icarus2
Registered Senior Member
V-2. Prove to the dark energy observation value(10[SUP]-47[/SUP] GeV[SUP]4[/SUP])
I’m sorry, I can’t English well.
Please, understand my insufficient explanation!
The value of observed dark energy has been proved in accordance with theoretical computation.
ρ_de = 1.38318 X 10^(-47)GeV^4 : Theoretical computation
ρ_obs = 10^(-47)GeV^4 : Observed value
Please have a look the attached proof and inspect the findings.
--- From a physics student
==============
In negative mass(minus mass) hypothesis,
Dark energy is corresponding to that plus potential term in total potential energy.
U_+ lus potential $$( + \frac{{GMm}}{r}) $$, U_ - :Minus potential $$( - \frac{{GMm}}{r}) $$, U_T :Total potential
*Potential energy between positive mass and positive mass has - value:$$U = \frac{{ - G(m_ +) (m_ +) }}{r} = 1U_ - $$
*Potential energy between negative mass and positive mass has + value:$$U = \frac{{ - G( - m_ - )(m_ +) }}{r} = 1U_ + $$
*Potential energy between negative mass and negative mass has - value:$$U = \frac{{ - G( - m_ - )( - m_ - )}}{r} = 1U_ - $$
When the number of negative mass is n[SUB]-[/SUB] , and the number of positive mass is n[SUB]+[/SUB] , total potential energy is given as follows.
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}+\sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})}
$$
$$
U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) + (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79)
$$
In equation (79)
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
From analysis of V-1, V-5,
If U[SUB]T[/SUB] ≥ 0, n[SUB]-[/SUB] ~ n[SUB]+[/SUB], Thus
Define, n[SUB]-[/SUB] = n[SUB]+[/SUB] = n = 10[SUP]80[/SUP] ( 10[SUP]80[/SUP] is about total proton number of our universe),
Observed ordinary matter density is about hydrogen atom 1ea/5m[SUP]3[/SUP] ,
So, m[SUB]+[/SUB] = m[SUB]p[/SUB] , m[SUB]-[/SUB] = 5m[SUB]+[/SUB],(because that dark matter has a five times ordinary matter)
m[SUB]p[/SUB] = proton mass(1.67 x 10[SUP]-27[/SUP] kg),
$$
\bar r_{ - + }= (13.7Gyr/2)=6.85Gyr=6.48065 X 10^{25}m $$
( r[SUB]-+[/SUB] must be calculated accurately, but that’s value(mean distence) has a (light velocity) X (universe’s age ~ half universe’s age); ct0~ct0/2)
$$
U_{de} = (5n^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) = 5n^2 \frac{{(6.67 \times 10^{ - 11} )(2.7889 \times 10^{ - 54} )}}{{6.48065 \times 10^{25} }} J
$$
U[SUB]de[/SUB] = (5n[SUP]2[/SUP] ) X 2.87039 X 10[SUP]-90[/SUP] J
1J = 1kg(m/s)[SUP]2[/SUP] = 6.25 X 10[SUP]18[/SUP] eV
U[SUB]de[/SUB] = (5n[SUP]2[/SUP] ) X 1.79399 X 10[SUP]-71[/SUP]eV= (5n[SUP]2[/SUP] ) X 1.79399 X 10[SUP]-80[/SUP] GeV
U[SUB]de[/SUB] of particle 1ea = U[SUB]de[/SUB]/2n
$$
\frac{{U_{de} }}{{total particle }} = \frac{{U_{de} }}{{2n}} = \frac{{(5n) \times 1.79399 \times 10^{ - 80} GeV}}{2}
$$
$$
\rho _{de} = \frac{2}{{5m^3 }} \times (U_{de} /2n) = \frac{{(n) \times 1.79399 \times 10^{ - 80} GeV}}{{m^3 }}
$$
2 particles (ordinary matter and dark matter) contained in 5m[SUP]3[/SUP], so 2 multiply.
Planck unit transformation by( 1GeV[SUP]-1[/SUP] = 1.975 X 10[SUP]-14[/SUP] cm )
1cm = 0.5063 X 10[SUP]14[/SUP] GeV[SUP]-1[/SUP]
$$
\rho _{de} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{cm^3 }} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{(0.5063 \times 10^{14} GeV^{ - 1} )^3 }}
$$
$$
\rho _{de} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{(0.1297) \times 10^{42} GeV^{ - 3} }} = (10^{80}) \times 1.38318 \times 10^{ - 127} GeV^4
$$
Insert to n = 10[SUP]80[/SUP]
ρ[SUB]de[/SUB] = 1.38318 X 10[SUP]-47[/SUP] GeV[SUP]4[/SUP]
Observation value is ρ[SUB]obs[/SUB] ~ 10[SUP]-47[/SUP] GeV[SUP]4[/SUP] (http://en.wikipedia.org/wiki/Cosmological_constant)
ρ[SUB]de[/SUB] ~ ρ[SUB]obs[/SUB] : "EUREKA"
In QFT(Quantum Field Theory), the energy density of the vacuum is estimated as 10[SUP]70[/SUP]GeV[SUP]4[/SUP], which is about 10[SUP]117[/SUP] orders of magnitude large than the observation value 10[SUP]-47[/SUP] GeV[SUP]4[/SUP].
Therefore,
You can see that negative mass hypothesis how to close to the observation and the universe.
from,
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
Origin of dark energy is particle not pressure or constant energy, because that n- and n+ are number of particle.
Also, because that gravity is repulsive, it is strongly suggest that negative mass is exist.
U[SUB]de[/SUB] is right. It means that U[SUB]T[/SUB] is also right.
Thus, from the analysis of U[SUB]T[/SUB],
It means that analysis of the inflation, fine tuning problem, decelerating and accelerating expansion, future of our universe are right.
You will know the mean which magnitude of dark energy is proved.
It is saying that cosmological constant has not existed and dark energy has not come from vacuum energy.
Definitely, it is against to the ΛCDM model.
For all that, why did the ΛCDM model show a similar result?
We can find out if we look at the total potential energy section, but if we look at the total potential energy (78) equation, (79) equation, it is in a form of :
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}+( \sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})} ) ---(78)
$$
U =(positive potential term) + (negative potential term)= Λ + (ordinary gravitation potential)
Which the positive term played an independent potential role as Λ.
---Icarus2
Hypothesis of Dark Matter and Dark Energy with Negative Mass :
http://vixra.org/abs/0907.0015
I’m sorry, I can’t English well.
Please, understand my insufficient explanation!
The value of observed dark energy has been proved in accordance with theoretical computation.
ρ_de = 1.38318 X 10^(-47)GeV^4 : Theoretical computation
ρ_obs = 10^(-47)GeV^4 : Observed value
Please have a look the attached proof and inspect the findings.
--- From a physics student
==============
In negative mass(minus mass) hypothesis,
Dark energy is corresponding to that plus potential term in total potential energy.
U_+ lus potential $$( + \frac{{GMm}}{r}) $$, U_ - :Minus potential $$( - \frac{{GMm}}{r}) $$, U_T :Total potential
*Potential energy between positive mass and positive mass has - value:$$U = \frac{{ - G(m_ +) (m_ +) }}{r} = 1U_ - $$
*Potential energy between negative mass and positive mass has + value:$$U = \frac{{ - G( - m_ - )(m_ +) }}{r} = 1U_ + $$
*Potential energy between negative mass and negative mass has - value:$$U = \frac{{ - G( - m_ - )( - m_ - )}}{r} = 1U_ - $$
When the number of negative mass is n[SUB]-[/SUB] , and the number of positive mass is n[SUB]+[/SUB] , total potential energy is given as follows.
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}+\sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})}
$$
$$
U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) + (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79)
$$
In equation (79)
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
From analysis of V-1, V-5,
If U[SUB]T[/SUB] ≥ 0, n[SUB]-[/SUB] ~ n[SUB]+[/SUB], Thus
Define, n[SUB]-[/SUB] = n[SUB]+[/SUB] = n = 10[SUP]80[/SUP] ( 10[SUP]80[/SUP] is about total proton number of our universe),
Observed ordinary matter density is about hydrogen atom 1ea/5m[SUP]3[/SUP] ,
So, m[SUB]+[/SUB] = m[SUB]p[/SUB] , m[SUB]-[/SUB] = 5m[SUB]+[/SUB],(because that dark matter has a five times ordinary matter)
m[SUB]p[/SUB] = proton mass(1.67 x 10[SUP]-27[/SUP] kg),
$$
\bar r_{ - + }= (13.7Gyr/2)=6.85Gyr=6.48065 X 10^{25}m $$
( r[SUB]-+[/SUB] must be calculated accurately, but that’s value(mean distence) has a (light velocity) X (universe’s age ~ half universe’s age); ct0~ct0/2)
$$
U_{de} = (5n^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) = 5n^2 \frac{{(6.67 \times 10^{ - 11} )(2.7889 \times 10^{ - 54} )}}{{6.48065 \times 10^{25} }} J
$$
U[SUB]de[/SUB] = (5n[SUP]2[/SUP] ) X 2.87039 X 10[SUP]-90[/SUP] J
1J = 1kg(m/s)[SUP]2[/SUP] = 6.25 X 10[SUP]18[/SUP] eV
U[SUB]de[/SUB] = (5n[SUP]2[/SUP] ) X 1.79399 X 10[SUP]-71[/SUP]eV= (5n[SUP]2[/SUP] ) X 1.79399 X 10[SUP]-80[/SUP] GeV
U[SUB]de[/SUB] of particle 1ea = U[SUB]de[/SUB]/2n
$$
\frac{{U_{de} }}{{total particle }} = \frac{{U_{de} }}{{2n}} = \frac{{(5n) \times 1.79399 \times 10^{ - 80} GeV}}{2}
$$
$$
\rho _{de} = \frac{2}{{5m^3 }} \times (U_{de} /2n) = \frac{{(n) \times 1.79399 \times 10^{ - 80} GeV}}{{m^3 }}
$$
2 particles (ordinary matter and dark matter) contained in 5m[SUP]3[/SUP], so 2 multiply.
Planck unit transformation by( 1GeV[SUP]-1[/SUP] = 1.975 X 10[SUP]-14[/SUP] cm )
1cm = 0.5063 X 10[SUP]14[/SUP] GeV[SUP]-1[/SUP]
$$
\rho _{de} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{cm^3 }} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{(0.5063 \times 10^{14} GeV^{ - 1} )^3 }}
$$
$$
\rho _{de} = \frac{{(n) \times 1.79399 \times 10^{ - 86} GeV}}{{(0.1297) \times 10^{42} GeV^{ - 3} }} = (10^{80}) \times 1.38318 \times 10^{ - 127} GeV^4
$$
Insert to n = 10[SUP]80[/SUP]
ρ[SUB]de[/SUB] = 1.38318 X 10[SUP]-47[/SUP] GeV[SUP]4[/SUP]
Observation value is ρ[SUB]obs[/SUB] ~ 10[SUP]-47[/SUP] GeV[SUP]4[/SUP] (http://en.wikipedia.org/wiki/Cosmological_constant)
ρ[SUB]de[/SUB] ~ ρ[SUB]obs[/SUB] : "EUREKA"
In QFT(Quantum Field Theory), the energy density of the vacuum is estimated as 10[SUP]70[/SUP]GeV[SUP]4[/SUP], which is about 10[SUP]117[/SUP] orders of magnitude large than the observation value 10[SUP]-47[/SUP] GeV[SUP]4[/SUP].
Therefore,
You can see that negative mass hypothesis how to close to the observation and the universe.
from,
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
Origin of dark energy is particle not pressure or constant energy, because that n- and n+ are number of particle.
Also, because that gravity is repulsive, it is strongly suggest that negative mass is exist.
U[SUB]de[/SUB] is right. It means that U[SUB]T[/SUB] is also right.
Thus, from the analysis of U[SUB]T[/SUB],
It means that analysis of the inflation, fine tuning problem, decelerating and accelerating expansion, future of our universe are right.
You will know the mean which magnitude of dark energy is proved.
It is saying that cosmological constant has not existed and dark energy has not come from vacuum energy.
Definitely, it is against to the ΛCDM model.
For all that, why did the ΛCDM model show a similar result?
We can find out if we look at the total potential energy section, but if we look at the total potential energy (78) equation, (79) equation, it is in a form of :
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}+( \sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})} ) ---(78)
$$
U =(positive potential term) + (negative potential term)= Λ + (ordinary gravitation potential)
Which the positive term played an independent potential role as Λ.
---Icarus2
Hypothesis of Dark Matter and Dark Energy with Negative Mass :
http://vixra.org/abs/0907.0015
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