Imagine a spherical volume of space and time representing a photon as 60 iterations that reduce, via sphere eversion, by one planck length at a time:
x^2+y^2+z^2-n=4/3pix^3
x/((2(3^59))=1.616229e-35
x=1.616229e-35*((2(3^59)))
x=4.56758796e-7
x^2+y^2+z^2+n=(4/3*pi*4.56758796e-7)^3
x+y+z=square root(7.0036847e-18-n)
x+y+z=7.0036847e-18^(1/2)-n^(1/2)
n^(1/2)=7.0036847e-18^(1/2)-(3(4.56758796e-7))
n=7.0036847e-18^(1/2)-(3(4.56758796e-7))^2
8x-6y-4z-7=2.6445699e-9
8x-6y-4z=2.6445699e-9+7
4x-3y-2z=(2.6445699e-9+7)/2
4x-3y-2z=3.50000000132
S(0)=x^2+y^2+z^2+4x-3y-2z-3.50000000132=0
S(60)=x^2+y^2+z^2+4x-3y-2z-(3.50000000132)/(2*(3^59))=0
S(60)=x^2+y^2+z^2+4x-3y-2z-1.2384658e-28=0
V©=x/((2(3^59))=1.616229e-35
x=1.616229e-35*((2(3^59)))
x=4.56758796e-7
4.56758796e-7/(2(3^59))=1.616229e-35
1.616229e-35=squareroot((1.0545718e-34*6.67408e-11)/c^3)
2.6121962e-70=(1.0545718e-34*6.67408e-11)/c^3
c^3=(1.0545718e-34*6.67408e-11)/2.6121962e-70
c=cuberoot((1.0545718e-34*6.67408e-11)/2.6121962e-70)
c=299792379.714
Now we suppose matter is trapped light. You can fit 8 spheres around the surface of a sphere, if these 9 spheres
represent a charged particle with 9 times the mass in 1/9th the volume of
your original photon than you can repeatedly perform these 9-fold compressions 28 more times before you exceed the planck mass. I calculated
that the entire 29th sphere would be <lp:
Since it’s charge must compress the photon by 1 planck length per planck
time for it to travel at c, the photon mass can be expressed by the quotient of
radii between a photon & planck length -> (7e-7/2)/(2(3^60))=4.1282194e-
36. Viz a viz, the photon density of elementary particle (EP) 1 is
4.9320464e-36/(4/3pi(7e-7)^3)=3.388006e-17 kg/m^3. Ergo, the particle
density of EP 28 is 4.9320464e-36 x 9^28/(4/3pi(1.6e-
35)^3)=1.4848022e+96 kg/m^3 this pretty much checks out as the densest
possible EP before you get a black hole planck particle.
In this theory the interior of the planck particle is virtually hollow, having less quantum fluctuations than the vacuum of space allowing for c to be increased, it's the photon sphere in which we get the gamma wavelength and where gravity is strongest as everything within the planck particle is pushed about the BH horizon - to prove this:
Black hole evaporation will be used to find the 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 protons do not have λmax = vacuum density, 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 proton’s collective
micro-BH quasars 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)))((4π)(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 ✓
So we rescale the photon after the final transformation.
v(QE)=X/(2(3^59))=1.616229e-35/(9^28)
X=(2(3^59))*(3.0882513e-62)
X=8.7276366e-34
8.7276366e-34/(2(3^59))=3.0882513e-62
3.0882513e-62=squareroot((1.0545718e-34*6.67408e-11)/c^3)
9.537296e-124=(1.0545718e-34*6.67408e-11)/c^3
c^3=(1.0545718e-34*6.67408e-11)/9.537296e-124
c=cuberoot((1.0545718e-34*6.67408e-11)/9.537296e-124)
v(QE)=cuberoot((1.0545718e-34*6.67408e-11)/9.537296e-124)
v(QE)=1.946917e+26 m/s
There was no big bang, and there will be no big bounce. We live in a black hole. The horizon is about 14 billion light years away, that's where we see the picture of the Cosmic Microwave Background at the edge of the quasar around our black hole. Our black hole was built from planck particles in a super cosmos that were about about 7e-7 meter across, that's about the length of the black holes that are merging with ours to produce Dark Energy. Our cosmos is collapsing via black hole evaporation, by the planck length which is the source of Dark Matter. That is why the universe has more Up Quarks than Down Quarks, and why there's not much anti-matter in the cosmos.
Also interesting to note, that when one travels even at 7e-7 meters every 10^-10 seconds as is the case beyond our cosmic event horizon, the photons there that we see as the CMBR are 38000 light years across so even if they weren't gamma shifted about a quasar they're still not going to register like normal light to our instruments.
So about the horizon of the Cosmic Black Hole we add S(0) about the circumference, which is around 44 billion light years worth of them, to the edge of the x&z-axes per planck time - and of course we are subtracting by S(60) per planck time of the photon particle:
Ex.) Plot2(S(0))=(x^2+(3.50000000132^2-1.2384658e-28^2))+y^2+(z^2+(3.50000000132^2-1.2384658e-28^2))+(4x+3.50000000132-1.2384658e-28))-3y-2z(+3.50000000132-1.2384658e-28)-(3.50000000132-1.2384658e-28)=0
I am not refuting that the formation of a black hole looks like a spontaneous expansion resembling the big bang in the center of a very large star in a very time dilated space that has gone supernova, but this big bang will be very cold and blue suns will become the SMBHs at the center of galaxies when they run out of fuel.
