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Using positive pressure in cosmological equations

  1. Aug 15, 2013 #1
    The Friedmann solution(s) to the GR cosmological equations (without Lambda) assume pressure=0. What happens if we let p>0 ?

    It seems to me that there is no continuous solution to the cosmological equations in that case. Yet, if we include electro-magnetic radiation (such as MBR) in the energy density, p must be > 0 as the Universe expands.
     
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  3. Aug 15, 2013 #2

    George Jones

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    Why not?

    Yes, for a radiation (photon gas) dominated universe, ##p = \rho/3##.

    More generally, it is often assumed that the cosmological fluid has equation of state ##p = w \rho##, with ##w=-1## for dark energy, ##w=0## for (dust) matter, and ##w=1/3## for radiation.
     
  4. Aug 16, 2013 #3
    Unfortunately one of the best textbooks covering the FLRW metrics is copywrited so I can't link it here. That book is Barbera Rydens "Introductory to cosmology"

    In that particular textbook she does a brilliant and often cited job of using the FLRW metrics to describe the universe in single and multi component universes.
    This included matter only, Lambda only, radiation only etc.
    She then proceeded to describe the numerous multi component universes.

    I've read numerous textbooks and articles on cosmology and to date none of them did as thorough if a job in this particular manner.
    The FLRW metric is more than capable in that manner.
    If your interested I highly recommend that particular text. As its also one of the most straight forward and easy
    to understand books that I am
    familiar with.

    Unfortunately I an working from my phone so it would be too painful to post her exampled as well as too lengthy.
     
  5. Aug 16, 2013 #4

    Chalnoth

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    As others have mentioned, this is the case for radiation.

    What it does is it causes the expansion of the universe to decelerate more rapidly, while at the same time the material with p > 0 dilutes more rapidly than matter (with p=0). This means that the period of time where the positive pressure stuff was relevant was the very early universe: very early-on, the universe was dominated by radiation, and was decelerating very rapidly. Later, the radiation diluted away, leaving most of the energy density of the universe dominated by matter which experiences no pressure on cosmological scales. The deceleration continued, but slowed.

    More recently, the matter has diluted to the point that dark energy (with negative pressure) has come to dominate, and the expansion has transitioned from decelerating to accelerating.
     
  6. Aug 21, 2013 #5
    The current estimated density ##\rho## of matter in the universe is about ##.05\rho_{critical}## . Suppose ##\rho_r## is the current radiation density of MBR. Both ##\rho## and ##\rho_{critical}## vary as ##a^{-3}## but ##\rho_r## varies as ##a^{-4}## where ##a## is the scale factor. Consequently as ##a## gets smaller, for some value of ##a=a_0##, ##\rho+\rho_r=\rho_{critical}## and for ##a<a_0##, ##\rho+\rho_r>\rho_{critical}##. The three distinct Friedmann solutions depend on only one of ##\rho+\rho_r<\rho_{critical}##, ##\rho+\rho_r=\rho_{critical}## or ##\rho+\rho_r>\rho_{critical}## being true throughout the history of the Universe.
     
  7. Aug 21, 2013 #6

    George Jones

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    Because the critical density is not constant, this isn't true. It is easy to show that the critical density evolves in time in such a way that ##\left( \rho +\rho_r \right) - \rho_{critical}## never changes sign.
     
  8. Aug 22, 2013 #7
    I didn't actually say the critical density is constant. I said it varies as ##a^{-3}##. For the cosmological equation with ##k=0 ##, ##(\dot a/a)^2=8 \pi G \rho /3 ##.
    The solution of this differential equation is ##a=a_0 t^{2/3} ## if ##t=(6 \pi G \rho)^{-1/2} ##. So, ##(\dot a/a)^2=(2/3t)^2=8 \pi G \rho /3 ## which implies ##\rho \propto t^{-2} ## and since ##a=a_0 t^{2/3} ##, ##\rho \propto a^{-3} ##. But in this case, ##\rho=\rho_{critical} ##. (Note: the ##a_0## used here is not necessarily the same as that used in post 5)
     
    Last edited: Aug 22, 2013
  9. Aug 24, 2013 #8

    George Jones

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    Sorry, I missed this; still, it is not true in general that critical density "varies as ##a^{-3}##."

    Exactly. This gives that the general form for critical density is

    $$\rho_{critical} = \frac{3}{8\pi G} \left(\frac{\dot{a}}{a} \right)^2 .$$

    This is true for pressure-free matter (dust), but ##t=(6 \pi G \rho)^{-1/2}## is not true for radiation, or for a mixture of radiation and dust, or when the cosmological constant is non-zero, ... .

    Assuming that a continuum approximation is valid on a large scale, matter in the universe is modeled by a perfect fluid with density ##\rho\left(t\right)## pressure ##p\left(t\right)##. An assumption of spatial homogeneity and spatial isotropy leads to an FLRW metric with scale factor ##a\left(t\right)##.

    To solve for the three functions ##\rho\left(t\right)##, ##p\left(t\right)##, and ##a\left(t\right)##, three equations are needed. One equation is the equation that, above, you called "the cosmological equation", which follows from Einstein's equation with a FLRW used for the left side, and the stress-energy tensor for a perfect fluid used for the right side. Einstein's equation leads also to a second independent equation, but, here, it is more useful to use an equivalent independent equation,

    $$\dot{\rho} + \left(\rho + \frac{p}{c^2}\right) \frac{3\dot{a}}{a} = 0.$$

    The left side of EInstein's equation is divergence-free, and, consequently, so is the right. Taking the divergence of the stress-energy tensor for a perfect fluid leads to the second equation. The third equation is an equation of state that relates density and pressure. An often used equation of state is (with ##c## restored) ##p = w\rho c^2##, with ##w## constant. As I said in my first post, radiation has ##w = 1/3##.

    Using the equation of state ##p = w\rho c^2## in the second equation gives ##\rho \propto a^{-3\left(1 + w\right)}##. Using this and ##w = 1/3## in the second equation leads, for radiation, to ##a \propto t^{1/2}## and ##\rho = \rho_{critical} \propto a^{-4}##.

    Thus, for radiation (and in general)

     
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