Does This Cosmological Model Re-Collapse?

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ajclarke
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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 what's the significance between the second two which allows me to isolate the value corresponding to collapse.

Cheers.
Adam
 
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What 'math' are you following through with?
 
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.
 
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.
 
zhermes said:
What 'math' are you following through with?

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.

cristo said:
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?

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

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

Chalnoth said:
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.

I'm uncertain as to how that determines which of the two remaining parameters is the recollapsing universe?
 
ajclarke said:
I'm uncertain as to how that determines which of the two remaining parameters is the recollapsing universe?
If the derivative of the Hubble parameter is negative, then it's recollapsing.
 
Don't you have Ωtotal0 equal to 3.01, instead of unity?
 
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.