The absolute electron mass in kg

[amoram]

So the energy-pound of the particle named electron is exactly 9 times
the dimension 10^16(existing in the transversal area if 1 m^2 existing
every s^2) x 10^16 (the same "charge" counted now how mass in the
direction of the mass flow) gets the 10^32 of the 4th power of the
10^8 dimension of the light. These 10^32 are (m/s)^4 that, in its 1/4
as "the quantity of the unitary potential presence", are m/s.
All the fourth dimension of the dimension 10^8 of the electromagnetism
is so 10^32 m/s, where the unitary mass in line is of 1 dm = m/10. So
we must divide for 10 as m, as s. The "time" 1/10 of the second
becomes the quantity 10^32 x 1/10, while m/10 gets 1 kg. So 9 x 10^31
electron energy masses are 1 kg.
This confirms the 9 exact energies as an unit, also of the unitary
electron particle, and this at dimension of 10^-31 kg.

We obtain the electron mass through this numerical calculus, referred
to 1 dm^3 of water equal to 1 kg:
900/100 x 10^-30 x 10^-1 kg = 900/100 (the energies of 10^-2 a.u.m.)
having 10^-30 x 10^-1 kg of mass.
900/100 are the 10^3 particles of matter in 1 a.u.m. divided in the
10^-6 of the matter-antimatter complex.
When these 10^3 are referred to 1 a.u.m., we have 540 particles (how
the unitary intensity of the Candle, 540 x 10^12 cycles/s) how the
"real matter" 500 (as 1/2 of 10^3) +40 (the real cycle), summed to 500
(the antimatter) -40 (that negative reality of the negative time). In
this reference to the a.u.m., the number of the mass precises the
10^-3 unit of the positive volume containing 10^3 units how the
540/10^6 complexes particle of 1 a.u.m., that is 0,000540 a.u.m.

But here we have to refer to the a.u.m. that contains 10^6 complex
particles, expanded in the 6 coordinate lines +x +y +z -x -y -z, while
the kg gravitation happens in only one. So 0,000540, divided by the 6
directions, is 0,00009 in one, while the 10^2 of its section are
0,009.
In this way we have the 9 "third minutes" corresponding to the time of
54/100 of second minutes that goes on simultaneously in the 6 positive-
negative directions of the complex space-time-mass. Where the time
advances in 6 directions in the 0,54 s of the time minimum of the
Sidereal Year (365,25 days, 9 prime, 9 second and 0,54 second), the
"third time" is 9 too, because 9 prime and 9 second (with a dimension
6x10 of unitary relative difference) already both respect the
expansion 6 in the cycle 10. On the contrary, 0,54 second minutes, to
respect the same cycle and to count in "third minutes", must divide
itself by 6/100, referring the 6 quantities +0,09x +0,09y +0,09z
-0,09x -0,09y -0,09z = 54/100, to the total 100 "red colored"
quantities of the area 10x10 of the flow. So 54/6 x 10^-2 x 10^2 = 9
"third units" = "third minutes", and 10^-3 = 0,001 how 0,000999 times
0,0000001 is presented in 9 "prime, second and third" minutes... in
which 0,001 (the 10^-3 unit of the mass 10^3) is all divided in third
minutes how 1 +9 +90 +900 units 10^-3 equal to 10^3 third minutes.

The same happens passing from the 0,00054 a.u.m, to the 0,009 energies
of the cycle of ten 0,0009.
A ideal balance measures as 100 units of pound this mass. The other
900/1.000 of the 1.000 thousandth masses of the electron are green
colored and are positioned on the first red level of the mass, and add
all how "pound added in gravitational energy". In fact the ideal
balance doesn't touch this 900 cubes, but accuses its gravitational
energy added to the only 100 known by the balance, because, caused by
this energy added in pound, the balance measures all the 1.000/1.000
of the particle 10^-1 of the u.a.m.
So, when in the electron mass, we put in 900/100 of energy, we
consider only the green colored quantity of the movement of the red
presence of the masses. In fact this 100 red colored masses are in
gravitational free fall. The situation is this other, in which I show
the 100 masses in the high position.
This free fall is acted by the electron masses always collocated at
distance 10, how a body in free circular orbit that is in eternal fall
but is always collocated to the same distance (1/10 of the a.u.m.
cube) because the tangential escape is always equal to the
simultaneous fall, and the result of this eternal fall is the
curvature of the tangent line in a circular form indicating the free
eternal fall. Every turn of the orbit is how the unitary cycle of this
fall that we have divided in 10 units. So the electron (red) mass,
that is always at 1/10 of the 10 distance of the unitary side of the
cubed u.a.m., in its real movement is only the green part that doesn't
appear, but is.
The division 900/100 = 9 reduces to one alone the 100 masses of the
red transversal plane, and the situation is this sequent, that
represents a pure line, because dividing a volume by its transversal
plane we have the only vertical line of the ideal fall of one only
particle of the u.a.m. equal in space at 10^-6 u.a.m. cube, having its
side 1 equal to 10 decimals.

I here show the green 9 decimals of the energy of the electron eternal
fall, since it is rotating around its nucleus.
We have these 9 decimal in space equal to 9 unitary energies, and use
the space to count the energy, because 1 energy is the movement of 1
decimal of the space 10 of all the line 1 of the side of the u.a.m.
cube.
Using the space to count the energy of the movement, and using the
same space to count the stopped position 1/10 of the unit of the
u.a.m. cube, this space is a pure line of 9/10 of the u.a.m. side,
where the unitary mass in line is collocated exactly at 10 decimals of
distance, and it is "pound" expressed in kg, the unit of the pound-
mass.

To calculate the quantity in kg, we must consider 10^10 u.a.m.
collocated one upon the other, in way that the distance was in space
dimension the same of 1 dm, the space reference of the kg cube.
10 u.a.m. cubes in line form a length of 1 Å (Angstrom, unit of the
atomic space) and 10^10 Å are 1 m. So the 1/10 of the Å is the space
reference of the u.a.m. and the same 1/10 of the m is the space
reference of the kg cube.
So it's evident that 10^10 a.u.m. cube in full linear sequence are the
1 dm of the kg cube.

This kg cube is a volume whose calculus is made in line: 1 dm x 1 dm x
1 dm.
If we put 10^10 u.a.m. sides, at the place of 1 dm we precise the same
length of 1 dm in the a.u.m. sides corresponding.
So the dm^3 (equal to 1 kg when is 1 dm^3 of H2O) is 10^10 x 10^10 x
10/10 = 10^30 a.u.m. sides of 10/10 = 1 kg.
To precise the electron quality of 9 energies of the particle that is
1/10 in line of 10/10 of a.u.m., we have to introduce the 9 energies
ad the 10 positions of the side of the 10^3 cube containing the 1.000
particles of the a.u.m., that are 10 in line.
In this way, 10^30 u.a.m. in line, referred to the 10 lines (energy +
mass) of the electron eternal gravitational fall, becomes the quantity
of 9 energies of 10^30 x 10 electrons (1 red + 9 green) = 1 kg. If
10^31 electrons are equal to 1 kg, 10^-31 kg is 1 electron having 9
energies. So - since the energy and the mass is the same, because it
is the energy of the mass 1, we don't precise the quality of the 9,
and count 9 masses in movement, only the green colored. and finally 9
x 10^-31 kg is the energy-mass of 1 electron.

WE CAN ALSO REASON IN THIS OTHER WAY, to have exactly 9 times 10^31
kg.
The mass 1 is the cycle 10 that has lost 9 unitary spaces, to can be
amassed in 1 only space.
So 1 mass, in space, is worthed a -9 quantity of space, lost in the
amassing.
When we measure this -9 of space we use 0,109389754 quantities kept
away (so subtracted) from -9, so they are negative ones, are -0,109.
The practical result of this subtraction to -9, a negative quantity
is: -9 -0,109 = -9,109.
This mass, for us that consider positive the space, is negative one
because it is an amassing, but we, for the 3th Dynamic principle
"action-reaction", perceive +9,109 (as reaction) the action -9,109 of
the unit mass counted. In fact we measure the mass through the
opposition to the same mass going down, really acted in up by a
balance.
On the contrary the expansion is positive, is +9 times 1 and the
subtraction of the TIE to c subtracts a quantity to the positive 9,
that is reduced to 8,98755...

How is it possible to unify this electron mass 9,109389754 with its
8,98755... expansion, when in a case the tie is added and in the other
it is subtracted, in way that afterwards the numbers are different
ones?

The lack in UNIFICATION is evident one!


E (energy in power) can be equal to two opposite -9 and +9 that summed
annihilate these two opposite works of the amassing and of the
expanding, or can present the Expansion as 9/1 (a 1 expansed 9 times)
and the amassing as the inverse 1/9 (of the 9 amassed in 1). In this
way mc^2 is in number 1/9 x 9/1 = 1 E.
The two oppositions -9 and +9 and 9 and 1/9, are the same thing.
10^+9 and 10^-9 is the absolute case of the index +9 and -9, and are
the two inverse 1.000.000.000/1 and 1/1.000.000.000.

Why today the electron appears to have the mass of 9,109389754 x10^-31
kg? This happens how a consequence of our unitary measurement, that
put in unitary geometrical TIES that are extraneous to the quantity
measured. Here are all this TIES at every dimension referred to the
unitary meter:

10^-1 is the unit of 10^1. It is the action 1/10 that shows the
reaction 10/1 of the unitary cycle of the space. The tie is 1/10, so
it is 0,1.
10^-2 is the unit of 10^2. It is the transversal area xy = 10x10, of
the flow of the mass that flows only in z direction. So we ave the
tie 0 at dimension 10^-2, that is 0,00.
10^-3 is the unit of 10^3. The time 1/10 is entered in joke and the
energy 3x3 of the transversal area xy, now can work in direction z, as
the tie 9 that, at the dimension 10^-3 is worthed 0,009.
10^-4 is the unit of 10^4. The volume 10^3 is in full real movement
and shows the tie 3 on the base of its numerical cycle. At the
dimension 10^-4 the tie 3 is 0,0003.
10^-5 is the unit of 10^5. This, referred to all the 10^10 Angstrom
that form 1 m, is the dimension (10^10)^1/2 of the electric flow. 10^3
x 10^5 counts all the masses electro-magnetics, and are 10^8, whose
tie is the 8 on the base of its numerical cycle 10. At the dimension
10^-5 the tie 8 is 0,00008.
10^-6 is the unit of 10^6 (all the complex space 10^3 x 10^3). The
unit of the mass volume 10^3, multiplied by all the line 10^6, gets
the tie 10^9, of the 9 on its decimal base. At the dimension 10^-6 the
tie 9 is 0,000009.
10^-7 is the unit of 10^7. All the space in 10^10, divided bu the 10^3
unitary masses get the number 10^7, whose tie is the 7 on the base of
its numerical cycle 10. At the dimension 10^-7 the tie 7 is
0,0000007.
10^-8 is the unit of 10^8, that is all the length electromagnetic
having the absolute area 10^2 in 10^10. This precise calculus 10^8 :
10^3 = 10^5 quantifies all the masses electrical of the electron. 5 is
the tie on the base of its numerical cycle 10. At the dimension 10^-8
the tie 5 is 0,00000005.
10^-9 is the unit of 10^9, =that is all the space runned by the cycle
10, in 10^10 (the usual Angstroms = 1 m). The calculus 10^9 : 10^5 =
10^4, numbers the electrical masses. Th tie is 4 on the base 10 of its
numerical cycle. At the dimension 10^-9 the tie 4 is 0,000000004.
The sum of all the ties ad every dimension is:
0,1
0,00
0,009
0,0003
0,00008
0,000009
0,0000007
0,00000005
0,000000004
==========
0,109389754

The Physicians cannot have more numbers than 9,109389754, because by
the cycle 10 can be obtained till 10 decimal numbers. A eleventh
decimal number exists, but "isn't yet present" for the reasons told by
Planck.


9 is the true, and 0,109389754 is the geometrical tie that if isn't
cut away seems a real increment of the mass.
If I put in a geometrical tie, how a lever, I can have more power.
If I put 1 kg on a lever multiplying for 2 the capacity of work... (of
work by 1/2), and on another that multiply for 3 this power... I have
always 1 kg, and not 2 or 3...
So the Physicians that affirm that the electron mass contains also the
geometric contribute similar to a space-temporal lever, are doing an
horrible mistake.