icarus2 Posted November 1 Author Share Posted November 1 (edited) 3.2.3.1. Why quantum fluctuations do not return to "nothing" and form the universe The existing model of the birth of the universe from nothing claims that the universe can be born from quantum fluctuations. However, the quantum fluctuations we know should return to "nothing" after a time of Δt. The existing model of the birth of the universe from nothing do not provide a reason or mechanism for the universe to be formed without quantum fluctuations returning to "nothing". Therefore, in the case where the universe is born from quantum fluctuations, a mechanism is needed that allows the quantum fluctuations to exist and not return to "nothing". Mechanism-1. If the total energy of the system, including the gravitational potential energy, is 0 or very close to 0. Quote If, Δt=t_p, ΔE=(5/6)m_pc^2, The total energy of the system is 0. In other words, a mechanism that generates enormous mass (or energy) while maintaining a Zero Energy State is possible. If ΔE_T --> 0, Δt --> ∞ Δt where quantum fluctuations exist can be very large. In other words, Δt can be larger than the current age of the universe, and these quantum fluctuations can exist longer than the age of the universe. Since the second mechanism changes the state of quantum fluctuations, it is thought that Δt does not necessarily have to be greater than the age of the universe. If we express the gravitational potential energy in the form including ΔE, If R = cΔt/2 ΔE_min means the minimum energy fluctuation that satisfies the equation ΔE≥hbar/2Δt. If Δt=t_P, U_gp ≥ -(3/5)ΔE_min Therefore, in this case, we must consider gravitational potential energy or gravitational self-energy. Therefore, If ΔE_T --> 0, Δt --> ∞ . Now, let's look at the approximate Δt that can be measured with current technology in the laboratory. We can see that gravitational potential energy term is very small compared to ΔE and can be ignored. In the case of a spherical uniform distribution, the total energy of the system, including the gravitational potential energy, is Therefore, we can see that the negative gravitational potential energy is very small in the Δt (much longer than the Planck time) that we observe in the laboratory, so the total energy of the system is sufficient only by ΔE excluding the gravitational potential energy, and the lifetime of the virtual particle is only a short time given by the uncertainty principle. If Δt>>t_P, Δt≥hbar/2ΔE_T ~ hbar/2ΔE Since E_T has some finite value other than 0, Δt cannot be an infinite value, but a finite value limited by ΔE_T. However, in the early universe, a relatively large Δt is possible because ΔE_T goes to zero, and as time passes and the range of gravitational interaction expands, if the surrounding quantum fluctuations participate in the gravitational interaction, an accelerated expansion occurs. Mechanism-2. Accelerated expansion due to negative energy or negative mass state In short, According to the uncertainty principle, it is possible to change (or create) more than (1/2)m_pc^2 energy during the Planck time, If an energy change above (5/6)m_pc^2 that is slightly larger than the minimum value occurs, the total energy of the mass-energy distribution reaches negative energy, i.e., the negative mass state, within the time Δt where quantum fluctuations can exist. However, since there is a repulsive gravitational effect between negative masses, the corresponding mass distribution expands instead of contracting. Thus, the quantum fluctuations generated by the uncertainty principle cannot return to nothing, but can expand and create the present universe. * Motion of positive mass due to negative gravitational potential energy, The force exerted by a negative (equivalent) mass on a positive mass is a repulsive (anti-gravity) force, so the positive mass accelerates and expands. The gravitational force acting between negative masses is attractive(m>0, F= - G(-m)(-m)/r^2 = - Gmm)/r^2), but since the inertial mass is negative in the case of negative mass, the gravitational effect is repulsive(m>0, F= (-m)a, a = - F/m ). So the distribution of negative energy or the distribution of negative equivalent mass is inflated. In a state of uniform energy distribution, when time passes, the radius of gravitational interaction increases. In this case, the mass energy increases in proportion to M, but the size of the gravitational potential energy increases in proportion to M^2/R. Therefore, since the negative gravitational potential energy increases faster than the positive mass energy, the phenomenon of accelerated expansion can occur. By combining mechanisms 1 and 2, we can simultaneously explain the existence of a universe born from quantum fluctuations without returning to "nothing", and the problem of inflation in the early universe. Edited November 1 by icarus2 Link to comment Share on other sites More sharing options...
swansont Posted November 1 Share Posted November 1 5 hours ago, icarus2 said: The existing model of the birth of the universe from nothing claims that the universe can be born from quantum fluctuations. However, the quantum fluctuations we know should return to "nothing" after a time of Δt Why? What requires that it be “nothing”? 5 hours ago, icarus2 said: If R = cΔt/2 How was this chosen? What if R were much larger? Link to comment Share on other sites More sharing options...
icarus2 Posted November 2 Author Share Posted November 2 11 hours ago, icarus2 said: 3.2.3.1. Why quantum fluctuations do not return to "nothing" and form the universe The existing model of the birth of the universe from nothing claims that the universe can be born from quantum fluctuations. However, the quantum fluctuations we know should return to "nothing" after a time of Δt. The existing model of the birth of the universe from nothing do not provide a reason or mechanism for the universe to be formed without quantum fluctuations returning to "nothing". Therefore, in the case where the universe is born from quantum fluctuations, a mechanism is needed that allows the quantum fluctuations to exist and not return to "nothing". Mechanism-1. If the total energy of the system, including the gravitational potential energy, is 0 or very close to 0. If ΔE_T --> 0, Δt --> ∞ Δt where quantum fluctuations exist can be very large. In other words, Δt can be larger than the current age of the universe, and these quantum fluctuations can exist longer than the age of the universe. Since the second mechanism changes the state of quantum fluctuations, it is thought that Δt does not necessarily have to be greater than the age of the universe. If we express the gravitational potential energy in the form including ΔE, If R = cΔt/2 ΔE_min means the minimum energy fluctuation that satisfies the equation ΔE≥hbar/2Δt. If Δt=t_P, U_gp ≥ -(3/5)ΔE_min Therefore, in this case, we must consider gravitational potential energy or gravitational self-energy. Therefore, If ΔE_T --> 0, Δt --> ∞ . Now, let's look at the approximate Δt that can be measured with current technology in the laboratory. We can see that gravitational potential energy term is very small compared to ΔE and can be ignored. In the case of a spherical uniform distribution, the total energy of the system, including the gravitational potential energy, is *I think there was a calculation error, so I just corrected the calculation. The core argument is the same. 3.2.3.1. Why quantum fluctuations do not return to "nothing" and form the universe The existing model of the birth of the universe from nothing claims that the universe can be born from quantum fluctuations. However, the quantum fluctuations we know should return to "nothing" after a time of Δt. The existing model of the birth of the universe from nothing do not provide a reason or mechanism for the universe to be formed without quantum fluctuations returning to "nothing". Therefore, in the case where the universe is born from quantum fluctuations, a mechanism is needed that allows the quantum fluctuations to exist and not return to "nothing". Mechanism-1. If the total energy of the system, including the gravitational potential energy, is 0 or very close to 0. If ΔE_T --> 0, Δt --> ∞ Δt where quantum fluctuations exist can be very large. In other words, Δt can be larger than the current age of the universe, and these quantum fluctuations can exist longer than the age of the universe. Since the second mechanism changes the state of quantum fluctuations, it is thought that Δt does not necessarily have to be greater than the age of the universe. If we express the gravitational potential energy in the form including ΔE, If R = cΔt/2 Therefore, in this case, we must consider gravitational potential energy or gravitational self-energy. Therefore, If ΔE_T --> 0, Δt --> ∞ . Now, let's look at the approximate Δt that can be measured with current technology in the laboratory. We can see that gravitational potential energy term is very small compared to ΔE and can be ignored. In the case of a spherical uniform distribution, the total energy of the system, including the gravitational potential energy, is Therefore, we can see that the negative gravitational potential energy is very small in the Δt (much longer than the Planck time) that we observe in the laboratory, so the total energy of the system is sufficient only by ΔE excluding the gravitational potential energy, and the lifetime of the virtual particle is only a short time given by the uncertainty principle. If Δt>>t_P, Δt≥hbar/2ΔE_T ~ hbar/2ΔE Since E_T has some finite value other than 0, Δt cannot be an infinite value, but a finite value limited by ΔE_T. However, in the early universe, a relatively large Δt is possible because ΔE_T goes to zero, and as time passes and the range of gravitational interaction expands, if the surrounding quantum fluctuations participate in the gravitational interaction, an accelerated expansion occurs. Mechanism-2. Accelerated expansion due to negative energy or negative mass state In short, According to the uncertainty principle, it is possible to change (or create) more than (1/2)m_pc^2 energy during the Planck time, If an energy change above (5/6)m_pc^2 that is slightly larger than the minimum value occurs, the total energy of the mass-energy distribution reaches negative energy, i.e., the negative mass state, within the time Δt where quantum fluctuations can exist. However, since there is a repulsive gravitational effect between negative masses, the corresponding mass distribution expands instead of contracting. Thus, the quantum fluctuations generated by the uncertainty principle cannot return to nothing, but can expand and create the present universe. * Motion of positive mass due to negative gravitational potential energy, The force exerted by a negative (equivalent) mass on a positive mass is a repulsive (anti-gravity) force, so the positive mass accelerates and expands. The gravitational force acting between negative masses is attractive(m>0, F= - G(-m)(-m)/r^2 = - Gmm)/r^2), but since the inertial mass is negative in the case of negative mass, the gravitational effect is repulsive(m>0, F= (-m)a, a = - F/m ). So the distribution of negative energy or the distribution of negative equivalent mass is inflated. In a state of uniform energy distribution, when time passes, the radius of gravitational interaction increases. In this case, the mass energy increases in proportion to M, but the size of the gravitational potential energy increases in proportion to M^2/R. Therefore, since the negative gravitational potential energy increases faster than the positive mass energy, the phenomenon of accelerated expansion can occur. By combining mechanisms 1 and 2, we can simultaneously explain the existence of a universe born from quantum fluctuations without returning to "nothing", and the problem of inflation in the early universe. =============== 5 hours ago, swansont said: Why? What requires that it be “nothing”? How was this chosen? What if R were much larger? I will write the answer to this question after I do what I have to do first. Within 1 day~ Link to comment Share on other sites More sharing options...
icarus2 Posted November 3 Author Share Posted November 3 On 11/2/2024 at 6:27 AM, swansont said: Why? What requires that it be “nothing”? How was this chosen? What if R were much larger? 1. If the quantum fluctuations that occur do not ultimately return to nothing but continue to exist, a situation will arise where some particles are born in the vacuum and continue to exist, which will break the law of conservation of energy. If this phenomenon has already been discovered in many laboratories, it would be an important issue and would have been reported. Also, isn't Δt generally treated as the lifetime of quantum fluctuations? 2. R=ct/2 This model uses the uncertainty principle and the concept of gravitational potential energy (or gravitational self-energy). From the uncertainty principle, the physical quantity that can be treated as the range of mass distribution seems to be Δx = 2R. Since the transmission speed of gravity is known to be the speed of light, it is thought that the model can be established as 2R = cΔt or R = cΔt from gravity. However, in the case of R = cΔt, -cΔt, 0, and +cΔt exist based on the center, but it seems difficult to see that -cΔt on the left enters into a gravitational interaction with +cΔt on the right. For this reason, I thought 2R=cΔt was a better rough approximation. However, it doesn't have to be completely accurate. It's the same reason why we don't use exactly ΔxΔP≥hbar/2 in problems where we apply the uncertainty principle, but rather use an estimate like ΔxΔP ~ hbar. In this problem, I think we can set up the following three models in relation to the distribution range of mass or energy. R=cΔt/2 : transmission range of gravity, R=Δx/2 : the uncertainty principle, R=λ/2=h/2p=h/2Δp : Interaction range from matter wave concept + uncertainty principle 1)R=cΔt/2 2)R=Δx/2 3)R=λ/2=h/2p=h/2Δp In all three cases, when Δt is near t_P, the total energy of quantum fluctuations can approach 0, and thus Δt can become very large (as large as the age of the universe). On the other hand, when Δt>>t_P, the total energy ΔE_T of quantum fluctuations cannot approach 0, and therefore has a relatively short finite time Δt. Since R is included in the denominator of the gravitational potential energy, when R increases while M is constant, the absolute value of the negative gravitational potential energy term decreases. It becomes difficult to achieve zero energy, and it also becomes difficult for the accelerated expansion of the mass distribution to occur. Link to comment Share on other sites More sharing options...
swansont Posted November 3 Share Posted November 3 12 hours ago, icarus2 said: 1. If the quantum fluctuations that occur do not ultimately return to nothing but continue to exist, a situation will arise where some particles are born in the vacuum and continue to exist, which will break the law of conservation of energy. If this phenomenon has already been discovered in many laboratories, it would be an important issue and would have been reported. Also, isn't Δt generally treated as the lifetime of quantum fluctuations? Interesting that you mention gravitational potential energy below but ignore it here. The requirement from the uncertainty principle is a return to zero energy (or ground state energy), not nothingness. So if the positive energy of the fluctuation is balanced by the negative gravitational potential energy, there is no violation. 12 hours ago, icarus2 said: 2. R=ct/2 This model uses the uncertainty principle and the concept of gravitational potential energy (or gravitational self-energy). From the uncertainty principle, the physical quantity that can be treated as the range of mass distribution seems to be Δx = 2R. Since the transmission speed of gravity is known to be the speed of light, it is thought that the model can be established as 2R = cΔt or R = cΔt from gravity. However, in the case of R = cΔt, -cΔt, 0, and +cΔt exist based on the center, but it seems difficult to see that -cΔt on the left enters into a gravitational interaction with +cΔt on the right. For this reason, I thought 2R=cΔt was a better rough approximation. What if expansion of space is not limited by c? (which it isn’t) Link to comment Share on other sites More sharing options...
Mordred Posted November 3 Share Posted November 3 (edited) The biggest problem here is that the universe can never have a state of quantum nothingness. Trying to solve for the quantum harmonic oscillator action can never be solved using any classical Newtonian method. I had already included the relevant equations previously to the quantum harmonic oscillator I was hoping you would have looked into them but apparently not. As this coincides with another thread on the harmonic oscillator vacuum catastrophe I'm going to port a worked solution showing the zero point energy ground state \[E_b=\sum_i(\frac{1}{2}+n_i)\hbar\omega_i\] where n_i is the individual modes n_i=(1,2,3,4.......) we can identify this with vacuum energy as \[E_\Lambda=\frac{1}{2}\hbar\omega_i\] the energy of a particle k with momentum is \[k=\sqrt{k^2c^2+m^2c^4}\] from this we can calculate the sum by integrating over the momentum states to obtain the vacuum energy density. \[\rho_\Lambda c^2=\int^\infty_0=\frac{4\pi k^2 dk}{(3\pi\hbar)^3}(\frac{1}{2}\sqrt{k^2c^2+m^2c^4})\] where \(4\pi k^2 dk\) is the momentum phase space volume factor. the effective cutoff can be given at the Planck momentum \[k_{PL}=\sqrt{\frac{\hbar c^3}{G_N}}\simeq 10^{19}GeV/c\] gives \[\rho \simeq \frac{K_{PL}}{16 \pi^2\hbar^3 c}\simeq\frac{10^74 Gev^4}{c^2(\hbar c)^3} \simeq 2*10^{91} g/cm^3\] compared to the measured Lambda term via the critical density formula \[2+10^{-29} g/cm^3\] method above given under Relativity, Gravitation and Cosmology by Ta-Pei Cheng page 281 appendix A.14 (Oxford Master series in Particle physics, Astrophysics and Cosmology). That's the calculations that led to the famous vacuum catastrophic. A large part of the corrections is renormalization methods for the reduced Hamilton and the phase summation using Pauli-Villars regularization. Renormalization procedures are not trivial unfortunately the mathematics gets quite complex Here is a paper for renormalizing the harmonic oscillator https://arxiv.org/abs/1311.6936 However a common method is dimensional regularization take the expression \[\int\frac{d^4p}{(p^2+m^2)^n}\] dimensional regularization replaces the integer with \[\int\frac{d^dp}{(p^2+m^2)^n}\] now what this allows us to do is eliminate divergences caused by integer vales of d which can be eliminated by non integer values of d (d being the power Edited November 3 by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Monday at 04:11 AM Share Posted Monday at 04:11 AM (edited) On 11/1/2024 at 9:35 AM, icarus2 said: 3.2.3.1. Why quantum fluctuations do not return to "nothing" and form the universe The existing model of the birth of the universe from nothing claims that the universe can be born from quantum fluctuations. However, the quantum fluctuations we know should return to "nothing" after a time of Δt. The existing model of the birth of the universe from nothing do not provide a reason or mechanism for the universe to be formed without quantum fluctuations returning to "nothing". Therefore, in the case where the universe is born from quantum fluctuations, a mechanism is needed that allows the quantum fluctuations to exist and not return to "nothing". Mechanism-1. If the total energy of the system, including the gravitational potential energy, is 0 or very close to 0. If ΔE_T --> 0, Δt --> ∞ Δt where quantum fluctuations exist can be very large. In other words, Δt can be larger than the current age of the universe, and these quantum fluctuations can exist longer than the age of the universe. Since the second mechanism changes the state of quantum fluctuations, it is thought that Δt does not necessarily have to be greater than the age of the universe. If we express the gravitational potential energy in the form including ΔE, If R = cΔt/2 ΔE_min means the minimum energy fluctuation that satisfies the equation ΔE≥hbar/2Δt. If Δt=t_P, U_gp ≥ -(3/5)ΔE_min Therefore, in this case, we must consider gravitational potential energy or gravitational self-energy. Therefore, If ΔE_T --> 0, Δt --> ∞ . Now, let's look at the approximate Δt that can be measured with current technology in the laboratory. We can see that gravitational potential energy term is very small compared to ΔE and can be ignored. In the case of a spherical uniform distribution, the total energy of the system, including the gravitational potential energy, is Therefore, we can see that the negative gravitational potential energy is very small in the Δt (much longer than the Planck time) that we observe in the laboratory, so the total energy of the system is sufficient only by ΔE excluding the gravitational potential energy, and the lifetime of the virtual particle is only a short time given by the uncertainty principle. If Δt>>t_P, Δt≥hbar/2ΔE_T ~ hbar/2ΔE Since E_T has some finite value other than 0, Δt cannot be an infinite value, but a finite value limited by ΔE_T. However, in the early universe, a relatively large Δt is possible because ΔE_T goes to zero, and as time passes and the range of gravitational interaction expands, if the surrounding quantum fluctuations participate in the gravitational interaction, an accelerated expansion occurs. Mechanism-2. Accelerated expansion due to negative energy or negative mass state In short, According to the uncertainty principle, it is possible to change (or create) more than (1/2)m_pc^2 energy during the Planck time, If an energy change above (5/6)m_pc^2 that is slightly larger than the minimum value occurs, the total energy of the mass-energy distribution reaches negative energy, i.e., the negative mass state, within the time Δt where quantum fluctuations can exist. However, since there is a repulsive gravitational effect between negative masses, the corresponding mass distribution expands instead of contracting. Thus, the quantum fluctuations generated by the uncertainty principle cannot return to nothing, but can expand and create the present universe. * Motion of positive mass due to negative gravitational potential energy, The force exerted by a negative (equivalent) mass on a positive mass is a repulsive (anti-gravity) force, so the positive mass accelerates and expands. The gravitational force acting between negative masses is attractive(m>0, F= - G(-m)(-m)/r^2 = - Gmm)/r^2), but since the inertial mass is negative in the case of negative mass, the gravitational effect is repulsive(m>0, F= (-m)a, a = - F/m ). So the distribution of negative energy or the distribution of negative equivalent mass is inflated. In a state of uniform energy distribution, when time passes, the radius of gravitational interaction increases. In this case, the mass energy increases in proportion to M, but the size of the gravitational potential energy increases in proportion to M^2/R. Therefore, since the negative gravitational potential energy increases faster than the positive mass energy, the phenomenon of accelerated expansion can occur. By combining mechanisms 1 and 2, we can simultaneously explain the existence of a universe born from quantum fluctuations without returning to "nothing", and the problem of inflation in the early universe. have you ever thought to look into zero energy universe ? https://arxiv.org/abs/gr-qc/0605063 you may find this helpful still looking if I still have a copy of one of the earlier Universe from nothing papers I once had a copy of Edited Monday at 04:27 AM by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Monday at 05:58 AM Share Posted Monday at 05:58 AM (edited) found a fundamental flaw with your total energy equation You have only calculated the Potential energy term and didn't include any term for the kinetic energy. The primary mistake results from just using e=mc^2 instead of the full energy momentum relation. You should be using this as your basis equation for the field \[E=\sqrt{(pc)^2+(m_o c^2)^2}\] the \(\frac{3}{5}\frac{GM^2}{R}\) is just potential energy it doesn't include any momentum its just from the mass term. This equation is accurate for the potential energy but not as the total energy your subtracting two equivalent terms by the operation \[E_t=mc^2-\frac{3}{5}\frac{GM^2}{R}=0\] would be the result it can only give zero the second term is a derivative of the first term. So the result can ONLY be zero. Under sphere treatment as the sphere expands via Hubble constant you do not have the equations of motion just the mass term Is that your intent ? if so what gives rise to any later energy content ? If not and if you like I can provide the kinetic energy term as a Newtonian treatment which you seemingly prefer for an expanding Universe. however if it is your intent then it would the equivalent of zero particles for any kinetic energy term as the result would always be zero nor would you have a mass to begin with and no remaining kinetic energy term to drive expansion which does not make any sense whatsoever. A total energy balance of zero is an empty universe with no component to drive expansion. Doesn't work You need to incorporate the kinetic energy side of the gravitational field as a result of expansion as well for total energy (hint the kinetic term can be derived using Hubble constant) if you want I will post how upon request. I know a Newtonian treatment for that. The essence however is that as the universe expands driven by the equations of state the kinetic term becomes involved as required to maintain a homogeneous and isotropic mass distribution. It is that term that is missing. Edited Monday at 11:33 PM by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Monday at 11:45 PM Share Posted Monday at 11:45 PM (edited) Here is how total energy can be calculated it took me a bit to find a methodology that matches your \[\frac{3}{5}\frac{GM^2}{R}\] relation The author also has that same relation but for the potential energy here is a dissertation paper where the author has the identical term for potential energy. The kinetic energy term is derived under Newtonian treatment "Total Energy of the Freidmann-Robertson_Walker Universes " by Maria Mouland https://core.ac.uk/download/pdf/30798226.pdf specifically arriving at \[\frac{3}{10}MH^2R^2\] The inclusion of both terms is what you need for total energy She shows further down how the two terms arrive at equation 1.28 \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] which is a well known relation used in the critical density formula. (proof of methodology ) it is a matter only treatment which is fine w=0 as the critical density formula is also a matter only treatment. If I recall correctly Peebles also had a similar treatment in one of his Introductory to cosmology textbooks. I do know he had the 3/5 relation in his density contrast term of his Large scale structure of the Universe article. Anyways hopefully you find the above article useful Edited Tuesday at 01:13 AM by Mordred Link to comment Share on other sites More sharing options...
icarus2 Posted Tuesday at 08:59 AM Author Share Posted Tuesday at 08:59 AM (edited) On 11/11/2024 at 2:58 PM, Mordred said: found a fundamental flaw with your total energy equation You have only calculated the Potential energy term and didn't include any term for the kinetic energy. The primary mistake results from just using e=mc^2 instead of the full energy momentum relation. You should be using this as your basis equation for the field E=(pc)2+(moc2)2−−−−−−−−−−−−√ the 35GM2R your subtracting two equivalent terms by the operation Et=mc2−35GM2R=0 would be the result it can only give zero the second term is a derivative of the first term. So the result can ONLY be zero. Under sphere treatment as the sphere expands via Hubble constant you do not have the equations of motion just the mass term Is that your intent ? if so what gives rise to any later energy content ? If not and if you like I can provide the kinetic energy term as a Newtonian treatment which you seemingly prefer for an expanding Universe. however if it is your intent then it would the equivalent of zero particles for any kinetic energy term as the result would always be zero nor would you have a mass to begin with and no remaining kinetic energy term to drive expansion which does not make any sense whatsoever. A total energy balance of zero is an empty universe with no component to drive expansion. Doesn't work You need to incorporate the kinetic energy side of the gravitational field as a result of expansion as well for total energy (hint the kinetic term can be derived using Hubble constant) if you want I will post how upon request. I know a Newtonian treatment for that. The essence however is that as the universe expands driven by the equations of state the kinetic term becomes involved as required to maintain a homogeneous and isotropic mass distribution. It is that term that is missing. As I explained before, in our time, when using the expression of rest mass, the expression “m_0” was mostly used. This is the notation of Marion, a representative classical mechanics book, and also the notation of Beiser’s book, which is used as a modern physics textbook. It also appears in the expression you wrote. And, you can see it in my previous paper. https://www.researchgate.net/publication/263468413_Is_the_State_of_Low_Energy_Stable_Negative_Energy_Dark_Energy_and_Dark_Matter Mc^2 is the total energy of objects with positive energy, and M is the relativistic mass or equivalent mass. On 11/11/2024 at 1:11 PM, Mordred said: have you ever thought to look into zero energy universe ? https://arxiv.org/abs/gr-qc/0605063 you may find this helpful still looking if I still have a copy of one of the earlier Universe from nothing papers I once had a copy of As I explained before, I do not think that the total energy of the universe is zero energy. I claim that the total energy of the observable universe is in a negative energy state. I claim that the universe is experiencing accelerated expansion because the negative energy of the observable universe is greater than the positive energy. I also think that this is the hidden truth behind standard cosmology, and that the mainstream is missing it because of their preconceptions about “negative mass”. 1.The first result of Friedmann equation for accelerated expansion was negative mass density Nobel lecture by Adam Riess : The official website of the Nobel Prize Refer to time 11m : 35s ~ https://www.nobelprize.org/mediaplayer/?id=1729 ===== Negative Mass? Actually the first indication of the discovery! *This is a phrase that appears on Adam Riess' presentation screen. ===== HSS(The High-z Supernova Search) team : if Λ=0, Ω_m = - 0.38(±0.22) : negative mass density SCP(Supernova Cosmology Project) team : if Λ=0, Ω_m = - 0.4(±0.1) : negative mass density *This value is included in a paper awarded the Nobel Prize for the discovery of the accelerated expansion of the universe. In the acceleration equation, (c=1) (1/R)(d^2R/dt^2) = -(4πG/3)(ρ+3P) In order for the universe to expand at an accelerated rate, the right side must be positive, and therefore (ρ+3P)must be negative. In other words, a negative mass density is needed for the universe to expand at an accelerated rate. They had negative thoughts about negative mass (negative energy). So, they discarded the negative mass (density). They corrected the equation and argued that the accelerated expansion of the universe was evidence of the existence of a cosmological constant. However, the vacuum energy model has not succeeded in explaining the value of dark energy density, and the source of dark energy has not yet been determined. They introduce negative pressure, which hides the negative mass density in the negative pressure, but this does not mean that the negative mass density has disappeared. ρ_Λ + 3P_Λ = ρ_Λ + 3(-ρ_Λ) = - 2ρ_Λ If we expand the dark energy term, the final result is a negative mass density of -2ρ_Λ. 2. Logical structure of the standard cosmology We need to look at the logic behind the success of standard cosmology. Let's look at the equation expressing (ρ+3P) as the critical density of the universe. In the second Friedmann equation (c=1), (1/R)(d^2R/dt^2) = -(4πG/3)(ρ+3P) Matter + Dark Matter (approximately 31.7%) = ρ_m ~ (1/3)ρ_c Dark energy density (approximately 68.3%) = ρ_Λ ~ (2/3)ρ_c (Matter + Dark Matter)'s pressure = 3P_m ~ 0 Dark energy’s pressure = 3P_Λ = 3(-(2/3)ρ_c ) = -2ρ_c ρ+3P≃ ρ_m +ρ_Λ +3(P_m +P_Λ)= (1/3)ρ_c +(2/3)ρ_c +3(−2/3)ρ_c= (+1)ρ_c + (-2)ρ_c = (−1)ρ_c ρ+3P ≃ (+1)ρ_c + (-2)ρ_c = (−1)ρ_c The logic behind the success of the ΛCDM model is a universe with a positive mass density of (+1)ρ_c and a negative mass density of (-2)ρ_c. So, finally, the universe has a negative mass density of “(-1)ρ_c”, so accelerated expansion is taking place. The current universe is similar to a state where the negative mass (energy) density is twice the positive mass(energy) density. And the total mass of the observable universe is the negative mass state. 3. So, what can correspond to this negative mass density? When mass or energy is present, the negative gravitational potential energy(gravitational self-energy or gravitational binding energy) produced by that mass or energy distribution can play a role. The gravitational potential energy model is a model in which +ρ(Positive energy density) and -ρ_gp(Gravitational potential energy density) exist, and shows that |-ρ_gp| can be larger than +ρ. [Result summary] Use a critical density ρ_c=8.50x10^-27[kgm^-3] At R=16.7Gly, (-M_gp)c^2 = (-0.39)Mc^2 |(-M_gp)c^2| < (Mc^2) : Decelerating expansion period At R=26.2Gly, (-M_gp)c^2 = (-1.00)Mc^2 |(-M_gp)c^2| = (Mc^2) : Inflection point (About 5-7 billion years ago, consistent with standard cosmology.) At R=46.5Gly, (-M_gp)c^2 = (-3.04)Mc^2 |(-M_gp)c^2| > (Mc^2) : Accelerating expansion period Even in the universe, gravitational potential energy (or gravitational action of the gravitational field) must be considered. And, in fact, if we calculate the value, since gravitational potential energy is larger than mass energy, so the universe has accelerated expansion. The gravitational potential energy model explains decelerating expansion, inflection points, and accelerating expansion. As the universe ages, the range of gravitational interactions increases, and thus accelerated expansion appears to have occurred because the negative gravitational potential energy increased faster than the positive mass energy. As the universe grows older, the range R of gravitational interaction increases. As a result, mass energy increases in proportion to M, but gravitational potential energy increases in proportion to -M^2/R. Therefore, gravitational potential energy increases faster. Therefore, as the universe ages and the range of gravitational interaction expands, the phenomenon of changing from decelerated expansion to accelerated expansion occurs. https://www.researchgate.net/publication/360096238_Dark_Energy_is_Gravitational_Potential_Energy_or_Energy_of_the_Gravitational_Field ----------- The accelerating expansion mechanism or inflation mechanism 1. Let’s assume that individual quantum fluctuations are born from zero energy. In this figure, the total energy of the individual quantum fluctuations that are born is 0. In figure (b), you can think of each point as an individual quantum fluctuation. For the first Δt time, it only interacts with itself gravitationally. However, as time passes, the radius of the gravitational interaction will grow to R1, R2, and R3. That is, more quantum fluctuations will enter the range of gravitational interaction, and accordingly, the total energy will change. If the mass-energy within the radius R_1 interacted gravitationally at t_1 (an arbitrary early time), the mass-energy within the radius R_2 will interact gravitationally at a later time t_2. As the universe ages, the mass-energy involved in gravitational interactions changes, resulting in changes in the energy composition of the universe. The total energy E_T of the system is In the case of a uniform distribution, comparing the magnitudes of mass energy and gravitational potential energy, it is in the form of -kR^2. That is, as the gravitational interaction radius increases, the negative gravitational potential energy value (Absolute value) becomes larger than the positive mass energy. Thus, the universe enters a period of accelerating expansion and inflation. 2.Even for a single quantum fluctuation, accelerating expansion can occur immediately, depending on the size of ΔE that changes during Δt Edited Tuesday at 09:15 AM by icarus2 Link to comment Share on other sites More sharing options...
Mordred Posted Tuesday at 01:05 PM Share Posted Tuesday at 01:05 PM (edited) That's clearly not shown in your mathematics The article I linked shows precisely how \[\frac{3}{5}\frac{GM}{R}\] is derived using an expanding volume. So your expression gives a balance of precisely zero at all times. If that is not your intent then that expression is wrong for your purposes. The relation above is specifically a derivative of the mass density evolution for an expanding volume clearly shown in the article I linked. specifically \[E_t=MC^2-\frac{3}{5}\frac{GM}{R}=0\] Now your telling us that your applying negative mass which under GR itself is considered impossible. If you ever applied a force to a negative mass and Newtons laws to negative mass you have inverse relations Specifically the more force applied the less displacement you get. All you need to get expansion is to have a kinetic energy term to exceed the positive energy term as per the scale field equation of state used in Cosmology. In essence pressure the energy density relations. That is how the FLRW metric works Is this paper not suppose to show how a universe from NOTHING is possible ? Is that not the claim of the paper ? the term mass energy if garbage by the way. mass is resistance to inertia change if an object has mass it has resistance to inertia change or acceleration. its already been mentioned energy DOES NOT exist on its own. It is a property just as mass is a property Your attempt to treat mass energy as something seperate from mass is simply wrong by any definition of mass. The definition for mass should make the inverse relations to force obvious had you considered the definition of mass with regards to negative mass. And please don't tell me your going to disagree with the very definition of mass or energy (ability to perform work) that has been around since the 16th century and how they apply to Newtons laws of inertia If you do then you better get busy rewriting the entirety of every physics theory since then. Edited Tuesday at 01:19 PM by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Tuesday at 07:28 PM Share Posted Tuesday at 07:28 PM (edited) I should have added "Where is this negative mass you claim exists ? No standard model particle has negative mass and even DM doesn't match the characteristics of negative mass. DE matches the characteristics of negative pressure but not a matter field. Just a side note \(E^2=(pc)^2+(m_o c^2)^2\) is the energy momentum relation if you place a minus sign in front of the RHS your not stating negative mass your stating negative momentum. Not even antiparticles has negative mass nor momentum. So the questions still remains name any particle with negative mass or momentum ? Good luck on that question I certainly know I can't think of any particle with negative coupling strength can you ? You may want to rethink your other article as well as this one. That's just a strong suggestion though Particularly since in your E- and E+ momentum that would be the equivalent to two seperate fields one collapsing while the other is expanding. If you don't believe that statement try using the stress energy momentum tensor with negative momentum with the Einstein field equations. Would not the result be two opposite curvature terms? Or did you even consider that problem ? When you apply thermodynamics to that, one field would be cooling down while the other is heating up. Not what we see is it ? What is obvious to me is that you never considered your proposals beyond Rudimentary Newtonian treatment nor considered the ramifications under a field treatment such as under GR. Let alone the absolute lack of any standard model particle to meet your criteria. If only negative mass was that easy we would already have an Alcubierre warp drive Edited Wednesday at 02:19 AM by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Wednesday at 02:28 AM Share Posted Wednesday at 02:28 AM (edited) Don't worry I fully expect you to search through internet looking for every possible instance of negative mass treatments they always boil down to not understanding how that negative mass is applied to a global vs local field treatment. What you are describing is a global treatment not local to a non zero global baseline. Example a solid state lattice network with negative and positive holes. Global energy value with negative local energy compared to the global energy value. If you had a non- zero baseline that would run counter to a Universe from nothing.... I will state this once again Energy is the ability to perform work a property... it does not exist on its own. Mass is resistance to inertia change hence the momentum terms.... As we are describing a Universe any theory must work as a field treatment such as GR. (Good luck on this point). For Newtonian case try it as a momentum vector field not simply scalar lol it would get trickier when you apply contravariant terms and anti- commutations... Lol after all I should be able to throw any SM particle into this universe and its subsequent equations of motion will work under multi particle field treatment. Hence all the above statements Edited Wednesday at 03:07 AM by Mordred Link to comment Share on other sites More sharing options...
icarus2 Posted 20 hours ago Author Share Posted 20 hours ago On 11/12/2024 at 10:05 PM, Mordred said: That's clearly not shown in your mathematics The article I linked shows precisely how 35GMR specifically Et=MC2−35GMR=0 Now your telling us that your applying negative mass which under GR itself is considered impossible. If you ever applied a force to a negative mass and Newtons laws to negative mass you have inverse relations Specifically the more force applied the less displacement you get. All you need to get expansion is to have a kinetic energy term to exceed the positive energy term as per the scale field equation of state used in Cosmology. In essence pressure the energy density relations. That is how the FLRW metric works Is this paper not suppose to show how a universe from NOTHING is possible ? Is that not the claim of the paper ? the term mass energy if garbage by the way. mass is resistance to inertia change if an object has mass it has resistance to inertia change or acceleration. its already been mentioned energy DOES NOT exist on its own. It is a property just as mass is a property Your attempt to treat mass energy as something seperate from mass is simply wrong by any definition of mass. The definition for mass should make the inverse relations to force obvious had you considered the definition of mass with regards to negative mass. And please don't tell me your going to disagree with the very definition of mass or energy (ability to perform work) that has been around since the 16th century and how they apply to Newtons laws of inertia If you do then you better get busy rewriting the entirety of every physics theory since then. The gravitational self-energy (or gravitational binding energy) equation is in most classical mechanics books, so there is no need to derive it separately. https://en.wikipedia.org/wiki/Gravitational_binding_energy The article you linked to is a mainstream model, and is different from my model. That model omits the gravitational self-energy that M itself has. In the following article, I will roughly explain the acceleration equation according to my model. On 11/13/2024 at 4:28 AM, Mordred said: I should have added "Where is this negative mass you claim exists ? No standard model particle has negative mass and even DM doesn't match the characteristics of negative mass. DE matches the characteristics of negative pressure but not a matter field. On 11/13/2024 at 11:28 AM, Mordred said: Don't worry I fully expect you to search through internet looking for every possible instance of negative mass treatments they always boil down to not understanding how that negative mass is applied to a global vs local field treatment. What you are describing is a global treatment not local to a non zero global baseline. Example a solid state lattice network with negative and positive holes. Global energy value with negative local energy compared to the global energy value. If you had a non- zero baseline that would run counter to a Universe from nothing.... I will state this once again Energy is the ability to perform work a property... it does not exist on its own. Mass is resistance to inertia change hence the momentum terms.... As we are describing a Universe any theory must work as a field treatment such as GR. (Good luck on this point). For Newtonian case try it as a momentum vector field not simply scalar lol it would get trickier when you apply contravariant terms and anti- commutations... Lol after all I should be able to throw any SM particle into this universe and its subsequent equations of motion will work under multi particle field treatment. Hence all the above statements People generally react that way when they encounter something different from their existing paradigm. One such object is “negative mass.” But don’t worry! There is a word prepared for such people. Equivalent, Effective Please think of it as negative equivalent mass, effective negative mass, not negative mass. In this model, particles, stars, and galaxies all have positive mass and positive energy. These positive masses create gravitational potential energy with a negative value. Therefore, the system, including the potential energy, can reach a negative energy state. However, the average density of negative gravitational potential energy is very low. Therefore, individual particles, stars, and galaxies are still in a state of positive mass. The universe (system) that includes them all is in a state of negative energy, and therefore, only the observable universe has a state of negative equivalent mass. Since individual particles and galaxies are still in a positive mass state, so if you're overreacting to the term "negative mass," please feel safe. ------------ Based on the gravitational potential energy model, the acceleration equation is derived as follows: The Friedmann equation can be obtained from the field equation. The basic form can also be obtained through Newtonian mechanics. If the object to be analyzed has spherical symmetry, from the second Newton’s law, R(t) = a(t)R By adding pressure, we can create an acceleration equation. Let’s look at the origin of mass density ρ here! What does ρ come from? It comes from the total mass M inside the shell. The universe is a combined state because it contains various matter, radiation, and energy. Therefore, the total mass m^* including the binding energy must be entered, not the mass “2m” in the free state.“m^∗ = 2m + (−m_gp)”, i.e. gravitational potential energy must be considered. We may need to correct the value for some reason, so let's calculate it by inserting the correction factor β(t). −ρ_gp is the equivalent mass density of gravitational potential energy, Instead of mass density ρ, we have (ρ) + (ρ_gp). So, Dark energy term and Cosmological constant term is *There is no cosmological constant in the gravitational potential energy model. However, since the existing standard model was built based on the cosmological constant model, it is expressed in the form of a cosmological constant term for comparison. If we obtain the cosmological constant and dark energy terms through this model, Please look at the case when there was no correction factor at all, that is, the pure predicted value. β=1 You can see that it is almost identical to the current cosmological constant and dark energy density. In other words, the gravitational potential energy model is 10^120 times more accurate than the vacuum energy model! (It is close to the dark energy density value.) In addition, since it shows that the dark energy density is a function of time, it is different from the ΛCDM model and can be verified. The current standard cosmology is the ΛCDM model. This model explains the dark energy problem by introducing a entity that has positive energy density and exerts negative pressure. However, cracks in this ΛCDM model have been increasing recently, and accordingly, the time is approaching when other possibilities and interpretations should be seriously examined. The ΛCDM model consists of Λ(Lambda) and CDM(Cold Dark Matter), but it has not succeeded in explaining the source of Λ, which is the core, and the vacuum energy, which was a strong candidate, has an unprecedented error of 10^120 between the theoretical value and the observed value. CDM has also not been discovered despite continued experiments. In addition, particle accelerator experiments, which are a completely different approach from WIMP detectors, have not found a candidate for CDM. Furthermore, the existence of the Hubble tension problem has recently become clear, and in January 2024, the DES (Dark Energy Survey) team raised the possibility that the cosmological constant model may be wrong, and in April 2024, the DESI (Dark Energy Spectroscopic Instrument) team announced results suggesting that the cosmological constant model may be wrong. Since it is only the first year of results, we will have to wait a little longer for the final results. 1)Dark Energy Survey team https://noirlab.edu/public/news/noirlab2401/?lang The standard cosmological model is known as ΛCDM, or ‘Lambda cold dark matter’. This mathematical model describes how the Universe evolves using just a few features such as the density of matter, the type of matter and the behavior of dark energy. While ΛCDM assumes the density of dark energy in the Universe is constant over cosmic time and doesn’t dilute as the Universe expands, the DES Supernova Survey results hint that this may not be true. An intriguing outcome of this survey is that it is the first time that enough distant supernovae have been measured to make a highly detailed measurement of the decelerating phase of the Universe, and to see where the Universe transitions from decelerating to accelerating. And while the results are consistent with a constant density of dark energy in the Universe, they also hint that dark energy might possibly be varying. “There are tantalizing hints that dark energy changes with time,” said Davis, “We find that the simplest model of dark energy — ΛCDM — is not the best fit. It’s not so far off that we’ve ruled it out, but in the quest to understand what is accelerating the expansion of the Universe this is an intriguing new piece of the puzzle. A more complex explanation might be needed.” 2)Dark Energy Spectroscopic Instrument team https://arstechnica.com/science/2024/04/dark-energy-might-not-be-constant-after-all/#gsc.tab=0 "It's not yet a clear confirmation, but the best fit is actually with a time-varying dark energy," said Palanque-Delabrouille of the results. "What's interesting is that it's consistent over the first three points. The dashed curve [see graph above] is our best fit, and that corresponds to a model where dark energy is not a simple constant nor a simple Lambda CDM dark energy but a dark energy component that would vary with time. The gravitational potential energy model creates a repulsive force on a cosmic scale, and it also explains the dark energy density value, so people better than me need to study this model more seriously. Dark Energy is Gravitational Potential Energy or Energy of the Gravitational Field Link to comment Share on other sites More sharing options...
Mordred Posted 19 hours ago Share Posted 19 hours ago (edited) Yes I know the first equation is just the potential energy. However under mainstream physics so is the invariant rest mass given by E=mc^2. So my argument still stands your subtracting potential energy from potential energy giving the result of zero. However in your article you claim that it is total energy and not potential energy and are ignoring kinetic energy terms for an expanding volume. Just because your model isn't mainstream doesn't mean it shouldn't match mainstream physics when it comes to the rudimentary physics under Newton treatment. (For the record stating someone's conjecture isn't mainstream is one of the most common excuses used on this forum) it's a very common and lame argument. Potential and kinetic energy relations has been mainstream since Newton if it can stand that test of time applicability then obviously it's a very robust and accurate methodology. You most certainly will not solve the cosmological constant nor any equation of state without including kinetic energy terms it's literally impossible as those terms involve kinetic energy for its momentum equivalence. If your theory has any motion or expansion of a matter field you need PE and KE. Hopefully you dropped negative mass as that has numerous ramifications you haven't considered. Edited 19 hours ago by Mordred Link to comment Share on other sites More sharing options...
icarus2 Posted 19 hours ago Author Share Posted 19 hours ago (edited) 27 minutes ago, Mordred said: Yes I know the first equation is just the potential energy. However under mainstream physics so is the invariant rest mass given by E=mc^2. So my argument still stands your subtracting potential energy from potential energy giving the result of zero. However in your article you claim that it is total energy and not potential energy and are ignoring kinetic energy terms for an expanding volume. Just because your model isn't mainstream doesn't mean it shouldn't match mainstream physics when it comes to the rudimentary physics under Newton treatment. (For the record stating someone's conjecture isn't mainstream is one of the most common excuses used on this forum) it's a very common and lame argument. Potential and kinetic energy relations has been mainstream since Newton if it can stand that test of time applicability then obviously it's a very robust and accurate methodology. You most certainly will not solve the cosmological constant nor any equation of state without including kinetic energy terms it's literally impossible as those terms involve kinetic energy for its momentum equivalence. If your theory has any motion or expansion of a matter field you need PE and KE. Hopefully you dropped negative mass as that has numerous ramifications you haven't considered. I can't understand what you're saying. In E_T=Mc^2 + (-3/5)GM^2/R, you seem to be saying that the Mc^2 term is also a potential energy term~ What does this mean? Are you thinking that I tried to write the E=E_k + E_p term, but made a mistake? Edited 18 hours ago by icarus2 Link to comment Share on other sites More sharing options...
Mordred Posted 18 hours ago Share Posted 18 hours ago (edited) Of course it is its the rest mass\invariant mass The momentum is p \[E=p+m^2\] using normalized units where c=1. The kinetic energy term isn't mc^2 its the momentum given by p If your using just PE your system is static with no particle motion. Edited 18 hours ago by Mordred Link to comment Share on other sites More sharing options...
icarus2 Posted 18 hours ago Author Share Posted 18 hours ago What did you read in my writing that made you think I tried to write E=E_k + E_p? Link to comment Share on other sites More sharing options...
Mordred Posted 18 hours ago Share Posted 18 hours ago (edited) Why would I think that when I'm referring to the first equation on your article. Where you have total energy but only included potential energy terms That's where the error lies your total energy formula is wrong. If the first formula is wrong any later formulas are likely incorrect as well. Edited 18 hours ago by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted 16 hours ago Share Posted 16 hours ago (edited) This is the equation you have but every term is just the potential energy without the kinetic energy term \[{E_T} = \sum\limits_i {{m_i}{c^2}} + \sum\limits_{i < j} { - \frac{{G{m_i}{m_j}}}{{{r_{ij}}}}} = M{c^2} - \frac{3}{5}\frac{{G{M^2}}}{R}\] Every term above is just potential energy so your subtracting PE from PE which is automatically zero. The article I included shows this relation. the Kinetic energy term \[\frac{3}{10}MH^2R^2\] arises from expansion and the movement of the mass distribution for an expanding volume. and the combination of the two terms for total energy in the Article and not your equation gives us the energy density we see in the FLRW metric \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] which will work even for all three curvature terms by the inclusion of k. Your method doesn't work to give the correct energy density Study the method in this article its why I included it to begin with. https://core.ac.uk/download/pdf/30798226.pdf The author show this is correct by the following \[E_t=E_k+E_p=\frac{3}{10}MH^2R^2-\frac{3}{5}\frac{GM}{R}\] \[=\frac{3M}{10}(H^2R^2-\frac{2GM}{R})\] \[=\frac{3M}{10}(H^2R^2-\frac{2G}{R}\frac{4\pi}{3}R^3\rho\] \[=\frac{3MR^2}{10}(H^2-\frac{8\pi G}{3}\rho)\] requiring E to be constant=conserved mass is already conserved in the above \[H^2-\frac{8\pi G}{3}\rho=\frac{10E}{3M}\frac{1}{R^2}\rightarrow H^2-\frac{10E}{3M}\frac{1}{R^2}=\frac{8\pi G}{3}\rho\] where \[H=\frac{1}{R}\frac{DR}{dt}=\frac{\dot{R}}{R}\] which is the velocity time derivative for expansion. With curvature under GR and k for curvature= constant \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] this is section 1.24 to 1.28 this above correctly includes the KE and PE terms for mass distribution of an expanding volume. I could tell you didn't really read the article as you missed the opening statement. "I will look at ’the total energy of the universe’. This is an interesting issue, because if the energy of the universe turns out not to be conserved, it will be in conflict with our common understanding of energy. So intuitively we expect the energy of the universe to be constant. Furthermore, if this total energy is constant and zero, it means that ’creating’ a universe does not require any energy. Such a universe could then, in principle, just ’pop up’ from nothing. Our universe is dominated by a so-called cosmological constant, or vacuum energy. It has the property that the energy density is constant in volume, so when the universe expands, the total amount of vacuum energy increases. Where does this new energy come from? One might immediately think that it could be energy from other components in the universe that is converted into vacuum energy. But it turns out not to be that simple, since the vacuum energy increases also for universe models which contain vacuum energy only. For flat (Minkowski) spacetimes, a global energy conservation law can be set up without problems. But we know that we need the general theory of relativity to describe the universe more realisticly. General relativity deals with curved spacetimes, and then it is in general not possible to set up a global law for conservation of energy." in essence the paper is an examination of the plausibility and consequences of Universe from Nothing based models She states energy conservation can work in the Newtonian case however will not work in curved spacetimes. This is an identical problem that plagued the zero energy universe model. The model only worked for flat spacetime. It is those curvature terms that will kill your model as well as your simply looking at the Newtonian case without examining curved spacetimes. We do not live in a critically dense universe our universe has a small curvature term that only approximates flat. The paper clearly looks at the issue of Lambda aka the cosmological constant. The very problem of Lambda being constant is something of a large issue with conservation of mass energy. The Newtonian method will not address this as shown in the paper and please don't claim your two papers solves that issue. Your not even close to solving that issue Edited 14 hours ago by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted 13 hours ago Share Posted 13 hours ago As an assist here is a recent Universe from nothing model based of the wavefunction of the universe Wheeler DeWitt. https://arxiv.org/abs/1404.1207 Might help Link to comment Share on other sites More sharing options...
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