Planck constant "h" "h" is a physical constant of elementary unit of action. It relates energy amount and wave of a quantum frecuency, is time by energy. h = ( √ 10 / 9 )( 2pi )( 10^{-n}U ) = 6,6230588.10^{-n}U "h" is the interaction between fields diameters ( 9 and 10U ) of space time, the factor "time" is represented in the equation by ( 2pi ). The relationship between this constant "h" and the universal background radiation "CMB" is evident since the difference among the wavelength in which the cosmic microwave background offers its highest black-body energy intensity and the reduced Planck constant ( h/2pi ) is exactly 31 orders of magnitude, total 10^{31}U Universal background radiation "CMB"= 0,001054 mU. Reduced Planck constant ( h/2pi ) = 1,054.10 ^{-34}mU. Multiplying the reduced Planck constant h / 2pi. ( 1,054.10^{-34}U ) by "2pi" we will have "h" and doing the same with the wavelength in which the cosmic microwave background offers its highest black-body energy intensity ( 0,001054 U ) by "2pi" will result an "inflated" Planck constant with total precision in "31" orders of magnitude 10^{31}U. |
The Neutron and Space Time
The Space-Time is composed of geometric units (triangles). To form a Neutron some of these units contract and with a mechanical coupling increase their density and create a particle. 10^{-n}U = NoE . h Space-Time has a energy density of 10^{-n}U per basic unit. When a Neutron is formed the triangles that make up cosmos, contract, increase their density and energy transforming the "basic unit" in "unit of action". In case of Neutron primary space-time energy passes from 10^{-n} to "h" the Planck constant, equal to 6,623.10^{-n}U. The difference between these two positions is the Neutron mass energy. m . c^{2} = 1,5099.10^{-n} 10^{-n}U / h = No . c^{2} Neutron mass energy at rest is the difference between the Planck constant and basic unit space-time energy |
The Proton and Space Time
The primary energy of space-time passes from 10^{-n}U to the gravity quantum "Gq" equal to 6,666.10^{-n}U . Where "Gq" is a constant related to gravity and slightly larger than the Planck constant. The difference among these two positions. 10^{-n}U / Gq . is the Proton mass energy ( m . c^{2}U )10^{-n}U / Gq = P+ . c^{2}. The slight mass difference between Neutron and Proton is the discrepancy of two constants " h " and " Gq ". Join equations of these particles and the resultant already incorporates Gravity constant " G ". 10^{-n}U = ( h ) ( No ) / ( G ) ( P+ ). |
A simple coupling between bricks of the Cosmos. The rotation centers approach of equilateral triangles 3U and 2U, causes readings such as Gravity constant, Proton mass, Proton mass energy, etc.
The gravity quantum "Gq" = 2 / 3. 10^{n} = 6,66.10^{-n} U. Proton mass = 10 ^{-n} / 3 . 2 =1,66.10^{-n} U. Proton mass energy = 3 / 2. 10^{n} = 1,5 .10^{-n} U. Newton constant= 2 / 3. 10^{n} = 6,66.10^{-n} U Wien constant = √ 3 / ( 2 . 3 . 10^{n} ) = 0,00288 U. |
These readings have their origin in the diameters of two space fields 9U and 10U, cause lectures like Neutron mass, Neutron mass energy, Planck constant, etc. Here readings are displayed in a perfect decimal line.
Balance diameter = √ 10 / 9 = 1,05409 U. 10^{-n} / Neutron mass energy . 2 pi = 1,05409.10^{-n}U. CMB highest black-body energy intensity = ( √ 10 / 9 ) ( 10^{-n} ) = 1,05409.10^{-n}U. Neutron mass . 2 pi . 10^{-n}= 1,05409.10^{-n} U. Reduced Planck constant = (h / 2pi) = 1,05409.10^{-n } U. |
Quarks and Space-time According to standard particle model , Quark UP + 2 / 3 and DOWN - 1 / 3 are fundamental constituents of atomic nuclei. Proton contains two UP + 2 / 3 quarks and one DOWN - 1 / 3 . Neutron contains one UP + 2 / 3 quark andtwos DOWN - 1 / 3 quarks We can clearly see the drawings corresponding to proton and neutron what we call Quarks, positive and negative charges are well defined with their true meaning. |
Cosmic Microwave Background CMB
It is the main support of Big Bang theory, in perfect agreement whith black-body energy distribution at 2,73 Kelvin temperature, According toWiens´ law a black body with K grade offers its max. photon energy intensity at wavelength of 0,00288m.U( √ 3 ) / ( 3 . 2 . 10^{n}) = 0, 00288 m. U. Then apply Wiens´ law for 2,73 Kelvin max wavelength energy intensity 0,00288 / 2,73 K = 0,001054 m. U. = ( √ 10 / 9 ) ( 10 ^{-n} ) It result from relation between two diameters 9U and 10U. |