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Solutions of Friedmann Equations

  1. Mar 17, 2007 #1
    Hello everybody,

    could somebody tell me how to get the scale factor as a function of time [R(t)] from the Friedmann Equations for a simple dust, pressureless universe when k != 0, [tex]\Lambda[/tex] = 0.

    Mathematica 5.2 doesn't want to give me any solution and wherever I searched the best thing I got was the parametric solution in terms of [tex]\Theta[/tex]:

    [tex]\noindent\(\pmb{R[\Theta ]=\alpha (1-\text{Cos}[\Theta ]);}\\[/tex]
    [tex]\pmb{t=\beta (\Theta -\text{Sin}[\Theta ]);}\)[/tex]

    for a closed universe and:

    [tex]\noindent\(\pmb{R[\psi ]=\gamma (\text{Cosh}[\psi ]-1);}\\[/tex]
    [tex]\pmb{t=\delta (\text{Sinh}[\psi ]-\psi );}\)[/tex]

    for an open universe.

    Where [tex]\noindent\(\pmb{\alpha , \beta , \gamma , \delta }\)[/tex] are some constants...

    Everywhere I also found plots of R vs. t for various parameters but without showing the solution for R(t).

    I guess I just need to somehow invert expressions for time...

    Thanks for help.

    Last edited: Mar 18, 2007
  2. jcsd
  3. Mar 18, 2007 #2


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    I changed a theta to a psi. Is this what you intended?
    my apologies if this is wrong, or isnt what you meant.

    It seems to me that you are asking if someone can express psi analytically as a function of t. Offhand I don't see how to do this.
    Last edited: Mar 18, 2007
  4. Mar 18, 2007 #3
    Simply, say, I want to make a plot of evolution of the scale factor R vs. t. Such plots are probably in every book which has something to do with cosmology.

    To do this I need R as a function of t, unfortunatelly have no idea how to get it.

    When k = 0, it's a simple differential equation solved on a piece of paper in a few moments, but what about k != 0 (open/closed universe)... :shy:

    Just to be complete, I am adding the Friedmann Equation I am talking about:

    [tex]\noindent\(\pmb{\left(\frac{dR[t]}{dt}\right)^2-\frac{8 \pi \rho _0 G}{3 R[t]}=-k c^2}\)[/tex]

    where [tex]\rho _0[/tex] is the density as measured today...
    Last edited: Mar 18, 2007
  5. Mar 18, 2007 #4
    Sometimes parametric equation like the one you have can't be transformed into simple explicit function...
    This maybe just the case.
    Still you can plot it , shouldn't be very hard.
  6. Mar 18, 2007 #5
    Thanks, you confirmed what I was thinking about.

    I guess this is what I don't know to do so I need R in terms of t... :blushing:
  7. Mar 18, 2007 #6
    on a second look I think you can express t as function of r
    You have r as a function of psi, you can write psi=acosh((r+gamma)/Gamma)
    and then replace psi in the second equation containing psi and t.
    just came trough my mind , try it.
    Last edited: Mar 18, 2007
  8. Mar 18, 2007 #7

    I doubt the result could be solved explicitly for R...

    Right now I've got three books in Astrophysics/Cosmology here, every of them provides the parametric solutions (above) and clearly explains how to get the age of the universe from it (expressing parameters in terms of [tex] \noindent\(\pmb{\Omega _0}\) [/tex] and so on)... But none of them explains how they got to graphing R(t) vs t... :mad:
  9. Mar 18, 2007 #8
    Graphing parametric equations is something I took in my calculus course in my first year in college...
    It shouldn't be that hard, I suppose you can learn how to graph it, by looking into a calculus book...

    try this link to draw: http://www.ies.co.jp/math/java/calc/sg_para/sg_para.html [Broken]
    Last edited by a moderator: May 2, 2017
  10. Mar 18, 2007 #9
    Thanks a lot, you opened my eyes... For some reason I didn't see the simple solution of plotting it as a parametric equation where x = t[[tex]\Theta[/tex]] and y = R[[tex]\Theta[/tex]]...

    I was probably expecting something more complicated...
    Last edited by a moderator: May 2, 2017
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