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.

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On The Gravitational Nature Of Time

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

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Thank you

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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!

Also, please keep in mind that while this may not be as responsive as the Fluster website, it was written in the very early stages of a markdown like syntax with the same transpiler that powers Fluster. The Conundrum transpiler will continue to improve daily, with blogging support being a major priority.

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 comes from the fine-structure constant).

derivation	function	% error
cosmological velocity from local gravitational parameters	`vbarCosmological`	Difficult to assess
kinematic velocity from local gravitational parameters	`vbarKinematic`	0.442258391884371
Earth's roational velocity via α	`rotationVelocityFromDivergence`	0.00677756400547747
Solar-Galactic Radius from visible mass	`galacticRadiusPolynomial`	0
α (solar)	`alphaSolar`	0.00342749503676447
α (Earth)	`omega` , `alphaFactor`	0.0253043303850287
α (orbital velocity)	`solarMetricVelocity` , `vOrbit`	0.054466549481003
α (from derived velocity)	`vbarAlphaEquivalence`	0.0761820393770567
J ≈ Φ	`currentEquivalent`	0.213633171310522
h from α	`omega` , `alpha`	0.0253887188634106
h/(2 - α) from ω	`omega` , `alpha`	0.00342749503676447
Φ mass density approximation	`massDensityEquivalence`	0.863214423082925

Error All error percentage calculations were done with the following function. See the Fluster & Conunundrum repository for a complete


.ipynb

notebook and/or install the


alpha_omega_gravity

python package.

 percent_diff.py

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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 wll satisfy 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 centripital influence, and Astropy provided values (equatorial) for everything else.

The Lighthouse & The Clocktower

An opening thought experiment

Consider two observers, $A$ and $B$ . Let $A$ and $B$ agree that $B$ should travel between two arbitrary points in space², $p_1$ and $p_2$ at an agreed upon velocity $v$ .

At $p_2$ place a time keeping device that remains visible from $p_1$ . Allow that $A$ remains at rest relative to $p_1$ , and very near $p_1$ while $B$ goes in motion (at an inertial pace) between $p_1$ and $p_2$ .

At what time does $A$ observe $B$ reach $p_2$ ? If the value is not $d/v$ , but $\gamma d/v$ , then $v \neq v_\prime$ , unless $d \to d\gamma$ . The only possible solution that allows for Einstein's additional factor of $\gamma$ while allowing the observer in an unchanged state to observe the kinetic effects of observer in motion is that the distance between $p_{1}$ and $p_{2}$ must elongate by a factor of $\gamma$ .

Is this not 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 $p_{1}$ and $p_{2}$ will be further seperated, if everything dilates, not just the space in-between galaxies. If we presume that this dilation Einstein applied to time may applied to space in the coordinate system of the observer in relative motion, than 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.

Does this not make more sense? A Universe driven by motion as a driving rate of change instead of an abstract quantity that we've never directly measured, time?

Our Cosmological Velocity

Let's consider that this dilation of space might give rise to our gravitational acceleration, where $p_1$ 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 $g$ with respect to $R$ .

\delta = \frac{d}{dR} - G \frac{M}{R^2} = 2 G \frac{M}{R^3}

Hint The negative sign here is applied to

g

since it is space that is moving, not the observer. This gives the observe a negative net

\dot{x}

despite the fact that

\Delta E_{\text{kinetic}} = 0

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

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

If we solve $\gamma$ for $v$ , we find:

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

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

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

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:

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

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

\Delta \tau = \frac{\int_p \dot{x}}{\int_0^v \dot{x}}

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Notation

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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 $\int_p$ 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 $t$ .

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

This gives:

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

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

\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{\dot{x}}{\hat{x}}

Notation Here and throughout the rest of this note,

\dot{x}

does not indicate a derivative with respect to time, but rather a fundamental, driving rate of change.

\hat{x}

is a notation used to demonstrate the magnitude equivalent of

1

for that velocity, making

\hat{x} = v

in units of distance. I'm hesitant to say

\hat{x} = v~1t

, because while mathematically acccurate, 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

t

If we substitute this result back into equation 4 we find:

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}

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

Error Margin:

0.43\%-0.92\%

From Orbital Parameters

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

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

F_{\text{centripital}} = F_{\text{gravity}}

We can solve this trigonometrically.

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

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

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

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

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

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

Pluging this value in to the equation we used to find $\delta$ gives:

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

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 $dt = 1$ and solving for $\dot{x}$ . 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 $B$ orbits $A$ at an agreed upon velocity $v_o$ , then this velocity would be decreased by a factor of $dt$ in the coordinate system of $A$ due to the decreased spatial density of $\mathbf{S}_B$ , while our relativistically corrected orbital velocity equation becomes as follows:

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

When we solve for $R$ and plugin our galactic parameters here on Earth, we find a radius of $8.13~ Kpc$ , perfectly on top of observatonal values, without any additional unseen mass!

Electromagnetism

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

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

Not only does this accomodate existing experimental validations of SR & GR, it does so while accomodating the original relativistic experiment: Michelson & Morely. 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 $\delta$ is this determinant, we should then take it's cube root:

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

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

2 - \alpha \pm 0.025\%

Giving a final equation of:

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

Solving this equation for $\alpha$ gives:

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

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

Final Error:

0.025\%

Take a second to recogize the fact that in the exact function we predicted to find a unifying quantity, the single most unifying electromagnetic property arises. This is not a dimensional coincidence.

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

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 $\text{spinor}=\left( \frac{1}{2} \right)^{1/3}$ , or the same quantity before the torsion applied by $\alpha$ and the dilation applied by $ds$ .

Plugging this back into equation 19 gives:

\left(\frac{1}{2 - \alpha}\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\%

Demonstrating that this function is more than just a coincidence.

Error

1 \pm \alpha

with an error of

0.00347\%

and

0.003427\%

respectively.

Significance of

\alpha

If you're unfamiliar with physics, let me take a second to break down how profound this is. That's an electromagnetic constant, comprised of multiple electromagnetic constants defined in terms of gravitational properties in what certainly looks like a recursive function.

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

\alpha = \frac{1}{4 \pi}\frac{e^2}{\hbar}\sqrt{\frac{\mu_0}{\epsilon_0}}

Component	What it does
$1/4\pi$	A consequence of the spherical geometry
$\epsilon_0$	The electric constant
$\mu_0$	The magnetic constant
$e$	The fundamental charge
$\hbar$	The momentum carrying quantity of a massless photon.
$\frac{1}{\sqrt{\mu_0 \epsilon_0}}$	The speed of light, and therefore the speed of causality.

That's basically the relationship of all of electromagnetism described in one function. Don't worry though, we can keep going! Now that we've derived a relationship to $\alpha$ , we can take advantage of the electromagnetic equations. Let's take advantage of the fact that the electromagnetic tensor is undefined at $\Delta t = 0$ . 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:

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

Since

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

then

\nabla \cdot \mathbf{B} = \delta

and

\nabla \cdot \mathbf{E} = c \delta = 921.116 ~\text{meters}

If we then multiply equation 26 by our spinor approximation, converting this linear spatial dilation to rotation, we find:

\frac{1}{2 - \alpha} c \delta = 462.244~ \text{meters}

That's our rotational velocity on Earth's equator $\pm -0.006778\%$ ! Off by a total of 0.03 meters on a radius the size of Earth's!

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Dimensionality



As in traditional models,

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

This model uses a velocity that is

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

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

\dot{x}~ ds = \text{distance}

In the same manner that

vt = \text{distance}

Velocity as a function of $\alpha$

Let's consider equation 19 once again. If we re-arrange this equation such that $ds$ is isolated on one side, we find:

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

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

\left(\alpha \left( 2 - \alpha \right) \right)^3 = ds^3 = \delta

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

\left(\alpha \left( 2 - \alpha \right) \right)^3 = \frac{ds^3}{dx^{\prime}} = \frac{\delta}{c} \quad \Rightarrow \quad v = c \sqrt{\left(\alpha \left( 2 - \alpha \right) \right)^3} = 525,693 ~ \text{meters}

This is our original velocity $\pm 0.0415\%$ . One can very logically make an argument for the small-angle approximations being a cause for coincidence here, but as a function of $\alpha$ ?

Displacement Current Equivalent

Then, consider the electromagnetic displacement current $\mathbf{{J}}$ where

\mathbf{{J}} = \rho_e v_0

Since, in electromagnetism:

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

Then

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

\mathbf{{J}} \propto 2 \frac{GM_\oplus}{R_{\oplus}^3} c \sqrt{1 - \frac{1}{\left(1 + \frac{g}{R}\right)^2}} = 1.6145 ~\frac{\text{m}^2}{\text{s}}

Note: There is a radius here on Earth's surface where this value is precisely $\Phi$ , the Golden Ratio.

Earth's Orbital Velocity From $\alpha$

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

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

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

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

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:

\frac{v_{g}}{v_{\text{orbit}}} = \eta \frac{\mathbf{AU}}{R}

Solving this for $\eta$ and plugging in our parameters gives:

\eta = \frac{\sqrt[3]{GM_\odot}}{2 \pi \mathbf{AU}^2} T_{\text{year}} R_{\oplus} = \alpha \pm 0.05443\%

Mass Density Approximation

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

\beta_0^2 = 1 - \frac{1}{ \left(1 + \frac{g}{R}\right)^2 } \approx \left( \alpha \left( 2 - \alpha \right) \right)^3

Then considering that

\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4\pi R^3}

and the fact that we defined

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

Then

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

and

\rho = \frac{3}{4\pi} \left(\frac{\delta}{2G}\right) = \frac{3 \delta}{8 \pi G}

And given the relationship that

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

and

\beta^2 \approx \left(\alpha \left(2 - \alpha\right)\right)^3

then finally

\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}

This is our observed mass density, $m_{0}$ ...

\rho = \frac{M_{\oplus}}{\frac{4}{3} \pi R^3} \pm 0.85\%

That's an error of under 1% for an approximation, that's using approximations and transcental quantities, demonstrating the wide application of this model.

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 $t - n$ and try to kill your grandfather. He won't be there anymore as he's now at $t$ , 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 existince of an electric monopole. Where do you think that 'sink and faucet' model is funelling 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.
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.

After all of this talk of divergence, it's important to understand the relationship between this geometry, the proposed spatial dilation, and the power 3 in our 3 dimensional coordinate system. If we recall in slight reworking of equation 19 :

R \alpha = \Gamma \sqrt[3]{2GM_\oplus}

Where

\Gamma = \frac{1}{2 - \alpha}

Is it a coincidence that $R \alpha^3$ gives the unit of measure, the cubit used by our most ancient civilizations? I'll leave that for you to ponder... With ongoing world events, maybe we'll soon know. At least we know how they got here.

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

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.

Assuming a uniform gravitational field across the path of travel.