Under active development

This blog is being built along with a blogging framework for the entire Conundrum ecosystem.

This is taking significantly more time than building a simple blogging website, but when this is complete in a couple months, all users will be able to publish their notes as a blog, independent of any specific service, including Fluster.

On The Gravitational Nature Of Time

A sample note & a novel interpretation of relativity yielding multiple directly observed quantities to well within measurement error.
Thank you

Thank you for taking the time to read this. My story is posted elsewhere, so to keep things short, this is the entire reason I built Fluster & Conundrum in the first place. Almost 5 years ago I realized Einstein made an assumption that made far more sense before the observations that give us the Big Bang, and after playing around with the math I realized I can derive an absolutely staggering amount of physics from a single, mathematically equivalent modification to Einstein's model. I quit my job, became homeless, and over the course of this pursuit built the original version of Fluster for my own personal use.

I'll do my best to convince the math types among you (this article does assume some basic understanding of physics and relativity) by the end of this sample note that Einstein made a mistake1, but regardless of your opinions on the model, I hope you enjoy Fluster and find it as useful as I have. Thanks for checking it out!

Included Derivations

This table is generated directly from Astropy and Scipy . All functions reflect names in the alpha_omega_gravity python package, because after 5 years of being homeless I think I earned the right to be grandiose. (α\alpha comes from the fine-structure constant).
Notebook There's now an interactive notebook published as a standalone application available for Mac and as a Jupyter notebook.
Derivation Function % Error
cosmological velocity from local gravitational parameters vbarCosmological Difficult to assess
kinematic velocity from local gravitational parameters vbarKinematic 0.442258391884371%0.442258391884371\%
Earth's rotational velocity via α\alpha rotationVelocityFromDivergence 0.00677756400547747%0.00677756400547747\%
Solar-Galactic Radius from visible mass galacticRadiusPolynomial 0.0%0.0\%
α\alpha (solar) alphaSolar 0.00342749503676447%0.00342749503676447\%
α\alpha (Earth) omega , alphaFactor 0.0253043303850287%0.0253043303850287\%
α\alpha (orbital velocity) solarMetricVelocity , vOrbit 0.0544665494810030%0.0544665494810030\%
α\alpha (from derived velocity) vbarAlphaEquivalence 0.0761820393770567%0.0761820393770567\%
JΦJ \approx \Phi currentEquivalent 0.213633171310522%0.213633171310522\%
hh from α\alpha omega , alpha 0.0253887188634106%0.0253887188634106\%
h/(2α)h / (2 - \alpha) from ω\omega omega , alpha 0.00342749503676447%0.00342749503676447\%
Φ\Phi mass density approximation massDensityEquivalence 0.863214423082925%0.863214423082925\%
GG as a function of α\alpha and ρ\rho -- 0.07618221222642045%0.07618221222642045\%
Cummulative magnetic diverence at R=ΦR = \Phi -- 0.0080514014152570250.008051401415257025%
Error All error percentage calculations were done with the following function. See the Fluster & Conundrum repository for a complete .ipynb notebook and/or install the alpha_omega_gravity python package.
percent_diff.py
def percentDiff(observed, derived):
    return (observed - derived) / observed * 100
    

Also, keep in mind that if we wanted to be selective we can choose a radius here on Earth's surface that will satisfy many of the above equations perfectly, giving an error of 0. The radii that were used were whichever was most fitting for the job... the mean radius for calculating volume, the polar radius when spin was involved to minimize centripetal influence, and Astropy provided values (equatorial) for everything else.

The Lighthouse & The Clocktower

An opening thought experiment

Consider two observers, AA and BB. Let AA and BB agree that BB should travel between two arbitrary points in space2, p1p_1 and p2p_2 at an agreed upon velocity vv.

At p2p_2 place a time keeping device that remains visible from p1p_1. Allow that AA remains at rest relative to p1p_1, and very near p1p_1 while BB goes in motion (at an inertial pace) between p1p_1 and p2p_2.

At what time does AA observe BB reach p2p_2? If the value is not d/vd/v, but γd/v\gamma d/v, then vvv \neq v_\prime, unless ddγd \to d\gamma. The only possible solution that allows for Einstein's additional factor of γ\gamma while requiring that v=vv=v_\prime in the coordinate system of the observer who's kinetic state remains unchanged, and that both AA and BB agree upon the time at which BB reaches p2p_2 is that ddγd \to d\gamma in the coordinate system of BB.

To summarize, we have 3 possibilities:

  1. Every single experimental validation of special and general relativity was flawed. This is statistically almost impossible.
  2. BB returns from p2p_2 with a photo of BB at p2p_2 that disagrees with the photo AA took of BB at p2p_2. There is nothing in physics that prohibits extending the time periods and distances to produce such absurdities as BB wearing a completely different shirt.
  3. dd elongates by a factor of γ\gamma. We might then presume that this dilation is the source of the equivalence principle, and watch how the physics just lines up.

If you think about it, isn't this inline with what we already believe? Do current models not describe cosmic inflation as a time dependent property? Does more time not pass for the observer in relative motion? Of course p1p_{1} and p2p_{2} will be further separated, if everything dilates, not just the space in-between galaxies. If we presume that this dilation Einstein applied to time may be applied to space in the coordinate system of the observer in relative motion, then we can describe this spatial dilation as the mechanism to the equivalence principle and combine cosmic inflation, gravitational acceleration and time itself into a single process as observed from different reference frames!

For the physics professionals: Yes, this model violates the Lorentz transformations, but it does so while satisfying every single experimental validation of either SR or GR. Don't stop reading yet, it's just about to get good.

Time Gravity Equivalence

Let's recognize the two primary properties that time occupies:

  • A 4th displacement coordinate
  • A driving rate of change

By attributing the 4th displacement coordinate to the density axis that arises from this geometry and the driving rate of change to motion, not time, we can remove our temporal axis entirely! Time becomes nothing more than a mathematical tool... a ratio of other changing properties, all in response to the primary rate of change: motion.

Of course we need to change the dimensionality of much of our physics, but as I will demonstrate here... the math remains remarkably consistent.

Our Cosmological Velocity

Let's consider that this dilation of space might give rise to our gravitational acceleration, where p1p_1 becomes akin to the cosmological rest frame. If we allow that this velocity gives rise to spatial dilation, which then gives rise to the gravitational acceleration that we feel, we should examine the derivative of gg with respect to RR.

Plot taken from Determining the motion of the Solar system relative to the cosmic microwave background using Type Ia supernovae by C. Gordon et. al.
δ=ddRg=2GMR3 \delta = \frac{d}{dR} g = 2 G \frac{M}{R^3}
1

If we take the average of this integral, we find a scalar product that can be used in the same manner as γ\gamma:

1R0R02GMR3=1+gR \frac{1}{R} \int_0^{R_0} 2 G \frac{M}{R^3} = 1 + \frac{g}{R}
2

If we solve γ\gamma for vv, we find:

vγ=c11γ2 v_\gamma = c \sqrt{1 - \frac{1}{\gamma^2}}
3

Substituting the value from equation 2 for γ\gamma we find:

vγ=c11(1+gR)2=526.6 km s1 v_\gamma = c \sqrt{1 - \frac{1}{\left(1 + \frac{g}{R}\right)^2}} = 526.6~ \text{km}~ \text{s}^{-1}
4

This value coincides closely with direct observation, but due to the wide error margins inherent in these types of observations, we're going to have to do better. Don't worry... we can.

Our Kinematic Velocity

Let's consider the fact that Earth kind of has two velocities, our kinematic velocity (the motion through space), and our cosmological velocity (the motion through space plus the motion of space). Since we just derived our cosmological velocity, we should now remove the temporal axis, giving us a velocity within a contained 3-dimensional 'slice'. This gives a velocity relative to the 'dough' in the common raisin bread analogy of cosmic inflation.

Let's first recognize a time dependent quantity as a dimensionless ratio of distances:

dx+dsdxΔt \frac{dx + ds}{dx} \propto \Delta t
5

If we consider that time has no driving properties anymore, we should then turn our attention to motion. Since this model proposes that dsds (cosmic inflation and now gravitational acceleration) is time, we can find this time dependent quantity as follows:

Δτ=px˙0vx˙ \Delta \tau = \frac{\int_p \dot{x}}{\int_0^v \dot{x}}
6
Notation
Introducing the notion of 'proportionally bound' derivatives and integrals

This notation is likely confusing, so let me clarify. To satisfy the properties time occupies, this quantity is integrated in a manner that is unique, keeping the denominator fixed at 1 unit of distance (the velocity in units of distance) while the integral p\int_p is meant to indicate a 'proportional' integral, or an integral that goes on indefinitely as 'time' progresses.

This gives us a dimensionless quantity that is distance/distance, of a magnitude that is equivalent to tt in this form, and a magnitude equivalent to 12t2\frac{1}{2}t^2 when integrated.

Let's then re-examine equation 4 and it's derivation. Since dxdx in equation 4 is per unit time, we should then divide by vγv_\gamma and multiply by our time equivalent quantity from equation 6 .

This gives:

0R0x[2GMR31vγτ] dR dτ \int_0^R \int_0^x \left[ -2 G \frac{M_\oplus}{R_\oplus^3} \frac{1}{v_\gamma} \tau \right]~ dR ~ d\tau
7

And following the rules of integration defined above, we find:

0R0x[δ1vγτ] dR dτ=1+12GMR31vγx˙2x^ \int_0^R \int_0^x \left[ \delta \frac{1}{v_\gamma} \tau \right]~ dR ~ d\tau = 1 + \frac{1}{2} \frac{G M_\oplus}{R^3_\oplus}\frac{1}{v_\gamma} \frac{\dot{x}^2}{\hat{x}}
8
Notation Here and throughout the rest of this note, x˙\dot{x} does not indicate a derivative with respect to time, but rather a fundamental, driving rate of change. x^\hat{x} is a notation used to demonstrate the magnitude equivalent of 11 for that velocity, making x^=v\hat{x} = v in units of distance. I'm hesitant to say x^=v 1t\hat{x} = v~1t, because while mathematically accurate, I feel that it hides underlying physics, or rather invents a derivative that is not an inherent part of the system only to remove it by multiplying by tt.

If we substitute this result back into equation 4 and solve for Δt=1\Delta t = 1 we find:

vγ=c11(1+12GMR3)2=371,567 km s1 v_\gamma^\prime = c \sqrt{1 - \frac{1}{\left( 1 + \frac{1}{2} G \frac{M}{R^3} \right)^2}} = 371,567 ~ \text{km} ~ \text{s}^{-1}
9

Observational ranges for this value that we can observe much more reliably is between 368,145369,943 km s1368,145-369,943~ \text{km}~ \text{s}^{-1}, giving us a margin of error of:

Error Margin: 0.43%0.92%0.43\%-0.92\%

Curvature

Note here, that as we integrate τ\tau, RR should be elongated by dRdR, or R=R+gΔtR^\prime = R + g \Delta t. However, this is obviously not what we observe due to our co-motion along our temporal axis. In other words, this integral is applied as a density gradient, inducing a curvature of precisely 1R\frac{1}{R} in the coordinate system of the observer in co-motion.

This means that a linear path in the coordinate system of the 'oven' may be an accelerated trajectory in the coordinate system of the 'dough' due to this intrinsic curvature, should that intrinsic curvature be smooth, continuous, and in our case... radial. This is what allows this steady rate of spatial dilation to produce the appearance of acceleration for a body with no known direct physical interaction; as gravity converts space to time, so too it converts x˙\dot{x} to x¨\ddot{x}. This axis acts as a spatial density gradient in the coordinate system of the observer in co-motion who is experiencing these dilatory effects of time, inducing this 1R\frac{1}{R} behavior, but is expressed as Euclidean distance in the coordinate system of the 'oven'.

doven=Cx˙ϕ(x,y,z) d_{\text{oven}} = \int_C \dot{x} \phi_{\left( x,y,z \right)}
10

Where ϕ\phi represents this spatial density. ϕ=1\phi = 1 at RR and scales according to dr\uparrow \uparrow dr where drdr is the radial distance from Earth's surface radius.

This in turn maps x˙x¨\dot{x} \mapsto \ddot{x}, remedying the asymmetry inherent in this model while resolving the issue of terminating geodesics in GR. Blackholes are now nothing more than a mirage in the distance that dissipates as an observer approaches that body's gravitational field.

Interestingly, if we take Earth's radius super-factorial, we find:

αηR=αη(1+3+6+10R)=AU \alpha \eta \uparrow \uparrow R = \alpha \eta \left( 1 + 3 + 6 + 10 \dots R_{\oplus}\right) = \mathbf{AU}
11

Then η1.014\eta \approx 1.014, approximating Earth's eccentricity.

super_factorial.js
let total = 0;
let n = 0;
let R = 6378100;
const ASTRONOMICAL_UNIT = 149597870700;

do {
  total += R - n;
  n += 1;
} while (n < R);

console.log(`
Total: ${total}
AU / Fundamental Radius: ${
    // Prints 0.007354830889518064
    ASTRONOMICAL_UNIT / total
    }
`);
From Orbital Parameters

Let's validate this model by testing this relationship one more time.

Consider the diagram below. If this model is correct, we can improve the derivation of our orbital behavior. Instead of the following:

Fcentripital=Fgravity F_{\text{centripital}} = F_{\text{gravity}}
12

We can solve this trigonometrically.

R02+x˙2dt2=(0R02GMR3)2=(R0+g)2 R_0^2 + \dot{x}^2 dt^2 = \left(\int_0^{R_0} 2 G \frac{M}{R^3}\right)^2 = \left(R_0 + g\right)^2
13

Solving this equality for dtdt gives only one non-zero solution:

dt=2GMR3G2M2R4x˙2 dt = - \frac{2 G M R^{3}}{G^{2} M^{2} - R^{4} \dot{x}^{2}}
14

If we substitute in x˙\dot{x} for it's Newtonian value, we find:

dt=R03(112GMR03)=1.00000077 dt = R_0^3 \left(\frac{1}{\frac{1}{2} GM - R_0^3}\right) = 1.00000077
15

As relativity predicts this value does not give 1, but a magnitude slightly higher as γ>1\gamma \gt 1.

Plugging this value in to the equation we used to find δ\delta gives:

v=c11dt2=373,453 km s1 v = c \sqrt{1 - \frac{1}{dt^2}} = 373,453 ~ \text{km} ~ \text{s}^{-1}
16

This value again coincides closely with our velocity observed relative to the CMB dipole, once again laying credibility to the model we are proposing... gravitational acceleration is a velocity dependent property.

In the relativistic frame

As we used the Newtonian derived orbital parameters in the previous section to derive a relativistic velocity correction in the form of an increase in elapsed 'time', we can now derive a relativistic equation for a body in stable orbit by setting dt=1dt = 1 and solving for x˙\dot{x}. This gives a model in the coordinate system of the observer in motion, as he travels at the 'agreed upon velocity' (any velocity multiplied by a scalar of 1).

If we extend our lighthouse & clocktower example to an orbital frame of reference, where BB orbits AA at an agreed upon velocity vov_o, then this velocity would be decreased by a factor of dtdt in the coordinate system of AA due to the decreased spatial density of SB\mathbf{S}_B, while our relativistically corrected orbital velocity equation becomes as follows:

GMRGM(GM+2R3)R2 \sqrt{\frac{GM}{R}} \quad \to \quad \frac{\sqrt{G M \left(G M + 2 R^{3}\right)}}{R^{2}}
17

When we solve for RR and plugin our galactic parameters here on Earth, we find a radius of 8.13 Kpc8.13~ Kpc, perfectly on top of observational values, without any additional unseen mass!

In the plot above, 'standard' is our standard orbital velocity equation (GM/R)1/2\left( G M / R \right)^{1/2}, 'derived' is the equation we derived in equation 17 , and 'local' points to our Sun using (R,vobserved)\left(R_{\oplus}, v_{\text{observed}}\right) where vobservedv_{\text{observed}} is the velocity found by A. Eilers and colleagues, 229,000 m s1229,000 ~ \text{m} ~ \text{s}^{-1} (The Circular Velocity Curve of the Milky Way from 5 to 25 kpc). The actual value we obtain is 230,145.52 m s1230,145.52 ~ \text{m} ~ \text{s}^{-1}, without any unseen mass.

Electromagnetism

Consider the determinant of a 3-dimensional, diagonal matrix, representing the change in volume of the coordinate system.

s=ω000ω000ω =ω3 s = \left|\begin{matrix} \omega & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 &\omega\ \end{matrix}\right| = \omega^3
18

Not only does this accommodate existing experimental validations of SR & GR1, it does so while accommodating the original relativistic experiment: Michelson & Morely23. The only direction they didn't evaluate was along the radial vector, precisely the direction described by this determinant.

If we consider that the magnitude of δ\delta is this determinant, we should then take it's cube root:

ω=δ3=2GMR33=0.01453 1s \omega = \sqrt[3]{\delta} = \sqrt[3]{\frac{2 G M_\oplus}{R_\oplus^3}} = 0.01453 ~ \frac{1}{\text{s}}
19

Woah... that number looks familiar that's 12α±0.39%\frac{1}{2} \alpha \pm 0.39\%! As a matter of fact, when you solve for the 2 , the solution is actually:

2α±0.025% 2 - \alpha \pm 0.025\%
20

Giving a final equation of:

α=12α2GMR33±0.025% \boxed{\alpha = \frac{1}{2 - \alpha} \sqrt[3]{2 \frac{G M_\oplus}{R_\oplus^3}} \pm 0.025\%}
21

Solving this equation for α\alpha gives:

α=1±1(2GMR3)1/3=1.9927 or 0.0072953 \alpha = 1 \pm \sqrt{1 - \left(\frac{2 G M_\oplus}{R_\oplus^3}\right)^{1/3}} = 1.9927 \text{ or } 0.0072953
22

That's almost precisely the value we were looking for when we add these values, and almost exactly α\alpha again when we subtract them, giving very Fibonacci-like vibes.

Final Error: 0.025%0.025\%

Using our solar parameters, equivalent to spinor=(12)1/3\text{spinor}=\left( \frac{1}{2} \right)^{1/3}

Let's consider the fact that we have a set of unique properties for two known bodies, us, and the sun. First, we know that all of our observational 'constants' are accurate here. Second, we know that the Sun is the dominant gravitational object in the vicinity. Let's take advantage of this fact by presuming that the lower limit of this 'spinor' approximation is found at the Sun.

Let's consider this function where the spinor=(12)1/3\text{spinor}=\left( \frac{1}{2} \right)^{1/3}, or the same quantity before the torsion applied by α\alpha and the dilation applied by dsds.

Plugging this back into equation 21 gives:

\limΓ(12)1/32GMR33=GMR33=α±0.0034% \lim_{\Gamma \to \left(\frac{1}{2}\right)^{1/3}} \sqrt[3]{2 \frac{G M_\oplus}{R_\oplus^3}} = \sqrt[3]{\frac{GM_\odot}{R_\odot^3}} = \alpha \pm 0.0034\%
23

Where

Γ=12α \Gamma = \frac{1}{2 - \alpha}
24

Demonstrating that this function is more than just a coincidence.

Error 1±α1 \pm \alpha with an error of 0.00347%0.00347\% and 0.003427%0.003427\% respectively.

Integrating Magnetic Divergence

Let's now take a second to recognize that to find this quantity

α(2α)=ddRg3 \alpha \left(2 - \alpha\right) = \sqrt[3]{\frac{d}{dR} g}
25

We first took the inverse of divergence. Let's now cube it to bring it back up to scale.

(α(2α))3=ddRg=2GMR3 \left( \alpha \left( 2 - \alpha \right) \right)^3 = \frac{d}{dR}g = 2 \frac{G M_{\oplus}}{R_{\oplus}^3}
26

Or expanded, this gives

ddRg=2GMR3=B=e6(μ0ϵ0)32(2e2μ0ϵ02h)38h3 \frac{d}{dR}g = 2 G \frac{M}{R^3} = \nabla \cdot \vec{B} = \frac{e^{6} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}} \left(2 - \frac{e^{2} \sqrt{\frac{\mu_{0}}{\epsilon_{0}}}}{2 h} \right)^{3}}{8 h^{3}}
27

If we then slide the RR_{\oplus} over to the over side to integrate magnetic divergence according to RR, we find:

0RB=R4(e6(μ0ϵ0)324h3+3μ0e10(μ0ϵ0)3264ϵ0h53μ02e816ϵ02h4μ03e12256ϵ03h6)=1.272122073168672×1021 \int_0^R \nabla \cdot \vec{B} = R^{4} \left(\frac{e^{6} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}}}{4 h^{3}} + \frac{3 \mu_{0} e^{10} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}}}{64 \epsilon_{0} h^{5}} - \frac{3 \mu_{0}^{2} e^{8}}{16 \epsilon_{0}^{2} h^{4}} - \frac{\mu_{0}^{3} e^{12}}{256 \epsilon_{0}^{3} h^{6}}\right) = 1.272122073168672 \times 10^{21}
28

That's

Φ×1042±0.00805% \sqrt{ \Phi \times 10^{42}} \pm 0.00805\%
29

Where the above plot illustrates the intersection of this function with

(R,Φ×1042) \left(R_{\oplus}, \sqrt{ \Phi \times 10^{42}}\right)
30
Significance of α\alpha

If you're unfamiliar with physics, let's take a second to break down what we're doing here.

This is what α\alpha looks like completely written out:

α=14πe2μ0ϵ0 \alpha = \frac{1}{4 \pi}\frac{e^2}{\hbar}\sqrt{\frac{\mu_0}{\epsilon_0}}
31
Component What it does
1/4π1/4\pi A consequence of the spherical geometry
ϵ0\epsilon_0 The electric constant
μ0\mu_0 The magnetic constant
ee The fundamental charge
\hbar The momentum carrying quantity of a massless photon.
1μ0ϵ0\frac{1}{\sqrt{\mu_0 \epsilon_0}} The speed of light, and therefore the speed of causality.

That's basically the relationship of all of electromagnetism described in one function, now defined in terms of gravitational parameters. Don't worry though, we can keep going! Now that we've derived a relationship to α \alpha, we can take advantage of the electromagnetic equations. Let's take advantage of the fact that the electromagnetic tensor is undefined at Δt=0\Delta t = 0. If electromagnetism can not exist without time, maybe time is masking a subset of electromagnetic properties.

Electric Field Divergence Equivalent

Let's start by assuming the following identity:

E=c(B) \nabla \cdot E = c \left(\nabla \cdot B\right)
32

Since

S=δ \nabla \cdot \mathbf{S} = \delta
33

then

B=e6(μ0ϵ0)32(2e2μ0ϵ02h)38h3δ \nabla \cdot \mathbf{B} = \frac{e^{6} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}} \left(2 - \frac{e^{2} \sqrt{\frac{\mu_{0}}{\epsilon_{0}}}}{2 h} \right)^{3}}{8 h^{3}} \approx \delta
34

and

E=ce6(μ0ϵ0)32(2e2μ0ϵ02h)38h3=921.116 meters \nabla \cdot \mathbf{E} = c \frac{e^{6} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}} \left(2 - \frac{e^{2} \sqrt{\frac{\mu_{0}}{\epsilon_{0}}}}{2 h}\right)^{3}}{8 h^{3}} = 921.116 ~\text{meters}
35

If we then multiply equation 35 by our spinor approximation, converting this linear spatial dilation to rotation, we find:

E=e6(μ0ϵ0)32(2e2μ0ϵ02h)2(μ0ϵ0)128h3=462.244 meters=dθdtRTday ±0.006778% \nabla \cdot \mathbf{E} = \frac{e^{6} \left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{\frac{3}{2}} \left(2 - \frac{e^{2} \sqrt{\frac{\mu_{0}}{\epsilon_{0}}}}{2 h} \right)^{2}}{ \left( \mu_0 \epsilon_0\right)^{\frac{1}{2}} 8 h^{3}} = 462.244~ \text{meters} = \frac{d \theta}{dt} R_\oplus T_{\text{day}} ~ \pm 0.006778\%
36

That's our rotational velocity on Earth's equator ±0.006778%\pm -0.006778\%! Off by a total of 0.03 meters (1.18 inches) for a quantity that takes into account a radius the size of Earth's and a period as long as our day!

Dimensionality

As in traditional models,

x˙=v=[dxdt] \dot{x} = v = \left[\frac{dx}{dt}\right]
37

This model uses a velocity that is

x˙=v=[dxds] \dot{x} = v = \left[\frac{dx}{ds}\right]
38

When multiplied by our 'time' equivalent ds=δds = \delta, this gives

x˙ ds=distance \dot{x}~ ds = \text{distance}
39
In the same manner that vt=distancevt = \text{distance}

Velocity as a function of α\alpha

Let's consider equation 21 once again. If we re-arrange this equation such that dsds is isolated on one side, we find:

α(2α)=ds=δ1/3 \alpha \left( 2 - \alpha \right)= ds = \delta^{1/3}
40

Cubing δ\delta to bring the units of time back to a power of 1 gives:

(α(2α))3=ds3=δ \left(\alpha \left( 2 - \alpha \right) \right)^3 = ds^3 = \delta
41

If we consider that all velocities are proportional in magnitude to cc, we should not look for vv, but for β2\beta^2. This then gives:

(α(2α))3=β02±0.08% \left(\alpha \left( 2 - \alpha \right) \right)^3 = \beta_0^2 \pm 0.08\%
42

One can very logically make an argument for the small-angle approximations being a cause for coincidence here, but as a function of α\alpha?

Expanding this using our earlier derivation of our cosmological velocity gives

(α(2α))3=11(1GMR3)2 \left(\alpha \left( 2 - \alpha \right) \right)^3 = 1 - \frac{1}{\left(1 - G \frac{M_\oplus}{R_\oplus^3}\right)^2}
43

And with some simple algebra, we find:

G=R3M(11(α(2α))31)±0.076% \boxed{ G = \frac{R^3_\oplus}{M_\oplus} \left(\sqrt{\frac{1}{1 - \left(\alpha \left(2 - \alpha\right)\right)^3}} - 1\right) \pm 0.076\% }
44

Displacement Current Equivalent

Let's now consider the electromagnetic displacement current J\mathbf{{J}} where

J=ρev0 \mathbf{{J}} = \rho_e v_0
45

Since, in classical electromagnetism:

E=ρeϵ0 \nabla \cdot E = \frac{{\rho_e}}{{\epsilon_0}}
46

Then

Jδv0 \mathbf{{J}} \propto \delta \cdot v_0
47

Or

J2GMR3c11(1+GMR3)2=1.6145 m2s \mathbf{{J}} \propto 2 \frac{GM_\oplus}{R_{\oplus}^3} c \sqrt{1 - \frac{1}{\left(1 + G\frac{M_\oplus}{R_\oplus^3}\right)^2}} = 1.6145 ~\frac{\text{m}^2}{\text{s}}
48
Note: There is a radius here on Earth's surface where this value is precisely Φ\Phi, the Golden Ratio.

Earth's Orbital Velocity From α\alpha

Consider the radius-independent magnitude of our metric for the dominant body in our system, the Sun:

vg=GM3 v_{g} = \sqrt[3]{GM_\odot}
49

Considering that nothing in this model is proportional to time, we should consider this model as proportional to dxdx, or orbital velocity.

vgvorbit=171.25209 \frac{v_{g}}{v_{\text{orbit}}} = 171.25209
50

As this relates the dilation of space to the velocity relative to the underlying coordinate system, we should then consider the relationship of static space, and guess what...

vgvorbit=αAUR \boxed{ \frac{v_{g}}{v_{\text{orbit}}} = \alpha \frac{\mathbf{AU}}{R_\oplus} }
51

Or if approximating as a perfectly circular orbit:

AUα=GM3Tyear2πAUR±0.014% \mathbf{AU}\alpha = \sqrt[3]{G M_\odot} \frac{T_{\text{year}}}{2 \pi \mathbf{AU}} R_\oplus \pm 0.014\%
52

Mass Density Approximation

Finally, let's wrap this all up with one last approximation. Consider the following, that we've derived above:

β0211(1+gR)2(α(2α))3 \beta_0^2 \approx 1 - \frac{1}{ \left(1 + \frac{g}{R}\right)^2 } \approx \left( \alpha \left( 2 - \alpha \right) \right)^3
53

Then considering that

ρ=MV=M43πR3=3M4πR3 \rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4\pi R^3}
54

and the fact that we defined

δ=2GMR3 \delta = 2 \frac{G M_\oplus}{R_{\oplus}^3}
55

Then

MR3=δ2G \frac{M}{R^3} = \frac{\delta}{2G}
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and

ρ=34π(δ2G)=3δ8πG \rho = \frac{3}{4\pi} \left(\frac{\delta}{2G}\right) = \frac{3 \delta}{8 \pi G}
57

And given the relationship that

δΦv0 \delta \approx \frac{\Phi}{v_0}
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and

β2(α(2α))3 \beta^2 \approx \left(\alpha \left(2 - \alpha\right)\right)^3
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then finally

ρ=3(δR32G)4πR3=3δ8πG=ΦGcα(2α)=5561.006gm3 \rho = \frac{3 \left( \frac{\delta R^{3}}{2G} \right)}{4 \pi R^{3}} = \frac{3 \delta}{8 \pi G} = \frac{\Phi}{G c \alpha \left(2 - \alpha\right)} = 5561.006 \frac{\text{g}}{m^3}
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This is our observed mass density, m0m_{0}...

ρ=M43πR3±0.85% \rho = \frac{M_{\oplus}}{\frac{4}{3} \pi R^3} \pm 0.85\%
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That's an error of under 1% for an approximation, that's using approximations and transcendental quantities, demonstrating the wide application of this model.

Results

If we omit this last derivation of ρ\rho, admitting that it is merely an approximation and the derivation of our cosmological velocity, as an accurate estimate is so difficult to obtain, we achieve an average error of 0.0842%0.0842\%, across more than a dozen functions. We've defined GG as a function of mass density and electromagnetic parameters, we've derived both our kinematic and cosmological velocities, and most importantly... we found α\alpha as a function of gravitational parameters in multiple, separately defined instances... all using the same principle: dd elongates by a factor of γ\gamma in the reference frame of BB, and this is the source of the equivalence principle.

Supportive Arguments

  • No terminating geodesics, and no more blackholes. Singularities become nothing but a mirage in the distance that vanishes as you approach that gravitational field.
  • The Bullet Cluster observations are heavily indicative of a repulsive gravity model.
  • This notion of time paints an entirely different picture of chronology, where there is no limitation on backwards time travel, but a limitation on affecting one's own past. Go back to tnt - n and try to kill your grandfather. He won't be there anymore as he's now at tt, if he's still around. Don't get me started on the potentials around massless conciousness in this buoyant, 4-dimensional environment.
  • Machian inertia. This model concludes that the Universe is almost precisely as Mach had predicted, with inertia being comprised of the sum of this radiating force for all other bodies in the Universe.
  • The entire equivalence principle. We've all become used to it so we overlook just how profound it is, but it's literally assuming what I'm saying is true mathematically and then just pretending it's not in reality.
  • The lack of a magnetic monopole, or rather, the existence of an electric monopole. Where do you think that 'sink and faucet' model is funneling charge to? Charge is just space flowing through a hole in an equi-temporal plane, while charged bodies are just bodies with a higher tendency towards one side of this plane due to the buoyant forces along this density axis.
  • Similarly, the induced 1R\frac{1}{R} curvature resolves the paradox produced by the equivalence principle, where a charged particle placed on the surface of a gravitating body does not radiate as if it were in an accelerated frame. This happens because an 'accelerated' frame in the coordinate system of the 'dough' is a linear velocity in the coordinate system of the 'oven'. I intentionally won't use inertial here to describe this path, as the Machian Universe produced by this model does not conserve energy and momentum across this density axis. This apparent violation of the conservation laws follows directly from purely Newtonian mechanics in a dilatory environment.
    • This is not a fluke in the math or a mathematical bandaid, this is a core piece of this geometry: This 4th axis is described entirely by spatial density.
  • The Fermi paradox. There's an entirely other dimension that life might exist across that we have been so far unable to transcend.
  • Relative motion. Think about it. Do we currently calculate all motion as being truly relative, or do we not already calculate motion as relative to the underlying gravitational acceleration and then lie to undergrads about it? The tensors of GR mask this fact to an extent, but reduce them to their calculus equivalents and we are doing precisely this.
If they say to you, "Is it you?" say 'We are it's children'. If they ask of you, 'What is the sign of your father in you?' say to them 'It is motion and rest'.
Nag Hamadi Codex

Footnotes

1.
The logical question here is well, why not just publish it in peer review? The only possible mathematical solution indicates that the Earth is expanding with the Universe. Einstein himself couldn't get a model like this past our current peer review culture.
2.
Assuming a uniform gravitational field across the path of travel.
3.
Interestingly, Michelson & Morely actually used the very same instrument in the very same year to measure both the proposed dilation of space and α\alpha! They just didn't find any dilation of space because they explicitly excluded the direction influenced by gravity.
Fluster