Ph 444
Solutions for Numerical Assignment 1
1. The particle horizon at the time
t
is calculated by taking each interval of proper
distance,
cdt
′
, covered by a photon between
t
′
and
t
′
+
dt
′
, increasing it by the expan-
sion of the universe between
t
′
and
t
,
a
(
t
)
/a
(
t
′
), and adding up the intervals between
zero and
t
:
d
hor
(
t
) =
a
(
t
)
integraldisplay
t
0
cdt
′
a
(
t
′
)
.
(1)
Changing variables in the integral to
a
′
and using
da
′
=
˙
a
′
dt
,
H
= ˙
a/a
, and Ryden
equation 6.6 for
H
:
d
hor
(
t
)
=
a
(
t
)
integraldisplay
a
(
t
)
0
cda
′
a
′
˙
a
′
=
a
(
t
)
integraldisplay
a
(
t
)
0
cda
′
(
a
′
)
2
H
(
a
)
(2)
=
ca
(
t
)
H
0
integraldisplay
a
(
t
)
0
da
′
(
a
′
)
2
(Ω
r,
0
/
(
a
′
)
4
+ Ω
m,
0
/
(
a
′
)
3
+ Ω
Λ
,
0
+ (1
−
Ω
0
)
/
(
a
′
)
2
)
1
/
2
(3)
=
c
(1 +
z
)
H
0
integraldisplay
1
/
(1+
z
)
0
da
′
(Ω
r,
0
+ Ω
m,
0
a
′
+ Ω
Λ
,
0
(
a
′
)
4
+ (1
−
Ω
0
)(
a
′
)
2
)
1
/
2
(4)
The last step uses
a
(
t
) = 1
/
(1 +
z
).
The dimensionless integral in equation 4 does not have a simple closed form. It
can be expressed in terms of exotic special functions, but this is not of much use
in obtaining numerical values.
I wrote a simple program to evaluate the integral
using trapezoidal rule integration with 10
4
equal-size steps in
a
′
(
i.e.
, multiplying the
integrand evaluated at the middle of the interval by the interval width and summing
the contribution from all of the intervals). I chose the number of steps by increasing
the number until the result stopped changing.
The value of the integral,
I
, for
Ω
r,
0
= 8
.
4
×
10
−
5
, Ω
m,
0
= 0
.
27, Ω
Λ
,
0
= 0
.
73, and
z
= 1090 is
I
= 6
.
697
×
10
−
2
.
Since
c/H
0
= 4220 Mpc for
H
0
= 71 km/s/Mpc,
d
hor
(
z
= 1090) = 0
.
2592 Mpc. This
small size reflects the young age of the universe at the time of decoupling, about
300,000 yrs.
The interval of comoving coordinate that corresponds to the size of the horizon at
z
= 1090 is
r
=
d
hor
(
z
= 1090)
/a
(
z
= 1090). The distance today that corresponds to
this interval of comoving coordinate is just
a
0
r
=
d
hor
(
z
= 1090)(
a
0
/a
(
z
= 1090)) =
d
hor
(1 +
z
) = 283 Mpc. Thus, any structure imposed on the size of the horizon at the
time of decoupling will show up at large physical sizes today, though this size is still
much smaller than the particle horizon today.
For
z
= 1090, it is a good approximation to ignore the Ω
Λ
,
0
(
a
′
)
4
and (1
−
Ω
0
)(
a
′
)
2
terms in the denominator of the integrand since they are smaller by a factor of at
least 10
3
compared to the Ω
m,
0
a
′
term, even at the upper limit of the integration.