Problem #1 [Structure Formation I: Radiation Era]

Roger Griffith
Astro 161
hw. # 9
Proffesor Chung-Pei Ma
Problem #1 [Structure Formation I: Radiation Era]
In class we derived the time evolution of density pertubations in non-relativistic matter δ m
(above the Jeans length) in the matter dominated era. Here we will find the behavior of δ m in
the radiation dominated era. Assume a flat ,Ω Λ = 0 universe for simplicity.
(a). Assume that the pertubation in radiation, δ r , is negligable. Show that the linearized
evolution equation for δ m can be written as
d 2 δ m
dy 2 + 2 + 3y dδ m
2y(1 + y) dy − 3
2y(1 + y) δ m = 0 (1)
where y ≡ ρ m /ρ r = a/a eq is the equality time.
The linearized equation that we need to solve is given as
We know that
¨δ m + 2ȧ
a ˙δ m − 4πGρδ m = 0
we also know
ẏ
y = ȧ
a
δ m = δ m
˙ δ m = dδ m
dy ẏ
¨ δ m = d2 δ m
dy 2 (ẏ)2 + dδ m
dy ÿ
(
ρ = ρ m + ρ r = ρ m 1 + ρ ) (
r
= ρ m 1 + 1 )
ρ m y
4πGρ = 3 2
the double derivative of the scale factor is given as
ä = ÿ = − 4π (
3 Gρ(1 + 3w)y = −4π 3 Gρ 1 + 2 )
y = − y + 2
y 2y(y + 1) (ẏ)2
putting all this together we find an equation of the form
(ẏ
y
) 2
y
y + 1
which yields
d 2 δ m
dy 2 (ẏ)2 + dδ m
dy ÿ + 2 dδ m
y dy (ẏ)2 − 4πGρδ m = 0
d 2 δ m
dy 2 − y + 2 dδ m
2y(y + 1) dy + 2 dδ m
y dy − 3
2y(y + 1) δ m = 0
d 2 δ m
dy 2 + 2 + 3y dδ m
2y(y + 1) dy − 3
2y(y + 1) δ m = 0
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(b). At late time (y ≫ 1), compare your solutions for δ m with those derived in class.
if y ≫ 1 the universe becomes matter dominated and the above expression becomes
d 2 δ m
dy 2
+ 3 dδ m
2y dy − 3
2y 2 δ m = 0
if we assume that this has a power solution of the form
δ m ∝ y α
dδ m
dy = αyα−1
d 2 δ m
dy 2
= α(α − 1)yα−2
we find that
which can be factored as
α 2 + 1 2 α − 3 2 = 0
(
(α − 1) α + 3 )
= 0
2
giving the two solutions as
α = 1 α = −3/2
and so we can see that
δ + m ∝ y and δ− m ∝ y−3/2
and we find a growing solution and a decaying solution. The two solutions from class are
δ + m ∝ a δ − m ∝ a −3/2
and so we know that when y ≫ 1 implies ρ m ≫ ρ r and we recover the solutions from class.
(c). Verify that δ m ∝ y+2/3 is a solution to the equation in part (a) in general. Can δ m grow
much in the radiation dominated era?
we can verify this solution by doing
δ m = y + 2 3
˙ δ m = 1
¨ δ m = 0
and plugging this into the differential equation yields
(
2 + 3y
2y(y + 1) − 3
y + 2 )
2y(y + 1) 3
= 0
0 = 0
thus, this is a solution to the differential equation.
We know that for radiation dominated era y ≪ 1 and since the solution for δ + m
can see it does not grow very much during the radiation dominated regime.
is linear we
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Problem #2 [Structure Formation II: Ω m ≠ 1 Models]
In class we derived the growths of linear density pertubations in matter for cosmological models
with different Ω m (assuming Ω Λ = 0).
(a). Plot the growing solution δ m (a) on a log-log scale for a=0.001 to 1 for four cosmological
models: Ω m = 0.01,0.1,0.3,and 1. Normalize all your curves to δ m (a = 0.001). Make
sure to plot all curves on a single figure so you can compare them.
The solutions to the linear equations governing the growth of density pertubation for the different
cosmological models are given as
δ + m ∝ a Ω 0,m = 1
δ + 3sinhθ(sinhθ − θ)
m ∝
(coshθ − 1) 2 − 2 Ω 0,m < 1
Ω 0
a(θ) =
2(1 − Ω 0 ) (coshθ − 1) Ω 0,m < 1
From these solutions we can solve θ in terms of a and plug into the growing solution, i.e
[ ]
a(θ)2(1 −
θ ′ = cosh −1 Ω0 )
+ 1
Ω 0
thus δ + m is given by δ + m ∝ 3sinhθ′ (sinhθ ′ − θ ′ )
(coshθ ′ − 1) 2 − 2
the plot is given as
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(b). How does the growth of pertubations depend on Ω 0,m ?
We can see that the growth of the fluctuations depend very strongly with Ω 0,m .
For Ω 0,m = 0.01 We can see that the growth of structure decreases very rapidly as z → 0, where
the slope of the curve aproaches 0 today. This seems very unlikely, due to the fact that we
live in a universe filled with galaxies.
For Ω 0,m = 0.10 we can see that increasing the mass density by one order of magnitude, causes
the growth of structure to continue as z → 0 and we can see that the more matter you put
into the universe the longer the growth of the fluctuations (over densities).
For Ω 0,m = 0.30 we can see that this is very similar to the last curve, except that we can see
now that it is approaching δ + m ∝ a.
For Ω 0,m = 1.00 we can see that the growth of structure is a constant
If the four models you plotted all produce the same level of pertubations today (that matches
the observed number density of galaxies, for example), how do you expect (qualitatively) the
amount of structures to differ in the four models at redshift z = 3?
To try to understand this question we can make the same plot as before, except that we must
normalize all the models to a(θ) = 1.00
1.000
δ m
+
vs a
0.100
δ m
+
(θ)
0.010
Ω m
0.001
0.001 0.010 0.100 1.000
a(θ)
thus from this plot we can see that at z = 3 the pertubations must have grown a lot faster in
the past for Ω 0,m < 1, with increasing growth as Ω 0,m → 0. This must have happened in order
for us to see the structures we see today.
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Problem #3 [Particle Horizon and the Horizon Problem]
The particle horizon at cosmic time t is the physical distance that light has traveled since t = 0;
it defines the maximum distances for causal communication:
Z r
d H (t) = a(t)
0
dr ′
√
1 − kr ′2 = a(t) Z t
0
cdt ′
a(t) ′ (2)
The integral can be solved exactly in the matter-dominated era for Ω Λ = 0
⎧
[ ]
√ c
(1 + z) ⎪⎨
−1 cosh −1 1 + 2(1−Ω 0,m)
H 0 1−Ω0,m (1+z)Ω
Ω
0,m
0,m < 1
2c
d H (z) =
H 0
(1 + z) −3/2 Ω 0,m = 1
[
]
⎪⎩ √ c
(1 + H 0 Ω0,m −1 z)−1 cos −1 1 − 2(Ω 0,m−1)
(1+z)Ω
Ω
0,m
0,m > 1
(a). Verify that eq. (2) indeed gives the expression above for Ω 0,m = 1 model. (Extra credit:
verify the other two models.)
We know that
Z t cdt ′
d H (t) = a(t)
0 a(t) ′ where a(t) ∝ t 2/3 a(t) ′ ∝ (t 2/3 ) ′
and this equation becomes
but we know that
Z t
d H (t) = ct 2/3 (t ′ ) −3/2 dt ′ = 3ct 2/3 t 1/3 = 3ct
0
a ∝ t2/3
t 2/3
0
putting this all together we find
t ∝ t 0 a 3/2 t 0 = 2
3H 0
a = 1
1 + z
t ∝ 2
3H 0
(1 + z) −3/2
thus we derive
d H (z) = 2c
H 0
(1 + z) −3/2
(b). What is the present horizon size (in Mpc) if Ω 0,m = 0.3,1, and 3?
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For the present horizon size we must set z = 0 into the above expressions. For Ω m = 0.3 we use
[
]
c
d H (z = 0) = √ cosh −1 2(1 − 0.3)
1 + = 8.7 Gpch −1
H 0 1 − 0.3 0.3
For Ω m = 1 we use
For Ω m = 3 we use
d H (z = 0) =
d H (z = 0) = 2c
H 0
= 6.0 Gpch −1
[
c
√ cos −1 1 −
H 0 3 − 1
]
2(3 − 1)
= 4.0 Gpch −1
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(c). Show that at high redshifts 1 + z ≫ Ω −1
0,m
, a good approximation for all three cases is given
by
2c
d H (z) ≈ √ (1 + z) −3/2
Ω0,m H 0
(you may find the identity cosh −1 x = ln(x + √ x 2 − 1) useful.)
For Ω 0,m < 1 we will use
d H (z) =
c
H 0
√
1 − Ω0,m
(1 + z) −1 cosh −1 (1 + α)
where α = 2(1 − Ω 0,m)
(1 + z)Ω 0,m
using the above relationship for cosh −1 we find
√
cosh −1 (1 + α) = ln[1 + α + (1 + α) 2 − 1] = ln[1 + α + √ 2α]
but we know that
√
2α > α
and we can just write
and we know that the Taylor expension for
so we now get
so now the distance horizon becomes
d H (z) ≈
cosh −1 (1 + α) = ln[1 + √ 2α]
ln(1 + x) ≈ x
cosh −1 (1 + α) ≈ √ 2α
c
H 0
√
1 − Ω0,m
(1 + z) −1√ 2α ≈
6
2c
√
Ω0,m H 0
(1 + z) −3/2

For Ω 0,m > 1 we will use
d H (z) =
c
√
H 0 Ω0,m − 1 (1 + z)−1 cos −1 (1 + α) where α = − 2(Ω 0,m − 1)
(1 + z)Ω 0,m
we know that the Taylor expansion for cos(θ) is given by
( )
cos(θ) = 1 − θ2
2 + ... → θ ≈ cos−1 1 − θ2
2
letting
we find
α = − θ2
2
θ = cos −1 (1 + α) ≈ √ −2α
and since we know what α is we can just rewrite this as
d H (z) =
which is what we had before, i.e
c
H 0
√
Ω0,m − 1 (1 + z)−1√ 2α
d H (z) ≈
2c
√
Ω0,m H 0
(1 + z) −3/2
For Ω 0,m = 1 we get the same thing as before, which is
d H (z) =
2c
√
Ω0,m H 0
(1 + z) −3/2 = 2c
H 0
(1 + z) −3/2
(d). Show that the comoving size of the particle horizon at the time of photon decoupling can
be written as λ dec = β(Ω 0,m h 2 ) −1/2 , and compute the value of β in Mpc.
We know that
λ dec = d H(z)
a dec
=
this can be written in as
λ dec =
2c
√
Ω0,m H 0
(1 + z dec ) −3/2 (1 + z dec ) =
2c sMpc
100 km √ 1 + z dec
(Ω 0,m h 2 ) −1/2 = β(Ω 0,m h 2 ) −1/2 where β =
we also know that
thus we find that β is
z dec = 1089
β ≈ 180 Mpc
2c
√
Ω0,m H 0
√ 1 + zdec
2c sMpc
100 km √ 1 + z dec
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(e). Estimate the angular size (in degrees) subtended by λ dec on the sky today. This is called
the horizon problem. Why is this a problem?
The angular size subtended in the sky is given by
θ =
λ dec
d H (z = 0)
for Ω 0,m = 1
so we find
θ = (1 + z dec ) −1/2 = .030 radians = 1.72 degrees
This is a problem because it states that things are causally connected by 1.7 degrees in the
sky, but yet we see that the CMB is very uniform at scales larger than this, which begs the question:
How do particles (which are not causally connected) have such a uniform temperature?
This is a problem that Inflation theory has been able to answer.
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