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Universe density in Hyperbolic Universe

  1. May 17, 2015 #1
    I want to know Universe density according to this equation( ##k=-1##) ?

    ##H^2(t)-8πρG/3=-k/a^2(t)##
    ##ρ_U=ρ_m+p_r##
    ##ρ_U##=Universe density
    ##ρ_m##=Matter density
    ##p_r##=Radiation density
     
  2. jcsd
  3. May 17, 2015 #2

    wabbit

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    With three terms (##1/a^2, 1/a^3, 1/a^4##) the differential equation for ##a(t)## is not solvable analytically, you only have an implicit form in terms of an integral which cannot be expressed in terms of usual functions - you can obtain this form by substituting ##\frac{\dot a}{a}## for ##H##, you get something of the form ## \int F(a)da=t-t_0 ##
     
    Last edited: May 17, 2015
  4. May 17, 2015 #3
    Then What should I do ? Is there's any way to escape this situation ?

    In ##k=-1## → ##Ω_0<1## Is this can help us ?
     
  5. May 17, 2015 #4

    wabbit

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    It depends what you're looking for. If you're brave enough I think you can express the integral in terms of standard elliptic integrals, which are available in most mathematical packages. But the easy route is just to integrate numerically, this works fine.

    ## \Omega_0<1 ## doesn't help - but the issue is not a hard one, the only thing is, the solution does not have a nice form. It is perfectly computable numerically though.
     
  6. May 17, 2015 #5
    I am in high school so I did not understand a single word what you're saying
     
  7. May 17, 2015 #6
    I cant do that myself Should I open a new thread or somebody can do that ?
     
  8. May 17, 2015 #7

    wabbit

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    Sorry about that. Very impressive, when I was in high school in didn't know anything at all about the Friedman equation or about General Relativity !

    To find ## a(t) ## (and therefore ## \rho(t) ## ), you need to program two things :
    1. The numerical calculation of the integral ##G(a)=\int F(a)da##
    2. The numerical solution of the equation ## G(a)=t ##

    This is a rather tough with high school tools though, you definitely need calculus. And the answer is a computer program, not a formula.

    What are you trying to achieve exactly, is your question about the qualitative behaviour of the solution ? There are some things you can tell without going through a complete solution.
     
    Last edited: May 17, 2015
  9. May 17, 2015 #8

    Chalnoth

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    Too many parameters, too few equations. Also, it depends a bit upon your conventions. There are two commonly-used conventions:
    1. ##k = {-1, 0, 1}##. With this convention, the scale parameter ##a## becomes the radius of curvature.
    2. ##a(now) = 1##. This is usually the easiest. With this convention, ##k## can take on any number, and represents the amount of spatial curvature today (when ##a = 1##).

    One way to make things easier to deal with is to use the concept of density fractions. With density fractions, I can rewrite:

    [tex]H^2(t) = {8\pi G \over 3} \left(\rho_m + \rho_r\right) - {k \over a^2}[/tex]

    as:

    [tex]H^2(t) = const \left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2}\right)[/tex]

    Here I've introduced a constant on the left, and a series of ##\Omega## constants. The equation becomes simplest if we require than when ##a = 1##, ##\Omega_m + \Omega_r + \Omega_k = 1##. In this case, when ##a = 1##, we get:

    [tex]H^2(now) = const[/tex]

    Where this is the same constant that goes out in front, so our constant is just ##H_0^2##, and the Friedmann equation becomes:

    [tex]H^2(t) = H_0^2 \left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2}\right)[/tex]

    So now we have four unknowns, but only one equation. That's not enough to give an answer unless three of the unknowns are set. What's done in practice is to take some measurable quantity which is a function of ##H(t)##:

    [tex]F(data) = f(H(t))[/tex]

    We then take lots of data points, and use computer simulations to figure out which choices for the constants ##H_0, \Omega_m, \Omega_r, \Omega_k,## and ##\Omega_\Lambda## fit the data most closely (that last one stems from the cosmological constant).
     
  10. May 17, 2015 #9

    Chalnoth

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    I don't think this is exactly what you've asked for, but this website calculates a bunch of different quantities based upon a variety of different inputs:
    http://www.astro.ucla.edu/~wright/CosmoCalc.html
     
  11. May 17, 2015 #10
    In our universe ##H_0## is known 70 and ##Ω_m=0.05## ... so we can change this things isnt it.I mean this quantites describes known universe.So Can I put ##H_0=65## to fit the data ?
     
  12. May 17, 2015 #11

    Chalnoth

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    ##\Omega_m## is not 0.05. This parameter includes both normal matter and dark matter.

    Planck's 2015 results report the best-fit estimates of these parameters are:
    ##H_0 = 68##km/sec/Mpc
    ##\Omega_m = 0.31##

    Anyway, if you tried to fit the data with these parameter values, but ##\Omega_\Lambda = 0##, you'd find pretty quickly that the data doesn't fit at all.
     
  13. May 18, 2015 #12
    I am trying to find density of universe ) If I put matter density 0.31 then whats the meaning to find density.We have already decided it.Is there any calculator which modified by 2015 plank results.
     
    Last edited: May 18, 2015
  14. May 18, 2015 #13

    ChrisVer

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    You can't just compute it... you have to make experiments in order to decide its value and try to fit the data... It's a free parameter.
     
  15. May 18, 2015 #14
  16. May 19, 2015 #15
    Ok,I understand the idea.But I want to ask something.My question is simple so I thought that theres no need to open a new thread and its related this question.
    ##H^2(t)-8ρπG/3=-k/a^2(t)## (Lets suppose a(t)=1) then
    ##H^2(t)-8ρπG/3=-k## In equations ##k## can be -1,0,1.
    Then If ##k<0## then ##-k>0## then
    ##H^2(t)-8ρπG/3>0 ##→Hyperbolic Universe. This means ##Ω_k>0## I mean If ##k## negative ##Ω_k## must be positive isnt it ? I am confused here.
    In cosmology calculator says ##Ω_k=1-Ω_m-Ω_Λ##

    (Its look like hyperbolic universe I am trying to understand)
    Thanks
     
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