icarus2
Registered Senior Member
Relations between radius of universe and dark energy density
fig11. Relations between radius of universe and dark energy density
* mass density of ordinary matter = 1 proton/5m[sup]3[/sup]
* Proton mass= 1.67264 X 10[sup]-27[/sup]kg
* G =6.6726 X 10[sup]-11[/sup] m[sup]3[/sup]/s[sup]2[/sup]kg
* 1J = 6.242 x 10[sup]18[/sup] eV
* $$n_-=n_+=n, m_-=(23.3/4.6)m_+=(5.06522)m_p$$
* $$\bar r_{ - + } = \frac{R}{{3.27273}}$$
* If $$\bar r_{ - + } = \frac{R}{{2.17879}}$$ $$(U_T \approx 0)$$, dark energy density has a 1/3 smaller than $$\bar r_{ - + } = \frac{R}{{3.27273}}$$.
In a WMAP, observed value $$\Lambda = 2.14 \pm 0.13 \times 10^{ -3} eV $$
Dark energy density :
$$
\rho _{de} = 2.09 \times 10^{ - 47} [_{ - 0.465}^{ + 0.557} ]GeV^4
$$
Ridius of the Universe :
$$
R_{UNI} = 96.76[_{ - 11.44}^{ + 12.13} ]Gly = 85.32 \sim 108.89Gly
$$
( If $$\bar r_{ - + } = \frac{R}{{2.17879}}$$,
$$
R_{UNI} = 118.8[_{ - 14.0}^{ + 14.9} ]Gly = 104.8 \sim 133.7 Gly
$$)
From Neil J. Cornish,
the universe' radius is at least 24Gpc(78Gly).
(2003, Neil J. Cornish, "Constraining the Topology of the Universe", http://arxiv.org/abs/astro-ph/0310233v1 )
[Proof]
In negative mass hypothesis, dark energy is corresponding to that positive potential term in total potential energy.
*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_- , and the number of positive mass is n_+ , total potential energy is given as follows.
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}
$$
$$
+\sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})}---(78)
$$
$$
U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) $$
$$
+ (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79)
$$
In equation (79)
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
If radius of the universe is 60Gyr, ordinary matter density is about proton 1ea/5m[sup]3[/sup]. So, m[sub]+[/sub] = m[sub]p[/sub],
$$
m_ - = km_ + \simeq (\frac{{23.3}}{{4.6}})m_ + = (5.06522)m_p$$
(because that dark matter has about (23.3/4.6) times ordinary matter in WMAP)
From equation (95)
$$
\bar r_{ - +} =(60Gyr/3.27273)= 1.73447 X 10^{26}m
$$
From analysis of V-5,
If $$U_T \ge 0$$, $$n_ - \approx n_ + $$, Therefore, Define, $$n_ - = n_ + = n $$
$$
V = \frac{{4\pi R^3 }}{3} = \frac{{4\pi \times (5.67648 \times 10^{26} )^3 }}{3} = 7.66171 \times 10^{80} m^3
$$
$$
n = \frac{{\rho V}}{{m_p }} = \frac{{(1m_p /5m^3 )V}}{{m_p }} = 1.53234 \times 10^{80}
$$
( 10[sup]80[/sup] is about total proton number of our universe).
$$
U_{de} = (kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }})
$$
$$
U_{de} = (5.06522)n^2 \frac{{(6.6726 \times 10^{ - 11} )(2.79772 \times 10^{ - 54} )}}{{1.73447 \times 10^{26} }}J
$$
$$
U_{de} = (n^2 ) \times 5.45168 \times 10^{ - 90} J = 1.28009 \times 10^{71} J
$$
1J = 1kg(m/s)[sup]2[/sup] = 6.242 X 10[sup]18[/sup] eV
$$
U_{de} = 7.99031 \times 10^{89} eV
$$
$$
\rho _{de} = \frac{{U_{de} }}{V} = \frac{{7.99031 \times 10^{89} eV}}{{7.66171 \times 10^{80} m^3 }} = \frac{{1.04289 \times 10^{ - 6} GeV}}{{cm^3 }}
$$
Planck Unit transformation(1cm =0.5063 x 10[sup]14[/sup]GeV[sup]-1[/sup] )
$$
\rho _{de} = \frac{{1.04289 \times 10^{ - 6} GeV}}{{1.29784 \times 10^{41} GeV^{ - 3} }} = 0.80355 \times 10^{ - 47} GeV^4
$$
$$
\rho _{de} = 0.80355 \times 10^{ - 47} GeV^4
$$
Observation value is $$\rho _{obs} \approx 10^{ - 47} GeV^4 $$
If R=90Gyr, $$\rho _{de} = 1.808 \times 10^{ - 47} GeV^4 $$(refer to fig11).
$$
\rho _{de} \approx \rho _{obs}
$$
[Proof end]
In 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].
Dark energy density :
$$
\rho _{de} = 2.09 \times 10^{ - 47} [_{ - 0.465}^{ + 0.557} ]GeV^4
$$
Ridius of the Universe :
$$
R_{UNI} = 96.76[_{ - 11.44}^{ + 12.13} ]Gly = 85.32 \sim 108.89Gly
$$
=========
Hypothesis of Dark Matter and Dark Energy with Negative Mass :
http://vixra.org/abs/0907.0015
fig11. Relations between radius of universe and dark energy density
* mass density of ordinary matter = 1 proton/5m[sup]3[/sup]
* Proton mass= 1.67264 X 10[sup]-27[/sup]kg
* G =6.6726 X 10[sup]-11[/sup] m[sup]3[/sup]/s[sup]2[/sup]kg
* 1J = 6.242 x 10[sup]18[/sup] eV
* $$n_-=n_+=n, m_-=(23.3/4.6)m_+=(5.06522)m_p$$
* $$\bar r_{ - + } = \frac{R}{{3.27273}}$$
* If $$\bar r_{ - + } = \frac{R}{{2.17879}}$$ $$(U_T \approx 0)$$, dark energy density has a 1/3 smaller than $$\bar r_{ - + } = \frac{R}{{3.27273}}$$.
In a WMAP, observed value $$\Lambda = 2.14 \pm 0.13 \times 10^{ -3} eV $$
Dark energy density :
$$
\rho _{de} = 2.09 \times 10^{ - 47} [_{ - 0.465}^{ + 0.557} ]GeV^4
$$
Ridius of the Universe :
$$
R_{UNI} = 96.76[_{ - 11.44}^{ + 12.13} ]Gly = 85.32 \sim 108.89Gly
$$
( If $$\bar r_{ - + } = \frac{R}{{2.17879}}$$,
$$
R_{UNI} = 118.8[_{ - 14.0}^{ + 14.9} ]Gly = 104.8 \sim 133.7 Gly
$$)
From Neil J. Cornish,
the universe' radius is at least 24Gpc(78Gly).
(2003, Neil J. Cornish, "Constraining the Topology of the Universe", http://arxiv.org/abs/astro-ph/0310233v1 )
[Proof]
In negative mass hypothesis, dark energy is corresponding to that positive potential term in total potential energy.
*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_- , and the number of positive mass is n_+ , total potential energy is given as follows.
$$
U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})}
$$
$$
+\sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})}
+ \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})}---(78)
$$
$$
U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) $$
$$
+ (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79)
$$
In equation (79)
$$
E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})
$$
If radius of the universe is 60Gyr, ordinary matter density is about proton 1ea/5m[sup]3[/sup]. So, m[sub]+[/sub] = m[sub]p[/sub],
$$
m_ - = km_ + \simeq (\frac{{23.3}}{{4.6}})m_ + = (5.06522)m_p$$
(because that dark matter has about (23.3/4.6) times ordinary matter in WMAP)
From equation (95)
$$
\bar r_{ - +} =(60Gyr/3.27273)= 1.73447 X 10^{26}m
$$
From analysis of V-5,
If $$U_T \ge 0$$, $$n_ - \approx n_ + $$, Therefore, Define, $$n_ - = n_ + = n $$
$$
V = \frac{{4\pi R^3 }}{3} = \frac{{4\pi \times (5.67648 \times 10^{26} )^3 }}{3} = 7.66171 \times 10^{80} m^3
$$
$$
n = \frac{{\rho V}}{{m_p }} = \frac{{(1m_p /5m^3 )V}}{{m_p }} = 1.53234 \times 10^{80}
$$
( 10[sup]80[/sup] is about total proton number of our universe).
$$
U_{de} = (kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }})
$$
$$
U_{de} = (5.06522)n^2 \frac{{(6.6726 \times 10^{ - 11} )(2.79772 \times 10^{ - 54} )}}{{1.73447 \times 10^{26} }}J
$$
$$
U_{de} = (n^2 ) \times 5.45168 \times 10^{ - 90} J = 1.28009 \times 10^{71} J
$$
1J = 1kg(m/s)[sup]2[/sup] = 6.242 X 10[sup]18[/sup] eV
$$
U_{de} = 7.99031 \times 10^{89} eV
$$
$$
\rho _{de} = \frac{{U_{de} }}{V} = \frac{{7.99031 \times 10^{89} eV}}{{7.66171 \times 10^{80} m^3 }} = \frac{{1.04289 \times 10^{ - 6} GeV}}{{cm^3 }}
$$
Planck Unit transformation(1cm =0.5063 x 10[sup]14[/sup]GeV[sup]-1[/sup] )
$$
\rho _{de} = \frac{{1.04289 \times 10^{ - 6} GeV}}{{1.29784 \times 10^{41} GeV^{ - 3} }} = 0.80355 \times 10^{ - 47} GeV^4
$$
$$
\rho _{de} = 0.80355 \times 10^{ - 47} GeV^4
$$
Observation value is $$\rho _{obs} \approx 10^{ - 47} GeV^4 $$
If R=90Gyr, $$\rho _{de} = 1.808 \times 10^{ - 47} GeV^4 $$(refer to fig11).
$$
\rho _{de} \approx \rho _{obs}
$$
[Proof end]
In 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].
Dark energy density :
$$
\rho _{de} = 2.09 \times 10^{ - 47} [_{ - 0.465}^{ + 0.557} ]GeV^4
$$
Ridius of the Universe :
$$
R_{UNI} = 96.76[_{ - 11.44}^{ + 12.13} ]Gly = 85.32 \sim 108.89Gly
$$
=========
Hypothesis of Dark Matter and Dark Energy with Negative Mass :
http://vixra.org/abs/0907.0015