Ryden - Galaxies/Cosmology Problems with Solution | PHYS
133
Physics
University of California - Santa Barbara
6 pag.
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Problem 1
“Big Crunch”
Ryden 6.5
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) =
t(θ) =
1 Ω0
(1 − cos θ)
2 Ω0 − 1
1
Ω0
(θ − sin θ)
2H0 (Ω0 − 1)3/2
where θ is allowed to run from zero to 2π. The idea here is to solve for t0
by setting a = 0, finding θ0 , and plugging that value into the time relation.
So:
2
a(θ0 ) = 1 ⇒ θ0 = arccos
−1
Ω0
sin(θ0 ) =
2 q
1 − Ω0
Ω0
Ω0
1
2
2 q
t0 = t(θ0 ) =
−
1
−
arccos
1 − Ω0
2H0 (Ω0 − 1)3/2
Ω0
Ω0
(If you don’t know how I got sin(θ0 ), ask me in office hours - it’s a very
useful trick.)
tcrunch occurs at θ = 2π:
tcrunch =
Ω0
π
H0 (Ω0 − 1)3/2
So the time until crunch is tcrunch − t0 :
tremaining =
1
2
Ω0
1 q
1
1 − Ω0 − arccos
−1
π
+
3/2
H0 (Ω0 − 1)
Ω0
2
Ω0
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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 =
Ω0
Ω0 − 1
1+z =
a(t0 )
a(te )
a(t0 )
Ω0 − 1
−1=
−1=
amax
Ω0
zmin =
− Ω10 = 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 ) =
t(θmax ) =
Ω0
⇒ θmax = π
Ω0 − 1
Ω0
π
2H0 (Ω0 − 1)3/2
tL = t0 − t(amax ) =
tL =
1
Ω0
H0 (Ω0 −1)3/2
1
2
arccos
2
Ω0
−1 −
1
Ω0
√
1 − Ω0 −
Problem 2
Cosmic Time - Redshift Relation
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π
2
This problem is very similar to the “lookback time” problem in last week’s
homework; parts i and iii can practically be copied from it. I will quickly
recap the solutions for these parts, and then move on to part ii.
For the lookback time problem, we had to use a cosmic time - redshift
relation. For the case of a flat universe with a single component and 1 ≥
w > −1, the solution was given by Ryden equations 5.51 and 5.48:
t0
1+z =
te
t0 =
2/(3+3w)
2
H −1
3(1 + w) 0
Putting everything in terms of H0 and z, recalling that w is roughly zero
for matter and is 1/3 for radiation, we find for parts i and iii respectively,
te =
te =
1
(1 + z)−2
2H0
2
(1 + z)−3/2
3H0
Plugging in H0−1 = 14 Gyr and z = 7, we find:
te = 1.1 × 102 Myr (radiation case)
te = 4.1 × 102 Myr (matter case)
To compute te for the benchmark model, we will need some means of
integrating numerically. You may use the website listed in the problem, or
you can use some program (Mathematica, Matlab, whatever) to perform the
integration yourself. If you are one of the intrepid souls who attempts this
feat, here’s a conceptual guide on how to do it:
Cosmic time is
te =
=
Z te
t=0
dt =
Z z(te )
z(t=0)
Z a(te )
a(0)
da Z a(te ) da Z a(te ) da
=
ȧ =
ȧ
a(0) a
a(0) aH(a)
a
Z ze
−dz(1 + z)−2
dz
=
−
=
(1 + z)−1 H(z)
∞ (1 + z)H(z)
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te =
Z ∞
ze
dz
(1 + z)H(z)
where one would obtain H(z) by a rearrangement of the Friedmann equation. So far, these equations hold true for any universe. We now specialize
them to the Benchmark Model:
H2
Ω0,m
= 3 + Ω0,Λ ⇒
2
H0
a
te =
1 Z∞
dz
q
−1
H0 ze (1 + z) Ω0,m
+ Ω0,Λ
a3
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 6.6
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
ȧ2 = H02 1 − ΩΛ,0 + a2 ΩΛ,0
recalling that H = ȧ/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, ȧ must be zero and ä > 0. Find the a for which ȧ = 0:
0 = 1 − ΩΛ,0 + a2bounce ΩΛ,0 ⇒
abounce =
s
1−
1
ΩΛ,0
Check that ä = 0 at this scale factor by differentiating the first equation
we wrote down:
äȧ = H02 ΩΛ,0 ȧa ⇒
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ä = H02 ΩΛ,0 a
Since this satisfies the requirement ä > 0, we have proved abounce =
q
1−
1
ΩΛ,0
Now, from the last equation, ä = H02 ΩΛ,0 a, we know that a must follow
the form
q
q
a = c1 exp(H0 ΩΛ,0 t) + c2 exp(−H0 ΩΛ,0 t)
We can find the constants c1 and c2 by requiring that at tbounce , a = abounce
and ȧ = 0. This is made much easier if we shift to a variable t − tbounce :
q
q
a = c3 exp(H0 ΩΛ,0 (t − tbounce )) + c4 exp(−H0 ΩΛ,0 (t − tbounce ))
a(tbounce ) = abounce =
s
1−
1
= c3 + c4
ΩΛ,0
ȧ(tbounce ) = 0 = c3 − c4 ⇒
1
c3 = c4 = abounce ⇒
2
q
q
1
a = abounce exp(H0 ΩΛ,0 (t − tbounce )) + exp(−H0 ΩΛ,0 (t − tbounce )) =
2
q
a = abounce cosh(H0 ΩΛ,0 (t − tbounce ))
Finally, we can find t0 by setting a = 1 and solving. We can then subtract
tbounce to find how much time has elapsed since then:
t0 − tbounce =
√ 1
Ω0 H0
cosh−1
q
Ω0
Ω0 −1
Problem 4
Distances
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.
Ryden 7.3
A very, very good equation to memorize or otherwise keep readily available is Ryden 7.38:
dL
1+z
If you know any one of these measures - angular diameter distance, proper
distance, or luminosity distance - you instantaneously know the other two as
a result.
We can snatch the relation between proper distance and redshift for a
flat, single-component universe from chapter five:
dA (1 + z) = dp (t0 ) =
2 c
1 − (1 + z)−(1+3w)/2
H0 1 + 3w
From this we immediately have:
dp (t0 ) =
dL (t0 ) =
dA (t0 ) =
c
2
(1 + z) 1 − (1 + z)−(1+3w)/2
H0 1 + 3w
c
2
1 1 − (1 + z)−(1+3w)/2
H0 1 + 3w 1 + z
We find the redshift for which dA (t0 ) has a maximum by taking the derivative and setting it equal to zero:
1
d
1 1 + 3w
−(1+3w)/2
dA (t0 ) = 0 = −
(1+z)−(1+3w)/2−1 ⇒
1
−
(1
+
z)
+
dz
(1 + z)2
1+z 2
1=
3 + 3w
(1 + z)−(1+3w)/2 ⇒
2
zmax =
2
3+3w
−2/(1+3w)
−1
Obtain the corresponding value of dA by plugging this back in. After
simplifying we get:
dA,max (t0 ) =
c
H0
2
3+3w
2/(1+3w)+1
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