How dark energy make universe to inflate?

[sweet springs]

Hi,
How dark energy make universe inflate? As it is positive energy, it seems to make universe contract by its gravity.
Regards.

 

[friend]

If dark energy is made of particles AND is responsible for space itself expanding (or contracting), then it seems there must be some interaction between space and particle, that particles can convert to space and visa versa.

 

[Chalnoth]

Hi,
How dark energy make universe inflate? As it is positive energy, it seems to make universe contract by its gravity.
Regards.

This stems from the Friedmann equations. The first of which is this:

H^2 = \rho

...where I have ignored constants for simplicity. H is the Hubble parameter:

H = {1 \over a} {da \over dt}

...while \rho is the energy density of the universe.

To give you an idea of how this works, let's take a universe where there is nothing but normal matter (or dark matter, take your pick). In such a situation \rho \propto 1/a^3. That is, the density of the matter decreases as as the volume of the universe increases. So with a matter-dominated universe we have:

H(a)^2 = {H_0^2 \over a^3}

Here I have introduced the Hubble constant, which is the value of the Hubble parameter now, when a = 1. If we substitute the definition of the Hubble parameter, some quick calculation leads to:

{da \over dt} = H_0 a^{-1 \over 2}

With a little bit of calculus, we can then integrate this equation to get how the scale factor a depends upon time:

a(t) \propto t^{2 \over 3}

Note that this power of time is smaller than one: this means that over time, the scale factor increases less and less. The universe decelerates. This is exactly what we expect from gravity, and is, in fact, the same whether we calculate based on General Relativity of Newtonian gravity. This should, I hope, make some degree of sense.

By contrast, we can take the simpler case where the energy density is a constant:

H(a)^2 = H_0^2
H(a) = H_0
{1 \over a} {da \over dt} = H_0
{da \over dt} = H_0 a

This is a classic differential equation: the rate of change in the scale factor is proportional to the scale factor. This is exponential growth. So as the universe expands, the scale factor increases, and the rate of change of the scale factor also increases! Thus, from the exact same law of gravity that predicts that a universe that dilutes will slow down, we end up with a universe that expands faster and faster if we just don't let the energy density dilute as it expands.

 

[sweet springs]

Hi.

Thank you so much, Chalnoth.
As for dark energy, density is conserved and total energy is not conserved, I see.

Some explanations refer to pressure of dark energy. Can someone tell me how pressure work in the theory ?

Regards.
I

 

[Chalnoth]

Hi.

Thank you so much, Chalnoth.
As for dark energy, density is conserved and total energy is not conserved, I see.

Yes. This article on energy conservation in General Relativity may interest you:
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/GR/energy_gr.html

Some explanations refer to pressure of dark energy. Can someone tell me how pressure work in the theory ?

Regards.
I

Well, one way to look at it is that the pressure is the reason why energy isn't conserved. Imagine, if you will, that we place some dark energy in a closed box. This dark energy, due to its nature, will place a negative pressure on the sides of the box. That is, it pulls the sides of the box inward in proportion to its energy density.

Let's say that this box, to begin with, has a volume of a^3: it's a cube with sides length a. Now, what happens if we increase the size of the box? Let's take the simplest case, where we only increase the size of the box in one direction, so that the new volume is a^2(a+b), where here we've lengthened the box in one direction by b. To do this, however, we had to apply a force to the side of the box, a force equal to the pressure times the area. Applying this force to cause the box to expand means that we did work on the box, increasing its energy. The amount that the energy increased is the force times the distance, which is equal to the pressure times the area of the side times the distance it increased, or:

\Delta E = \rho a^2b

...where I have substituted the energy density for the pressure, since they have the same magnitude. It should be no surprise, then, that this is exactly the energy increase in the dark energy inside the box from the increase in size: since the energy density of the stuff in the box stays the same, and the volume increases by a^2 b.

So we can understand the energy increase in the dark energy as coming from the force which is required to expand it. In an expanding universe, this force is gravity, and we can understand the added energy for dark energy as coming from gravitational potential energy.

 

[sweet springs]

Hi Chalnoth.
Thanks again.
As for dark energy dE= roh dV=-pdV p<0, I see.

Let me show further question.

1. One should interprete that the inflation in one of two ways ( roh or p ) or that generated dark energy has pressure p (roh and p) ?

2. What direction does pressure p have ? Does it lie in the universe or perdinicular to it ?

Regards.

 

[Chalnoth]

1. One should interprete that the inflation in one of two ways ( roh or p ) or that generated dark energy has pressure p (roh and p) ?

For inflation/dark energy, \rho = -p (or close to it).

2. What direction does pressure p have ? Does it lie in the universe or perdinicular to it ?

Pressure is omni-directional. There is no notion of a direction "outside" the universe in General Relativity.

 

[sweet springs]

Hi.
Thank you so much. I summarize my understanding. Dark energy has uniform energy density roh and pressure p=-roh direction of which is omni-directional and inward to any point considered.

I've read dark energy cause effective repulsive force. I imagine that any distance between a pair of gravitationally interacting bodies increases by the inflation. Prolonged distance r weakens gravitational force of -r^-2 effectively. Does it sound good ?

Regards.

 

[Chalnoth]

Hi.
Thank you so much. I summarize my understanding. Dark energy has uniform energy density roh and pressure p=-roh direction of which is omni-directional and inward to any point considered.

I've read dark energy cause effective repulsive force. I imagine that any distance between a pair of gravitationally interacting bodies increases by the inflation. Prolonged distance r weakens gravitational force of -r^-2 effectively. Does it sound good ?

Regards.

I don't think so. I seem to recall that in Newtonian gravity, it has the effect of adding a repulsive force proportional to the distance between two objects. That is:

F = (G m_1 m_2)\left({-1 \over r^2} + {r \Lambda}\right)\hat{r}

I'm having a hard time looking it up online, though, so I could be wrong here. But I think it's roughly correct.

 

[George Jones]

Yes. I also posted something similar in

https://www.physicsforums.com/showthread.php?p=2799641#post2799641.

I don't know of any online sources, but it is given in the books:

General Relativity: An Introdution for Physicsists by Hobson, Efstathiou, and Lasenby;

Gravitation: Foundations and Frontiers by Padmanabhan.

 

[Chalnoth]

Thanks! I see that I at least got the form correct, provided that the spherical case translates to the two particle case in the obvious way.

 

[Ich]

IMHO, the first Friedmann equation is not a good start for the problem. Better use the second, which explicitly states that deceleration = density plus three times pressure. It's the negative pressure that drives the acceleration, not density being constant. The latter is just a consequence of the continuity equation for p=-rho.
I just see that I got a sign wrong in the thread George Jones linked to, read "rho + 3p" in https://www.physicsforums.com/showthread.php?p=2799641#post2799641".

 

[sweet springs]

Hi.
Thanks George for showing a "springy" repulsive force by cosmic constant.
For tasting it better, let me ask how cosmic constant relates to dark energy density and its pressure ?
Regards.

 

[sweet springs]

Hi.
Einstein equation is
Gμν = 8πG/c^4 (Tμν+ ρgμν) where rho is energy density of dark energy, or
Gμν + Λgμν = 8πG/c^4 Tμν
so Λ = - 8πG/c^4 ρ

If Λ is negative, it causes attractive force.
Where did I make a mistake ?
Regards.

 

[Chalnoth]

Hi.
Einstein equation is
Gμν = 8πG/c^4 (Tμν+ ρgμν) where rho is energy density of dark energy, or
Gμν + Λgμν = 8πG/c^4 Tμν
so Λ = - 8πG/c^4 ρ

If Λ is negative, it causes attractive force.
Where did I make a mistake ?
Regards.

Er, that first equation is just wrong, namely:
G_{\mu\nu} = {8\pi G \over c^4} \left(T_{\mu\nu} + \rho g_{\mu\nu}\right)

The second equation is the correct form of Einstein's equations.

The energy density \rho is part of the stress energy tensor T_{\mu\nu}, not separate from it.

 

[sweet springs]

Hi, Chalnoth. Thanks for your patience.

Er, that first equation is just wrong, namely:
G_{\mu\nu} = {8\pi G \over c^4} \left(T_{\mu\nu} + \rho g_{\mu\nu}\right)

The second equation is the correct form of Einstein's equations.

The energy density \rho is part of the stress energy tensor T_{\mu\nu}, not separate from it.

So in the correct formula Gμν + Λgμν = 8πG/c^4 Tμν, cosmological constant is in LHS and dark energy is included in Tμν in RHS. They are two independent quantity. George taught us that positive Λ brings spring like repulsive force but none about effect of dark energy. Am I right ?
Regards.

 

[Chalnoth]

So in the correct formula Gμν + Λgμν = 8πG/c^4 Tμν, cosmological constant is in LHS and dark energy is included in Tμν in RHS. They are two independent quantity. George taught us that positive Λ cause spring like repulsive force but none about effect of dark energy. Am I right ?
Regards.

You specifically meant \rho_{\Lambda}, then? In that case, you have to be a bit more careful and look at the specific way in which you would construct a stress-energy tensor that has a constant energy density in time and space.

You're right that you pick up one minus sign by moving the \Lambda term to the other side of the equation, but you can't simply equate that to energy density.

 

[George Jones]

For inflation/dark energy, \rho = -p (or close to it).

Hi, Chalnoth. Thanks for your patience.So in the correct formula Gμν + Λgμν = 8πG/c^4 Tμν, cosmological constant is in LHS and dark energy is included in Tμν in RHS. They are two independent quantity. George taught us that positive Λ cause spring like repulsive force but none about effect of dark energy. Am I right ?
Regards.

Dark energy that satisfies the condition that Chalnoth gave is indistinguishable from a positive cosmological constant. To see why, take the cosmological constant term to the other side of Einstein's equation.

I think that it will take a quantum theory of gravity to disentangle things.

[edit]Didn't see Chalnoth's previous post.[/edit]

 

[sweet springs]

Hi.

Dark energy that satisfies the condition that Chalnoth gave is indistinguishable from a positive cosmological constant. To see why, take the cosmological constant term to the other side of Einstein's equation.

I read you mean Gμν = 8πG/c^4 Tμν - Λgμν = 8πG/c^4 (traditional Tμν + Tμν of dark energy )
where Tμν of dark energy = (8πG/c^4)^-1 ( - Λgμν )
Here we see T00 of dark energy, dark energy density, is negative. I expect it is ρ as we named in this thread and positive.
What's wrong ?
Regards.

 

[Chalnoth]

Hi.


I read you mean Gμν = 8πG/c^4 Tμν - Λgμν = 8πG/c^4 (traditional Tμν + Tμν of dark energy )
where Tμν of dark energy = (8πG/c^4)^-1 ( - Λgμν )
Here we see T00 of dark energy, dark energy density, is negative. I expect it is ρ as we named in this thread and positive.
What's wrong ?
Regards.

That depends upon your convention for the signature of the metric. Make sure you've properly kept track of all of the indices.

 

[sweet springs]

That depends upon your convention for the signature of the metric. Make sure you've properly kept track of all of the indices.

I confirm your suggestion. From
T_μν of dark energy = (8πG/c^4)^-1 ( - Λg_μν )
By contraction
Tμ_μ of dark energy = (8πG/c^4)^-1 ( - Λgμ_μ)
So in (+---)　way in conformal system,
ρ - 3p = (8πG/c^4)^-1 (- Λ)4
As ρ=-p>0 for dark energy,
4ρ = - 4 (8πG/c^4)^-1 Λ < 0 ?
I think something is wrong in my tensor calculation.

Or am I confusing sign of Λ ?
I order sign of Λ to be positive for the inflating universe with acceleration, its definition is Gμν + Λgμν = 8πG/c^4 Tμν or Gμν = 8πG/c^4 Tμν + Λgμν ?

Regards.

 

[sweet springs]

PS
Formula (20) of the paper http://arxiv.org/PS_cache/astro-ph/pdf/0207/0207347v2.pdf applies the definition Gμν = 8πG/c^4 Tμν + Λgμν
Formula (24) shows spring like repulsive force.
Regards.

 

[Chalnoth]

PS
Formula (20) of the paper http://arxiv.org/PS_cache/astro-ph/pdf/0207/0207347v2.pdf applies the definition Gμν = 8πG/c^4 Tμν + Λgμν
Formula (24) shows spring like repulsive force.
Regards.

As an aside, you should really take advantage of the forum's Latex capabilities. Just enclose the Latex code in [noparse][/noparse] tags.

Anyway, I'd have to look into it a bit more, but it is possible that there is a sign convention issue going on here. In any case it is clear that a cosmological constant that equates to positive energy density produces accelerated expansion.

 

[sweet springs]

I've learned a lot in this thread. Thanks.
Is it right to say that accelerating inflating universe and spring like repulsive force comes from minus sign of invariant contraction of dark energy tensor, i.e. rho + 3p < 0 in comparison to invariant contraction of energy tensor of usual and dark matters, i.e. energy density + 3 * pressure > 0.
Regards.

 

[Chalnoth]

I've learned a lot in this thread. Thanks.
Is it right to say that accelerating inflating universe and spring like repulsive force comes from minus sign of invariant contraction of dark energy tensor, i.e. rho + 3p < 0 in comparison to invariant contraction of energy tensor of usual and dark matters, i.e. energy density + 3 * pressure > 0.
Regards.

Yup.

 

[sweet springs]

Yup.

Thanks.

 

[sweet springs]

Hi.

The weak-field limit of Einstein's equation with cosmological constant/dark energy \Lambda leads to a modified "Poisson" equation,

\nabla^2 \Phi = 4 \pi G \rho - \Lambda c^2.

For a spherical mass M, the divergence theorem applied to the above gives

\vec{g} = \left(-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3} r \right) \hat{r}.

The second term is a "springy" repulsive term for positive \Lambda.

Distance between any two bodies increase with acceleration by expansion of universe with dark energy.
I interpret term + \frac{c^2 \Lambda}{3} r \right) \hat{r} express this accelerating increase of the distance or generation of new space between.
Is it OK?
Regards.

 

[sweet springs]

Hi.

Thanks.

I resume again here. In GR not only energy but momentum (pressure) take part in evolution of universe. Ordinary plus energy and extraordinary minus pressure of dark energy work out virtually repulsive force and accelerating expansion of universe.
Thank you.

 

[Chalnoth]

Hi.


I resume again here. In GR not only energy but momentum (pressure) take part in evolution of universe. Ordinary plus energy and extraordinary minus pressure of dark energy work out virtually repulsive force and accelerating expansion of universe.
Thank you.

Largely, yes. However, I'd like to nitpick a little here.

Pressure and momentum are different things. Energy density is the time-time component of the stress-energy tensor. Pressure makes up the diagonal space-space elements. The off-diagonal space-space elements are twisting forces (shear). The space-time components of the stress-energy tensor is the momentum in different directions.

One way to think about it is that talking about the momentum of a chunk of matter, if it is non-zero, means that chunk of matter as a whole is moving. The pressure, on the other hand, talks about how much the constituent elements of that chunk of matter are pushing against one another.

When we're talking about a fluid, we can consider the fluid being inside a closed box. In that situation, the pressure is the force per unit area on the sides of the box. A positive pressure pushes outward, while a negative pressure pulls inward.

For normal matter and photons, we can understand pressure as being related to the individual momenta of the constituent particles: the pressure comes from the particles bouncing off the sides of the box. This description doesn't make sense with regard to dark energy, however.