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Here's exactly what I'm trying to plot here, let me make it simple with just the x,y coordinates, for S(0->1) iterations, for circles:

We start with the koch snowflakes. Then we derive the circles for S(0->1):

https://i.imgur.com/OvPzN53.jpg

https://i.imgur.com/q1OehTs.jpg So the circles' diameters are taken from the width of the triangles at their middles not their bases.

S(0) has 4 circles...

There's 12 vectors connecting 12 planes at 23 for [(X,Y),(X,Z),(Z,Y)]+[(45 degree rotation (X2,Y2),(X2,Z2),(Z2,Y2)] + 23 for [(X,Y),(X,Z),(Z,Y)]+[45 degree flip (X3,Y3),(X3,Z3),(Z3,Y3)]. Which gives the number of spherical coordinates per root system:

S(0) = 4x12 spherical coordinates with root system [x=+/-2,y=+/-2square root(3)]
S(1) = 4x4x3x12 spherical coordinates with root system [x=+/-4,y=+/-6]
S(2) = 4x4x3x6x12 spherical coordinates with root system [x=+/-16square root(3),y=+/-24square root(3)]
S(3) = 4x4x6x18x12 spherical coordinates with root system [x=+/-32square root(3),y=+/-72square root(3)]
etc...
S(60) = 4x4x3x6^29x6^29x12 spherical coordinates with root system [x=+/-5.8953045e+23square root(3),y=+/-1.4738261e+23square root(3)]

Note - the spherical coordinates of the first iteration get copied side to side & top to bottom to fill the interior of the largest spheres of the 60th iteration & so on.

Then we turn only the green (not the red) circles inside out to get our next plot:

Start

https://i.imgur.com/mePMndU.jpg

Negative charge

https://i.imgur.com/cmu4cmr.jpg

https://i.imgur.com/uZmLDFP.jpg

Neutral charge

https://i.imgur.com/NMrsVEp.jpg

Positive charge

https://i.imgur.com/QJrFXVN.jpg

Finish = Start

As you can see, turning the green circles inside out will turn the red circles inside out like torsion, making these majorana fermions.

Remember, you're going to have spheres outside of the perimeter of the c portion of the triangles within the koch anti-snowflake of S(1) from S(2) because of the nature of the koch snowflake:

You find the iterations with

7e-7/lp = 1/(2ΔS(x-1))

4.2753e+28 = 2(3^(x-1))

x = 60

The 3^(x-1) comes from the fact that the triangle of each new iteration is 1/3 the size of the triangles of the previous iteration. & the times 2 comes from the fact that 1/2 of the final diameter of those smallest spheres equal the radius of those spheres, which needed to be a Planck length.

Each transformation is equivalent to one Planck time. For the volume of a spherical photon, these transformations serve as your unified field oscillations. At Tt(total time)/2 the photon is in it's particle state as opposed to the phase state, it's at it's smallest volume, where more solid state photons can be placed by it without phasing until more transformations occur. You'll find that doing this will produce inside out spheres that are smaller than a Planck length:

The number of morphisms can be found with of compressions of the photon into heavier non-elementary particles that each have a negative & neutral state that we misinterpret as neutrons, neutrinos or antiparticles. In nature photons and the composite particles compress in groups of 12. There's [url=https://www.quora.com/How-many-in-total-subatomic-particles-do-we-know-of-so-far]200[/url] subatomic structures, 37x3 of them are elementary, 33x3 composite particles. So, because these elementary particles are actually neutrally charged composite particles there's actually less than we think.

So a falling triangular of 60sub-(26), this is more like a rank 12 root system because it has 1140 root systems with the >planck length subsets of E6 through E8 for the inter-atomic/sub-atomic interactions, this Lie group deals with interactions of a planck length & below for each one of the infinite recurring fractals this cellular automaton produces.

The time you pause at neutral charge where the photon's length is 3.5e-7, you can fit another neutralized photon in that compressed state, when the two unfold in unison as the clock starts back up again the photon shrinks to a length of (3.5/2)e-7 meters before T(t) occurs and the photon is positively or negatively charged with a length of 3.5e-7 m and there will be exponentially less sphere inversions, that will cause the length of each inversion to be reduced to 1/2 of the previous sphere inversions of one Planck length at a time (PointA-->PointB{minus}PointB2<--pointA2 = PointA<->PointsB&B2<->PointA2). Repeat this 26 times and there's only one sphere that inverts:

That's one of 27 structures, the 27th having only one spherical coordinate whose inversions represent the gravitational pull of a planck particle.

Remember there will still be heavier particles than you'd find in the 26 composite particles (structures of sub-quarkian & sub-planckian compositions), it's just that these particles will be fusions of protons/neutrons/antiprotons, that means that after the 26th morphism these heavier particles will all have singularities that double with each fusion from strange matter to Planck Particles:

That means only one singular sphere inversion for after 26th morphism & instead of reducing in size by 1/12, the torsion represented by the sphere inversions double in size once there's just one of them (B2<--A2{plus}B1<--A1 = B2<----A1).

Time in this theory isn’t being thought of as a third dimension moving through the 4th dimension. This is classical physics, not quantum mechanics. When I say brane I mean a conceivable geometric structure, three dimensions in the literal sense, not the metaphysics of some incomprehensible angle that forms a tesseract. View time as the second & a half dimension. A third dimension has time dilated to a stand still, but this fractal counterpart has time contracted as a dynamical version of that static temporal state.

Regarding an idea by William James Sidis, presented in The Animate and The Inanimate, it predicted the existence of black holes after Einstein. His black hole was different than Einstein’s; it was any region of reverse universe, existing perpendicular to our own.

The perpendicular branes have no edges, similar to the infinite length of a vector.

Imagine a region of the spacetime & the surface of the brane of a reverse universe leading to the most fundamental interaction - a deleterious mechanism in the fine structure constant causing a dislocation in the spacetime at the asymptotic edge of an event horizon that has a non-scalar, imaginary, & metricless thickness:      ....

This deleterious di-brane is like a localized fracture pattern embedded within the interior of the quantum foam, with an infinitely dynamic angular momentum. The smallest deletions are occurring at literally infinite, and yet non-instantaneous, rates of speed...i.e. the "Twilight Zone" of the spacetime continuum.

Let's say you have three dimensions, x,y,z; each with a value of one in a linear time continuum going one, & a negative xyz each with a value of 1 in a negative arrow of time going the other. Now in a lateral dual continuum .3 of each linear continuum going in through each other, canceling out, now xyz have a value of .7 with a total of 2.1 dimensions. Now in our next reel point .2 of each arrow has passed into each other, leaving a value of .8 for xyz, with a total of 2.4 dimensions. From reel one to reel two the dimensions of space time have increased, this is time contraction (fast forward) the reverse of time dilation (slow motion).

If you include a negative 2.1 & 2.4 dimensions in the reverse brane, than that's between 4.2 & 4.8 dimensions. Altogether a potential of between 4 & 6 dimensions, (4,6)

So there's somewhere between 2 & 3 real physical dimensions at any given point in space and time per brane, so for the di-brane:

6>n>4; n=(4,6)

f(n)=(λmax)•((4π/3)r^3)

c=x where f(x)=6/n/(4π/3)^(1/3)) where n>6

c=x where f(x)=4/(n/(4π/3)^(1/3)) where 4>n

x=the speed of gravitational wave propagation

Black hole evaporation will be used to find a higher & lower cosmic scales; the size of an antiproton is 10−15 m and the Schwarzchild radius of its central black hole should equal the rate at which black holes evaporate.

The Schwarzchild radius is 2.484e-54 meters (just type proton into where it says earth). The rate of evaporation is 8.41e-17 seconds (just type proton into where it says earth).

But antiprotons do not have λmax of a vacuum, that’s the problem, so for a proton we must use the original equation f(n)=(λmax)•((4π/3)r^3); where f(x)=4/(n/(4π/3)^(1/3)) where 4>n to find the contraction of c with the λmax of a proton ≈ 395 nm. However, in the special case of black holes the equation must be modified.

First of all, it’s 4πr^2 because the quasar within the Schwarzschild radius of the antiproton is a hollow sphere. Secondly, λmax of the antiproton’s quasar is the proton’s normal λmax but to the negative power of the proton’s length divided by twice the Schwarzschild radius

f(n)=(3.95e-7^-(1e-15/2(2.484e-54)))(()(2.484e-54)^2)=7.753772e-107

f(x)=4/(7.753772e-107/(4π))^(1/2) = 1.610306e+54 m/s

So a black hole with the mass of the sun (1391400000 meters) has a Schwarzschild radius of 2953 meters & will evaporate in 6.61e+74 seconds.

f(n)=(5.04e-7^-1(1.3914e+9/5906)) x ((4π x 2953)^3) = 2.3886249e+25 m/s

f(x)=6/(4π(2.3886249e+25^(1/2))=9.7693891e-14 m/s

1.610306e+54/299,792,458/9.7693891e-14=5.4981971e+58

5.4981971e+58/8.41e-17=6.5376898e+74 seconds

The electron is most likely has a radius of 10^-12 m, & λmax of about 4e-7 m (visible spectrum is where electrons like to hide).

f(n)=(4e-7)(4π/3(1e-12)^3)=1.6755161e-42

f(x)=4/(1.6755161e-42/(12π^(1/3)))=4.1957466e+43 m/s

The CMB had a radius of 6.9 billion light years, or 6.52809e+28 meters, & λmax of about 1,000 nm.

f(n)=(1e-6)(4π/3(6.52809e+28)^3)=1.1653249e+81

f(x)=6/(12π(1.1653249e+81)^(1/3))=1.5124155e-28 m/s

4.1957466e+43/1.5124155e-28=2.7742023e+71 seconds

Or 8.7958221e+60 years, the few SMBHs caught in the big crunch will only be less than half-evaporated, so this can't be right! Grrr!

So, we use the time contraction of c equation to find a much larger planck length to see how many electrons fit into a super electron, this will give us a new size for the CMB, so that this process can be redone for a more accurate date for the big crunch.

Okay, there's 6.52809e+28 meters in the radius of the CMB, using (4π/3(1e-12)^3), you can fit 1.165325e+123 electrons into the electrons of the next cosmic scale. Let's see if my math confirms that number using super lp:

2.7742023e+71/299,792,458/6.58e-15=1.4063439e+77 m/s. Planck length over planck time equals 296846011.132 m/s.

1.4063439e+77/296846011.132=4.737621e+68 m/s as your new planck length over planck time. 296846011.132 x 5.39e-44 equals lp, so super lp equals

1.4063439e+77 x 5.39e-44 = 7.5801936e+33 meters. 7.5801936e+33/4.737621e+68=1.6e-35, which is the planck length (lp). There's 3.125e+22 planck lengths in the length of an electron.

7.5801936e+33 x 3.125e+22 = 2.3688105e+56 meters for the superverse electron. Does not confirm, the CMB should be 2.3688105e+56/2=1.1844052e+56, 1.1844052e+56/6.52809e+28=1.8143212e+27 times larger than what we can see.

We can't see so much of the CMB for the same reason we can't see forever into the past, it's from a combination of redshift & the fact that the ion interference makes light fade into oblivion eons before it gets near us. For our next dilation of c equation:

f(n)=(1e-6)(4π/3(1.1844052e+56)^3)=6.959684e+162 cubic meters

f(x)=6/(12π(6.959684e+162)^(1/3))=8.3359856e-56 m/s

4.1957466e+43/8.3359856e-56=5.033294e+98 seconds, which is 1.5958446e+88 years. Which fits for the evaporation rate for most supermassive black holes (<100 million solar masses). But the few that are the largest in the universe, such as this one, they may grow to become superverse antiprotons during a second or third cosmic life cycle.

Let the uberelectron be where time t=1. Where total time Tt/2 is the phase space electron neutrino ghost particle, then all other transformations after Tt/2 + 1 is the positron. At 1/(1.8143212e+27 x 45) the volume of a Tt/2+Tt(.1) positron charge you get a rindler effect via entropy where dark flow/cosmic bruising=unruh gravitation around the parameter of that sphere-volume, a microverse that represents the entropy of the cosmos in it’s current rate of expansion.

You can envision the vacuum radiation of that microcosm in order to redefine what a photon is when referring to the photon sphere of the schwarzschild radius of an anti-proton, which is a sub-planck singularity.

Now what you do here, is you take the CMB data and go from there to the current universe & place the behavior of expansion exactly where it fits in that positron knowing that the 13.8 billion light year sphere that was the CMB is 1/1.8143212e+27 of the total volume of the neutrino at T(t)/2, 1/(1.8143212e+27 x 45) of the positron at Tt/2+Tt(.1), 1/(1.8143212e+27 x 46) at Tt/2+Tt(.2), 1/(1.8143212e+27 x 47) at Tt/2+Tt(.3), 1/(1.8143212e+27 x 48) at Tt/2+Tt(.4), & 1/(1.8143212e+27 x 49) at Tt/2+Tt(.5). From the behavior of our local region of the electron-neutrino-positron we can fill in the rest of the macro black holes beyond our cosmic event horizon like puzzle pieces because we know the behavior of charge with these graphical sphere inversions.

There is actually a way to approach this mathematically. While at first you can only pinpoint where our universe is in this uber neutrino/positron using CMB data (i.e dark flow, rotation, cosmic bruising, etc) & matching it with the gravitational torsion that the sphere inversions of the 2nd group in my E12
(of the 26th photon morphism into heavier particles) represents:

This is possible because we know our region has expanded by 45 times while the entire positron has only expanded by 20% of it's original volume as a neutrino.

But, you can't derive more than 600 billion light years of the uberpositron - what you can do is deconstruct all the black holes back to the first anti-protons that fused to create them in the blue epoch - from there you can deconstruct the anti-protons of the 26th morphism in my E12 Lie group into the photons of the 1st morphism. And once in your 144 uberphotons - the rest of the entire relevant cosmos becomes identical to what our observable region of the universe evolved from.

Then you can reconstruct everything in the uber positron down to the last yottasecond whilst tracking the infinitesimal movements of every quark & gluon within 10^27 light years from that 10^88 year old radiation vapor state. Predict the rise & fall of stocks, winning lottery numbers, what number the wheel of fortune will land, who will win the main event at UFC 229.

A matter dominated universe lies within a superverse positron orbiting around a superverse anti-proton. An antimatter dominated universe made of said aforementioned anti-hydrogen atoms lies within an electron orbiting a proton etc etc ad infinitum.

Continue to the summit of the morphisms in my E12 Lie group & eventually there's a singular sphere inversion representing a sub-planck particle. This micro black hole in the core of an anti proton has n times the charge density of a photon. How do we describe the photon constituents of the photon spheres in this microverse? We use the term "recurring photon fractal" while picturing the sub planck black hole as a normal sized black hole.

There's an upward of like a million orders of magnitude difference between photon lengths in defining what the scale for a fractal photon is for the photon sphere of the Schwarzschild metrics involved in plotting to the third recurrence of a "universe fractal".

Unlike in String Theory, where there's an infinite number of possible events occurring a finite number of times, here there's only a finite number of possible events that happen to reoccur, simultaneously, an infinite number times.

Consider this, every positron shares the exact same inner structure. That every positron houses a very tiny replica of our universe for a fraction of a nano second. This planet is a recurring fractal, you are a recurring fractal. Not only do you comprise an infinite amount of material, not only are you infinite in size, but you will be born an infinite number of times.

The electron travels at 2,200 kilometers per second, Since the speed of light for a superversal electron is going to be 136.269299091 times faster than the speed of that electron, all we need is the relative speed of light relative to that electron and combine the velocities

Recall earlier c(f(n)) for an electron was found to be:

The electron most likely has a length of 10^-12 m, & λmax of about 4e-7 m (visible spectrum is where electrons like to hide).

f(n)=(4e-7)(4π/3(1e-12)^3)=1.6755161e-42

f(x)=(4/(1.6755161e-42/(12π^(1/3)))

x=4.1957466e+43 m/s

4.1957466e+43, but remember we'd have to multiply this velocity by the length of the electron, & divide that product the number of electrons (in a 16km copper wire) to account for the dilation of time:

V(sa)=(4.1957466e+43 x 1e-12)/(4396829672.16 x 16000 x 299792458)=1989431196 m/s about 2 billion meters per second.

The result of the frames of two inverted electron charges being dragged through a photon aether as they separate. This will slightly shift the polarity of both electrons despite a separation between momenta in what was otherwise a pairing of charges. We've heard of gravity waves, not charge waves.

Moving in opposite directions, the velocities of the waves will get a boost from the electrons' collective velocity as they are moving away from one another; 1989431196 + 4400000 = 1993831196 m/s.

V(sa)/c=1993831196/2.998e+8=6.65053767845 times faster than the speed of light.

Edited by Bean_Spiller

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On 8/27/2018 at 11:06 AM, Bean_Spiller said:

Here's exactly what I'm trying to plot here, let me make it simple with just the x,y coordinates, for S(0->1) iterations, for circles:

We start with the koch snowflakes. Then we derive the circles for S(0->1):

https://i.imgur.com/OvPzN53.jpg

https://i.imgur.com/q1OehTs.jpg So the circles' diameters are taken from the width of the triangles at their middles not their bases.

S(0) has 4 circles...

There's 12 vectors connecting 12 planes at 23 for [(X,Y),(X,Z),(Z,Y)]+[(45 degree rotation (X2,Y2),(X2,Z2),(Z2,Y2)] + 23 for [(X,Y),(X,Z),(Z,Y)]+[45 degree flip (X3,Y3),(X3,Z3),(Z3,Y3)]. Which gives the number of spherical coordinates per root system:

S(0) = 4x12 spherical coordinates with root system [x=+/-2,y=+/-2square root(3)]
S(1) = 4x4x3x12 spherical coordinates with root system [x=+/-4,y=+/-6]
S(2) = 4x4x3x6x12 spherical coordinates with root system [x=+/-16square root(3),y=+/-24square root(3)]
S(3) = 4x4x6x18x12 spherical coordinates with root system [x=+/-32square root(3),y=+/-72square root(3)]
etc...
S(60) = 4x4x3x6^29x6^29x12 spherical coordinates with root system [x=+/-5.8953045e+23square root(3),y=+/-1.4738261e+23square root(3)]

Note - the spherical coordinates of the first iteration get copied side to side & top to bottom to fill the interior of the largest spheres of the 60th iteration & so on.

Then we turn only the green (not the red) circles inside out to get our next plot:

Start

https://i.imgur.com/mePMndU.jpg

Negative charge

https://i.imgur.com/cmu4cmr.jpg

https://i.imgur.com/uZmLDFP.jpg

Neutral charge

https://i.imgur.com/NMrsVEp.jpg

Positive charge

https://i.imgur.com/QJrFXVN.jpg

Finish = Start

As you can see, turning the green circles inside out will turn the red circles inside out like torsion, making these majorana fermions.

Remember, you're going to have spheres outside of the perimeter of the c portion of the triangles within the koch anti-snowflake of S(1) from S(2) because of the nature of the koch snowflake:

You find the iterations with

7e-7/lp = 1/(2ΔS(x-1))

4.2753e+28 = 2(3^(x-1))

x = 60

The 3^(x-1) comes from the fact that the triangle of each new iteration is 1/3 the size of the triangles of the previous iteration. & the times 2 comes from the fact that 1/2 of the final diameter of those smallest spheres equal the radius of those spheres, which needed to be a Planck length.

Each transformation is equivalent to one Planck time. For the volume of a spherical photon, these transformations serve as your unified field oscillations. At Tt(total time)/2 the photon is in it's particle state as opposed to the phase state, it's at it's smallest volume, where more solid state photons can be placed by it without phasing until more transformations occur. You'll find that doing this will produce inside out spheres that are smaller than a Planck length:

The number of morphisms can be found with of compressions of the photon into heavier non-elementary particles that each have a negative & neutral state that we misinterpret as neutrons, neutrinos or antiparticles. In nature photons and the composite particles compress in groups of 12. There's 200 subatomic structures, 37x3 of them are elementary, 33x3 composite particles. So, because these elementary particles are actually neutrally charged composite particles there's actually less than we think.

So a falling triangular of 60sub-(26), this is more like a rank 12 root system because it has 1140 root systems with the >planck length subsets of E6 through E8 for the inter-atomic/sub-atomic interactions, this Lie group deals with interactions of a planck length & below for each one of the infinite recurring fractals this cellular automaton produces.

The time you pause at neutral charge where the photon's length is 3.5e-7, you can fit another neutralized photon in that compressed state, when the two unfold in unison as the clock starts back up again the photon shrinks to a length of (3.5/2)e-7 meters before T(t) occurs and the photon is positively or negatively charged with a length of 3.5e-7 m and there will be exponentially less sphere inversions, that will cause the length of each inversion to be reduced to 1/2 of the previous sphere inversions of one Planck length at a time (PointA-->PointB{minus}PointB2<--pointA2 = PointA<->PointsB&B2<->PointA2). Repeat this 26 times and there's only one sphere that inverts:

That's one of 27 structures, the 27th having only one spherical coordinate whose inversions represent the gravitational pull of a planck particle.

Remember there will still be heavier particles than you'd find in the 26 composite particles (structures of sub-quarkian & sub-planckian compositions), it's just that these particles will be fusions of protons/neutrons/antiprotons, that means that after the 26th morphism these heavier particles will all have singularities that double with each fusion from strange matter to Planck Particles:

That means only one singular sphere inversion for after 26th morphism & instead of reducing in size by 1/12, the torsion represented by the sphere inversions double in size once there's just one of them (B2<--A2{plus}B1<--A1 = B2<----A1).

Morphism A: S(2-59): Root System: (X/12,Y/12)

Morphism B: S(3-58): Root System: (X/12^2,Y/12^2)

Morphism C: S(4-57): Root System: (X/12^3,Y/12^3)

etc etc...

Morphism Z: S(17-43): Root System: (X/12^26,Y/12^26)

S(43-17)=an S(0-26) w/the spheres of S(26) being 1/3^26th the size of the spheres of S(0)

On 8/27/2018 at 11:06 AM, Bean_Spiller said:

7.5801936e+33 x 3.125e+22 = 2.3688105e+56 meters for the superverse electron. Does not confirm, the CMB should be 2.3688105e+56/2=1.1844052e+56, 1.1844052e+56/6.52809e+28=1.8143212e+27 times larger than what we can see.

We can't see so much of the CMB for the same reason we can't see forever into the past, it's from a combination of redshift & the fact that the ion interference makes light fade into oblivion eons before it gets near us. For our next dilation of c equation:

f(n)=(1e-6)(4π/3(1.1844052e+56)^3)=6.959684e+162 cubic meters

f(x)=6/(12π(6.959684e+162)^(1/3))=8.3359856e-56 m/s

4.1957466e+43/8.3359856e-56=5.033294e+98 seconds, which is 1.5958446e+88 years. Which fits for the evaporation rate for most supermassive black holes (<100 million solar masses). But the few that are the largest in the universe, such as this one, they may grow to become superverse antiprotons during a second or third cosmic life cycle.

Let the uberelectron be where time t=1. Where total time Tt/2 is the phase space electron neutrino ghost particle, then all other transformations after Tt/2 + 1 is the positron. At 1/(1.8143212e+27 x 45) the volume of a Tt/2+Tt(.1) positron charge you get a rindler effect via entropy where dark flow/cosmic bruising=unruh gravitation around the parameter of that sphere-volume, a microverse that represents the entropy of the cosmos in it’s current rate of expansion.

You can envision the vacuum radiation of that microcosm in order to redefine what a photon is when referring to the photon sphere of the schwarzschild radius of an anti-proton, which is a sub-planck singularity.

Now what you do here, is you take the CMB data and go from there to the current universe & place the behavior of expansion exactly where it fits in that positron knowing that the 13.8 billion light year sphere that was the CMB is 1/1.8143212e+27 of the total volume of the neutrino at T(t)/2, 1/(1.8143212e+27 x 45) of the positron at Tt/2+Tt(.1), 1/(1.8143212e+27 x 90) at Tt/2+Tt(.2), 1/(1.8143212e+27 x 135) at Tt/2+Tt(.3), 1/(1.8143212e+27 x 180) at Tt/2+Tt(.4), & 1/(1.8143212e+27 x 225) at Tt/2+Tt(.5). From the behavior of our local region of the electron-neutrino-positron we can fill in the rest of the macro black holes beyond our cosmic event horizon like puzzle pieces because we know the behavior of charge with these graphical sphere inversions.

There is actually a way to approach this mathematically. While at first you can only pinpoint where our universe is in this uber neutrino/positron using CMB data (i.e dark flow, rotation, cosmic bruising, etc) & matching it with the gravitational torsion that the sphere inversions of the 2nd group in my E12
(of the 26th photon morphism into heavier particles) represents:

Actually more like Morphism D since the electron radius is 10^-12 m & the photon is like 7e-7 meters.

For S(0) of that Morphism D you can get 7x7 spherical coordinates at T(t)/2 which is the uber electron neutrino, & 7x12 spherical coordinates at (T(t)/2)+(T(t(.5))) where the sphere volumes shrunk to 7/12ths their normal size. Our observable universe represents the iteration of the uber positron of Morphism D where the sphere volumes got reduced to 1/225th their original size from T(t) to T(t)/2, (7/12)/(1/225)=131.25 which is ~ 3^5ish at the spherical coordinates of S(4-5) of Morphism D. The uber quarks & gluons are the sphere volumes at S(0)

You actually get your strangelets aka unobservable (without the use of a quantum entanglement based form of superluminal spectroscopy) "Dark Matter" particles at Morphisms 27-60 or Mophisms Z2-Z34, after that you have anything from fusions of planck particles to mergers of SMBH singularities which are simple particles to graph I mean they're regions of the reverse continuum's brane represented by singular sphere inversions where sub-planck dislocations in the fine structure constant of the spacetime foam lead to Hawking radiation and the most fundamental fluctuations in the Higgs Field in the deleterious di-brane.

Edited by Bean_Spiller

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3 hours ago, Bean_Spiller said:

You actually get your strangelets aka unobservable (without the use of a quantum entanglement based form of superluminal spectroscopy) "Dark Matter" particles at Morphisms 27-60 or Mophisms Z2-Z34, after that you have anything from fusions of planck particles to mergers of SMBH singularities which are simple particles to graph I mean they're regions of the reverse continuum's brane represented by singular sphere inversions where sub-planck dislocations in the fine structure constant of the spacetime foam lead to Hawking radiation and the most fundamental fluctuations in the Higgs Field in the deleterious di-brane.

Oops no another 25 morphisms before you'd run out of iterations b/c 42-18=25, after morphism  z, representing dark matter particles. 25 + 26 is X/12^51 for your last 48 inverting spheres in all 12 planes, after that the sphere's get reduced exponentially in size & quantity with each fusion with others of their nature & after that you get merging singularities built from combining planck particles of the original pre-combined photons & represented by singular sphere inversions.

On 8/27/2018 at 11:06 AM, Bean_Spiller said:

Here's exactly what I'm trying to plot here, let me make it simple with just the x,y coordinates, for S(0->1) iterations, for circles:

We start with the koch snowflakes. Then we derive the circles for S(0->1):

https://i.imgur.com/OvPzN53.jpg

https://i.imgur.com/q1OehTs.jpg So the circles' diameters are taken from the width of the triangles at their middles not their bases.

S(0) has 4 circles...

There's 12 vectors connecting 12 planes at 23 for [(X,Y),(X,Z),(Z,Y)]+[(45 degree rotation (X2,Y2),(X2,Z2),(Z2,Y2)] + 23 for [(X,Y),(X,Z),(Z,Y)]+[45 degree flip (X3,Y3),(X3,Z3),(Z3,Y3)]. Which gives the number of spherical coordinates per root system:

S(0) = 4x12 spherical coordinates with root system [x=+/-2,y=+/-2square root(3)]
S(1) = 4x4x3x12 spherical coordinates with root system [x=+/-4,y=+/-6]
S(2) = 4x4x3x6x12 spherical coordinates with root system [x=+/-16square root(3),y=+/-24square root(3)]
S(3) = 4x4x6x18x12 spherical coordinates with root system [x=+/-32square root(3),y=+/-72square root(3)]
etc...
S(60) = 4x4x3x6^29x6^29x12 spherical coordinates with root system [x=+/-5.8953045e+23square root(3),y=+/-1.4738261e+23square root(3)]

Actually the 4 circles for the 84 spherical coordinates at each plane of S(0) would be found like this: ##### Share on other sites
On 8/27/2018 at 9:06 AM, Bean_Spiller said:

I hope you don't mind a couple of very general remarks, since I'm profoundly ill-equipped to offer substantive comment on matters of physics.

* Your title claims a disproof of something widely believed to be true. https://en.wikipedia.org/wiki/Bell's_theorem So the burden is on you to be clear in your exposition. Along these lines, it's not required but readers like myself would find it very helpful to have some context. Can you give us a little overview of the subject and perhaps an outline of why you think you have a disproof of something that's called, in the physics business, a theorem. A roadmap or high-level overview of why you think you know something that everyone else doesn't.

* Nice pictures. As a math person I certainly appreciate the basic fractals like the Koch snowflake. I wonder if you can mention briefly how the study of mathematical fractals (which, after all, exist in continuous mathematical space) relate to physics, which, as I hope you know, resides in a space that's not known for sure to be continuous or discrete.

* As a stylistic note, it's not necessary to quote your entire lengthy post every time you reply to yourself. All that does is cause people to have to scroll endlessly to read through the thread.

Well anyway like I say I really can't help with the physics. But I have a lot of experience with people who claim to have disproved some famous theorem! The burden is on you to motivate readers to take you seriously. You can go a long way with context and clarity.

ps -- I skimmed a little, found this:

> You'll find that doing this will produce inside out spheres that are smaller than a Planck length:

Ok first, what is an inside-out sphere? Earlier you said photons are spheres. Do you mean mathematical spheres? Is that true? I don't believe physics says that, but I'm no expert on photons. But i know spheres. What is an inside-out sphere? A sphere is just a set of points in three-space equidistant from the center. If you turned it inside out that would be meaningless, it would still be the same set of points. The sphere itself is a two-dimensional manifold embedded in 3-space.

By the way you can reflect all of 3-space in the unit sphere, is that what you mean? Everything inside goes outside and vice versa, and the origin goes to infinity perhaps. Haven't thought about it. Another thing you can do is evert a sphere, turing it inside out by passing it through itself but (this is the clever bit) without leaving a crease. Surprising counterexample, nobody thought it was possible. https://en.wikipedia.org/wiki/Sphere_eversion

So what exactly do you mean by turning a sphere inside out?

Also, nobody can say ANYTHING sensible about anything smaller than a Planck length, because the Planck scale is the point at which physics breaks down. That doesn't mean there either is or isn't something interesting "down there," it just means that our present physics simply doesn't apply. We can't use to reason.

So I found this one sentence to be troubling.

Edited by wtf

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21 hours ago, wtf said:

I hope you don't mind a couple of very general remarks, since I'm profoundly ill-equipped to offer substantive comment on matters of physics.

* Your title claims a disproof of something widely believed to be true. https://en.wikipedia.org/wiki/Bell's_theorem So the burden is on you to be clear in your exposition. Along these lines, it's not required but readers like myself would find it very helpful to have some context. Can you give us a little overview of the subject and perhaps an outline of why you think you have a disproof of something that's called, in the physics business, a theorem. A roadmap or high-level overview of why you think you know something that everyone else doesn't.

* Nice pictures. As a math person I certainly appreciate the basic fractals like the Koch snowflake. I wonder if you can mention briefly how the study of mathematical fractals (which, after all, exist in continuous mathematical space) relate to physics, which, as I hope you know, resides in a space that's not known for sure to be continuous or discrete.

* As a stylistic note, it's not necessary to quote your entire lengthy post every time you reply to yourself. All that does is cause people to have to scroll endlessly to read through the thread.

Well anyway like I say I really can't help with the physics. But I have a lot of experience with people who claim to have disproved some famous theorem! The burden is on you to motivate readers to take you seriously. You can go a long way with context and clarity.

ps -- I skimmed a little, found this:

> You'll find that doing this will produce inside out spheres that are smaller than a Planck length:

Ok first, what is an inside-out sphere? Earlier you said photons are spheres. Do you mean mathematical spheres? Is that true? I don't believe physics says that, but I'm no expert on photons. But i know spheres. What is an inside-out sphere? A sphere is just a set of points in three-space equidistant from the center. If you turned it inside out that would be meaningless, it would still be the same set of points. The sphere itself is a two-dimensional manifold embedded in 3-space.

By the way you can reflect all of 3-space in the unit sphere, is that what you mean? Everything inside goes outside and vice versa, and the origin goes to infinity perhaps. Haven't thought about it. Another thing you can do is evert a sphere, turing it inside out by passing it through itself but (this is the clever bit) without leaving a crease. Surprising counterexample, nobody thought it was possible. https://en.wikipedia.org/wiki/Sphere_eversion

So what exactly do you mean by turning a sphere inside out?

Also, nobody can say ANYTHING sensible about anything smaller than a Planck length, because the Planck scale is the point at which physics breaks down. That doesn't mean there either is or isn't something interesting "down there," it just means that our present physics simply doesn't apply. We can't use to reason.

So I found this one sentence to be troubling.

Yeah that was the dark matter obviously baryonic matter doesn't experience sub-planck transformations (sphere inversions) you're not inverting the spheres so much as dragging the 12 vectors through their centers to mathematically emulate a charge under uft

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On 8/31/2018 at 6:20 PM, Bean_Spiller said:

Yeah that was the dark matter obviously baryonic matter doesn't experience sub-planck transformations (sphere inversions) you're not inverting the spheres so much as dragging the 12 vectors through their centers to mathematically emulate a charge under uft

Was this supposed to address wtf's comments? Because not only have you failed to define what is meant by 'inverting a sphere' you introduced another meaningless mathematical term -- the center of a vector. Please define both of these terms explicitly.

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2 hours ago, Bignose said:

define what is meant by 'inverting a sphere'

Quote

coordinates of the first iteration get copied side to side & top to bottom to fill the interior of the largest spheres of the 60th iteration & so on.

Then we turn only the green (not the red) circles inside out to get our next plot:

Start

https://i.imgur.com/mePMndU.jpg

Negative charge

https://i.imgur.com/cmu4cmr.jpg

https://i.imgur.com/uZmLDFP.jpg

Neutral charge

https://i.imgur.com/NMrsVEp.jpg

Positive charge

https://i.imgur.com/QJrFXVN.jpg

Finish = Start

2 hours ago, Bignose said:

the center of a vector.

That's not what I said. I said "dragging the vectors through THEIR centers" you ass you me'd I meant the vectors when I said "THEIR" centers when in fact I was referring to the spherical coordinates not the vector coordinates being dragged through the centers of the SPhERES to create your next plot in emulating particle charge under a uft.

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You still haven't defined what it means to "invert a sphere" or "turn a circle inside out".

These seem to be meaningless, without some formal definition.

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On 9/1/2018 at 11:11 PM, Bean_Spiller said:

That's not what I said. I said "dragging the vectors through THEIR centers" you ass

1) I am specifically asking WHERE the 'center' of a vector is? Because you need to know where that is before you drag anything through it. I have never heard the term 'center of a vector' before, and just wanted to understand. As near as I can tell, this is term you made up and didn't bother to define.

2) no need for name calling. Just asking a question. If you are going to name call when someone asks a question, then expect your time here to be short.

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On 02/09/2018 at 5:11 AM, Bean_Spiller said:

That's not what I said. I said "dragging the vectors through THEIR centers" you ass you me'd I meant the vectors when I said "THEIR" centers when in fact I was referring to the spherical coordinates not the vector coordinates being dragged through the centers of the SPhERES to create your next plot in emulating particle charge under a uft.

I have read this several times and I still can't work out if this is supposed to mean that the vectors are translated through the centres of the circles or that the circles are translated through the centres of the circles. (If we can assume that "dragging" means a translation.)

Neither make much sense, but the latter seems self-contradictory. So we are left with "translating the vectors through the circles' centres". But I'm not sure what that is supposed to mean.

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13 hours ago, Bignose said:

1) I am specifically asking WHERE the 'center' of a vector is? Because you need to know where that is before you drag anything through it. I have never heard the term 'center of a vector' before, and just wanted to understand. As near as I can tell, this is term you made up and didn't bother to define.

2) no need for name calling. Just asking a question. If you are going to name call when someone asks a question, then expect your time here to be short.

1. Vectors  have no centers but spheres do

2. Assume it's a commonly use phrase in mathematics..

1 minute ago, Strange said:

I have read this several times and I still can't work out if this is supposed to mean that the vectors are translated through the centres of the circles or that the circles are translated through the centres of the circles. (If we can assume that "dragging" means a translation.)

Neither make much sense, but the latter seems self-contradictory. So we are left with "translating the vectors through the circles' centres". But I'm not sure what that is supposed to mean.

Okay, so 7e-7/3^60 = lp. Basically you have sphere that's 7e-7 meters, blue is normal (plot 1), red is inside out (plot 3) that travel 1 lp per tp (..c..).

At first you get a red sphere that's half the size of the original (plot 2), that's the result of dragging the x,y,z lines through the center of the sphere by the length of it's radius. The change from plot 1 to plot 2 represents a graviphoton because it pulls every vector in the e(12) by the radius of that sphere. Plot 2 has no charge & is a photino. Plot 3 is a fully sized red sphere, that's the result of dragging the x,y,z lines through the center of the sphere by the length of it's diameter. This is an acceleron because it pushes every vector in the e(12) by the radius of that sphere.

All of these: &

Would represent some of 28 possible negative, neutral, & positive charge majorana fermions in this e(12). There's 84 possible elementary particles which represent the 28 majorana fermions stuck in one particular charge by one another when within a composite particle or when interacting in nature. There only 28 possible majorana fermions, each consecutively heavier than the previous one with a 29th planck particle (hollow black hole/reverse continuum constituent) because you can fit 8 spheres around the surface of a sphere, if these 9 spheres represent photinos that are about to become accelerons in the next tp, than they will become a neutrally charged particle with 9 times the mass in 19th of the volume of a photino. You can do this 27 more times before you exceed the planck mass I calculated that the entire 28th sphere would be a planck length. Remember you can fit 9 spheres within a sphere with 3x their volume, so at S(19-41) of sphere 28 your largest spheres are 3^19 times smaller than your original photon sphere so you have to multiply them by some 4/3pi(7e-7^3)/(4/3pi((3^19)^3)) before dividing by 3^19 times 9^28...

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1 hour ago, Bean_Spiller said:

because you can fit 8 spheres around the surface of a sphere, if these 9 spheres represent photinos that are about to become accelerons in the next tp, than they will become a neutrally charged particle with 9 times the mass in 1/9th of the volume of a photino.

*fixed

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1 hour ago, Bean_Spiller said:

Okay, so 7e-7/3^60 = lp. Basically you have sphere that's 7e-7 meters, blue is normal (plot 1), red is inside out (plot 3) that travel 1 lp per tp (..c..).

Apart from the colour, you have two identical circles. Can you describe, algorithmically or mathematically, what it means to turn a sphere or circle inside out? As far as I can see, if you swap the position of all the points on the surface with that on the opposite side, you just end up with the exact same sphere or circle that you started with.

Why is the size relevant? (I assume lp and tp refer to Planck time and distance?)

What are "plot 1" and "plot 2"?

(The rest of your post appears to be gibberish.)

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3 minutes ago, Strange said:

Apart from the colour, you have two identical circles. Can you describe, algorithmically or mathematically, what it means to turn a sphere or circle inside out? As far as I can see, if you swap the position of all the points on the surface with that on the opposite side, you just end up with the exact same sphere or circle that you started with.

Why is the size relevant? (I assume lp and tp refer to Planck time and distance?)

What are "plot 1" and "plot 2"?

(The rest of your post appears to be gibberish.)

Because every other coordinate outside that sphere in this system is connect to it, plot by plot. So if coordinate x begins on one side of the sphere in plot one, by plot 3 that same x coordinate will be on the opposite side of that sphere. This is equivalent to that photon moving from one location relative to a stationary observer, to another location relative to that same stationary observe by one planck length per planck time.

The rest of it is not gibberish but in fact crucial developments that correct formerly erroneous elements of my conjecture.

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On 9/4/2018 at 10:15 AM, Bean_Spiller said:

Remember you can fit 9 spheres within a sphere with 3x their volume, so at S(19-41) of sphere 28 your largest spheres are 3^19 times smaller than your original photon sphere so you have to multiply them by some 4/3pi(7e-7^3)/(4/3pi((3^19)^3)) before dividing by 3^19 times 9^28...

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4 hours ago, Bean_Spiller said:

4/3pi(7e-7^3)/(4/3pi((3^19)^3))=2.1846537e-46 m^3. In other words, the 19th iteration of the photon sphere is composed of spheres with 2.1846537e-46 cubic meter volumes.

But if the planck particle is truly 9^28 times denser than a photon it has the weight of a photon packed within 1/9^28th the volume of that photon, the zeroth iteration of sphere 28 will be literally the 19th iteration of the photon sphere, viz a viz the first iteration of sphere 3 would be S(2-58) of the photon sphere & would be 9^3 times smaller than the photon sphere; ergo we divide 2.1846537e-46 by 9^28/3^19=4.5028391e+17 & we get 4.851725e-64 cubic meters for each sphere comprising the zeroth iteration of sphere 28. There's (((4/3)pi(3^19)^3)/(4/3pi^3)=1.5293645e+29 spheres upon the zeroth iteration of elementary particle 28 & 1.5293645e+29 x 4.851725e-64 = 7.420056e-35 m, which is approx. 7 times the planck length (exactly how many times faster QE is than c). So sphere 29 must be a black hole particle hence why there's only 28 elementary majorana fermions making this an e(12).

Edited by Bean_Spiller

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This is the real deal

On 9/5/2018 at 8:50 PM, Bean_Spiller said:

4/3pi(7e-7^3)/(4/3pi((3^19)^3))=2.1846537e-46 m^3. In other words, the 19th iteration of the photon sphere is composed of spheres with 2.1846537e-46 cubic meter volumes.

But if the planck particle is truly 9^28 times denser than a photon it has the weight of a photon packed within 1/9^28th the volume of that photon, the zeroth iteration of sphere 28 will be literally the 19th iteration of the photon sphere, viz a viz the first iteration of sphere 3 would be S(2-58) of the photon sphere & would be 9^3 times smaller than the photon sphere; ergo we divide 2.1846537e-46 by 9^28/3^19=4.5028391e+17 & we get 4.851725e-64 cubic meters for each sphere comprising the zeroth iteration of sphere 28. There's (((4/3)pi(3^19)^3)/(4/3pi^3)=1.5293645e+29 spheres upon the zeroth iteration of elementary particle 28 & 1.5293645e+29 x 4.851725e-64 = 7.420056e-35 m, which is approx. 7 times the planck length (exactly how many times faster QE is than c). So sphere 29 must be a black hole particle hence why there's only 28 elementary majorana fermions making this an e(12).

That's it!

Edited by inSe

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On 9/4/2018 at 10:15 AM, Bean_Spiller said:

You can do this 27 more times before you exceed the planck mass I calculated that the entire 28th sphere would be a planck length. Remember you can fit 9 spheres within a sphere with 3x their volume, so at S(19-41) of sphere 28 your largest spheres are 3^19 times smaller than your original photon sphere so you have to multiply them by some 4/3pi(7e-7^3)/(4/3pi((3^19)^3)) before dividing by 3^19 times 9^28...

This was actually close mp is 2e-9 kg, 4/3pi(7e-7)^3/(4/3pi(7e-7/3^19)^3=1.5700429e+27, 1.5700429e+27/(3^19 x 9^28)=2.58117479e-9

Also I can plot/graph the internal dynamics of all 84 elementary particles by compressing the photon: Edited by Pock Suppet

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3 hours ago, Pock Suppet said:

This was actually close mp is 2e-9 kg, 4/3pi(7e-7)^3/(4/3pi(7e-7/3^19)^3=1.5700429e+27, 1.5700429e+27/(3^19 x 9^28)=2.58117479e-9

Note: 7e-7/1.5700429e+27= 4.4584769e-34, the energy of 5.2334763e+26 photons is almost compressed into a planck length.

3 hours ago, Pock Suppet said:

Also I can plot/graph the internal dynamics of all 84 elementary particles by compressing the photon: ##### Share on other sites This topic is now closed to further replies.

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