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Problem 1 “Big Crunch” Ryden 6. Section 6.1 of Ryden gives us the information we need to do this problem. How long until the Big Crunch? In equations 6.17 and 6.18 we are given a scale factor - time relation in a rather unusual form:
a(θ) =
(1 − cos θ)
t(θ) =
2 H 0
(Ω 0 − 1)^3 /^2
(θ − sin θ)
where θ is allowed to run from zero to 2π. The idea here is to solve for t 0 by setting a = 0, finding θ 0 , and plugging that value into the time relation. So:
a(θ 0 ) = 1 ⇒ θ 0 = arccos
)
sin(θ 0 ) =
√ 1 − Ω 0
t 0 = t(θ 0 ) =
2 H 0
(Ω 0 − 1)^3 /^2
( arccos
) −
√ 1 − Ω 0
)
(If you don’t know how I got sin(θ 0 ), ask me in office hours - it’s a very useful trick.) tcrunch occurs at θ = 2π:
tcrunch = π H 0
(Ω 0 − 1)^3 /^2
So the time until crunch is tcrunch − t 0 :
tremaining =
H 0
(Ω 0 − 1)^3 /^2
( π +
√ 1 − Ω 0 −
arccos
))
Minimum blueshift This part of the problem is ambiguously worded, but is intended to mean, what is the blueshift (i.e. redshift) of the most blueshifted object visible? Just as an object appears redshifted if the universe was smaller when it emitted light, an object is blueshifted if the universe was larger when it emitted light. The most blueshifted object is one that emitted when the scale factor a(t) was at a maximum. Ryden gives us this maximum scale factor:
amax =
1 + z = a(t 0 ) a(te)
zmin = a(t 0 ) amax
− (^) Ω^10 = zmin
Lookback time to the most redshifted object This can be solved in a manner similar to the first part of the question.
amax = a(θmax) =
⇒ θmax = π
t(θmax) =
π 2 H 0
(Ω 0 − 1)^3 /^2
tL = t 0 − t(amax) =
tL = (^) H^10 (Ω 0 −^ Ω^0 1) 3 / 2
( (^1) 2 arccos^
( (^2) Ω 0 −^1
) − (^) Ω^10
1 − Ω 0 − π 2
)
Problem 2 Cosmic Time - Redshift Relation
te =
∫ (^) ∞ ze
dz (1 + z)H(z) where one would obtain H(z) by a rearrangement of the Friedmann equa- tion. So far, these equations hold true for any universe. We now specialize them to the Benchmark Model:
H^2 H 02
Ω 0 ,m a^3
te =
H 0 −^1
∫ (^) ∞ ze
dz (1 + z)
√ (^) Ω 0 ,m a^3 + Ω^0 ,Λ Setting the limiting value ze to 7, you are now ready to integrate. My Mathematica code for doing this is attached. As you will see, the age of the universe for the Benchmark Model is te =. Problem 3 “Big Bounce” Ryden 6. One way to start this problem is to take the matter + curvature + lambda model given in section 6.3 and set Ωm, 0 = 0. You can then rearrange equation 6.34 to read
a˙^2 = H 02
( 1 − ΩΛ, 0 + a^2 ΩΛ, 0
)
recalling that H = ˙a/a. The “big bounce” that the problem speaks of is a point in time when a contracting universe reaches its smallest allowed scale factor, turns around, and begins expanding. When this “big bounce” occurs, ˙a must be zero and ¨a > 0. Find the a for which ˙a = 0:
0 = 1 − ΩΛ, 0 + a^2 bounceΩΛ, 0 ⇒
abounce =
√ 1 −
Check that ¨a = 0 at this scale factor by differentiating the first equation we wrote down:
¨a a˙ = H 02 ΩΛ, 0 aa˙ ⇒
¨a = H 02 ΩΛ, 0 a
Since this satisfies the requirement ¨a > 0, we have proved abounce =
√ 1 − (^) Ω^1 Λ, 0. Now, from the last equation, ¨a = H 02 ΩΛ, 0 a, we know that a must follow the form
a = c 1 exp(H 0
√ ΩΛ, 0 t) + c 2 exp(−H 0
√ ΩΛ, 0 t) We can find the constants c 1 and c 2 by requiring that at tbounce, a = abounce and ˙a = 0. This is made much easier if we shift to a variable t − tbounce:
a = c 3 exp(H 0
√ ΩΛ, 0 (t − tbounce)) + c 4 exp(−H 0
√ ΩΛ, 0 (t − tbounce))
a(tbounce) = abounce =
√ 1 −
= c 3 + c 4
a ˙(tbounce) = 0 = c 3 − c 4 ⇒
c 3 = c 4 =
abounce ⇒
a =
abounce
( exp(H 0
√ ΩΛ, 0 (t − tbounce)) + exp(−H 0
√ ΩΛ, 0 (t − tbounce))
)
a = abounce cosh(H 0
√ ΩΛ, 0 (t − tbounce))
Finally, we can find t 0 by setting a = 1 and solving. We can then subtract tbounce to find how much time has elapsed since then:
t 0 − tbounce = √Ω^10 H 0 cosh−^1
(√ (^) Ω 0 Ω 0 − 1
)
Problem 4 Distances
