The Birth Mechanism of the Universe from Nothing and New Inflation Mechanism!

[icarus2]

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

[swansont]

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?

[icarus2]

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~

[icarus2]

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.

[swansont]

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)

[Mordred]

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

[Mordred]

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 November 11 by Mordred

[Mordred]

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 November 11 by Mordred

[Mordred]

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 November 12 by Mordred

[icarus2]

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 November 12 by icarus2

[Mordred]

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 November 12 by Mordred

[Mordred]

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 November 13 by Mordred

[Mordred]

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 November 13 by Mordred

[icarus2]

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

 

[Mordred]

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 November 14 by Mordred

[icarus2]

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 November 14 by icarus2

[Mordred]

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 November 14 by Mordred

[icarus2]

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

[Mordred]

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 November 14 by Mordred

[Mordred]

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 November 14 by Mordred

[Mordred]

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

[icarus2]

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).

 

[Mordred]

Sigh your grasping the importance of the KE term. 

Treat your model above as a toy universe. Then ask two fundamental questions.

1) What prevents your toy universe from collapsing under its own self gravity ?

2) what expansion rate would it have ?

Let me know when you want to address those two questions.

Also further consider the zero point energy value you have is the vacuum catastrophe value.

Do you really want your model to match the worse mistake in QM predictions  ?

Do you not think it may be better to match the Observed vacuum energy density of the cosmological constant instead of the vacuum catastrophe ?

I will let you think about those questions 

I also don't agree with your gravitational self energy apply Newtons Shell theorem to the Earth case and not just two seperste hemispheres you must look at all contributions of gravitational  force not just two points.

Specifically sum of all forces from every coordinate in the Earth's shell.

If you have a uniform mass distribution then any point  can be treated as a center of mass. The sum of all force vectors using Newtons gravitational law will be zero for gravity

g=0 in any uniform mass distribution it is non uniform mass distribution that leads to gravity hence under GR the acceleration is handled through the curvature terms.

Nice try but Newtons Shell theorem is basic classical physics why didn't you consider that ? On a multi-point analysis ie sum of forces surrounding any arbitrary CoM in a uniform mass distribution ?

Of course you could have just asked yourself why does gravitational force vary with the 1/r^2 relation in regards to your proton example as you didn't include a distance between any points you used. 

How do you have a gravitational self energy that doesn't decrease the greater the distance. That should have been obvious  just as obvious that every other point in the Earth's volume would also have influence 

Edited Saturday at 04:32 PM by Mordred

[icarus2]

20 hours ago, Mordred said:

I also don't agree with your gravitational self energy apply Newtons Shell theorem to the Earth case and not just two seperste hemispheres you must look at all contributions of gravitational  force not just two points.

Specifically sum of all forces from every coordinate in the Earth's shell.

If you have a uniform mass distribution then any point  can be treated as a center of mass. The sum of all force vectors using Newtons gravitational law will be zero for gravity

g=0 in any uniform mass distribution it is non uniform mass distribution that leads to gravity hence under GR the acceleration is handled through the curvature terms.

Nice try but Newtons Shell theorem is basic classical physics why didn't you consider that ? On a multi-point analysis ie sum of forces surrounding any arbitrary CoM in a uniform mass distribution ?

Of course you could have just asked yourself why does gravitational force vary with the 1/r^2 relation in regards to your proton example as you didn't include a distance between any points you used. 

How do you have a gravitational self energy that doesn't decrease the greater the distance. That should have been obvious  just as obvious that every other point in the Earth's volume would also have influence 

 

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.

 

20 hours ago, Mordred said:

1) What prevents your toy universe from collapsing under its own self gravity ?

 

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.

On 11/16/2024 at 4:09 PM, icarus2 said:

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.

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.

On 11/12/2024 at 5:59 PM, icarus2 said:

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.

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~

 

20 hours ago, Mordred said:

Also further consider the zero point energy value you have is the vacuum catastrophe value.

Do you really want your model to match the worse mistake in QM predictions  ?

Do you not think it may be better to match the Observed vacuum energy density of the cosmological constant instead of the vacuum catastrophe ?

 

Are you not reading my writing?

On 11/14/2024 at 8:21 PM, icarus2 said:

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.

 

Dark Energy is Gravitational Potential Energy or Energy of the Gravitational Field

 

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.

 

20 hours ago, Mordred said:

2) what expansion rate would it have ?

 

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

 

Edited Sunday at 09:25 AM by icarus2

[Mordred]

It doesn't take a genius to figure out in order to describe how a uniform mass distribution can expand by one basic equation that exists in the FLRW metric.

The equation already exists for that and that equation uses both PE and KE and works with the Newtons Shell theorem.

Obviously you don't want to include the necessary KE component.

Trying to use just PE is insufficient.  That equation is the scalar field equation of state .

Of course I'm going to follow examples pioneered by others I know  they work I would be foolish not to. Those examples provide key lessons your choosing to ignore.

The article I provided clearly demonstrates that it requires both PE and KE terms to arrive at the first FLRW metric equation.

If you choose to ignore that detail that's your choice but the method in article works.

Yet you choose to ignore it. So I can't really help you as you would rather try to reinvent physics that works in favor of your model that you would not be able to calculate an expansion rate or even answer the question 

1) what prevents your universe from collapsing under self gravity

2) what expansion rate will it have

So good luck to you.

You don't even have a means to derive a pressure term to determine vacuum energy you require KE vs PE for that so your on your own.

Quite frankly it's foolish to ignore 

Kinetic energy =energy due to motion is required for pressure which is required for vacuum energy. That is a very basic classic physics lesson your choosing to ignore.

 

It's the commonly used method by any professional physicist because it works. All you need to do is study the equations for a piston to recognize those pressure equations uses both Potential and kinetic energy.

So why you choose to ignore that lesson I have no idea. The equations of state which relate the pressure terms to energy density uses the same hydrodynamic relations from classical physics lessons such as that piston example. 

So your blooming right I will rely on the work pioneered by 100's of years of collective research. I know those methods work as opposed to trying to determine pressure terms for a vacuum as opposed to your method. Bloody right I will favor the standard method over yours.

\[\omega=\frac{P}{\rho}\]

pressure to energy density relation above.

scalar field equation of state 

\[\omega=\frac{\frac{1}{2}\dot{\phi}^2-V(\phi)}{\frac{1}{2}\dot{\phi}^2+V(\phi)}\]

scalar field equation of state has both pressure and energy density. When the kinetic energy exceeds the pressure you get the negative pressure relation w=-1 for an incompressible fluid and only the w=-1 value is the only value that gives a vacuum pressure that is constant.

that is the lesson you choose to ignore There is VERY good reasons why the FLRW metric uses Pressure to energy density relations and those relations involve both Kinetic energy and potential energy. NOT JUST POTENTIAL ENERGY.

https://en.wikipedia.org/wiki/Equation_of_state_(cosmology)

there is good reasons those equations of state are used by the professional community. Reasons you choose to ignore lol if you really want to understand that last equation you can literally track it back to Bernoulli's Principle in fluid mechanics. It employs the same principles

However as the FLRW metric applies GR its better to use the Einstein Field equations.

which quite frankly Your method is not compatible with as you are ignoring the Stress energy momentum tensor.

\[T^{\mu\nu}=(\rho+P)\mu^\mu\mu^\nu+pg^{\mu\nu}\]

this is the equation that those equations of state derived from via Raychaudhuri equations. So Yes I choose the GR method over your any day. I know they work and how they were derived starting from classical physics

As an FYI for yourself and other readers one can use the Canonical formalism with the above (QFT)

the action for a minimally coupled scalar fields in spacetime is

\[S=\int d^4x \sqrt{-g}(\frac{R}{2k^2} +\mathcal{L}_m+\mathcal{L}_\phi\]

Where R is the Ricci scalar \(\sqrt{-g}\) is the metric determinant \(\mathcal{L}_m\) is the matter field Langrangian for matter fields.

scalar field 

\[\mathcal{L}_\phi=-\frac{1}{2}g^{\mu\nu}\partial _\mu\phi\partial_\nu\phi-V(\phi)\]

where \(V\phi\) is the scalar field potential Terms previous is the kinetic energy terms.

gives Klein Gordon equation

\[\square\phi-V_\phi=0\]

where \(V_phi=\frac{\partial V}{\partial\phi}\) and \(\square\phi=\nabla_\mu\nabla^\nu \phi\)

variation with respect to the gravitational metric \(g_{\mu\nu}\) yields

\[R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=k^2(T_{\mu\nu}+T^\phi_{\mu\nu})\]

leading to the canonical treatment of the energy momentum tensor

\[T^\phi_{\mu\nu}=\partial_\mu\phi \partial_\nu \phi-\frac{1}{2}g_{\mu\nu}(\partial\phi)^2-g_{\mu\nu}V(\phi)\]

and \(\partial\phi^2=g_{ab}\partial^a\phi\partial^b\phi\)

so using the above with k=0 for flat and equation of state \[p=\omega\rho\] with FLRW metric

\[ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2\]

one can derive the Freidmann equations and acceleration equation which is quite lengthy as you must go through the Ricci tensor with all the relevant Christoffels as very members understand Christoffel symbols I will skip that portion.

to arrive at 

\[3H^2=k^2(\rho+\frac{1}{2}\dot{\phi}+V(\phi)\]

\[2\dot{H}+3\dot{H}^2==k^2(\omega\rho+\frac{1}{2}\dot{\phi}^2-V(\phi))\]

gives energy density as

\[\rho_\phi=\frac{1}{2}\dot{\phi}^2+V(\phi)\]

\[p-\phi=\frac{1}{2}\dot{\phi}^2-V(\phi)\]

which gives the mathematical proof of the scalar field equation for the FLRW metric above. Clearly demonstrates the relevance of kinetic energy and potential energy now doesn't it. You ask why I will use the work of other professional Physicists over yours there is your answer. The GR method as well as QFT method both work with the FLRW metric for all equations of state including the scalar field equation of state.

Edited Sunday at 04:41 PM by Mordred