Questions about Negative Pressure and Vacuum Energy

[darkdark10]

Currently, dark energy is described as a being that exerts a negative pressure while having a positive energy density.

$${\rho _\Lambda } + 3{P_\Lambda } = {\rho _\Lambda } + 3( - {\rho _\Lambda }) = - 2{\rho _\Lambda }$$

However, there seems to be a problem with the negative pressure assertion.

In the book(An Introduction to Modern Astrophysics P1161)

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.

Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy

1. Sign of the negative pressure
In general relativity, pressure is equivalent energy density of the kinetic energy.
In the acceleration equation, 3P(c=1) has the idea of an equivalent mass 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.
$$K = \frac{1}{2}m{v^2} < 0$$
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,
$$P = \frac{1}{3}(\frac{{{v^2}}}{{{c^2}}})\rho = \omega \rho$$
In the case of matter, v << c, So $$P = \frac{1}{3}{(\frac{v}{c})^2}\rho \simeq 0$$
In the case of radiation, v=c, So $$P = \frac{1}{3}{(\frac{v}{c})^2}\rho = \frac{1}{3}\rho$$
However, in the case of dark energy $$P = \frac{1}{3}{(\frac{v}{c})^2}\rho = - \rho$$
$$v = (\sqrt 3 c)i$$
We need energy density with imaginary super-luminous speed.

Ideal gas state equation applies to "massless ~ no upper limit'' particles, and the velocity ranges from "0 ~ c'' are all included. If we seriously consider the physical presence of P= -ρ = - 3((1/3)ρ), we will find that there is a serious problem.

It is very strange. So, let's look again at the logic we're convinced of negative pressure.
dU=dQ-dW, if dQ=0, dU=-dW=-PdV
dU=-PdV : if dU=ρdV, P=-ρ

I thought about whether the logic using dU=-PdV is correct.
1) The argument regarding negative pressure is an inverted explanation
Pressure P = equivalent energy density of kinetic energy
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''. Can we be sure that this logic is correct?

2) ρ_Λ+ 3P_Λ = ρ_Λ + 3(-ρ_Λ) =-2ρ_Λ
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_Λ+ 3P_Λ =-2ρ_Λ. In other words, it suggests that the object (or energy density) has a gravity with a negative mass density of -2ρ_Λ. This is different from a vacuum with a positive energy density +ρ_Λ, which we think of.

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

I am not sure if this equation(dU=-PdV) 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.

Q1. Is it possible to have negative kinetic energy while having positive mass density?

Q2. Does the dU=-PdV equation hold for a system in which energy is not conserved?

Q3. Does vacuum energy with positive energy density ρ produce negative pressure?
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 volume?

Q4. Can't there be a vacuum energy density with P=0 ~ (1/3)ρ?

 

[PeterDonis]

there seems to be a problem with the negative pressure assertion.

No, there isn't. The vacuum has to be Lorentz invariant. That requires that the vacuum have the equation of state $p = - \rho$.

Your objections are based on misconceptions:

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

No, it isn't.

We need energy density with imaginary super-luminous speed.

No, we don't. The energy density of the vacuum is not moving. I don't know where you are getting your "ideal gas state equation" in terms of $v / c$, but it doesn't apply to the vacuum.

Both mass density ρ and pressure P are sources of gravity.

Yes, but that doesn't mean energy density and pressure are the same thing, which is what you're basing your reasoning on.

in the case of vacuum energy and the cosmological constant, energy conservation does not hold.

This is correct for the meaning you are assigning to "energy conservation". It's also not a problem, since GR does not require that "energy conservation" in your sense must hold. The only "energy conservation" that must hold in GR is that the covariant divergence of the stress-energy tensor is zero. The stress-energy tensor of the vacuum is $\Lamba g_{\mu \nu}$, with $\Lambda$ a constant, and the covariant divergence of that is zero.

Q1. Is it possible to have negative kinetic energy while having positive mass density?

No, but that has nothing to do with vacuum energy. See above.

Q2. Does the dU=-PdV equation hold for a system in which energy is not conserved?

Not necessarily, but that's not a problem. See above.

Q3. Does vacuum energy with positive energy density ρ produce negative pressure?

Yes. See above for the equation of state of vacuum energy.

Q4. Can't there be a vacuum energy density with P=0 ~ (1/3)ρ?

No.

 

[darkdark10]

In the book(An Introduction to Modern Astrophysics - Bradley W. Carroll and Dale A.Ostile)

Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy

Isn't the pressure P in the acceleration equation the equivalent energy density of kinetic energy?

 

[George Jones]

In the book(An Introduction to Modern Astrophysics - Bradley W. Carroll and Dale A.Ostile)

View attachment 302768
Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy

Isn't the pressure P in the acceleration equation the equivalent energy density of kinetic energy?

Dark energy is not caused by a song on the Mason Williams Phonograph Record.

 

[PeterDonis]

Isn't the pressure P in the acceleration equation the equivalent energy density of kinetic energy?

I already answered this. No.

 

[PeterDonis]

Pressure P = equivalent energy density of kinetic energy
P/c^2 = equivalent mass density of kinetic energy

No. I'm not familiar with the textbook you referenced, but I would say the passage you quoted is badly worded. The "equivalent mass of the particle's kinetic energy" goes in the mass density (or energy density, depending on which units you use) $\rho$, not the pressure $P$. The pressure $P$ is the force per unit area exerted by one parcel of fluid on another (assuming we are talking about a perfect fluid). Pressure does appear in the stress-energy tensor and in the "source of gravity" term in the Friedmann equation, but it is not the same as "equivalent energy density of kinetic energy"--kinetic energy goes in $\rho$, not $P$, as above.