Roger Griffith
Astro 161
hw. # 5
Proffesor Chung-Pei Ma
Problem #1 [Ω in Photon Background Radiation]
The present temperature of the cosmic background photons is known to exquisite precision
T0 = 2.725 ± 0.0001(1σ) Kelvin (Mather et al 1999).
(a). Calculate the present number density of photons (in cm−3 ) in this radiation. (Don’t
worry about the error bars.)
we know that the number density for bosons is given by
! "3
gb · ζ(3) k3 3
gb
kT
=
nb = 2 ζ(3)
T
π
!c
π2 !3 c3
where
gb = 2
ζ(3) = 1.202
T = 2.725 Kelvin
therefore we find the number density of these photons to be
nb =
2.2404 k3
· (2.725)3 = 411 cm−3
π2 !3 c3
(b). Calculate Ω0,γ , the present ratio of the mass-energy density in photons to the critical
density. (Express your answer in terms of the Hubble parameter h.) Is the universe today
radiation or matter dominated?
the mass-energy density for bosons is given by
εγ = αT 4
α=
π 2 gb k 4 π 2 k 4 4
=
T
30 !3 c3 15 !3 c3
ρc =
3H 2 c2
8πG
The ratio Ω0,γ is given by
εγ
8π3 k4 GT 4
π2 k4 4 8πG
8 π3 k4 G(2.725 K)4
T ·
=
=
Ω0,γ =
=
ρc 15 !3 c3
3H 2 c2
45
45H 2 c5
H 2 c5 !3
where
H0 = 3.241 × 10−18 hs−1
thus we find
1
Ω0,γ = 2.47 × 10−5h−2
From this results we can see that the universe is matter dominated because Ω0,m # Ω0,γ .
Problem #2 [Ionizing Photons]
Suppose the photon temperature T of a background distribution such that kT $ Q, where Q =
13.6 eV is the ionization energy of hydrogen.
(a). What fraction f of the blackbody photons are energetic enough to ionize hydrogen?
The distribution function for bosons is given as
4πgb
nγ =
(2π!)3
Z ∞
0
p2
ecp/kt − 1
if we let
E = cp
dp =
dp
dE
c
if
kT $ Q
we find
nγ =
4πgb
(2π!)3
Z ∞
Q
E 2 e−E/kT dE
using integration by parts we find
nγ = [0 − (−e−Q/kT kT (2k2 T 2 + 2kQT + Q2 ] =
4πgb −Q/kT
[e
kT [2k2 T 2 + 2kQT + Q2 ]
(2π!c)3
and the total number density is given by
! "3
kT
gb
nγT 2 ζ(3)
π
!c
therefore the fraction f is fiven by
f=
nγ
e−Q/kT [2k2 T 2 + 2kQT + Q2 ]
=
nγT
2k2 T 2 ζ(3)
(b). As we will discuss later, the cosmic microwave background photons came from the last
scattering surface when the universe was at T ≈ 3700 K. What is the numerical value of f at
T ≈ 3700 K? What is f today?
2
for
T = 3700 Kelvin
kT = .3188 eV
and f is found by using
e−Q/kT [2k2 T 2 + 2kQT + Q2 ]
= 2.35 × 10−16
2k2 T 2 ζ(3)
and for the value today is given by
f (3700) =
kT = 2.35 × 10−4 eV
T = 2.725 Kelvin
f (2.725) =
e−Q/kT [2k2 T 2 + 2kQT + Q2 ]
≈0
2k2 T 2 ζ(3)
Problem #3 [Gamow’s Prediction of the CMB]
A fascinating bit of cosmological history is that of George Gamow’s prediction of the Cosmic
Microwave Background in 1948. (Unfortunately, his prediction was premature; by the time the
CMB was actually discovered in the 1960’s, his prediction had fallen into obscurity.) Let’s see
if you can reproduce Gamow’s line of argument. Gamow knew that nucleosynthesis must have
taken place at a temperature Tnuc ≈ 109 K, and that the age of the universe is currently t0 ≈ 10
Gyr.
Assume that the universe is flat and contains only radiation. With these assumptions, what
was the energy density ε at the time of nucleosynthesis? What was the Hubble parameter H
at the time of nuceosynthesis? What was the time tnuc at which nucleosynthesis took place?
What is the current temperature T0 of the radiation filling the universe today? If the universe
switched from being radiation-dominated to being matter-dominated at a redshift zr,m > 0, will
this increase T0 for fixed values of Tnuc and t0 ? Explain your answer.
The energy density is given by
ub =
π 2 gb k 4 T 4 π 2 k 4 T 4
=
30 !3 c3
15 !3 c3
for Tnuc = 109 K we find
uNC =
π2 [1.381 × 10−23JK−1 ]4 [109 K]4
π2 k 4 T 4
=
= 7.54 × 1020 Jm−3
3
3
−34
3
8
−1
3
15 ! c
15[1.055 × 10 Js] [3 × 10 ms ]
The Hubble parameter H at the time of nucleusynthesis is given by
2
HNC
=
8πG
uNC
3c2
therefore H is given as
HNC =
#
8πG
uNC = .0022 s−1
2
3c
3
tnuc can be found by using
tnuc =
1
= 230.91s
2HNC
for T we use
T (t) =
!
" $ %1/2
$ t %1/2
tp
45
p
Tp
≈ 0.61 Tp
2
32π
t
t
where Tp = 1.4 × 1032 K is the Planck temperature and t p ∼ 5 × 10−44 s−1 is the Planck time
and t0 = 10 Gyr = 3.15 × 1017 s
thus
! "1/2
tp
= 34.024 K
T0 = 0.61 Tp
t0
To explain why the temperature of the matter dominated universe is less than the temperature
of the radiation dominated universe we can use
T∝
and since we know
1
a(t)
a(t)r ∝ t 1/2
a(t)m ∝ t 2/3
so for any time t we know that
a(t)r < a(t)m
plugging this into the temperature equation we find that
TrD > TmD
thus we just showed that the temperature for radiation dominated universe is greater that the
temperature of a matter dominated universe. It follows from the simple analogy that the bigger
the volume the lower the temperature. Since a(t) for a matter dominated universe is greater than
a(t) for radiation dominated we know that the volume of a matter dominated universe is greater
than for a radiation dominated universe, and thus the temperature is lower.
4