Συζήτηση χρήστη:Nalxhal

TWO EXPERIMENTS TO EXAMINE THE UNIFIED HYPOTHESES

Author : Alexandris Nikos

http://profiles.yahoo.com/nalxhal http://profiles.pathfinder.gr/nalxhal

ABSTRACT

We can close this open presentation with an experimental proposes. This start with a philosophical book (1999-2006, ISBN 960-8353-85-8) with the main conclusion about the existence of two dimension of time and extra dimension of space. Papers, articles in an open way include a mathematical proceeding with calculations between hypothesis and the acceptable parameters of unified theory as mass plank, temperature of plank and length of plank. Then we found the connection with real world as Avogadro number, proton, electron and fine structure of electron. We propose more hypotheses to search this relation and some ways to prove them, like neutrino prediction, length of Fermi, cmb radiation and nuclear particles. To close this discussion we must to propose an experimental application of these ypotheses and metrical analysis.

MAIN ARTICLE

We must to remember some interest functions and relations of two papers All metrical analysis describe little black holes LBH

meg = 4,66.10-9.kg = 2.61x1018.GV/C2 of 5.1a), function 108 of paper: m=e/k > 0 e : electron charge , Ke: Coulomb constant , k =(G/Ke)1/2 = (4πGε0)1/2 =8,6164×10-11 C/Kg , G: gravity constant , function (106) ,π=3.14..,c:velocity of light ,λplank: length of plank , h: plank constant , le:length of charge 5.29x10-11.m , Na:avogadro’s number , kb: Boltzman’s constant . paper : http://www.wbabin.net/science/alexandris.pdf one very important constant π*= 3,1598199 with units (76)

                h2.c2.G / Ke3. kbl 4 = π*4   

Temperature in a unified area

T =(n1/n2)-3/2. N-3/2.1,085×1016.Q/ (λ.lc)1/2 , λ = lc , a = ap (102)

For n1 =10 , n2 =12 , N=1 , Q=e we have

T.lg = 5,755×10-3.m.K , double Wien’s constant (171)

Using the law of Stefan-Boltzman , function 102 or 171 for an appropriate length and N we have results in agreement while Wien’s law cannot . Wien’s law is a case without gravity but has gravity . Gravity affects to Wien’s law and calculations are in disagreement with the law of Stefan - Boltzman the function 102 also gives the Tplank for lplank without sqrt(2π). The Wien’s constant is not valid.

Mass plank calculation

mc =c2.λPlanck / 2π.k2. Kel =2,17671.10-8 kg = MPlanck > 0 (117)

Expotential geometry of Interactions :

(E/l)c2.(E/l)G1. (E/l)*4 = (E/l)e3 .(E/l)T 4 (150) or forces fc2.fG1. f*4 = fe3 .fT 4 (154)

Gaus geometry of energies So the sum of energies is not valid, but the sum of the squares of the energies multiplied by the square of the number of the spatial dimensions of each corresponding interaction are:

                                        Εcge2 = -4.Ec2- Eg2 +9.Ee2                                              (179)
     Ec =-3.h.c/ lc ,  Eg= -3.G.m2 / lg ,   Ee=3.(Ke.(k.m)2 /lc   ,  lc = λPlanck                (180)
  

lg=lc.sqrt(2π) From (158) function : m = mcge = 1,7209×10-7.kg or energy 1,5467×1010.J and this mass comes from function 155 : fc2.fG1= fe3 the relation of forces : electromagnetic , gravitational and electrical force . We rename the Εcge =6,181595963x1010.J to Εsqrt =sqrt( -4.Ec2- Eg2 +9.Ee2 ) We must replace the type error Εsqrt =6,5963x10-14.J

Function 175 : From 162,172 could be : Εcge = A175/sqrt(2). lc , A175=2,5041x10-24.J.m/kg2 Εsqrt / Εcge = 1,97 or Function 177 : for A177 sqrt(2). lc , A177=2,383x10-24.J.m/kg2 , n=10 , we have :Εsqrt / Εcge = 2,07 , thaus Εcge = 2.A/sqrt(2). lc or Εcge = A/(sqrt(2)/2). lc = A/(sqrt(2)/2). lc = A/cos45o. lc so Εcge = A/cos45o. lc

Avogadro number                  
                          meg.le2 .NA-2 = MPlanck.lg2                                                 
       le=5,291×10-11.m ,  meg >  0,  MPlanck>  0                                                (194)

This meg is of hypothesis 5.1a)

Proton mass

Empirical form of angular momentum that is relevant to the mass of proton.

                (NA.mp. (meg/(2π)1/2)) 1/2 .c. λplank = p.10.h,  p = 1,0065     
                                                    meg >  0                                                        (195)

This (meg/(2π)1/2 is meg of hypothesis 5.1.b mp=mass of proton, lg (length of gravity interaction)=λPlanck.(2.π)1/2.

From article : http://www.wbabin.net/science/alexandris5.pdf

Moles and fine structure of electron

Na.mp =A.meg and

A=(Eplank2/ Emeg2).100.( 2π ) 1/2 Na.me = 10.(E1/E2).meg Eplank2/ Emeg2=Je/h=21,80995 ap/a=2π or ap=2π.a , a=1/137,035 , ap : fine structure of proton

So we have two new energies : 1) Ecge and 2) Emcge Emcge 1.5467E+10.J Emeg 4.1890E+08.J Eplank 1.9563E+09.J Ecge , sum 6.18159E+10.J

1th Experiment

Verification of hypothesis lg=sqrt(2π).lc . This hypothesis said that some electromagnetic waves or all related with a gravitational wave. Two lights like laser cross the space between earth and a planet or satellite in a distance of sun. The sun effects with gravity. The two lights meet each other in a different phase. If gravity wave existed, the frequency c/lc and the lines of lights will change.

2th experiment

Astronomical observation of unified points. The unified points must have density of energy in accordance with length of light in two laws: 1) Stefan Boltzman law 2) law of Wien with double constant functions 102 , 171 and length of gravity lg 3) lengths of first paper λ = 2πlc , ypothesis 4 and lg = sqrt(2π).lc , ypothesis 5 http://www.wbabin.net/science/alexandris.pdf We observe and calculate the areas of sky in the same way of CMB prediction Modified law of Wien in function 171 constant171=2.constant of Wien's constant CMB radiation ( http://hyperphysics.phy-astr.gsu.edu/hbase/bkg3k.html )has energy P/S = 1,9X10-3.w/m2 ,

this value in the law of Stefan-Boltzman ( function 166 ) P/S=σ.T4, 

gives 13,52K Then in modified law of Wien ( in function 171 ) T.lg=5,755×10-3.m.K so lg = 0,42mm λ=2πlc , lg=sqrt( 2π ).lc so λ=sqrt(2π).lg =1,052mm , the cosmic microwave background radiation (CMB). This length λ is length of meg or the length of remnant force as we extracted in http://www.wbabin.net/science/alexandris9.pdf So CMB radiation is the radiation of meg = e/k First proceeding in this solution was in below paper But we must fix the type errors: http://www.wbabin.net/science/alexandris5.pdf

• All presentation in : http://www.wbabin.net , list of authors