Pressuron's phenomenology

Transcription

1 Pressuron's phenomenology Olivier Minazzoli Aurélien Hees

2 The Pressuron : - scalar particle - couples non-minimally to both curvature and matter But - is not sourced by pressure-less fields!!! (weird isn't it?!) 2

3 Basics of the pressuron S = d 4 x g(φ R ω(φ) Φ ( σ Φ) 2 [Minazzoli, Hees, 2 Φϵ) Phys. Rev. D, 88, issue 4 (2013)] Effective Lagrangian for perfect fluid : L m = ϵ= c 2 ρ P (ρ) ρ d ρ with σ (ρu σ )=0 [Minazzoli, Harko, Phys. Rev. D, 86, issue 8 (2012)] [Minazzoli, Phys. Rev. D, 88, issue 2 (2013)] T αβ = 2 g δ( g ϵ) δ g αβ =[ϵ+p ]U α U β + P g αβ 3

4 Basics of the pressuron S = d 4 x g(φ R ω(φ) Φ ( σ Φ) 2 [Minazzoli, Hees, +2 Φϵ) Phys. Rev. D, 88, issue 4 (2013)] 2 ω+3 Φ 2 Φ+ ω, Φ Φ ( Φ)2 = 1 Φ ( T +ϵ ) T = ϵ+3p 2ω+3 Φ 2 Φ+ ω, Φ Φ ( Φ)2 = 3P Φ 4

5 A simple way to understand the mechanism : the dust case 5

6 the dust case S = d 4 x g (Φ R Z (Φ)( Φ) 2 )+ A Φ m A d τ A Conformal transformation g αβ =Ω 2 g αβ S = d 4 x g ( Φ Ω 2 R F (Ω,Φ,( Φ) 2, ( Ω) 2 )) + A Φ Ω m A d τ A 6

7 the dust case S = d 4 x g (Φ R Z (Φ)( Φ) 2 )+ A Φ m A d τ A Going to the Einstein frame g αβ =Φ g αβ S = d 4 x g ( R Z (Φ)( Φ) 2 )+ A m A d τ A 7

8 Dust field in Einstein frame In other words : In the dust case, the pressuron reduces to «veiled» general relativity cf. Deruelle & Sasaki S = d 4 x g (Φ R Z (Φ)( Φ) 2 )+ A Φ m A d τ A S = d 4 x g ( R Z (Φ)( Φ) 2 )+ A m A d τ A 8

9 Dust field in Einstein frame Or equivalently : S = d 4 x g (h 2 R Z (h)( h) 2 +V (h))+ A h m A d τ A S = d 4 x g ( R Z (h)( h) 2 +Ṽ (h))+ A m A d τ A 9

10 Dust field in Einstein frame Brans-Dicke : S = d 4 x g (Φ R Z (Φ)( Φ) 2 )+ A m A d τ A S = d 4 x g ( R F (Φ)( Φ) 2 )+ m A A Φ d τ A Pressuron : S = d 4 x g (Φ R Z (Φ)( Φ) 2 )+ A Φ m A d τ A S = d 4 x g ( R F (Φ)( Φ) 2 )+ A m A d τ A 10

11 But Pressuron no longer «veiled» general relativity when there is Pressure! 2ω+3 Φ 2 Φ+ ω, Φ Φ ( Φ)2 = 3P Φ Scalar field couples to matter via pressure 11

12 Phenomenology : basics Weak field : Cosmology : PN parameters = 1 Same trajectories as GR at 1.5PN level No Nordtvedt effect at 1.5PN Light wave trajectory different from GR at 2PN Gravitational redshift different from GR at 10-6 relative level Cosmologically quickly converges toward GR in matter era (no coupling) Decouples dynamically in radiation era (Damour & Nordtvedt like) (ie. ω(φ) Big) Cannot explain DE without potential or Λ 12

13 Why is it interresting? 1/ Unusual phenomenology 2/ One of the possible solutions to the presence of very light coupled scalar fields (while we live in a GR-like wolrd). 13

14 What remains to do Strong field : Pressuron should appear in regimes with high-pressure, how about black holes? To be done (several difficulties) 14

15 What remains to do Strong field : Pressuron should appear in regimes with high-pressure, how about black holes? To be done (several difficulties) Link to microphysics : what coupling effectively leads to S m d 4 x g Φρ? in progress Dilaton (multiplicative) partial decoupling in general (still UFF violation but weaker) Higgs-like coupling 15

16 Thank you for your attention References : - Minazzoli & Hees, Physical Review D, vol. 88, Issue 4 (2013) - Minazzoli, Physics Letters B, Volume 735 (2014) - Minazzoli & Hees, Physical Review D, Volume 90, Issue 2 (2014) About microphysics : - Minazzoli & Hees (or switched around), to be submited soon Anatidaephobia : The fear that one is being constantly watched by a duck. 16

17 Visit Scholarpedia.org Peer-reviewed totally free encyclopedia (already contributed : 12 Nobel laureates, 4 Fields medalists, 8 Dirac medalists, Etc.) 17

18 Perfect fluid Lagrangian L m = αϵ+β P γ γ α+β T μ ν =(ϵ+p)u μ U ν + P g μ ν S m d 4 x g L m Degenerate : one can take any value of {α, β} S m d 4 x f (Φ) g L m No longer degenerate : only one value of {α, β} Which one? Dust case : L m = c 2 ρ α=1, γ=1 β=0 L m = ϵ Only possible choice of effective Lagrangian 18

19 Basics of the pressuron S = d 4 x g(φ R ω(φ) Φ ( σ Φ) 2 +2 Φ L m ) Lets consider a dust (pressure-less) field : [Minazzoli, Hees, Phys. Rev. D, 88, issue 4 (2013)] T μ ν = i μ i δ( x i )u i μ u i ν L m = i μ i δ( x i ) μ i =m i /(u 0 g ) m i Conserved mass R μ ν Φ( = 1 T μ ν 1 T) 2 gμ ν + 1 Φ ( μ ν Φ g μ ν σ σ Φ ) +ω( 2) Φ μ Φ ν Φ 1 2 gμ ν ( α Φ) 2ω+3 Φ σ σ Φ+ ω, Φ Φ ( σ Φ) 2 = 1 Φ ( T L m ) 19

20 Damour & Polyakov dilaton S = d 4 x g B(ϕ)[ 1 α' (R+4 2 ϕ 4( ϕ) 2 ) k 4 F 2 Ψ D Ψ...] [Damour and Polyakov, Nucl. Phys. B, 423 (1994)] S = d 4 x g(φ R ω(φ) Φ ( σ Φ) 2 +2Φ n L m ) With n=1 20

21 Electromagnetism does not take the form of a perfect fluid L EM =F 2 [Peter and Uzan, Primordial cosmology (2009) Oxford U. Press] T αβ =(ρ+p)u α u β +P g αβ +(u α q β +q α u β )+π αβ Heat vector flux Viscous shear tensor Photons : not perfect fluid 21