16-5-2014see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-2014-c8-1 Galaxy Formation Summary • We can’t say from equilibrium physics what the structure is of a galaxy • We don’t know what a galaxy consists of Dark matter (90% ?) some stars, gas (10%?) we don’t even know where a galaxy stops ! Why are galaxies like we see them ? How is the structure of galaxies determined ? 16-5-2014see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-2014-c8-2 the bad news equilibrium physics does NOT fix galaxy structure (No HR diagram for galaxies as for stars) the good news equilibrium physics does NOT fix galaxy structure galaxy structure is determined by GALAXY FOR- MATION, i.e., the process by which galaxies formed The big question in galaxy research is that of GALAXY FORMATION
How does the univers evolve ? It was very smooth atearly times. Below shows the images of the cosmic mi-crowave background as obtained by the COBE satellite
The top image is as observed, the second image is athigher contrast, and the third is (after subtraction ofcontamination, but the Milky Way is still there) thereal picture of the high redshift universe.
The terms here can be easily identified: on the left isthe kinetic energy, on the right is the gravitationalenergy, and the integration constant. The total en-ergy is given by
E = 12 (r)2 −
GM
r= c
Now write M = 43πr3ρ
12 (r)2 =
G 43πr3ρ
r+ c
= G 43πρr2 + c
(
r
r
)2
= 83πGρ +
c
r2
(6)This is the final equation of motion. Remember, how-
ever, that ρ is not a constant, it varies like ∝ r−3
The left term is special, since it is equal to H2, whereH is the Hubble “constant”. The consequence is,that the hubble constant is not constant !
Notice that E is the energy of the sphere, normalizedto r = ∞. The value of E will determine the evolu-tion of the sphere. The equations above imply that
E = c
We have different types of models:
• E = c = 0. The total energy is zero. The gravita-tional and kinetic energy compensate each other.The expansion halts at t = ∞. Require thatr = tα, we find for α: α = 2/3. Hence, thesolution is r = t2/3
• E = c < 0 Gravitational energy dominates. Theuniverse will halt, and collapse again !
• E = c > 0 Kinetic energy dominates. The universewill keep expanding. Gravitational energy will be-come less and less important, and at some phaseexpansion will be at a constant rate.
We can rewrite the Energy criterium as a density cri-terium. If c = E = 0, we define a critical density ρcfrom the last equation:
We know galaxies with redshifts of 5 and higher - i.e.,the universe was 6 times smaller, and 63 =216 timesdenser, when that light was emitted than it is now...
We generally use redshift to indicate the relative size ofthe universe, because it is so easy to use.
How can we derive time of emission from red-shift ?
for Ω = 1 universe
r ∝ t2/3
Redshift is defined by
(1 + z) =λobs
λemit=
robs
remit=
(
tobs
temit
)2/3
Hence
temit =tobs
(1 + z)1.5
Take MicroWaveBackground: z = 1000, tobs = 13 109
year, hence temit = 4 105 year.
Cosmological constant
In recent years, we have found very strong evidencefor a cosmological constant “Λ”. This term can enter
equation (6) through an effect of general relativity:
(
r
r
)2
= 83πGρ +
c
r2+
Λ
3
It can also enter through an effect of (particle) fieldtheories (changing the redshift dependence of ρ) Thecosmological constant influences the universe only inthe late stage. In early phases all terms in the equationabove are much larger than currently, except for thecosmological constant term, which does not dependon redshift. Hence its effect will be minimal at higherredshifts (in practice, z > 1). For convenience, weignore the cosmological constant in the following.
Homework assignments:1) What is the dimension of H0 ? What is the value ofH0 in cgs units ? (cm, gram, seconds).2) Calculate the value of the critical density of the uni-verse in solar mass per cubic Mpc. Assume H0 = 70km/s/Mpc.3) Calculate the age of the universe at redshifts of1,2,3,4,5,6,7,8,9,10. Assume the current age is 13 Gyrand Ω = 1. How long has the light travelled which wasemitted at z=1 ?4) CCDs are often used to take spectra of galaxies. Athigh redshift, the most prominent line is that of hydro-gen: Lyman alpha at 1216 A. CCDs are sensitive to1 micron. Out to what redshift can we used CCDs todetect galaxies ?
Now we know that ρ ∝ r−3, hence ρr2 ∝ r−1 ∝ 1 + zIn short, the left term is proportional to 1/(1 + z),and gets smaller and smaller with increasing redshiftz. Hence the term on the right also gets smaller as1/(1 + z). Now write Ω = 1 + δΩ. Hence to first order
1
Ω− 1 = −δΩ
And since 1Ω − 1 evolves like (1/(1 + z))
δΩ ∝ 1/(1 + z)
Hence, with increasing redshift, Ω gets closer and closerto 1As a consequence, Ω was very, very close to 1 athigh redshifts, independent of the current value !
How do we now form a galaxy ?
Answer: look back in the distant past (very high red-shift). Ω was very close to 1 at that time. Assumethat in some volume, the density ρ was enhancedby a minute fraction δρ. Since Ω was almost 1 ,
the smallest δρ is high enough to push the local Ωabove 1. This local volume has Ω higher than 1, andthe total energy is lower than 0. In short, it will notkeep expanding like the rest would (if, e.g., Ω ≤ 1).Hence it will collapse at some time, and will form agalaxy. Or a cluster...
Ω will be close to 1 in the distant past of our universe,independent of the current value. This is put graphi-cally below
0 2 4 6 8 100
0.5
1
1.5
2
z
0 200 400 600 800 10000
0.5
1
1.5
2
z
Now, take the universe at a nominal redshift of say,z = 106. We know that Ω is very close to 1. As-sume that for some reason or another, density fluctu-ations are present in this universe:
δ = ρ/ρ − 1
where ρ is the mean density of the universe at thatepoch, which we take to be the critical density.
We now wish to understand how a overdensity δ > 0evolves with time.
It turns out that the overdensity grows with expansionlike
We can prove this by considering a homogeneous sphereat z = 106 with overdensity δ. The sphere is embed-ded in a Ω = 1 universe.
The particles in the sphere do not feel anything fromthe outside universe. Hence the sphere will evolvelike it is a “separate universe”, with Ω = (1 + δ).Since a universe with Ω > 1 will collapse at sometime, the sphere will collapse at some time tcollapse.Before that, the density of the sphere will evolve like
1
Ω(t)− 1 =
1Ω0
− 1
1 + z(t)=
1Ω0
− 1
1/r(t)= (
1
Ω0− 1)r(t)
where r is the “radius” of the universe (expansion pa-rameter). Take at z = 106 : Ω0 = 1 + δ0, then wefind
Hence, the density contrast of the sphere increases lin-early with expansion radius. As a result, if a fluctua-tion was present of 10−6 at z = 106, it would havegrown to a fluctuation of δ = 1 at z = 0 under itsown gravity.
This mechanism is the basic mechanism toform galaxies
In detail, what happens is the following:
The sphere will collapse, and start oscillating (if we ig-nore the material just outside of the sphere). In reality,the sphere will have internal density fluctuations, and itwill settle to an equilibrium structure, with a radius ofabout half the “maximum expansion ” radius.
This radius is often called the “virialization radius” ,rvir. The sphere will obtain this radius at the first col-lapse time, which is also called the “virialization time”(or “formation time”)By comparing the maximum expansion radius of thesphere to the “normal expansion” radius of the uni-verse (with Ω = 1) at the virialization time, we canderive:
rmax =1
14 (12π)2/3
rΩ=1(tcollapse) (≈ 0.36rΩ=1(tcollapse))
This is derived using analytical solutions for the expan-sion of the universe. Since rvir = 1/2rmax
rvir =1
12 (12π)2/3
rΩ=1
The relative density of the sphere, compared to therest of the universe, is simply given by the ratio(rΩ=1/rvir)
3, since the mass of the sphere is con-served, but the density is increased compared to theΩ = 1 universe since the mass is put in a smaller den-sity structure.Hence
ρvir
ρ(universe)(z = zvir)= (rΩ=1/rvir)
3
= (1/2(12π)2/3)3 = 18π2 = 178
This makes a very specific prediction for the density ofobjects (galaxies, clusters, etc : If a galaxy forms at aredshift zform, it will have a density which is 178 timeshigher the density of the universe at zform.
After the galaxy has formed, it will remain the same,whereas the universe will keep expanding. Hence, thedensity contrast will increase with time
ρvir
ρ(universe)= 178 ∗ (r/rform)3
= 178 ∗ ((1 + zform)/(1 + z))3 = 178 ∗ (t/tform)2
This can be used as a simple recipe: we now measurethat galaxies have an overdensity of about 105 insidethe optical radius. This part would be formed at aredshift of
(1 + zform) = (105/178)1/3 = 8
The galaxy is much bigger, however, than the opticalradius. The halo has a density profile which goes likeρ ∝ r−2. The average density goes down like r−2, andthe density contrast will be a lot smaller if we take thehalo into account. If we assume that the halo extendsto 100 kpc (10 times further), the density will be lowerby a factor of 100, and the formation redshift will be
(1 + zform) = (103/178)1/3 = 1.8
Hence, the fact galaxies have halos has a very impor-tant consequence for galaxy formation: it makes thembigger, have lower mean density, and thereby formmuch later !At maximum expansion, the region has an overdensityof about 5, this increases very rapidly to 178 in thenext half of the total collapse time.
Homework assignment:5) Calculate the luminosity overdensity of a typicalgalaxy inside 10 kpc. Assume that the luminosity func-tion implies a luminosity density of 0.01 L∗ per Mpc3,where L∗ is the typical luminosity of a galaxy.6) Estimate when this galaxy would have formed usingthe equations given above.
The fact that galaxies follow these relations is remark-able, because it is not at all necessary for galaxies tohave such a relation from “simple” dynamics.
The relations are even more remarkable since the pre-diction is for the mass, not the light, of galaxies tofollow such a relation. As a consequence, there is alikely, simple relation between mass and light
Mhalo ∝ Lstars???
The explanation for this relation is usually the follow-ing: Each dark matter halo has some fraction f of itsmass in baryons (hydrogen mainly). These baryons sinkto the center, and form stars. The light of the stars isproportional to the mass in baryons, which is propor-tional to the mass in dark matter. This simple explana-tion might just work...
Homework Assignment:7) If the luminosity of a galaxy is proportional to circu-lar velocity to the power 4, how does the surface den-sity scale with mass ? Assume that the mass-to-lightratio is constant as a function of mass.
In practice, we have to do computer simulations tostudy the formation of galaxies. It is clear that the trueinitial conditions are not like overdense homogeneousspheres. In practice, the initial fluctuations are verycomplex
δ = δ(~x)
are expected to be irregular on all scales. For most ini-tial conditions, the overdensities at the smallest scaleshave the highest < δ >, and these will collapse first.Then larger structures will collapse.These type of formation theories are called “hierarchi-cal”, because the smaller structures will form first, andlarger structures will form later.
On the next page, we show some outputs from com-puter simulations. The first shows the evolution ofdark matter, then galaxies are added, and finally galaxyages are shown.
Galaxies are put into the halos of the simulationswith a simple recipe. They trace the large scalestructure very well, and look like the observed uni-verse
Now galaxy ages are shown in a small section of thesimulation which forms into a cluster at z=0. Bluegalaxies are young, red ones are old. The clusterscontain galaxies with old ages (at z=0)
