G a la x ies - Jacobs University Bremen · G a la x ies 1 .1 S ta rs 1 .1 .1 T h e m a g n itu d e...

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Chapter 1 Galaxies 1.1 Stars 1.1.1 The magnitude system Flux A star is first of all characterized by its luminosity, L, which may lead to a detectable energy flux, f , at the position of the Earth. Distant stars, i. e. all stars except the Sun, are that far away that they can be considered as point sources. Hence, the flux can written as f = L 4πr 2 . (1.1) Distance measures For distances, r, astrophysics has developed its own unit system. For dis- tances in the solar system it is convenient to use the average distance between the Sun and the Earth, one Astronomical Unit (AU), as base, 1 AU = 1.496 × 10 13 cm. (1.2) The distance of objects beyond the Solar System in usually expressed in ‘Parsec’. This measure is derived from trigonometric method to derive the distance of an object. Astronomers realised that while the Earth moves around the Sun some of the stars on the sky oscillate a bit around their 1

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Chapter 1

Galaxies

1.1 Stars

1.1.1 The magnitude system

Flux

A star is first of all characterized by its luminosity, L, which may lead to adetectable energy flux, f , at the position of the Earth. Distant stars, i. e.all stars except the Sun, are that far away that they can be considered aspoint sources. Hence, the flux can written as

f =L

4! r2. (1.1)

Distance measures

For distances, r, astrophysics has developed its own unit system. For dis-tances in the solar system it is convenient to use the average distance betweenthe Sun and the Earth, one Astronomical Unit (AU), as base,

1 AU = 1.496 ! 1013 cm. (1.2)

The distance of objects beyond the Solar System in usually expressed in‘Parsec’. This measure is derived from trigonometric method to derive thedistance of an object. Astronomers realised that while the Earth movesaround the Sun some of the stars on the sky oscillate a bit around their

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1.1. STARS

Figure 1.1: Trigonometric parallax: the angle between a star close by anda very distant star at the sky varies over the course of the year. The am-plitude of this oscillation is called ‘parallax’. Since also the solar sytem, thestar and the fix star move along their orbits in the galaxy, the a drift maysuperimposed.

average position compared to majority of stars. The angle, ", between anoscillating star and other ‘fix’ stars, Sfix, varies with time, t, by

" = p Sfixsin(2! t/yr) + "0.

If we choose a star in the plane of Earth orbit we obtain the maximumamplitude, p, which is called parallax. It measures the distance to staroscillating on the sky,

p = arctan

!1AU

r

"

"1AU

r. (1.3)

The maximum variations of those oscillating stars on the sky is about onearc second, hence we are allowed to neglect the arctan. We can use theparallax to introduce a new distance measure: A star with a parallax of1 arc sec is defined to be at the distance of one Parsec (pc),

1 pc =1AU

1!!= 3.086 ! 1018 cm. (1.4)

Luminosity - distance degeneracy

Eq. (1.1) immediately reveals one of the fundamental di!culties in identify-ing the nature of an astrophysical object: We can only measure the flux of a

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1.1. STARS

source, to derive its luminosity we have to know its distance and vice versa.History has repeated itself in this respect:

• Galaxies has been observed for centuries, e. g. Charles Messier (1730-1817) compiled a catalogue of ‘nebulae’ while searching for comets.Only few people, e. g. Immanuel Kant (1724-1804), speculated at thistime that some of the nebulae reside outside of the Milky Way. Notuntil the observations of Edwin Hubble it was generally accepted thata lot of those nebulae are separate galaxies. He was able to identifysome Cepheid variables.

• Some hydrogen clouds have velocities much higher than the typicalvelocities in the Milky Way disk (High velocity clouds). It is stillunclear if these clouds are part of the disk or if they reside in thegalactic halo.

• Gamma Ray Burststo be added

Apparent magnitude

Astronomers still use a rather old-fashioned unit system to represent theflux of a star: the apparent magnitude. Modern, well calibrated, etc. in-struments might prompt the flux in ergs/s. However, in the beginning as-tronomers classified stars by the naked eye. The Greeks already introduceda system with six luminosity classes. The most luminous stars on the skydefine the first class. Later on the star Vega in the constellation Lyra wasselected as reference star, and it is still used for it. This reveals also thestrength of the apparent magnitude system: Its definition includes a pre-scription how everybody can calibrate its photometric system, i.e. his fluxmeasurement. Our senses work in a logarithmic manner, hence, the appar-ent magnitude represents the logarithm of the flux. More precisely, it isdefined by

mA # mB = #2.5 log10

!fA

fB

"

. (1.5)

This equation tells us how the flux and the magnitude of two stars, A andB, are related. With the additional definition that the flux arriving fromVega, fVega, has zero apparent magnitude,

mVega = 0 (1.6)

we can fix also the zero point of the apparent magnitude scale.

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1.1. STARS

Absolute magnitude

We can use Eq. (1.1) and substitute the flux in Eq. (1.5)

mA = mB # 2.5 log10

!LA

LB

"

+ 5 log10

!10 pc

rB

"

+

# $% &

=:MA

5 log10

!rA

10 pc

"

,

where additionally a distance scale, namely 10 pc, is introduced. The firstthree terms at the r.h.s. are defined as absolute magnitude of the star A. It isimportant to notice that it does only depend on the LA – beside propertiesof the reference star B, mB, LB and rB. Thus, while the apparent magni-tude is a measure for the flux arriving at Earth, the absolute magnitude ismeasure for actual luminosity of the star. A more practical approach fordetermining the absolute magnitude is to substract the distance modulus

from the apparent magnitude,

M = m # 5 log10

!r

10 pc

"

. (1.7)

The magnitude system can also be used to classify extended objects likegalaxies: You can give the flux, integrated over the entire area, of an object,hence you can also give the flux arriving within a certain opening angle. So,don’t wonder when you encounter mag/arcsec2.

Flux in a band - Filters

to be added

Colors

to be added

1.1.2 Stellar evolution

Gravitational instability

The rather smoothly distributed gas in a galaxy in transferred into a veryclumpy structure when star formation sets in. The driving force the con-traction is mainly gravitation, while the gas pressure helps to erase densitydi"erences. We wish to elaborate what conditions are necessary to allowgravitation to form a clumps from a homogeneous distribution. Our ev-eryday life experience tells us that perturbations in the air propagate anddissipate: The fate of sound waves.

The conclude for the conditions which allow perturbations to grow weconsider a very artificial situation and make some bold simplifications. How-ever, a much more sophisticated treatment would lead to the same conclu-sion. We assume that there is a homogeneous gas sphere with density #. We

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1.1. STARS

subdivide the sphere into an outer, ’shell’ and an inner, ’core’ part. Bothcore and shell are assumed to have the mass M . The core attracts the shellgravitationally,

Fgrav "GM2

r2, (1.8)

where we have used simply the radius r of the spherical separating surfacebetween the shell and the core as distance. Note, we wish to derive herequalitative result, hence we don’t care for numerical constants of the order ofunity (one of which would introduced by a proper calculation of the force).The pressure, p, of the core gas pushes the shell outwards with a force of

Fpress = p 4!r2 . (1.9)

We have an equilibrium situation if the pressure force balances the gravi-tational force exactly. Everything would stay in place. Let’s assume now,that for some reason the shell is slightly contracted. The separating surfaceis at radius r # dr. Let’s assume that the outer shell contracts slightly. Weget the following modification of the forces. For the gravitational attractionis increased by

dFgrav =2GM2

r3dr. (1.10)

On the other hand also the pressure is increased

dFpress =

'dp

dr4!r2 + p 4! 2r

(

dr. (1.11)

For the derivative of the pressure we can write

dp

dr=

dp

d#

d#

dr=

dp

d#

d

dr

3M

4!r3= #v2

s9M

4!r4,

where v2s denotes the speed of sound of the gas. The ratio of the first and

the second term in the brackets is

1

2

dp

dr

r

p.

If the pressure is a power-law function of the radius, p $ r! (at least locallysuch a relation can be found), the expression above reduces to "/2. Fromthe result for dp/dr we can conclude that " is #4 plus some dependenceof the sound speed. Hence, the second term in the brackets leads to thesubtraction of a fraction of the first term. Since we are not interested innumerical constants of the order of unity we simply neglect the second term.Finally, we can write for the change of the pressure force

dFpress = v2s

9M

r2dr. (1.12)

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1.1. STARS

We have now to decide whether the equilibrium situation, dFgrav =dFpress, is stable or not. It is unstable if the gravitational attraction in-creases faster than repulsive gas pressure,

dFgrav

dFgrav=

2GM

9v2s r

=8!

27

G#r2

v2s

"G#r2

v2s

> 1 (unstable). (1.13)

Let us consider this result in a more general situation as introduced above:consider an infinite almost homogeneous gas distribution with small per-turbation on it. Only those perturbation will grow, which meet the cri-terion above, Eq. (1.13). Small perturbations grow like sound waves (i.e.the equilibrium situation is stable), while large perturbations give rise for agravitational collapse. The dividing length scale is called Jeans length,

$J = vs

)

1

G#. (1.14)

Jeans –or gravitational– instability is the main driving agent for the forma-tion of any structure in the Universe, from clusters of galaxies to stars.

Black body radiation

The light is emitted from the photosphere and spectral intensity #($, T ) isbasically distributed according to Planck’s law of radiation

#($) d$ =8!hc

$3

1

ehc / kB"T # 1d$. (1.15)

Recall some features of this equation: The maximum shifts antiproportionalto temperature

T$max = 0.29 cm K, (1.16)

and the total luminosity, integrated over the entire spectrum, from the stellarsurface with radius r is

L = 4!r2 %SB T 4 (1.17)

(Stefan-Boltzmann-law) with %SB = 5.67 ! 10"5 erg cm"2 K"4 s"1. Hence,surface temperature and radius of star determine its luminosity.

Stellar evolution

A-B protostellar evolution:gas cloud contracts due to self-gravitythe gas is heated by compressionshort phase % 50 Myr for 1 M# star.

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1.1. STARS

Figure 1.2: Path of a star in the color magnitude diagram (CMD).

B ’zero age main sequence’ (ZAMS)central hydrogen burning startsinitial mass function (IMF) gives probability distribution of stellarmasses

p(M) $ M"2.35 0.1 ! M ! 100.0

(Salpeter IMF)

BC main sequence:the central nuclear burning process

4H+ + 2e" & He++ + 2 &

replaces radiative losses, quasi-stable phase in stellar lifethe time in this phase depends on the mass of the star

tMS % 10Gyr

!M

M#

""2.5

CD Hydrogen shell burning startsstar expands, outer shells coolstar becomes a red giant

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1.1. STARS

DE Helium contractiongradual transition of central helium core into liquid phase releasesadditional gravitational energy

E helium core ignitesexplosive burning phase which leads to stellar winds

F horizontal branch starouter shells are partly removedremaining envelop is highly transparenthence, light is emitted from a hot core region

G asymptotic giant branch (AGB)hydrogen and helium burn in shellsshells are not convective stable, pulses occur which lead to stellar windsand then to planetary nebularsurounding gas is enriched by metalls

H White dwarfall envelops are ejecteda hot, degenerated Helium core remains which cools very slowly (dueto small radius)

Late, explosive stages

Some stars may su"er an explosive burning phase as final evolution step.Those stars appear for a brief period as very luminous object. Such anevolution appears in particular for very massive stars.

• Wolf Rayet stars% 50 M#

outer shells are expelled during early evolutionstar appears as a hot, luminous, young object

• Supernovaeclassified by spectra

Ib, Ic, II: core collapse SNiron core becomes gravitationally unstable and collapses to a neu-tron star or black hole

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1.1. STARS

Figure 1.3: Spectra of di"erent type stars.

gravitational energy is released and leads to a shock front pushingouter shells

Ia: explosive burning of a white dwarfin a binary system mass is transferred from a companion to awhite dwarftransferred material may be burned in an explosive processno hydrogen lines since possibly only C, Si is transferred, othershell are already expelledall Ia SN show similar light curves, which can be ’rescaled’.The so-called Phillips relation connects the maximum B-bandmagnitude with the dimminhg rate. The latter is measured bythe magnitude decrement after 15 days

Mmax(B) = #21.726 mag + 2.698#m15(B). (1.18)

Classification of stellar spectra

Looking at the spectra of stars the shift of the maximum of emission can beseen clearly. Also a system of absorption lines can be seen clearly. They are

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1.1. STARS

Figure 1.4: The color-maginude diagram as obtained by Hipparcos.

caused by Hydrogen line transitions

$n$m =c

R%(n"2 # m"2)

with R% = 3.3 ! 1015 s"1. Transitions from the ground state n = 1 arecalled the Lyman series, those from the first excitet state belong to theBalmer series. The Balmer lines are in the visible range of the spectrum.

H! 656 nmH# 486 nmH$ 434 nmH% 410 nmH& 397 nmHcontinuum 370 nm

Balmer lines are only present under certain density/temperature con-ditions: is the temperture too low, almost all hydrogen atoms are in theground state, if the tempertaure is too high all atoms are in higher states orthe atom is fully inonized. In both cases there are too few atoms in the firstexcited state to cause a significant absorption. Hence, the relative strenghtof these lines is a good measure for the surface temperature.

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1.1. STARS

Figure 1.5: The luminosity function of stars in the local neighborhood.[Reid, Gizis, Hawley (2002) AJ 124]

O blue HeII emission + absorption25000 K

B blue HeI11000 K

A blue Maximum of Balmer lines7500 K

F blue-white metallic lines become noticable6000 K

G white-yellow neutral metallic lines5000 K

K orange-red metallic lines dominate3500 K

M red Molecular bands TiO VO

1.1.3 Stars in the color magnitude diagram

Hipparcos mission

CMD covers 10 mag, if mass range 0.2 - 60 -¿ L % L#(M/M#)4.0

The Hipparcos satellite (High Precision Parallax Collection Satellite,1989-1993) has measured the luminosities (detection limit % 11mag) and

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1.1. STARS

M L/L# M/M# tMS/Gyr $/(10 pc)3mag $corr

15 1.2 ! 104 10.4 0.03 12.5 3.0 ! 103

10 1.2 ! 102 3.3 0.51 7.5 1.0 ! 102

5 1.2 ! 100 1.0 9.07 3.5 3.50 1.2 ! 10"2 0.3 161.2 0.2 0.2

parallaxes (angular resolution: % 1 milliarcsec) of about 120,000 stars. Oneof the most simple representations of the obtained data set is the luminosity

function

$(M) =# stars in the interval

*

M # 12dM, M + 1

2dM+

volume dM. (1.19)

For the Hipparcos mission the volume is restricted by angular resolution ofthe telescope. Therefore, for all magnitude intervals the volume occupiedby the stars included in the data set is the same. The resulting sample isvolume limited, i.e. no stars are too faint to be detected. In contrast, thenumber of stars in a sample may be restricted by the detection limit of thetelescope. Eq. (1.1) tells us immediately that with increasing luminosity ofa star it can be detected at larger distance. In this case the sample is fluxlimited and the observable volume depends on the absolute magnitude.

Fig. 1.5 shows the luminosity function obtained by Hipparcos. One canuse it to obtain a rough estimate for the initial mass function (IMF) of thestars produced in the solar neighborhood. The luminosity function does notimmediately tell the IMF, since massive stars have a short life time whilethat of small stars exceeds even the age of the Universe. Therefore, allsmall stars are still present in the Milky, regardless when they have beenformed. We can correct for the loss of massive stars by assuming a constantstar formation rate over time. Furthermore, we assume that the age of theMilky Way equals half of the age of the Universe, i.e. t0 % 7 Gyr. Hence, ifthe life time of a star is less than the age of the Universe, only the fractiontMS/t0 is still present.

Open clusters

to be added

Clobular clusters

to be added

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1.2. MANY-BODY DYNAMICS

1.2 Many-body dynamics

1.2.1 Virial theorem

Newton’s law

Two point masses, " and ', are gravitationally attracted according to New-

ton’s law

F ! =d

dt(m!v!) = #

Gm!m#|x! # x#|3

(x! # x#). (1.20)

If the system consists of many point masses, many bodies, we have to sumup the force causes by all neighbors

d

dt(m!v!) = #

,

!! !="

Gm!m#|x! # x#|3

(x! # x#). (1.21)

Gravitational interaction is a long-range force, i.e. it decreases only propor-tional to the square of the distance. In a homogeneous matter distributionthe mass in a spherical shell around a given point x! is # 4!r2 #r. Themass in the shell increases with the same power of r as the force decreases,therefore a nearby and a very distant shell lead to the same contribution.Moreover, for gravity there is no shielding as for electric charges.

Gravitational potential

The force can be computed –for a conservative system– from the gravita-tional potential $ by

d

dt(m!v!) = #m!'!$(x!), (1.22)

where the index " could also be omitted. Comparing Eq. (1.22) and Eq. (1.21)we find that the potential can be computed in a many-body system by

$(x!) = #,

!! !="

Gm#|x! # x#|

, (1.23)

since

'!1

|x! # x#|= #

x! # x#

|x! # x#|3x# (= x!. (1.24)

Continuum description

Before we continue discussing many body systems we write down the def-inition of the potential for a continuous matter distribution. Substituting

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1.2. MANY-BODY DYNAMICS

m & #d3x we can rewrite Eq. (1.23)

$(x) = #-

G #(x!)

|x # x!|d3x! x! (= x. (1.25)

The force acting on a test body with mass m becomes

F (x) = #m'$(x) = #m

-

d3x! G #(x!) (x # x!)

|x # x!|3. (1.26)

The continuum description is advantageous when the number of point massesis very large and the motion of a each body is governed by the distributionof the particles as a whole, i.e. the interaction with immediate neighbors isnegligible. We will discuss below under which circumstances these conditionsare fulfilled.

Poisson equation

The integral equation (1.25) can be transferred into a di"erential equationby applying the Laplace operator

'2x$(x) = #

-

d3x! G #(x!) '2x

1

|x # x!|# $% &

=0 for x" &=x

x! (= x. (1.27)

One may be tempted to assume that the r.h.s. is zero, but it turns out thatthis not true when treating the singularity x! # x = 0 properly. To do so,we assume that the density, #(x), is homogeneous within a small sphere, S&,with radius ( and center x. Furthermore, we apply the divergence theorem

-

V'2f d3x =

-

'V'f d2x, (1.28)

where f has to su!ciently ‘regular’ within the arbitrary volume V . Itssurface is denoted by )V . Thus, the volume integral of the divergence of fcan be replaced by an surface integral of its gradient. We rewrite Eq. (1.27)

'2x$(x) " #G #(x)

-

'S#

'x1

|x # x!|d3x!.

Eq. (1.24) tells us that the integrand is constant at the surface )S& and hasthe value #1/(2. Note, that the normal vector of a surface element is an-tiparallel to the direction of the centrally orientated force, this leads to anadditional minus sign. So, we can simply multiply this value by the surfacearea 4!(2. Finally, we obtain the Poisson equation

'2x$(x) = #4!G #(x). (1.29)

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1.2. MANY-BODY DYNAMICS

In general, it may be di!cult to find a solution for the Poisson equation.For complex matter distributions one has to rely on numerical methods.Poisson’s equation can be substantially simplified if the matter distributionis spherical.

Spherical symmetry

Inside a spherical, homogeneous shell of the force acting on a test body van-ishes, since the contributions from two opponent cones cances out. Hence,the potential inside the shell is constant. In general, inside a sphericallysymmetric mass distribution the force is zero since it can be considered tobe composed of individual shells.

Outside a spherically symmetric mass distribution we integrate first theforce, F , over a spherical surface with radius r

-

'Sr

F d2x = #G

-

'Sr

d2x

-

Sr

d3x! #(x!) (x # x!)

|x # x!|3.

Reconsidering the steps we have done we find that the r.h.s. simplifies to4!GM , where M is the total mass inside the sphere with radius r

M(< r) =

-

Sr

d3x!#(x!).

For spherical symmetry the surface integral over the force is 4!r2 F , whereF = |F |. Hence, we can write for the force outside a spherical mass distri-bution

F = #GM(< r)

r3r. (1.30)

For spherical symmetry the gradient simplifies to )/)r, hence Eq. (1.22) cansimply integrated and we obtain for the potential

$(r) = #GM(< r)

r. (1.31)

Note, the potential is normalised in a way that $(r & )) & 0.When there is mass inside and outside the force can be still computed

according to Eq. (1.30). In contrast for the potential the mass beyond the

radius in questions has to be taken into account. It can be easily shown that

$(r) = #'

GM(< r)

r+ 4!G

- %

rdr! #(r!)r!

(

(1.32)

leads with F = #)$/)r to the force derived above.

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1.2. MANY-BODY DYNAMICS

Single-body energy conservation

We resume the discussion of a many-body system. First, we multiplyEq. (1.20) by v!,

v! ·d

dt(m!v!) = #m!v! ·'$(x!) (1.33)

If the potential is independent from time we can write

d

dt$(x) = v! ·'$(x!).

Thus, Eq. (1.33) becomes

d

dt

.1

2m!v

2 + m! $(x!)

/

= 0. (1.34)

This results tells us simply, that the energy of a particle moving in a constantpotential has a constant total energy,

E =1

2mv2

# $% &

EK

+ m$(x)# $% &

EP

, (1.35)

where EK denotes the kinetic energy and EP the potential energy. A testbody can only leave the potential well if E > 0, therefore the escape speed

at a given position is

v2e(x) = #2$(x). (1.36)

Many-body energy conservation

We can also work out the energy conservation for the energy conservationfor a system of point masses. To this end we multiply Newton’s equation ofa many-body system (1.21) by v! an sum over all "

,

!

v! ·d

dt(m!v!) =

d

dtEK = #

,

",!! !="

Gm!m#|x! # x#|3

(x! # x#) · v! (1.37)

Consider the r.h.s. of the equation. We could simply exchange " and ' sincewe sum over both of them. This operation would not change the result. Sowe just do it and add both versions. Taking care for the sign in the bracketswe find

2d

dtEK = #

,

",!! !="

Gm!m#|x! # x#|3

(x! # x#) · (v! # v#). (1.38)

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1.2. MANY-BODY DYNAMICS

This equation can be finally rewritten as

d

dt

0

12EK #

1

2

,

",!! !="

Gm!m#|x! # x#|

3

45 = 0 (1.39)

The second term, i.e. the potential energy of a many-body system, can berepresented in several ways, either for discrete particles

EP = #1

2

,

",!! !="

Gm!m#|x! # x#|

=1

2

,

!

m!$(x!) (1.40)

or for a continuous matter distribution

EP =1

2

-

d3x #(x)$(x). (1.41)

We have found that in many-body system the energy is conserved when onlythe gravitational interaction between the particles is present. Unfortunately,this does neither tell how the energy is distributed among the particles norwhat fraction is stored as kinetic energy. Here the virial will theorem help.

Virial theorem

We start again from Newton’s law for a many-body system, Eq. (1.21). Theforce derived by this equation includes only the gravitational interactionswithin the many-body system. For completeness we consider additionallyexternal forces, F ext

! , whose origin is irrelevant here. Now we proceed in asimilar way as above, but this time we multiply Newton’s law by x! andsum over ",

,

!

x! ·d

dt(m!v!) = #

,

",!! !="

Gm!m#|x! # x# |3

(x!#x#) ·x!+,

!

F ext! ·x!. (1.42)

This time the l.h.s. is a bit more di!cult to interpret. It can be rewrittenby

,

!

x! ·d

dt(m!v!) =

1

2

,

!

d2

dt26

m!x2!

7

#,

!

m!v2! =

1

2

d2I

dt2# 2EK , (1.43)

where we have used the definition of the moment of inertia

I =,

!

m! x! · x!. (1.44)

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1.2. MANY-BODY DYNAMICS

We follow again the steps we have made to derive the energy conservation ofa many-body system: exchange " and ' in Eq. (1.42) and add the result tothe equation itself. In the resulting equation we can simplify the expression

#,

",!! !="

Gm!m#|x! # x#|3

(x! # x#) · (x! # x#) = #,

",!! !="

Gm!m#|x! # x#|

= 2EP . (1.45)

Hence, we can rewrite Eq. (1.42)8

1

2

d2I

dt2

9

# 2 *EK+ = *EP + +,

!

:

F ext! · x!

;

, (1.46)

where we have applied the time average operation,

*"+ =1

T

- T

0dt ". (1.47)

For any given time the second derivative of the moment of inertia can belarge, hence, we couldn’t conclude for the partition between kinetic and po-tential energy. When the many-body system reaches an equilibrium state,the derivative should vanish, at least in average. We call this process virial-

ization. Therefore a virialized system obeys the relations

2 *EK+ + *EP + +,

!

:

F ext! · x!

;

= 0, (1.48)

the viral theorem. It is of great use, since it allows us in combination withthe energy conservation, Eq. (1.39) to derive the amount of kinetic energyin a system for which we know the total energy, if there is no external forcedisturbing the system.

A mass estimator

Let us consider a very simplified picture for an star cluster: It may composedas a homogeneous sphere with constant number density inside and no starsoutside,

n(r) =

'

n0 : r < R0 : elsewhere

(1.49)

The stars in this cluster may have the mass m and move with velocity V ,randomly directed. For these assumptions the kinetic energy is easily derived

EK =1

2

,

!

mv2 =1

2

4

3!R3 n0 m 3v2

z =3

2Mv2

z , (1.50)

where M is the total mass. For the spherical model the velocities do nothave a preferred direction, therefore we have used v2 = v2

x + v2y + v2

z = 3v2z .

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1.2. MANY-BODY DYNAMICS

To derive the potential energy needs a little bit more e"orts. We computeit according to the continuous description given in Eq. (1.41). To use thisformula we have to determine the potential $(r), to which end Eq. (1.31)can be utilized,

$(r) = #G4

3!n0mr3 1

r= #

4!G

3n0mr2 for r < R. (1.51)

We can restrict ourselves to the discussion inside the sphere, i.e. r < Rsince outside the density is zero, hence these regions will not contribute tothe potential energy. Inserting the expressions for potential and density intoEq. (1.41) gives

EP = #1

24!

- R

0dr r2 n0m

4!G

3n0mr2 = #

8!2 G

15n2

0m2R5

= #3

10

GM2

R(1.52)

Finally, we assume that the cluster is virialized and there are no externalforces, hence

2EK = #EP

3Mv2z =

3!2

10

GM2

R.

We solve for the mass and obtain

M = 10v2zR

G. (1.53)

Thus, if we are able to measure the extension of a star cluster and thevelocity dispersion in one direction we can conclude for the mass of thecluster. More realistic mass distributions would only change the numericalconstant in front of the r.h.s., namely % 3.

1.2.2 Two-body relaxation

Strong encounters

When a mass, m, passes another, M , close by the gravitational attraction de-flects the trajectories of both particles. We assume here that M , m, henceonly the smaller mass moves along a curve. We wish to distinguish betweenstrongly and weakly bent. When the mass, m, gains as much gravitationalenergy as it had kinetic energy originally the trajectory is significantly af-fected. The closer m passes by the stronger the trajectory will be bent. Weuse the energy criterion to define below which distance m will be stronglybent:

|EP |max # EorgK

GmM

r#

1

2mv2.

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1.2. MANY-BODY DYNAMICS

Hence, we define the strong encounter radius

rs =2GM

v2. (1.54)

Consider the simple model for a globular cluster above. The average velocityis 10km s"1, and stars may have in average solar mass. The resulting strongencounter radius is 9 ! 10"5 pc. On the other hand, the number density ofstars is 3 ! 104 pc"3, therefore the average separation is much larger thanthe strong encounter radius. However, from time to time a strong encounterwill happen. We estimate how frequent this happen for a single star. Onecan consider the area !r2

s as cross section for a strong encounter. The areacovered by N stars is

N !r2s . (1.55)

If a star moves with velocity v with respect to all other stars we can definea volume by

Avt, (1.56)

where A is an arbitrary area perpendicular to the velocity v, and t is thetime since the star cluster has been formed. The sum of the cross sectionsof all stars in this volume is

Avt n !r2s , (1.57)

where n is the number density of stars. If the sum of all cross section is alarge as the area, A, it is very likely that the particle su"ers at least onestrong encounter while moving through this volume. Hence, the time afterwhich an encounter should happen is

ts =1

! v n r2s

=v3

4!Gn M2(1.58)

= 4 ! 1012 yr< v

10 km s"1

=3!

m

M#

""2 !n

pc"3

""1

,

where we have used the definition of the strong encounter radius for the laststep. As a result of a strong encounters the trajectories of stars becomevirtually random, the star cluster undergoes relaxation. For the samplecluster above the relaxation time is 1.4 ! 108 yr. Hence, globular clustersare rather relaxed systems.

Weak encounters

If a star passes another with a minimum distances larger the strong en-counter radius it will still deflected by a small angle. To determine thisangle we integrate the force acting on the mass, m. Since the deflection is

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1.2. MANY-BODY DYNAMICS

tiny we can assume for the force calculation that the trajectory is straight.The force is given by

F =GnM

r3r, (1.59)

where r is the vector from M to m

r =

!r'r(

"

=

!b

vt

"

. (1.60)

Thus, the perpendicular force is given by

F' =GmMb

-b2 + v2t2

3 . (1.61)

Integration along the trajectory gives the gained perpendicular velocity

m#v' =

- %

"%dt F' =

2GmM

bv. (1.62)

Therefore, we can write for the deflection angle

" "#v'

v=

2GM

bv2(1.63)

where we have used that " is a small angle. The total deflection can be con-sidered as a sequence of random deflections with di"erent impact parameter,b, caused by the neighbours along the trajectory. For the sum of all deflec-tions with the same impact parameter we obtain the total perpendicularvelocity

v2'db = #v2

' 2!bdb vt n (1.64)

Taking into account the contributions from all possible impact parametersthe perpendicular velocity becomes

v2' =

- bmax

bmin

db8!G2 M2 nt

bv=

8!G2 M2nt

vln

!bmax

bmin

"

(1.65)

To determine the ratio % = bmax/bmin we make the following assumptions

bmin: the strong encounter radius rs

bmax: the extension of the globular cluster

Similar to strong encounters the initial parameters of the trajectory becomeunimportant when the deflection becomes large, v' % v. Using this condi-tion we define the relaxion time according to weak encounters

tweak =v3

8!G2 M2 n ln%=

1

2 ln%ts. (1.66)

Hence, weak encounters are more e!cient as the strong encounters andgovern therefore the relaxation time of a cluster.

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1.3. DISK GALAXIES

Figure 1.6: Hubble sequence with examples: M87 (E1), M74 (Sc), M51 (Sc),ngc1300 (SBb), ngc613 (SBc), ngc1365 (SBc).

1.3 Disk galaxies

The Hubble sequence

Galaxies have shapes from virtually spherical to thin and disky. The mor-phology is described in the Hubble sequence.

1.3.1 Building blocks for disk galaxies

Disk galaxies have some di"erent constituents:

nucleus:Central, very compact, star cluster

bulge:A rather massive, spherical assembly of stars in the centre

bar:Some galaxies show an elongated, bar-like, assembly of stars

disk:Consists of stars, gas, and dust, there is a thin and a thick disk

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1.3. DISK GALAXIES

halo:Spherical distribution of stars in which the disk is embedded in, con-sists basically of dark matter

open clusters:Young star clusters in the disk

globular clusters:Old star clusters in the halo

Exponential disk

The distribution of stars in the disk of the Milky Way can be well describedby a double-exponential function

nP(R, z) = nP0 exp

'

#R

hR

('

#|z|hP

z

(

, (1.67)

where hz is the scale height and hR the scale length. The scale height and thedensity constant, n0, are di"erent for di"erent stellar populations. Moreover,considering G and K stars, i.e. looking at a rather old star population,the vertical distribution is best described by the sum of two exponentialfunctions, with scale heights of about 330 and 1500 kpc. In contrast youngstars follow a single, thin disk with a scale height about 250 kpc. Thisindicates that stars are formed in the thin disk and by two-body interactionstheir velocity dispersion is increased.

1.3.2 Galactic rotation

Looking face-on at a disk galaxy the projected star density, and thereforealso the surface luminosity, &, becomes a single exponential function

&(R) = &0e"R/hR . (1.68)

We can determine the total luminosity inside the radius R by integration

L(< R) =

- R

0dR 2!R &(R)

= 2!&0 h2R

- R/hR

0d* * e"(. (1.69)

Integration by parts leads to

L(< R) = 2!&0 h2R

>

#* e"( # e"(?R/hR

0

= 2!&0 h2R

.

1 #R

hRe"R/hR # e"R/hR

/

. (1.70)

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1.3. DISK GALAXIES

Figure 1.7: Rotation curve for an exponential disk.

Let us consider the questions how the shape of the disk can be sustained. Thegravitational attraction towards the galactic centre has to be compensated.In the disk the rotation of the stars leads to balancing centrifugal force. Fora stable configuration we demand

F centri = F grav (1.71)

mv2

R=

GmM(< R)

R2.

To determine the gravitational force at radius R we have used the approxi-mation that the mass distribution is spherical (only in this case we can usethe r.h.s. expression). For a disk galaxy this is clearly a very bold assump-tion, however, a more detailed analysis would lead to similar conclusions.We assume therefore in the following that the circular velocity is given by–even if the disk is the dominant structure–

v2circ(R) =

GM(< R)

R. (1.72)

Using the surface brightness of an exponential disk, Eq. (1.70), and assuminga constant mass-to-light ratio, M/L, we can write for the circular velocity

v2circ(R) = 2!G&0

M

LhR

.hR

R#

hR

Re"R/hR # e"R/hR

/

. (1.73)

Fig. 1.7 indicates that for an exponential disk the circular velocity shouldbe maximal at 1.8 scale lengths and should decrease for larger radii.

Oort’s constant

How can we determine the circular velocity of the Sun? This is not a trivialtask at all, since we cannot observe the ‘rest frame’ of the Milky Way. What

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1.3. DISK GALAXIES

Figure 1.8: The Sun and another star moving on circular orbits.

we can determine instead are radial and proper motions of stars and gasclouds in the galactic plane. Fortunately, from these informations we candeduce the circular velocity of the Sun.

Consider Fig. 1.8. The di"erence between the velocity of the Sun andanother star, both moving on circular orbits, is given by

v) # v# = ')R)

!cos +

# sin +

"

# '#R#

!1

0

"

. (1.74)

For the position of the star we can write

R) = R)

!sin +

cos +

"

=

!D sin l

R# # D cos l

"

. (1.75)

We can use this last equation to eliminate R) cos + and R) sin + from theequation for the relative velocity,

v) # v# =

!R#(') # '#) # D') cos l

#D') sin l

"

. (1.76)

In order to apply the formalism to observed data we should decompose thevelocity into its radial and its proper motion part. We do this by projectingthe relative velocity. The radial velocity becomes

vr = (v) # v#) ·!

sin l

# cos l

"

= R# (') # '#) sin l, (1.77)

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1.3. DISK GALAXIES

and the proper motion (tangential velocity) can be written as

vt = (v) # v#) ·!

cos l

sin l

"

= R# (') # '#) cos l # D'). (1.78)

Hence, if know the radius, R#, and the angular velocity, '#, of the Sun wecan deduce these properties for another star when we are able to measureits relative velocity with respect to us. However, the angular velocity ofthe Sun is something we want to know, therefore we have to proceed in ouranalysis. Considering only close-by stars we can approximate

') # '# =d'

dR

@@@@R#

(R) # R#) =

.1

R

dv

dR#

v

R2

/

R#

(R) # R#). (1.79)

Remember, if stars move on circular orbits, the balance between gravita-tional and centrifugal forces determine its velocity. Therefore the angularspeed, ' = v/R, is a well defined function of the radius. We introduce nowthe Oorts constants

A := #1

2

A

dv

dR

@@@@R#

#v#R#

B

(1.80)

B := #1

2

A

dv

dR

@@@@R#

+v#R#

B

. (1.81)

In particular the di"erence and the sum of the Oorts constants lead to simpleproperties of the galactic rotation, namely the angular velocity and the sheararound the Sun,

A # B =v#R#

= '# (1.82)

#(A + B) =dv

dR

@@@@R#

. (1.83)

Using the definition of the Oorts constants –and the approximation for ')#'#– we can write for the radial and the tangential velocity

vr " #2A(R) # R#) sin l (1.84)

vt " #2A(R) # R#) cos l # ')D. (1.85)

For small distances, i.e. |R) # R#| . D we can use the following approxi-mations: R) #R# " #D cos l and ') " '# = A#B. Hence, we can finallywrite

vr " AD sin 2l (1.86)

vt " AD cos 2l + BD, (1.87)

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1.3. DISK GALAXIES

Figure 1.9: Rotation curve of the Milky Way

where we have used 2 cos2 l # 1 = cos 2l. Thus, plotting the radial velocityand the proper motion of stars with similar distance, D, as a function of thelongitude, l, allows to derive the Oorts constants A and B, and hence theangular velocity of the Sun and the local shear. Observations give

A = 14.4 ± 1.2 km s"1 kpc"1 (1.88)

B = #12.0 ± 2.8 km s"1 kpc"1. (1.89)

1.3.3 Rotation curve

The circular velocity v(R) is called rotation curve of a galaxy. Consider aline-of-sight starting from the Sun’s position inwards, i.e. #90 < l < 90. Itis possible to measure radial velocities along such a line-of-sight using atomicor molecule emission lines in the regime of radio wavelengths. For opticalemission the extinction due to interstellar dust is too strong. Moreover, itis advantageous to measure the radial velocity of molecular clouds, insteadof stars, since they haven’t been subject to relaxation, i.e. they have arather low velocity dispersion. The 21 cm radiation caused by hyperfinesplitting of neutral hydrogen (i.e. the interaction between an electron andthe magnetic moment of the nucleus) is used to measure the velocity of themolecular clouds. Along a line-of-sight the Sun-centred galactic longitude,l, is constant, hence we can rewrite Eq. (1.77)

vr(D) = R#('(D) # '#) sin l. (1.90)

It is reasonable to assume, that ' = v/R is monotonically decreasing out-wards, at least in the solar neighbourhood where we already expect v(R) to

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1.3. DISK GALAXIES

be decreasing. Therefore, '(D)#'# is maximal when the line-of-sight hasits smallest distance to the galactic centre. For this particular point the po-sition vector of the molecular cloud, RMC, is perpendicular to the directionof the line-of-sight, + = 90 # l. Hence, we can write

R'MC = R# cos(90 # l) = R# sin l. (1.91)

Thus, from the galactic longitude we can conclude for the radius of themolecular cloud with the maximal radial velocity along the line-of-sight.Inserting Eq. (1.91) into Eq. (1.90) and using ' = v/R we find for circularvelocity of the cloud with maximal radial velocity

v(R) = vmaxr +

R'MC

R#v#. (1.92)

See Fig. 1.3.4 for the distribution of radial velocities of molecular clouds inthe Milky Way. Fig. 1.10 shows the rotation curve of the Milky Way. Oneof the most important results is that the rotation curve does not show anysign of decrease for large radii. This is not consistent with a circular velocityinduced by the exponential dist, therefore, additional (‘dark’) matter haveto exist in the halo of the galaxy.

1.3.4 Spiral structure

The winding problem

From the pictures of spiral galaxies one may get the impression that spiralarms are rather solid and stable over a long time. As a consequence thespiral arms would rotate as entity around the galactic centre. To test thehypothesis of such a ‘solid’ spiral arms let us consider how an idealisedarm evolves. It may have a straight shape in the beginning, with the Sunat the radius R# and another star at half of the Sun’s radius. The anglebetween the Sun and the star, +, (seen from the galactic centre) is zero inthe beginning. If we assume a flat rotation curve, as indicated for the MilkyWay, how long does it take until the angle + becomes 2!? We can write forthe angle

+ = (') # '#) t. (1.93)

We solve this equation for t and set + = 2!. Moreover, we use ' = v/R,R) = R#/2, and we assume flat rotation curve, v = const. = v#. Hence, weget for the time

t2) =2!

') # '#=

2!R#

v#. (1.94)

For the parameters of the Sun, v# = 250 km s"1 and R# = 8 kpc, we obtain

t2) " 0.2 Gyr. (1.95)

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1.3. DISK GALAXIES

If the present configuration of the Milky Way is much older as this time,which is for instance indicated by the vertical structure of the disk, the armsshould from very tight spirals. A similar argument should be also valid forother galaxies. In contrast such very tight spirals are seldom observed. As aconsequence we have to conclude that spiral arms are not long-living, stableentities.

Epicycles

In the preceding sections we have assumed that stars move on perfect circlesaround the galactic centre. This is not necessarily the case. Stars mayadditionally oscillate around the circular orbit, around the guiding centre.Considering the radius of a star, R, we can write

R(t) = Rgc + r0 cos(,t + +0), (1.96)

where Rgc is the radius of the guiding centre, r0 is amplitude of the oscilla-tion, , the frequency of the oscillation, and + the phase of the oscillation att = 0. We will show below that such harmonic oscillations exists and that, is well defined function of the radius.

Radius equation

A star with mass M may orbit in a spherical gravitational potential $(R).The acceleration is given by the gradient of the potential, see Eq. (1.22).

#'$ = #!)x

)y

"

$ =

!vx

vy

"

=

!x

y

"

, (1.97)

where we have considered only the galactic plane, x-y. To get an equationfor the radius R we introduce a circular coordinate system, x = R cos + andy = R sin +, hence the trajectory of the star is given by the functions R(t)and +(t). We insert the circular coordinate system into Eq. (1.97) and getfor the x-component:

#)x$ = x = R cos + # R+ sin + # R+2 cos +. (1.98)

We are free to choose the coordinate system in a way that for t = 0 the staris located at the x-axis, hence )x = )/)R, cos + = 1, and sin + = 0. As aresult, we obtain an equation for the radius

#)$

)R= R # R+2. (1.99)

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1.3. DISK GALAXIES

E!ective potential

The angular momentum of the star with velocity v and relative position tothe centre of the rotation x is defined by L(x) = M x ! v. A torque r ! Fmay change the angular momentum according to L = x!F. However, in aspherically symmetric system the direction of the gravitational force #'$ isalways radial. Therefore, F is parallel to x, no torque occurs, and the angular

momentum is conserved. We denote with Lz that component of the angularmomentum which is perpendicular to the x-y-plane, i.e. perpendicular tothe plane of motion. Lz can be written in circular coordinates

Lz = MR R+ = MR2+. (1.100)

This provides a relation between + and R, and allows us to eliminate + inEq. (1.99)

R = #)$

)R+

L2z

M2R3

= #)

)R

'

$ +L2

z

2M2R2

(

(1.101)

= #)$e!

)R,

where we define the e!ective potential which depends of course on the radiusR but also on the angular momentum, Lz, of the star

$e!(R,Lz) = $(R) +L2

z

2M2R2. (1.102)

In general, we expect the e"ective potential to have a minimum at a certainradius. At the position of the minimum the first derivative has to vanish.From Eq. (1.101) we obtain for the minimum

0 =)$e!

)R

@@@@min

=)$

)R

@@@@Rmin

#L2

z

M2R3min

=)$

)R

@@@@Rmin

# Rmin+2(Rmin). (1.103)

This restores the condition for a circular orbit, F grav = F centri, see Eq. (1.31)and Eq. (1.71). Therefore, the minimum in the e"ective potential reflects acircular motion with radius Rmin, i.e. with radius Rgc as denoted here inthe context of epicycles.

Perturbed trajectory

We consider now a trajectory which deviates only slightly from the guidingcentre orbit, as indicated by Eq. (1.96). Eq. (1.99) describes the evolution of

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1.3. DISK GALAXIES

the radius. Since we allow only small deviation from the guiding centre wecan expand the e"ective potential into a Taylor series around Rgc

$e!(R) = $e! |Rgc+

)$e!

)R

@@@@Rgc

(R # Rgc) +1

2

)2$e!

)R2

@@@@Rgc

(R # Rgc)2 + ... .

From Eq. (1.103) we know that the second term on the r.h.s., i.e. the linearterm )R$e! , vanishes. The evolution of the radius is given by the quadraticterm

R =d2

dt2(R # Rgc) = #

)$e!

)R" #

)2$e!

)R2(Rgc) (R # Rgc) .

We have added on the l.h.s. the constant radius Rgc, which vanishes whenperforming the time derivative. We have now obtained a linear di"erentialequation of the type r = #,2r for the di"erence r = R # Rgc.

Epicyclic frequency

The solution for this ordinary di"erential equation is a harmonic oscillationof r, see Eq. (1.96), with epicyclic frequency ,, which abbreviates the secondderivative

,2(Rgc) =)2$e!

)R2(Rgc) =

)

)R

!)$

)R#

L2z

M2R3

"@@@@Rgc

=)2$

)R2

@@@@Rgc

+3L2

z

M2R4

@@@@Rgc

=)2$

)R2

@@@@Rgc

+ 3'2(Rgc) .

The epicyclic frequency for each radius Rgc depends only on the properties'(Rgc), which is given by the rotation curve, and )2

R$ of the guiding centreorbit. We can also characterize the potential by the '(Rgc) of circular orbits.We use to this end Eq. (1.103) and set + = '. Hence, we can write

)2$

)R2

@@@@Rgc

=)

)R

)$

)R

@@@@Rgc

=)

)R

6

R'27@@@@Rgc

.

Therefore the epicyclic frequency is given by

,2(R) =)(R'2)

)R+ 3'2(R) =

1

R3

)

)R

6

R4'27

,

where we have omitted the index ‘guiding centre’.

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1.3. DISK GALAXIES

Figure 1.10: Pattern speed

Pattern speed

We have seen above that for each radius the frequency of an perturbation ofthe circular orbit is determined by the local potential and angular momen-tum. With this knowledge about the epicyclic frequency we return to thediscussion of the spiral structure. In general, an epicyclic motion does notlead to a closed orbit. But a rotating observer exists for which the epicyclictrajectory of a star appears to be closed. Assume an observer rotates withangular speed 'P with respect to the Milky Way. The time of one epicycleis

Te =2!

,. (1.104)

During this time the star reached an angle #- = 'Te with respect to thegalactic rest frame. The rotating observer sees the angle

#-P = #-# 'P. (1.105)

We can choose the angular speed, 'P, in a way that the angle #-P isrationale fraction of 2!

m#-P = n 2!. (1.106)

This means that for the rotating observer the trajectory is closed after nfull orbits and the star has fulfilled m epicycles during this time. InsertingEq. (1.106) into Eq. (1.105) and taking into account that #-/Te = ' we find

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1.3. DISK GALAXIES

'P = ' #n

m,. (1.107)

We can interpret the result as follow: The observer in the galactic rest framesees the closed orbit of the (as derived for the rotating observer) rotatingwith 'P. In the rotating frame a ‘pattern’ is generated which rotates forthe rest frame observer as a whole. It turns out that 'P is varies much lesswith radius than ' itself, for a suitable choice of n and m. This defines thenumber of spiral arms which are seen. Therefore, spiral arms are density

waves in the galactic disk.

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Figure 1.11:Molecular cloudsin the galactic plane.

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Chapter 2

Structures in the Universe

2.1 Local Group

Our companions

Chemical enrichment history

The metallicity of the stellar populations and of the gas (if present) in galax-ies sheds some light on the conversion history from gas to stars. Basically,all elements heavier than Helium (‘metals’ in astronomers jargon) are pro-duced in stars and partly mingled with interstellar gas when massive starsexplode. The metal budget in a galaxy is in general complex since a lotprocesses a"ect the matter circuit, see Fig. 2.2.

We wish to estimate by the help of a toy model the metal enrichmenthistory of the Milky Way. In general, the abundance of an element is givenwith respect to the solar abundance

[A/H] = log10

'(# atoms of A / # atoms of H))(# atoms of A / # atoms of H)#

(

. (2.1)

He we have used the hydrogen number density for reference, however, anypair of elements may be used. A more rough approach is to give the metal-

licity, i.e. the mass fraction locked in metals, Z.

Let us now establish a toy model for the evolution of the metallicity of agalaxy according to the baryon budget shown in Fig. 2.2. To keep the modelas simple as possible we have to use several drastic simplifications. We thedenote the entire gas mass in the galaxy by g and all in stars and stellarremnants by s. The cumulative mass in locked in all elements heavier maydenoted as gZ , hence, the metallicity becomes Z = gZ/g. With the help ofthe following assumptions we can construct the enrichment history of thegalaxy

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2.1. LOCAL GROUP

Figure 2.1: Galaxies in the local group

Figure 2.2: Galactic baryon budget

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2.1. LOCAL GROUP

i. Closed-box model: There is no mass exchange with the environmentof the galaxy, i.e. no accretion, merging or stripping takes place. Asa result, mass can only be exchanged between the gas and the starreservoir

ds = #dg (2.2)

• Instantaneous recycling: In general the feedback of metals at a giventime, t depends on the yield, p, i.e. the fraction of matter convertedinto metals in a star, furthermore on the star formation rate, (, atthe time t # .(m)/, where .(M) is the life-time of a star with massM , and on the initial mass function, $. The metal feedback withinthe time interval dt at the time t is given by the convolution

-

dM p(M)(M # Mremn(M)) ((t # .(M)) $(M) ! dt, (2.3)

where Mremn is the mass of the stellar remnant. A large fraction ofmetals are produced in the very massive stars, which life-time muchshorter than the age of the Milky Way. Therefore, we assume thatthere are only two types of stars, those which explode immediatelyafter forming and those which life forever. The amount of metalsreturned from stars in the time interval dt becomes

pds, (2.4)

where all factors are incorporated in the yield, p.

• Instantaneous mixing: The metals in gas reservoir are always wellmixed.

With the help of those assumption the increase of metals in the gasreservoir can be written as

dgZ = #Zds + pds. (2.5)

For the change of the metallicity in the gas we obtain

dZg = d

!gZ

g

"

=1

gdgZ #

gZ

g2dg (2.6)

=pds # Zg(ds + dg)

g. (2.7)

For a closed-box model this reduces to

dZg = #pdg

g, (2.8)

which can be integrated

Zg(t) = Zg(0) + p lng(0)

g(t). (2.9)

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2.1. LOCAL GROUP

This result reflects that in a closed-box scenario the continuous consumptionof the gas reservoir entails its enrichment with metals. At a given time, t,newly formed stars have the metallicity of the gas reservoir. Hence, all starsformed before t have a lower metallicity. The mass in these stars is just theconsumed gas mass

s(< Zs) = g(0) # g(t) = g(0)

'

1 # expZs # Z(0)

p

(

. (2.10)

Comparison with the metallicity distribution in the Milky Way allows toestimate the yield

pMilkyWay % 0.5 # 0.7Z#, (2.11)

where the value of the disk is lower. This may possibly attributed to theoutflow of enriched gas, since the potential well in the disk is less deep. Asa result the e"ective yield is reduced. For the average metallicity of stars avalue of 0.7Z# has been found.

Even if in general the distribution of metallicities is well reproduced withthose values for the yield, for low-metallicity stars a discrepancy occurs. Forthe ratio between metal-poor and average stars we find

s(< 0.25Z#)

s(< 0.7 Z#)=

1 # exp(#0.25Z#/p)

1 # exp(#0.7 Z#/p)" 0.5. (2.12)

Thus half of the stellar mass should be found in metal-poor stars. It hasbeen found (first for G-dwarfs) that the fraction is significantly smaller, G-

dwarf problem. A possible solution is, that the Milky Way has been buildfrom gas which has been partly enriched in progenitor galaxies.

Dynamical friction

When a satellite galaxy moves the halo of a host it is decelerated. Thereason for this is that the mass in the halo gets accelerated. To evaluate thee"ect we redo the steps of the computation of the weak encounter time. Weinclude now that also the deflecting mass gets accelerated. We have foundthat one encounter leads to a perpendicular velocity of

#v' =2Gm

bv, (2.13)

see Eq. (1.62). Note, m denotes here the mass of the scattering centre.Momentum is conserved, therefore

M #V' = m v'. (2.14)

The total energy in the perpendicular motions become

#Ekin' =1

2M(#V')2 +

1

2m(#v')2 (2.15)

=2G2

b2V(Mm(m + M). (2.16)

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2.2. GALAXY GROUPS

The total energy (/ + 0) is conserved, therefore the parallel must decrease.Initially there is only energy in the parallel motion, hence, we can derive thechange in velocity, #V(,

1

2MV 2

( =1

2M

6

V( + #V(

72+ #MV 2

' + #mv2'. (2.17)

Computing the square, considering only changes in first order, i.e. neglectingthe term with #V 2

( , leads to

#V 2( = #

2G2m(M + m)

b2V 3(

. (2.18)

To obtain the total decrease of V( we have to sum up all interactions withneighbours. We proceed similar to the discussion of weak encounters and in-tegrate over all neighbours in a cylinder around the trajectory. The decreaseof velocity becomes

dV( = #nV(dt

- bmax

bmin

db 2!b2G2m(M + m)

b2V(. (2.19)

Hence, the dynamical friction becomes

dV(

dt= #

4!G2 nm(M + m)

V 2(

lnbmax

bmin(2.20)

2.2 Galaxy groups

Galaxy groups are assemblies of a few (roughly up to twenty) large diskgalaxies. They may accompanied by a large number of smaller galaxies(which are in fact hard to observe). Consider for example the Hicksongroup 51. It consists of seven galaxies with a radial velocity dispersionof %r = 470 km s"1. The core radius for this group is estimated to berc = 370 kpc (as far as this is possible for such a small number of objects).We can use the virial theorem to estimate the mass of the group

Mgroup %3%2

r rc

G" 6 ! 1013 M# (2.21)

In contrast, for the luminosity of all galaxies a value of LB = 2 ! 1011 L#

has been found. There is about hundred times more ‘gravitational active’mass than ‘luminous’ mass. The galaxies seem to be embedded in a darkhalo. Galaxy groups are believed to be ‘young’ on cosmological time scales.For the Hickson group 51 we know the velocity dispersion, if we can also

39

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2.2. GALAXY GROUPS

estimate the matter density of the dark halo we can infer the dynamicalfriction time scale. A simple estimate for the density is

# =3 Mgroup

4!r3c

= 1.8 10"26 g cm"3 = 0.01 cm"3 mH. (2.22)

Hence, the dynamical friction time scale is of the order of

tdyn"frict = v

@@@@

dv

dt

@@@@

"1

=v3

4!G2 nm (M + m) ln%= 0.6 Gyr. (2.23)

Hickson groups are compact in general, therefore for these groups the dy-namical friction time is the shortest. These groups couldn’t exist for the ageof the Universe.

Tully-Fisher relation

How can we estimate the distance of galaxies? We will see in the cosmologychapter that the redshift of absorption or emission lines is a good indicator.However, to get the redshift one needs a high resolution spectrum. Alterna-tively it has been found that the circular velocity of disk galaxies is relatedto its luminosity. In Sec. 1.3.2 we have derived an expression for the circu-lar velocity of an exponential disk, see Eq. (1.73). The rotation curve hasa maximum at R/hR " 1.8, see Fig. 1.7. The expression in the bracketsin Eq. (1.73) takes its maximal values % 0.3 at this point. Hence, we canconclude for the maximal circular velocity

v2circ,max = 2!&0

M

LhR 3.0. (2.24)

From Eq. (1.70) we see that the total luminosity is

Ltot = L(R & )) = 2!&0 h2R (2.25)

Replacing hR in Eq. (2.24) (since we cannot measure the scale length whenwe do not know the distance of the galaxy) we can write for the maximalcircular velocity

v4max $

!M

L

"2

&0 Ltot. (2.26)

This relation is known as Tully-Fisher relation . The theoretical derivationis rather weak, since it is not obvious that the mass-to-light ratio, M/L andthe central surface brightness is universal for all disk galaxies. However, theobservational evidence is overwhelming, that disk galaxies follow in generalsuch a relation, e.g. in H-band (it is advantageous to use red or infra-red,since in this bands the luminosity is less a"ected by ongoing star formation)

LH

3 ! 1010 LH,#=

!vmax

196 km, s"1

"3.8

(2.27)

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2.3. CLUSTERS OF GALAXIES

2.3 Clusters of galaxies

Central dominant galaxy (cD)

Clusters of galaxies contain significantly more galaxies, up to % 1000. Incontrast to groups the centre of a clusters typically hosts a very massive,elliptical galaxy. One can easily imagine that those galaxies are formed bythe continuous merging of disk galaxies, as expected for compact groups.The surface luminosity profile of usually approximated by a Sercic formula

&(R) = &(Re) exp

C

#b

A!

R

Re

" 1n

# 1

BD

, (2.28)

where n is typically found to be % 4 (de Vaucouleur-law). Re denotes thee"ective radius for which half of the light comes from inside. This definitionfixes b

b = 2n # 0.327. (2.29)

The total luminosity is given by

Ltot =

- %

0dR 2!R&(R) " 22.7&(Re)R2

e. (2.30)

The radius is di!cult to measure, hence we replace it using again the virialtheorem

Ltot $ Mtot $ Re%2r . (2.31)

As a result we obtain the Faber-Jackson relation (the equivalent of the Tully-Fisher relation)

Ltot $%4

r

&(Re). (2.32)

Unfortunately, it turns out that elliptical galaxies scatter broadly aroundthis relation. However, to some extent even elliptical galaxies follow a scalingrelation. The reason that &(Re) is universal (to some extent) reflects thatelliptical galaxies have similar formation histories.

Intra-cluster medium

The space between the galaxies in a cluster is filled by the intra-clustermedium (ICM). The temperature of it can be estimated from the velocitydispersion of the galaxies. In clusters the dispersion can be about %r %1000 km s"1. Hence, the temperature of the ICM is

T %mH %2

r

kB= 1.2 ! 108 K. (2.33)

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2.3. CLUSTERS OF GALAXIES

It is important to notice, that the gas also cools. Hence, one might besurprised that the ICM stays at such a high temperature since it looses con-stantly energy. The main cooling mechanism is free-free emission (bremsstrahlung).An electron that is deflected in the vicinity of a hydrogen ion, radiates someenergy away. The energy loss rate per volume for a fully ionised hydrogengas, i.e. ne = nH, in thermal equilibrium is

%free"free = %0

!T

K

" 12 < ne

cm"3

=2, (2.34)

with %0 = 1.4 ! 10"27 erg cm3 s"1. We can estimate the cooling time

tcool =specific internal energy

energy loss rate=

3/2 (n2e + nH) kBT

%0T 1/2n2e

=3kBT 1/2

%0ne(2.35)

Assuming a temperature of 108K and as an conservative estimate a hydrogennumber density two orders of magnitude smaller as the number densityderived for the group, i.e. ne = 10"4 cm"3 we find a cooling time of

tcool % 940 Gyr. (2.36)

Thus, cooling in the ICM is very ine!cient. Only in the very centre of thecluster, where the density is much higher the cooling time is about the ageof the Universe. The best approach to observe the bremsstrahlung emissionis to use the x-ray band. A lot of clusters and their internal structures havebeen detected using x-ray telescopes. Since the ICM is optically thin forx-ray photons, internal dense structures (e.g. shock fronts or cold fronts)can be seen.

2.3.1 Gravitational lensing

Lensing by a point mass Clusters of galaxies are large the most mas-sive objects in the Universe. The resulting gravitational potential a"ectsalso the light from galaxies behind the clusters. In Sec. 1.2.2 we have shownthat deflection angle by a weak encounter does not depend on the mass ofthe deflected object. This suggest that even the trajectories of photons,which have zero rest mass, get deflected in a gravitational well. Generalrelativity leads to the result that the deflection angle, ", is even by a factorof two large than the result in Eq. (1.63), namely

" =4GM

c2

1

*, (2.37)

where * is the minimal distance between light ray and bending mass, M ,and c is the speed of light. Consider the configuration of a star, a light

42

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2.3. CLUSTERS OF GALAXIES

dL

dLS

dS

Source Image

Lens

!

"

#

x

y

$

Figure 2.3: Geometry of a lensing situation.

bending galaxy and an observer as shown in Fig. 2.3. We wish to computethe direction of the image, +, by the help of Eq. (2.37). To this end we usethe identity

y = x + (y # x).

For small angles we can substitute y = dS+, x = dS', and y # x = dLS".Inserting these expressions and using Eq. (2.37) leads to

+ = ' +dLS

dS

4GM

c2

1

*.

From Fig. 2.3 we read again * = dL+. Moreover, we define the Einstein an-

gle, +E, which summarises the geometry of the lensing situation

+E =dLS

dS dL

4GM

c2. (2.38)

To obtain the direction to the image, +, we have to solve the quadratic equa-tion

+2 # '+ # +2E = 0.

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2.3. CLUSTERS OF GALAXIES

Hence, in general there are two solutions

+± =1

2' ±

)

1

4'2 + +2

E. (2.39)

• ' = 0We expect a circular image since the source, the bending mass andthe observer are aligned. No plane in which the bending takes placeis determined by this configuration. We see an Einstein ring withopening angle +E.

• ' (= 0We obtain two pictures, one within the (hypothetical) Einstein ringand one outside.

Lensing by a distributed mass We generalize the lensing by singlemass. To this end we assume that mass is distributed in a plane (thin lens

approximation) with surface mass density &(!!). Integrating over all masselements, d2!! &(!!), we obtain the total deflection angle

" =4!

G

-

d2!! &(!!)(! # !!)

|! # !!|2. (2.40)

Note, also the opening angle is vector-like, since we have to consider thedirection of the bending by each mass element, d2! &(!). We have addthese contributions in a two-dimensional vectorial manner. The fundamentaldrawback of Eq. (2.40) is we do not know the position, !, where the lightray crosses the lens plane. A general aim of lensing studies is to derive theunderling mass distribution. However, with only one lensed image we cannothope to solve the equation for &(!!) since very di"erent mass distributionmay lead to the same result.

Assuming spherical symmetry allows to derive at least the enclosed mass.One can show that the integral along a circular line with radius, R, can besubstantially simplified

-

circled2!!

(! # !!)

|! # !!|2=

C

0 : |!| < R!

|!|2 : |!| > R(2.41)

We can apply this formula, since in a spherical symmetric configuration thesurface density along a circular line is constant. Eq. (??) indicates that onlythe mass inside the radius at which a light ray crosses the lens plane con-tributes to the deflection,

" =4!

G

M(< *)

*. (2.42)

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2.3. CLUSTERS OF GALAXIES

"

!

Figure 2.4: Direction of the image, +, versus direction of the original source, ' fora point mass (&(< *) $ 1/+2, dashed line) and for a projected Plummer spheredistribution (&(< *) $ 1/(1.0 + +2)2, solid line). For an o"-axis galaxy behind thesmoothly distributed mass of a galaxy cluster we expect three images (dotted line).

Similar to the procedure for a point mass we can replace " by the help ofEq. (2.42). As a result we obtain

+ = ' +dLS

dS

4GM(< *)

c2

1

*.

We solve for ' and rearrange the terms around M(< *) using * = dL+,

' = +

'

1 #dLS dL

dS

4!G

c2

M(< *)

!*2

(

.

We introduce the critical surface density

&crit =dS

dLS dL

c2

4!G. (2.43)

As a result we obtain again a quite simple lens equation

' = +

'

1 #&(< *)

&crit,

(

(2.44)

with &(< *) = M(< *)/!*2. Three images are obtained if somewhere&(< *) > &crit, see Fig. 2.4

Magnification @@@@

d'image

d'original

@@@@=

@@@@

+

'

d+

d'

@@@@

(2.45)

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2.4. THE LARGE SCALE STRUCTURE

Figure 2.5: Schechter function

2.4 The large scale structure

It is obviously interesting to map the entire distribution of galaxies in theUniverse. First of all it provides a luminosity function for galaxies.

Galaxy luminosity function

To construct the luminosity function we count galaxies in luminosity bins#L. It has been found that the best fit to the distribution is a so-calledSchechter function

$(L) =$)

L)

!L

L)

"!

exp

'

#L

L)

(

, (2.46)

with parameters

L) " 1010 h"2 L#

$) " 10"2 h3 Mpc

" " #0.7 to # 1.1.

Note, using the Schechter function the total number of galaxies in the Uni-verse is not finite, but the luminosity is. The value L) defines ‘big’ galaxies.Galaxies with luminosity below 0.1L) are called dwarf galaxies.

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Chapter 3

Cosmology

3.1 Newtonian cosmology

Hubble expansion

The cosmological principle: Viewed on su"ciently large scales, there

are no preferred directions or preferred places in the Universe.

More mathematically speaking: The Universe is homogeneous and isotropic.Galaxy surveys indicate that on scales beyond % 100 Mpc this hypothesisis reasonably fulfilled. We axiomatically state that properties derived onsuch large scales, e.g. the average density, are representative for the entireUniverse.

Hubble’s law: Galaxies seem to move away from us. Hubble realized thatthere is a linear relation between radial velocity, ur, and distance, D,

ur = H0 D. (3.1)

The Hubble constant H0 is commonly reduced to a dimensionless factor h

H0 = h 100 km s"1 Mpc,

which allows you to proceed in a calculation and to insert not until theend the preferred value for h. A simple example: the Hubble time H"1

0gives the age for a constantly expanding Universe –see paragraph ‘comoving

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3.1. NEWTONIAN COSMOLOGY

coordinates’ below. So we write

H"10 = 9.77h"1 Gyr, (3.2)

i.e. we have to divide 9.77Gyr by h to obtain the actual Hubble time.However, it is convenient to leave h in the equation. The combination ofseveral recent observations indicates

h = 0.7 ± 0.01,

so we are quite convinced that we know the true Hubble constant.

Comoving coordiantes: One distinguishes between distance-dependentHubble flow and additional peculiar motions. To separate these two we in-troduce the comoving coordinate system:

r(t) = a(t)x(t) a(t0) = 1, (3.3)

where a(t) is called scale factor or expansion parameter. The index 0 indi-cates the present-day value of a quantity. Computing the time-derivativeleads to total relative velocity, u, between two points separated by r

u = r = a x + a x

substitute x by Eq. (3.3)

u =a

ar + v,

where v denotes the peculiar velocity, v = ax. On average peculiar motionsshould cancel out,v = 0, hence comparison with Hubble’s law, Eq. (3.1), in-troduces the time-dependent Hubble constant

H(t) =a

a. (3.4)

The observed Hubble constant quantifies the expansion rate today, H0 = a0.If the Universe expanded with a constant rate the Hubble time 1/H0 wouldgive the age of the Universe. For completeness, we write down the acceler-ation in comoving coordinates

r = ax + 2ax + ax. (3.5)

In the following we consider an isotropic, homogeneous Universe. Any pe-culiar velocity, v, would give a preferred direction, and the Universe wouldnot be homogeneous anymore. Hence, throughout this section we assumethat there are no peculiar velocities, x = x = 0.

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3.1. NEWTONIAN COSMOLOGY

Figure 3.1: Hubble’s original law, published 1929, PNAS 15. Note the Hubbleconstant % 530 km s!1 Mpc, also the ‘non-cosmological’, small maximum distanceof 2 Mpc.

Cosmological redshift

Due to the Hubble expansion distant galaxies move relative to us. Whenwe detect the light of those galaxies we have to be aware of the fact, thatthe received radiation has wavelength $r which is di"erent from that ofthe emitted radiation, $e, due to the Doppler e"ect. For non-relativisticvelocities the shift of the wavelength amounts to

$r

$e= 1 +

ue # ur

c,

where ue is positive when the source departs from the receiver. Hence, wecan write for the relative shift between two neighbouring points

d$

$=$r # $e

$e=

ue # ur

c= #

du

c.

Since we have assumed that relative velocities are solely introduced by theHubble flow we can substitute the relative velocity by du = Hdr. By thehelp of Eq. (3.4) we can write

d$

$= #

a

a

dr

c= #

da

a

1

c

dr

dt=

da

a,

where we have used that light travels with dr/dt = #c (minus since thelight travels towards the observer, i.e. r gets smaller). Integration leads toln$ = # ln a + const., such that we can write

$r

$e=

ar

ae.

The redshift gives the relative shift of the wavelength, z = ($r # $e)/$e.Therefore, a photon emitted when the scaling factor of the Universe was ae

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3.1. NEWTONIAN COSMOLOGY

has the cosmological redshift when observed today (ar = 1)

z + 1 =1

ae, (3.6)

Note, this result indicates that the wavelength behave as a comoving quan-tity. If a spectrum allows us to derive the redshift, we immediately knowthe expansion factor at emission time. In order to translate this informa-tion into a time and a spatial distance, we have to know more about thecosmology we live in. Clearly, peculiar motions that are superimposed onthe Hubble flow are additional sources of redshift.

Friedmann equation

For now, we assume that the expanding Universe can be described with stan-dard Newtonian dynamics. Consider a spherical homogeneous distributionof matter and a probe at the outer radius of the sphere, r. When the sphereexpands or contracts exactly as the probe moves, i.e. the outer radius equalsalways the position of the probe, it does not feel that the mass distributionchanges. The potential is equivalent to a mass concentration in the centre,see Sec. 1.2.1. If the probe is only accelerated by the mass distribution thetotal energy of the probe, Ekin + Epot, is conserved. Hence we can write

u2

2#

GM

r= const. .

with M = (4/3)! #r3 we can write

u2

2#

4!G

3# r2 = const. .

Inserting comoving coordinates leads to (the are no peculiar velocities)

a2x2

2#

4!G

3# a2x2 = const. .

Renaming the constant, finally gives the Friedmann equation

a2 =8!G

3# a2 # Kc2. (3.7)

Remarks:

• Don’t take the derivation above too serious. However, a proper deriva-tion using General Relativity leads to the same final equation.

• Pay attention when comparing with other textbooks: there is nocanonical naming for the constant K, sometimes c2 is included, some-times not.

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3.1. NEWTONIAN COSMOLOGY

• Remember the cosmological principle: None of quantities in the Fried-mann equation is allowed to depend on spatial coordinates (Homo-geneity).

• The dimension of K is 1/length2.

The Friedmann equation relates the evolution of the scale factor, i.e. a, tothe total density in the Universe, # and the constant K. The derivation usingGeneral Relativity indicates that # reflects the relativistic energy density

#c2 =

- %

0dp f(p) pc, (3.8)

where p is the relativistic momentum and f(p) the momentum distributionfunction. In other words, the driving agent for expansion is the relativisticenergy density of the Universe.

If K = 0 we have an exceptional situation. We may ask what conditionsare needed to achieve it. Obviously, if the density equals the critical density,

#crit =3H2

8!G, (3.9)

the constant K is zero. Since K does not depend on time, it has be zeropermanently.

Fluid equation

First law of thermodynamics states energy conservation. Basically, no en-ergy is exchanged with any heat reservoir while the Universe expands (onlyduring phase transition latent heat may be released). Thus the Universeexpands adiabatically

dU = #P dV, (3.10)

where P is the pressure, V the volume and U the relativistic energy content,U = #c2V . For an expanding cube, V = (aL0)3, we can write

d(#c2a3) = #P da3.

Applying the product rule leads to

a3 d(#c2) + #c2 3a2 da = #P 3a2 da.

Re-arranging the terms and dividing by dt leads to the fluid equation

d#

dt+ 3

a

a

!

#+P

c2

"

= 0. (3.11)

In general, a particular type of a fluid (we just avoid here the term ‘matter’

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3.1. NEWTONIAN COSMOLOGY

since it is used below) is specified by its equation-of-state

P = w #c2, (3.12)

i.e. the constant w defines the type of the fluid. The fluid equation –we re-move dt again– gives

d#

#= #3 (1 + w)

da

a.

Integration leads to ln # = #3(1 + w) ln a + const., hence

# $ a"3(1+w).

We consider now three ideal cases.

Radiation (0 = 1/3): The ideal case of fully relativistic particles. Pho-tons may serve as prototype. The energy of an photon is h&. Assume thata volume L3 are photons with number density n. The energy density inthe volume is n h&. The momentum of a photon is h&/c. The pressure isthe momentum transfer, 2h&/c, times the number of incidents L2 c#t n/6,divided by time interval, #t and surface L2. Hence, the pressure becomesn h&/3. Therefore, pressure and energy density of a radiation are related by

Pr =1

3#rc

2

In this case integration of the fluid equation leads to

#r $ a"4.

The number density dilution due to the expansion and the redshift of therelativistic particles a"ects the energy density.

Dust (w = 0): The ideal case of non-relativistic matter, called ‘dust’ orsimply ‘matter’. Here, the pressure is much smaller than the energy density,hence, we assume w = 0. From the fluid equation we obtain that the densityscales according to

#m $ a"3,

which is just the matter dilution due to the expansion.

Vacuum energy (w = #1): Quantum mechanics tells us that the spacevacuum has a constant, non-zero vacuum energy density, #" = const. If weexpand a given volume the internal energy, U , increases, thus the pressurehas to be negative, see Eq. (3.10). Therefore, the equation-of-state is

P" = ##"c2.

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3.1. NEWTONIAN COSMOLOGY

Dimensionless density parameters

The cosmic fluid may comprise the three ideal fluids introduced above. It isconvenient to give the density of each component with respect to the criticaldensity, see Eq. (3.9).

'i =#i

#crit=

8!G #i

3H2,

where i denotes dust, radiation, or vacuum energy (m,r,%). These densityparameters specify the Universe we live in, they are typically given for thepresent time t0. Only if stated explicitly, we consider their time-dependence.Using the scaling with a for each cosmic fluid component we can express thetotal density by the help of the density parameters

#tot(t) =,

i#i(t)

= #crit,0

,

ia"3(1+wi) #i,0

#crit,0(3.13)

= #crit,0

'

'r1

a4+ 'm

1

a3+ '"

(

.

We can also express the constant K of the Friedmann equation in terms ofthe density parameters and the Hubble constant. To this end, we note thatK is constant with time. Thus, it is su!cient to derive K for one particulartime, e.g. t0. We solve the Friedmann equation for K and insert the densityas give in Eq. (3.14)

Kc2 = H20 ('tot # 1),

(remember t = t0, i.e. a = 1) where we have used

'tot = 'm + 'r + '".

This implies that the constant K depends on the total energy density in theUniverse:

K > 0 for 'tot > 1

K = 0 for 'tot = 1

K < 0 for 'tot < 1.

It is remarkable that we need just four numbers to determine the Universewe live in, namely H0, 'm, 'r, and '".

Age of the Universe: We can rewrite Friedmann’s equation in terms ofthe four parameters defining the Universe

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3.1. NEWTONIAN COSMOLOGY

!a

a

"2

= H20

'

'r1

a4+ 'm

1

a3+ '" + (1 # 'tot)

1

a2

(

. (3.14)

Now, the age of the Universe can be obtained after separation of the vari-ables

H(a) =a

a=

1

a

da

dt& dt =

1

aH(a)da

and subsequent integration.

The acceleration equation

Acceleration equation: Di"erentiate the Friedmann equation with re-spect to time, i.e.

2aa =8!G

3

E

#a2 + #2aaF

We can eliminate the time derivative of the density, #, by using the fluidequation. This leads to the acceleration equation

a

a= #

4!G

3

'

# +3p

c2

(

. (3.15)

The latter has at least a formal similarity to the result we expect from New-ton’s law:

r = #GM

r2= #

4!

3#r3 G

r2.

Hence the Newtonian result is

a

a= #

4!G

3# (Newtonian).

Comparison shows that our relativistic treatment of energy density providesalso the pressure as source of gravity.

Deceleration parameter: The change of the expansion rate is quantifiedby the deceleration parameter

q = #1

H2

a

a.

For the present time, t = t0, we can write the acceleration equation as

a

a(t0) = #

4!G

3

,

i#i0 (1 + 3wi) = #

1

2H0

,

i'i (1 + 3wi).

Hence the deceleration parameter can be written as

q0 = #1

H0

a

a(t0) =

1

2

,

i'i (1 + 3wi). (3.16)

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3.1. NEWTONIAN COSMOLOGY

For decades, this seemed to be the most promising method to derive thematter density in the Universe, especially since a pure matter-dominatedUniverse was assumed. With the advent of new observations, this picturehas changed.

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3.2. GEOMETRY OF THE SPACE-TIME

3.2 Geometry of the Space-time

Robertson-Walker metric

Metric: In general, geometry can be defined by a local metric tensor gµ*

ds2 =,

µ*gµ*dxµdx* ,

where ds is the invariant line element, i.e. it has the same length for all possi-ble observers and in any coordinate system. For instance, in two dimensionsPythagoras theorem states

ds22D = dx2 + dy2,

i.e. the metric tensor is

gµ* =

!

1 00 1

"

.

(We are not going to use this tensor formalism further.) You may choosea di"erent coordinate system in two dimensions, such as polar coordinates.The invariant line element becomes now

ds22D = dr2 + r2d+2.

Curvature: We introduce the curvature by considering equilateral trian-gles. The legs are in general geodesics, i.e. they are shortest connectionbetween two points. For any triangle in the 2D geometry introduced abovethe sum of all inner angles is 180*.

Now, let us consider an equilateral triangle on the surface of a sphere.Here the geodesics are great circles. Therefore, the legs of the triangle lieon great circular arcs. The sum of the inner angles is larger than 180*.This is easiest seen by considering a triangle that includes one pole and theequator. Since all angles now have 90* the sum is 270*. Choosing di"erentside-lengths we see that the sum varies with the length.

We call a space curved when triangles have an inner angle sum di"erentfrom 180*. The symmetry of the sphere entails that the curvature is thesame everywhere.

The independent line element on the surface is

ds22D = d*2 + dL2

= d*2 + R2 sin2(*/R)d+2,

see Fig. 3.3.Demanding constant curvature everywhere allows only one alternative

solution –beside the flat geometry–, namely the hyperbolic space. Here, theinner sum is smaller than 180* and the invariant line element is given by

ds22D = dr2 + R2 sinh2(r/R)d+2.

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3.2. GEOMETRY OF THE SPACE-TIME

Figure 3.2: Triangle on a spherical surface, composed by great circular arcs. Theinner sum of the triangle is larger than 180".

Robertson-Walker metric: Similar to the two dimensional curved space,there are general solution in three dimensions for a curved space with con-stant curvature everywhere –we have to demand this since we have to obeythe cosmological principle:

ds23D = d*2 + f2

K(*)E

d+2 + sin2 + d-2F

.

(Hyperspherical coordinates). Where the function fK(*) is given by

fK(*) =

G

H

I

1/-

K sin(-

K*) K > 0* for K = 01/-#K sinh(

-#K*) K < 0

, (3.17)

where R = 1/J

|K| is the curvature radius.

The equation above describes the geometry of a curved space, withconstant curvature everywhere as demanded by the cosmological principle.Curved spaces may have features which are less intuitive. For instance, ina closed space we reach the original point if we go constantly in the samedirection. On the sphere, we see that it means we follow just great circles.In three dimensions it is di!cult to imagine, but the same would happen.

We combine the description of the space with the idea of the Hubbleflow introduced in the preceeding section. All distances are scaled with thetime-dependent expansion factor a(t). In addition, we demand that the in-variant line element is Lorentz invariant, i.e. we have to combine space andtime according to special relativity

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3.2. GEOMETRY OF THE SPACE-TIME

Figure 3.3: Coordinate system on the surface of a sphere: R is the radius of thesphere, * is the length of the arc, l is the projection of R on the plane defines bythe circle of all points with variable + and constant arc length *. For a small d+ thearc length along this circle is given by dL = l d+ = R sin(") d+ = R sin(*/R) d+.

c2d.2 = c2dt2 # a2(t)E

d*2 + f2K(*)d'

F

, (3.18)

where d' = d+2 + sin2 + d-2. This so-called Robertson-Walker metric is themost general solution for a Lorentz-invariant space-time with isotropic cur-vature. It is a solution of the Einstein equations

Gµ* + %gµ* = #8!G

c2Tµ* ,

where Gµ* is the Einstein tensor, gµ* the metric, % is Einstein’s cosmolog-ical constant and Tµ* is the stress tensor. It is beyond the scope of thisintroduction to derive and explain Einstein’s equations, hence we do notfurther explore their realm. However, we have to import one crucial result:Einstein’s equation links the energy constant K introduced in the last sec-tion to the curvature of the space, the constants which both are deliberatelylabelled K are identical. Thus, the energy density determines, both,

• the dynamics of the expansion parameter a(t) and

• the curvature of the space.

Light rays

For light the invariant line element d. is zero. Let us assume that an observeris placed in the centre of the coordinate system and receives signals fromdistant sources. For this particular choice, light rays travels along a radialline of the coordinate system.and we have d' = 0. Hence we can write

cdt = a(t)d*.

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3.2. GEOMETRY OF THE SPACE-TIME

Integration relates the comoving distance with time between emitting andreceiving the light signal

*(re) =

- tr

te

cdt

a(t).

Alternatively, we can relate the comoving distance to the expansion factor

*(re) =

- ar

ae

c da

a a(a)=

- ar

ae

c da

a2 H(a). (3.19)

Since this relation is quite important we should examine it more closely.Typically, we are the observer and we see the light from a distance source.In this case the limits are ar = 1 and ae = 1/(1 + ze). When we know thecosmology, the integral provides the comoving distance. For small distanceswe can Taylor-approximate the integrand around a = 1:

1

aa=

1

aa

@@@@a=1

#a2 + a

a2a3

@@@@a=1

(a # 1) + O((a # 1)2)

For the second term we used the quotient rule and the identity d/da =(1/a)d/dt. In the approximation above we insert the Hubble constant, H0,and the deceleration parameter, q0,

1

H0{1 + (1 # q0)(1 # a)} + O((a # 1)2).

Integration from ae to a = 1 gives

c

H0

'

(1 # ae) +1

2(1 # q0)(1 # ae)

2

(

+ O((a # 1)3). (3.20)

The first term resembles Hubble law: the comoving distance * is proportionalto the redshift ze = (1 # ae)/ae " 1 # ae. The second term reflects that thecomoving distance depends also on the derivative of a. Let us compare twosituations: Light travels from ae to ar in a universe with a > 0 and witha < 0. For the former the scaling of the universe was at any time smallerthan for the latter. Therefore, for the former the comoving distance betweenemission and receiving is larger. Eq. (3.20) reflects this fact, since a $ #q0,see Eq. (3.16). When we find a method to determine *(ae), we can derivethe deceleration parameter.

Mapping a time interval: Consider the following situation: A lightemitting source sends a signal at t = te, which is received at another placeat tr. After a very short time, i.e. at te + #te a second signal is emittedwhich is received at tr +#tr. Both source and receiver are in rest, therefore

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3.2. GEOMETRY OF THE SPACE-TIME

1

a

1t/t0

"

"

a > 0

a < 0

Figure 3.4: Depending on the second derivative, a, the same interval in a reflectsa di"erent time interval, t.

the comoving distance between them is constant. Assuming again that thelight travels along radial coordinate lines we can write

- tr

te

cdt

a(t)= * =

- tr+#tr

te+#te

cdt

a(t)"- tr

te

cdt

a(t)+

c #tra(tr)

#c #tea(te)

Therefore the time between sending the two signals is di"erent from the timebetween receiving the two according to

#tr#te

=a(tr)

a(te).

This is closely related to an earlier result: Regarding the light signal as awave than the frequency changes when traveling through the Universe, i.e.the light is redshifted.

Angular diameter distance: We can derive the distance of an objectwhen we know its extension and when we can measure its apparent openingangle. Let’s put the observer in the centre of the coordinate system andan object with linear extension S at distance * where ‘extension’ means thesize perpendicular to the line-of-sight. Assume the dL is small compared todistance between source and observer. We use the definition of the invariantline element in 3D and set ds3D = l and d* = d- = 0, hence

dL = fK(*) d+.

The observed opening angle d+ is now related to the linear extension of adistant object by the function fK . That is precisely how we have defined it.

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3.2. GEOMETRY OF THE SPACE-TIME

In an expanding space we have to take into account that the process isnot instantaneous: There is an emission time te and a receiving time tr. Inthe meanwhile the light rays –which allows us to measure the linear sizeof the object– travel along radial coordinate lines and the space expands.Hence the proper extension dL at emission time is obtained by multiplyingby a(te). We define the angular diameter

D! :=dL

#+= fK(*) a(te).

Luminosity distance: We can infer the the distance of an object byknowing its luminosity L since the flux F decreases with distance

F =L

4!d(Euklidian space).

Put a light source in the centre of a comoving coordinate system. After timetr # te the light has reached a comoving distance of *. The essential pointhere is that the radius of the sphere is given by fK(*) a(and not * as onemight naively think). Hence, the physical sphere at tr is

4! f2K(*)a2(tr).

Assume the luminosity gives the energy radiated per time interval #te by thesource. The flux which arrives at the observed is a"ected by the expansiontwofold:

• photons are redshifted

• the time interval has to be mapped

These are two di"erent e"ects –even if they look at the first glance quitesimilar. In summary the flux arriving at the observer is altered by

F & Fno"redshifta2(te)

a2(tr).

So we have

F =a2(te)

a2(tr)

L

4! f2K(*)a2(tr)

.

Since we are usually the observer we can set a(tr) = 0. An analogy to theEuclidian result we can define the luminosity distance as DL =

J

L/4!fr.The luminosity distance in an expanding curved space becomes

DL =fK(*)

a(te)

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3.2. GEOMETRY OF THE SPACE-TIME

Figure 3.5: The hunt for high redshift supernovae.

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3.2. GEOMETRY OF THE SPACE-TIME

The supernovae project

Consider the luminosity distance DL: If we know the luminosity of an objectthe measured flux, i.e. the observed magnitude gives us the luminosity dis-tance. On the other hand DL depends only on the redshift via fK(*)/a(te).Therefore if we can determine DL for a large redshift range we may restrictcosmological parameters, since we can restrict fK(*). The relation can beobtained numerically by integrating *(ae) and choosing the appropriate fK .However, for small distances 1# ae . 1 we can make the following assump-tions:

• The curvature has negligible e"ect, fK(*) = *

• Use the Taylor approximation for *(ze), see above.

• Dividing *(ze) by ae results in the luminosity distance

DL =c

H0

'(1 # ae)

ae+

1

2(1 # q0)

(1 # ae)2

ae

(

• Considering only term of O(1 # ae)2 we can approximate

(1 # ae)2

ae"

(1 # ae)2

a2e

• From the definition of the redshift we obtain

(1 # ae)

ae= ze

Thus we can approximate the luminosity distance for ’small’ redshifts by

DL =c

H0

'

ze +1

2(1 # q0)z

2e + O(z3

e )

(

.

According to the definition of luminosity distance we see, that redshift anda known luminosity of an objects leads to an expectation of for the observedflux, i.e. the magnitude of the object. When we find systematic deviationswe can constrain constrain cosmological parameters as done by the highredshift supernova project.

Source count

Assume we have an isotropic distribution of unchangeable objects over theuniverse. We can determine how many objects we should find up to a givenredshift ze. The density may n0. Let’s compute the volume in comovingspace, here the density does not vary with time. So we have to derive what is

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3.2. GEOMETRY OF THE SPACE-TIME

the volume we see if we include everything with z < ze. Simply 4/3!*3(ze)?This expression would miss the curvature of the space. Consider again the2D case: choose a distance * and connect all points which have this distance.The resulting circle has the length 2!fK(*). In 3D this means the surfaceof a shell of a given xi is

4!f2K(*).

We had this result already found when we discussed the luminosity distance:the size of a sphere of light that traveled for a given time. We obtain thetotal volume when we integrate over all distances

N(< ze) = 4!

- ((ze)

0d*! f2

K(*!).

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3.3. THERMAL HISTORY OF THE UNIVERSE

3.3 Thermal history of the Universe

The cosmic microwave background

Temperature fix point The cosmic microwave background (CMB) is ex-ceptional isotropic over the entire sky. It is very unlikely that a combinationof local sources result in such a evenly distributed radiation. In contrast weassume that the early Universe was very homogeneous, therefore releasingthermal radiation at this times would naturally explain the isotropy.

The spectrum of the CMB perfectly follows the Planck distribution,hence its temperature can be precisely determinded, see Fig 3.6,

TCMB = 2.725 ± 0.001 K.

Momentum space distribution We generalise the Planck spectrum givenin Sec ??. The thermal distribution of particles in momentum space is

f(p) dp =g

2!2 !3

p2 dp

exp{(E(p) # µ)/kBT} ± 1,

where µ is the chemical potential and g the statistical weight. The energyis in general given by the relativistic expression

E2(p) = m20c

4 + p2c2,

where m0 is the rest mass.

Number of photons For photons we have: ‘-’ since they are bosons,statistical weight of g = 2 since they may have two helicities, E = pc,p = h&/c since photons are always relativistic, and µ = 0. The numberdensity of photons is given by

n$(T ) =

- %

0dp f(p) =

k3BT 3

!2 !3c3

- %

0

y2 dy

ey # 1=

2 1(3)

!2

!kBT

!c

"3

, (3.21)

where 1(x) is the zeta-function, 1(3) " 1.202. Since we know the tempertureof the CMB, we can directly obtain the number density of photons in themicrowave background

nCMB$ (t0) " 400 cm"3.

Baryon-to-photon ratio We now compare the average baryon numberto the photon number density in the Universe. The term ‘baryons’ basi-cally refers to nuclei, i.e. it includes all ‘normal’ matter as stars, gas anddust. The mass contribution of electrons, which are not baryons but leptons,is negligible. Hence, we subdivide the matter content –remember ‘matter

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3.3. THERMAL HISTORY OF THE UNIVERSE

Figure 3.6: The black-body spectrum of the cosmic microwave background.

means in this context non-relativistic– in the Universe into a ‘baryonic’ anda ‘dark’ component

'm = 'b + 'dm.

For the baryonic matter we can compute the number density assuming thatit mainly consists of non-relativistic protons and neutrons. We introducehere the mass of a baryon mb which is simply the mass of a proton

mb = mp.

For the number density of baryons we can write

nb(t0) ='b #crit

mb" 1.12 ! 10"5 'b h2 cm"3.

We can express the baryon number density in units of the photon numberdensity as derived from the CMB

nb = 2 n$ = 2.81 ! 10"8 'b h2 n$ ,

where 2 is the baryon-to-photon ration n$/nb. Even if the energy densityof the CMB is low compared to that of matter, photons vastly outnumberthe baryons. We will see below that the baryon-to-photon ration is a crucialparamter for the primordial nucleosynthesis.

Expansion cooling

First law of thermodynamics According to the first law the change ofinternal energy U in a Volume V is given by

dU = TdS # pdV + µdN,

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3.3. THERMAL HISTORY OF THE UNIVERSE

where S is the entropy, µ and N are the chemical potential and the par-ticle number, respecively. If more than one species are present we have tosum over all terms µidNi. We consider only reversible processes, hence dUand TdS are total di"erentials. The entropy may change, e.g., if particlespecies annihilate and heat the remaining plasma. Otherwise, the entropyis constant and the Universe expands adiabatically, dS = 0.

Radiation energy density We have already seen for photons that theyare redshifted by the expansion of the Universe, cf. Sec 3.1. We now derivehow the temperature evolves. First we determine the energy density of aphoton-‘fluid’, which is given by the integral

#c2 =

- %

0dp f(p) pc =

k4BT 4

!2 !3 c3

- %

0

y3 dy

ey # 1= arad T 4.

The integral gives !4/15, hence the radiation density constant is

arad =!2 k4

B

15 !3 c3.

Radiation The energy density of radiation in a volume V = (aL0)3 is

Ur = arad T 4 a3L30,

where the index r means radiation. The pressure of radiation is #c2/3. Foradiabatic expansion we obtain with the help of the first law of thermody-namics

d(T 4a3) = #1

3T 4da3.

Using di"erentiation rules leads to

a3 4T 3 dT + T 4 3a2 da = #1

3T 4 3a2 da.

Separation of variables results in

dT

T= #

da

a,

and integration gives finally

Tr(a) =a0

aTr0.

We have added here the index r to the temperature, since this result is onlyapplicable for radiation. Clearly, we have reproduced the result we aleadyobtained for the redshift of the photons.

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3.3. THERMAL HISTORY OF THE UNIVERSE

Matter The internal energy of a mono-atomic gas in a volume V = (aL0)3

is

Um =3

2nb kBT (aL0)

3.

The density evolves with nb = nb0/a3 and the pressure is given by p =nbkBT . Hence, we get for adiabatic expansion

3

2dT =

T

a3da3.

Following similar steps as given in the paragraph above we obtain the tem-perature evolution of matter

Tm(a) =<a0

a

=2Tm0.

Radiation matter equality Today the energy density of radiation ismuch less than the energy density of matter. However, since the energydensity of radiation increases significantly with the radiation temperaturethe was a time in the past when the two densities were equal. The time atwhich the two energy densities were equal,

#m(aeq) = #r(aeq),

can be estimated by

'm1

a3eq

= 'r1

a4eq

.

Hence

aeq ='r

'm.

We have neglected here the neutrino content in the Univere and that in thetime from aeq to adec the photon temperature is governed by the mattercontent. Before aeq the Universe is radiation dominated.

Which component does determine the temperature evolution at aeq? Ra-diation or matter? We compare to this end the thermal energy content in avolume (aL0)3

Um

Ur=

3nbkBT

2 aradT 4.

Inserting arad, substituting nb = 2n$ , and replacing n$ leads to

Um

Ur=

451(3)

!32.

Since 2 is very small the thermal energy content in matter is negligible. Aslong as radian and matter in thermal contact always radiation will dominate.(Note, we should actually compare the specific heats.)

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3.3. THERMAL HISTORY OF THE UNIVERSE

Time of recombination

Since at earlier times the average temperature of radian and matter washigher there must have been an epoch at which both were coupled to eachother. At this time hydrogen was ionized and the photons were scattered bythe electrons and hydrogen ions. At the moment when hydrogen becomesneutral photons start to travel freely. This transition is called recombination

or decoupling. The CMB is assumed to be the photonic remainder of thistransition.

We now work out the following line-of-reasoning: The density of iniosizedhydrogen atoms, nHII = xnH, depends on the temperature T . Assuming afraction x therefore allows us to derive the temperature at with radiationdecouples and, since we know how radiation evolves with expansion, we canfix the expansion factor of decoupling adec.

First we derive the number of particles ni from the distribution functionin momentum space f(p). The binding energy of hydrogen is

BH = (mHII + me)c2 # mHIc

2 = 13.6 eV,

we expect that recombination happens at kBT % BH. Since rest-masses ofelectron, protons and hydrogen atom are much higher we derive the numberdensity for non-relativistic particles

E(p) =E

m20c

4 + p2c2F 1

2 = m0c2

C

1 +

!pc

m0c2

"2D 1

2

= m0c2

C

1 +1

2

!pc

m0c2

"2

+ ...

D

" m0c2 +

1

2

p2

m0,

where the last term is nothing else than the kinetic energy of the particle.The integral becomes now

n =g

2!2!3exp

'

#m0c2 # µ

kBT

(- %

0

p2 dp

exp{p2/2m0kBT}, (3.22)

where we have already neglected the ±1 in the nominator, since it is muchsmaller than exp(mc2/kBT ). With

K%0 dyy2 exp(y2) =

-!/4 the integration

gives

n =g

!3

'm0kBT

2!

( 32

exp

'

#m0c2 # µ

kBT

(

. (3.23)

So we have obtained the familiar Boltzmann distribution. To get rid of therest-masses we form the ratio

1

nH

nenHII

nHI=

x2

x # 1=

1

nH

'mekBT

2!!2

( 32

exp

'

#B

kBT

(

.

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Here we have used for the factor in front mHI " mHII, the chemical potentialcancel out

µe + µHII = µHI,

and the statistical weights are

ge = gHII = 2, gHI = 1.

Up to here we have worked out a relation between temperature and ion-ization fraction. Now we solve for the temperature. Let’s introduce someabbreviations

BH

kBT=

BH a

kBTCMB= k1 a,

'2!!2

meB

( 32

= V1.

This simplifies the temperature–ionization fraction relation

x # 1

x2=

nH(t0)

a3V1 (k1adec)

2/3 exp{k1adec}.

Taking the logarithm gives

adec =1

k1

'

ln

!x # 1

x2

"

# ln(nH(t0)V1) + 3 ln(adec) #3

2ln(k1adec)

(

Using numerical values

nH(t0)V1 " nb(t0)V1 = 7.40 ! 10"29 'bh2, k1 = 5.85 ! 104

and demanding an ionization fraction x allows us to solve the equationrecursively. For x = 0.1 we get adec = 8.0 ! 10"4. Thus, the redshift ofdecoupling is about

zdec % 1000.

Freeze-out The derivation above is based on the assumption that neutraland ionized hydrogen is always kept in equilibrium. This is only true whenthe time-scale for the reaction

HI + 0 1& HII + e" + 0,

–photoionisation is dominant here– is much shorter than the expansion time-scale. We can formalise this a bit. Recall that number density, cross sectionand mean free path are related by n%l = 1. The mean free path is andthe relative velocity between projectile and target gives how much time ittakes that for one target atom a reaction occur, #t = l/v. The inverse isthe reaction rate RAB = n%v (for a reaction A + B). If we have particles

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3.3. THERMAL HISTORY OF THE UNIVERSE

with di"erent velocities we have to average over all velocities including aprobability distribution f(v)

RAB(T ) = nB *v%+ = nA1

K%0 dv f(v)

- %

0dv f(v) v %(v, T ).

Note, f(v) denotes here a general distribution function, not necessarilynormalised by

K

dvf(v) = 1. Let’s make an assumption how everythingscales when the universe expands. The density goes with nB $ a"3. Since*v%+ depends on the temperature which in turn depends on a, we assume*v%+ $ a"* . From a given time to the future the total number of reactionsis

N =

- %

tdt RAB =

- %

a

da

aH(a)RAB(a).

Let’s assume also H scales also according to a power-law H(a) $ a"µ

N 2R(a)

H(a)

!1

a

"µ"3"* - %

ada aµ"1"3"* =

1

3 + & # µ

R(a)

H(a).

Therefore the ratio R(a)/H(a) (forget about the small contribution from&, µ) tells us in how many reactions the particle A will take part in thefuture of time a. If the ratio is much less than one basically no reaction willhappen, this type of reaction freezes out.

We can compute the freeze out time for the photo-ionization. The cross-section is given by Kramer’s semi-classical rule, the momentum distributionof photons is given in the beginning of this section, and we know how thephoton density evlves with expansion. The integral has to be performednumerically. It shows that recombination freeze out at x % 10"5.

Decoupling sophisticated A few items should be added, since decou-pling is even more complicated than discussed above

• The momentum distribution of photons is not thermal: recombinationprovides photons precisely with ionization energy. It is very di!cult todestroy those photons. This is done by recombination with two-photonemission.

• Even after freezing out recombination photons are still scattered byionized hydrogen and electrons. The freeze out of photon scatteringtakes place somewhat later.

Big bang nucleosynthesis

The screenplay Let’s summarize the major steps in nucleosynthesis:

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3.3. THERMAL HISTORY OF THE UNIVERSE

Figure 3.7: The evolution of elements.

• Neutrons and protons are in equilibrium by weak interactions, e.g.

p + e+ 3 n + &.

According to the Boltzmann distribution the ratio between the numberdensity of neutrons and protons is determined by their mass di"erence

nn

np= exp

'

#mn # mp

kBT

(

.

• Weak interactions freeze-out.

• Neutrons decay freely.

• Chemical elements, i.e. H, He, etc., are more or less in equilibrium.The small baryon-to-photon, 2, leads to an e!cient photo-dissociationof all elements heavier than hydrogen.

• When the temperature decreases photo-dissociation becomes less e!-cient. Helium can be formed, the neutron abundance decreases rapidlysince they are consumed to from helium.

• Nuclear reactions freeze out, the final abundance pattern survives.

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3.3. THERMAL HISTORY OF THE UNIVERSE

Figure 3.8: Final abundances

Equilibrium abundances We wish to derive an expression for the equi-librium distribution of nuclei. Again, the Boltzmann distribution is theappropriate tool. Integrating the Boltzmann distribution gives the num-ber density of particles, see Eq. (3.23). To simplify the derivation above wefactorise the expression

n =g

!3

'm0kBT

2!

( 32

# $% &

F

exp

'

#m0c2

kBT

(

# $% &

G

exp

kBT

(

# $% &

H

(3.24)

For a nuclei, X, composed by Z protons and (A # Z) neutrons, where Z isthe proton number and A the mass number, the chemical potential is givenby the sum of that of protons and neutrons

µX = Zµp + (A # Z)µn.

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3.3. THERMAL HISTORY OF THE UNIVERSE

Therefore, we can write for the factor H of the number density

HX = HZp HA"Z

n .

The binding energy of a nucleus is given by

BX = Zmp + (A # Z)mn # mX.

As a result we can write for the factor G

GX = GZp GA"Z

n exp

'BX

kBT

(

.

The number density of the element X can now be related to the numberdensity of protons and neutrons

nX = FXGXHX = FX GZp GA"Z

n exp

'BX

kBT

(

HZp HA"Z

n

=FX

FZp FA"Z

nnZ

p nA"Zn exp

'BX

kBT

(

, (3.25)

where we have used for the last step HG = n/F . The mass fraction, X, ofan element X is given by the ratio

XX =AX nX

nb.

By the help of Eq. (3.25) we obtain

XX =AX

nb

FX

FZp FA"Z

nnA

b XZp XA"Z

n exp

'BX

kBT

(

,

note Ap = An = 1. We can finally eliminate the baryon density, nb, byusing the baryon-to-photon ratio, 2. The photon density is fully determinedby the temperature, see Eq. (3.21). As a result we can finally write for themass fraction of the element X

XX = AX FXF"Ap nA"1

$ 2A"1 XZp XA"Z

n exp

'BX

kBT

(

, (3.26)

where we have used Fp " Fn. The first terms, F and n$ , a given by thetemperature. Most importantly, the formation of elements is strongly ham-pered by photo-dissociation, what is indicated by smallness of 2A"1. Notuntil the temperature dependent first term can compensate for the small2A"1 elements can be build. Thus, we draw the following conclusions

• Photo-dissociation hampers formation of elements beyond hydrogen,as indicated by the smallness of 2A"1

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• The larger the baryon-to-photon ratio, 2, the earlier elements, in par-ticular helium, can be formed, the more neutrons are preserved fromdecaying. As a result the higher is the mass fraction of helium. Note,helium has by far the highest binding energy of light nuclei, thereforeit is the most abundant element beside hydrogen.

• The higher the baryon content in the universe, 'b the later the nu-clear reactions will freeze out. At this times the neutron density issmall compared to the hydrogen and helium density since only a fewfree neutron remain. Solving Eq. (3.26) for Xn with constant Xp andXHe we see that the final neutron abundance decreases with decreasingtemperature. Thus, when the freeze-out takes place later, the temper-ature of the cosmic plasma is lower, less neutrons remain, and henceless deuterium is produced. Deuterium shows the opposite behaviorthan helium, see Fig. 3.8.

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