Cosmology Solutions
Useful constants:
1AU = 1.50 × 1011 m
Msun = 1.99 × 1030 kg
1pc = 206, 265AU
mproton = 1.67 × 10−27 kg
G = 6.67 × 10−11 kg −1 m3 s−2
1. Cosmic density
(a) Calculate the energy density of particles today (giving your answer in SI units)
Calculate the energy density of particles today (giving your answer in SI units) and
hence calculate the number density of the particles today if the “particles” are:
Since H0 given to 2 sig fig answers should be given to same precision
H0 = 72kms−1 M pc−1
=
=
(t0 ) =
=
=
(1)
72
s−1
8
× 1.49 × 10 km/AU
(2)
106 AU/M pc
206265 ×
2.3 × 10−18 s−1
3c2 H02
8πG
3(3 × 108 ms−1 )2 (2.3 × 10−18 s−1 )2
(8π(6.67 × 10−11 m3 kg −1 s−2 )
8.5 × 10−10 kgm−1 s−2 (2sf )
(3)
(4)
(5)
(6)
(b) Hence calculate the number density of the particles today if the “particles” are:
i. Hydrogen atoms (giving your answer in atoms m−3 ).
ii. Stars, the same mass as our Sun (giving your answer in stars kpc−3 ).
iii. Galaxies, the same mass as the Milky Way (giving your answer in galaxies
M pc−3 ).
i. Hydrogen atoms (giving your answer in atoms m−3 ).
n(t0 ) =
(t0 )
mc2
(7)
Assume
m(H) = mp = 1.7 × 10−27 kg
8.5 × 10−10 kgm−1 s−2
nH (t0 ) =
(1.7 × 10−27 kg)(3 × 108 ms−1 )2
= 5.6atoms m−3 (2sf )
Cosmology - 1
(8)
(9)
(10)
ii. Stars, the same mass as our Sun (giving your answer in stars kpc−3 ).
mSun = 2 × 1030 kg
(8.5 × 10−10 kgm−1 s−2
nstars (t0 ) =
(2 × 1030 kg)(3 × 108 ms−1 )2
= 4.7 × 10−57 m−3
−3
= 137stars kpc
(2sf )
(11)
(12)
(13)
(14)
iii. Galaxies, the same mass as the Milky Way (giving your answer in galaxies
M pc−3 ).
m(M W ) = 5.8 × 1011 Msun
(15)
−3
137starskpc
× (103 M pc/kpc)3 M pc−3 (16)
ngalaxies (t0 ) =
11
5.8 × 10 stars/M W
= 0.24galaxies M pc−3 (2sf )
(17)
2. Observing distance galaxies {Expanded from Harwit: Astrophysical Concepts}
This question deals with what limits the observation of distant galaxies - is it their size,
brightness or whether the universe has been around long enough for light from them to
have reached us? The smallest extragalactic sources resolved with currently available
telescopes are of the order of 0.05” of arc in diameter. While the faintest observable
objects are of about 30th magnitude.
(a) Find the distance to
i. the farthest resolved galaxies
Careful with units here - don’t forgot to change arcsec to radians
Either calculate distance in same units as diameter (in which case you need to
change angle in arcsec to RADIANS).
Distance, D, to galaxy in kpc is related to diameter ,d, in kpc and resolution, θ
in radians.
10 1o 2πrad
60” 600 360o
2.4 × 10−7 rad
d(kpc)
θ(radians)
44
2.4 × 10−7 rad
1.81 × 108 kpc
θ(rad) = 0.05”
(18)
=
(19)
D(kpc) =
=
=
= 181Gpc
(20)
(21)
(22)
(23)
Or use analogy to trigonometric parallax to use angle in arcseconds directly
(BUT remember diameter is then measured in AU and distance is in pc!).
Cosmology - 2
Distance, D, to galaxy in pc is related to diameter ,d, in AU and resolution, θ,
in arcsec. In Euclidean
d(AU )
θ(arcsec)
4.4 × 104 pc × 206265AU/pc
=
0.05”
= 1.82 × 1011 pc
D(pc) =
= 182Gpc
(24)
(25)
(26)
(27)
ii. The faintest obesrvable galaxies
Let magnitude be mgal , and FM 31/gal be the observed flux from M31/ galaxy,
FM 31
Fgal
2
LM 31 Dgal
= 2.5 lg
(since F = L/4πD2 )
2
Lgal DM
31
mgal − mM 31 = 2.5 lg
(28)
(29)
Since we are assuming they have the same luminosity
mgal
Dgal
= 4.4 + 5 lg
778kpc
!
(30)
finally we want the galaxy to be the faintest, so mgal = 30 so that
Dgal
30 − 4.4 = 5 lg
778kpc
!
→ Dgal = 103Gpc
(31)
(b) Giving a sensible value of H0 , what would you estimate the size of the observable
horizon to be today?
H0 = 72kms−1 M pc−1
c
dHubble =
H0
3 × 105 kms−1
=
72kms−1 M pc−1
= 4.2Gpc(2 sig f ig)
(32)
(33)
(34)
(35)
Note that this would imply that the galaxies we are interested in are receding from
us at superluminal speeds (if Hubble’s law holds to those distances)
H0 =
72
2.06265 ×
106 AU/M pc
× 1.49 × 108 km/AU
Cosmology - 3
s−1
(36)
= 2.3 × 10−18 s−1
(37)
v = H0 D
(38)
−18 −1
9
11
−1
(2.3 × 10 s )(118 × 10 pc × 206265AU/pc × 1.49 × 10 ms )
=
(39)
3 × 108 ms−1
= 83 × 108 ms−1
(40)
= 27.8c
(41)
In Special Relativity it is not possible to accelerate matter to the speed of light or
beyond. However in General Relativity, in which space is expanding, it is possible
for matter to move away from us at superluminal speeds.
To get some idea of this, consider two ants moving on the surface of a balloon
initially 2cm apart from one another, with the ants having a maximum speed of
1cms−1 (as an analogy to the speed of light). If the balloon is not being blown up
then the ants could approach each other and meet 1 second later. However if the
ballon is being blown up, then the rate of increase in the distance separating the
ants is in no way related to their maximum walking speed, but rather this rate of
expansion of space (the balloon’s surface) which is not related in any way to their
maximal speed.
(c) So what is the main limitation on distant galaxy observation?
Based on the (simplistic) analysis above, the main limitation on observing these distant galaxies, would be that the galaxies lie outside the observable horizon (roughly
taken to be ∼ the Hubble length c/H0 ). We’ve not taken into account the fact that
the Hubble parameter H(t) varies over time
3. Cosmological and kinematical redshift
Consider light from a galaxy at cosmological redshift, zcosmo , and with a peculiar recession velocity (the difference between the total velocity and the Hubble expansion at the
position of the galaxy), vpec , purely along the line of sight.
(a) Show that the combined redshift due to these two effects (kinematical and cosmological) is
(1 + ztot ) = (1 + zcosmo )(1 + zkin )
(42)
where the kinematical redshift, zkin , is related to the peculiar velocity through the
Doppler formula. [Hint: compare with situation for an observer close to the source].
In a static universe the observed redshift of light from a galaxy receding with velocity vpec will be observed with wavelength
λreceived,static = (1 + zkin )λemitted
vpec
(vpec c)
zkin ≈
c
Cosmology - 4
(43)
(44)
This redshift is independent of the distance of the observer from the source, only
the relative velocity. In a non-expanding universe this would be the only effect
observed. The redshift caused by the expansion of the universe is then an additional,
separate effect distorting the photon already stretched by the peculiar velocity. This
effect is dependent on how much the universe has expanded as the light travels to
the observer from the source.
λreceived,expanding = (1 + zcosmo )λreceived,static
= (1 + zcosmo )(1 + zkin )λemitted
λreceived,expanding
(1 + ztotal ) =
λemitted
= (1 + zcosmo )(1 + zkin )
(45)
(46)
(47)
(48)
You can visualize this separation of the 2 redshifts by imagining an observer right
next to the source, which is receeding from them at speed vpec then the expansion
of the universe will not alter his observed wavelength of light. This redshifted light
as it passed to a second observer on earth today will be cosmologically redshifted
with respect to the first observer in the form above.
(b) Assuming H0 = 72M pc−1 kms−1 and a typical galactic peculiar velocity of 500
kms−1 :
i. At what distance does the cosmological redshift exceed that caused by the
peculiar motion?
ii. Comment on whether peculiar velocities are going to be an important consideration for measurements of the universe’s expansion using galaxies i) within
our ‘Local Group’ of galaxies, ii) in galaxies within the Coma cluster. [Feel
free to ‘Google’ to get any necessary information!]
Assuming Hubble’s law vcosmo = H0 d for inferred cosmological recession ve locity, vcosmo at a distance d then for vcosmo > vpec
d >
vpec
500km/s
≈
≈ 6.94M pc
H0
75M pc−1 km/s
(49)
So this would be outside our ‘Local Group’ of galaxies (diameter ∼ 3M pc) but
still within the ‘Local Supercluster’ and closer than the Coma cluster of galaxies
(∼ 100M pc away). Peculiar velocities therefore would be an issue when using
Local Group galaxies but not a dominant effect on cosmological scales e.g. to
Coma. Even so they would remain a source of statistical error.
4. Curvature {Slightly modified from Ryden Ch. 3 Qu. 3.2 and 3.3}
Consider yourself a 2-dimensional being living on the surface of the Earth, modeled as a
perfect sphere radius R = 6371km. In this 2D world, light travels on the surface of the
Earth.
Cosmology - 5
(a) The metric is
2
2
2
2
ds = dr + R sin
r
dθ2
R
(50)
(b) Sketch the variation of the angular width dθ you measure for an object of length
ds = 10km( R) (lying perpendicular to your line of sight), as its distance away
from you, r, varies. Clearly give a value on the sketch for any minimum or maximum value of dθ in arcseconds. Since object is lying perpendicular to your line of
sight (i.e. wholly in the θ direction) dr = 0
ds = R sin
dθ(rad) =
r
dθ
R
(51)
10km
1
6371km sin(r(km)/6371km)
(52)
As r increases from 0 to πR/2 the angular diameter decreases, this is because the
distance separating the two lines of sight to the ends of the object first increase as
you move towards the equator, and then decrease as you move towards the opposite
pole. As you get to the pole all lines of sight converge to a single point i.e. all lines
of sight will be able to see the opposite pole, and the object is of infinite angular
extent (the concept of angular diameter is ill defined).
d!
d!min =5.40’ (3sf)
d!min
0.
0.5
Cosmology - 6
1
r/"R
(c) If you draw a circle of radius r on the Earth’s surface, the circle’s circumference is
r
C = 2πR sin
.
(53)
R
If you could measure distances with an error of ±1m, how large a radius would a
circle you draw on the Earth’s surface have to be to convince yourself that the Earth
is spherical rather than flat? Assume that R is exact to the nearest meter, and give
your answer to the nearest meter.
r
≥ 1m
R
r
10−3
, (x = )
x − sin x ≥
2π(6371)
R
Cf lat − Csphere = 2πr − 2πR sin
(54)
(55)
Either solve by iteration or use small angle approximation for sin x,
x3
(f or x 1)
3!
!1/3
6 × 10−3
x ≥
2π(6371)
x ≥ 5.31197 × 10−3
sin(x) ≈ x −
r ≥ 33.843km (to nearest m)
(56)
(57)
(58)
(59)
5. Gravitational lensing
The angle by which a photon is deflected by a compact mass M , at an impact parameter
b is given by
4GM
c2 b
α =
(60)
A light ray just grazes the surface of a neutron star (M = 3.0 × 1030 kg, R = 1.2 ×
104 km). Through what angle α is the light ray bent by the gravitational lensing? Give
you answer in arcminutes.
4GM
(rad)
c2 R
= 7.41 × 10−4 rad
α =
0
= 2.55
(61)
(62)
(63)
6. Measuring dark matter from the dynamics of a cluster of galaxies
It is 1933 and your professor, Fritz Zwicky, has just given you some new observations
of a cluster of galaxies for which he wants you to do a quick analysis to estimate the
cluster’s mass based on their dynamics. He has measured the radial velocities (i.e. along
the line of sight) of twelve galaxies:
Cosmology - 7
1110 km/s 890 km/s 1140 km/s 860 km/s
750 km/s 1250 km/s 1320 km/s 670 km/s
810 km/s 1190 km/s 590 km/s 1410 km/s
He reports that the galaxies all have luminosities of about 1011 Lsun , and says it is reasonable for you to assume that the stars in the galaxies all have about the same mass
and luminosity as the sun (we know this is not true, but on average it is a reasonable
approximation).
(a) Briefly describe how one might measure the radial velocity of a galaxy.
(b) Calculate an estimate of the 1D velocity dispersion of the galaxies.
(c) Based on these observations, and treating the galaxies as if they were uniformly
distributed in a spherical cluster of constant density with radius R = 300kpc (a
back of an envelope assumption), show that the virial theorem is not satisfied for
this system.
(d) If the virial theorem were not satisfied, (qualitatively & briefly) what would happen
to the cluster of galaxies in the future? If you can, give a rough time scale for your
answer.
(e) Zwicky suggests that there might be some form of unseen dark matter that is responsible for making the virial theorem hold. How much more dark matter is required?
You can assume that the dark matter is evenly distributed and has the same kinetic
energy per unit mass as the galaxies and that Mtotal = f Mluminous , where f is a
constant. Give your answer in terms of f .
7. Nuclear energy
Suppose the Sun was made up of 100% carbon-12 and that energy was produced through
a reaction
12
C +12 C →24 M g.
(64)
The atomic weight of 12 C is 12 and 24M g is 23.985.
(a) How much energy, in eV , is released per carbon nucleus in this reaction?
Change in binding energy, ∆E, is given by
∆E = [mass(24 M g) − 2 × mass(12 C]c2
(65)
= −(23.985 − 24)931.46M eV
(66)
= −14.0M eV
(67)
Negative because energy is released.
Cosmology - 8
(b) Assuming the Sun could burn 10% of the Carbon, how long would its lifetime be?
Change in Binding energy per mole of Magnesium formed
∆E = 1.4 × 107 eV × 1.6 × 10−19 J/eV × 6 × 1023 nuclei/mol
12
= 1.3 × 10 Jmol
−1
(68)
(69)
Let lifetime be T , and total energy emitted from burning 10% of the carbon be,
Etot ,
Etot
Lsun
= 3.8 × 1026 W
T ∼
Lsun
(70)
(71)
To get Etot , note that 2 carbon nuclei are required to produce 1 magnesium nuclei,
0.1 × Msun
× 1.3 × 1012 Jmol−1
2 × 12g)
2 × 1029 kg
=
× 1.3 × 1012 Jmol−1
−3
24 × 10 kg
= 1.1 × 1043 J
Etot =
(72)
(73)
(74)
so
1.1 × 1043 J
3.8 × 1026 J/s
= 2.9 × 1016 s
T =
= 930 million years
Cosmology - 9
(75)
(76)
(77)