icarus2 Posted November 27, 2009 Share Posted November 27, 2009 (edited) Proof of that value of observed dark energy(10-47GeV4) 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. ρde = (0.134 ~ 3.34) X 10-47GeV4 : Theoretical computation ρobs = 10-47GeV4 : Observed value Please have a look the attached proof and inspect the findings. ============== *Potential energy between positive mass and positive mass has - value:[math]U = \frac{{ - G(m_ +) (m_ +) }}{r} = 1U_ - [/math] *Potential energy between negative mass and positive mass has + value:[math]U = \frac{{ - G( - m_ - )(m_ +) }}{r} = 1U_ + [/math] *Potential energy between negative mass and negative mass has - value:[math]U = \frac{{ - G( - m_ - )( - m_ - )}}{r} = 1U_ - [/math] When the number of negative mass is n- , and the number of positive mass is n+ , total potential energy is given as follows. [math] U_T = \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})} [/math] [math] +\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) [/math] [math] U_T = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) [/math] [math] + (\frac{{n_ - (n_ - - 1)}}{2}(\frac{{ - Gm_ - m_ - }}{{\bar r_{ - - } }}) + \frac{{n_ + (n_ + - 1)}}{2}(\frac{{ - Gm_ + m_ + }}{{\bar r_{ + + } }})) ---(79) [/math] 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. [math] E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) ---(80) [/math] If radius of the universe is 60Gyr, ordinary matter density is about proton 1ea/5m3. So, m+ = mp, [math] m_ - = km_ + \simeq (\frac{{23.3}}{{4.6}})m_ + = (5.06522)m_p [/math] (because that dark matter has about (23.3/4.6) times ordinary matter in WMAP) From equation (95) [math] \bar r_{ - + } = \frac{R}{{2.17879}} ---(95) [/math] [math] \bar r_{ - +} =(60Gyr/2.17879)=2.60533 \times 10^{26}m [/math] From analysis of V-5, If UT ≥ 0, n - ~ n+, Therefore define, n - = n+ = n [math] 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 [/math] [math] n = \frac{{\rho V}}{{m_p }} = \frac{{(1m_p /5m^3 )V}}{{m_p }} = 1.53234 \times 10^{80} [/math] ( 1080 is about total proton number of our universe). [math] U_{de} = (kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) ---(81) [/math] [math] U_{de} = (5.06522)n^2 \frac{{(6.6726 \times 10^{ - 11} )(2.79772 \times 10^{ - 54} )}}{{2.60533 \times 10^{26} }}J [/math] [math] U_{de} = (n^2 ) \times 3.62940 \times 10^{ - 90} J = 8.52207 \times 10^{70} J [/math] 1J = 1kg(m/s)2 = 6.25 X 1018 eV Ude = 5.31948 X 1089 eV [math] \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 }} [/math] Planck Unit transformation(1cm =0.5063 x 1014GeV-1 ) [math] \rho _{de} = \frac{{6.94294 \times 10^{ - 7} GeV}}{{1.29784 \times 10^{41} GeV^{ - 3} }} = 5.34961 \times 10^{ - 48} GeV^4 [/math] ρde = 0.534961 X 10-47GeV4 Observation value is [math]\rho _{obs} \approx 10^{ - 47} GeV^4 [/math] If R=90Gyr, ρde = 1.203 X 10 -47GeV4 (refer to fig11). [math] \rho _{de} \approx \rho _{obs} [/math] [Proof End] In Quantum Field Theory, the energy density of the vacuum is estimated as 1070GeV4, which is about 10117 orders of magnitude large than the observation value 10-47GeV4. Therefore, You can see that negative mass hypothesis how to close to the observation and the universe. [math] E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) [/math] 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. Ude is right. It means that UT is also right. Thus, from the analysis of UT, 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. ========= If the proof above is right, please read another section of the article. Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Merged post follows: Consecutive posts mergedV-2-2)Proof from the gravitational self-energy In case of, mass M and mass distribution is 0 ≤ r ≤ R and mass density is ρ. fig10 caption : Gravitational self-energy Gravitational self-energy of the universe [math] U_S = - \frac{3}{5}\frac{{GM^2 }}{R} [/math] 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-, and the number of positive mass is n+. The other side, Us is total potential energy when all particles are positive mass. Therefore, Us is total potential energy when dark energy term has an opposite sign. General gravitational potential defined, [math] U_{gp} = - ((\frac{{n_ - (n_ - - 1)}}{2}\frac{{Gm_ - m{}_ - }}{{r_{ - - } }}) + (\frac{{n_ + (n_ + - 1)}}{2}\frac{{Gm_ + m{}_ + }}{{r_{ + + } }})) [/math] Therefore, US = - Ude + Ugp ---(88) From analysis of V-5, If UT ≥ 0, [math] \frac{{U_{\max } }}{{U_{GeneralParticle} }} = \frac{{nU}}{{n(2n - 1)U}} = \frac{1}{{(2n - 1)}} \approx \frac{1}{{2 \times 10^{80} }} [/math] We know that Umax ≥ UT = Ude + Ugp [math] \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} }} [/math] Therefore, [math] U_{gp} \approx - U_{de} ---(89) [/math] Substitution equation (89) in equation (88 ) [math] U_{de} \approx - \frac{{U_S }}{2} [/math] Finally, [math] U_{de} \approx - \frac{1}{2}U_s = \frac{3}{{10}}\frac{{GM^2 }}{R} ---(91) [/math] (This equation can be applied at UT ≥ 0) i) In case of n- = n+ =n, m - = km+ (k ≥ 1) Total mass [math]M = (n_ - \times m_ - ) + (n_ + \times m_ + ) = (k + 1)nm_p [/math] [math] U_{de} = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R} ---(92) [/math] Equation (81) = Equation (92) [math] (kn^2 )(\frac{{Gm_p^2 }}{{\bar r_{ - + } }}) = \frac{{3(k + 1)^2 n^2 }}{{10}}\frac{{Gm_p^2 }}{R} [/math] [math] \bar r_{ - + } = \frac{{10k}}{{3(k + 1)^2 }}R [/math] If [math] k=\frac{{(dark\_matter)}}{{(ordinary\_matter)}} \approx \frac{{23.3}}{{4.6}} = (5.06522)[/math] [math] \bar r_{ - + } = \frac{R}{{2.17879}}---(95) [/math] From eq(91), [math] U_{de} = \frac{3}{{10}}\frac{{GM^2 }}{R} = \frac{3}{{10}}\frac{{G(5.06522 + 1)^2 n^2 m_p^2 }}{R} = (11.036068n^2 )\frac{{Gm_p^2 }}{R} [/math] [math] U_{de} = (5.06522n^2 )\frac{{Gm_p^2 }}{{(\frac{R}{{2.17879}})}} ---(96) [/math] Eq(96) is same to the (81) Therefore, Dark Energy define equation. [math] E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }})=\frac{3}{{10}}\frac{{GM^2 }}{R} [/math] Used to the gravitational self-energy, dark energy value of today is explained. Therefore, it means that dark energy is gravitational potential energy. 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 Merged post follows: Consecutive posts mergedV-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 = 1 proton/5m3 * Proton mass= 1.67264 X 10-27kg * G =6.6726 X 10-11m3/s2kg * 1J = 6.242 x 1018 eV comment-1. Radius of the universe is at least 78.2 billion light-years(24Gpc) wide. (2003, Neil J. Cornish, "Constraining the Topology of the Universe", http://arxiv.org/abs/astro-ph/0310233v1 ) comment-2. This is the Ned Wright BEST FIT volume In his recent article (January 2007) Wright gives a version of the universe that he calls best fit (to 4 or 5 different datasets) and Omega = 1.011 According to Smoot's notes the radius of curvature R is equal to the current Hubble radius divided by the square root of (Omega - 1) and the square root of 0.011 is around 0.105, So you divide the Hubble radius 13.8 billion LY by 0.105 and you get R = 130 billion LY. - 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(ΩM = - 0.38( ± 0.22)), using field equations which do not have the cosmological constant. ΩM = - 0.38( ± 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-3) - 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. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Edited November 28, 2009 by icarus2 Consecutive posts merged. Link to comment Share on other sites More sharing options...
Airbrush Posted November 29, 2009 Share Posted November 29, 2009 Interesting stuff, but the math is beyond me. Anyone want to explain what he is saying in 50 words or less? It would seem like the effect of dark energy is opposite to matter, so call it "negative mass". Why not? And not to be confused with antimatter. Link to comment Share on other sites More sharing options...
icarus2 Posted November 29, 2009 Author Share Posted November 29, 2009 (edited) Interesting stuff, but the math is beyond me. Anyone want to explain what he is saying in 50 words or less? It would seem like the effect of dark energy is opposite to matter, so call it "negative mass". Why not? And not to be confused with antimatter. Thanks! Airbrush! I'm sorry, I can't English well. So, My explain is short. Negative mass is not same to the antimatter. Antimatter has a value of positive mass. I explained property of the negative mass in chapter II. [ Abstract ] From the observance of the HSS team in 1998, they gained the mass density of the negative( ΩM = -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_M 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 ([math]10^{-47}GeV^{4}[/math]) 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 Edited November 29, 2009 by icarus2 Link to comment Share on other sites More sharing options...
michel123456 Posted November 30, 2009 Share Posted November 30, 2009 If I understand well, what you are saying is that we are actually observing a repulsive force in the universe. When you consider this force as repulsive gravity, it must been radiated by negative mass, and when you imput these hypotheses into the equations, everything fits well. Right? Link to comment Share on other sites More sharing options...
icarus2 Posted November 30, 2009 Author Share Posted November 30, 2009 If I understand well, what you are saying is that we are actually observing a repulsive force in the universe. When you consider this force as repulsive gravity, it must been radiated by negative mass, and when you imput these hypotheses into the equations, everything fits well.Right? Thanks! michel123456! That’s Right! In my proof, Dark energy is gravitational potential energy has a positive value. Therefore it suggested that negative mass is existed! Assuming that negative mass and positive mass were born together at the beginning of universe(in energy and momentum conservation state), everything fits well. We studied the negative energy state at Modern physics and Quantum Mechanics Please read to the below explain. And then, please think about the means of that negative mass has a harmonic oscillation at the maximum point. ---------- II-3. Negative mass is stable at the maximum point! therefore the catastrophe to energy level of minus infinity never happens 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. 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, [math] F(x) = F(0) + \frac{x}{{1!}}F^{'} (0) + \frac{{x^2 }}{{2!}}F^{''} (0) + \frac{{x^3 }}{{3!}}F^{'''} (0) + \cdots [/math] [math] + \frac{{x^n }}{{n!}}F^{(n)} (0) + \cdots \\ [/math] 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 [math]{x^2}[/math] and higher powers of x. We have, therefore, the approximate relation [math] F(x)=+kx [/math] The force is always the opposite directed toward the unstable equilibrium position(the origin), the derivative F'(0) is positive and therefore k is a positive constant. [math] - m\ddot x = + kx [/math] [math] \ddot x + \omega _0^2 x = 0 [/math] [math] (\omega _0^2 = \frac{k}{m}) [/math] 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. [math] \vec F = - \nabla U [/math] [math] U = - \frac{1}{2}kx^2 [/math] [math] {E_ -} = T + U = - \frac{1}{2}m\dot x^2 - \frac{1}{2}kx^2 \\ = - \frac{1}{2}m\omega _0^2 A^2 \\ [/math] In phase space [math] \frac{{x^2 }}{{(\frac{{ - 2E_ - }}{k})}} + \frac{{p^2 }}{{( - 2mE_ - )}} = 1 [/math] This equation is ellipses equation, because total energy [math]E_- < 0 [/math] 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 Link to comment Share on other sites More sharing options...
Arch2008 Posted December 1, 2009 Share Posted December 1, 2009 http://vixra.org/disclaimer Link to comment Share on other sites More sharing options...
michel123456 Posted December 1, 2009 Share Posted December 1, 2009 Icarus, I have to admit that I have not the knowledge to criticise your thinking. Wonderful work as it looks to me the same admirable as a magic presentation: I understand nothing. But i will try to dicuss it. First of all, I don't know how you will avoid circular reasonning in your calculations. It looks to me that your input is the same as your conclusion. I may be wrong on this comment, just make attention to it. Then if you want scientists look in their telescopes in search of negative mass, you should describe its properties. Some time ago I went on a search upon this subject, and there are a few things to find. It came to my understanding that negative mass, if it exists, is not so strange as it could appear at first sight. For example, (be carefull, I am far from an expert), it appears that negative mass is connected with negative energy through the e=mc^2 equation. On some sites (this is not an adequate reference) it is claimed that the "real" equation is e=+- mc^2, where the negative sign has been discarded as illogical. On some other sites, it is claimed that negative energy is related to time, and to serious search about time travel ( sorry if this comment will propulse this thread in the speculations waste basket, i don't want that to happen). In my understanding (it is less than a reference) negative mass is simply regular mass traveling backwards in time. But that would take a while to explain, surely in the waste basket. Link to comment Share on other sites More sharing options...
icarus2 Posted December 1, 2009 Author Share Posted December 1, 2009 (edited) Icarus, I have to admit that I have not the knowledge to criticise your thinking. Wonderful work as it looks to me the same admirable as a magic presentation: I understand nothing. But i will try to dicuss it. First of all, I don't know how you will avoid circular reasonning in your calculations. It looks to me that your input is the same as your conclusion. I may be wrong on this comment, just make attention to it. Then if you want scientists look in their telescopes in search of negative mass, you should describe its properties. Some time ago I went on a search upon this subject, and there are a few things to find. It came to my understanding that negative mass, if it exists, is not so strange as it could appear at first sight. For example, (be carefull, I am far from an expert), it appears that negative mass is connected with negative energy through the e=mc^2 equation. On some sites (this is not an adequate reference) it is claimed that the "real" equation is e=+- mc^2, where the negative sign has been discarded as illogical. On some other sites, it is claimed that negative energy is related to time, and to serious search about time travel ( sorry if this comment will propulse this thread in the speculations waste basket, i don't want that to happen). In my understanding (it is less than a reference) negative mass is simply regular mass traveling backwards in time. But that would take a while to explain, surely in the waste basket. Thanks! Michel~ I appreciate to your opinion. I’m sorry. I can’t English well. So, my answer is short and insufficient. Proof 1) is not same to proof 2) [math] E_{de}=U_{de} = (n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) -(80) [/math] Eq(80) can be applied at the whole time of universe. [math] U_{de} \approx - \frac{1}{2}U_s = \frac{3}{{10}}\frac{{GM^2 }}{R} -(91) [/math] But, Eq (91) can be applied only UT near to 0. Proof 1) has started from analysis for individual particle. "When the number of negative mass is n-, and the number of positive mass is n+ , total potential energy is given as follows." But, mean distance r-+ has not getting from equation (80) Therefore mean distance must be get from other method. Eq(81) = Eq(92) It is process for getting the mean distance. Negative mass originated from relativistic energy eq. [math] E^2 = p^2 c^2 + (m_0 c^2 )^2 [/math] [math] E = \pm \sqrt {p^2 c^2 + (m_0 c^2 )^2 } [/math] Therefore, [math] E = - \sqrt {p^2 c^2 + (m_0 c^2 )^2 } = - m_ - c^2 [/math] It is a solution also. However, In the modern physics and quantum mechanics, [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.] So, the solution of negative energy trashed away. And DIRAC applied the solution of negative energy to antiparticles. But, antiparticle's mass has a positive value. Finally, antiparticle is not negative mass particle. But, I fine out very important property of negative mass. It is that negative mass is stable at the maximum point, so the catastrophe to energy level of minus infinity never happens. Therefore, at this time, [math] E = - \sqrt {p^2 c^2 + (m_0 c^2 )^2 } = - m_ - c^2 [/math] Negative energy(mass) solution must be revived. My article has a many explains for negative mass.(20 Pages) I explained that properties of the negative mass. II-1-2)The law of motion of negative mass and positive mass. II-1-3)The law of motion of negative mass and negative mass. II-2. Negative mass cannot form the structure greater than atom. II-3. Negative mass is stable at the maximum point. II-4. The difference in mass when creating the pair of negative mass and positive mass. And I explained that collision of Bullet Cluster Among them, I will explain that the law of motion of negative mass and positive mass. Also, I will explain that collision of the Bullet Cluster. Again, I apologize for my English ability. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Merged post follows: Consecutive posts merged II. Property of negative mass 1-1)The law of motion of positive mass and positive mass ---------- +m1 ------ +m2 Fig01. Positive mass +m1 and positive mass +m2 (initial velocity =0, m1 >0, m2 >0) [math] m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r [/math] [math] \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r [/math] [math] m_2 \vec a_2 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r [/math] [math] \vec a_2 = - G\frac{{m_1 }}{{r^2 }}\hat r [/math] 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 - r, so the distance between two objects are reduced gradually. Merged post follows: Consecutive posts merged1-2)The law of motion of negative mass and positive mass ---------- -m1 ------- +m2 fig02. Negative mass - m1 and positive mass +m2 (initial velocity =0, m1 >0, m_2 >0) [math] - m_1 \vec a_1 = - G\frac{{( - m_1 )(m_2) }}{{r^2 }}\hat r [/math] [math] \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r [/math] [math] + m_2 \vec a_2 = - G\frac{{( - m_1 )(m_2) }}{{r^2 }}\hat r [/math] [math] \vec a_2 = G\frac{{m_1 }}{{r^2 }}\hat r [/math] 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). The type of force is repulsion, so the potential energy has positive value. ==> 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. Merged post follows: Consecutive posts merged1-3)The law of motion of negative mass and negative mass --------- - m1 ------ - m2 fig03. Ngative mass - m1 and negative mass - m2 (initial velocity =0, m1>0, m2> 0) [math] - m_1 \vec a_1 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r [/math] [math] \vec a_1 = + G\frac{{m_2 }}{{r^2 }}\hat r [/math] [math] - m_2 \vec a_2 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r [/math] [math] \vec a_2 = + G\frac{{m_1 }}{{r^2 }}\hat r [/math] 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. 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 Edited December 1, 2009 by icarus2 Consecutive posts merged. Link to comment Share on other sites More sharing options...
michel123456 Posted December 2, 2009 Share Posted December 2, 2009 icarus, I really try to follow. Sorry for this silly question, but you wrote [math] m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r [/math] I thought it was Where F=m a of course. Link to comment Share on other sites More sharing options...
icarus2 Posted December 2, 2009 Author Share Posted December 2, 2009 icarus, I really try to follow.~~ Strictly speaking, Your expression is wrong. [math] m_1 \vec a_1 = - G\frac{{m_1 m_2 }}{{r^2 }}\hat r [/math] It is a spherical coordinate expression. It is correct expression. [math] \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r [/math] G>0, m2>0, r>0. The direction of acceleration is in the direction of - r, so the distance between two objects are reduced gradually. It means that force is attraction. Strictly speaking, your expression is [math] \vec F_1 = + G\frac{{m_1 \times m_2 }}{{r^2 }}\hat a_x [/math] [math] \vec F_2 = - G\frac{{m_1 \times m_2 }}{{r^2 }}\hat a _x [/math] [math]\hat a_x [/math] is unit direction vector. So, [math] \vec F_1 = - \vec F_2 [/math] [math] |\vec F_1 | = |\vec F_2 | = G\frac{{m_1 \times m_2 }}{{r^2 }} [/math] [math]F_1 = F_2 = G\frac{{m_1 \times m_2 }}{{r^2 }}[/math] is only magnitude of force. Link to comment Share on other sites More sharing options...
michel123456 Posted December 5, 2009 Share Posted December 5, 2009 O.K. for your last post. Digestion will take time for the rest. I think you have to make a search in another direction, in reality, the same direction: try to prove that the notion of mass (rest mass) is time dependent. I do not use the term "frame dependent", because we know that rest mass is not frame dependent. When we accept, as posed in wikipedia, that "For a system of particles going off in different directions, the invariant mass of the system is the analog of the rest mass, and is the same for all observers", we are speaking of observers in the same time (the present), observers in the past & observers in the future, traveling smoothly in Time. Time is considered as invariable. Let's suppose it is not: maybe you will see negative mass rise from the equations. Merged post follows: Consecutive posts mergedBack to the topic. Icarus wrote:"II-2. Negative mass cannot form the structure greater than atom." What makes you think that? Link to comment Share on other sites More sharing options...
icarus2 Posted December 6, 2009 Author Share Posted December 6, 2009 (edited) Icarus wrote:"II-2. Negative mass cannot form the structure greater than atom." What makes you think that? Thanks michel~ II-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 [math]\vec F=-Q®\hat r[/math], Q® is the positive function of distance r, thus nuclear force is in the form of attraction worked in the direction of -[math]\hat r [/math]. Here, for the force worked on negative mass –m, [math] \vec F=-m\vec a=-Q®\hat r [/math] [math] \vec a=\frac{{Q®}}{m}\hat r [/math] 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 Merged post follows: Consecutive posts mergedV-6. Colliding with Bullet Cluster As shown in the recent observation of dark matter, dark matter seems not to interact with each other when colliding with galaxy clusters, and it can be predicted with essential characteristics of negative mass. Uniformly distributed negative mass receives attractive effect from massive positive mass, so dark matter which has negative mass is clustered around galaxy because of attraction of galaxy. If the type of force worked between negative mass when colliding with galaxy clusters is repulsive, and negative mass is distributed almost uniformly without forming massive mass structure, the shape of dark matter is not distorted or transformed even if dark matter(negative mass) and dark matter(negative mass) pass through similar area, that is, they will seem not to interact each other. Also there is repulsive effect between dark matters that are made up with negative mass, thus there will be almost no direct collision as well. Repulsive effect that is the nature between negative mass as described above provides the proper description that existing dark matters do not interact each other, although they interact gravitationally with positive mass. As the factor that breaks the uniform distribution of negative mass, First, negative mass receives the attractive effect from massive positive mass, thus for the distribution of negative mass near massive positive mass such as galaxy or galaxy cluster, the density of negative mass is higher as it is closer to galaxy or galaxy cluster, and is lower as it is farther. Second, If positive mass(like galaxy cluster) that has strong gravity or interstellar cloud that has positive mass pass through existing area that negative mass is distributed, negative mass can be disappeared when meeting positive mass or it can be drawn owing to attractive effect of massive positive mass at this moment, so there can be the area that negative mass, namely, dark matter is not uniformly distributed Fig20. Collision of Bullet Cluster. We can see that ordinary matters(red color) are close to each other, and dark matters(blue color) are on the far side. (http://arxiv.org/abs/astro-ph/0608407v1) Fig23. MACS J0025.4-1222 Cluster. We can see that ordinary matters(yellow line) are close to each other(center), and dark matters(red line) are on the far side. (http://arxiv.org/abs/0806.2320v2) ========== i)Positive mass, positive mass : attractive ii)Negative mass(dark matter), negative mass(dark matter) : repulsive iii)Massive positive mass, negative mass : attractive iv)Result : At least, from 3 characteristics above, we can predict that positive mass and positive mass are arranged on the close side, and negative mass(dark matter) and negative mass(dark matter) are arranged on the far side. Can we explain the phenomenon that dark matter is arranged on the far side and visible positive matter is arranged on the close side in Fig.20 with other dark matters? Negative mass shows the result matched with the phenomena, and it means that it is necessary to perform more strict simulation with negative mass. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Edited December 6, 2009 by icarus2 Consecutive posts merged. Link to comment Share on other sites More sharing options...
icarus2 Posted December 15, 2009 Author Share Posted December 15, 2009 Fig23. MACS J0025.4-1222 Cluster. We can see that ordinary matters(yellow line) are close to each other(center), and dark matters(red line) are on the far side. (http://arxiv.org/abs/0806.2320v2) ========== i)Positive mass, positive mass : attractive ii)Negative mass(dark matter), negative mass(dark matter) : repulsive iii)Massive positive mass(Galaxy and Galaxy Cluster), negative mass : attractive ==> 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. iv)Result : At least, from 3 characteristics above, we can predict that positive mass and positive mass are arranged on the close side, and negative mass(dark matter) and negative mass(dark matter) are arranged on the far side. Link to comment Share on other sites More sharing options...
icarus2 Posted January 7, 2010 Author Share Posted January 7, 2010 (edited) 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/5m3 * [math]n_-=n_+=n, m_-=(23.3/4.6)m_+=(5.06522)m_p[/math] * [math]\bar r_{ - + } = \frac{R}{{3.27273}}[/math] * If [math]\bar r_{ - + } = \frac{R}{{2.17879}}[/math] [math](U_T \approx 0)[/math], dark energy density has a 1/3 smaller than [math]\bar r_{ - + } = \frac{R}{{3.27273}}[/math]. In a WMAP, observed value [math]\Lambda = 2.14( \pm 0.13) \times 10^{ -3} eV [/math] Dark energy density : [math] \rho _{de} = 2.09 \times 10^{ - 47} [_{ - 0.465}^{ + 0.557} ]GeV^4 [/math] Ridius of the Universe : [math] R_{UNI} = 96.76[_{ - 11.44}^{ + 12.13} ]Gly = 85.32 \sim 108.89Gly [/math] ( If [math]\bar r_{ - + } = \frac{R}{{2.17879}}[/math], [math] R_{UNI} = 118.8[_{ - 14.0}^{ + 14.9} ]Gly = 104.8 \sim 133.7 Gly [/math]) ========= Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Edited January 7, 2010 by icarus2 Link to comment Share on other sites More sharing options...
pywakit Posted January 7, 2010 Share Posted January 7, 2010 (edited) icarus2 writes: V-2-2)Proof from the gravitational self-energy In case of, mass M and mass distribution is 0 ≤ r ≤ R and mass density is ρ. fig10 caption : Gravitational self-energy Gravitational self-energy of the universe 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-, and the number of positive mass is n+. The other side, Us is total potential energy when all particles are positive mass. Therefore, Us is total potential energy when dark energy term has an opposite sign. General gravitational potential defined, Therefore, US = - Ude + Ugp ---(88) From analysis of V-5, If UT ≥ 0, We know that Umax ≥ UT = Ude + Ugp Therefore, Substitution equation (89) in equation (88 ) Finally, (This equation can be applied at UT ≥ 0) i) In case of n- = n+ =n, m - = km+ (k ≥ 1) Total mass Equation (81) = Equation (92) If From eq(91), Eq(96) is same to the (81) Therefore, Dark Energy define equation. Used to the gravitational self-energy, dark energy value of today is explained. Therefore, it means that dark energy is gravitational potential energy. 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 It is impossible for me to comprehend your equations. However, if I did not know better ( do I? ) it would seem you have just proved the existence of Dark Energy. Would the same equations apply if Dark Energy was actually the energy of space itself? Edited January 7, 2010 by pywakit Link to comment Share on other sites More sharing options...
icarus2 Posted January 20, 2010 Author Share Posted January 20, 2010 (edited) It is impossible for me to comprehend your equations. However, if I did not know better ( do I? ) it would seem you have just proved the existence of Dark Energy. Would the same equations apply if Dark Energy was actually the energy of space itself? I’m sorry. I can’t English well. So discussion by English is difficult. I’m very sorry. (_ _); Would the same equations apply if Dark Energy was actually the energy of space itself? Case by case~ At first, definition of energy of the space itself is needed. Strictly speaking, gravitational self-energy is not the energy of the space itself. But, if energy of the space is gravitational potential energy, gravitational self-energy method can be applied. Though energy of the space is not same to the gravitational potential energy, spherical integral shape is useful. [math]dU_s = - \frac{{GM'dm}}{r}[/math] [math]M' = \frac{4}{3}\pi r^3 \rho[/math], [math]dm = \rho r^2 dr\sin \theta d\theta d\phi[/math] [math] dU_s = - G\frac{{4\pi \rho ^2 }}{3}r^4 dr\sin \theta d\theta d\phi [/math] [math] 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 } [/math] [math] U_s = - G\frac{{(4\pi \rho )^2 }}{3}(\frac{1}{5}R^5 ) [/math] [math] U_S = - \frac{3}{5}\frac{{GM^2 }}{R} [/math] “The hypothesis of dark matter and dark energy with negative mass” insists that dark matter is not different from dark energy each other, dark matter is the effect of centripetal force by negative mass out of galaxy, and dark energy is that positive gravitational potential term in total potential energy, and total gravitational potential energy has positive value at now. So universe has accelerating expansion. [math] E_{de}= \sum\limits_{i,j}^{i = n_ - ,j = n_ + } {(\frac{{Gm_{ - i} m_{ + j} }}{{r_{ - + ij} }})} =(n_ - \times n_ + )(\frac{{Gm_ - m_ + }}{{\bar r_{ - + } }}) [/math] other expression, [math]E_{de}= - \frac{{U_S }}{{(a + 1)}} = k_c \frac{{GM^2 }}{{R}}[/math] (k_c is function of the time, at now k_c =0.45062) Cosmological constant and dark matter with positive mass are not observed as yet. Accelerating expansion and gravitational effect(centripetal force effect and gravitational lensing effect) are only observed. We observed not positive mass but gravitational effect. Ω_M = - 0.4( ± 0.1) SCP team : Presentation at the January 1998 Meeting of the American Astronomical Society, (referenced in Perlmutter et al., B.A.A.S., v. 29, no. 5, p. 1351, 1997) http://www-supernova.lbl.gov/public/papers/poster198preprint.pdf <= 7P Ω_M = - 0.38( ± 0.22) HSS Team : http://arxiv.org/abs/astro-ph/9805201 - 14P ----------- HSS team and SCP team had observed our universe, however they got a negative mass density from inspected field equation during 70years. They are very confused. So, they insert Einstein’s cosmological term(Λ). But…, Although they can be explained that universe is accelerating expansion by insert Λ, they make two new problems at least. Finally, problem has not solved but increased. What is the origin of cosmological constant? And, what is the reason of that cosmological constant has a special value? We tried to explain with vacuum energy but it was failed. The energy density of the vacuum is estimated as [math]10^{70}GeV^4[/math], which is about [math]10^{117}[/math] orders of magnitude large than the observation value [math]10^{-47}GeV^4[/math]. Also, we have must explained dark matter. However, if negative mass density is exist, accelerating expansion is explained rightly. And I have assert that negative mass can be explained for the dark matter, decelerating expansion and accelerating expansion, fine tuning problem of mass density, inflation mechanism, the reason of that dark energy seems to has a small and constant value(like cosmological constant Λ), collision of Bullet cluster. This problem is related to the field equation. ********************************************** (Einstein, Direc, Hubble,...; Inspected field eq. during 70 years.) vs ( New field eq. supported by 60~70%(?) of relation branch physists during 10 years.) ********************************************** =========== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Edited January 20, 2010 by icarus2 Link to comment Share on other sites More sharing options...
pywakit Posted January 20, 2010 Share Posted January 20, 2010 (edited) Let me ask the question a different way. ( sorry about the english ) I have talked about this before on another thread. Could dark energy be a PROPERTY of space rather than a seperate force acting upon space? Is it possible that space is PULLING galaxies apart, rather than PUSHING them apart? I think it could be possible. I also think that space does not like energy in the form of mass/gravity. The 'lines' of space work toward being perfectly straight, while mass bends the lines. I also think it is very possible that space's energy is the mechanism that 'artificially' creates a limit to the velocity of matter, and photons. Are there NEW measurements being made that will ( or could possibly ) correctly quantify space's energy quotient, if any? Thank you. Edited January 20, 2010 by pywakit Link to comment Share on other sites More sharing options...
Martin Posted January 20, 2010 Share Posted January 20, 2010 (edited) http://vixra.org/disclaimer Icarus, I want to convey congratulations on this interesting thread discussing your highly speculative theory. I appreciate your bringing it here to SFN, and the large amount of work put into graphic illustration, and the Latex typography. Because the work is so highly speculative, it belongs in the Speculations forum. I will move it there. This does not indicate disapproval. I encourage you to continue discussing and explaining your interesting idea. The Astronomy/Cosmology forum is more focused on questions and answers regarding mainstream A/C. ==================== I don't want to get into a discussion of your ideas, myself. There are several people who are already interested and it's up to them to ask questions. Nevertheless, I did want to toss in a comment---and to mention what seems like a peculiarity of your "Laws of Motion" in post #8. It seems to require a distinguished frame of reference simply to be able to state your laws of motion. Or am I missing something? I see no Newtonian "equal and opposite reaction" assumption regarding forces. Would momentum be conserved in a universe which obeyed your laws? Edited January 20, 2010 by Martin Link to comment Share on other sites More sharing options...
icarus2 Posted January 21, 2010 Author Share Posted January 21, 2010 (edited) Icarus, I want to convey congratulations on this interesting thread discussing your highly speculative theory.I appreciate your bringing it here to SFN, and the large amount of work put into graphic illustration, and the Latex typography. Because the work is so highly speculative, it belongs in the Speculations forum. I will move it there. This does not indicate disapproval. I encourage you to continue discussing and explaining your interesting idea. The Astronomy/Cosmology forum is more focused on questions and answers regarding mainstream A/C. ==================== I don't want to get into a discussion of your ideas, myself. There are several people who are already interested and it's up to them to ask questions. Nevertheless, I did want to toss in a comment---and to mention what seems like a peculiarity of your "Laws of Motion" in post #8. It seems to require a distinguished frame of reference simply to be able to state your laws of motion. Or am I missing something? I see no Newtonian "equal and opposite reaction" assumption regarding forces. Would momentum be conserved in a universe which obeyed your laws? Would momentum be conserved in a universe which obeyed your laws? To Martin Negative mass has obey to the preexistence physics laws(Newton’s law, energy conservation law, momentum conservation law, …) 1-2)The law of motion of negative mass and positive mass ---------- -m1 ------- +m2 fig02. Negative mass - m1 and positive mass +m2 (initial velocity =0, m1 >0, m_2 >0) [math] - m_1 \vec a_1 = - G\frac{{( - m_1 )(m_2) }}{{r^2 }}\hat r [/math] [math] \vec a_1 = - G\frac{{m_2 }}{{r^2 }}\hat r [/math] [math] + m_2 \vec a_2 = - G\frac{{( - m_1 )(m_2) }}{{r^2 }}\hat r [/math] [math] \vec a_2 = G\frac{{m_1 }}{{r^2 }}\hat r [/math] [math]\vec P_T = ( - m_1 \vec v_1 ) + ( m_2 \vec v_2 ) = 0[/math] [math]m_1 \vec v_1 = m_2 \vec v_2 [/math] [math]\vec v_1 ,\vec v_2 [/math] is same direction.(In eq.,viewpoint is the spherical coordinate : [math]\hat r[/math] ) in the spherical coordinate [math]\vec a_1 = - \hat r , \vec a_2 = + \hat r[/math] in the rectangular coordinate [math]\vec a_1 = + \hat x , \vec a_2 = + \hat x[/math] for the simple analysis, if m1=m2 Acceleration a1=a2, therefore v1=v2, In the Rectangular coodinate system It is correct.(momentum conservation) ---------- - m1 ------ - m2 fig03. Ngative mass - m1 and negative mass - m2 (initial velocity =0, m1>0, m2> 0) [math] - m_1 \vec a_1 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r [/math] [math] \vec a_1 = + G\frac{{m_2 }}{{r^2 }}\hat r [/math] [math] - m_2 \vec a_2 = - G\frac{{( - m_1 )( - m_2 )}}{{r^2 }}\hat r [/math] [math] \vec a_2 = + G\frac{{m_1 }}{{r^2 }}\hat r [/math] In this case, momentum conservation is valid. [math]\vec P_T = ( - m_1 \vec v_1 ) + ( - m_2 \vec v_2 ) = 0[/math] [math]m_1 \vec v_1 = - m_2 \vec v_2 [/math] , because, let's view to the acceleration term. [math]\vec a_1 [/math] , [math]\vec a_2 [/math] is opposite direction(spherical coordinate +r direction). So, velocity direction is opposite direction. in the spherical coordinate [math]\vec a_1 = + \hat r , \vec a_2 = + \hat r[/math] in the rectangular coordinate [math]\vec a_1 = - \hat x , \vec a_2 = + \hat x[/math] For the energy conservation and momentum conservation, Refer to below wiki, http://en.wikipedia.org/wiki/Negative_mass#Negative_mass Negative mass has obey to the preexistence physics laws(Newton’s law, energy conservation law, momentum conservation law, …) My argument never speculative. I mentioned many methods or calculations for the investigation. But peoples have not try to see it. I graduated in physics of the university. My grade of Classical Dynamics I, II is all A+(total 50 person, only two person is A+) I don't claim to be a bright man. What I want to say is this : At least, I’m not speculative. I love to the physics. I’m disappointed your decision. But, I can’t English well. So, never more say. Have a nice day! Edited January 21, 2010 by icarus2 1 Link to comment Share on other sites More sharing options...
icarus2 Posted February 4, 2010 Author Share Posted February 4, 2010 [ Description of cosmic decelerating expansion and accelerating expansion ] fig08. caption : Decelerating expansion and accelerating expansion of the universe Fig.8 shows current cosmic expansion. Currently expansion velocity of our universe was reduced owing to attraction of matter, but it has increased again for last 7 billion years. At the beginning of universe, most of negative energy was converted to negative mass owing to the characteristics of negative energy and positive energy pair-created, whereas some positive energy was exist as radiant energy(pair annihilation of matter and antimatter, radiation) and bond energy, So, if we induce how many terms of potential energy survive when the number of negative mass is greater than that of positive mass, Focus on the number of terms of remaining potential energy rather than the concrete value of potential energy, *Potential energy between positive mass and positive mass has – value: U = ([math]\frac{{ - Gm_ + m_ + }}{r}[/math]) = 1U- *Potential energy between negative mass and positive mass has + value: U = ([math]\frac{{ - G( - m_ - )(+ m_ + ) }}{r}[/math]) = 1U+ *Potential energy between negative mass and negative mass has – value: U = ([math]\frac{{ - G( - m_ - )( - m_ - )}}{r}[/math]) = 1U- (if, U+ = - U-) 1-1)When the number of negative mass is 2, and the number of positive mass is 1 [math] U_T = 2(\frac{{ - G( - m_ - )(m_ +) }}{r}) + ( - \frac{{G( - m_ - )( - m_ - )}}{r}) [/math] = 2U+ + 1U- = 1U+ 1-2)When the number of negative mass is 3, and the number of positive mass is 1 UT = 3U+ + 3U- = 0 1-3)When the number of negative mass is 4, and the number of positive mass is 1 UT = 4U+ + 6U- = - 2U+ 1-4)When the number of negative mass is 4, and the number of positive mass is 2 UT = 8U+ + 7U- = 1U+ 1-5)When the number of negative mass is n-, and the number of positive mass is n+, total potential energy is given as follows. [math] U_T = (n_ - \times n_ + )U_ + + (\frac{{n_ - (n_ - - 1)}}{2}U_ - + \frac{{n_ + (n_ + - 1)}}{2}U_ - ) ---(77) [/math] equation (77) is expressed as follows more strictly [math] 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} }})} [/math] [math] 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_{ + + } }})) [/math] Calculate the value of U- = - U+, n- = 10, n+ = (1 -10) with equation (77) when there is a difference in the number between negative mass and positive mass to examine changes, ( n- , n+ ) (10,1) UT = 10U + + 45U - = - 35U + (10,2) UT = 20U + + 46U - = - 26U + (10,3) U T = 30U + + 48U - = - 18U + (10,4) U T = 40U + + 51U - = - 11U + (10,5) U T = 50U + + 55U - = - 5U + (10,6) U T = 60U + + 60U - = 0 (10,7) U T = 70U + + 66U - = 4U + (10,8) U T = 80U + + 73U - = 7U + (10,9) U T = 90U + + 73U - = 9U + (10,10) U T = 100U + + 90U - = 10U + 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. fig17 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(or amount) of negative mass to the number(or amount) of positive mass, the total potential energy has 0. ==> 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(or amount) of positive mass less than criticle number(amount), 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)(UT = nU+). ==> Final state of the universe. From eq.(72) ([math] U_n = \sum\limits_{i = 1}^{n^2 } {U_{ + i} } + \sum\limits_{j = 1}^{n(n - 1)} {U_{ - j} } = nU_+ [/math]) The analysis above can describe that current universe expanded with decelerated speed approximately 7 billion years ago, whereas it expanded with accelerated speed in last 7 billion years, and most of negative energy was converted to negative mass owing to the characteristics of negative energy and positive energy pair-created at the beginning of universe, whereas some positive energy was exist as radiant energy(pair annihilation of matter and antimatter, radiation) and bond energy, which does not have positive matter and mass. At this time, it expanded with decelerated speed because total potential energy had negative value, then positive matter gradually increased as the universe was cooled down and currently it entered in accelerating expansion which total potential energy of universe has positive value after passing through the point that total potential energy of universe is zero. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass http://vixra.org/abs/0907.0015 Link to comment Share on other sites More sharing options...
icarus2 Posted May 26, 2010 Author Share Posted May 26, 2010 (edited) [ Dark matter ] Calculating centripetal force in galaxy from negative mass distributed out of galaxy fig11 caption : Revolution velocity of galaxy. The revolution velocity through the total quantity of matter that can be observed is A, while the actual observed value is B. We observed not positive mass but gravitation effect. That is, universe has not said that dark matter is positive mass. This study mentioned the effect of centripetal force described previously, and the effect of centripetal force of negative mass with the description of dark matter. The researcher suggests that it is possible to prove the existence of negative mass with the comparison between observed value and revolution velocity of star according to the distance in galaxy which is calculated from negative mass distributed out of galaxy. fig12 caption : Negative mass effect on arms of galaxy. The actual observed spiral galaxy is on the left. Since repulsive effects occur between negative masses in Newton’s dynamics, negative masses will be distributed all over space because it cannot form large mass structures like stars. Negative mass within the galaxy is cancelled out by attraction from large positive mass during the galaxy formation process. Furthermore, the space, other that the galaxy, will maintain the distribution state of negative mass. When negative mass(negative energy) is distributed uniformly, if we examine the area of galaxy diameter working the gravity on the basis of mass m for the analysis, there is no negative mass in the space of spiral space on the left, so negative mass can exist as much as the size of black and white on the right. negative mass on the right works the force on positive mass m on the origin to the left(antigravity;positive mass moves after receiving the force that becomes farther from negative mass), and this direction will be in keeping with the central direction of gravity of real galaxy. [math] V®^2 = GM®/r [/math] [math] M® = M®_{visible} + M®_{dark} [/math] From the calculation of centripetal force in Newton dynamics, the important matters to calculate gravity effect of negative mass are the size and model of empty space of negative mass by real galaxy on the left, in case of spiral galaxy or oval galaxy, it may be close to oval if we consider the fact that most of mass is placed on galaxy core, the rotation of galaxy, and the gravitational radius of galaxy core, thus we regard empty space of negative mass as spherical or oval, and perform calculation to compare with observed value. fig13 caption : Effective negative mass. The negative mass distribution, where gravity is applied without cancellation of the oblique field located on the right side. [math] \frac{{mv^2 }}{{r_ + }} = \frac{{GM_ + (r_ + )m}}{{r_ + ^2 }} + \frac{{GM_ - (r_ - )m}}{{r_ - ^2 }} ---(93) [/math] ([math]M_ + (r_ + ) > 0, M_ - (r_ - ) > 0[/math]) Adding the term of centripetal force by negative mass to the term of centripetal force by positive mass as described above, we compare revolution velocity of star and the density of negative mass, then expand this to apply to other galaxies, and we can inspect if the interpretation of dark matter distributed out of galaxy is proper or not. The distribution of negative mass that remains without being offset in Fig. 13 is crescent-shaped, so it seems to be difficult to calculate, but if we assume that the white empty space on the left is full with negative mass and positive mass at the same density, here negative mass is uniformly distributed over the whole area, so the effect of negative mass on mass m is 0. Remaining positive mass is distributed over the white area on the left at the density of negative mass, and the gravity that uniformly distributed positive mass works on positive mass m place on radius r is worked only by the distribution of mass within radius r. fig14 caption : Equal to the degree of gravity that is applied on the equally distributed positive mass within the radius r of the left side. The gravitational effect from negative mass, which functions at mass m is equal to the gravitational effect from positive mass within radius r. Therefore, the effect of negative mass that remains out of galaxy without being offset can make it approximate to the gravity generated by the distribution of positive mass within the radius R in galaxy [math] M_ - = \rho _ - ®V = \rho _ - ®(\frac{{4\pi r^3 }}{3}) [/math] In equation(93) [math] \frac{{mv^2 }}{{r_ + }} = \frac{{GM_ + (r_ + )m}}{{r_ + ^2 }} + \frac{{GM_ - (r_ + ^{} )m}}{{r_ + ^2 }} [/math] [math] v = \sqrt {\frac{{G(M_ + + M_ - )}}{{r_ + }}} = \sqrt {\frac{{G(M_ + + \rho _ - (r_ + )V)}}{{r_ + }}} [/math] [math] v = \sqrt {\frac{{GM_ + }}{{r_ + }} + \rho _ - (r_ + )(\frac{{4\pi r_{_ + }^2 }}{3})} ---(97) [/math] The mass density of collisionless isothermal sphere is given as [math]\rho ® = \frac{{\sigma ^2 }}{{2\pi Gr^2 }}[/math] in galaxy dynamics. [math]\sigma ^2 [/math] is velocity dispersion. [math] M_{dark}=M_{negative}= \frac{{2\sigma ^2 }}{{3G}}r ---(98) [/math] Equation (97) is applied to arbitrary spherical or oval galaxy, and in case of normal shaped spiral galaxy, the distance which spherical approximation can be applied from the center of galaxy will be effective. So if we consider that the distribution of negative mass out of galaxy comes under isothermal, the effect of mass in galaxy by the distribution of negative mass out of galaxy is proportional to r as shown in equation (98), and this is in keeping with observed dark matter. Positive mass dark matter distributed in galaxy does not form the spherical distribution, and it is broken owing to the gravity of galaxy core, gravitational action with stars, and rotation of galaxy, if its shape is not very intentionally assumed. Furthermore, there are various heat sources in galaxy according to position, so it cannot be regarded as collisionless Isothermal. However, the effect of centripetal force that negative mass works from the exterior of galaxy can keep the spherical distribution of mass, and make the mass effect that increases linearly in accordance with distance r as seen above because it receives less influence from elements such as the gravity of galaxy core, or rotation of galaxy. Also ideal state of collisionless can be achieved by negative mass, because negative mass has repulsive effect each other. At the solar system, non-observation of dark matter is explained. In negative mass hypothesis, dark matter exists at the exterior of galaxy, so dark matter is not observed at the solar system. It described that the quantity of negative mass decided at the beginning of universe can explain the effect of centripetal force in galaxy(V-4), repulsion effect on cosmic expansion(V-3), decelerating expansion at the first half, and accelerating expansion(V-1) of universe at the second half(7 billion years ago) at the same time. Therefore, it is necessary to try to calculate and observe negative mass more strictly, laying aside the abstract aversion of negative mass. Please review to the fig12, fig13, fig14. And then, please understand that the gravitational effect from negative mass(negative energy) is equal to the gravitational effect from positive mass within radius r. ======== Hypothesis of Dark Matter and Dark Energy with Negative Mass : http://vixra.org/abs/0907.0015 Edited May 26, 2010 by icarus2 Link to comment Share on other sites More sharing options...
icarus2 Posted September 20, 2010 Author Share Posted September 20, 2010 (edited) modify the equation (97), (98) [math] v = \sqrt {(\rho _ + (r_ + ) + \rho _ - (r_ + ))\frac{{4\pi Gr_ + ^2 }}{3}} [/math] The mass density of collisionless isothermal sphere is given as [math]\rho ® = \frac{{\sigma ^2 }}{{2\pi Gr^2 }}[/math] in galaxy dynamics. [math]\sigma ^2 [/math] is velocity dispersion. [math] M_{dark} = M_{negative} = \int {\rho ®dV} = \frac{{2\sigma ^2 }}{G}r [/math] Motion of the negative mass and dark matter -1 Motion of the negative mass and dark matter -2 ========= Negative mass, Dark matter, Dark energy, CC Problem! http://www.bautforum.com/showthread.php/105870 Hypothesis of dark matter and dark energy with negative mass : http://vixra.org/abs/0907.0015 Edited September 20, 2010 by icarus2 Link to comment Share on other sites More sharing options...
vuquta Posted September 21, 2010 Share Posted September 21, 2010 I read your proof and conclusions that your theoretical value is at the pobs. How does your value and the mainstream cause the universe to expand outwards. modify the equation (97), (98) [math] v = \sqrt {(\rho _ + (r_ + ) + \rho _ - (r_ + ))\frac{{4\pi Gr_ + ^2 }}{3}} [/math] The mass density of collisionless isothermal sphere is given as [math]\rho ® = \frac{{\sigma ^2 }}{{2\pi Gr^2 }}[/math] in galaxy dynamics. [math]\sigma ^2 [/math] is velocity dispersion. [math] M_{dark} = M_{negative} = \int {\rho ®dV} = \frac{{2\sigma ^2 }}{G}r [/math] Motion of the negative mass and dark matter -1 Motion of the negative mass and dark matter -2 ========= Negative mass, Dark matter, Dark energy, CC Problem! http://www.bautforum.com/showthread.php/105870 Hypothesis of dark matter and dark energy with negative mass : http://vixra.org/abs/0907.0015 How do you create negative mass? Link to comment Share on other sites More sharing options...
dragonstar57 Posted September 26, 2010 Share Posted September 26, 2010 could someone dumb this down by a factor of 10000000000000^0000000000000 Link to comment Share on other sites More sharing options...
icarus2 Posted September 26, 2010 Author Share Posted September 26, 2010 (edited) I read your proof and conclusions that your theoretical value is at the pobs. How does your value and the mainstream cause the universe to expand outwards. How do you create negative mass? I’m sorry. I can’t English well. My native language is not English. So, my expression is short and insufficient. ();; Again, I’m sorry. 1. How does your value and the mainstream cause the universe to expand outwards. fig10. caption : Expansion of the universe by negative mass distribution. The force, which is applied to the galaxy and galaxy cluster and located within radius R from the center of a 3-dimensional space, when assuming equal distribution of negative mass from repulsive effects between negative masses across space is valid. As considered the problem of the gravity at the point placed on arbitrary radius R of spherical object like the earth that mass is uniformly distributed, the gravity by mass distribution out of radius R is offset each other, and there is only the gravity effect by mass within radius R. If negative mass was born at the beginning of universe together with positive mass, it should be uniformly distributed in universe by repulsion between negative mass. Of course, there can be difference in negative mass density owing to the local distribution of massive positive mass. For the gravity effect that almost uniformly distributed negative mass has on positive mass(galaxy or galaxy cluster) located in radius R from the center of universe(3-dimensional center), negative mass within radius R only has the gravity effect on positive mass located within radius R, and at this time, positive mass is accelerated in the direction of begin far away from negative mass. The direction that positive mass becomes farther from negative mass is in keeping with the direction of cosmic expansion, and also negative mass continuously works the force, so finally positive mass is gradually accelerated. As examined in Chapter II, in the description of the force of positive mass and negative mass, the force is repulsion(the direction of +R), and integrated potential energy is positive value. Therefore, the assumption of the existence of negative mass can suggest the explanation of repulsive effect required for cosmic expansion, and the accompanying positive dark energy. "The hypothesis of dark matter and dark energy with negative mass" insists that dark matter is not different from dark energy each other, dark matter is the effect of centripetal force by negative mass out of galaxy, and dark energy is that positive potential term in total potential energy, and total potential energy has positive value at now. So universe has accelerating expansion. Accordingly, if we calculate the distribution and mean density of negative mass that describes rotation velocity within galaxy, and explain accelerating expansion of current universe with the size of potential energy that has positive value by this negative mass, we can prove the hypothesis that negative mass is the origin of dark matter and dark energy. ============= 1) Calculation by physicists. Refer to below paper : HSS(The High-z Supernova Search) team : Brian P. Schmidt, et al. : "Observational Evidence from Supernovae for an Accelerating Universe and a Cosmological Constant" http://arxiv.org/abs/astro-ph/9805201 14P, 26~29lines. If Λ=0, Ω_m = - 0.38(±0.22) : negative mass density SCP(Supernova Cosmology Project) team : S.Perlmutter et al.: "Cosmology from Type Ia Supernovae" http://arxiv.org/abs/astro-ph/9812473 7P If Λ=0, Ω_m = - 0.4(±0.1) : negative mass density From the observance of the HSS team and SCP team in 1998, they gained the mass density of the negative(HSS: Ω_M = -0.38( ± 0.22), SCP: Ω_M = -0.4( ± 0.1)), using field equations which do not have the cosmological constant.(It is inspected Field equation during 70 years.) In they thought, the quantity of the mass couldn't be negative value. So the value was discarded. We have to know that not the field equation has disposed the value, but our thought disposed that value. Also, negative mass is stable at the maximum point. Therefore the catastrophe to energy level of minus infinity never happens So, negative mass can be exist. 2) My calculation, refer to the first posting. http://www.scienceforums.net/topic/43190-dark-energy-dark-matter-fine-tuning-problem-negative-mass/page__view__findpost__p__509927 The value of observed dark energy has been proved in accordance with negative mass hypothesis. ρ_de = (0.201 ~ 5.02) X [math]10^{-47}GeV^4[/math] : Theoretical computation(Refer to recently paper ) ρ_obs = [math]10^{-47}GeV^4[/math] : Observed value ============== 2. How do you create negative mass? Initial energy value of universe 1) It looks more natural when an initial energy value of universe is 0. Therefore, negative energy is needed to offset positive energy of matters and in my opinion, only with negative gravitational potential energy, it is difficult to offset positive energy. - The present theory for the universe creation from nothing Argue that E_T = 0 = (+E) + (-E) = 0 = (Σ+mc^2) + (Σ-GMm/r) = 0 - But, in my opinion, (Σ+mc^2) > |(Σ-GMm/r)|, so (Σ+mc^2) + (Σ-GMm/r) ≠ 0 In my think, E_T = 0 = (+E) + (-E) = 0 = (Σ+ m+c^2) + (Σ - m_c^2) + (Σ U) = 0 (m+ : positive mass / m_ : negative mass / U : gravitational potential energy) 2) We observe the universe today. If we run all the particles positions backwards, they appear to be packed together at infinite density. Maybe, early universe has a density of black hole. In generally, particles cannot escape from the black hole. However, present theories have explains to the situations with ambiguous method. But, Assuming that negative mass(energy) and positive mass(energy) were born together at the beginning of universe, If we run all the particles positions backwards, they appear to be packed together at infinite density(each positive mass and negative mass). But, Universe has not a density of black hole. So universe can be expanded. The reason why we have not found negative mass on the earth until now is that negative mass exists as the state of mass when it was born without forming massive mass structure which can be easily measured owing to basic characteristics of negative mass(repulsive effect) as described above, and that negative mass is not observed because negative mass which exists around massive positive mass such as earth or the solar system receives attractive effect from massive positive mass, and it was disappeared long time ago when forming galaxy, even if it existed at the beginning of universe. Negative mass within the galaxy is cancelled out by attraction from large positive mass during the galaxy formation process. Furthermore, the space, other that the galaxy, will maintain the distribution state of negative mass.(view to the simulation video) I’m very sorry. I can’t English well. So I have a difficulty in English expression. ========= Negative mass, Dark matter, Dark energy, CC Problem! http://www.bautforum.com/showthread.php/105870 Hypothesis of dark matter and dark energy with negative mass : http://vixra.org/abs/0907.0015 Edited September 26, 2010 by icarus2 Link to comment Share on other sites More sharing options...
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