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How dark energy make universe to inflate?

  1. Sep 11, 2010 #1
    Hi,
    How dark energy make universe inflate? As it is positive energy, it seems to make universe contract by its gravity.
    Regards.
     
  2. jcsd
  3. Sep 11, 2010 #2
    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.
     
  4. Sep 12, 2010 #3

    Chalnoth

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

    [tex]H^2 = \rho[/tex]

    ...where I have ignored constants for simplicity. [itex]H[/itex] is the Hubble parameter:

    [tex]H = {1 \over a} {da \over dt}[/tex]

    ...while [itex]\rho[/itex] 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 [tex]\rho \propto 1/a^3[/tex]. That is, the density of the matter decreases as as the volume of the universe increases. So with a matter-dominated universe we have:

    [tex]H(a)^2 = {H_0^2 \over a^3}[/tex]

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

    [tex]{da \over dt} = H_0 a^{-1 \over 2}[/tex]

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

    [tex]a(t) \propto t^{2 \over 3}[/tex]

    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:

    [tex]H(a)^2 = H_0^2[/tex]
    [tex]H(a) = H_0[/tex]
    [tex]{1 \over a} {da \over dt} = H_0[/tex]
    [tex]{da \over dt} = H_0 a[/tex]

    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.
     
  5. Sep 13, 2010 #4
    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
     
  6. Sep 13, 2010 #5

    Chalnoth

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    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

    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 [itex]a^3[/itex]: it's a cube with sides length [itex]a[/itex]. 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 [itex]a^2(a+b)[/itex], where here we've lengthened the box in one direction by [itex]b[/itex]. 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:

    [tex]\Delta E = \rho a^2b[/tex]

    ...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 [tex]a^2 b[/tex].

    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.
     
  7. Sep 14, 2010 #6
    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.
     
  8. Sep 14, 2010 #7

    Chalnoth

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

    Pressure is omni-directional. There is no notion of a direction "outside" the universe in General Relativity.
     
  9. Sep 15, 2010 #8
    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.
     
  10. Sep 15, 2010 #9

    Chalnoth

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    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:

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

    I'm having a hard time looking it up online, though, so I could be wrong here. But I think it's roughly correct.
     
  11. Sep 15, 2010 #10

    George Jones

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  12. Sep 15, 2010 #11

    Chalnoth

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    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.
     
  13. Sep 15, 2010 #12

    Ich

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    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".
     
    Last edited by a moderator: Apr 25, 2017
  14. Sep 15, 2010 #13
    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.
     
    Last edited: Sep 15, 2010
  15. Sep 17, 2010 #14
    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.
     
  16. Sep 17, 2010 #15

    Chalnoth

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

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

    The energy density [itex]\rho[/itex] is part of the stress energy tensor [itex]T_{\mu\nu}[/itex], not separate from it.
     
  17. Sep 18, 2010 #16
    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 brings spring like repulsive force but none about effect of dark energy. Am I right ?
    Regards.
     
    Last edited: Sep 18, 2010
  18. Sep 18, 2010 #17

    Chalnoth

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    You specifically meant [itex]\rho_{\Lambda}[/itex], 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 [itex]\Lambda[/itex] term to the other side of the equation, but you can't simply equate that to energy density.
     
  19. Sep 18, 2010 #18

    George Jones

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    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]
     
  20. Sep 18, 2010 #19
    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.
     
  21. Sep 18, 2010 #20

    Chalnoth

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