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Calcluating the Hubble Radius for an open universe?

  1. Sep 6, 2010 #1
    Given the following parametric form of the Friedmann Equation for an open, dust-filled (matter-dominated) universe:
    [tex]a(x)={a_0 \Omega \over 2(1-\Omega)}(cosh(x)-1)[/tex]

    [tex]t(x)={\Omega \over 2 H_0 (1- \Omega)^{3/2}}(sinh(x)-x)[/tex]

    I am trying to calculate the Hubble Radius, R=c/H(t) where H(t)=(da/dt)/a and have come up with the following solution:

    [tex]{da \over dt} = {da \over dx} {dx \over dt}[/tex]

    [tex]{da \over dx}={a_0 \Omega \over 2(1-\Omega)}sinh(x)[/tex]

    [tex]{dt \over dx}={\Omega \over 2H(1- \Omega)^{3/2}}(cosh(x)-1)[/tex]

    [tex]{da \over dt}=Ha_0(1-\Omega)^{1/2}coth({x \over 2})[/tex]

    Integrate to find a:

    [tex]a=Ha_0(1-\Omega)^{1/2}coth({x \over 2})t[/tex]

    [tex]{{da \over dt} \over a}={1 \over t}[/tex]

    Therefore, the Hubble Radius:

    [tex]R_H={c a \over ({da \over dt})}=ct[/tex]

    I just can't seem to find any references to this solution, so I don't know if it is actually correct.
    Can anyone confirm?
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2

    Chalnoth

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    Could you please enclose the latex in [noparse][tex][/tex][/noparse] tags?

    e.g.

    [noparse][tex]{dy \over dx} = x[/tex][/noparse]
     
  4. Sep 7, 2010 #3
    Sorry, fixed now.
     
  5. Sep 7, 2010 #4

    Chalnoth

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    Thanks!

    Unfortunately, I don't think it's correct. The step after "Integrate to find a:" appears to be wrong. Basically, x and t are not independent variables, so you can't integrate a function which includes x over t without first substituting x as a function of t.

    Edit: Looks like a potential solution may be:

    [tex]\int f(x)dt = \int {f(x) \over {dx \over dt} } {dx \over dt} dt = \int {f(x) \over {dx \over dt} } dx[/tex]
     
    Last edited: Sep 7, 2010
  6. Sep 7, 2010 #5
    Thanks, I thought that might have been the case. Any ideas how to get a solution for a then?
     
  7. Sep 7, 2010 #6

    Chalnoth

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    Heh, you replied too fast :) I think the edit in my above post may be a way to go.
     
  8. Sep 7, 2010 #7
    Tricky. Thanks for your help!
     
  9. Sep 7, 2010 #8

    Chalnoth

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    Heh, just realized that doesn't actually get you anywhere, because it just gives a(x), which you already have!

    I also think I've noticed another mistake, where you say da/dt is proportional to the hyperbolic cotangent. I don't think this is the case, as it looks like you've dropped the -1 in dt/dx.

    Anyway, you should be able to get the Hubble radius as a function of x no problem. Getting it as a function of t analytically may not be possible.
     
  10. Sep 8, 2010 #9
  11. Sep 8, 2010 #10

    Chalnoth

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    Oh, I see, it's the hyperbolic cotangent of x/2, not x. That makes sense now. For some reason I missed the division by 2 before. Obviously they're using some interesting trigonometric identities to get that answer.
     
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