Hypothesis of Dark Matter and Dark Energy with Negative Mass :
V-2-2)Proof from the gravitational self-energy
In case of, mass M and mass distribution is [tex] 0 \le r \le R[/tex]
and mass density is [tex]\rho [/tex]
In case of, mass M and mass distribution is [tex] 0 \le r \le R[/tex]
and mass density is [tex]\rho [/tex]
fig10
caption : Gravitational self-energy
Gravitational self-energy of the universe
[tex]
U_S = - \frac{3}{5}\frac{{GM^2 }}{R}
[/tex]
A coefficient 3/5 is constant for geometric shape of the universe.
At this time, let's analyze to the relation between total potential energy and gravitational self-energy.
Equation (79) is total potential energy when the number of negative mass is [tex]n_-[/tex], and the number of positive mass is [tex]n_+[/tex].
The other side, [tex]U_s[/tex] is total potential energy when all particles are positive mass. Therefore, [tex]U_s[/tex] is total potential energy when dark energy term has an opposite sign.
General gravitational potential defined,
[tex]
U_{gp} = - ((\frac{{n_ - (n_ - - 1)}}{2}\frac{{Gm_ - m{}_ - }}{{r_{ - - } }}) + (\frac{{n_ + (n_ + - 1)}}{2}\frac{{Gm_ + m{}_ + }}{{r_{ + + } }}))
[/tex]
Therefore,
[tex]
U_S = - U_{de} + U_{gp} ---(88)
[/tex]
From analysis of V-5,
If [tex]U_T \ge 0[/tex],
[tex]
\frac{{U_{\max } }}{{U_{GeneralParticle} }} = \frac{{nU}}{{n(2n - 1)U}} = \frac{1}{{(2n - 1)}} \approx \frac{1}{{2 \times 10^{80} }}
[/tex]
We know that [tex]U_{max } \ge U_T = U_{de} + U_{gp} [/tex]
[tex]
\frac{{U_T }}{{U_{GeneralParticle} }} = \frac{{U_{de} + U_{gp} }}{{U_S }} = \frac{{U_{de} + U_{gp} }}{{U_{de} + ( - U_{gp} )}} \le \frac{1}{{2 \times 10^{80} }}
[/tex]
Therefore,
[tex]
U_{gp} \approx - U_{de} ---(89)
[/tex]
Substitution equation (89) in equation (88)
[tex]
U_{de} \approx - \frac{{U_S }}{2}
[/tex]
Finally,
[tex]
U_{de} \approx - \frac{1}{2}U_s = \frac{3}{{10}}\frac{{GM^2 }}{R}
[/tex]
(This equation can be applied at [tex]U_T \ge 0[/tex])
i) In case of [tex]n_- =n_+ =n[/tex], [tex]m_ - = km_ + (k \ge 1)[/tex]
Total mass [tex]M = (n_ - \times m_ - ) + (n_ + \times m_ + ) = (k + 1)nm_p [/tex]
[tex]
U_{de} = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R} ---(92)
[/tex]
Equation (81) = Equation (92)
[tex]
(kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R}
[/tex]
[tex]
\bar r_{ - + } = \frac{{10k}}{{3(k + 1)^2 }}R
[/tex]
If [tex] k=\frac{{(dark\_matter)}}{{(ordinary\_matter)}} \approx \frac{{23.3}}{{4.6}} = (5.06522)[/tex]
[tex]
\bar r_{ - + } = \frac{R}{{2.17879}}
[/tex]
Used to the gravitational self-energy, dark energy value of today is explained. Therefore, it means that dark energy is gravitational potential. Also, because that gravitational potential is plus value, it is strongly suggested that negative mass is exist.
V-2-3) Relations between radius of universe and dark energy density
caption : Gravitational self-energy
Gravitational self-energy of the universe
[tex]
U_S = - \frac{3}{5}\frac{{GM^2 }}{R}
[/tex]
A coefficient 3/5 is constant for geometric shape of the universe.
At this time, let's analyze to the relation between total potential energy and gravitational self-energy.
Equation (79) is total potential energy when the number of negative mass is [tex]n_-[/tex], and the number of positive mass is [tex]n_+[/tex].
The other side, [tex]U_s[/tex] is total potential energy when all particles are positive mass. Therefore, [tex]U_s[/tex] is total potential energy when dark energy term has an opposite sign.
General gravitational potential defined,
[tex]
U_{gp} = - ((\frac{{n_ - (n_ - - 1)}}{2}\frac{{Gm_ - m{}_ - }}{{r_{ - - } }}) + (\frac{{n_ + (n_ + - 1)}}{2}\frac{{Gm_ + m{}_ + }}{{r_{ + + } }}))
[/tex]
Therefore,
[tex]
U_S = - U_{de} + U_{gp} ---(88)
[/tex]
From analysis of V-5,
If [tex]U_T \ge 0[/tex],
[tex]
\frac{{U_{\max } }}{{U_{GeneralParticle} }} = \frac{{nU}}{{n(2n - 1)U}} = \frac{1}{{(2n - 1)}} \approx \frac{1}{{2 \times 10^{80} }}
[/tex]
We know that [tex]U_{max } \ge U_T = U_{de} + U_{gp} [/tex]
[tex]
\frac{{U_T }}{{U_{GeneralParticle} }} = \frac{{U_{de} + U_{gp} }}{{U_S }} = \frac{{U_{de} + U_{gp} }}{{U_{de} + ( - U_{gp} )}} \le \frac{1}{{2 \times 10^{80} }}
[/tex]
Therefore,
[tex]
U_{gp} \approx - U_{de} ---(89)
[/tex]
Substitution equation (89) in equation (88)
[tex]
U_{de} \approx - \frac{{U_S }}{2}
[/tex]
Finally,
[tex]
U_{de} \approx - \frac{1}{2}U_s = \frac{3}{{10}}\frac{{GM^2 }}{R}
[/tex]
(This equation can be applied at [tex]U_T \ge 0[/tex])
i) In case of [tex]n_- =n_+ =n[/tex], [tex]m_ - = km_ + (k \ge 1)[/tex]
Total mass [tex]M = (n_ - \times m_ - ) + (n_ + \times m_ + ) = (k + 1)nm_p [/tex]
[tex]
U_{de} = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R} ---(92)
[/tex]
Equation (81) = Equation (92)
[tex]
(kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R}
[/tex]
[tex]
\bar r_{ - + } = \frac{{10k}}{{3(k + 1)^2 }}R
[/tex]
If [tex] k=\frac{{(dark\_matter)}}{{(ordinary\_matter)}} \approx \frac{{23.3}}{{4.6}} = (5.06522)[/tex]
[tex]
\bar r_{ - + } = \frac{R}{{2.17879}}
[/tex]
Used to the gravitational self-energy, dark energy value of today is explained. Therefore, it means that dark energy is gravitational potential. Also, because that gravitational potential is plus value, it is strongly suggested that negative mass is exist.
V-2-3) Relations between radius of universe and dark energy density
fig11
caption : Relations between radius of universe and dark energy density
* mass density of ordinary matter = [tex]1 proton/5m^3[/tex]
* Proton mass= 1.67264X [tex]10^{-27}[/tex]kg
* G =6.6726X [tex]10^{-11} m^3/s^2[/tex]kg
* 1J = 6.242 x [tex]10^{18}[/tex] eV
caption : Relations between radius of universe and dark energy density
* mass density of ordinary matter = [tex]1 proton/5m^3[/tex]
* Proton mass= 1.67264X [tex]10^{-27}[/tex]kg
* G =6.6726X [tex]10^{-11} m^3/s^2[/tex]kg
* 1J = 6.242 x [tex]10^{18}[/tex] eV
========
Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
댓글
댓글 쓰기