Non-unifying Geometrized Newton-Cartan Gravity

[Strange]

13 minutes ago, inSe said:

The length of the base (c), which would equal the length of it's sine

The length if a side is not equal to a sine.

Sines do not have a "length".

Quote

of the isoscoles right and left triangles made from cutting every equilateral triangle in the Koch snowflake/antisnowflake of S(0) & S(1) in half.

All the triangles are similar (technical term, that; you might need to look it up) and so the angles, and therefore the sines will be the same for all of them.

Quote

It clearly shows why the radius of the spheres of the 60th iteration of the sphere fractal equals a Planck length, given the length of what's being graphed equals the length of a photon.

You might think it shows that, but it is gibberish, not mathematics.

Also, what is the "length of a photon"?

However, it seems that 3⁶⁰ * Planck length (which is what you mean, expressed very much more simply) is the wavelength of red light. (Not that this means much, whatever number it came to could be the wavelength of a photon).

[inSe]

On 5/23/2018 at 6:49 AM, Mordred said:

 if your using Mandelbrot then you should have recursive functions. This includes the recursive functions of the Koch snowflake.

ie [latex] D=\frac{log m}{log f}

D=logmlogf

Alright, do you understand that I'm making 8 Koch snowflakes and also the inner 8 Koch antisnowflakes for each iteration of S(1->60)? There's 8 of each because not only am I inverting the two dimensional fractal x,y plane, but I'm also inverting another two dimensional fractal the x,z plane.

If you've got that down, do you understand that isn't the final fractal I use? The final fractal I use are spheres from the Koch snowflake/antisnowflakes which I've explained how to derive (using the c (as in abc) of the right and left isoscoles triangles made from cutting the equilateral triangles of S(0->60) in half.

Thats the fractal I need the thread scheduling codes to input into mathetica for. I still don't understand how to apply the recursive functions for S(0->60) iterations of THAT spherical fractal I derive for thread scheduling

  [latex] D=\frac{log m}{log f}   

 

17 minutes ago, Strange said:

The length if a side is not equal to a sine.

Sines do not have a "length".

All the triangles are similar (technical term, that; you might need to look it up) and so the angles, and therefore the sines will be the same for all of them.

You might think it shows that, but it is gibberish, not mathematics.

Also, what is the "length of a photon"?

However, it seems that 3⁶⁰ * Planck length (which is what you mean, expressed very much more simply) is the wavelength of red light. (Not that this means much, whatever number it came to could be the wavelength of a photon).

7e-7 meters. It does show that. It's as simple as using 7e-7/1.6e-35 in order to find out how many iterations one needs to do. 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 x2 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 

I just need to know where to graphicalle place some (n)e+28 spheres of the 60th iteration, plus all the spheres of the previous 59 iterations in order to make the 3D plot of the inner behavior of my version of the photon using these transformations

to construct each new consecutive plot, until the entire photon gets turned inside out.

This illustrates a photon going from negative charge, to neutral charge at T(t)/2, to positive charge.

Edited May 31, 2018 by inSe

[swansont]

No progress on making a prediction based on your conjecture?

[inSe]

On 5/23/2018 at 6:49 AM, Mordred said:

 if your using Mandelbrot then you should have recursive functions. This includes the recursive functions of the Koch snowflake.D=logmlogf

I can't figure out the recursive thread scheduling input for mathematica unless you go into maximum detail.

But first you need to know exactly what I'm trying to plot here, I feel I've done a poor job in explaining it so far so let me make it simpler with just the x,y coordinates as opposed to the x,y & x,z coordinates, for S(0->1) iterations as opposed to S(0->60) iterations, for circles as opposed to spheres:

We start for the koch snowflakes

Then we derive the circles

Then we turn the blue and red circles only inside out to get our next plot. You'll notice that doing this is equivalent to a negative charge, and it will distort the shape of the black circles into a phi-like shape. Doing this will eventually reduce the size of the entire graph to 1/2 of it's original size and what the graph looks like there is it's neutral charge. Do this enough times and the graph will be the same as when it started, but inside out (positive charge):

Now imagine doing that for the x,z coordinates that cross through the x,y coordinates, for S(0->60) iterations of the koch snowflake, for spheres. Instead of two transformations, you'll end up with n number of transformations (n number of planck times), the first 1/2n-1 of them will plot the negative charge of the photon, the 1/2nth iteration will plot the photon in it's neutrally charged state at 3.5e-7 meters in diameter as opposed to 7e-7 meters in diameter (existing for 1 planck time), the final 1/2-1 iterations will plot the positive charge of the photon. Instead of 48 circles at the final iteration, you'll have (8xn)e+28 spheres upon the final iteration.

Then for 33 compressions of neutral photon states to create heavier particles, you're representing the fusion of (n)e+19 photons. A process that gives you 33 negatively charged particles, 33 positively charged particles, & 33 neutrally charged particles, the building blocks for a periodic table of elements that's based on a new certainty principle, with non-instantaneous, information transmitting ESPs, bridged by microGW charges. This process I've explained here:

 

On 5/23/2018 at 5:48 AM, inSe said:

Finding the right online programs for fractal analysis wouldn't take long to get the centers of the spherical coordinates for those 60 iterations (so that the inversive transformations of that 3D graph of the inside of a photon can be expressed with equations).

Basically what those equations would entail, according to this classification of a TSVF, that CAN be distinguished from the standard quantum interpretation; is that a photon's charge = M + (8A + lp x n)

n = f(m)

f(m) = number of transformations (in Planck units) that have occurred in the graph between t(1) & T(t)/2:

The area belonging to the (7e-7)^3 meter volme of the triangle_S(0) from the S(60) Koch antisnowflake from which the spherical coordinates for graphing the inner structures of a photon are derived 

The mass based off of the midpoint for T(t) in the repeated addition of photon masses in the topological transformations representing 33 compressions of the photon. The time you pause at neutral charge where the photon's length is 3.5e-7, you can fit another neutralized photon in that compacted state, when the two unfold in unison as the clock stars 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 the topological transformations for each inversion to be reduced to 1/2 of the previous inversive transformation representing t(Planck time) that was one Planck length at a time. The second time, however,  well you get the idea. Repeat it 33 times and there's only one sphere that inverts. The photon mass in kilograms equals the length of that singular coordinate transformation, which is (n)e-54 m for the schwarzchild radius of the proton.

For an anti-photon every positive value in that equation becomes negative.

Time dilation can be seen as t + n, w f(n)=M (of the photon) - M (of any particles which are heavier than a photon, an atom, a collection of atoms, or even a stellar mass black hole)

This also defined rainbow gravity, the notion that photons of different wavelengths experience gravity differently based on the color spectrum of light visibility.

 

1 hour ago, swansont said:

No progress on making a prediction based on your conjecture?

Am I not allowed to ask for help?

Edited May 31, 2018 by inSe

[inSe]

Lol, a few last corrections:

1 hour ago, inSe said:

Now imagine doing that for the x,z coordinates that cross through the x,y coordinates, also you're going to rotate the whole graph of x,y & x,z horizontally about the x, y, & z axes to fill the three dimensions, this 2D plot is actually going to be done 6-fold, for S(0->60) iterations of the koch snowflake, for spheres. Instead of two transformations, you'll end up with n number of transformations (n number of planck times), the first 1/2n-1 of them will plot the negative charge of the photon, the 1/2nth iteration will plot the photon in it's neutrally charged state at 3.5e-7 meters in diameter as opposed to 7e-7 meters in diameter (existing for 1 planck time), the final 1/2-1 iterations will plot the positive charge of the photon. Instead of 48 circles at the final iteration, you'll have (24xn)e+28 spheres upon the final iteration.

Then for 33 compressions of neutral photon states to create heavier particles, you're representing the fusion of (n)e+19 photons. A process that gives you 33 negatively charged particles, 33 positively charged particles, & 33 neutrally charged particles, the building blocks for a periodic table of elements that's based on a new certainty principle, with non-instantaneous, information transmitting ESPs, bridged by microGW charges. This process I've explained here:

 

Finding the right online programs for fractal analysis wouldn't take long to get the centers of the spherical coordinates for those 60 iterations (so that the inversive transformations of that 3D graph of the inside of a photon can be expressed with equations).

Basically what those equations would entail, according to this classification of a TSVF, that CAN be distinguished from the standard quantum interpretation; is that a photon's charge = M + (24A + lp x n)

n = f(m)

f(m) = number of transformations (in Planck units) that have occurred in the graph between t(1) & T(t)/2:

The area belonging to the (7e-7)^3 meter volme of the triangle_S(0) from the S(60) Koch antisnowflake from which the spherical coordinates for graphing the inner structures of a photon are derived 

The mass based off of the midpoint for T(t) in the repeated addition of photon masses in the topological transformations representing 33 compressions of the photon. The time you pause at neutral charge where the photon's length is 3.5e-7, you can fit another neutralized photon in that compacted state, when the two unfold in unison as the clock stars 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 the topological transformations for each inversion to be reduced to 1/2 of the previous inversive transformation representing t(Planck time) that was one Planck length at a time. The second time, however,  well you get the idea. Repeat it 33 times and there's only one sphere that inverts. The photon mass in kilograms equals the length of that singular coordinate transformation, which is (n)e-54 m for the schwarzchild radius of the proton.

For an anti-photon every positive value in that equation becomes negative.

Time dilation can be seen as t + n, w f(n)=M (of the photon) - M (of any particles which are heavier than a photon, an atom, a collection of atoms, or even a stellar mass black hole)

This also defined rainbow gravity, the notion that photons of different wavelengths experience gravity differently based on the color spectrum of light visibility.

 

 

Edited May 31, 2018 by inSe

[inSe]

3 hours ago, inSe said:

 

That's totally wrong, it's actually going to look more like this for S(0->1):

Now graphing it in three dimensions, you have that for (xy), you have the same coordinates as xy for (xz), but also for (yz). Everything gets copied 3-fold.

Edited June 1, 2018 by inSe

[swansont]

12 hours ago, inSe said:

 

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Am I not allowed to ask for help?

!

Moderator Note

Of course. If you e.g. need help with math, ask a math question. But do so in another thread, and not in the context of making claims you can't back up. 

This topic is closed until such time as you can make predictions from the conjecture.