative argument instead of the second, which under literature is

□^2 = ∂^2_x + ∂^2_y + ∂^2_z - ∂^2_0

And it is from here I will apply my knowledge on heat conduction for a Riemannian manifold and later transpose the above equation into curved space. To take it into curved space we need to use the specific language of general relativity which defined the operator as

□^2_μv = g_μv [∇, ∇] = ∂_μ∂_ν − Γ^σ_μν ∂_σ

Which uses the connection, the main correction factor that took us out of Newton Ian thinking;

∇ = ∂ + Γ

The full relationship found from general relativity is

g^μv R_μv = g^μv [∇_μ, ∇_v]

= g^μv ∇_μ (∂_v) = g^μv (∂_μ∂_v + Γ^σ_μv ∂_σ)

= - 1/√-g ∂_μ (g^μv √-g ∂_v)

The Survival Probability and Collapse of Ricci flow

We start with the new Ricci flow interpretation

∂_0 g_μν( R) = - k (2 ∂_0 T_μν - ∂_0 g_μν T)

This form of the equation on the right hand side could be interpretated from a difference in entropy as;

S(p, q) = Σ_i p_i(log_2 p_i - log_2 q_i)

Giving the geometry a mathematical property of statistical survival. Doing so is not difficult, you just rewrite the entropy in terms of energy and interpret the energy through the presence of a stress energy tensor, which typically has units of energy density. If gravity has a relationship to a statistical nature, it can be assigned a wave function and in such cases, geometric collapses are allowed to happen as Anandan has shown in his own work. I extended those idea to Mandlestram-Tamm inequalities. Entropy is defined the usual way

S = Σ_i |c_i|^2 log_2 |c_i|^2

Or as

S = sum 1/N ln(1/N) = ln(1/N)

The energy density related to entropy is then

ρ = ln(1/N) nkT

And a stress energy is obtained simply as

T = 1/2 ρc

where here k is not the Einstein constant but the Boltzmann constant and n is the particle number density. It is thermal in description but due to equipartition the kinetic energy is roughly equivalent

1/2 mv^2 = 3/2kT

In the low energy regime.

And moving particles have further consequence to the model concerning velocity pressures.

Using

∂_0 g_μν(R, t) = - k (T_μν - 1/2g_μν T)

We can form;

∂_0 g_μν (ψ, ψ') = - k Σ ψ (log_2 T_μν ψ - 1/2 log_2 T_μν ψ' )

With g_μν units of time and inverse length squared. Just not using usual notation to stop it looking crowded

This flow now has a statistical nature brough to the geometric evolution. In a collapsed situation upon the square of the wave function we expect the amplitude to form as;

S = Σ |ψ|^2 log_2 |ψ|^2

And the observed difference

S(ψ,ψ') = Σ |ψ|^2 (log_2 |ψ|^2 - log_2 |ψ' |^2)

Known as a Shannon entropy. In this form it has an identical structure to

S(p, q) = Σ_i p_i(log_2 p_i - log_2 q_i)

The whole point of giving the metric a collapse probability is so that the Ricci flow can be modelled under survival probabilities.

Let's cover some basics first, Einstein's equations are;

G_μv = - 2R_μν + 2/n R g_μν = g_μν k T

Where

T = 1/2 ρcu

k = 8πG/c^4

And the scalar curvature is

R = g^μv ∇_μ∇_v

The curvature tensor is

R_μv = g_μv R = ∇_μ∇_v = ∇_μ (∂_v φ) = ∂_μ∂_v + Γ^σ_μv ∂_σ

Generally will not commute even in classical physics,

[∇_μ, ∇_v] = (∂_μ + Γ_μ) (∂_v + Γ_v)

With g having units of inverse length squared and one inverse of time as found in the Ricci flow allows us to write it as

g^μv □_μv = ∇^μ ∂_μ √-g ∇^v ∂_v

□_μv = [∇^μ, ∇^v] = (∂_μ + Γ_μ)√-g(∂_v + Γ_v)

= √-g (∂_μ ∂_v + ∂_μ Γ_v + Γ_μ ∂_v + Γ_μ Γ_v)

The torsion typically arises from the last terms (which are under commutation laws)

[Γ_μ, Γ_v] = Γ_μ Γ_v - Γ_v Γ_μ

The Ricci flow is

∂_0 g^μv / ∂ √-g = - 2 √-g (∂_0 g^μv / ∂ g) = 4 √-g (R^μv / ∂g)

While a geometric flow is does not preserve the volume so we write,

∂_0 g_μν(t) / ∂ √-g = - √-g (4R_μν + R g_μν) / ∂g

But here is the interesting thing that leads us to a new equation for the Ricci flow where we notice the renormalized case is identical to the Einstein equations of the form

G_μν = R_μν - D/2Rg_μν

When you plug in for the spacetime dimension D=4 you get 4R_μν - 2Rg_μν and when divided through by four we get R_μν - 1/2Rg_μν. From here we remind ourselves we can rewrite the factor as the stress energy tensor

G_μν = R_μν - 1/2 g_μν (-kT)

Allowing us to define

R_μν = k (T_μν - 1/2 g_μν T)

This means the Ricci flow has to be associated to an energy flow, we don't need to meddle about now with the volumetric case because we will find the ordinary Ricci flow will suffice,

∂_0 g_μν(t) = - 2R_μν

And redefine under the equivalent form of the curvature tensor using the equation we stated before

R_μν = k (T_μν - 1/2 g_μν T)

And the Ricci flow becomes a stamement about the flow contributed by the stress energy tensor and is attributed to the manifolds geometry evolution as,

∂_0 g_μν(t) = - k (2T_μν - g_μν T)

Since the Ricci flow is a heat flow, there may be a universal application for the formula in regards that the stress energy tensor is applied to all types of fields in nature. A heat flow under the stress energy will require a proper meld of the right physics. If we make all metric terms equal in dimensions would help keep in clearer, involving some small modifications,

∂_0 g_μν( R) = - k (2 ∂_0 T_μν - ∂_0 g T)

The right hand side can be reinterpreted into four dimensions with first derivatives in □ giving with c =1. And to get the four dimensional case as you will know simply requires replacing the complete 3-space operator ∇ for its punctual cousin □ sticking to first order equations in the process. Three dimensional case

∂_0 g_μν( R) = - 2k ∇ T_μν

= - 2k (∂_x T_μν + ∂_y T_μν + ∂_z T_μν)

And also for calculation the squared form of this

g_μν ∂^0 R^μν( R) ∂_0 R g_μν = - 2k g_μν [∇, ∇] T_μν T^μν

= - 4k^2 (∂_x T_μ + ∂_y T_μ + ∂_z T_μ)(∂_x T_v + ∂_y T_v + ∂_z T_v)

No one taught me how to do this, but it is perfectly acceptable it seems to interpret a volumetric flow in this way. And the four dimensional case is thus a modification of;

∂_0 g_μν(vR) = - 2 □ kT_μν

= - k (2 ∂_x T_μν + 2 ∂_y T_μν + 2 ∂_z T_μν - ∂_0 g_μν T)

The d'Alemertian operator is inside this last equation where g_μν(R, t) just means a metric with Riemannian curvature dimension (typically) of inverse length squared so g(t) would be concerned with time alone ang g(x... N) same idea here but is related to the dimensions position length derivative alone. Here we would relate any metric without a dependent term as dimensionless for clarity.

We are allowed to implicate the d'Alemertian because of the presence of the time derivative, a feature which del operator lacks.

∂_x + ∂_y + ∂_z - ∂_0

Which is a low limit relativistic case and for this we had to choose then first derivative argument instead of the second, which under literature is

□^2 = ∂^2_x + ∂^2_y + ∂^2_z - ∂^2_0

And it is from here I will apply my knowledge on heat conduction for a Riemannian manifold and later transpose the above equation into curved space. To take it into curved space we need to use the specific language of general relativity which defined the operator as

□^2_μv = g_μv [∇, ∇] = ∂_μ∂_ν − Γ^σ_μν ∂_σ

Which uses the connection, the main correction factor that took us out of Newton Ian thinking;

∇ = ∂ + Γ

The full relationship found from general relativity is

g^μv R_μv = g^μv [∇_μ, ∇_v]

= g^μv ∇_μ (∂_v) = g^μv (∂_μ∂_v + Γ^σ_μv ∂_σ)

= - 1/√-g ∂_μ (g^μv √-g ∂_v)

The Survival Probability and Collapse of Ricci flow

We start with the new Ricci flow interpretation

∂_0 g_μν( R) = - k (2 ∂_0 T_μν - ∂_0 g_μν T)

This form of the equation on the right hand side could be interpretated from a difference in entropy as;

S(p, q) = Σ_i p_i(log_2 p_i - log_2 q_i)

Giving the geometry a mathematical property of statistical survival. Doing so is not difficult, you just rewrite the entropy in terms of energy and interpret the energy through the presence of a stress energy tensor, which typically has units of energy density. If gravity has a relationship to a statistical nature, it can be assigned a wave function and in such cases, geometric collapses are allowed to happen as Anandan has shown in his own work. I extended those idea to Mandlestram-Tamm inequalities. Entropy is defined the usual way

S = Σ_i |c_i|^2 log_2 |c_i|^2

Or as

S = sum 1/N ln(1/N) = ln(1/N)

The energy density related to entropy is then

ρ = ln(1/N) nkT

And a stress energy is obtained simply as

T = 1/2 ρc

where here k is not the Einstein constant but the Boltzmann constant and n is the particle number density. It is thermal in description but due to equipartition the kinetic energy is roughly equivalent

1/2 mv^2 = 3/2kT

In the low energy regime.

And moving particles have further consequence to the model concerning velocity pressures.

Using

∂_0 g_μν(R, t) = - k (T_μν - 1/2g_μν T)

We can form;

∂_0 g_μν (ψ, ψ') = - k Σ ψ (log_2 T_μν ψ - 1/2 log_2 T_μν ψ' )

With g_μν units of time and inverse length squared. Just not using usual notation to stop it looking crowded

This flow now has a statistical nature brough to the geometric evolution. In a collapsed situation upon the square of the wave function we expect the amplitude to form as;

S = Σ |ψ|^2 log_2 |ψ|^2

And the observed difference

S(ψ,ψ') = Σ |ψ|^2 (log_2 |ψ|^2 - log_2 |ψ' |^2)

Known as a Shannon entropy. In this form it has an identical structure to

S(p, q) = Σ_i p_i(log_2 p_i - log_2 q_i)

The whole point of giving the metric a collapse probability is so that the Ricci flow can be modelled under survival probabilities.

**Edited by Dubbelosix, Yesterday, 11:42 AM.**

### #4

Posted Yesterday, 12:16 PM

I have plans to take these ideas now further as the flow geometry will involve the drag as it has been under my model, sorry I cannot express the Tex code at this moment, but for now this will have to do which highlights the derivation of drag into three dimensions.

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 (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:p

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

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...Reynolds_number)

https://en.wikipedia...rag_coefficient (https://en.wikipedia...rag_coefficient)

**Drag Coefficient in Three Dimensions**

$\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 $

and/or

$\frac{mv^2}{n^2fx^3} + \frac{mv^2}{n^2fy^3} + \frac{mv^2}{n^2fz^3} = - \frac{A_b}{A_f} \rho v^2 = - \frac{A_b}{A_f} T^{00} $

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

**The Drag Coefficient Explains the Cosmological Constant as a Thrust**

Using the derivation in previous work for the drag in three dimensions, it was found can be formulated as:

$\frac{1}{f}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - \frac{A_b}{A_f} \cdot \rho v^2 = - \frac{A_b}{A_f} \cdot T^{00} $

Solving for the stress energy alone we have

$ \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - T^{00} $

(concentrate on the sign, because this does not indicate a negative energy per se, we will see how the sign is connected specifically to the definition of the cosmological impetus $\Lambda$)

Einstein did think the cosmological constant as an independent parameter, but as we know today, it is inherently linked to the energy of the vacuum - in the mechanical explanation I have provided in my own work, it arises as a ''pressure difference'' and that slightly more momentum was owed to the expansion of the universe rather than it collapsing. Instead of writing that avenue out, all we need for now is to write the energy-momentum component as:

$T^{00} = - g^{00}\ \frac{\Lambda c^4}{8 \pi G}$

And the density as

$\rho_{vac} = \frac{\Lambda c^2}{8 \pi G}$

So the first equation can be formulated as

$- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = \frac{\Lambda c^2}{8 \pi G} = \rho_{vac} v^2 $

Which relates the cosmological constant as a ''thrust.. "

And the second replacement for the stress energy:

$ - \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00} $

The conclusion here is that the cosmological constant is a type of cosmic thrust!

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 (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:p

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

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...Reynolds_number)

https://en.wikipedia...rag_coefficient (https://en.wikipedia...rag_coefficient)

**Drag Coefficient in Three Dimensions**

$\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 $

and/or

$\frac{mv^2}{n^2fx^3} + \frac{mv^2}{n^2fy^3} + \frac{mv^2}{n^2fz^3} = - \frac{A_b}{A_f} \rho v^2 = - \frac{A_b}{A_f} T^{00} $

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

**The Drag Coefficient Explains the Cosmological Constant as a Thrust**

Using the derivation in previous work for the drag in three dimensions, it was found can be formulated as:

$\frac{1}{f}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - \frac{A_b}{A_f} \cdot \rho v^2 = - \frac{A_b}{A_f} \cdot T^{00} $

Solving for the stress energy alone we have

$ \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - T^{00} $

(concentrate on the sign, because this does not indicate a negative energy per se, we will see how the sign is connected specifically to the definition of the cosmological impetus $\Lambda$)

Einstein did think the cosmological constant as an independent parameter, but as we know today, it is inherently linked to the energy of the vacuum - in the mechanical explanation I have provided in my own work, it arises as a ''pressure difference'' and that slightly more momentum was owed to the expansion of the universe rather than it collapsing. Instead of writing that avenue out, all we need for now is to write the energy-momentum component as:

$T^{00} = - g^{00}\ \frac{\Lambda c^4}{8 \pi G}$

And the density as

$\rho_{vac} = \frac{\Lambda c^2}{8 \pi G}$

So the first equation can be formulated as

$- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = \frac{\Lambda c^2}{8 \pi G} = \rho_{vac} v^2 $

Which relates the cosmological constant as a ''thrust.. "

And the second replacement for the stress energy:

$ - \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00} $

The conclusion here is that the cosmological constant is a type of cosmic thrust!

### #5

Posted Yesterday, 12:17 PM

The final discoveries which used the stress energy will become a focal point to constructing the new set of equations I have in mind which the Ricci flow and the drag are related phenomenon, so more to come pretty soon.