Structure formation. Yvonne Y. Y. Wong Max-Planck-Institut für Physik, München

Transcription

1 Structure formation Yvonne Y. Y. Wong Max-Planck-Institut für Physik, München

2 Structure formation... Random density fluctuations, grow via gravitational instability galaxies, clusters, etc. Initial perturbations (could have) come from quantum fluctuations during inflation.

3 Computational tools: N-body simulations: Represent matter content with a sample of massive particles and watch them move under the influence of gravity. Used in the study of small-scale structures, e.g., dark matter halo profile. Perturbation theory: On scales of O(100) Mpc, the universe is almost homogeneous. Small fluctuations can be tracked with perturbation theory. Used in large-scale structure calculations.

4 Plan... Newtonian dynamics in cosmology. Fluid dynamics. Structure formation in: A CDM-dominated universe. A HDM-dominated universe. A mixed CHDM universe.

5 Useful references... E. Bertschinger, Cosmological dynamics, astro-ph/ F. R. Bouchet, Introductory overview of Eulerian and Lagrangian perturbation theories, astro-ph/ J. Binney and S. Tremaine, Galactic dynamics, Princeton University Press, Princeton NJ, S. Dodelson, Modern cosmology, Academic Press, San Diego CA, 2003.

6 Newtonian dynamics in cosmology... On subhorizon scales and for nonrelativistic matter, Newtonian gravity suffices. Newton's law: d 2 r i d t 2 = g i ; g i = j 1 G m j r i r j In the continuum limit r i r j 3 d 3 r G r ', t r r ' r r ' 3 Define Newtonian potential: g r ρ = Density r r 2 r =4 G Poisson equation

7 Comoving coordinates... Factor out the expansion. Comoving spatial coordinates: Conformal time: x= r a t t = 0 dt ' a t Comoving (or peculiar) velocity: or v d x d = d r dt 1 a d = dt a t d a dt r = H(t), Hubble expansion What we measure

8 Newton's law in comoving coordinates... Define: x, = x, Homogeneous Perturbation Then Newton's law becomes: d 2 x d ȧ 2 a d x d x d ȧ a d = ' with 2 =4 G a 2 2 '=4 G a 2 Note definitions / ; / x

9 First term on the RHS: Since 2 =4 G a 2 = 4 3 G a2 x Last term on the LHS: From Friedmann equation: x d d 1 a d a d = 4 3 G a2 x Finally ' Damping term due to expansion d 2 x d ȧ d x 2 a d = 2 =4 G a 2 x, Potential due to perturbations only

10 Eulerian fluid dynamics... Consider a nonrelativistic collisional gas (i.e., rapid interactions establish local thermal equilibrium). Fluid equations: Continuity Euler 3 ȧ v =0 a v v v ȧ a v= 1 P Pressure ; x = x, v= v x, = x, P=P x,

11 Define x, 1 x,, i.e., factor out the mean behaviour. Also use = 0 a 3 Then: 3 ȧ a v=0 Continuity equation (alternative notation) [ 1 v ]=0

12 For large-scale structure, we are interested in small amplitude fluctuations. Linearise fluid equations: [ 1 v]=0 v v v ȧ a v= 1 P v 0 v ȧ a v 1 P Keep terms up to first order in small quantities δ, v, φ, P.

13 For a collisional gas, we have an equation of state: P=P, S S = Specific entropy e.g., an ideal, nonrelativistic, monoatomic gas: T = Temperature T ds =d 3 2 P P d 1 Hence, the pressure term in the Euler equation: 1 P=c 2 s 2 3 T S ; c 2 s 5 3 P c s = Sound speed

14 Finally, the full set of linearised fluid equations: Continuity equation: v=0 Euler equation: v ȧ a v= c 2 s 2 3 T S Poisson equation: 2 =4 G a 2

15 Isentropic fluctuations... No entropy gradient: S=0 Natural outcome of inflation followed by thermalisation. Euler v ȧ a v= c 2 s 2 3 T S = 0 Combine Continuity, Euler, and Poisson into a 2 nd order DE: ȧ a =4 G a 2 c s 2 2 Driven, damped oscillator Damping Driving force

16 ȧ a =4 G a 2 c s 2 2 Perform Fourier Transform: x, = d 3 k k, exp i k x ȧ a =4 G a 2 cs 2 k 2 k J 2 k 2 c s2 where k J 4 G a 2 c s 2 k J = Comoving Jeans wavenumber

17 In the large k limit k k J ȧ a k 2 k J 2 c s 2 =0 Positive δ Damped oscillations: Density fluctuations do not grow.

18 In the small k limit k k J ȧ a Negative k 2 k J 2 c s 2 =0 Neglect for now the damping term: 2 =0; 2 k J 2 k 2 c s 2 Solution: = A exp B exp A exp Decays, 0 Fluctuations grow! Including damping term generally causes growth to be slower than exponential (example later).

19 Physically, we have a competition between gravity and pressure. k k J Large k: damped oscillations k k J Small k: density perturbations grow Pressure wins Gravity wins Define two time-scales: Gravitational time-scale: Sound-crossing time-scale: G ~ 4 G a 2 1/ 2 S ~ k c s 1 Jeans wavenumber is defined as k when: G = S

20 Cold dark matter... Pressureless, collisionless, i.e., no interaction to establish local thermal equilibrium, but... Perturbation evolution can be described by fluid equations with: c s 2 =0 ȧ a =4 G a 2 Same for all values of k! i.e., no acoustic oscillations for CDM.

21 Assume flat, CDM-dominated universe. Using: a t 2/ 3 2 ; 4 G a 2 = 3 ȧ 2 2 a We get: 2 = 6 2 Meszaros equation Solutions: Growing Decaying a In a CDM-dominated universe, perturbations grow like the scale factor for all k.

22 Multi-component fluid... Only the Poisson equation requires modification: Hence: 2 =4 G a 2 i i i x, Sum over all clustering fluids i ȧ a i =4 G a 2 j j j c s 2 k 2 i

23 Hot dark matter... Collisionless, and with large velocity dispersion. This means: Shear, i.e., nonzero off-diagonal entries in the stress-energy tensor. ρ and P alone not enough to describe this! Need to consider evolution of the phase space density, f x, p,

24 Use the Vlasov equation (or collisionless Boltzmann equation): D/Dt = Liouville operator D f D f d x d f x d p d f p =0 Note definitions: p=a m x x, = g 2 3 m a 3 d 3 p f p e.g, relativistic Fermi-Dirac distribution: 1 f FD p = f FD p = 1 exp p/t 0 T 0 = Present day temperature

25 Substitute: d x d = p a m ; d p d =am d 2 x d ȧ a d x = a m d f d x d f x d p d f p =0 We get: Vlasov f p am f x a m f p =0 Poisson 2 =4 G a 2 i i i x,

26 We are interested in small perturbations. Define: f x, p, = f 0 p f 1 x, p, Linearised Vlasov equation: f 1 p a m f 1 x a m f 0 p 0 Transform to super-comformal time, s= d /a ; ds=d /a f 1 s p m f 1 x a2 m f 0 p 0

27 To solve, perform Fourier Transform: f 1 x, p, s f 1 k, p, s ; x i k ; x, s k, s f 1 s i k p m f 1 i ma 2 k f 0 p 0 Solution: = 0 if initial phase space is unperturbed f 1 k, p,s = f 1 k, p,s ini e i k. p s s ini /m m i k f s 0 p s ini ds' a 2 s' k, s e i k p s s ' / m

28 We are interested in the total perturbation in the neutrino gas: k, s k, s = m a 3 d 3 p f 1 k, p, s = 1 d 3 p s ma 3 d 3 p f 0 p n f 1 k, p, s 0 s k,s = k 2 s ini d s' a 2 s' k, s' s s' F[ k s s' m ] where F q 1 n 0 d 3 p e i p q f 0 p Relativistic Fermi-Dirac F q = 4 1 n 1 n 3 3 n=1 n 2 q 2 T Maxwell- Boltzmann f 0 p =exp p /T 0 1 F q = 1 q 2 T 2 0 2

29 s k, s = k 2 s ini d s' a 2 s' k, s' s s' F[ k s s' m ] where F q 1 n 0 d 3 p e i p q f 0 p But what does this tell us? Can we rewrite this in a way comparable to the fluid equations?

30 Yep. Here's the trick: Rewrite: s k,s = m k s ini ds' a 2 s' k, s' [q F q ]; q k s s' m Differentiate wrt s: s s k, s = k ds' a 2 s' k, s' d [q F q ] s ini dq Differentiate wrt s again: 2 s k, s = k 2 a 2 s k, s k 3 2 m s ini s ds' a 2 s' k, s' d 2 [q F q ] 2 dq Note that F q=0 =1

31 Observe that: If d 2 qf d q 2 = A d qf dq B qf 2 s =C then 2 s D This is true if: F q =exp q T 0 Recall: F q 1 n 0 d 3 p e i p q f 0 p f 0 p = n p2 2 T 0 2 T 0 Unperturbed phase space density

32 So we have: F q =exp q T 0 2 s 2 T k 0 2 m d 2 d q 2 qf = 2 T 0 s T k 0 m 2 = k 2 a 2 s k, s d dq qf T 0 2 qf Transforming back to conformal time (s τ) and using the Poisson equation, we get: ȧ a 2 k c k 2 c 2 =4 G a 2 i i i ; c T 0 m a cf fluid equation i ȧ a i k 2 c s2 i =4 G a 2 j j j Neutrino thermal velocity

33 Suppose we have only hot dark matter: ȧ a 2 k c k 2 c 2 4 G a 2 =0 Solution in the small k limit: Third term: Second term: k 2 c 2 4 G a 2 k c 1 ȧ 2 a = G a2 cf Jeans criterion ȧ a 4 G a 2 0 At small k, HDM evolve exactly like CDM.

34 Solution in the large k limit. Are there acoustic oscillations? k 2 c 2 4 G a 2 k c 1 ȧ 2 a = G a2 2k c k 2 c 2 0 cf fluid equation i ȧ a i k 2 c s 2 i 0 Suppose c ν is constant. Then:, 1 exp k c, 2 exp k c Both decay At large k, HDM cannot support acoustic oscillations. Perturbations do not grow either.

35 What does this mean physically? Gravitational time-scale: Free-streaming time-scale: G ~ 4 G a 2 1/ 2 FS ~ k c 1 If FS G k k FS or What is k FS? 4 G a 2 2 c Structures cannot form c ~ T 0 ma =50 1 z ev km s 1 m FS 2 /2 =0.4 1 /2 1 z 1 ev k FS m h 1 Mpc c c Using, we get: 2 FS, max =32 1/ ev m 1 /2 h 1 Mpc In pure HDM cosmology, there are no structures below λ FS,max.

36 Mixed cold and hot dark matter... Small k limit: CDM HDM c ȧ a ȧ a c 4 G a 2 c c 0 4 G a 2 c c 0 CDM and HDM evolve in the same way if =0 = c =0 Expected from inflation Flat, matter-dominated universe: = c a

37 Large k limit: HDM ȧ a 2k c k 2 2 c 4 G a 2 c c =0 For k c 1, we have a steady-state solution: 1 f k 2 /k 2 f k FS c FS k 2 2 c k 2 FS 4 G a 2 2 m /c f / m m c If f ν << 1 HDM perturbations at large k can grow if there are growing CDM perturbations, but are always somewhat suppressed.

38 What about the CDM perturbations in the large k limit? c ȧ a c 4 G a 2 c c 0 Highly suppressed Recall the Meszaros equation for flat, CDM-dominated universe: c 2 c 6 2 c =0 c 2 a For mixed CHDM at large k: c 2 c f c =0 c f 1 /2 e.g., f =0.1 c 1.88 The presence of HDM causes CDM perturbations at large k to grow at a slower rate.

39 δ cdm Initial time... δ cdm Some time later... CDM-dominated CHDM k k The presence of HDM suppresses the growth of CDM perturbations at large k.

40 Back to cold dark matter... We have been using collisional fluid equations to describe CDM. Can this be justified from the collisionless Boltzmann equation? Vlasov Continuity Euler f p am f x?? a m f p =0 3 ȧ a v=0 v v v ȧ a v= Yes, by taking velocity moments of the Vlasov equation.

41 Take zeroth moment by integrating over phase space: m a 3 d 3 p f ma 3 d 3 p p am f x ma 3 d 3 p a m f p =0 First term: m a 3 d 3 p f = 3 ȧ a Second term: m a 3 Third term: x d 3 p v f = v ; v d 3 p v f d 3 p f ; p=a m v d 3 p p f = f d 2 S 0 Since f(p) 0 for large p values 3 ȧ a v =0 Zeroth moment Continuity

42 First moment: m a 3 d 3 p v j f m a 3 d 3 p v j First term: m a 3 Second term: p i a m d 3 p v i f = v i v i 4 ȧ a v i f ma 3 d 3 f p a m v x j =0 i x i p i m a 3 Third term: d 3 p v x i v j f = v i x i v j ; i v i v j d 3 p v i v j f d 3 p f Use divergence thereom and then integration by parts: d 3 p p j f p i = d 3 p p j p i f = ij a 3 / m ma 3 d 3 p p j x i f p i = x j

43 Collect all terms: v j 3 ȧ a v i x v j v i v j i x i ȧ a v j v i v j =0 x i x j = 0 from the continuity equation Define stress tensor: 2 ij v i v i v j v j = v i v j v i v j v j v v j i ȧ x i a v j 1 x j First moment (Almost) Euler ij x i =0 Stress tensor has off-diagonal entries But CDM has negligible velocity dispersion, i.e., σ ij = 0. Pressureless fluid description is good.

44 In general, there is an infinite number of moments and just as many differential equations to describe their evolution: Boltzmann hierarchy. The integrated Vlasov equation for the ith moment generally depends on the (i+1)th moment. To truncate the hierarchy, we need: Relation between the highest and some lower moments, e.g., equation of state for a perfect fluid. and/or Higher order dependence on a small quantity, e.g., CDM velocity.