Problem Set 1 Solutions
Michael Eastwood
[email protected]
1.
MIT and Caltech
Consider three galaxies in a straight line with
1 Mpc separating adjacent galaxies.
According to the Hubble law, MIT and Caltech
are moving apart at a rate of
If the Hubble law is linear (ie. ṙ ∼ r, for
the sake of convenience we’ll set constants of
proportionality to 1), the middle galaxy observes the two other galaxies to be moving
away from each other at a speed 2r. However, the key is that these two galaxies also
measure that they’re moving away from each
other at a speed 2r. This means Hubble’s law
is consistent with homogeneity, and because
we can pick the line to be in any direction,
Hubble’s law is consistent with isotropy.
v = H0 d
= 70 km s−1 Mpc−1 × 1.4 × 10−16 Mpc
= 9.8 × 10−15 km s−1
= 0.31 mm yr−1 .
Google tells me that the continents are drifting
due to plate tectonics at speeds of several cm yr−1 ,
so this expansion rate is small even compared to
continental drift.
However, MIT and Caltech are not actually
moving away from each other due to the expansion of space. This is because the Earth is held
together by gravitational forces. Thus the Earth is
held together at a fixed physical size (see the next
question!) and does not expand with the universe.
2.
If Hubble’s law was instead quadratic (ie. ṙ ∼
r2 ), the middle galaxy again observes the two
other galaxies to be moving away from each
other at a speed 2r2 . However this time, these
two galaxies each measure that they’re moving
away from each other at a rate 4r2 and hence
homogeneity is violated.
Co-Moving and Physical Distances
(b) The mean density of matter in the universe is
The physical distance is the distance one would
measure between two points on a t = constant slice
of the universe. The co-moving distance is the distance that one would measure between those same
two points if you made the measurement today.
That is, the physical distance is normalized to
be the co-moving distance today. A structure that
is expanding with the rest of the universe has a
fixed co-moving size. On the other hand, a bound
object that is held together and does not expand
with the universe has a fixed physical size.
The mathematical relationship between the
physical distance dphys , co-moving distance dco
and the redshift z is
dphys =
3.
dco
.
1+z
3H02
8πG
= 2.3 × 10−30 g cm−3 .
ρm = Ωm
The mean density of baryons in the universe
is
3H02
ρb = Ωb
8πG
= 4.0 × 10−31 g cm−3 .
The factor by which the mass interior to the
sun’s orbital radius in the Milky Way exceeds
the mean density of matter in the universe is
(1)
2 × 1044 g
4
3
3 π(8 kpc)
Homogeneity and Isotropy
(a) Hubble’s law needs to be the same at every point and every direction in the universe.
·
1
∼ 1.4 × 106 .
ρm
The factor by which the mass of baryons interior to the sun’s orbital radius in the Milky
1
but we can set the constant to zero by requiring a(T = 0) = 0. Hence
Way exceeds the mean density of baryons in
the universe is
2
3
× 2 × 1044 g
4
3
3 π(8 kpc)
·
a(T ) = T .
1
∼ 5.3 × 106 .
ρb
(4)
(c) The form of the FRW metric we assumed is
The baryons are more concentrated than the
dark matter because gas clouds can collide,
lose energy and settle into the bottom of a
gravitational potential well. Dark matter,
on the other hand, is collision-less, so that
it doesn’t dissipate energy in collisions, and
hence can’t be as “clumpy” as the baryons
can. Furthermore, dark matter can’t cool radiatively like baryonic matter can.
ds2 = −dT 2 + a2 (T ) dχ2 + sinh2 χ dΩ2 .
(5)
The Minkowski metric (in spherical coordinates) is
ds2 = −dt2 + dr2 + r2 dΩ2 .
(6)
(c) The factor by which the mass in the vicinity
of the Milky Way and Andromeda exceeds the
mean density of matter in the universe is
If these two metrics are equivalent, the coefficients of dΩ2 must be equal. Additionally, we
require −dT 2 + a2 (T )dχ2 = −dt2 + dr2 .
2 × 4 × 1045 g 1
·
∼ 28 .
4
3
ρm
3 π(1 Mpc)
The first requirement tells us r = a(T ) sinh χ.
Substituting this into the second requirement
(and using the form for a(T ) derived in the
previous part) gives
(d) The universe is actually isotropic on the scales
where voids, filaments, and galaxy clusters can
be averaged down. Voids tend to be about
10 to 100 Mpc in size, and if you’re familiar
with baryon acoustic oscillations (BAO), this
creates an excess of galaxies clustering on 150
Mpc scales. However, these are the largest
structures in the universe, so if you average
over volumes > (200 Mpc)3 , the universe looks
homogeneous and isotropic.
4.
−dT 2 + T 2 dχ2 = −dt2 + sinh χ dT
2
+ T cosh χ dχ ,
which simplifies to
dt2 =
cosh χ dT + T sinh χ dχ
2
.
Therefore the transformation we are looking
for is
GR, the Metric, and Expanding Space
(a) Because we have chosen Ω0 = 0, the total
density of the universe is less than the critical density (3H02 /8πG) as long as H0 6= 0.
This means that we should use the FRW metric assuming negative (open) curvature.
r(T, χ) = T sinh χ ,
(7a)
t(T, χ) = T cosh χ .
(7b)
(d) For a fixed χ, we can divide Equations 7a and
7b to obtain
(b) Under the assumptions we have made (Ω0 = 0,
k = −1), the Friedmann equation reduces to
H 2 = a−2 .
r(t) = t tanh χ .
(8)
Therefore an observer at fixed χ appears to
move away from the origin at a constant speed
in Minkowski space.
(2)
Recalling that H = ȧ/a we have ȧ(T ) = 1 so
that
a(T ) = T + constant ,
(3)
(e) The space-time is exactly flat. However, the
T = constant slice has a negative curvature.
2