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_+ Please Register or Log in to view the hidden image!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
- Negative mass has observed, but it was trashed away - From the observance of the HSS team(The High-z Supernova Search team) in 1998, they gained the mass density of the negative(\(\Omega _M = -0.38( \pm 0.22)\)), using field equations which do not have the cosmological constant. \( \Omega _M = -0.38( \pm 0.22) \) HSS team : http://arxiv.org/abs/astro-ph/9805201 :14P 26, 29 lines - Negative mass is stable at the maximum point!(Refer to II-5) - As examined in the question of Harmonic oscillation, in case of positive mass, a point of minimum value which energy is the lowest is stable. However, in case of negative mass(minus mass), stable equilibrium is a point of maximum value, not a point of minimum value. So negative mass is toward a point of maximum value to be stable. In the world of positive mass, ground state is a point that energy is low, but in case of negative mass, ground state is a point that energy is the highest. Accordingly, in the world of negative mass, energy level is filled from the highest to the lowest, and stable state means the highest energy state, so the catastrophe to energy level of minus infinity never happens even if negative mass spontaneously emits energy. ---Icarus2
Property of negative mass -1 1.Extended Newton's motion law When an object with mass of +m1 is away from an object with mass of +m2 by distance r, the force worked between two objects is described as following type. \( \vec F = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r \) When an object with mass of -m[sub]1[/sub] is away from an object with mass of -m[sub]2[/sub] by distance r, what type does the force worked between two objects have? Many people think “the force decides motion of object in gravity”, but actually the acceleration decides the motion of object in gravity, and also plays a role of deciding the moving direction of object which is at the state of stop. We can set up following dynamic equation to describe the motion of object. That is, the gravity created on mass m[sub]1[/sub] by mass m[sub]2[/sub] is expressed as follows: \( \vec F_1 = m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r \) \( m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r \) \( \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r \) As we can see in the equation above, the term of acceleration remains only because mass m[sub]1[/sub] is erased from both terms. Now the equation of motion means the equation of acceleration, not the equation of force, the acceleration provides information of motion direction, and decides the direction of motion.
The law of motion of positive mass and positive mass 1-1)The law of motion of positive mass and positive mass Please Register or Log in to view the hidden image! ----- r ----- Please Register or Log in to view the hidden image! +m1 ---------- +m2 caption : Positive mass +m[sub]1[/sub] and positive mass +m[sub]2[/sub] (initial velocity =0, m_1 > 0, m_2 > 0) \( m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r \) \( \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r \) \( m_2 \vec a_2 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r \) \( \vec a_2 = - G\frac{{m_1 }}{{r^2 }}\hat r \) Positive mass and positive mass : The force worked between positive mass is attraction, and two objects move toward the center of mass. The force is attraction, thus their potential energy has negative value. The direction of acceleration is in the direction of \(-\hat r\), so the distance between two objects are reduced gradually.
1-2)The law of motion of negative mass and positive mass 1-2)The law of motion of negative mass and positive mass Please Register or Log in to view the hidden image! ----- r ----- Please Register or Log in to view the hidden image! - m1 ---------- +m2 caption : Negative mass -m1 and positive mass +m2 (initial velocity = 0, m1 > 0, m2 > 0) \( - m_1 \vec a_1 = - G\frac{{( - m_1 )m_2 }}{{r^2 }}\hat r \) \( \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r \) \( + m_2 \vec a_2 = - G\frac{{( - m_1 )m_2 }}{{r^2 }}\hat r \) \( \vec a_2 = G\frac{{m_1 }}{{r^2 }}\hat r \) Negative mass and positive mass : Negative mass is accelerated in the direction of positive mass, and positive mass is accelerated in the direction to be far away from negative mass. The direction of acceleration a1 worked on negative mass –m1 is - r, so -m1 moves in the direction of reducing distance r, and the direction of acceleration a2 worked on positive mass +m2 is + r, so positive mass +m2 is accelerated in the direction that distance r increases, namely the direction of being far away from negative mass. If the absolute value of positive mass is bigger than that of negative mass, they will meet within finite time(attractive effect), and if the absolute value of positive mass is smaller than that of negative mass, the distance between them will be bigger, and they cannot meet(repulsive effect). ==>Uniformly distributed negative mass receives attractive effect from massive positive mass(Galaxy and Galaxy cluster), so dark matter which has negative mass is clustered around galaxy because of attraction of galaxy. The type of force is repulsion, so the potential energy has positive value. ---Icarus2 Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
The law of motion of negative mass and negative mass 1-3)The law of motion of negative mass and negative mass Please Register or Log in to view the hidden image! ----- r ----- Please Register or Log in to view the hidden image! - m1 -------- - m2 caption : Negative mass - m1 and negative mass - m2 (initial velocity =0, m1 > 0, m2 > 0) \( - m_1 \vec a_1 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r \) \( \vec a_1 = + G\frac{{m_2 }}{{r^2 }}\hat r \) \( - m_2 \vec a_2 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r \) \( \vec a_2 = + G\frac{{m_1 }}{{r^2 }}\hat r \) Negative mass and negative mass: Both two objects are accelerated in the direction of +r which extends distance r, so as time passes, the distance between them is greater than initially given condition, and the force between them is attraction, but the effect is repulsive. The force is attraction(-Gm1m2/r^2), thus the potential energy between them has negative value. If negative mass and positive mass were born together at the beginning of universe, positive mass has attractive effect each other, so it forms star and galaxy structure now, but negative mass has repulsive effect each other, so they cannot make massive mass structure like star or galaxy. ----------- Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
Negative mass is stable at the maximum point! 3. Negative mass is stable at the maximum point! Nature prefers stable state, and has the tendency to go to stable state. Additionally, this can be expressed in another way that nature prefers low energy state, and has the tendency to go to low energy state. Such an idea is frequently used as a logic which denies the existence of negative mass. That is, if there is negative mass and negative energy level, negative mass spontaneously emits energy to be stable, and goes to energy state of minus infinity, so finally it is confronted by catastrophe. Is it right? In case of positive mass, stable state means low energy state, therefore it is not necessary to divide which one nature prefers among two states(stable state and low energy state). By the way, does stable state mean low energy state also in case of negative mass? We can get an answer, if we examine Harmonic oscillation. Please Register or Log in to view the hidden image! fig04 caption : When there is negative mass in potential which has a point of maximum value and a point of minimum value. We begin by considering the oscillatory motion of a particle that is constrained to move in one dimension. We assume that there exists a position of unstable equilibrium for the particle and we designate this point as the origin. Restoring force is in general some complicated function of the displacement and perhaps of the particle's velocity or even of some higher time derivative of the position coordinate. We consider here only case in which the restoring force F is a function only of the displacement F(x) can be expanded in a Taylor series, \(F(x) = F(0) + \frac{x}{{1!}}F^{'} (0) + \frac{{x^2 }}{{2!}}F^{''} (0) + \frac{{x^3 }}{{3!}}F^{'''} (0) + \cdots\) \( + \frac{{x^n }}{{n!}}F^{(n)} (0) + \cdots \\ \) Since the origin is defined to be the equilibrium point, F(0) must vanish, Then, if we confine our attention to displacements of the particle that are sufficiently small, we can neglect all terms involving x[sup]2[/sup] and higher powers of x. We have, therefore, the approximate relation F(x)=+kx The force is alaways the opposite directed toward the unstable equilibrium position(the origin), the derivative F[sup]'[/sup](0) is positive and therefore k is a positive constant. \( - m\ddot x = + kx \) \( \ddot x + \omega _0^2 x = 0 \) \( (\omega _0^2 = \frac{k}{m}) \) Please Register or Log in to view the hidden image! This form of differential equation is the same as that of particle which has positive mass. But we have to notice that positive mass carries out harmonic oscillation on a point of minimum value, whereas negative mass carries out harmonic oscillation on a point of maximum value. Additionally, restoring force is +kx at this time. \( \vec F = - \nabla U \) \( U = - \frac{1}{2}kx^2 \) \( {E_ -} = T + U = - \frac{1}{2}m\dot x^2 - \frac{1}{2}kx^2 \\ = - \frac{1}{2}m\omega _0^2 A^2 \\ \) In phase space \( \frac{{x^2 }}{{(\frac{{ - 2E_ - }}{k})}} + \frac{{p^2 }}{{( - 2mE_ - )}} = 1 \) This equation is ellipses equation, because total energy E[sub]-[/sub] < 0 As examined in the question of Harmonic oscillation, in case of positive mass, a point of minimum value which energy is the lowest is stable. However, in case of negative mass, stable equilibrium is a point of maximum value, not a point of minimum value. So negative mass is toward a point of maximum value to be stable, not a point of minimum value which energy is low. In the world of positive mass, ground state is a point that energy is low, but in case of negative mass, ground state is a point that energy is the highest. Accordingly, in the world of negative mass, energy level is filled from the highest to the lowest, and stable state means the highest energy state, so the catastrophe to energy level of minus infinity never happens even if negative mass spontaneously emits energy. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
Icarus; I have a pet theory that dark energy is simply gravity that has travelled around a 'wraparound universe' (4D topology), therefore acting as a force of repulsion. Does this kind of imagery fit with your theory?
I’m sorry. I can’t English well. Though, I don’t able to see your all idea. But, your thought is not same to my negative mass hypothesis. Your theory need to the extra dimension (for wrap). So, total dimensions are over 5D. Also, wrap around universe --> therefore acting as a force of repulsion. How acting repulsion? ======== For my negative mass hypothesis : Negative mass is real particle at 4D(x,y,z,t). I have not inserts anything for negative mass. Negative mass has obey to the preexistence physics laws(Newton’s law, energy conservation law, momentum conservation law, relativistic energy eq., 4Dimension, also observation results, …)
The Shape of the Universe. The top part of the diagram on the right i.e. the sphere. If a graviton particle (which I believe can be represented by a spinning helix) wraps around the universal sphere then it will appear in the opposite direction and have an attractive force away from the originating mass i.e. indistinguishable from a force of repulsion. Look at the top motion diagram of the archimedes screw in Wikipedia Archimedes screw. The red ball represents the direction of force. Imagine that the screw wraps around the universe. The red ball is now moving AWAY from the initial object mass which radiated the graviton i.e. now a force of repulsion.
I'm sorry, I can't English well So, My answer is short. Read to the classical dynamics(the gravitational potential of a spherical shell) or shell theorem. In the shell theorem (http://en.wikipedia.org/wiki/Shell_theorem) Thick shells part ================ Now consider a spherically symmetric shell of finite thickness, with inner radius R_a and outer radius R_b. The behavior entirely inside or outside the shell is no different than for a thin shell, but what is the force felt by an observer somewhere within the shell (i.e. R_a < r < R_b)? inside the inner shell ( r ≤ R_a), F_r =0 ================ It is means that gravitational force is zero inside the inner spherical shell. Therefore, the outer mass distribution of your universe has not role the repulsive force.
Negative mass cannot form the structure greater than atom! 2. Negative mass cannot form the structure greater than atom As examined the equation of motion for negative mass, it is marked in form of F= - ma (m>0), when attraction is applied together with nuclear force(when usually nuclear force is attraction, but has the form of repulsive core, and assuming nuclear force has the form of \(\vec F=-Q(r)\hat r\), Q(r) is the positive function of distance r, thus nuclear force is in the form of attraction worked in the direction of -\(\hat r \). Here, for the force worked on negative mass –m, \( \vec F=-m\vec a=-Q(r)\hat r \) \( \vec a=\frac{{Q(r)}}{m}\hat r \) The term of acceleration is positive, so the effect of increasing distance r, namely repulsive effect appears. This means that negative mass cannot form the structure like atom(massive nucleon, baryon, particle consists of multi elementary particle), because nuclear force has not binding negative mass when it is applied to negative mass. Also, gravity has not binding negative mass(repulsive) Additionally, for the problem of mesons that mediates nuclear force or weak interaction, if there is no meson that delivers strong interaction or weak interaction, it is doubtful if strong interaction or weak interaction can be worked or not. For example, nucleon must have internal structure including meson or quark, but in case of negative mass, nuclear force is repulsive, so it cannot have the internal structure of nucleon from the beginning. That is, there is a great possibility that negative mass cannot include meson or quark which has negative mass in nucleus. The fact that it cannot make nucleon means that it is impossible to form massive mass structure like a star in addition to atomic structure. This provides proper explanation of the fact that negative mass is not seen as it has visible massive mass structure. Also generally it satisfies the nonbaryonic matters required for dark matter. If negative mass was born at the beginning of universe, there is higher possibility that it exists until now as a certain basic state born at the beginning of universe, and that it does not have strong interaction like nuclear force, weak interaction, and electromagnetic interaction(neutral or has not internal structure(positive mass elementary particle has a ±1/2, ±1/3, 2/3 charge)). This point is keeping with current characteristics required for dark matter. If negative mass and positive mass were born together at the beginning of universe, positive mass has attractive effect each other, so it forms star and galaxy structure now, but negative mass has repulsive effect each other and nuclear force cannot form nucleons by binding negative mass, so they cannot make massive mass structure like star or galaxy. If dark matter is negative mass, non-observation of dark matter star and galaxy can be explained. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
More strictly Calculation [ More strictly Calculation ] Proof of that value of observed dark energy(10^(-47)GeV^4) 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 = (0.134 ~ 3.34) 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. ============== U_+ Please Register or Log in to view the hidden image!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_- , and the number of positive mass is n_+ , total potential energy is given as follows. \( U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})} \) \( +\sum\limits_{i,j,i > j}^{i,j = n_ - } {(\frac{{ - Gm_{ - i} m_{ - j} }}{{r_{ - - ij} }})} + \sum\limits_{i,j,i > j}^{i,j = n_ + } {(\frac{{ - Gm_{ + i} m_{ + j} }}{{r_{ + + ij} }})}---(78) \) \( U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) \) \( + (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79) \) V-2-1)Proof from the definition equation of dark energy [ Proof Start! ] In equation (79), Dark energy is corresponding to that plus potential term in total potential energy. \( E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) ---(80) \) If radius of the universe is 60Gyr, ordinary matter density is about proton 1ea/\(5m^3\). So, \(m_ + = m_p \), \( m_ - = km_ + \simeq (\frac{{23.3}}{{4.6}})m_ + = (5.06522)m_p \) (because that dark matter has about (23.3/4.6) times ordinary matter in WMAP) From equation (95) \( \bar r_{ - + } = \frac{R}{{2.17879}} ---(95) \) \( \bar r_{ - +} =(60Gyr/2.17879)=2.60533 X 10^{26}m \) From analysis of V-5, If \(U_T \ge 0\), \(n_ - \approx n_ + \), Therefore, Define, \(n_ - = n_ + = n \) \( V = \frac{{4\pi R^3 }}{3} = \frac{{4\pi \times (5.67648 \times 10^{26} )^3 }}{3} = 7.66171 \times 10^{80} m^3 \) \( n = \frac{{\rho V}}{{m_p }} = \frac{{(1m_p /5m^3 )V}}{{m_p }} = 1.53234 \times 10^{80} \) ( \(10^{80}\) is about total proton number of our universe). \( U_{de} = (kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) \) \( U_{de} = (5.06522)n^2 \frac{{(6.6726 \times 10^{ - 11} )(2.79772 \times 10^{ - 54} )}}{{2.60533 \times 10^{26} }}J \) \( U_{de} = (n^2 ) \times 3.62940 \times 10^{ - 90} J = 8.52207 \times 10^{70} J \) \( 1J = 1kg(m/s)^2 = 6.25 \times 10^{18} eV \) \( U_{de} = 5.31948 \times 10^{89} eV \) \( \rho _{de} = \frac{{U_{de} }}{V} = \frac{{5.31948 \times 10^{89} eV}}{{7.66171 \times 10^{80} m^3 }} = \frac{{6.94294 \times 10^{ - 7} GeV}}{{cm^3 }} \) Planck Unit transformation(1cm =0.5063 x\(10^{14}GeV^{-1}\) ) \( \rho _{de} = \frac{{6.94294 \times 10^{ - 7} GeV}}{{1.29784 \times 10^{41} GeV^{ - 3} }} = 5.34961 \times 10^{ - 48} GeV^4 \) \( \rho _{de} = 5.34961 \times 10^{ - 48} GeV^4 \) Observation value is \(\rho _{obs} \approx 10^{ - 47} GeV^4 \) If R=90Gyr, \(\rho _{de} = 1.203 \times 10^{ - 47} GeV^4 \)(refer to fig11). \( \rho _{de} \approx \rho _{obs} \) [Proof End] In Quantum Field Theory, the energy density of the vacuum is estimated as \(10^{70}GeV^4\), which is about \(10^{117}\) orders of magnitude large than the observation value \(10^{-47}GeV^4\). Therefore, You can see that negative mass hypothesis how to close to the observation and the universe. \( E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) \) From equation (80), 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 suggested that negative mass is exist. \(U_{de}\) is right. It means that \(U_T\) is also right. Thus, from the analysis of \(U_T\), 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 =(positive potential term) + (negative potential term)= Λ + (ordinary gravitation potential) Which the positive term played an independent potential role as Λ. ========= Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 ---Icarus2
Abstract! [ Abstract ] From the observance of the HSS team in 1998, they gained the mass density of the negative( Ω[sub]M[/sub] = -0.38( ± 0.22)), using field equations which do not have the cosmological constant. The quantity of the mass couldn't be negative value in they thought, the value is trashed away. We have to know that not the field equation has disposed the value, but our thought disposed that value. In the world of positive mass, ground state is a point that energy is low, but in case of negative mass, ground state is a point that energy is the highest. Accordingly, in the world of negative mass, energy level is filled from the highest to the lowest, and stable state means the highest energy state, so the catastrophe to energy level of minus infinity never happens even if negative mass spontaneously emits energy. Assuming that negative mass exists, Newton's Law of motion was derived in between negative and positive masses and also between negative and negative masses. As a method for proving the existence of negative mass, an explanation on the revolution velocity of the galaxy through negative mass has been presented. In this process, the existence of spherical mass distribution was given; furthermore, explanation was done using this, to show observation results where dark matter effect through negative mass is proportional to distance r. If Ω[sub]M[/sub] is -0.38, universe's age is 14.225 Gyr. It is in the range estimated by other observations. Assuming that negative mass and positive mass were born together at the beginning of universe, it satisfies the various problems that previous dark matter and dark energy possess, such as, [ Dark matter ] - Centripetal force effects of galaxy and galaxy clusters from previous dark matters, - Mass effects that is proportional to the distance r, - Low interaction between dark matter when collision occurs between dark matter. [ Dark energy ] : Repulsive force needed for expansion, dark energy that has positive values, [ Fine tuning problem of mass density ] : The reason of that mass density close to the critical mass density. [ Cosmological Constant Problem ] - The reason of that dark energy seems to has a small and non-zero value. - Phase transition problem of dark energy [ Others ] - Collision of Bullet cluster, - Deceleration expansion and acceleration expansion of universe, - Age of the universe with negative mass density - Size of the universe [ Proof of that observed dark energy value ] : Dark energy observation value (\(10^{-47}GeV^{4}\)) As a result, the necessity of observation focusing on exact computation and detection of negative mass is stated. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
Fine tuning problem and inflation mechanism Fine tuning problem and inflation mechanism Pair creation of negative mass and positive mass seems to provide the proper explanation of homogeneity, local non-homogeneity, and flatness of our universe without separate assumption with basic characteristics of Newton dynamics. With reference to the flatness of universe, the calculation of cosmic critical density with previous positive mass leads to the problem that present density should be close to critical density, the cosmic density at the beginning should be close to critical density, and initial condition should be set delicately. However, according to the hypothesis of negative mass, potential energy of universe is not single potential when there is only positive mass(potential that has both + and -), and the density of universe close to critical density was from basic mechanism of pair creation, namely 1:1 correspondence of negative mass to positive mass, and form of potential energy. 5-2) Inflation mechanism - Inflation start - If we consider that inflation started at the point of time that there were large-scale pair creations of negative mass and positive mass after Big Bang, not at the point of time that the gravity was separated, the number of positive mass and negative mass might be born together at this time, so here also the term of positive potential energy by equation (72)(nU_+(if pair creations are \(10^{80}\), \(U_T = 10^{80} U_+\) survives), and this can provide the start power of inflation. - Inflation finish - If positive mass is converted to radiant energy(pair annihilation of matter and antimatter, radiation) or energy when strong interaction(Some mass changed to the bond energy. Therefore, positive mass defect has come.), weak interaction, and electromagnetic force are separated after the start of inflation, the number of positive mass falls down below critical ratio which total potential energy is 0 in (V-1.), so at this time, inflation also is naturally finished. Reference-1 =========== Potential energy when there are generally n pairs of negative mass and positive mass : \( U_n = \sum\limits_{i = 1}^{n^2 } {U_{ + i} } + \sum\limits_{j = 1}^{n(n - 1)} {U_{ - j} } = nU_+ ---(72) \) If n pairs of negative mass and positive mass are created, generally the number of term of positive potential energy is n^2, the number of term of negative potential energy is n(n-1), so the number of term of positive potential energy is greater than that of term of negative potential energy by n. Accordingly, even if the absolute value of negative mass is the same as that of positive mass at the beginning of universe, the universe has the value of positive potential energy, and expands. =========== Reference-2 =========== Calculate the value of \(U_- = - U_ + \), \(n_ - = 10,n_ + = 1 - 10\) with equation (79 ) when there is a difference in the number between negative mass and positive mass to examine changes, We can see the change in total potential energy in accordance with the difference in the number of negative mass and positive mass from 10 samples above. Please Register or Log in to view the hidden image! fig09 caption : Potential energy from ratio of negative mass and positive mass - The tendency of total potential energy in accordance with the number of negative mass and positive mass - i)In Critical ratio of the number of negative mass to the number of positive mass, the total potential energy has 0. --> When the universe is flat. (ex. \((n_ - ,n_ + )=(10,6))\) Here, we note that total potential energy does not have 0 when general matter comes under 60% of dark matter. 60% is the proportion by assuming that all terms of potential energy are identical, and prescribing that the number of \(n_-\) is 10. ii)If number of positive mass less than critical number, total potential energy has negative value. --> Cosmic decelerating expansion iii)If number of positive mass over than criticle number, total potential energy has positive value. --> Cosmic accelerating expansion iv)The value of total potential energy increases as the number of positive mass approaches to the number of negative mass. v)If the number of pair of negative mass and positive mass is n, n terms of positive potential remains as shown in equation (72)(\(U_T = nU_+\)). --> Final state of universe. =========== 5-3) Fine tuning problem of mass density and cosmological constant Universe mass density is same critical mass density, that is correspond with total potential energy=0 \( U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) \) \(+ (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }})+ \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79 ) \) I can't seek exactly solution for total potential energy equation. So I consider that these three potential can be the same U, and I will seek to the whole tendency. Maybe, consider of uniformly distribution, it is near to stable state when three value of potential energy are almost in same level. Define, \(x = n_ - ,y = n_ + \), if \( U_{ - + } = - U_{ - - } = - U_{ + + } = U\) Equation(79) is \( U_T = \frac{{2xy - x^2 + x - y^2 + y}}{2}U \) If x=y=n, \( U_{\max } = nU \) If, \(x \to \infty ,y \to \infty ;n_{h}(t) = \frac{y}{x}\) = \(\frac{{n_ + }}{{n_ - }}\) \( U_T = - \frac{{x^2 }}{2}[\{ 1 + (\frac{y}{x}){}^2 - \{ \frac{{2y}}{x} + \frac{1}{x} + \frac{y}{{x^2 }}\} ]U \) \( U_T \simeq - \frac{{x^2 }}{2}[\{ 1 + (\frac{y}{x}){}^2 - \{ \frac{{2y}}{x} + 0 + 0\} ]U \) \( U_T = - \frac{{x^2 }}{2}(1 - \frac{y}{x})^2 U \) \( U_T = - \frac{{n_ - ^2 }}{2}(1 - n_h (t))^2 U ---(105 ) \) y \( \ne x \), at most the section, Total potential energy is proportion square term of negative mass number(similar positive mass number) Divide \(U_{max}\) by equation(105 ), Ratio of potential size \( \frac{{U_{\max } }}{{|U_T| }} = \frac{{n_ - }}{{\frac{{n_ - ^2 }}{2}(1 - n_h (t) )^2 }} = \frac{2}{{n_ - ^{} (1 - n_h (t) )^2 }} \) - Cosmological constant problem - The reason of that cosmological constant has a small and non-zero value. \(n_ - \) is total number of negative mass. If \(n_ -\) is \(10^{80}\), we can know how the present potential values are smaller than the absolute value of minimum potential values of universe, how this present potential value is close to zero. It is means that present dark energy(cosmological constant) has a small positive value. Because that dark energy is gravitational potential energy(eq.(80)) of universe and then total potential energy has positive value at now. Reference-3 ========= \( E_{de}(t)=U_{de}(t) = (n_- (t) \times n_+ (t) )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + }(t) }}) ---(80) \) ========= Other process, total potential energy term by 2n general particles, has a \(\frac{{2n(2n - 1)}}{2}\)U. The other side, U_max(positive maximum potential energy by negative mass and positive mass) has a nU. So, if n = 10^80 \( \frac{{U_{\max } }}{{U_{GP} }} = \frac{{nU}}{{n(2n - 1)U}} = \frac{1}{{(2n - 1)}} \approx \frac{1}{{2 \times 10^{80} }} \) It is correspond with the ratio of flatness. \(U_{max}\) is that the upper limit of total potential energy. Please Register or Log in to view the hidden image! fig16 caption : Changes in universe's potential energy over time - Fine tuning problem of mass density - The reason of that mass density of universe close to critical mass density. About the fine tuning of mass density in the early universe, even though the mass density of present universe is equally same with the value of critical mass density, when negative mass and positive mass coexist, it doesn't mean that the density of early universe must be very close with the value of critical density. And also because negative and positive mass's rate is going close to 1, that present universe is almost at the same with critical density values. The mass density of universe close to critical mass density was from basic mechanism of pair creation, namely 1:1 correspondence of negative mass to positive mass, and form of potential energy. In fig.16, negative and positive mass created n pairs, and in this period that potential energy values in \(U_{\max } = nU_+ \). That cause accelerates expansion of universe(Bigbang or Inflation). Strong interaction, weak interaction and electromagnetic force are separate(some mass change to the bond energy) and also antimatter and matter generate pair annihilation in during inflation. So most of positive matter became radiation. Then total potential energy become negative value(result of V-1, fig.9). Therefore we call this time \(t_{if}\)(inflation finish time) As the universe is getting cooler, the radiation is getting lower and that radiation changed the matter with positive mass. And after values of positive mass became the critical ratio of negative mass, that potential energy becomes zero. We call this time by \(t_h\). According to hypothesis with negative mass, The last state of universe is when positive mass and negative mass almost the same and potential energy has the value of \(U_T = n(t)U(t)\) at this time. - Phase transition problem of dark energy - The reason of that dark energy have a very big positive value(Inflation Energy) in early universe and very small positive value(Cosmological Constant) at now. This value is strong power that can generate inflation in early universe. But roll of this value is very smaller than inflation period in the growth universe(mean distance r is very bigger than initial mean distance \(r_0\) and because of our universe have positive potential energy, so it's doing expansion now. If \(r_0 = l_p\) = 1.61624 X \(10^{-35}m\), \(r_{now}\) = \(ct_{0}/2 \sim ct_0 = (6.48065 \sim 12.9613) X 10^{25}\)m \(U_0 = (4.0097 \sim 8.0194) X 10^{60} U_{now}\), also \(\rho _{0} >> \rho _{now} \) \(\rho _{now} = 10^{ - 47} GeV^4 \) \(\rho _{\inf } \approx (10^{60} \sim 10^{243} ) \times \rho _{now} = (10^{13} \sim 10^{196} )GeV^4 \) It is very important. Because that \(U_{de}\) explain dark energy that very big positive value(Inflation Energy) in early universe and very small positive value(Cosmological Constant) at now. ================= Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
Proof from the gravitational self-energy 5-2-2)Proof from the gravitational self-energy In case of, mass M and mass distribution is \( 0 \le r \le R\) and mass density is \(\rho \) Please Register or Log in to view the hidden image! Fig-Gravitational self-energy Gravitational self-energy of the universe \(dU_s = - \frac{{GM'dm}}{r}\) \(M' = \frac{4}{3}\pi r^3 \rho $, $dm = \rho r^2 dr\sin \theta d\theta d\phi \) \(dU_s = - G\frac{{4\pi \rho ^2 }}{3}r^4 dr\sin \theta d\theta d\phi \) \(U_s = - G\frac{{4\pi \rho ^2 }}{3}\int\limits_0^R {r^4 d} r\int\limits_0^\pi {\sin \theta d\theta } \int\limits_0^{2\pi } {d\phi } \) \(U_s = - G\frac{{(4\pi \rho )^2 }}{3}(\frac{1}{5}R^5 )\) \( U_S = - \frac{3}{5}\frac{{GM^2 }}{R} \) 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 n[sub]-[/sub], and the number of positive mass is n[sub]+[/sub]. The other side, U[sub]s[/sub] is total potential energy when all particles are positive mass. Therefore, U[sub]s[/sub] is total potential energy when dark energy term has an opposite sign. General gravitational potential defined, \( U_{gp} = - ((\frac{{n_ - (n_ - - 1)}}{2}\frac{{Gm_ - m{}_ - }}{{r_{ - - } }}) + (\frac{{n_ + (n_ + - 1)}}{2}\frac{{Gm_ + m{}_ + }}{{r_{ + + } }})) \) Therefore, \( U_T = U_{de} + U_{gp} \) \( U_S = - U_{de} + U_{gp} \) From WMAP \( a = \frac{{|U_{gp} |}}{{U_{de} }} = (\frac{{23.9}}{{72.1}}) = 0.33148 \) \(U_{gp} = - aU_{de} \), \(U_S = - (a + 1)U_{de} \) Substitution U[sub]gp[/sub] in equation (89) Finally, \( U_{de} = - \frac{{U_S }}{{(a + 1)}} = (k_c)(\frac{{GM^2 }}{{R}}) \) i) In case of \(n_- = n_+ =n\),\( m_- = km_ + (k \ge 1)\) Total mass \(M = (n_ - \times m_ - ) + (n_ + \times m_ + ) = (k + 1)nm_p \) \( U_{de} = - \frac{{U_S }}{{(a + 1)}} = \frac{3}{{5(a + 1)}}\frac{{G((k + 1)nm_p )^2 }}{R} \) \( U_{de} = \frac{{3(k + 1)^2 n^2 }}{{5(a + 1)}}\frac{{Gm_p^2 }}{R} = (0.45062)(k + 1)^2 n^2 \frac{{Gm_p^2 }}{R} \) Equation (81) = Equation (93) \((kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ + - } }}) = \frac{{3(k + 1)^2 n^2 }}{{5(a + 1)}}\frac{{Gm_p^2 }}{R}\) \( \bar r_{ - + } = \frac{{5(a + 1)k}}{{3(k + 1)^2 }}R \) If \(k = \frac{{(dark_matter)}}{{(ordinary_matter)}} \approx \frac{{23.3}}{{4.6}} = (5.06522)\) (In WMAP) \( \bar r_{ - + } = \frac{R}{{3.27273}} \) 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. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015
