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Critical density: question correction

  1. Jul 23, 2015 #1
    Hope I'm in the right section for this question! In the big bang model, the expansion of the universe is slowed down by gravity. If there is enough matter in the universe, then the expansion can be overcome and the universe will collapse in the future. The density of matter that is just sufficient to eventually halt the expansion is called the critical density. The equation for the critical density is

    ρcrit = 3H₀²/ 8πG

    You can see that the critical density is proportional to the square of the Hubble constant — a faster expansion requires a higher density to overcome the expansion.


    We can calculate ρcrit by inserting the gravitational constant, G = 6.67 × 10-11 Nm2 / kg2, and adopting H0= 70 km/s/Mpc. We first convert the Hubble constant to metric units, H0= 2 × 10-18s-1. Now we can solve to get ρcrit = 3 × (2.1 × 10-18)2 / 8 × 3.14 × 6.7 × 10-11 = 7.9 × 10-27 kg/m3 ≈ 10-26 kg/m3. Equal to about five hydrogen atoms per cubic meter.

    With all that being said, can anyone tell me why;
    8π x G? Or why 3 × H²?

    In other words, I'm looking for an explanation as to why we are using particular numbers like 8π or 3H² to achieve density?
     
    Last edited: Jul 23, 2015
  2. jcsd
  3. Jul 23, 2015 #2
    From equations which describes Universe.

    Did you know friedmann equation
    ##H^2-8πGρ/3=-k/a^2(t)+Λc^2/3##
    If we assume ##k=0## (Which it is) and ##Λ=0## (We assume ##Λ=0## for simplicity) then we get
    ##H^2=8πGρ/3##
    then ##p=3H^2/8πG##
    These numbers derived from equations.Einstein Field Equations gives us friedmann equation.And friedmann equation gives us critical density equation
     
  4. Jul 23, 2015 #3
    Yes, thank you for your reply, I'm am aware of these calculations. My question can even be related to them if preferred. Question still stands how are the values 8π and 3 relevant to the equations parameters. They are acting as constants in themselves but for what purpose? Where's is the dimensionality in 8π other than to say it undergoes 4 rotations? Relatively so, what is the dimensionality of 3? In other words what are they explaining in the the equation?
     
  5. Jul 23, 2015 #4
    I meant to tag you for a response. Here we go
     
  6. Jul 23, 2015 #5

    PeterDonis

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    Staff: Mentor

    Note that this is only true if the cosmological constant is zero. If it's positive (which, according to our best current model, it is), a critical density universe will still expand forever. In fact, according to our best current model, our universe is at the critical density (if the density of dark energy, i.e., the cosmological constant, is included), and will expand forever.

    We're not. We're using them because they are needed to make the equations match reality.

    These are dimensionless numbers; they're not there to correct the units of anything. They're there because they have to be there to make the equations match reality.

    I'm not sure what you mean. The equation is supposed to describe reality, i.e., to describe the relationship between density and expansion rate (among other things) that we actually observe. But without the ##8 \pi## and the ##3## in there, the equation doesn't correctly describe that relationship.
     
  7. Jul 23, 2015 #6
    The derivation of Friedmann equation:
    ##1/2mV^2-mMG/r=U##
    ##V^2-2MG/r=2U/m##
    ##M=4/3πρR^3##, so
    ##H^2R^2-8πGR^2/3=-k##
    ##H^2-8πG/3=-k/R^2##

    ##k=-2U/m## U=Total energy

    This derivation is wrong but It can give you an idea about where this 8,π,G,3 come from.

    Or Lets think the other way ##p=3H^2/8πG##
    ##8πG/3=H^2/p##
    So these numbers are just constants to make ##H^2/p## same for all time(H and p is time dependet)
     
  8. Jul 23, 2015 #7


    Thanks for your replies these are all great answers! Bare with me I'm still highschool level. Although
    Very interesting!
    Is there any other way physicists can derive critical density without it being derived from the Friedman equations?
    Seems to me there must logically be another way to calculate to the density of the universe and arrive with the same value and units...
     
  9. Jul 23, 2015 #8

    PeterDonis

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    No.

    The actual density of the universe is not the same, conceptually, as the critical density. The actual density is whatever we measure it to be. The critical density is a theoretical quantity that is calculated from the Friedmann equations. Our best measurements indicate that, to within experimental error, the two are numerically the same, but that doesn't make them the same thing. It just means they happen to be equal numerically.
     
  10. Jul 23, 2015 #9
    Ahh yes Well noted. Interesting though... There numerical similarity describes what then ? If there separateness lies in their description, then there must be some sort of error in our collective methods and descriptions of such mechanics... Wouldn't you agree?
     
  11. Jul 23, 2015 #10

    PeterDonis

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    Um, that the actual density of the universe is the same as the theoretically calculated critical density? I'm not sure what you're trying to ask here.

    I don't know because I don't understand what you're asking.
     
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