

[Cosmology] Scale Factor Values

  1. Hello.

    I have been working through some questions and answers to do with cosmology. One of them asks you to consider a model where:

    [tex]\Omega_{MO}=3 [/tex]
    [tex]\Omega_{\Lambda O}=0.01 [/tex]
    [tex]\Omega_{RO}=0 [/tex]
    and asks you to show mathematically that the model re-collapses.

    Following through the math, I get three values of a: -14.87,1.51 and 13.36.

    Clearly the first can be disregarded and unphysical since a cannot be negative, but I can't decide whats the significance between the second two which allows me to isolate the value corresponding to collapse.

  2. jcsd
  3. What 'math' are you following through with?
  4. BillSaltLake

    BillSaltLake 184
    Gold Member

    If a is normalized time, then it may have zero diameter 14.87 time units in the past, first collapse 1.51 in the future, and a "recollapse" later. Not sure if that's correct though.
  5. cristo

    cristo 8,408
    Staff Emeritus
    Science Advisor

    What is the definition of [itex]\Omega_{s0}[/itex] for some species [itex]s[/itex]? What is [itex]\Omega_{\rm total 0}[/itex] in the universe you are studying?
  6. Chalnoth

    Chalnoth 5,243
    Science Advisor

    Make use of the second Friedmann equation to make sure that when [itex]H(a)[/itex] goes to zero, [itex]dH/da[/itex] is negative.
  7. I used the equation for the Hubble Parameter as a function of redshift, then changed this over to be a function of scale factor instead.

    [tex]\Omega_{total 0} = 1[/tex]

    I don't understand the first bitof the question I'm sorry.

    I'm uncertain as to how that determines which of the two remaining parameters is the recollapsing universe?
  8. Chalnoth

    Chalnoth 5,243
    Science Advisor

    If the derivative of the Hubble parameter is negative, then it's recollapsing.
  9. BillSaltLake

    BillSaltLake 184
    Gold Member

    Don't you have Ωtotal0 equal to 3.01, instead of unity?
  10. George Jones

    George Jones 6,388
    Staff Emeritus
    Science Advisor
    Gold Member

    Use the second derivative test from elementary calculus. [itex]a\left(t\right)[/itex] has a local maximum at [itex]t = t_1[/itex] if [itex]da/dt \left(t_1 \right) = 0[/itex] and [itex]d^2 a/dt^2 \left(t_1 \right) < 0[/itex]. To find [itex]d^2 a/dt^2 [/itex], differentiate the Friedmann equation.

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