Problems with negative pressure and vacuum energy

[icarus2]

From the second Friedmann equation or acceleration equation,

[math] \frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - (\frac{{4\pi G}}{3})(\rho + 3P)[/math]

Since ρ is the mass density in the acceleration equation, the pressure term P has a dimension of energy density (c=1). Current acceleration equation inevitably require negative mass density. Without negative (gravitational) mass density, it is impossible to create acceleration expansion. The mainstream produces acceleration expansion by setting the pressure of the cosmological constant or vacuum energy to P=−ρ .

Currently, dark energy is described as a being that exerts a negative pressure while having a positive energy density. However, there is a serious problem when considering a physical existence with these characteristics.

ρ+3P=ρ+3(−ρ)=−2ρ

1) Sign of the negative pressure

Since the universe is treated as a fluid with uniform density, the pressure between regions is zero. In general relativity, pressure is equivalent energy density of the kinetic energy.

Quote

Note that the effect of the pressure P is to slow down the expansion (assuming P>0). If this seems counterintuitive, recall that because the pressure is the same everywhere in the universe, both inside and outside the shell, there is no pressure gradient to exert a net force on the expanding sphere. The answer lies in the motion of the particles that creates the fluid’s pressure. The equivalent mass of the particle’s kinetic energy creates a gravitational attraction that slows down the expansion just as their actual mass does. - An Introduction to Modern Astrophysics

In the acceleration equation, 3P has the idea of an equivalent energy density corresponding to the kinetic energy of the particle. So, assuming that the pressure P term has a negative energy density is same assuming that it has negative kinetic energy. In order to have negative kinetic energy, it must have negative inertial mass or imaginary velocity. But, because they assumed a positive inertial mass, it is a logical contradiction.

[math]K = \frac{1}{2}m{v^2} < 0[/math]

m<0 or v=Vi : negative mass or imaginary speed.

Negative mass contradicts the assumption of positive energy density, and energy density with imaginary speed is far from physical reality.

2) Size of the negative pressure

In the ideal gas state equation, we obtain,

[math]P = \frac{1}{3}(\frac{{{v^2}}}{{{c^2}}})\rho = \omega \rho[/math]

In the case of matter, v << c, So [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho \simeq 0[/math]

In the case of radiation, v=c, So [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = \frac{1}{3}\rho [/math]

However, in the case of dark energy [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = - \rho [/math]

a) Only considering the size (absolute value)

[math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = | - \rho |[/math]

[math]v = \sqrt 3 c[/math]

We need energy density with super-luminous speed.(FTL)

b) Considering even the negative sign

[math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = - \rho[/math]

[math]v = (\sqrt 3 c)i[/math]

We need energy density with imaginary super-luminous speed.

We can avoid this problem by assuming that cosmological constant or vacuum energy does not apply to the results (ideal gas state eq.) obtained through physics. However, ideal gas state equation applies to "massless ~ infinite mass'' particles, and the velocity ranges from "0 ~ c'' are all included. If we seriously consider the physical presence of P=-p=-3(p/3), we will find that there is a serious problem.

 

3) Incorrect application of dU = - PdV

In the equation dU = - PdV, researchers claim that negative pressure exists if the energy density remains constant as the volume increases. Is this claim correct?

Going to the root of the problem, the analysis of this formula seemed simple (dU=pdV), so I had the illusion that the logic of negative pressure was certain. But~

a) This explanation is an inverted explanation

Since pressure is a property of an object, pressure exists first, and because of this pressure, changes in internal energy according to volume change appear.

That is, since pressure is positive, if dV>0, then dU<0. Since the pressure is positive, if dV<0, then dU>0.

By the way, we use the logic "if dV>0, dU>0, then P<0''. How are you sure that this logic is correct?

b) ρ+3P=ρ+3(−ρ)=−2ρ

Mass density ρ and pressure P are properties of the object to be analyzed. Both mass density ρ and pressure P are sources of gravity.

It means that even if the region maintains a constant size without expanding or contracting, gravitational force is applied as much as ρ+3P .

In other words, it suggests that the object (or energy density) has a gravity with a negative mass density of −2ρ . This is different from a vacuum with a positive energy density ρ , which we think of.

c) dU = - PdV is the expression obtained when the law of conservation of energy is established

dU=dQ-dW, if dQ=0, dU=-dW=-PdV

However, in the case of vacuum energy and the cosmological constant, energy conservation does not hold. As the universe expands, the total energy in the system increases. Therefore, we cannot guarantee that dU = - PdV holds. dU = - PdV is an equation that holds when the energy of the system is conserved. However, researchers are applying this equation to vacuum energy or cosmological constant where the energy of the system is not conserved.

"In the equation dU = - PdV, even if the universe expands, if it has a uniform density, it should have a negative pressure...'' What's wrong here?

I am not sure if this equation holds even for negative pressure. However, although this equation holds even in the case of negative pressure, its interpretation is as follows.

This equation holds true when substances in radius r_1 expand from r_1 to r_2 (r_2 > r_1), and have the same uniform density in r_1 and r_2. In other words, it is argued that a negative pressure is required to create a uniform density effect only with the material present in radius r_1. But, vacuum energy is a form in which energy is newly generated by an increased volume. It is also energy that can be assumed to have an initial speed of 0 ~ c. Considering the positive energy density, pressure term seems reasonable to assume 0 ~ (1/3)p.

 

If the cosmological constant term is shifted to the right side of the field equation, it becomes a negative equivalent mass. It is not an object that has a positive energy density (positive inertial mass) and acts on a negative gravitational mass, but just a negative mass.

It is not necessary for an entity to exist that exerts a negative pressure while having a positive energy density. Only a negative mass component may exist, or a positive energy component may exist and a negative energy component may exist. And, the negative energy component may be the energy of the gravitational field or the gravitational potential energy.

 

Q1. Is it possible to have negative kinetic energy while having positive mass density? Is it possible for a physical entity to have an imaginary velocity while having a positive energy density?

Q2. [math]P =  - \rho  =  - 3(\frac{1}{3}\rho ) [/math]

Can such an entity (an entity with a pressure(kinetic energy component) three times greater than light) physically exist?

Q3. Is dark energy energy with FTL (Faster Than Light) or an imaginary FTL?

Q4. Does the dU=-PdV equation hold for a system in which energy is not conserved?

Q5. Even if the universe expands, If an entity with uniform mass (energy) density exists, it is said that it will have a negative pressure. Is this claim correct?
We can considered the vacuum energy density with P=0. However, in order to have a uniform energy density even when space expands, does it have to suddenly have negative pressure? Shouldn't it be newly created, with P=0, and filling the larger universe?

Can we not think of a vacuum energy density with P=0? Can't there be a vacuum energy density with P=0 ~ (1/3)p?

Even if vacuum energy is present, it is not certain whether it creates a negative pressure.

ρ+3P=ρ+3(−ρ)=−2ρ
There seems to be a serious problem with this argument.

Edited June 4, 2022 by icarus2

[swansont]

8 hours ago, icarus2 said:

In general relativity, pressure is equivalent energy density of the kinetic energy.

And in the quote you provided, it says P>0

So how do you get from that to saying it’s negative?

[NTuft]

From a simple viewpoint, if we are assuming a value for P>0, an increasing volume (i.e. universe expansion), and conservation of mass (amount of substance, n); from PV=nRT , the increasing V may necessitate an increasing temperature, or a decreased mass density \(p\) = mass/V , if we can equate n to density as mass/V; the mass density is now negative in change from what it was before the expansion in volume?

 

This brief exchange (link is to full thread) may be relevant:

Physics Forums - Negative kinetic energy?

Quote

[...]

For a particle trapped in a potential that is greater than its total energy, the physical intepretation is that it has a kinetic energy<0. Is that reasonable? Since its potential is goverened by the potential of the well it is in and is fixed. The only value that can change is kinetic energy. OR in that situation should be abolish the term kinetic energy and define something else?

[...]

Kinetic Energy is never negative. If a Particle is bound in some kind of potential its TOTAL ENERGY is negative (this is characteristic of a bound system, classical or quantum), but its KINETIC ENERGY is still positive.

[...]

Kinetic energy can be negative in GR.

[...]

In what context

[...]

write out the Hamiltonian constraint, the terms quadratic in the momentum are indefinite : 1/2 (P_c^c)^2 - P_ab P^ab. The ``potential'' term is the densitized Ricci curvature of the spacelike metric and can be positive and negative as well.

[...]

That points to this (which I am not through with)

Lecture I: Constrained Hamiltonian systems (Courses in canonical gravity) - Yaser Tavakoli, Dec. 15, 2014

 

I will look to read your paper, good work on presentation here.

Edited June 5, 2022 by NTuft

[icarus2]

On 6/5/2022 at 7:35 AM, swansont said:

And in the quote you provided, it says P>0

So how do you get from that to saying it’s negative?

Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy
So, in the Book(An Introduction to Modern Astrophysics   1161P) say that equivalent mass (density) of the particle's kinetic energy...  

When a new phenomenon is discovered, we try to explain the new phenomenon within the existing theoretical framework. The current negative pressure is also a problem caused by trying to explain within the existing theoretical framework.

The current cosmology is based on general relativity, among them, the 1915 field equation without cosmological constant and the field equation with cosmological constant in particular.

For the universe to accelerating expansion, the right side of Friedmann's equation must be positive, and therefore (p+3P)<0.

So, the researchers came up with the following logic:

p+3P=p+3(-p)=-2p

However, as raised in the main text, p+3P=p+3(-p)=-2p has serious problems and contradicts existing physics.

In my opinion,

Positive energy density and negative kinetic energy are logically contradictory. Moreover, p+3P=p+3(-p)=-2p It seems impossible for a physical existence that satisfies this relational expression. Nothing seems to have a kinetic energy component that is three times greater than that of light.

Even if vacuum energy with positive energy density p exists, the pressure P created by that vacuum energy is estimated to have a positive pressure of 0~(1/3)p. So, even if vacuum energy with positive energy density exists, it cannot create negative pressure.

There is no negative pressure, instead there is a negative gravitational potential energy by a positive energy density.

So how do you create a negative pressure component, that is, a repulsive force component?
After all, the root of the problem is Einstein's 1915 field equation.

Einstein's field equation was a huge success in weak gravitational field. However, there is a problem in strong gravitational field where the energy of the gravitational field must be taken into account. The singularity problem is strong evidence that Einstein's field equation is incomplete.

1. The fundamental principle of general relativity states that “all energy is a source of gravity”. However, the field equation created by Einstein did not fully realize this principle

The energy of the gravitational field must also function as a gravity source.
Einstein was also aware of this, and for over two years, beginning in 1913, he worked to formulate a field equation that included the energy-momentum of the gravitational field. However, because it was difficult to define the energy-momentum tensor of the gravitational field in general relativity, Einstein could not complete the field equation including the gravitational action of the gravitational field. So, we have a field equation involving only the energy-momentum tensor of matter. Because of the omission of the energy-momentum of the gravitational field, the singularity problem and the dark energy problem came into existence.

1)Gravitation and Spacetime (Book)

(1) According to (48) matter acts on the gravitational field (changes the fields), but there is no mutual action of gravitational fields on matter; that is, the gravitational field can acquire energy-momentum from matter, but nevertheless the energy-momentum of matter is conserved ({\partial _\nu }{T^{\mu \nu }} = 0). This is an inconsistency.

(2) Gravitational energy does not act as source of gravitation, in contradiction to the principle of equivalence. Thus, although Eq. (48) may be a fair approximation in the case of weak gravitational fields, it cannot be an exact equation.

The obvious way to correct for our sin of omission is to include the energy-momentum tensor of the gravitational field in T^{\mu \nu }. This means that we take for the quantity T^{\mu \nu } the total energy-momentum tensor of matter plus gravitation:

T^{\mu \nu }} = T_{(m)}^{\mu \nu } + {t^{\mu \nu }}

2)Explanation of GRAVITY PROBE B team

Do gravitational fields produce their own gravity?

Yes. A gravitational field contains energy just like electromagnetic fields do. This energy also produces its own gravity, and this means that unlike all other fields, gravity can interact with itself and is not 'neutral'. The energy locked up in the gravitational field of the earth is about equal to the mass of Mount Everest, so that for most applications, you do not have to worry about this 'self-interaction' of gravity when you calculate how other bodies move in the earth's gravitational field.

2. As the mass increases, the ratio of negative gravitational potential energy to mass energy increases

Wikipedia, Gravitational binding energy

Two bodies, placed at the distance R from each other and reciprocally not moving, exert a gravitational force on a third body slightly smaller when R is small. This can be seen as a negative mass (Negative mass - Wikipedia) component of the system, equal, for uniformly spherical solutions, to:

M_binding=-(3/5)(GM^2)/(Rc^2)

When two masses m are separated by a distance r,

● —- r —- ●

The total energy of the two particle systems is

In the dimension analysis of energy, E has kg(m/s)^2, so all energy can be expressed in the form of (mass) x (speed)^2. So, E=mc^2 (Here, m is not used as a word for rest mass) holds true for all kinds of energy. If we introduce the equivalent mass -m_gp for the gravitational potential energy,

That is, when considering the gravitational action of a bound system, not only the mass in its free state but also the binding energy term (negative gravitational potential energy) should be considered.

Look at the second term, the repulsive gravitational force term.

In general, gravitational self-energy is small compared to the mass energy, but as the mass increases, the ratio of negative gravitational potential energy to mass energy increases.

-M_gs = Equivalent mass of total gravitational potential energy (gravitational self-energy) of an object. Mass energy is proportional to M, whereas the gravitational self-energy is proportional to -M^2/R.

Moon’s -M_gs = (-2.0 x 10^-11)M_Moon
Earth's -M_gs = (- 4.17 x 10^-10)M_Earth
Sun's -M_gs = (- 1.27 x 10^-4)M_Sun
Black hole's -M_gs = (- 3 x 10^-1)M_Black-hole
*In the case of the Earth, the Sun and the Black hole, please do your own calculations. Then you will notice that this value increases considerably.

It can be seen that as the mass increases, the ratio of negative gravitational self-energy to mass energy increases. Therefore, as the mass increases, the gravitational action of the gravitational field must be taken into account. Since mass energy is proportional to M, whereas gravitational self-energy is proportional to -M^2/R, the greater the mass, the greater the proportion of negative gravitational self-energy.

So, now the question we have to ask is,

"What value would the negative gravitational potential energy in the case of a universe with a greater mass?"

3. In the universe, if we calculate the gravitational self-energy or total gravitational potential energy
The universe is almost flat, and its mass density is also very low. Thus, Newtonian mechanics approximation can be applied.
Since the particle horizon is the range of interaction, if we find the Mass energy (Mc^2) and Gravitational self-energy ((-M_gs)c^2) values at each particle horizon, Mass energy is an attractive component, and the equivalent mass of gravitational self-energy is a repulsive component. Critical density value p_c = 8.50 x 10^-27[kgm^-3] was used.

At particle horizon R=16.7Gly, (-M_gs)c^2 = (-0.39M)c^2 : |(-M_gs)c^2| < (Mc^2) : Decelerated expansion period
At particle horizon R=26.2Gly, (-M_gs)c^2 = (-1.00M)c^2 : |(-M_gs)c^2| = (Mc^2) : Inflection point (About 5-7 billion years ago, consistent with standard cosmology.)
At particle horizon R=46.5Gly, (-M_gs)c^2 = (-3.04M)c^2 : |(-M_gs)c^2| > (Mc^2) : Accelerated expansion period

Even in the universe, gravitational potential energy (or gravitational field’s energy) 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.
Gravitational potential energy accounts for decelerated expansion, inflection point, and accelerated expansion.

If we could create a new field equation that perfectly reflects the principle(All energy is a source of gravity, so, the energy of the gravitational field is also a source of gravity) of general relativity and find a solution to it, the dark energy problem will be solved.

The model of Gravitational Potential Energy

1) has a clear origin,

2) is a physical quantity required by the principle of general relativity,

3) has decelerated expansion, inflection point, and accelerated expansion period,

4) does not require fine-tuning,

5) does not have the problem of coincidence,

6) Since dark energy term is presented as a function of time, it is verifiable.

Λ(t) = (6πGR(t)ρ(t)/5c^2)^2

 

https://www.researchgate.net/publication/360096238_Dark_Energy_is_Gravitational_Potential_Energy_or_Energy_of_the_Gravitational_Field

 

Edited June 6, 2022 by icarus2

[swansont]

16 hours ago, icarus2 said:

Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy
So, in the Book(An Introduction to Modern Astrophysics   1161P) say that equivalent mass (density) of the particle's kinetic energy...  

When I look up the equations in wikipedia, there is an additional term that includes the cosmological constant

https://en.wikipedia.org/wiki/Friedmann_equations

The version you present is referred to as the "matter only" form

You note that the vacuum energy density is negative, which is the source of the negative pressure, and there is no kinetic energy associated with this. Vacuum energy and matter are two different things

 

[icarus2]

On 6/7/2022 at 7:50 PM, swansont said:

When I look up the equations in wikipedia, there is an additional term that includes the cosmological constant

https://en.wikipedia.org/wiki/Friedmann_equations

The version you present is referred to as the "matter only" form

You note that the vacuum energy density is negative, which is the source of the negative pressure, and there is no kinetic energy associated with this. Vacuum energy and matter are two different things

It's same the expression on the wiki. Since I only talked about the dark energy term in some equations, I omitted the notation a little, so there seems to be a misunderstanding.

Since vacuum energy or cosmological constant can also be expressed in terms of mass density [math]{\rho _\Lambda }[/math] and pressure [math]{P_\Lambda }[/math], standard cosmology introduces, So~

[math] \frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {\rho  + \frac{{3P}}{{{c^2}}}} \right) [/math] stands for

[math]\frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {{\rho _m} + {\rho _{rel}} + {\rho _\Lambda } + \frac{{3({P_m} + {P_{rel}} + {P_\Lambda })}}{{{c^2}}}} \right)[/math]

From the equation including the cosmological constant on the wiki,

[math]\frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {\rho  + \frac{{3P}}{{{c^2}}}} \right) + \frac{{\Lambda {c^2}}}{3} =  - \frac{{4\pi G}}{3}\left( {\rho  + \frac{{3P}}{{{c^2}}} - \frac{{\Lambda {c^2}}}{{4\pi G}}} \right)[/math]

If we express each component including,

[math]\frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {\rho  + \frac{{3P}}{{{c^2}}} - \frac{{\Lambda {c^2}}}{{4\pi G}}} \right) =  - \frac{{4\pi G}}{3}\left( {{\rho _m} + {\rho _{rel}} + {\rho _\Lambda } + \frac{{3({P_m} + {P_{rel}} + {P_\Lambda })}}{{{c^2}}}} \right) [/math]

Therefore, in the standard cosmology?, the dark energy term holds the following relation.

[math] - \frac{{\Lambda {c^2}}}{{4\pi G}} = {\rho _\Lambda } + \frac{{3{P_\Lambda }}}{{{c^2}}} [/math]

Standard cosmology or mainstream cosmology defines [math] \rho _\Lambda [/math] as

[math] {\rho _\Lambda } \equiv \frac{{\Lambda {c^2}}}{{8\pi G}} [/math]

[math] {P_\Lambda } =  - {\rho _\Lambda }{c^2} =  - \frac{{\Lambda {c^4}}}{{8\pi G}} [/math]

[math] - \frac{{\Lambda {c^2}}}{{4\pi G}} = \frac{{\Lambda {c^2}}}{{8\pi G}} + \frac{{3( - \frac{{\Lambda {c^2}}}{{8\pi G}}{c^2})}}{{{c^2}}} = \frac{{\Lambda {c^2}}}{{8\pi G}} - \frac{{3\Lambda {c^2}}}{{8\pi G}} =  - 2(\frac{{\Lambda {c^2}}}{{8\pi G}}) =  - 2{\rho _\Lambda }[/math]

If C=1 is set,

[math] - \frac{{\Lambda {c^2}}}{{4\pi G}} = {\rho _\Lambda } + \frac{{3{P_\Lambda }}}{{{c^2}}} [/math]

[math] - 2{\rho _\Lambda } = {\rho _\Lambda } + 3{P_\Lambda } = {\rho _\Lambda } + 3( - {\rho _\Lambda }) =  - 2{\rho _\Lambda }[/math]

In this article, I am only talking about the dark energy term, and point out the problem with the core logic of dark energy. So, I used this expression to simply express only the core.

[math] \rho  + 3P = \rho  + 3( - \rho ) =  - 2\rho [/math]

The mainstream argument for dark energy is that it has a positive energy density [math]+\rho _\Lambda [/math] , a negative pressure, that is, a negative kinetic energy component [math]-3\rho _\Lambda [/math], and finally a mass density of [math]-2\rho _\Lambda [/math].

 

Another result of standard cosmology

[math] \frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {{\rho _m} + {\rho _{rel}} + {\rho _\Lambda } + \frac{{3({P_m} + {P_{rel}} + {P_\Lambda })}}{{{c^2}}}} \right) [/math]

For simple analysis, applying c=1, [math] {\rho _{rel}} + \frac{{3{P_{rel}}}}{{{c^2}}} <  < {\rho _m} [/math],

[math]\frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}\left( {{\rho _m} + {\rho _\Lambda } + 3{P_\Lambda }} \right)[/math]

Current standard cosmology is, approximately [math] {\rho _m} \sim 31.7\% ,{\rho _\Lambda } \sim 68.3\% [/math] ([math]\rho _m[/math] is matter + dark matter)

[math]{\rho _m} \sim  \frac{1}{3}{\rho _c}[/math]

[math]{\rho _\Lambda } \sim \frac{2}{3}{\rho _c}[/math]

[math]{\rho _m} + {\rho _\Lambda } + 3{P_\Lambda } = \frac{1}{3}{\rho _c} + \frac{2}{3}{\rho _c} + 3( - \frac{2}{3}{\rho _c}) =  - {\rho _c}[/math]

Consequently, the result of standard cosmology is that the universe has a negative mass density [math] - {\rho _c}[/math] .
The current state of the universe is a form in which negative mass (density) twice that of positive mass exists. Since the negative mass density conflicts with their notion, they do not talk about the final result, but only talk about the slightly vague negative pressure and stop.

People say: Dark energy has a positive energy density and acts as a negative pressure. So, the accelerated expansion of the universe is explained. Wow!, problem solved. They call it "standard cosmology" as they like~

The introduction of negative pressure did not solve or explain the problem. It just feels like it's been solved. The introduced negative pressure conflicts with existing physics. (Existences with positive energy density and negative kinetic energy, super-luminous speed, imaginary speed~, whether dU=-PdV is applicable to a situation in which energy is not conserved, Minor error of 10^120...)

Quote

You note that the vacuum energy density is negative, which is the source of the negative pressure, and there is no kinetic energy associated with this.

No. I'm saying that even if vacuum energy (with [math] \rho _\Lambda [/math]) exists, it has no negative pressure. Therefore, the current mainstream model is wrong. And, gravitational potential energy or gravitational field's energy seems to be the source of dark energy.

Edited June 8, 2022 by icarus2