Download Atom (Single) with 3 Electrons and 3 Protons Or (Double) 6

Atom (Single) with 3 Electrons and 3 Protons
Or (Double) 6 Electrons and 6 Protons
Atom (Single) with 3 Electrons and 3 Protons
Or (Double) 6 Electrons and 6 Protons - conjugates
not shown
PS- designates pseudosphere. In this case we
can directly add or subtract the energies.
PS1 = PS3 energy at level 3
PS1+PS2 = 2PS1
Inclined cones, when straightened
will separate the projected circles.
3S3
PS1
Hyper - planes, each with its
own energy tube/string
Difference in energy
correspond to distance
3S2
2S2
2PS1
3S
1S2
2S1
2S
1S
1S orbital
2S orbital
3S orbital
Z
Z
30° degrees
14° degrees
11.5°
(11.36°)degrees
Note that the heptagon has n!/2n = 7!/14 = 360 possible configurations which fall into:
(n-1)! - (n-1) / 2n sets.
[ (7-1)! - (7-1) ] /2x7 = 51 configuration groups, thus deviding the circle into 7 degrees!
The 51 groups of single electron multiplied by 2 = 102 elements of the periodic table with 2 electrons assigned to each
orbital.
This amounts to 360 wave functions from n orbitals with single electron assigned to each orbital.
Arrange the table with double or single electron, and number of electrons and protons
External sphere has sunk in splitting the vertical axis (the particle and
anti-particle have rotated in opposite direction) giving us two projection
points.
when the vertical axis splits to 9.5°, the sphere has sunk in half way,
similar to a projectile on the face of the larger sphere.
At 11.5 degrees, the sphere has sunk in all the way inside the larger
sphere.
At 45° degrees the sphere is back down to its original sate.
14 and 13.25, there is 0.75 missing which has to do with whether we
take the sphere on the interior or exterior. In the figure to the right we
have four spheres 3 of which are inside the large sphere and one is on
the outside.
14 - 11.5 = 2.5
we had 13.25 ( 1 /2 (26.5) related to golden ratio)
14 - 13.25 / 13.25 = 0.0566 , 0.0566 x 2 = 0.1132
@14° degrees Four spheres
@9.5° degrees 3.5 Spheres
@11.5° degrees 3 Spheres
@45° degrees
one sphere
Particle in a Box Vs. Free Particle
S
S
Apparent and true location of the sphere is not
and the same. Sphere moves along the
diagonal from OS to OS.
S
1
2S
If we rotate the sphere 120° degrees on the sphere and 240°
degrees in space, the sphere will disappear!!!
Like the sine or cosine curve. Note they are out of phase by 2.
60°
r
Determine distance, rotation, and size of molecules.
F
r'
F'
45° r
r'
S
60°
F
120°
Complex plane C
moves up and down
forming the bell curve
Plane reverses by 180 at
each 90 degree turn on the
sphere
Parabolic form of the complex plane. Similar to the bell curve.
Conjugates not shown. When we move up the sphere with circles
parallel to the equator the complex plane bends with a single
curvature similar to the figure above. Double equal curvature
shown below, with a circular as opposed to an elliptical base, for
when the planes are not parallel to the equator.
C3
1
S
Pseudospheres inside shown
in green and red
S
C1
1
2 C1
C6
b
C1
potential energy function
G'
d γ
a
a
G''
E
G
F
F'
y
θ
r
γ
β
E
α
β
a
F''
D CF
CF' β = 30°
CF'' β = 60°
C1
height of
the box C
1
2
C1
a
a
If ds is kept the same length then the curvature will be equal and the
complex plane will curve into a sphere, all curvatures are equal.
To turn it into an ellipse, take two different lengths for ds projecting from
Z and Y or X, find the difference of the two circles and obtain an ellipse.
For example from the Z axis we reduce the mass to 21 C1 , while from the
X or Y axis reduce the mass to 34 C1.
ds
60°
OS
true path O
120°
2C3
1
2
7 particles of a single atom correspond to a heptagon
To picture the Oscillation and Rotation of the Molecule:
Take C1 (Sphere S) as the nuclei inside C7 as having what is called: Nuclear oscillation,
nuclear rotation, and nuclear spin.
Then C6 has electronic orbital motion and electronic spin as C7 rotates on C7 .
This gives five functions, one describing the orbital motion of the electrons the second
orientation of the electron spins, the third the oscillational motion of the nuclei, the fourth
the rotational motion of the nuclei, and the fifth the orientation of the nuclear spins. For a
detailed explanation see Introduction to quantum mechanics, article 43f, pg. 355
2C6
Conjugate projection curve of height of the complex
plane above C6 vs. diameter of the projected circle
Angles from the line of projection to C and C 0
remain equal from the 45° limit shown as 30°
C6
30°
C1
1
0
C3
This is easy to see. The difficulty is in seeing what the complex plane
does as N moves down on the sphere in a spiral fashion. We obtain a
spiral stair in reverse.
Take the north pole on the Y and Z or X and Z.
x
N
C4
ar
app
S
C7
Even more interesting is when we take N on three axes and create three
spirals on the sphere and relate this to the hypersphere.
a
OS
C6
th
pa
t
n
e
N
S
C5
a'
1
2
Z
ds
Behavior of the complex plane
during Loxodormic transformation
Free particle
The boundary of the top of the
rectangular or square box with
sides a and b corresponding to
ds (x), and ds (y) of the
projectile
OS
Plane inverts from zero to 1 and S (at zero)
rotates on S . Sphere will rotate about itself!
Spiral stair form of the
complex plane is an inverted
sphere.Conjugates, C4 and
C5 not shown
Mass reduced to zero at the
equator
60°
O
ds
Split open the complex plane by 120° degrees, in essence rolling S on S
by 60° degrees.
If we split open the complex plane by 180°, then S moves directly to S.
Say we split open the plane by 120°. Then projecting on the complex
plane as shown to the right, for an angle of 60° from S, will give us the
electronic orbital motion and spin of C6. The spin will be 180°, and C6
will reduce in size by one half.
1
2
120°
C1
S
North pole is taken at the Z axis for the figure to the right. The complex
plane relative to N moves up from C6 to C 0 .
If we move N along the Z axis down to 21 C1, then the complex plane will
move to C6(21). N , in general does not move along a straight line. It
moves down on S on a spiral. This is called Loxodormic transformation.
ar
app
ds
C6
Start at the north pole on the
sphere and move downwards on
a spiral
th
pa
t
n
e
The wave and its conjugate will have
different amplitude depending on whether S
moves straight up or on a curved path to S
S
0
OS
1
2
Rotate the force F and its perpendicular distance r to F' and
r'. In the 60° case, the sphere reduces to 21 its size and in
the 45° case it disappears!
parabolic form of the complex
plane.Conjugates C4 and C5
not shown - particle in a box
S
true
Radius of new sphere after rotation by 60° on the sphere and
120° in space is one half the original.
inversion shown as
rotation along the
diagonal
height of the
complex plane
above C6
2C6
N
Normalized wave function for C1. We can
normalize it for a unit sphere for convenience
30°
at Ch
4°
1
2 C1
at C0
8°
C0
Diameter of the projected cirlce
at C6
C6 (21)
C6
It will be more convenient to invert the
potential energy function, as shown in many
text books.
Plot of the height of the complex plane above C6 vs. diameter of the projected circle.
Axis of the diameter of the projected circle should move down to C0
Having this curve, we can determine the length of our segment on C1.
When we cross our vectors we obtain a rectangle, which
is the top of the box, on the sphere. what happens at the
boundary?
Rotate and enlarge the green rectangle.
For two directions this would give us the inclination of a plane to the sphere and
where we can derive its normal.
Plan and section shown on the
same diagram
Free particle - Take N on three different axes, X, Y, and Z, with three different
zpheres. Rotate each N on each axis down the axis of its sphere in an spiral fashion
towards the equator. Relate this to the hypercube.
This will give us the nine dimensional space.
If each axis represents a property, T, A, G for example, then performing this
operation on each sphere on the axis will give us a means to categorize the
properties like for example the human genome.
inversion shown as
rotation along the
diagonal
Form of the complex plane for a Free Particle - Quarks and Anti Quarks
For an analogy, rotating the north pole N down the sphere in a spiral
fashion is much like peeling an apple.
Next sphere if there are two
spheres joined at N
Sphere rolling down on the
complex plane
N
After peeling the surface we can:
Next we rotate this cylinder in space.
1. Stretch it, in the opposite direction of the "gravitational" pull, where
matter is trying to pull back together.
We can do multiple wraps at any hight above the equator we wish hence in
effefct peeling a string, at the speed of light, from a cylinder with a
diameter of our choice.
2. Reverse it hence inverting the sphere, which is what the complex plane
has done for us !
Spiral stair form of the
complex plane is an inverted
sphere. Conjugates, C4 and
C5 not shown
if the material is resisting
a tensile pull, it will result
in a Cosine curve
The size of the knife and the width of the peel vs the size of the sphere.
The smaller the width of the peel, the larger the sphere. Also the longer it
will take to go around a given sphere. The unit sphere is taken for
convenience.
Lets take the apple and peel it from the equator up ever so close to the
north pole and decrease the size of the knife or width of the peel so that
we have a string right before approaching the north pole. At this point we
are traveling with the speed of light.
If we keep rotating in a spiral fashion near the north pole, with a width
equal to a string, in effect we have our complex plane peeling a cylinder
with a diameter ever so smaller than the sphere.
After peeling, let the peeled sufrace hang in the air. If there was no
gravitational pull from the earth, the shape would remain like an apple,
however because of the downward gravitational pull the form will stretch.
Think of the force which has stretched the apple as an internal force which
is trying to stretch the sphere. The sphere will then resisti with an internal
compressive force trying for form itself in the shape of the apple. It will be
like we have put a compressive force on the material. If we reverse the
peeled surface, then it is as if we have put a tensile force on the material.
N
width
Equator
Changing the width of the peel, will result in a different Cosine curve,
longer or shorter, or one with a lower or higher amplitude.
S
5 Revolutions and 5 spheres
N
S
Given this one sphere, we can then generate an infinite amount of waves
or spheres depending on the width of the peel.
From infinite number of curves or wave functions
such as the one shown above we can identify
our curve.
Z
Four spheres in space will provide a convenient way of forming a volume.
Instead 2 spheres of 3 quarks and 3 antiquarks for a total of 6, we have four
spheres of 3 for a total of 12. The six tubes will increase to seven or 12, with
7x7=49 to 12x7=84 particles.
N
C1
y
We will have four hypercubes each with 256 constants, and the number of
sontatns will increase from 256 to 1024.
The figure to the right, with three spirals shown will have an average sum of a
fourth sphere not shown.
Our given ds, or length of the projectile, then fall in any category. If we are
also given the potential energy function and size of the sphere from which
it was obtained, we have an initial sphere say S which is then reduced in
size say to 21 S. The problem is solved for the unforced, undamped
oscillator. From the infinite spheres we are given the one. From the
function we know the size is reduced by 1/2.
Rotating the figure to the left in space, and hence our sphere shown on the
right is not the same thing as taking three different spheres with N taken on
each axes and performing the operation.
N
S
Last segment of the
four-vector a the boundary
will determine the
inclination of the third plane
N
Equator
S
Generate a cylinder of diameter d p , inside the sphere S. Place a sphere
with the diameter equal to d p , and place it on the equator. At the height
above the equator of sphere S decrease the width of the peel to a string
and wrap it around S. The complex plane will generate a cylinder at the
north pole of the sphere dp, 21 dp above the equator of S .
N
S
Equator
N
1 p
2d
S
dp
The height of the cylinder will equal to the diameter of the sphere dp.
Once the cylinder is generated, continue wrapping the string about the
cylinder to increase its length and guide it along any path in space.
S
Magnetic monopole
Pseudospheres inside shown
in green and red
There will be two such cylinders formed inside the sphere corresponding,
C1 and C4, and the two pseudospheres.
if we take C1 and C4 as composed of 3 Spheres each, call them quarks
and anti-quarks, then we have a total of six cyliners! or pseudospheres.
C1
If Seven particles are taken on each pseudosphere then this results in a
total of 42 particles, 21 for quarks and 21 for antiquarks.
For four spheres we would have 84. And the number of constants will
increase from 256 to 1024.
When C5 and C6 are tangent, then the three quarks and antiquarks merge
into one.
At this point, the complex plane should form a sphere.
If we take If we take C1 and C4 as composed of three spheres then C1,
and C4 will each have three hyper cubes, and instead of three surfaces we
would have 9 surfaces on each octant. If we take four spheres C1 to C4
each composed of 3 spheres, then we would have 12 planes.
In other words, we are simultaneously peeling the apple from the north
pole of the X, Y, and Z axes respectively, with different size kinfes and or
widths. The apple could then be composed of one sphere if the size and
width of the peel is taken equal in all directions and could be of different
sizes if the size and width of the peel in the three directions varries.
C6
C4
C5
Z
C7
N
N
C1
y
S
Magnetic monopole in three dimensions
and 3 embedded spheres would result in
a hypercube with 9 planes. Taking the
sum of the spheres as the fourth this will
result in a hypercube with 12 planes.
U
V
M
O
X
e
h
0
J
a
c
K
C2
P
g
b
C4
Residue energy surface given off if the column is
braced and the load keeps increasing.
C3
Q
d
C1 = C2 = C3 = C4 = 0.5 C5 = 0.5 C6
Magnetic Monopole
In comsology, residue from this surface is termed
Cosmic Microwabe Background Radiation by
Stephen Hawking
C1
i
Residue Surface generated
from the Magnetic Monopole
41.5°
Pure Electricity / Charge
No Magnetic field
f
Electric/Magnetic Field / Energy - Plastic stage
Electric/Magnetic Field / Energy - Elastic stage
W
N
R
S
Rotating this cylinder in three dimensional space and increasing its length
with the number of wraps, which will be many! we can in effect create
holes and go in and out of our sphere !!! This is referred to as a Magnetic
Monopole.
Lets call this the periodicity of the complex form.
S
N
L
C6
Magnetic Monopole
dp
0
4°
0
Equator
S
4°
N
1
2 dp
0
#2 Rays which stabilize the
surface
S
r
D
For three embedded spheres the hypercube would have 9 such planes in each
octant. And for four cubes there will be 12 such planes.
d
r
Magnetic monopole
#3 - rays which stabilize the
core.
d
Hyper Cube corresponding to a single sphere has three planes in each octant for
a total of 24 planes. Planes shown on one octant at h, P, and X, in the figure
above.
S
8°
T
± 90°
S
N
C1
C2
C3
C4
C5
Magnetic Monopoles
N
Electric/Magnetic Field / Energy - Elastic stage
Represents a solid as opposed to gas
Residue Surface generated
from the Magnetic Monopole
Residue energy surface given off if the column is
braced and the load keeps increasing. Compare
with 4°
Radii of Particles in the region of the
new potential energy surface should
be 40 times smaller than the ones
inside the quark
S
Electric/Magnetic Field / Energy - Plastic stage
Represents gas
New potential energy surface
Magnetic Monopole
The energy given off in the form
of heat when the column is loaded
up to the yeild point
Pure Electricity / Charge
No Magnetic field
New segment on the sphere
N
4°
dp
4°
Equator
S
0
N
1
2 dp
4° segment would result in a 100
fold increase in the diameter of
the sphere dp
0
S
8° segment would result in a 40
fold increase in the size of the
sphere dp
± 90°
Equator
N
0
51.5° - 52°
#2 Rays which stabilize the
surface
N
S
N
1 p
2d
Equator
S
dp
S
Magnetic Monopole
Dipole
Gradual change in shape of the magnetic dipole into an
elipse. To slow down or prevent this change in shape
one could build two accelerators above or below the
equator at 51.5 to 55 degrees.
D
Direction of fluid flow
and pressure
Monopole
#3 - rays which stabilize the
core.
d
r
d
Walls of the corpus
callosum
r
Centrifugal force perpendicular
to the direction of wrap
We obtain twice the information flow
S
8°
S
Inversion of the complex plane , Energy and Angular momentum (Copy)
Radius of new sphere after rotation by 60° on the sphere and 120° in space is
one half the original.
S
S
1
2S
If we rotate the sphere 120° degrees on the sphere and 240° degrees in space,
the sphere will disappear!!!
60°
r
Like the sine or cosine curve. Note they are out of phase by 2.
r'
F'
Rotate the force F and its perpendicular distance r to F' and
r'. In the 60° case, the sphere reduces to 21 its size and in
the 45° case it disappears!
S
60°
S
r'
F
F
Determine distance, rotation, and size of molecules.
45° r
Pseudospheres inside shown
in green and red
S
120°
ds
C1
C1
C6
C3
ds
S
C4
C5
Split open the complex plane by 120°
degrees, in essence rolling S on S by
60° degrees.
C7
Yin - Yang
Rotate the sphere 90 degrees on the sphere and we have opened up the complex plane by 180. Again, a
difference in phase of 2 (Bottom right)
In essence, rotate OA about O.
When we rotate 90° degrees in space and 45° degrees on the sphere, our sphere disappears, as its ratius
decreases to zero. At this point we reverse the plane.
Note: during rotation S and S reverse.
Take the bottom left figure. S rotates on S to zero. Then from zero to 1 /2 S rotates on S .
Also if the particles were to slide on one another and rotate, we would obtain the figure to the right. vise versa.
ar
p
p
a
Up
S
60°
O
0 - ready to except
another mollecule
0
Double inversion sum
1
S
1
S
0
1
2
45°
true
th
pa
t
n
e
S2
1
0
OS
1
2
inversion shown as rotation
along the diagonal
inversion shown as
rotation along the
diagonal
Apparent and true location of the sphere is not
and the same. Sphere moves along the
S
diagonal from OS to OS.
The complete picture
true
45°
O
O
O
A
A
A
120°
1
2
1
2
apparent
OS
1
S
1,0
Plane inverts from zero to 1 and S (at zero)
rotates on S . Sphere will rotate about itself!
OS2
1
2
0
1
2
60°
A
0
1
S
O
120°
S
OS
OS
1
1
n
re
a
p
p
a
th
a
p
t
S
0
60°
OS
true path O
inversion shown as
rotation along the
diagonal
90°
45°
A
O
A
Trigonal/Hexagonal - Elastic - Longer period
(wavelength)
Anisotropic - the two diagonals do not have the same
length and the material has different properties in two
directions
1
OS
OS
Cubic - Rigid - Shorter period
Isotropic - the two diagonals of the
square have the same length.
S
1
2
0
Plane inverts from zero to 1 and S (at zero) rotates on O
as S(0) has shrunk to zero.
Magnitude of the forces keeping the mollecule together is
equal to the diagonals, from the center of the sphere to
O. As the particel spins around itself, its plane is inverted
two times.
0
A
S
Wavelength
OS
S(0)
true path
apparent path
120°
OS
S(0)
Apparent and true location( at O) of the sphere
while the sphere may seem like it is traveling on a square, it is
actually going along the diagonal OS to O !!! Actually it travels
along the curved path shown bottom right.
O
S
OS
A
S
1
There will be a double inversion sum and cross product
pointing into the board. The figure shows how a sphere
would spin in three dimensions. This is the basic
principle behind Penrose diagrams.