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I Hubble parameter in terms of the scale factor

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  1. Aug 3, 2016 #1
    From Introduction to Cosmology by Matt Roos, he wanted to derive the Hubble parameter in terms of the scale factor. From the Friedmann's equation,

    ##\frac{R'^2 + kc^2}{R^2} = \frac{8πG}{3}ρ##

    The density parameter is ##~Ω(a) = \frac{8πG}{3H_o^2}ρ(a)~## and let ##~Ω_k = \frac{-kc^2}{H_o^2}##

    So, ##~H(a)^2 = H_o^2 ( Ω_k a^{-2} + Ω(a) )~~## where ##Ω(a) = Ω_m(a) + Ω_r(a) + Ω_λ(a)##
    m = matter, r = radiation, k = curvature, λ = cosmological constant

    It is also known that the density parameter evolves as (for example, a matter dominated universe)
    ##~Ω_m(a) = Ω_m \frac{H_o^2}{H^2} a^{-3}##

    The expression for the Hubble parameter in terms of the scale factor is,
    ##~H(a)^2 = H_o^2 ( Ω_m a^{-3} + Ω_r a^{-4} + Ω_k a^{-2} + Ω_λ ) ##

    But how is that possible? If I substitute the density parameters in terms of how they evolve, there will be an extra factor of ##~ \frac{H_o^2}{H^2}~##. Any comments?
     
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  3. Aug 3, 2016 #2

    Chalnoth

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    I think the problem here is that [itex]\Omega_m(a)[/itex] is quite different conceptually from [itex]\Omega_m[/itex]. The former is the density fraction over time, while the latter is the current density fraction.

    The density of matter over time, for example, is given by:
    [tex]\rho_m = {3H_0^2 \over 8\pi G}{\Omega_m \over a^3}[/tex]

    You can think of the [itex]\Omega_m[/itex] in this equation as providing a sort of "initial condition" for the density parameter, which then evolves in accordance with the conservation of stress-energy (which gives the [itex]a^{-3}[/itex] scaling).

    The density fraction over time ([itex]\Omega_m(a)[/itex]) compares that density over time to all of the other densities, providing an estimate of how relevant that particular component of the universe was to the expansion at any given point in time.
     
  4. Aug 3, 2016 #3

    Bandersnatch

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    I must admit I also don't get it. Why is there the ##\frac{H_0^2}{H^2}## at all? Shouldn't it be just ##\Omega_m(a)=\Omega_ma^{-3}##? Aftetr all, if you add the ##1/H^2## the matter density goes down slower than ##a^{-3}##, which is what I understood was supposed to happen in a purely matter dominated universe?
     
  5. Aug 3, 2016 #4

    Chalnoth

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    That wouldn't give you the density fraction over time. It'd give you a dimensionless matter density over time.
     
  6. Aug 3, 2016 #5
    But the current density in the book and other sources in the internet is given by
    ##\rho_m = {3H_0^2 \over 8\pi G}{\Omega_m}~##, so that ##Ω_m = \frac{ρ_m}{ρ_c}~## where ##ρ_c## is the critical density.

    It is the relationship of ##~Ω_m(a)~## and ##~Ω_m~## that should involve the scaling, say, ##~a^{-3}~## in the case of matter dominated universe.
     
  7. Aug 4, 2016 #6

    George Jones

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    It is conventional (but confusing) to drop a subscript in order to avoid notational clutter, i.e., this is shorthand for
    $$H(a)^2 = H_o^2 ( Ω_{m,0} a^{-3} + Ω_{r,0} a^{-4} + Ω_{k,0} a^{-2} + Ω_{λ,0} ).$$

    See the text between equations (1.3.34) and (1.3.135) in Baumann's notes.
     
  8. Aug 4, 2016 #7
    Yes, my question is the derivation Matt Roos did, I know that there should be a factor of ##\frac{H_o^2}{H^2}## in the evolution of the density parameter, but by plugging it in the corresponding equation to get
    $$H(a)^2 = H_o^2 ( Ω_{m,0} a^{-3} + Ω_{r,0} a^{-4} + Ω_{k,0} a^{-2} + Ω_{λ,0} ).$$
    it wouldn't turn out that way, as in my post.
     
  9. Aug 4, 2016 #8

    George Jones

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    Sorry, I misunderstood. Maybe I still misunderstand, but ...

    in a matter-dominated universe,
    $$
    H_0^2 \left[ Ω_{m,0} a^{-3} + \Omega_{k,0} a^{-2} \right] = H_0^2 \left[ \frac{H^2}{H_0^2} \Omega_m+ \frac{H^2}{H_0^2} \Omega_k \right] = H_0^2 \left[ \frac{H^2}{H_0^2} \Omega_m+ \frac{H^2}{H_0^2} \left( 1 - \Omega_m \right) \right],
    $$
    where the Omegas without zero subscripts are the varying Omegas.
     
  10. Aug 4, 2016 #9

    Jorrie

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    No because the critical density also evolves. If you want to check: the matter density at CMB (last scattering, a=1/1090) was ~0.75. This agrees with
    ##~Ω_m(a) = Ω_m \frac{H_o^2}{H^2} a^{-3}##.

    Figaro's substitution removes the ## a^{-3}## from the equation he has given for H(a), so it is not the same equation any more.

    PS. George has it right above. ;)
     
    Last edited: Aug 4, 2016
  11. Aug 4, 2016 #10
    I think there might be a possible error in Roos's book? Can you take the ratio of the arbitrary density to the present critical density and define it to be the arbitrary density parameter? I think it should not be the case, I think it should be,

    ##~Ω(a) = \frac{8πG}{3H^2}ρ(a) = \frac{ρ(a)}{ρ_c}~## where ##~ρ_c~## is the arbitrary critical density

    as opposed to

    ##~Ω(a) = \frac{8πG}{3H_o^2}ρ(a) = \frac{ρ(a)}{ρ_{c,o}}~## where ##~ρ_{c,o}~## is the present critical density
     
  12. Aug 4, 2016 #11

    Jorrie

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    Interestingly, the reason why this is not given in the books is probably because the substitution, after cancelling the Ho's, just gives you: ##H(a) = H##.
     
  13. Aug 4, 2016 #12
    Yes because the sum of the density parameters should be 1?
     
  14. Aug 4, 2016 #13

    Jorrie

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    Yes, by definition. This is why ##\Omega_k## is there.
     
  15. Aug 4, 2016 #14
    So how should I resolve the derivation? It really seems there is something wrong

    Presentation1.jpg

    In (4.13) ##Ω(a)## corresponds to the arbitrary density parameters which are related as I have stated, but there is a factor ##~\frac{H_o^2}{H^2}~## preventing me to get the correct expression.
     
  16. Aug 4, 2016 #15

    Chronos

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    how does this differ from what george jones offered?
     
  17. Aug 4, 2016 #16
    Plug in for example, ##~Ω_m(a) = Ω_m\frac{H_o^2}{H^2} a^{-3}~## there will be an extra ##~\frac{H_o^2}{H^2}##.
     
  18. Aug 4, 2016 #17

    Chronos

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    Pardon my density, but, i'm not seeing how you arrive at that conclusion.
     
  19. Aug 4, 2016 #18
    In a matter dominated universe, the energy density evolves as

    ##ρ_m(a) = ρ_m (\frac{a_o}{a})^3##

    The density parameter is defined to be,

    ##Ω_m(a) = \frac{8πG}{3H^2}ρ_m(a)~## and ##~Ω_m = \frac{8πG}{3H_o^2}ρ_m~##

    ##Ω_m(a) = \frac{8πG}{3H^2}ρ_m(a) = \frac{8πG}{3H^2}ρ_m (\frac{a_o}{a})^3 = \frac{8πG}{3H^2} \frac{3H_o^2}{8πG} Ω_m (\frac{a_o}{a})^3##

    Thus, ##~Ω_m(a) = Ω_m\frac{H_o^2}{H^2} a^{-3}~## where we set ##a_o = 1##
     
  20. Aug 4, 2016 #19

    Jorrie

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    Yes, but then the ##a^{-3}## disappears from the equation for H(a). I think you are simply doing the substitution the 'wrong way around'. You have to plug in ##\Omega_m##, not ##\Omega_m(a)##.
     
  21. Aug 4, 2016 #20
    No, in my post#14, equation 4.13 is given by

    ##H(a)^2 = H_o^2 (Ω_k a^{-2} + Ω(a) )~## where ##Ω_k = 1 - Ω_o##

    ##Ω(a) = Ω_m(a) + Ω_r(a) + Ω_λ(a)##

    Each evolves as

    ##Ω_m(a) = Ω_m\frac{H_o^2}{H^2} a^{-3}~~~~Ω_r(a) = Ω_r\frac{H_o^2}{H^2} a^{-4}~~~~Ω_λ(a) = Ω_λ\frac{H_o^2}{H^2} a^{0}##

    So,

    ##H(a)^2 = H_o^2 (Ω_k a^{-2} + Ω_m\frac{H_o^2}{H^2} a^{-3} + Ω_r\frac{H_o^2}{H^2} a^{-4} + Ω_λ\frac{H_o^2}{H^2})##

    There is the factor ##~\frac{H_o^2}{H^2}~## that should not be around to make it correct.

    Unless Roos made a mistake which I just noticed right now. In my post#14, just above eq 4.13, he substituted
    ##\frac{8πG}{3}~## by ##~Ω(a) H_o^2~## but I think it should be ##~Ω(a) H^2~## since everything is arbitrary, not the present value (this is my question to George Jones in my post#10), if it is the case then eq 4.13 should be

    ##H(a)^2 = H_o^2 Ω_k a^{-2} + H^2 Ω(a)~## then by substituting the corresponding density parameters
    ##H^2## will cancel with ##\frac{H_o^2}{H^2}~## leaving ##~H_o^2~## which will correspond to the correct expression.
     
    Last edited: Aug 4, 2016
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