I Variable Dark Energy: Would P Remain Constant?

[Ranku]

If dark energy density were to vary with time, in the equation of state w = p/ρ, would p remain constant and only ρ vary?

 

[Arman777]

I am not sure but here is my idea.
Let us take the equation,

$P=wε$ and $ε(a)=ε_0a^{-3(1+w)}$ so from here if you want to vary the dark energy first thing is $w≠-1$. So from the equation we can see that $w$ should be a constant. And then if $ε$ changes then $P$ should change to.

 

[PeterDonis]

If dark energy density were to vary with time, in the equation of state w = p/ρ, would p remain constant and only ρ vary?

Not if the equation of state remains the same, since it gives a fixed relationship between $p$ and $\rho$. I am not aware of any speculative models where the equation of state can vary, but that doesn't mean there couldn't be one. However, even in a model where the equation of state varied, it would have to be very fine-tuned to keep $p$ constant if $\rho$ changes.

 

[kimbyd]

If dark energy density were to vary with time, in the equation of state w = p/ρ, would p remain constant and only ρ vary?

It depends entirely upon how the relationship between pressure and density varied. Basically it follows conservation of the stress-energy tensor, which in turn behaves as if the co-moving volume in an expanding universe is an insulated box, and the dark energy does work on the box as the universe expands. Positive work decreases energy within the co-moving volume, negative work increases it.

 

[kimbyd]

I am not sure but here is my idea.
Let us take the equation,

$P=wε$ and $ε(a)=ε_0a^{-3(1+w)}$ so from here if you want to vary the dark energy first thing is $w≠-1$. So from the equation we can see that $w$ should be a constant. And then if $ε$ changes then $P$ should change to.

That equation only works if $w$ is a constant. If $w$ is not constant, then you have to do something a bit more complicated. It's not too terrible, though. The statement I wrote above works.

If you want to get really technical, the stress-energy conservation law applied to the FRW universe can be written as:

$$\dot{\rho} + 3 {\dot{a} \over a}\left(\rho + p\right) = 0$$

If you take $p = w\rho$ and $w =$ constant, then you get the equation Arman777 wrote above. If $w$ is not constant, it's more complicated.

 

[Arman777]

That equation only works if $w$ is a constant. If $w$ is not constant, then you have to do something a bit more complicated. It's not too terrible, though. The statement I wrote above works.

If you want to get really technical, the stress-energy conservation law applied to the FRW universe can be written as:

$$\dot{\rho} + 3 {\dot{a} \over a}\left(\rho + p\right) = 0$$

If you take $p = w\rho$ and $w =$ constant, then you get the equation Arman777 wrote above. If $w$ is not constant, it's more complicated.

That is the fluid equation, right.

I think the technical problem is can $w$ change ? Since if it changes it means that the property of the matter/material is also changing. Or I believe, at least there must be limits on this change.

 

[Ranku]

Not if the equation of state remains the same, since it gives a fixed relationship between $p$ and $\rho$. I am not aware of any speculative models where the equation of state can vary, but that doesn't mean there couldn't be one. However, even in a model where the equation of state varied, it would have to be very fine-tuned to keep $p$ constant if $\rho$ changes.

I am trying to understand this in terms of the recent paper which uses quasars as standard candles, and finds that dark energy density is increasing with time, i.e., $w$ <-1. In terms of $w$ = - P/ρ, to obtain increasing values of -1, with dark energy density increasing, we would need both P and ρ to increase, with P increasing at a greater rate than ρ. But can we not get the same effect of increasing values of -1 with P increasing and ρ remaining constant, or P remaining constant and ρ decreasing? So how do we determine which of these three possibilities are occurring?

 

[Ranku]

It depends entirely upon how the relationship between pressure and density varied. Basically it follows conservation of the stress-energy tensor, which in turn behaves as if the co-moving volume in an expanding universe is an insulated box, and the dark energy does work on the box as the universe expands. Positive work decreases energy within the co-moving volume, negative work increases it.

Positive work would be expansion of the universe, and negative work would be contraction of the universe?

 

[kimbyd]

That is the fluid equation, right.

I think the technical problem is can $w$ change ? Since if it changes it means that the property of the matter/material is also changing. Or I believe, at least there must be limits on this change.

Oh, definitely. Take normal matter, for instance. At high temperatures, it behaves like radiation with $w=1/3$. At low temperatures it has $w=0$.

Most alternative models of dark energy use scalar fields, and they basically never have constant $w$.

 

[kimbyd]

Positive work would be expansion of the universe, and negative work would be contraction of the universe?

No, that's not what I mean.

The pressure of the fluid exerts a force per unit area on the imaginary wall of the imaginary box. As the box expands with the universe, that force does work. If the imaginary force on the imaginary box is pushing outward, then it does positive work, causing a loss of energy of the fluid. If the imaginary force on the imaginary box is pulling inward, such that the force is in opposition to the motion, then it does negative work resulting in an increase of energy of the fluid.

 

[Ranku]

No, that's not what I mean.

The pressure of the fluid exerts a force per unit area on the imaginary wall of the imaginary box. As the box expands with the universe, that force does work. If the imaginary force on the imaginary box is pushing outward, then it does positive work, causing a loss of energy of the fluid. If the imaginary force on the imaginary box is pulling inward, such that the force is in opposition to the motion, then it does negative work resulting in an increase of energy of the fluid.

Is the imaginary force the force of the pressure?

 

[kimbyd]

That equation only works if $w$ is a constant. If $w$ is not constant, then you have to do something a bit more complicated. It's not too terrible, though. The statement I wrote above works.

If you want to get really technical, the stress-energy conservation law applied to the FRW universe can be written as:

$$\dot{\rho} + 3 {\dot{a} \over a}\left(\rho + p\right) = 0$$

If you take $p = w\rho$ and $w =$ constant, then you get the equation Arman777 wrote above. If $w$ is not constant, it's more complicated.

By the way, to explain this equation in a little bit more detail, you can understand each of the three terms with respect to the co-moving volume mentioned earlier. The first term is just the density changing over time. The second can be thought of as the volume changing as the box expands, which will change the total energy in the volume if density remains constant. The third term is the hypothetical pressure on each side of the box. For the latter two terms, the factor of 3 comes in because the box is expanding in three dimensions.

Just because I feel like delving into the math a bit, here's one way to understand it:

Imagine a cube. It has sides of length $a$. That cube is expanding with its center at the origin, which means that each face of the cube is moving outward at a speed of $\dot{a}/2$: if opposing faces move away from one another at $\dot{a}$, then they move away from the origin at half that speed.

Now, the rule we're going to use is that the fluid represented by $\rho$ and $p$ conserves energy in the sense that the change in energy of the fluid is equal to the work the fluid performs on this hypothetical box.

The change in energy for the fluid can be represented as:

$${dE \over dt} = {d \over dt} \rho V = V {d \rho \over dt} + \rho {dV \over dt}$$

The first term is easy: it's just $V \dot{\rho}$ by definition. But what's the second term? Well, $V = a^3$, so $\dot{V} = 3a^2\dot{a} = 3V \dot{a}/a$.

Finally, what about the work performed? Well, the pressure exerts a force per unit area on each side of the box. The work performed is the vector product of the force and the velocity of the wall of the box, multiplied by the number of faces of the box, i.e.:

$$W = 6 p a^2\dot{a} / 2 = 3V p {\dot{a} \over a}$$

Each term has a factor of $V$ in it, so we can divide by that that, which gives us the fluid equation:

$$\dot{\rho} + 3 {\dot{a} \over a} \rho + 3 {\dot{a} \over a} p = \dot{\rho} + 3 {\dot{a} \over a} \left(\rho + p\right)$$

You can verify I got the sign right by noting that a positive work means a drop in the total energy of the fluid, while a negative work increases the energy of the fluid.

Granted, this isn't a very rigorous calculation, and I'm not sure you can always do this kind of analysis. But the answer is consistent with the full math stemming from General Relativity in this specific case, so I'll take it.

 

[Arman777]

Most alternative models of dark energy use scalar fields, and they basically never have constant www.

Hmm that's interesting. I know that dark energy is generally called $w<-1/3$ but I never knew that the $w$ of dark energy can change in time.

 

[Arman777]

Is the imaginary force the force of the pressure?

Yes indeed. Look

The pressure of the fluid exerts a force per unit area on the imaginary wall of the imaginary box.

And what else it could be...?

Thats a nice derivation

 

[Ranku]

Yes indeed. Look

And what else it could be...?

I was trying to once again ascertain what was denied in the previous post.

 

[kimbyd]

Is the imaginary force the force of the pressure?

The force from the pressure, yes. It's imaginary because the box isn't actually there, so there's nothing for the pressure to push against.

 

[kimbyd]

Hmm that's interesting. I know that dark energy is generally called $w<-1/3$ but I never knew that the $w$ of dark energy can change in time.

Maybe. There aren't any really good models of dark energy. Except for the cosmological constant, which is on solid theoretical foundations but has a really really weird value, there hasn't yet been a model of dark energy that really makes theoretical sense.

 

[Arman777]

Maybe. There aren't any really good models of dark energy. Except for the cosmological constant, which is on solid theoretical foundations but has a really really weird value, there hasn't yet been a model of dark energy that really makes theoretical sense.

I see. Its interesting