Modern Cosmology. Tom Charnock

Transcription

1 Modern Cosmology Tom Charnock

2 Contents 1 Introduction Observable Features The Metric Hubble s Law Curved Space Generalisation Light and Redshift Geodesic Equations Einstein s Equations Energy Evolution Cosmological Solutions Observational Parameters Horizons and Distances in Cosmology Particle Horizon Event Horizon Luminosity Distance Angular Diameter Distance Thermal History Ideal (Bose and Fermi) Particles Relativistic Species Entropy of the Background Non-Relativistic (Massive) Species Several Relativistic Species (Degrees of Freedom) Temperature vs. Time Density of the Universe and Dark Matter Dark Matter Candidates Cosmological Perturbation Theory Newtonian Gravity Poisson Equation Fluid Equations Structure Formation Static Universe Expanding Universe Relativistic Perturbation Theory Scalar-Vector-Tensor Decomposition Perturbative Einstein Equations Observing Structure Cosmic Microwave Background Anisotropies Recombination Decoupling

3 Chapter 1 Introduction 1.1 Observable Features The observable path of the universe, in which the light has been able to travel to the Earth since the beginning of time, is 3000Mpc years cm. On scales greater than 100Mpc then the universe is homogeneous and isotropic such that there is no unique place and all features look the same in all directions. The evidence for homogeneity comes from the distribution of large scale structure and for isotropy it is the distribution of the cosmic microwave background (CMB) photons. The universe is expanding at speed v Hd where H is the Hubble constant and d is the distance between objects, such as galaxies. In fact the universe is also accelerating which indicates that there must be some form of dark energy. There is a uniform distribution of radiation at T 2.275K±10 5 K which is in the microwave region. Observations of the fluctuations in the CMB indicate that when the universe was roughly a thousandth of the size of today then the fluctuations were as small as The universe contains baryons, about 1 per every 10 9 photons. There is very little anti-matter. The baryonic matter is distributed with 75% hydrogen and roughly 24% helium. The rest is trace amounts of the heavier elements. These elements were not produced in the early universe but instead in stars or supernovae. Baryons make up about 4% of the total energy density and the rest is in dark stuff. Cold dark matter makes up roughly 23% of the universe and it is unknown non-electromagnetic interactions, non-relativistic particles. These particles are pressureless. Dark energy makes up the rest of the universe (about 73%). The dark energy has a negative pressure. 1.2 The Metric In two dimensional flat space then the distance is given by Pythagorus, s 2 x x 2 2. If the space is stretched over time such that it is expanding uniformly then the distance increases as s 2 a 2 (t) ( x x 2 2). a(t) gives the rate of expansion and x1 and x 2 are comoving coordinates. Since time is a coordinate in general relativity then spacetime coordinates are used. curvature and so the infinitesimal separation needs to be used: There can be ds 2 g µν dx µ dx ν g µν are the components of the metric tensor and is a function of the coordinates and so allows space to be curved. dx µ and dx ν are the infinitesimal coordinate separation. The greek indices take values µ 0, 1, 2, 3. The time coordinate is given by x 0 and the spatial coordinates are x i (x 1, x 2, x 3 ). Vectors are components with raised indices, A µ (A 0, A i ) with i 1, 2, 3. A µ and A µ are different objects, A µ is called a one-form. Changing between vectors and one-forms can be done using the metric tensor: A µ g µν A ν and A µ g µν A ν 2

4 g µν is the inverse metric defined by g µν g νβ δβ ν. This arises since raising the indices of the metric gives: g µν g µα g νβ g αβ So g νβ g αβ 1 when ν α and g νβ g αβ 0 when v α. g µν is symmetric and so it has 4 diagonal and 6 off-diagonal components. In special relativity then Minkowski spacetime is used where x µ (ct, x i ) and the metric in this case is: 1 η µν For an expanding universe then a scale factor a(t) can be included, so for a flat space time the metric is: 1 g µν a 2 (t) a 2 (t) a 2 (t) This means the line element is: 1.3 Hubble s Law ds 2 g µν dx µ dx ν c 2 dt 2 + a 2 (t) ( dx 2 + dy 2 + dz 2) ( ) 2 ( ) 2 invariant time + interval interval ( ) 2 scale factor ( ) 2 comoving interval Because the expansion is uniform then the physical distance can be written as: r(t) a(t) x(t) Where x(t) is the comoving distance. Differentiating this with respect to time gives: ṙ H(t)r(t) + a(t)ṫ Here H ȧ/a is the Hubble parameter. The aẋ term is the peculiar velocity, or the local dynamics. The Hr term is the dynamics due to the expansion of the universe. This term is key for cosmology since it shows how the expansion rate relates the separation and velocity of recession of distant galaxies. Hubble s law is: v H(t)r (t) (t) The convention is to use subscript 0 to denote todays values, so Hubble s parameter today is: H 0 100h km s 1 Mpc 1 h is a measure of the uncertainty and, as of 2012, is: h ± If h 0.72 then for a galaxy travelling at v exp 7200km s 1 then the separation between the earth and the galaxy is 100Mpc. Uncertainty in h feeds into all the cosmological estimations. To estimate the ages of the universe then: H h years This is not a bad estimate. The evolution of the scale factor, a(t), depends on the density of matter. 3

5 1.4 Curved Space Generalisation The cosmological principle implies that the universe is isotropic and homogeneous on large scales. This means that the spatial part of the metric must have constant curvature. The most general form of the three dimensional space with constant curvature (in spherical polars) is: ds 2 3 a 2 ( dr 2 1 Kr 2 + r2 (dϑ 2 + sin 2 ϑdφ 2 ) Here a 2 > 0 and from now on dω 2 dϑ 2 + sin 2 ϑdφ 2. K measures the curvature of space. If K > 0 then the space is spherical (or closed), if K 0 then space is flat and if K < 0 then space is hyperbolic (or open). The full spacetime metric is: ( ) dr ds 2 c 2 dt 2 + a 2 2 (t) 1 Kr 2 + r2 dω 2 This is the Robertson-Walker metric. Taking the spatial section and introducing a new variable dχ 2 dr 2 /1 K 2, then normalising K ±1, 0 and integrating gives: } χ arcsinhr K 1 0 χ χ r K 0 χ sin r K 1 0 χ π This means that the metric for the spatial part can be written as: Where S K (χ) is: ds 2 3 a 2 (dξ 2 + S 2 K(χ)dΩ 2 ) S K (χ) sinh χ K 1 S K (χ) χ K 0 S K (χ) sin χ K 1 For the three dimensional sphere (K 1) then the distance element on the surface of the two sphere with radius χ is: ds 2 a 2 sin 2 ξdω This is equivalent to the line element on a sphere of radius R a sin χ in flat three dimensional space. The surface area of the two dimensional surface is: S 2D (χ) 4πR 2 4πa 2 sin 2 χ As the radius χ increases then the surface area grows to a maximum of χ π/2 and then decreases until it vanishes at χ π. The volume is then obtained by integrating dv S 2D (χ)adχ: χ0 V (χ 0 ) 4πa 3 sin 2 χdχ 0 2πa 3 (χ sin 2χ 0 ) ) 1.5 Light and Redshift The line element in general for a Friedmann-Robertson-Walker universe is: ( ) dr ds 2 c 2 dt 2 + a 2 2 (t) 1 Kr 2 + r2 dω 2 If there are two galaxies A and B then the light emitted from B has a wavelength which could be changed by the expansion of the universe when observed at A. Light follows null geodesics ds 2 0. For a radial trajectory dω 0 so the change in time between emision and observation is: rb r A dr 1 Kr 2 tobs t emit c dt a(t) 4

6 A second peak of the wave is emitted at t emit + λ emit /c and is observed at t obs + λ obs /c and since the integral over r is just a constant then: tobs This reveals the result: t emit c dt a(t) v emit v obs tobs +λ obs /c t emit+λ emit/c a obs a emit c dt a(t) Since v emit /v obs 1 + z then observing shifts in spectral lines allows the relative size of the universe to be determined when the light was emitted. This is key for observational cosmology. 1.6 Geodesic Equations The geodesics is the trajectory in curved space. It is the generalisation of the straight line in Minkowski space in the absence of external forces, i.e. Newton s Law generalised to an expanding universe. Since time is a coordinates then a new evolution parameter is needed. For this role, λ is introduced, which increases monotonically along the particles path. The geodesic equation is: Γ µ αβ d 2 x µ dλ 2 are the Christoffel symbols which are defined as: Given the FRW metric with K 0 then: + dx α dx β Γµ αβ dλ dλ 0 Γ µ αβ 1 2 gµν (g αν,β + g βν,α g αβ,ν ) Γ 0 ij aa δ ij and Γ i 0j Γ i j0 a a δi j Where a da/cdt ȧ/c. All the other Christoffel symbols vanish. 1.7 Einstein s Equations Einstein s equation relate the components of the Einstein tensor (geometry) to the energy-momentum tensor (matter). The Einstein tensor is: R µν is the Ricci tensor and is symmetric given by: G µν R µν 1 2 g µνr R µν Γ α µν,α Γ α µα,ν + Γ α βαγ β µν Γ α βνγ β µα R is the Ricci scalar obtained by contracting the inverse metric with the Ricci tensor R g µν R µν. Einstein s equations are: G µν 8πG c 4 T µν Λg µν Λ is the cosmological constant and T µν is the energy-momentum tensor, which is symmetric. For a perfect isotropic fluid then the energy-momentum tensor is: T ν µ ϱc 2 p p p ϱ(t) is the energy density and p(t) is the pressure of the fluid. By substituting in the Christoffel symbols found in section 1.6 into the Ricci Tensor then it can be seen that: R 00 3a a R ij δ ij ( 2a 2 + aa ) 5

7 The Ricci scalar can also be found: R g µν R µν R 00 + R ii 6 ( a a + a 2 ( ) ) a 2 The Friedmann equation is the 00 component of Einstein s equations: R g 00R 8πG c 4 T 00 Λg 00 (ȧ a ) 2 8πG Λc2 ϱ(t) When considering curvature then the Christoffel symbols obtain curvature terms and so the Friedmann equation is: (ȧ ) 2 8πG Kc2 ϱ(t) a 3 a 2 + Λc2 3 Now considering the ij components of Einstein s equations gives the acceleration equation: ( ä a 4πG ϱ(t) + 3p(t) ) 3 c 2 + Λc2 3 The cosmological constant can be absorbed by making: ϱ ϱ + Λc2 8πG a and p p Λc2 8πG Λc 2 /8πG can be thought of as an energy density ϱ λ or a pressure p Λ so that p ϱc 2. This is important for the evolution of the universe and inflation since substituting p ϱc 2 into the absorbed acceleration equation shows that if ϱ + 3p/c 2 < 0 then ä > 0. The acceleration and Friedmann equations with the cosmological constant absorbed are: ä a 4πG 3 ( ϱ(t) + 3p(t) ) c 2 H 2 8πG Kc2 ϱ(t) 3 a 2 The critical density can now be defined as the density for which the universe is flat, i.e. K 0. This density is a function of time and is continually changing: The density parameter is then defined as: ϱ C (t) 3H2 (t) 8πG Ω ϱ(t) ϱ C (t) 8πGϱ 3H 2 The density today is ϱ Ω 0 h 2 kgm 3. The density is extremely low but in terms of different units then ϱ Ω 0 h 2 M Mpc 3. This is on the galactic scale and says that there is roughly solar masses in the galaxy which is approximately correct. If ϱ ϱ C then Ω 1 for a spatially flat universe. 6

8 1.8 Energy Evolution The energy-momentum conservation is given by: T µ ν;µ T µ ν,µ + Γ µ αµt α ν Γ α νµt µ α 0 When ν 0 then T µ 0,µ + Γµ αµt α 0 Γ α 0µT µ α 0, and assuming isotropy then T 0 i vanishes since there is no strain and thus placing in the Christoffel symbols for the FRW universe reveals: ϱ + 3ȧ a (ϱ + p ) c 2 0 This is the fluid equation. The pressure and energy density need to be related and is done so using the equation of state p p(ϱ). For most fluids (with no torsion) then p wϱ where w (γ 1) is a constant. The fluid equation has a solution: ϱ a 3(1+w) The Friedmann equation can be solved to give: a t 2/(3(1+w)) These can be used to find the time evolution versions of the energy density and scale factor. Matter Domination When w 0 then p 0 and since matter is a pressureless dust. When Ω m 1 then the universe is Einstein-de Sitter and the values of the energy density and the scale factor are: ( a0 ) ( ) 3 2/3 t ϱ m (t) ϱ m0 and a(t) a 0 a t 0 Radiation Domination Radiation has a w 1/3 and a universe with Ω r 1 is called a Tolman universe. The energy density and scale factor evolve as: ( a0 ) ( ) 4 1/2 t ϱ r (t) ϱ r0 and a(t) a 0 a t 0 Cosmological Constant For the cosmological constant then w 1 and so the scale factor has to be expanded as a Taylor series to find its evolution. If the universe has a cosmological constant then it is de Sitter and has energy density and a scale factor: Here H 8πGϱ 0 /3. ϱ Λ (t) ϱ 0 and a(t) a 0 e H(t t0) The total energy density is given by ϱ ϱ m + ϱ r + ϱ v where ϱ v is the vacuum energy density where w is not confined to 1 as for a cosmological constant. If w varies with time then it provides the force known as quintessence. The Friedmann equation can now be solved using the time evolution energy densities: H 2 8πG 3 (ϱ m + ϱ r + ϱ v ) Kc2 a 2 8πG 3 (ϱ m0 (1 + z) 3 + ϱ r0 (1 + z) 4 + ϱ v0 (1 + z) 3(1+w)) Kc2 a 2 8πG 3H0 2 H0 (ϱ 2 m0 (1 + z) 3 + ϱ r0 (1 + z) 4 + ϱ v0 (1 + z) 3(1+w)) Kc2 a 2 H 2 0 (Ω m0 (1 + z) 3 + Ω r0 (1 + z) 4 + Ω v0 (1 + z) 3(1+w)) Kc2 a 2 7

9 Since K is not observable then it should be eliminated by solving the Friedmann equations for the constants known today: Kc 2 a 2 0H 2 0 (Ω 0 1) a 2 0H 2 0 (Ω m0 + Ω r0 + Ω v0 ) Substituting this back into the Friedmann equation gives: ( ) H 2 H0 2 (1 + z) 2 Ω m0 ((1 + z) 1) + Ω r0 ((1 + z) 2 1) + Ω v0 ((1 + z) (1+3w) 1) This is very powerful since t(z) can be found by integrating and the integrals are numerically straight forward. 1.9 Cosmological Solutions The universe has lots of fluids and may not be flat. Conformal time can be introduced so that the propertime becomes a(t)τ. This conformal time, η is given by cddt a(η)dη so that: cdt η a(t) For c 1 then the solutions to different cosmologies can be calculated in terms of η. It is useful to note that a da/dη. The Friedmann equation is now written as: a 2 8πG 3 ϱa4 Ka 2 And the acceleration equation is: a 4πG 3 (ϱ 3p)a3 Ka For a single fluid, say radiation with p ϱ/3 then the acceleration equation is simply: a + Ka 0 This is simple to solve since it is of the form of a harmonic oscillator (depending on K). The solutions are: } a(η) c r sinh η K 1 0 η a(η) c r η K 0 a(η) c r sin η K 1 0 η π The second integration constant is fixed by a(0) 0. The physical time is then: t a(η)dη For the three different curvature situations then: t(η) c r (cosh η 1) K 1 t(η) c rη 2 K 0 2 t(η) c r (1 cos η) K 1 These are parametric solutions, i.e. a(η) and t(η). This give a(t) but not in such an accessible form, except for the trivial K 0 case. For dust then the same steps can be taken by now p 0. Some unusual solutions can be found. Take the radiation-λ solution (in K 0). the Friedmann equation for radiation-λ is: ( ( H 2 (z) H0 2 a0 ) ) 4 Ω r0 + ΩΛ0 a 8

10 Square rooting both sides and rearranging gives: Here α Ω Λ0 /Ω r0. Introducing y a 2 /a 2 0 then: Integrating this gives a solution to a(t): d ( ) ) 4 1/2 dt a2 2H 0 Ωr0 a 2 0 (1 + α 2 aa0 ẏ 2H 0 Ωr0 ( 1 + α 2 y 2) 1/2 ( ) 1/4 Ωr0 ( ) 1/2 a(t) a 0 sinh(2h 0 ΩΛ0t) Ω Λ0 Where a(0) 0 has been used as the boundary condition. At early times (t 1) then sinh t t so a(t) t 1/2 which is radiation dimension. At late times, t 1 then sinht e t /2 so a(t) e H0 ΩΛ0t which is an inflationary solution. This model could be a precursor before the onset of inflation in the early universe. For a matter-λ solution (a good way of describing the universe today since Ω Λ0 0.7 and Ω m and so are comparable) then when K 0 the Friedmann equation is: ( ( H 2 (z) H0 2 a0 ) ) 3 Ω m0 + ΩΛ0 a Now y (a/a 0 ) 3/2 and solving for a(t) gives: ( ) 1/2 ( ( Ωm0 3 a(t) a 0 sinh Ω Λ0 2 H 2/3 0 ΩΛ0 t)) Again, at early times then a(t) t 2/3 which is usual for matter domination and once again, for large t then sinh t e t and so a(t) e H0 ΩΛ0t which is inflationary or dark energy Observational Parameters Hubbles parameter (constant - which is not constant) - H 0 100hkms 1 Mpc 1 with h ± Deceleration parameter - q 0 ä(t 0 )/(a(t 0 )H 2 0 ). This is measurable directly on large scales. Type 1a supernovae show that q and so the observation show that the universe is accelerating Horizons and Distances in Cosmology The FRW universe metric is: ( dr ds 2 c d t 2 + a 2 2 (t) 1 Kr 2 + r62 ( dϑ 2 + sin 2 ϑdφ 2) ) A light ray travels from r 0 and t t em to r r 0 and t t along a radial null direction so that ds dϑ dφ 0. It is useful to use conformal time and dχ dr/ 1 Kr 2 so that the metric becomes: 0 dη 2 + dχ 2 Integrating through these gives ξ(η) ±η + constant. The solutions are straight lines at angle at 45 o in the η χ plane and is true of all geometries, K ±1, 0. 9

11 Particle Horizon For a universe of finite age then light can only travel a finite distance in that time. Moreover, since light has only travelled a finite distance then there is a finite volume of space within which signals could be received. The boundary of this volume is the particle horizon which is expected to be of order c age of the universe. The maximum distance that light can propagate is given by the difference in chi at the beginning and end of the universe. Given that χ(η) ±η+constant then: χ(η) η η i t At η events at χ > χ p (η) are inaccessible to an observer located at χ 0. It is usually alright to choose η i t i 0 especially when there is an initial singularity. The physical size of the particle horizon is: t i R(t) a(t)χ p cdt a a(t) t i t cdt a(t) This is the radius of the observable universe at some time t. It could be infinite or finite depending on the cosmology. If the cosmology is decelerating then R(t) is finite, i.e. a(t) t α (0 < α < 1) has a radius: R(t) ct 1 α This is finite at any given t. The Einstein-de Sitter universe has α 2/3 and so the particle horizon is at R(t) 3ct 0 > ct Event Horizon The event horizon can be thought of as the compliment of the particle horizon. It encloses a set of points from which signals sent at a given moment t(η) will never be received by an observer in the future. These correspond to the points: χ ev (t) η max η tmax t cdt a(t ) η max is the final moment of conformal time. The physical size of the event horizon is: R ev (t) ca(t) tmax t dt a(t ) If the universe expands forever then t max and when K 0, 1 then these are decelerating universes so that χ ev and R ev and so there is no event horizon. If the universe is accelerating then R ev is finite even for K 0, 1 and so there is an event horizon. For a flat de Sitter expansion, a(t) e Ht where H is a constant then: R ev (t) ce Ht ch 1 t dt e Ht This is a size of order the scale of the curvature of the universe Luminosity Distance In Minkowski space then the definition scale is no problem so if a source was emitting with absolute luminosity L s then the flux of light, F, received at a distance d is given by: F L s 4πd 2 10

12 In the expanding case then the true distance is not known and so the Minkowski result is used as the new definition of distance, called the Luminosity Distance: d 2 L An object has an absolute luminosity L s which is located at comoving distance χ S from an observer at χ 0. The energy of the light emitted within time interval t 1 is E 1 and it reaches the observer on a sphere of radius χ S as E 0 over a period t 0. E ν and since c is invariant then c ν 1 λ 1 ν 0 λ 0. Relating this all to the red shift gives: L s 4πF 1 + z λ 0 λ 1 ν 1 ν 0 t 0 t 1 E 1 E 0 The relation between the frequency and the time interval, ν 0 t 0 ν 1 t 1, has been used. The luminosities are: L s E 1 and L 0 E 0 t 1 E 1 These can be rearranged using the previous relations to give: L s L 0 (1 + z) 2 The two factors of (1 + z) arise from the fact that each photon loses energy and that the number of photons arriving each second decreases over time as the universe expands. Light travels along the χ direction (dϑ dφ ds 0) then: χ s χs 0 t0 0 dχ s cdt a(t) c z dz a 0 0 H(z ) This relation has been obtained from ż H(1 + z) and 1 + z a 0 /a. The area of a sphere at t t 0 is given by S 4π(a 0 S k (χ s )) 2 as seen on page 4 and from this result the observed flux is: F L 0 4π(a 0 S k (χ s )) 2 The luminosity distance d 2 L L s/f can now be found by substituting in relations to get: d L a 0 χ s (1 + z) The consequence of this relation is that distant objects appear further away than they really are because of redshift. Comparing the luminosity distance to the physical distance is: d phys a(t) a 0 χ S χs So d L d phys (1 + z) which means that for z 1 then d L d phys Angular Diameter Distance Objects of a given physical size, l are assumed to be perpendicular to an observers line of sight so that if they subtend an angle ϑ (which is small in astronomy) then the angular diameter distance is: d diam 11 0 l ϑ dχ

13 The line element in comoving coordinates is: ds 2 a 2 (η) [ dη 2 + dχ 2 + S k (χ)(dϑ + sin 2 ϑdφ 2 ) ] At the particle horizon then S k (χ p ) 2/(a 0 H 0 Ω 0 ), which is true for any value of Ω 0. A dust dominated universe is considered so that Ω 0 Ω m. When it comes to the angular diameter distance then only the redshift z of the galaxy when the light was emitted which is being detected. S K (χ emit (z)) 2 Ω 2 0 a 0H 0 (1 + z) ( Ω 0 z + (Ω 0 2) ( 1 + Ω0 z 1 Taking the large z limit then S K (χ p ) is recovered. The large z limit is further back in time. If there is an extended object of given transverse size l at a comoving distance χ emit then it can be aligned so that φ constant. Photons are emitted from both sides of the object at t emit and propagate along the radial geodesic and arrive at t 0. These photons arrive at an apparent angular separation ϑ. When the light is emitted then dt dφ dχ so the interval between emission events at the end points: l ds 2 a(t emit )S k (χ emit ) ϑ )) The angular diameter distance is therefore: d diam l ϑ a(t emit )S k (χ emit ) a 0 (1 + z) 1 S k (χ emit ) d L (1 + z) 2 This means that d L (1 + z) 2 d diam. This is true for all curved spaces. ϑ has some unusual features: ϑ l a(t emit )S k (χ emit ) l a(η 0 χ emit )S k (χ emit ) If an object is nearby such that χ emit η 0 then a(η 0 χ emit ) a(η 0 ) and also S k (χ emit ) χ emit for all curvature. It follows from these two results thats: ϑ l a(η 0 )χ emit l d phys This is as expected. Far away objects which are near the particle horizon then η 0 χ emit η 0 so a(η 0 χ emit ) a(η 0 ) and S k (χ emit ) S k (χ p ) 2/(a 0 Ω 0 H 0 ) constant. This means that: ϑ l a(η 0 χ emit ) As the object gets further away it looks as if it is getting bigger, i.e. ϑ increases with distance. As the object approaches the particle horizon then then image will cover the whole sky. The angular size of objects peak nearby but also far away. This can be understood because the universe was smaller in the past so a given physical sized object appears larger. For a matter dominated flat universe then the angular diameter distance peaks at a redshift of z 5/4. Using 1 + z a 0 /a(t emit ) then the angular diameter distance is: ϑ (1 + z)l a 0 S k (χ emit (z)) 12

14 χ emit can be written in terms of z using dz (1 + z)h(t)dt: χ emit (z) η 0 η emit t0 t emit cdt a(t) 1 a 0 z 0 cdz H(z) Also from this the age of the universe, depending on choice of cosmology, can be calculated: t(z) z dz (1 + z)h(z) A flat dust filled universe, where S k (χ emit ) χ emit and Ω 0 Ω m0, has H(z) H 0 (z)ω m0 (1 + z) 3/2 so: t(z) 2 1 and χ(z) 2c ( ) 1 1 3H 0 (1 + z) 3/2 a 0 H z and ϑ lh 0 2c (1 + z) 3/2 (1 + z) 1/2 1 Here ϑ has a minimum at z 5/4. This corresponds to the maxima in the angular diameter distance. 13

15 Chapter 2 Thermal History Assume there is a statistic equilibrium then there will be a number density n, an energy density ϱc 2 and a pressure p, all arising from a distribution function. One type of particles, species A, has a distribution function f A (p, t) where p is the 3-momentum. Different species interact with rate Γ(t) > H(t). The interactions are thought to be short range, so provide the mechanism to maintain thermal equilibrium. 2.1 Ideal (Bose and Fermi) Particles The equilibrium distribution function is: f A (p, t)d 3 p g A d3 p (2π) 3 e β(e A µ A) ± 1 β (k B T ) 1 where k B is Boltzmann s factor, g A is the spin degeneracy which gives the number of relativistic species at any given temperature T A. µ A is the chemical potential and E A (p0 p 2 c 2 + m 2 c 4. the + sign refers to fermions and the - sign to bosons. The number of particles is conserved so µ A 0. From the distribution function it follows that the number density is: n 1 3 f(p )d 3 p The energy density is: g (E 2 m 2 c 4 2π 2 c 3 3 mc e βe ± 1 EdE 2 ϱc 2 En 1 3 E(p )f(p )d 3 p And the pressure is given by: g (E 2 m 2 c 4 2π 2 c 3 3 mc e βe ± 1 2 E2 de p 1 3 mc 2 p c2 3E(p ) f(p )d3 p g (E 2 m 2 c 4 ) 3/2 6π 2 c 3 3 mc e βe de ±

16 2.1.1 Relativistic Species In the relativistic regime then mc 2 k B T. Introducing x βe then the number density is: n (Bosons) (Fermions) ( kb T c ( kb T c ( kb T c ) 3 g x 2 dx 2π e x ± 1 ) 3 gζ(3) π 2 ) 3 gζ(3) π 2 ( 1 3 ) n B T 3 The ζ function is given by: ζ(n) n1 m n 1 u n 1 du Γ(n) 0 e u 1 The fermions can be written as two species of bosons at different temperatures: 1 e x e x 1 2 e 2x 1 The CMB photons have a number density which can be found using the bosonic equation. Since T 2.725K and g 2 for photons then n γ m 3. The energy density for bosons is: And for fermions is: ϱ B c 2 π2 30c 3 3 g(k BT ) 4 ϱ F c ϱ Bc 2 These are both proportional to T 4. The pressure is given by p ϱc 2 /3. Since ϱ B T 4 and ϱ B a 4 for relativistic particles then it can be seen that the temperature is inversely proportional to the scale factor T 1/a Entropy of the Background Entropy and energy are both extensive quantities, so are proportional to the number of particles in the system, such that: S V S E and V V E V Starting with the first law of thermodynamics and substituting in de and ds gives: E E dt + T de T ds pdv V dv T ( S T dt + S V dv ) pdv This is true for arbitrary dt and dv and so collecting the terms gives: dt : E T T S T dv : S E + pv T 15

17 In the relativistic limit where k B T mc 2 then the entropy density is obtain by substituting in the results from section 2.1.1: s S V 4 ϱ B c 2 3 T 2π2 k B 45c 3 3 g(k BT ) 3 There would be a factor of 7/8 for fermions. The number density, n B T 3 for relativistic particles and since s scales in the same way then it also counts the number of particles. This is why cosmologists say that the ration n γ /n baryon is the entropy per baryon. n γ /n baryon 10 9 today Non-Relativistic (Massive) Species The non-relativistic (massive) species have mc 2 k B T. In this case the number density is: n g 3 ( mkb T 2π ) 3/2 e βmc2 It is Boltzmann suppressed. The energy density is ϱ mn and the pressure is p n(kt ). This is much less than the energy density so p 0. The baryon to photon ratio is η nγ/n baryon. The photon number is n γ 411cm 3 and the baryon number is n B ϱ B /m B Ω B ϱ c /m B. The critical density is ϱ C kgm 3 and the mass of a proton is m p g so that: ( η ΩB h 2 ) 0.02 This means there is of the order of one billion photons for every baryon. therefore n B 0.22m 3. The number of baryons is Several Relativistic Species (Degrees of Freedom) The g factor used throughout section 2.1 describes the number of species. If there is a collection of multiple species of relativistic particle each at equilibrium at a different temperature, T i, then the total energy is: ϱ Rel c 2 k4 B T γ 4 ( ) 4 Ti u 3 du 2π 2 3 c 3 g i T γ x i e u ± 1 i T γ is the temperature of the CMB photons. The total energy density can be written is a compact from by doing the integral in the limit x i 0 giving ζ(4): g is the effective number of degrees of freedom: g bosons ϱ Rel c 2 (kt γ) c 3 π3 g ( ) 4 Ti g i + 7 T γ 8 fermions ( ) 4 Ti g i T γ As the temperature of the background photons decreases the effective number of degrees of freedom in radiation also decreases since massive particles become non-relativity, m i c 2 > kt i. T 1MeV The only relativistic particles are the three neutrinos and the photon. The photon is a boson and has two polarisation states (two degrees of freedom) and the neutrinos are fermions with two degrees of freedom 16

18 (spin up and spin down). Neutrinos at this scale are decoupled from the thermal bath and are slightly colder T ν (4/11) 1/3 T γ. The effective number of degrees of freedom is: g ( 4 11 ) 4/3 1Mev < T < 100MeV 3.36 The electron and the positron have masses 0.5MeV and so both of these are relativistic. The difference between T γ and T ν is due to the electron-positron annihilation, but when the electrons and positrons are relativistic then this process shuts down and so T γ T ν. The electrons now need to be included in the calculation of the effective number of degrees of freedom. T 300GeV g ( ) At high energies such as at the electroweak unification at the scale of the standard model particles then g Temperature vs. Time For photons then the radiation density parameter is: Ω r ϱ r ϱ c π 2 30c 5 3 ϱ c g(kt ) h 2 The neutrino density parameter is: Ω ν 3 7 ( ) 4/3 4 8 Ω r Ω r h 2 If the only relativistic particles today are photons and neutrinos then the relativistic contributions are: Ω rel Ω r + Ω ν h 2 Since Ω mo 0.25 then all the relativistic particles are negligible compared to non-relativistic matter. The evolution of these quantities can be determined using ϱ rel a 4 and ϱ m a 3 : Ω rel Ω m Ω m0 h 2 ( a0 ) a Ω m0 h 2 (1 + z) 17

19 There is a value of z when Ω rel Ω m which is a period called equality. This redshift is at: ( ) 4 (1 + z eq ) 24074(Ω mo h 2 T0 ) 2.725K For T > T eq then the universe is radiation dominated and when T < T eq then the universe is matter dominated. Decoupling is a period when the formulation of the CMB occurred because of the photons decoupling from the electrons as atoms formed. (1 + z dec ) Density of the Universe and Dark Matter The critical density today is ϱ c (t 0 ) 1.88h kgm 3. For stars then Ω stars , for baryons from nucleosynthesis (tightest constraints) Ω B h For h 0.7 then Ω B 0.03 Ω stars which implies there is baryonic material in the universe which is not visible in stars. The dynamics of galaxies such as rotation curves give a density parameter Ω halo. Going further to clusters of galaxies and looking at the virial dynamics and peculiar velocities then Ω m0 0.3h 1/2. The current best bound on the contribution of matter today is Ω m There is also gravitational lensing of distant quasars which gives Ω m Numerical simulations can be done and tend to suggest that Ω Λ0 0.2 to get the models to work. All of these suggest that ω m0 < 1. Ω 0 is of the order unity and so there must be some dark energy out there. CMB anisotropies are used to probe the geometry of the universe and hence obtain a value of Ω 0. The WMAP7 data suggests that Ω K and so is consistent with a flat universe. When combined with the Hubble space telescope data then Ω K and Ω m ± Today it is believed that Ω K 0, Ω m from nucleosynthesis and so most of the matter is in dark matter. Ω Λ0 0.7 is the dark energy. More evidence of dark matter is present in the Bullet Cluster Dark Matter Candidates There are three basic types of dark matter candidates: Hot Dark Matter The particles decouple from the background radiation when they are relativistic. Their number density is n HDM n γ where the typical mass scale is m O(ev) such as neutrinos. Ω rel m ν. These are not currently favoured since free-streaming leads to erased structures on large scales. Weak Dark Matter These relic particles decouple early enough such that there is a similar abundance to photons but can be boosted by annihilations other than just through electron-positron annihilation. There are 100 distinct particles species which have m W DM 1 10keV. These fell out of favour because of free streaming but they have recently come back in vogue as a way of addressing dwarf galaxy problems in large scale structure formation. Cold Dark Matter The particles decouple when they are non-relativistic. The number density is therefore suppressed n CDM e βmc2. The freeze out temperature is O(MeV ) where the mass scales of the particles are m CDM 10GeV O(T ev ). The main candidates are called WIMPs (weakly interacting massive particles) and come from supersymmetry, i.e. neutralinos or gravitinos. There are also axions which are a class of particles invented to solve the ϑ problem in QCD, but can also be used as dark matter candidates. They have masses 10 6 ev < m Axion < 10 3 ev. Primordial black holes have a mass g m g where the lower bound comes from the black hole evaporation rate and the upper bound from microlensing. These have not been ruled out. Maybe dark matter and dark energy do not exist and what is really happening is a modification of gravity on large scales. These ideas still need to be explored. 18

20 Chapter 3 Cosmological Perturbation Theory The idea behind perturbation theory is to use Taylor expansion and use the leading orders in the theory. If there is a function at a point X 0, f(x 0 ) f 0 then a point close to this is given by f(x) f ) + f 1 X where f 1 X is small. The full set equations would have FWR background plus a small fluctuation and dropping second order (fluctutation) 2 terms. Units energy. Natural unit are used so that c k B 1. Using this all units can be measured in units of [Energy] [Mass] [Temperature] The unit of energy which will be used is the GeV J. 1GeV Mpc 1 1pc 3 light years 1kg GeV Mpc 1 1K GeV Mpc 1 1m GeV Mpc 1s GeV Mpc m 1 [Length] 1 [Time] Hubble s constant is: H 0 100hkm s 1Mpc 1 100h 103 m m Mpc 3.33h 10 4 Mpc 1 The current size of the universe is: 2.1h GeV 1 H Mpc 3000Mpc 3.1 Newtonian Gravity F Gm 1m 2 r 2 ˆr 19

21 To work out the dynamics of the universe then the force caused by each particle needs to be taken into consideration. This is obviously impossible. Tho rectify this then the particles are modelled as a fluid. If the total volume is V and there is a small volume δv which still contains a large number, N, of particles then the mass density is: Nm ϱδv lim δv 0 δv By doing this for many small volumes to make up the total universe then the energy density is obtained ϱ(t, x ). Each particle has its own velocity and so taking a small volume of fluid then average out the velocities of all the particles to get the fluid velocity ṽ(t, x ). The pressure can be defined in a similar way, ϱ(t, x ) Poisson Equation The force between the two regions is calculated using Newtons gravitation equation: δ 2 F Gδmδm i r r i 2 (ˆr ˆr i) This is now done for all of the particles effecting just the particle at r. The force density can also be defined f δf /δv : δf G i f δv δ m δ mi r r i 2 (ˆr ˆr i) Since ϱ(t, r )δv δm then: f G i ϱ(t, r )ϱ(t, r i) r r i 2 δv i (ˆr ˆr i ) Take the small volume limit so that δv d 3 r and the sum becomes an integral then: f Gϱ(t, r ) d 3 r ϱ(t, r ) r r 2 (ˆr ˆr ) This unit vector can be made into a scalar using the the gradient operator: r r x y z (x x ) 2 + (y y ) 2 + (z z ) 2 2(x x 1 ) 2 2(y y ) (x x ) 2 + (y y ) 2 + (z z ) 2 2(z z ) r r r ˆr ˆr This means that: ˆr ˆr ( ) 1 r r r r The force density can therefore be written as: ( ) 1 f(t, r ) Gϱ(t, r ) d 3 r ϱ(t, r ) r r 20

22 This is equivalent to Newtons second law, m ã f ϱ(t, r ) g. The acceleration due to gravity is: F ( ) 1 g(t, r ) G d 3 r ϱ(t, r ) r r The Newtonian potential is defined as φ(t, r ) and since dφ/dr then: φ(t, r ) G d 3 ϱ(t, r ) r r r This means that g φ. Taking the scalar product of this gives: g 2 φ G G ( ) 1 d 3 r ϱ(t, r ) 2 r r d 3 r ϱ(t, r )( 4πδ 3 (r r ) 4πGϱ(t, r ) This is the Poisson equation: 2 φ 4πGϱ This is the analogue of Newtons law for continuous media Fluid Equations The mass in a volume V is time dependent: m(t) d 3 rϱ(t, r ) The rate of decrease of this mass out of the volume is therefore: dm dt d 3 r ϱ t The change in mass is related to the flux of particles into or our of the surface of V. This depends on the velocity of the particles. The flux is maximal when the velocity vector is parallel to the normal to the surface of the volume, ds, and is zero when the flux is perpendicular. This suggests that the flux is proportional to ṽ d s. Using dimensional analysis then it can be seen that: [ṽ] LT 1 0 [ds ] L 2 [ṽ ds ] L 2 [ ] m t [ϱ] ML 3 L 4 ML 1 L 2 So the required quantity for the rate of change in mass is ϱṽ since its dimension is L ds 2. Integrating this over the surface gives: ϱṽ d S ds 3 r ϱ V t 21

23 Using Stokes theorem then: S ϱṽ ds V V d 3 r (ϱṽ) Substituting this into the previous equation gives: d 3 r ϱ t d 3 r (ϱṽ) Since this holds for any volume then: (ϱṽ) + ϱ t 0 Using the product rule it can be seen that this is just the conservation of mass equation: 0 D Dt ϱ + ϱ ṽ ϱ + ṽ ϱ + ϱ ṽ t Here D/Dt is the convective or substantial derivative. Newtons second law of motion can be used to get the rate of change of velocity. The second law written in terms of the force per unit volume is: f ϱ ã The velocity now depends not only on time but also on position and so the new form of ã needs to be found. Consider A(t, r ) then a small change in A is given by: δa A t δt + A x Now divide by δt and take the limit where t 0 so δt dt: da dt A A δx + δy + y z δz A t + A dx x dt + A dy y dt + A dz z dt A t + ṽ A This is just the convective derivative. Using this it can be seen that ã Dṽ/Dt and so the force density is: The force density due to gravity is: f ϱ ã [ ] ṽ ϱ t + ṽ ṽ f grav ϱ g ϱ φ The pressure is the hydrodynamical force and is given by the force per area. The force on a slab whose top has a point A and whose bottom has a point B is F z F A F B. The force at A is the pressure at A per area F A p A A p A X Y and the force of B is F B p B X Y. So F (p A p B ) X Y. For a small Z then the pressures at A and B can be Taylor expanded: The difference in the forces in the z direction is: p B p A + p z z F z p X Y Z z p z V 22

24 Repeating for the forces along the x and y directions reveals that the total F is given by: F F x + F y + F z ( p x p y p ) V z p V So the force density is just: The total force density is finally: f hydro p The Euler equation is then obtained: f ϱ φ p ṽ The three equations which describe fluids are: t [ ] ṽ ϱ t + ṽ ṽ + ṽ ṽ + p p + φ 0 0 ṽ p + ṽ ṽ + t p + φ 0 ϱ + ṽ ϱ + ϱ ṽ t 4πGϱ 2 φ There are three equations for four unknowns, ϱ, p, ṽ and φ. What is needed is an equation of state to relate p and ϱ. They are related by the speed of sound, c 2 s: In this case Eulers equation becomes: p c 2 s ϱ ṽ t + ṽ ṽ + c2 s ϱ + φ 0 ϱ These are impossible to solve analytically. Certain approximations need to be made. 3.2 Structure Formation Static Universe First assume that the fluid is homogeneous such that any terms are zero. This leaves ϱ/ t 0 so that ϱ ϱ is constant in time and space. The same is also true for the velocity. In fact a frame of reference can always be chosen where ṽ 0. Now assume that ϱ(t, x ) ϱ + δϱ where δϱ is some small fluctuation and the velocity is ṽ(t, x ) ṽ + δṽ δṽ. Define the density constant δ δϱ/ϱ. If δ > 0 then there is a concentration of particles and if δ < 0 then there are fewer particles that average. Plugging these terms into the fluid equations and ignoring any terms quadratic or higher gives: 0 δṽ t + c2 s δϱ + φ ϱ δṽ t + c2 s δ + φ 0 δϱ t + ϱ δṽ δ + δṽ 23

25 4πGϱ (1 + δ) 2 φ Only the small term of this equation is needed since the first term is large. Eliminating the velocity by taking the time derivative of the continuity equation and substituting in ṽ/ t gives: δ + ( ṽ) t δ + { c 2 s δ φ } δ c 2 s 2 δ 2 φ δ c 2 s 2 δ 4πGϱδ The first two terms are the components of a wave equation with speed c s. The last term is the gravitational term. Oscillations are expected. If the pressure dominates then there is no collapse but if gravity dominates then the fluid collapse into bound structures. Fourier transforms are given by: A(t, x ) Ã(t, k ) d 3 k eik x Ã(t, k ) (2π) 3 d 3 xe ik x A(t, x ) The Dirac δ-function can be defined using the Fourier transforms: d 3 k A(t, x ) eik x Ã(t, k ) (2π) 3 d 3 k (2π) 3 eik x d 3 x d 3 x e ik x A(t, x ) d 3 k eik (x x ) (2π) 3 A(t, x ) d 3 x δ 3 (x x )A(t, x ) A(t, x ) The Fourier transforms of the δ-function is: δ 3 (x x ) 1 d 3 k (2π) 3 eik (x x ) d 3 xe ik x δ3 (x ) The Fourier transforms can be used to make the partial differential equations for the density contrast into an ordinary differential equation. This is done by finding what 2 δ is in Fourier space: 2 d 3 k δ(t, x ) [ (2π) 3 2 e ik x ] δ(t, k ) d 3 (2π) 3 eik x ( k2 )δ(t, k ) The density contrast equation can be written in Fourier space as: ϱ + (k 2 c 2 s 4πGϱ)δ 0 24

26 This is simply harmonic motion with frequency: ω k 2 c 2 s 4πGϱ When ω 2 > 0 then there are two oscillatory solutions δ 0 cos(ωt) and δ 1 sin(ωt). When ω 2 < 0 then the solutions are exponential, δ 0 e ωt and δ 1 e ωt. δ 1 decays away quickly but δ 0 leads to exponential collapse. When ω 0 then c 2 sk J 4πGϱ so: k J 1 c s 4πGϱ k is a frequency and so a wavelength can be defined using λ 2π/k: λ J 2πc s 4πGϱ c s π Gϱ This is the Jeans length. If a collection of particles are in an area of side length L then when L > λ J then there is collapse and when L < λ J then the system is stable Expanding Universe These structure formation conditions are only true for a static universe. In an expanding universe then the background velocity cannot be zero. Now: v H(t)x Here H(t) is some function of time which describes the expansion of the universe. Now the fluid equations can be found is exactly the same way but using: v + Ḣx Hẋ 0 since à à for any vector Ã(t, x ). To see this take: ẋ x A x à A y and x (xyz) A z Now carry out the calculation: à à A x x x A y y y A z z z A x A y A z à So if à ṽ it can be seen that ṽ x ṽ and as such: v dx dt x t + ṽ x (0) + ṽ ṽ 25

27 Now plugging the background velocity, ṽ into the the continuity equation gives: ϱ t + ϱ ṽ ϱ + ϱ (Hx ) ϱ + ϱh x ϱ + 3Hϱ The Euler equation is now: 0 ṽ t + ṽ ṽ + φ Ḣx + H2 x + φ 0 This can now be substituted into Poisson s equation: φ (Ḣ + H2 )x 2 φ 3(Ḣ + H2 ) 4πGϱ The rate of change of H is therefore: Ḣ + H 2 + 4πGϱ 0 3 Now looking at small fluctuations again ϱ(t, x ) ϱ(t)+δϱ where δ δϱ/ϱ with ϱ ϱ(1+δ) and ṽ ṽ +ũ where ũ is small and ṽ H(t)x. The potential is also φ φ + δφ. This reveals the equations: Continuity Equation δ + ũ + Hx δ 0 Euler Equation ũ + H(ũ + x ũ) c 2 sδ δφ Poisson s Equation 2 δφ 4πGϱδ Again going to Fourier space makes analysis simpler, but because everything has time dependence a new comoving coordinate, r is introduced such that: dr dt r t + ṽ r r t + Hx r 0 An ansatz for would be x /a(t) so that: r r dr ȧ ( xã ) dt a 2 x + Hx x ã 0 ( ) ȧ a + H This means that H ȧ/a. The Fourier transforms are made using r : d 3 k δ(t, r ) eik r δ(t, k ) (2π) 3 d 3 k (2π) 3 e(ik x )/a δ(t, k ) 26

28 Now introducing a new coordinate for conveinience, ϑ ũ/a the liearised equations are: Continuity Equation δ + ik ϑ 0 Euler Equation + ϑ 2Hϑ ik [ c 2 s δ δφ ] a2 Poisson s Equation k 2 δφ 4πGϱa 2 δ The ϱ variable can be eliminated by multiplying the Euler equation by and substituted into the partial ik derivative with respect to time of the continuity equation: ( δ + 2H δ c 2 + s k 2 ) 4πGϱ δ 0 a 2 If H 0 and a 1 then the static case is recovered. In general then this equation is a damped harmonic oscillator equation where 2H δ is the damping term. The frequency is: ω 2 c2 sk 2 { ω 2 > 0 Stable System, Oscillates a 2 4πGϱ ω 2 < 0 Instability, Collapses Setting ω 0 means a Jeans length can be defined using: 4πGa k J 2 ϱ c 2 s 2π λ J λ J c s π a Gϱ When L > λ J then there is collapse under gravity. This Jeans length is time dependent and so a state which is initially stable can eventually begin to collapse, depending on a. For the collapsing case where ω 2 < 0 then c 2 sk 2 /a 2 is negligible and so the equation is: δ + 2H δ 4πGϱδ 0 In matter domination then ϱ a 3, a t 2/3 and H 2/(3t) so that: δ + 4 3t δ 2 3t 2 δ 0 The solution to this equation is of the form δ δ 0 t n so using n(n 1)t n 2 + 4/(3n)t n 2 2/3t 2 0 gives 3n 2 + n 2 0 which can be solved and gives solutions n 1 and n 2/3. When n 1 then δ t 1 and so decays and is unimportant. When n 2/3 then δ t 2/3. During the matter era then fluctuations grow as t 2/3 a. The collapse has slowed down in an expanding universe but collapse still occurs. In radiation domination then the density fluctuations in matter is given by: δ + δ t 0 This is because 4πGϱδ H 2 and a t. A trivial solution to this is when δ is constant. Another solution is δ δ 0 + δ 1 ln t and so collapse happens but extremely slowly. This is called the Mészáros effect. It means that any structure formation in the universe must have happen during matter domination really. 27

29 3.3 Relativistic Perturbation Theory Newtonian perturbation theory is good for pressureless matter for scales much smaller than the cosmological horizon. This means that it can be used to calculate structure on sub-horizon scales. Relativity is needed to study the CMB, inflation and dark energy. To start working out the relativistic perturbation theory then the metric is needed: g µν a 2 (η) [ dη 2 + γ ij dx i dx j] Here γ ij are the background variables which satisfy γ aj γ jb δb a. The overline is use to denote that it is in conformal time, η. The metric can be placed into the Einstein equation: G µ ν 8πGT µ ν This then gives the two Friedmann equations where H a /a and a a/ η. Small fluctuations can be added to g µν to give: g µν g µν + δg µν ( 1 δ ij ) ( δg00 δg + 0i δg ij ) δg 00 is a scalar, δg 0i is a three-vector and δg ij is a tensor. This is called the 3+1 decomposition. In general relativity then the physics cannot depend on the coordinates system and so four components of δg µν can be fixed. The metric has 10 components since it is symmetric. These can be reduced by fixing some of the modes Scalar-Vector-Tensor Decomposition Since δg oi is a vector then in can be denoted v i. This can then be written in terms of a scalar mode and a vector mode: i i v + ˆv i i v is the scalar mode and ˆv i is the pure vector mode. The tensor δg ij can be decomposed into: dg ij hγ ij + ( i j 13 ) 2 γ ij A + i ˆfi + j ˆfi + h T ij γ ij are the background variables and h is a scalar mode and A is another scalar mode where the third comes from making this term cancel out when contracting with γ ij. The vector modes have the condition that i ˆf i 0. The vector mode has two degrees of freedom since one is removed by this condition. The scalar fields have one degree of freedom each. The tensor has six degrees of freedom but this can be reduced by making it traceless, γ ij h T ij 0 and stating that its divergence is zero, i h T ij 0 and so the degrees of freedom are reduced to two meaning δg ij has the expected six degrees of freedom. Tensor modes are pure gravitational waves. In total δg µν has four scalar modes: δ 00, δ 0i i B, δg ij hγ ij + ( i j 13 ) 2 γ ij A There are also four vectors and two tensors. Two of the scalars and two of the vectors can be removed by ( covariance and ) so that there are now six degrees of freedom. The scalars which are fixed are δ 0i and i j 1/3 2 γ ij A. The effect of the vectors is so small that they are negligible so they can be ignored and so there are now four degrees of freedom. The tensors cannot be removed. The remaining scalar modes are: ( ) 2ψ(t, x ) δgµν scalar 2φ(t, x )γ ij 28

30 Here φ and ψ are gravitational potentials. The fluctuations in the energy momentum tensor can also be found in terms of Scalar-Vector-Tensor decomposition and gives: ϱδ δt µ [ (ϱ p) i u ] ν 1 ϱ 3 Πδi j + (1 + w)di j σ (ϱ p) i u is the velocity decomposed into a scalar mode and a vector mode where only the scalar mode is considered in δt µ ν. Π is the pressure contrast given by δp ϱπ. σ is the shear and D i j i j 1/3 2 δ ij. The four scalars are therefore δ, u, Π and σ and none of these can be fixed Perturbative Einstein Equations The four perturbative Einstein equations (with ϱδ the sum of all species, i.e. baryons, cold dark matter (CDM), photons, γ and neutrinos, ν) are: δg 0 0 : 2 2 φ 6H(φ + Hψ) 8πGa 2 ϱδ δg 0 i : 2(φ + Hψ) 8πGa 2 (ϱ + p)u δg i j : φ + Hψ + 2Hφ + (2H + H 2 + 1/3 2 )ψ 1/3 2 φ φ ψ 4πGa 2 ϱπ 8πGa 2 (ϱ + p)σ i j i j These equations replace Poisson s equation in Newtonian gravity. They can be simplified slightly by substituting in three times the δg 0 i equation into the δg0 0 term to get: 2 φ 4πGa 2 ϱ [δ + 3H(1 + w)u] This is very similar to the Newtonian gravity Poisson s equation but it is also sourced by the velocity. Π 0 for pressureless matter such as baryons, dark matter or cosmological constant and Π 1/3δ for radiation like photons or neutrinos. The shear σ 0 for baryons, cold dark matter or cosmological constant but is not for photons and neutrinos. It is still very small and so when ignoring it then φ ψ. The conservation of mass from Newtonian perturbation theory now becomes µ T µ ν. When ν 0 then the relativistic analogue of the continuity equation is: δ 3H(wδ Π) + (1 + w)( 2 u + 3φ ) Taking ν i then the relativistic analogue of Euler equation is: u H(1 3w)u w σ + ψ There is additional sourcing from the shear and time dependence of the potentials. Solutions can be found for the potential and the matter variables. Assuming that there is only one fluid then the Friedmann equation in conformal times is: 3H 2 8πGa 2 ϱ δ can be eliminated using the δg 0 0 equation and the δg i i equation to get: φ + 3H(1 + w)φ + wk 2 φ 0 This is valid for a single fluid with parameter w 0 on all scales. Starting with a given mode with wavenumber k then it can be determined whether it is a superhorizon or subhorizon mode. The horizon is given by 1/H and so a superhorizon mode occurs when k H and so the wavelength is the much greater than the length of the universe and subhorizon mode occurs when k H such that the wavelength is much less than the length of the universe. This mode is time dependent in an expanding universe and so at early times the mode may be superhorizon and become 29

31 subhorizon after H k. For superhorizon scales then the k 2 term is negligible so that: φ + 3H(1 + w)φ 0 There are two solutions, φ constant and φ A/a 3(1+w) dη. This can be integrated to get φ a. 3(1+w) This is a decaying mode and so only the φ constant needs to be considered. This means on superhorizon scales φ sup φ sup (k ). The superhorizon scale total density becomes: Substituting in φ sup gives: 6H(φ + Hφ) 8πGa 2 ϱδ 6H 2 φ sup 8πG2 ϱ 3H 2 δ Therefore δ sup 2φ sup. The density contrast is constant on superhorizon scales. The total velocity on superhorizon scales is the same and so after substituting in φ sup and δ sup then: Since H 2/(1 + 3w)η then: ũ sup ũ sup 2 φ sup 3(1 + w) H 1 + 3w 3(1 + w) φ supη These results give the conditions of curvature and adiabatic initial conditions. When η 0 then u sup 0 so all the particles at the start of the universe are comoving with the expanding universe. The fluctuations are regular as η 0. If w 0 (pressureless matter domination) then the solutions are the same as the superhorizon calculations even on sub horizon scales. On small scales (subhorizon) then the Newtonian calculations are good. During the matter era then φ constant so that: δ k 2 ũ and ũ Hũ + φ These are exactly the Newtonian equations of motion. For two fluids, radiation and matter, then the total density, total velocity and total pressure become: 2 2 φ 6H(φ + Hψ) 8πGa 2 [ϱ r δ r + ϱ m δ m ] 3H 2 (Ω r δ r + Ω m δ m ) 2(φ + Hψ) 8πGa 2 (ϱ r + ϱ m + p r + p m )u 3H 2 ( 4 3 Ω rδ r + Ω m δ m ) φ + Hψ + 2Hφ + (2H + H 2 + 1/3 2 )ψ 1/3 2 φ 4πGa 2 ϱ r δ r Π Only one fluid dominates at a given time and so the solution for φ for a single fluid is still valid. The adiabatic condition states that δ m 3δ r /4. In the radiation era then δ r 2φ sup and so δ m 3φ sup /2 and in the matter era then δ m 2φ sup and so δ r 8φ sup /3. Superhorizon Superhorizon Subhorizon Subhorizon Radiation Era Matter Era Radiation Era Matter Era φ constant constant oscillatory decay constant δ m constant constant constant+log term grow η 2 ũ m grow grow η decay grow η 30

32 φ δ m η hor1 k 1k 2 η hor2 k 2 k 1 η hor1 η hor2 Equality η This is only true for truly pressureless matter. The universe contains baryons, 75% H and 25% He. At high temperatures, i.e. in the early universe, then these are ionised and so are charged. This means that the baryons interact with photons via Compton scattering on the scale 1/m 2 p as do the electrons with scale 1/m 2 e. Baryons also interact with electrons via Rutherford scattering. This is called the tight-coupling regime. When the temperature drops then neutral atoms begin to form which do not interact with photons and so the Compton scattering switches off and so photons can travel without much interaction. This is called the free-streaming regime. In tight coupling then baryons have pressure because they are just a constituent of a photon-baryon fluid. Since δp c 2 sδϱ then there is a speed of sound c 2 s 0. In free-streaming then the baryons decouple and so there are two separated fluids and the speed of sound c 2 s 0. This means that the baryons become pressureless. The Jeans scale is: λ J c s π a Gϱ b (a) During tight-coupling then λ J 1/H and so is of the horizon scale. The baryons form waves and cannot form structures. During free-streaming then the Jeans length λ J 0 since c s 0 and as such structures can form. Where the horizon crossing occurs sets the phase of the oscillations: 31

33 δ b η hor1 η hor2 Equality η η dec Oscillations should be imprinted in the galaxy distribution but this is not seen on the predicted scale. This is because in the real universe there is cold dark matter which is completely pressureless since it does not interact electromagnetically. δ b Cold Dark Matter Baryons Equality η η dec The baryons fall in to potential wells, φ and so the oscillation pattern is mostly erased. Of course dark matter feels some back reaction so there can still be seen some small oscillations Observing Structure Using the galaxy distribution across the sky can give an idea of the structure. Galaxies in a volume formed of the some box of area da and some thickness given by the redshift, z, can 32

34 be counted to find a number density contrast: δ n (z, ˆn) N(z, ˆn) N(z) N(z) ˆn is the direction of of the galaxies, N(z, ˆn) is the number of galaxies in that direction and N(z) is the mean number of galaxies at redshift z. The galaxy mass density contrast in this region is given by δ s (z, ˆn) b 1 δ n where b 1 is a phenomenological number which depends on the type of galaxy being observed. What is really needed is δ g (η, r, ˆn) where r is the position of the galaxies. Since galaxies are in the past lightcone then η r. Now as the galaxies are being observed by redshift then a redshift space distortion needs to be taken into account. Galaxies have velocities which cause doppler shift. This means that the galaxies which are moving further away have a greater redshift and as such are closer than they appear and galaxies moving towards are blue shifted and so are really further away than they appear. The redshift is therefore given by: z z + δz H 0 r + ṽ ˆn H 0 Here s is the redshift vector: s r + ṽ ˆn H 0 As a vector then: ( s 1 + ṽ ˆn ) r H 0 r The volume should be the same whether in redshift space or normal space: What is found is that: n s d 3 s n r d 3 r [ n s n r 1 1 ] (ṽ ˆn) H 0 r The last term is small. The number density is therefore n r (r ) n(z) ( 1 + δg(r ) ) and so the mass density contrast of the galaxies is: δ s δ g (r ) 1 ( ) ṽ ˆn H 0 r Fourier transforming and looking in k-space gives: δ s (k ) δ g (k ) + µ2 H 0 k 2 v(k ) Where µ ˆk ẑ. The velocities appear to add to the density of the galaxies. Defining δ k 2 v allows a growth rate to be written: f d ln δ d ln a Now δ s b(1 + µ 2 β)δ(k ) where β f/b is the bias which is included since galaxies do not trace the density field exactly. The density contrast is then given by: δ(k ) δ 0 (k 0)T (k ) Here δ 0 (k ) is an initial condition and T (k ) is a transfer function which describes how to get from the initial conditions to todays density contrast. The transfer function contains the density of baryons and cold dark matter: δ(k ) δ 0 (k ) ( Ω b δ b (k ) + Ω CDM δ CDM (k ) ) δ(k ) is a random variable and so δ(k ) 0. The power spectrum is what is actually observed: P (k ) δ(k )δ(k ) P 0 (k )T 2 (k ) 33

35 Here P 0 (k ) is the initial power. The power spectrum is sketched as: P (k ) Subhorizon Superhorizon Baryon Oscillations k 3.4 Cosmic Microwave Background The cosmic microwave background is made of photons and was first observed (accidentally) by Penzias and Wilson for which they won a Nobel prize. The CMB has a current temperature T K as measured by the COBE satellite in There are s 1 cm 2 photons in the CMB. It is almost isotropic since it is the same temperature in all directions. The anisotropies are very small with a dipole on the scale of 10 3 and higher multipoles The CMB has a Planck spectrum which implies equislibrium. The intensity is: 4πν 3 I(ν) e 2πν/T 1 The power per unit area can be calculated: P σt 4 σ is a constant. The CMB is the best example of a Plank spectrum in the universe. The fact that the photons are in equilibrium means that they can be analysed using statistical mechanics. The Bose-Einstein distribution is a function of phase space F (t, x, k ). Since the universe is isotropic and homogeneous then the background is: 2 f(t, k ) ) (2π) (e 3 k /T 1 The energy-momentum tensor can be calculated using: T µ ν d 3 kf(t, x, k ) kµ k ν E For the Friedmann-Robertson-Walker universe then this is: T µ ν 2 k µ k ν 1 (2π) 3 E e k /T 1 34

36 The energy density is given by ϱ γ T 0 0: ϱ γ 2 (2π) 3 d 3 k k e k /T 1 8π (2π) 3 0 k 3 dk e k /T 1 T 4 π 2 0 π2 15 T 4 π2 T a 4 x3 dx e x 1 The pressure is obtained from T i j and is: pγ 1 3 ϱ γ Energy conservation is given by µ T µ ν 0 and in FRW is given by ϱ γ + 4Hϱ γ 0. This can be seen to be the result of the Boltzmann equation, which also gives the evolution of the temperature. For FRW then the Boltzmann equation is: So the partial derivatives are: f t Hk f k 0 f t e k/t k T (e k/t 1) 2 T 2 t f k ek/t 1 (e k/t 1) 2 T Substituting these into the Boltzmann equation gives: This implies that T T 0 /a Anisotropies T t + HT 0 The dipole is apparent on a scale of 10 3 and corresponds to v (627 ± 22)km s 1 which is the velocity the Earth is moving with respect to the CMB. Removing this dipole allows the higher multipole anisotropies to be seen. These multipoles are seen by observing in direction ˆn and then constructing T (ˆn) for all directions. The average is then T T (ˆn) and the temperature anisotropy is then given by: Θ(ˆn) T (ˆn) T T Θ(ˆn) is position dependent so Θ(ˆn) 0. Spherical harmonics can be used to describe the temperature anisotropy: Θ(ˆn) a lm Y lm (ˆn) lm Where Y lm (ˆn) is the spherical harmonic and l 0, 1, 2, 3, and m l,, l. a lm can be found using: a lm dωˆn Ylm(ˆn)Θ(ˆn) The spherical harmonic satisfies: [ 1 sin ϑ ϑ ( sin ϑ ) + 1 ϑ 2 sin 2 ϑ φ 2 35 ] Y lm l(l + 1)Y lm

37 Legendre polynomials can be related to the spherical harmonics. Legendre polynomials are P l (µ) where µ 1,, 1 and satisfy: (1 µ 2 ) d2 P l dµ 2 2µdP l dµ + l(l + 1)P l 0 Since for two directions ˆn and hatn then ˆn ˆn cos ϑ then it can be seen that: P l (cos ϑ) 4π Y 2l + 1 lm(ˆn)y lm (ˆn) For the CMB then Θ(ˆn) lm a lmy lm (ˆn) is used and the average between Θ(ˆn) and Θ(ˆn ) is taken: Θ(ˆn)Θ(ˆn ) C(Θ) lm m l m Y lm (ˆn)Y l m (ˆn ) a lm a l m If the universe is statistically isotropic then a lm a lm C lδ ll δ mm and so the correlator becomes: C(Θ) lm C l Y lm (ˆn)Y lm(ˆn ) 1 4π (2l + 1)C l l C l is the angular power spectrum where C 0 0 is the monopole, when l 1 then C 1 is the dipoles and then the anisotropies are given by the l 2 terms. Each l has 2l + 1 samples. 3.5 Recombination In tight-coupling then there is thermal equilibrium so that e + p H + γ. This means that: n e n p n H n(0) e n (0) p n (0) H Here n e and n p are the free electrons and protons and n (0) i n (0) i g i ( mi T 2π The Boltzmann equation for massive particles is: ( 1 d a 3 dt (a3 n e ) n (0) e n p (0) σv are the equilibrium number density: ) 3/2 e mi/t n H n (0) H This is zero at equilibrium. The electron ionisation fraction is: n en p n (0) e n (0) p ) X e n e n e + n H n p n p + n H X e 1 for the completely ionised case and X e 0 for the completely neutral case. When including the He then X e 1.15 when completely ionised. Using this it can be seen that: n e n p n H n2 e n H ( me T 2π ) 3/2 e (me+mp m H)/T 36

38 Here m e + m p m H is the binding energy ε H 13.6eV. By measurement n e /n H X e /(1 X e ) which reveals the Saha equation: Xe 2 ( ) 3/2 1 me T e ε H/T 1 X e n e + n H 2π n e + n H n p + n H is the baryon density, n b 10 9 T 3. Using this it can seen that Xe 2 /(1 X e ) diverges so X e 1. Beyond equilibrium then the Boltzmann equation needs to be solved: [ ( ) ] 3/2 1 d me T a 3 dt (a3 n e ) n b σv (1 X e ) e εh/t Xe 2 n b 2π Then use n e X e n b to get: β is the ionisation rate: dx e dt (1 X e )β X 2 e n b α (2) ( ) 3/2 me T β σv e ε H/T 2π α (2) is called the recombination rate which describes how electrons and protons form a neutral hydrogen: X e With Helium Free Electrons and Protons From Hydrogen Decoupling a Decoupling Decoupling occurs when n e σ T H. The number density of baryons is: n b ϱ b m b This means: n e σ T H 3H2 0 Ω 0b m b a 3 The Friedmann equation gives: H H 0 Ω0m a 3 σ 0Ω 0b h 2 X e a a eq a 37