icarus2

In the shell theorem, the force inside the sphere is 0. However, the gravitational potential energy inside the sphere is not 0. Therefore, the sum of the gravitational potential energies created by each sphere is not 0, and exists. That is the gravitational self-energy equation. The universe is a negative mass state, so it is a universe dominated by repulsion, not attraction. Therefore, it does not collapse due to self-gravity. In the early days of the universe, I showed that E_T can be 0. I also explained that E_T can be in a negative energy state less than 0. Since I proved Zero Energy, it is easy to infer that it creates a negative mass state. Let's assume that 10 Planck masses are created during the Planck time. It becomes a state of negative mass and negative energy. Since there is a repulsive gravitational effect between negative masses, the universe cannot contract but expands. Therefore, the universe exists without collapsing, and it can explain the accelerated expansion of the early universe. Even though a single quantum fluctuation is set to 0, as time passes, the surrounding quantum fluctuations participate in gravitational interactions, so the negative gravitational potential energy increases faster than the positive mass energy, so it can explain that the total energy of the system becomes a state of negative energy and expands. The above mechanism explains the mechanism of the early universe expanding without collapsing. Now, let's look at the current observable universe. If we calculate the gravitational potential energy of the current observable universe, we get a value of approximately -3.04 times the mass energy. The current gravitational potential energy of the universe is approximately 2.04 times greater than the mass energy. Therefore, the current universe is also in a negative mass state, and therefore, the universe does not collapse. This is also consistent with the hidden result of standard cosmology. Why doesn't the universe collapse? The universe is in a state of negative energy, so it didn't collapse because anti-gravity is stronger than gravity~ Are you not reading my writing? I have proven the value of dark energy density through my model, and In a rough estimate without correction factor, the cosmological constant value by the gravitational potential energy model is 2.45x10^-52m^-2, which is very close to the observed value. While the value of dark energy was proven through the gravitational potential energy model, this paper introduces quantum fluctuations, and one may wonder whether this also creates a vacuum catastrophe problem. Here are some points that are different from the mainstream vacuum catastrophe problem occurrence and my personal opinion. 1) I am negative about applying zero point energy derived from simple harmonic oscillation to the universe. In the SHO model of quantum mechanics, there is a potential V with a negative value, and the width of the well also exists as a function of the potential V, so in this case, zero point oscillation will exist, but I think there is no basis for the existence of a potential V with a negative value in the universe and for limiting the width of the well to the Planck length. Since the vacuum of the universe is likely to have similar energy values, I think there is a problem with assuming that there is a potential V with a negative value and that it will exist at every Planck length, as the mainstream estimates. 2) In the mainstream vacuum catastrophe problem, the mainstream does not consider gravitational potential energy, that is, gravitational self-energy at all, so the vacuum has an enormous energy density and this becomes a problem. However, in my model, even in the case of quantum fluctuations, there is a gravitational source ΔE and a time Δt for gravity to act, so I have said several times that gravitational self-energy should be considered. And, considering the gravitational self-energy, we proved that near t=t_P, the total energy, including the gravitational potential energy, and the total mass can take the value 0. That is, I showed that the energy density of the vacuum can be 0 or close to 0. Here is one possibility. If, Δt=t_p, ΔE=(5/6)m_P, (5/6)m_P is near m_P. 3) In the previous article 3.6(3.6. The Vacuum Catastrophe Problem or Cosmological Constant Problem), when the mainstream claims that the vacuum energy density exists, we calculated the size of the event horizon formed by the vacuum energy density. The event horizon created by the vacuum energy density within a radius R=l_P/2 is R_S=2.09 l_P, which is larger than the radius R of the mass or energy distribution, so the vacuum energy is trapped in the event horizon formed by its own mass, and therefore, it cannot exert a force on the outside and has the possibility of disappearing. 4) There may be an unknown reason that we have not found. In the case of the gravitational potential energy model, the mass M inside the shell is not a constant but a variable, and it is difficult to solve the problem because it includes M(t) --> M_(t) + ( -M_gp(t)). After thinking about it for a while, it seems that the total energy inside the shell is not conserved, and it seems difficult to derive the first Friedmann equation using the mechanical energy conservation equation. To derive this, we need a genius who can take a new path, not someone who follows the examples pioneered by others. If there is anyone among the readers who has the ability, please try to derive it. So, until a genius who can build such a new model appears, if we apply a temporary method, we can think of the following method. Both the existing Friedmann 1 and 2 equations include the same dark energy term (1/3)Λc^2. Therefore, we can put the dark energy term obtained from the second Friedmann equation into the first Friedmann equation as it is. I derived the dark energy term from the second Friedmann equation, and the dark energy value calculated using the derived dark energy term is very close to the observed value, and by adding a correction factor, we can make it match completely. Therefore, Dark Energy is Gravitational Potential Energy or Energy of the Gravitational Field

E_T = Mc^2 - (3/5)(GM^2/R) is not an equation used to describe the conservation of mechanical energy in an expanding universe. It is an equation close to rest mass energy. M can also be expressed as M_rel. The rest mass of an object is the total energy including the rest mass, kinetic energy, and potential energy (gravitational self-energy, electromagnetic self-energy, and potential energy related to the strong and weak forces) of the particles that make up the object, divided by c^2. In other words, the rest mass of an object includes not only the rest mass of the elements that make up the object, but also the potential energy between the elements that make up the object. Representative potential energies include gravitational potential energy and electrostatic potential energy. This can also be understood through the following figure. M is the combined state of dMs. Therefore, it has potential energy. If there is an object (mass or energy distribution) as above, let's divide this object in half virtually. Then, the left hemisphere A and the right hemisphere B will exist, and the centers of mass of the two hemispheres will exist. The centers of mass of the two hemispheres are separated by a distance r, and this r is not 0. Then, according to the definition of gravitational potential energy, in the equation U=-G(m_A)(m_B)/r, m_A(mass of hemisphere A), m_B(mass of hemisphere B) exist, and the distance r also exists between the two centers of mass. Therefore, gravitational potential energy U exists by itself. Since the differential concept of dividing all objects into the masses of the infinitesimals that compose them is established, gravitational potential energy exists between the infinitesimals that compose the objects. Therefore, in any entity or any energy distribution, we must consider gravitational self-energy(or - gravitational binding energy) and electrostatic self-energy when calculating the total energy. However, if we do not assume the existence of charge (if we simply define the energy distribution), we can exclude the electrostatic self-energy term. Also, since electrostatic self-energy is positive energy, it can be included in the equivalent mass M of positive energy. As a result, if we assume an entity with some energy that has a spherical uniform distribution, the total energy of the entity can be expressed as E_T=Mc^2 - (3/5)(GM^2/R). Strictly speaking, M is the total energy when the particles that make up the entity are in a free state. However, in most situations, the gravitational self-energy is small compared to the mass energy, so M_free ~ M_0 holds. 1) Mass energy and gravitational self-energy in the case of a proton Proton mass = 1.67x10^-27kg Proton radius = 0.84x10^-15m For example, in the case of a proton, the gravitational self-energy is very small compared to the mass energy, at the level of 10^-39. 2) Mass energy and gravitational self-energy in the case of the Earth In the case of the Earth, the gravitational self-energy of the Earth is also very small compared to the mass energy, at the level of 10^-10. Therefore, we have not considered this gravitational self-energy (the gravitational potential energy that the mass distribution of the object itself has) in most problems, and have forgotten its value. As can be seen from the calculation above, the negative gravitational self-energy is not always equal to the mass energy. However, there are situations where this gravitational self-energy becomes very important, and these situations are presumed to be the source of important problems in current physics and astronomy. 1) Black hole singularity problem 2) Dark matter and dark energy 3) Inflation mechanism 4) Birth of the universe from quantum fluctuations ... ========== Now, let's look at the birth process of a single quantum fluctuation covered in this paper. Let's assume that a single quantum fluctuation is born with the size of the Earth and the mass energy of the Earth. The energy ΔE=(M_E)c^2 remains, and therefore, it has a very short Δt. Therefore, it is difficult to explain the birth of the universe from nothing (zero energy) and its existence to the present. Let's assume that it was born with the size of a proton and the mass energy of a proton. The energy of ΔE=(m_p)c^2 remains, and therefore, it has a very short Δt. Therefore, it is difficult to explain the birth of the universe from nothing (zero energy) and its existence to the present. In both cases, ΔE=(M_E)c^2 and ΔE=(m_p)c^2 remain, and it is difficult to explain the birth from nothing. However, let's assume the following situation calculated in this study. 1)If, Δt=t_p, ΔE=(5/6)(m_p)c^2, 2)If, Δt=(3/5)^(1/2)t_P, ΔE=(5/12)^(1/2)(m_p)c^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. In addition, by the 2nd mechanism, since the negative gravitational potential energy exceeds the positive mass energy, the corresponding energy distribution has the characteristics of negative mass, and through this, it can explain the accelerated expansion of the early universe(Inflation) and the accelerated expansion of the current universe(Dark energy).

What did you read in my writing that made you think I tried to write E=E_k + E_p?

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?

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. 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 2)Dark Energy Spectroscopic Instrument team https://arstechnica.com/science/2024/04/dark-energy-might-not-be-constant-after-all/#gsc.tab=0 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

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. 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

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.

*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. =============== I will write the answer to this question after I do what I have to do first. Within 1 day~

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.

If the universe was nothing at the beginning, how did this state of nothing then create what is the universe? We don't yet know what the answer is. However, because it is an important issue, I am writing my personal opinion. Let's start with the following equation: A = A Before the universe was birthed, the concept of A did not even exist. This A is an concept created by intellectual creature called humanity, 13.8 billion years after the creation of the universe. Also, mathematical terms, including =, are concepts created by humans born after the birth of the universe. If we move "A" from the left side to the right side, 0 = A - A = 0 To make the idea clearer, let's express this a little differently. 0 = (+A) + (-A) = 0 This equation can be conceptually decomposed as "0", "0=(+A)+(-A)", "(+A)+(-A)=0", "0=0". 1)"0" : Something did not exist. Nothing state. 2)"0 = (+A) + (-A)" : (+A) and (-A) were born from "nothing". Or "nothing" has changed to +A and -A. Something state. 3)"(+A) + (-A) = 0" : The sum of (+A) and (-A) is still zero. From one perspective it's "something state", from another perspective it's still “nothing state”. 4)"0 = 0" At the beginning and end of the process, the state of “nothing” is maintained. 5) "B = 0 = (+A) + (-A) = 0" : The intelligent life form called humanity defines the "first nothing" as B. B may be total A, which is the sum of all A, or it may be a new notion. In other words, “nothing” can create something +A and something -A and still remain “nothing” state. And, the newly created +A and -A create new physical quantities and new changes. For example, in order for the newly created +A and -A to be preserved in space, a new relational equation such as a continuity equation must be created. ∂ρ/∂t + ∇·j=0 Let's look at how pair production occurs from photon (light). B = 0 = (+Q) + (-Q) = 0 The charge of a photon is zero. When photon do pair production, photon do not conserve charge by creating beings with zero charge, but by creating +Q and -Q to preserve zero. That is, in all cases, in all circumstances, in order to satisfy or maintain “nothing”, this equation of the form (+Q) + (-Q) = 0 must hold. This may be because "0" is not representative of all situations and is only a subset of (+Q) + (-Q) = 0. At the beginning and end of the process, the total charge is conserved, but in the middle process +Q and -Q are created. Due to the electric charge generated at this time, new concepts including electromagnetic fields and electromagnetic forces are needed. According to Emmy Noether's theorem, if a system has a certain symmetry, there is a corresponding conserved physical quantity. Therefore, symmetry and conservation laws are closely related. Conservation of spin, conservation of particle number, conservation of energy, conservation of momentum, conservation of angular momentum, conservation of flux... etc.. New concepts may be born from conservation laws like these. Let’s look at the birth process of energy. https://icarus2.quora.com/The-Birth-Mechanism-of-the-Universe-from-Nothing-and-New-Inflation-Mechanism E_T = 0 = (+E) + (-E) = Σmc^2 + Σ(-Gmm/r) = 0 “E_T = 0” represents “Nothing” state. Mass appears in “Σ(+mc^2)” stage, which suggests the state of “Something”. In other words, “Nothing” produces a negative energy of the same size as that of a positive mass energy and can produce “Something” while keeping the state of “Nothing” in the entire process (“E_T = 0” is kept both in the beginning of and in the end of the process). Another example is the case of gauge transformation for scalar potential Φ and vector potential A in electromagnetic fields. Φ --> Φ - ∂Λ/∂t A --> A + ∇Λ Maxwell equations of electromagnetism hold them in the same form for gauge transformation. After all, the existence of some symmetry or the invariance that the shape of a certain physical law must not change requires a gauge transformation, and this leads to the existence of new physical quantities (Λ, ∂Λ/∂t, ∇Λ) that did not exist in the beginning (Φ, A). This can be interpreted as requiring the birth of a new thing in order for the conserved physical quantity to be conserved and not change. The condition or state that should not change is what makes change. Why was the universe born? Why is there something rather than nothing? Why did the change happen? B = 0 = (+Q) + (-Q) = 0 E_T = 0 =(+E) + (-E) = Σmc^2 + Σ-Gmm/r = 0 ∂ρ/∂t + ∇·j=0 Φ --> Φ - ∂Λ/∂t A --> A + ∇Λ It changes, but does not change! It changes in order not to change! What does not change (B = 0) also creates changes in order not to change in various situations (Local, Global, phase transformation, translation, time translation, rotation transformation ...). This is because only the self (B) that does not want to change needs to be preserved. The change of the universe seems to have created a change by the nature of not changing. The universe created Something (space-time, quantum fluctuation, energy, mass, charge, spin, force, field, potential, conservation laws, continuity equation...) to preserve Nothing. By the way, as this something was born, another something was born, and the birth of something chained like this may still preserve the first "nothing", and in some cases, the first "nothing" itself may also have changed.

Mathematics began as human intellectual activity to describe phenomena that exist in nature and the universe. Therefore, mathematics and the universe have a very deep connection, and events that occur in the universe can generally be described through mathematics. However, in mathematics, if a proposition is true within a defined axiomatic system, then this proposition is true and has value. In other words, because mathematics reflects human intellectual activity, even things that are imaginary can become true. On the other hand, the universe is true to exist and has value. Therefore, mathematics and the universe are not completely the same thing.

Pressure, energy, and mass are all sources of gravity. Therefore, these terms enter the Friedmann equation and play the role of -(4πG/3)p, which is the gravitational effect. According to Birkhoff’s theorem, The first Friedmann equation without a cosmological constant is derived from general relativity, but there is also a form derived from the conservation of mechanical energy of Newtonian mechanics (dust model), and this is taught in major books. At this time, the terms entered are kinetic energy and gravitational potential energy. [math]\frac{1}{2}m{v^2}(t) - G\frac{{{M_r}m}}{{r(t)}} = E[/math] Therefore, the right-hand side term (8πG/3)ρ comes from the total mass M_r inside the shell. Friedmann's second equation without a cosmological constant is also derived from general relativity, but can also be derived from F=ma=-GMm/R^2. This also uses gravitational force. Because this is intuitive, if we look at this derivation process [math]ma = - \frac{{GMm}}{{{R^2}}}[/math] [math]m\frac{{{d^2}R(t)}}{{d{t^2}}} = - \frac{{GM(t)m}}{{R{{(t)}^2}}}[/math] [math]M(t) = \frac{{4\pi {R^3}(t)}}{3}\rho (t) = \frac{{4\pi {a^3}(t){R^3}}}{3}\rho (t)[/math] [math]m\frac{{{d^2}a(t)R}}{{d{t^2}}} = - Gm\frac{{\frac{{4\pi }}{3}{a^3}(t){R^3}\rho (t)}}{{{a^2}(t){R^2}}} = - Gm\frac{{4\pi }}{3}a(t)R\rho (t)[/math] [math]\frac{{\ddot a(t)}}{{a(t)}} = - \frac{{4\pi G}}{3}\rho (t)[/math] Adding pressure, we can create an acceleration equation. [math]\frac{{\ddot a}}{a} = - \frac{{4\pi G}}{3}(\rho + \frac{{3P}}{{{c^2}}})[/math] In other words, we can see that ρ(t) is the mass density inside the expanding shell, and the -(4πG/3)ρ term describes gravity or gravitational effects. It can be seen that the total mass M inside the shell is included in both Friedmann equations 1 and 2. This means that the equivalent mass of total energy is included. If matter and radiation have the same energy density ρ0, it is thought that they should be the same when entering the acceleration equation. In my opinion, We can put in the equivalent mass of the energy density of the radiation divided by c^2, and let the pressure of the radiation be 0, or since the rest mass of the radiation is 0, let the mass density be 0, and in the pressure term (1/3)ρ must be inserted. Photons are also included as relativistic particles. However, in this explanation, I tried to distinguish between relativistic particles that have rest mass and photons that do not have rest mass. For photons, which do you think is correct: 1ρ0 or 2ρ0 ?

Consider a situation where matter and radiation have the same energy density ρ_0. If we set c=1, mass density = energy density For matter, [math]{\rho _m} + 3P{}_m = {\rho _m} + 3(0) = {\rho _m} = {\rho _0}[/math] For radiation (photon), [math]{\rho _r} + 3P{}_r = {\rho _r} + 3(\frac{1}{3}{\rho _r}) = 2{\rho _r} = 2{\rho _0}[/math] Now if we think about the physical meaning of the two cases, This means that if matter and radiation have the same energy density ρ_0, the gravitational effect of the radiation is twice that of the matter. If photon and matter have the same energy density, does the photon exert twice the gravitational force (or effect) than the material exerts? Is this true? In my opinion, In the case of photon, it seems wrong to include both density in the (mass) density term and pressure term.

I think I didn't phrase the question appropriately. [math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}})[/math] [math]\rho + \frac{{3P}}{{{c^2}}} = ({\rho _m} + \frac{{3P{}_m}}{{{c^2}}}) + ({\rho _r} + \frac{{3P{}_r}}{{{c^2}}}) + ({\rho _\Lambda } + \frac{{3P{}_\Lambda }}{{{c^2}}})[/math] In this question the first and third terms are not of interest. What happens if you summarize(Energy density of photon ρ) the second term? When the energy density of a photon is ρ, I am curious as to whether its gravitational effect in the acceleration equation is 1ρ or 2ρ. *When considering the dimension, it must be divided by c^2, but this is not an important issue. [math]{\rho _r} + \frac{{3P{}_r}}{{{c^2}}} = [/math]

Gravitational effect and pressure of photons in Friedmann equation? [math] \frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}}) [/math] ρ is the unit of mass density. Pressure of matter : P=0 Pressure of radiation : P=(1/3)ρc^2 Regarding the source of pressure, there are the following expressions in major books: In special relativity, [math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math] For matter, [math]{E^2} = {({m_0}{c^2})^2}[/math] For, relativistic particle, [math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math] For, radiation or photon, [math]{E^2} = {(pc)^2}[/math] So, here's the question. For a relativistic particle with rest mass it would be: [math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math] In the case of relativistic particles, there is a pressure component due to kinetic energy or momentum, and this value cannot be ignored compared to mass energy. [math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _{{\rm{relativistic particle}}}} + \frac{{3{P_{{\rm{relativistic particle}}}}}}{{{c^2}}}) = - \frac{{4\pi G}}{3}({\rho _{{\rm{rel - par}}}} + \frac{{3(\frac{1}{3}{\rho _{{\rm{rel - par}}}}{c^2})}}{{{c^2}}}) = - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rel - par}}}})[/math] [math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rel - par}}}}) = - \frac{{4\pi G}}{3}(\frac{{2{\rho _{{\rm{rel - par}}}}{c^2}}}{{{c^2}}})[/math] When the energy density of matter is ρ, in the Friedmann equation, matter generates the gravitational effect of ρ. When the energy density of a relativistic particle is ρ, in the Friedmann equation, the relativistic particle generates a gravitational effect of 2ρ. But what about photons, which are not relativistic particles? Which expression is correct for photons that have no rest mass? 1) [math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}}) = - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rad}}}}) = - \frac{{4\pi G}}{3}(\frac{{2{\rho _{{\rm{rad}}}}{c^2}}}{{{c^2}}})[/math] Considering radiation(When limited to photons) with energy density ρ, it generates a gravitational effect equal to twice the energy density : 2ρ ? 2) [math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _{{\rm{radiation}}}} + \frac{{3{P_{{\rm{radition}}}}}}{{{c^2}}}) = - \frac{{4\pi G}}{3}(0 + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}}) = - \frac{{4\pi G}}{3}({\rho _{{\rm{rad}}}}) = - \frac{{4\pi G}}{3}(\frac{{{\rho _{{\rm{rad}}}}{c^2}}}{{{c^2}}})[/math] Considering radiation(When limited to photons) with energy density ρ, it generates a gravitational effect equal to 1 times the energy density : 1ρ ? 1) Is the expression correct? 2) Is the expression correct?