Infrared Effects at the BAO Scale

Transcription

1 Infrared Effects at the BAO Scale Gabriele Trevisan CosPA 2017 Senatore, Trevisan: arxiv , Scoccimarro, Trevisan: arxiv xxxx.xxxx

2 BAO in pictures Eisenstein et al

3 BAO in pictures Eisenstein et al

4 R = 150 Mpc

5 BAO in the sky Planck 2015 r drag = ± 0.49 Mpc

6 BAO in the sky Planck 2015 Anderson et al r drag = ± 0.49 Mpc D V (z =0.57) = 2056 ± 20 Mpc

7 Standard Perturbation Theory (SPT) ˆ + Ò [(1 + ) v] =0, Fluid Equations: ˆ v + H v + v Ò v + Ò =0 = 3 2 H2 m q n Perturbative solution: n ( k) k F n F n n linear( q) q 1

8 SPT 2-Point Function P (k) tree level = P lin P(k) [Mpc/h] tree-level k [h/mpc] Where is the BAO in Fourier space? The BAO signal is ~5% oscillation with freq. 1/150 Mpc

9 SPT 2-Point Function P lin P 22 P 13 P lin P (k) 1 loop = P lin P lin F 2 F 2 P lin F 3 P lin P(k) [Mpc/h] loop tree-level k [h/mpc] Perturbation Theory should recover the BAO, so the more loops the better, right?

10 SPT 2-Point Function SPT completely fails around the BAO scale r 2 ξ [Mpc/h] tree-level r [Mpc/h]

11 SPT 2-Point Function SPT completely fails around the BAO scale r 2 ξ [Mpc/h] loop tree-level r [Mpc/h]

12 SPT 2-Point Function SPT completely fails around the BAO scale r 2 ξ [Mpc/h] loop 1-loop tree-level r [Mpc/h]

13 Why SPT fails? The wiggly component (BAO) receive large infrared (IR-enhanced) contribution from loop integrals P w 1 loop(k) k 2 P w lin(k) = P w lin(k) s<. 1 BAO.p.k d3 p P lin (p) (2fi) 3 p 2

14 Why SPT fails? The wiggly component (BAO) receive large infrared (IR-enhanced) contribution from loop integrals P w 1 loop(k) k 2 P w lin(k) = P w lin(k) s<. 1 BAO.p.k d3 p P lin (p) (2fi) 3 p 2 and s< 1 Bad for doing PT! Better not to expand

15 To fix just resum Lagrangian PT (à la Zeldovich) short modes should not be resummed, calculations are more cumbersome, especially EFT, Fourier space is numerically more challenging IR-resummation (à la EPT) Senatore, Zaldarriaga 15 numerically more demanding Consistency relations (using EP) Baldauf et. al 15 split into smooth and wiggly components, NLO corrections Time-Sliced PT (à la QFT) Blas et al. 15 need to split into smooth and wiggly components manually check the IR-enhanced contribution resummation relies on separation of scales calculations are more cumbersome, especially NLO and UV

16 IR-resummation The idea is to resum IR displacement modes K 0 (k, q) exp k ik j A IR ij (q)+ i 6 k ik j k k B IR ijk(q) 6

17 IR-resummation The idea is to resum IR displacement modes K 0 (k, q) exp k ik j A IR ij (q)+ i 6 k ik j k k B IR ijk(q) 6 ~ s<

18 IR-resummation The idea is to resum IR displacement modes K 0 (k, q) exp k ik j A IR ij (q)+ i 6 k ik j k k B IR ijk(q) 6 ~ s<

19 IR-resummation The idea is to resum IR displacement modes K 0 (k, q) exp k ik j A IR (q)+ i ij 6 k ik j k k Bijk(q) IR 6 ~ s< For example at tree-level tree (r) = 1 A ij (r) 1/2 d 3 q E tree(q) exp (r q) ia 1 ij (r)(r q) j. ressions in Eqs. (31) and (33), is a -dependent Guassian-like kernel

20 IR-resummation with NLO terms r 2 ξ [Mpc/h] ξ linear ξ ZA 0 ξ 0-loop ξ 1-loop ξ 0-loop cubic ξ 1-loop cubic ξ 2-loop ξ 2-loop cubic r [Mpc/h] Senatore, Trevisan: arxiv

21 IR-resummation with NLO terms r 2 ξ [Mpc/h] ξ linear ξ 0-loop ξ 1-loop ξ ZA ξ 0-loop cubic ξ 1-loop cubic (ξ-ξ cubic )/ξ cubic (100) loop 1-loop ξ 2-loop ξ 2-loop cubic r [Mpc/h] 2-loop r [Mpc/h] Senatore, Trevisan: arxiv

22

23 Detection of the BAO in the 3-PF Gaztanaga et. al

24 Detection of the BAO in the 3-PF Gaztanaga et. al Slepian et al BAO scsle fittitg : Mitimsl mbdel Dsts; BAO mbdel Dsts; tb wiggle mbdel σ χ σ nσ α

25 Detection of the BAO in the 3-PF Gaztanaga et. al Slepian et al BAO scsle fittitg : Mitimsl mbdel Dsts; BAO mbdel Dsts; tb wiggle mbdel σ χ σ nσ α Analyses use P phys (k) =[P (k) P nw (k)] exp k 2 2 nl/2 + P nw (k), with P the linear theory power spectrum. P plugged into the tree-level SPT 3-PF is the n

26 SPT predictions for the 3-PF 1.5 Equilateral triangles ζ [Equilateral] ζ tree-level l [Mpc/h]

27 SPT predictions for the 3-PF B211 B411 B222 F2 104 II B321»»B411» I B Mpc6 D 10-6 Mpc6 D» 1000 B222 Bloop 100 B F D I B321 F3 F2 II B321 Btree» 1000 F B222»B411» B222 Figure 3: Tree-level and one loop-bispectra. F2 + 5 perm., F2 F4 II B321 = 6F2 (k2, k3 ) (k2 ) (k3 ) 1 Z PFlin 3 (k3, q, F2 q) (q) + 5 perm. q = F2 (k2, k3 ) (k2 )P13 (k3 ) + 5 perm., Z B411 F4 (q, q, k2, k3 ) (q) + 2 cyc. per = 12 (k2 ) (k3 ) q 0.20 II F3 the one-loop contribution to the power spectrum st Note that B321 reduces to the correlator h (3) P(1)lini, i.e. P13. Again, these integrals -1 can be divergent just of the one-loop power spectrum. An important part Dof this paper is dedic F2 that these divergences can be cancelled. In sum, the SPT bispectrum at the reads 100 I II BSPT (k1, k2, k3 ) = B112 + B222 + B321 + B321 + B II B321 1 B411 If properly the integrals in Eqs. (2.26), (2.27), (2.28) and (2.29 Figure 3: Tree-levelregularized, and one loop-bispectra. uated analytically for a power-law linear power spectrum (k) / k n in in Ref. [22]. For a more realistic CDM universe, these integrals have to + 5 perm., numerically since we do not have an analytic form (2.27) of the linear power spe Z Also, in this case we do not encounter formally divergent i present epoch. = 6F2 (k2, k3 ) modes (k2 )entering (k3 ) Fthe, q, q) P 5 perm.domination, 3 (k3horizon lin (q) + during radiation are suppressed. q = F2 (k2, k3 ) (k2 )P13 (k3 ) + 5 perm., 3 Z = 12 (k2 ) (k3 ) F4 (q, q, k2, k3 ) (q) + 2 cyc. perm. 12 (2.28) (2.29) q 0.1 G. Trevisan - NYU B F2 B411» F3 II B F2 321 II B Mpc6 D I B321 Btree 0.1 F4 Baldauf et.al F2 Btree B222 II Note that B321 reduces to the one-loop contribution to the power spectrum stemming from the correlator h (3) (1) i, i.e. P13. Again, these integrals can be divergent just as in the case I the one-loop power spectrum. An important part of this paper is dedicated to prove Bof 321 that these divergences can be cancelled. In sum, the SPT bispectrum at the one-loop level CosPA 2017

28 SPT predictions for the 3-PF B211 B411 B222 F2 F4 Baldauf et.al »»B411» Bloop 100 B F D Mpc6 D 10-6 Mpc6 D II B321 I B321 B ζ [Equilateral] 10-6 Mpc6 D» I B F Btree B222 F2 Btree F3» 100 B222»B411» B F2 B II B F2 321» F3 B222 Figure 3: Tree-level and one loop-bispectra. F2 + 5 perm., F2 F4 II B321 = 6F2 (k2, k3 ) (k2 ) (k3 ) 1 Z PFlin 3 (k3, q, F2 q) (q) + 5 perm. q = F2 (k2, k3 ) (k2 )P13 (k3 ) + 5 perm., Z B411 F4 (q, q, k2, k3 ) (q) + 2 cyc. per = 12 (k2 ) (k3 ) q 0.20 II F3 the one-loop contribution to the power spectrum st Note that B321 reduces to the correlator h (3) P(1)lini, i.e. P13. Again, these integrals -1 can be divergent just of the one-loop power spectrum. An important part Dof this paper is dedic F2 that these divergences can be cancelled. In sum, the SPT bispectrum at the reads I II BSPT (k1, k2, k3 ) = B112 + B222 + B321 + B321 + B B411 G. Trevisan - NYU II B321 1 II B321 II B I B321 Btree F2 If properly the integrals in Eqs. (2.26), (2.27), (2.28) and (2.29 Figure 3: Tree-levelregularized, and one loop-bispectra. uated analytically for a power-law linear power spectrum (k) / k n in in Ref. [22]. For a more realistic CDM universe, these integrals have to + 5 perm., numerically since we do not have an ,II analytic form (2.27) of the linear power spe Z present epoch. Also, in this case we do not encounter formally divergent i = 6F2 (k2, k3 ) modes (k2 )entering (k3 ) Fthe, q, q) P 5 perm.domination, 3 (k3horizon lin (q) + during radiation 321,I 411are suppressed. ζ tree-level ζ ζ ζ 1-loop ζ ζ q = F2 (k2, k3 ) (k2 )P13 (k3 ) + 5 perm., 3 Z = 12 (k2 ) (k3 ) F4 (q, q, k2, k3 ) (q) + 2 cyc. perm q l [Mpc/h] (2.28) 140 (2.29) II Note that B321 reduces to the one-loop contribution to the power spectrum stemming from the correlator h (3) (1) i, i.e. P13. Again, these integrals can be divergent just as in the case I the one-loop power spectrum. An important part of this paper is dedicated to prove Bof 321 that these divergences can be cancelled. In sum, the SPT bispectrum at the one-loop level (a) Equilateral triangles of side l. CosPA 2017(b)

29 IR-resummation for the 3-PF Similar to the resummation of the 2PF For example at tree-level IR+ tree ( x 1, x 2, x 3 )= d 6 r E tree(r)g(x, x r) Gaussian-like kernel induced by long displacement modes Scoccimarro, Trevisan: in preparation

30 IR-resummation for the 3-PF IR ζ [Equilateral] Similar to the resummation of the 2PF 1.5 For example at tree-level tree ( x 1, x 2, x 3 )= ζ tree ζ tree IR+ ζ 1-loop IR+ d 6 r E tree(r)g(x, x r) Gaussian-like kernel induced by long displacement modes l [Mpc/h] (a) Equilateral triangles of side. Scoccimarro, Trevisan: in preparation

31 Conclusions A simpler approx. for the IR-resummation Although IR-enhanced, IR mode-coupling leads to <1% in the 2-PF IR-resummation fixes also the 3-PF Tree-level and 1-loop already agree quite well Analysis of 2+3 PF may be an alternative to reconstruction

32 Thanks!

33 A brief excursus on Lagrangian PT (LPT) Instead of using comoving coordinates, use fluid coordinates x( q, t) = q + s( q, t)

34 A brief excursus on Lagrangian PT (LPT) Instead of using comoving coordinates, use fluid coordinates x( q, t) = q + s( q, t) and sum over all initial positions 1+ ( x, t) = d 3 q 3 D( x q s( q, t))

35 A brief excursus on Lagrangian PT (LPT) Instead of using comoving coordinates, use fluid coordinates x( q, t) = q + s( q, t) and sum over all initial positions 1+ ( x, t) = d 3 q 3 D( x q s( q, t)) P (k) = to obtain the 2-PF d 3 q 12 e i k q 12 e e i k ( s(q 1 ) s(q 2 )) f

36 A brief excursus on Lagrangian PT (LPT) The linear order solution for the displacement field is s(p) ƒ s 1 (p) =i p p 2 lin(p), and leads to the Zel dovich approximation:

37 A brief excursus on Lagrangian PT (LPT) The linear order solution for the displacement field is s(p) ƒ s 1 (p) =i p p 2 lin(p), and leads to the Zel dovich approximation: P (k) = d 3 q 12 e i k q 12 e 1 2 k ik j Ès i s j Í(q 12 ) Ès i s j Í where d 3 p P lin (p) (2fi) 3 p 2

38 Why SPT fails? Lets go back to the 1-loop expression P 1 loop (k) 1 2 pπ d 3 p (p k) 2 (2fi) 3 p 4 [P lin ( k p )+P lin ( k + p ) 2P lin ( k )] P lin (p).

39 Why SPT fails? Lets go back to the 1-loop expression P 1 loop (k) 1 2 pπ d 3 p (p k) 2 (2fi) 3 p 4 [P lin ( k p )+P lin ( k + p ) 2P lin ( k )] P lin (p). P lin à k n For a smooth component P lin (k) p2 k 2, Very long modes ( p π k) do not contribute to the loop

40 Why SPT fails? Lets go back to the 1-loop expression P 1 loop (k) 1 2 pπ d 3 p (p k) 2 (2fi) 3 p 4 [P lin ( k p )+P lin ( k + p ) 2P lin ( k )] P lin (p).

41 Why SPT fails? Lets go back to the 1-loop expression P 1 loop (k) 1 2 pπ d 3 p (p k) 2 (2fi) 3 p 4 [P lin ( k p )+P lin ( k + p ) 2P lin ( k )] P lin (p). P lin à sin(k/k osc ) For the BAO P w lin(k)(cos(p BAO ) 1) So there is an IR-enhancement for modes 1 BAO. p. k

42 IR-resummation LPT calculations involve the average of an exponential P (k) = d 3 q 12 e i k q 12 e e i k ( s(q 1 ) s(q 2 )) f which can be done as e e ik (q) f =expc Œ ÿ n=1 ( i) n n! È(k (q)) n Í c D = K(k, q),

43 IR-resummation LPT calculations involve the average of an exponential P (k) = d 3 q 12 e i k q 12 e e i k ( s(q 1 ) s(q 2 )) f which can be done as e e ik (q) f =expc Œ ÿ n=1 ( i) n n! È(k (q)) n Í c D = K(k, q), Once expanded to some order N in LPT = SPT

44 IR-resummation and the idea is to resum IR modes (~ s< ) X N means up to order N K 0 contains only IR-displacements K IR+ N = K 0 = Nÿ j=0 K K N R N j K j

45 IR-resummation and the idea is to resum IR modes (~ s< ) X N means up to order N K 0 contains only IR-displacements K IR+ N = K 0 = Nÿ j=0 K K N R N j K j P (k) = d 3 qe iq k to get* ÿ N j=0 R(k, q) N j E j(q) *we actually use a much simpler and intelligible approximation wrt the original paper which is parametrically justified

46 ( x 1, x 2, x 3 )= ( x 12 ) 2 ( x 13 )+Ò 1 i 2 ( x 12 )Ò i 2 ( x 13 )+Ò i 2 ( x 12 )Ò 1 i 2 ( x 13 ) Ò iò 1 j 2 ( x 12 )Ò i Ò 1 j 2 ( x 13 )+cyc.

47 Back-up 0.2 (100) imp )/ξ 2-loop imp (ξ-ξ 2-loop ξ linear ξ 0-loop ξ ZA ξ 0-loop cubic -0.3 ξ 1-loop ξ 1-loop cubic ξ 2-loop ξ 2-loop cubic r [Mpc/h]

48 1.10 SDSS MGS WiggleZ (DV/rdrag)/(DV/rdrag)Planck DFGS BOSS LOWZ BOSS CMASS z Fig. 14. Acoustic-scale distance ratio D V (z)/r drag in the base CDM model divided by the mean distance ratio from Planck TT+lowP+lensing. The points with 1 errors are as follows: green star (6dFGS, Beutler et al. 2011); square (SDSS MGS, Ross et al. 2015); red triangle and large circle (BOSS LOWZ and CMASS surveys, Anderson et al. 2014); and small blue circles (WiggleZ, as analysed by Kazin et al. 2014). The grey bands show the 68 % and 95 % confidence ranges allowed by Planck TT+lowP+lensing. D V (z) = " # 1/3 (1 + z) 2 D 2 cz A (z). H(z) nds in the figure show the ±1 and ±

49 1 loop TSPT P 1 loop w, ( ; k) hard = s 4 + s 3 s 3 + s 3 K 2 P IR res,lo+nlo s w, ( ; k) = k + w 4 + w 3 s C 2 + s 4 + K 3 s 3 s 3 + w 6 + w 5 s 3 + s 3 w 5 s 3 + K 2 K 2 + K 3 + s 3 K 2 + w 8 + w 7 s 3 + w w 4 + w 3 s 3 + w 3 K2 ( + The perturbative solution parametrically goes as s n s 1 n 1 1