Delta-gravity. Jorge Alfaro Pontificia Universidad Católica de Chile. Segundo Encuentro CosmoConce Concepción, March 16, Motivation...

Transcription

1 Delta-gravity Jorge Alfaro Pontificia Universidad Católica de Chile Segundo Encuentro CosmoConce Concepción, March 16, 2012 Table of contents Motivation Definition of Delta gravity Symmetries

2 Equations of motion Particle motion in the gravitational field Distances and time intervals Energy momentum tensor for a perfect fluid Friedman-Robertson-Walker metric Red Shift Distances Supernova Ia data The Newtonian limit Dark Matter Conclusions and open problems

3 Motivation Delta Gauge Theories,J. A, bv gauge theories, hep-th ,J. A and P. Labraña,Phys. Rev. D 65, (2002) Classical equations of motion are satisfied in the full Quantum theory They live at one loop. All higher loop corrections are strictly zero. They are obtained through the extension of the former symmetry of the model, introducing an extra symmetry(δ), which is formally obtained as the variation of the original symmetry. General Relativity without matter and without a cosmological constant is finite at one loop, G. thooft and M. Veltman, One-loop divergencies in the theory of gravitation, Annales de l I.H.P, section A, tome 20, number 1 (1974), pag J.A., P. González and R. Avila, A Finite Quantum Gravity Field Theory Model, gr-qc , Class. Quantum Grav. 28 (2011) J.A. Delta Gravity and Dark Energy, gr-qc , Phys. Lett B709(2012)101. 3

4 Definition of Delta gravity We use the metric convention (-+++). κ= 8πG c 4 and κ 2 is an arbitrary constant. S(g, g,λ)= d d 1 x g ( 2κ R + L M) + κ 2 [(R µν 1 ] 2 g µν R )+ κ T µν g g µν d d x κ 2 κ g (λ µ;ν + λ ν;µ )T µν d d x (1) Symmetries Action (1) is invariant under the following transformations(δ), δg µν = g µρ ξ ρ ρ 0,ν + g νρ ξ 0,µ + g µν,ρ ξ ρ 0 = ξ 0µ;ν + ξ 0ν;µ δg µν (x) = ξ 1µ;ν + ξ 1ν;µ + g µρ ξ ρ ρ ρ 0,ν + g νρ ξ 0,µ + g µν,ρ ξ 0 ρ δλ µ = ξ 1µ + λ ρ ξ 0,µ +λ µ,ρ ξ 0 (2) 4

5 From now on we will fix the gauge λ µ = 0. This gauge preserves general coordinate transformations but fixes completely the extra symmetry with parameter ξ 1µ. Equations of motion The equation of motion for g µν is: S γσ + 1 (Rg γσ g µν R σγ g µν ) gσγ ( g g µν ν ),µ + g gσγ ( g g ασ g µν g µν σ ),α = κ δt µν g µν (3) g δg γσ where S σγ = (U σβγρ + U γβσρ U σγβρ ) ;ρβ U αβγρ = 1 2 [ g γρ ( g βα 1 ] 2 gαβ g g µν µν ) (4) The equation of motion for g µν is Einstein equation: ( R µν 1 ) 2 g µνr + κt µν = 0 (5) 5

6 Covariant derivatives as well as raising and lowering of indices are defined using g µν. Notice that outside the sources(t µν = 0), a solution of (3) is g µν = λg µν, for a constant λ. We will have g µν = g µν, assuming that both fields satisfy the same boundary conditions. 6

7 Particle motion in the gravitational field The action of a particle in a gravitational field is 1 2κ d n y g R m The variation in x µ produces the geodesic equation. dt g µν ẋ µ ẋ ν Instead in δ gravity we have: κ 2 d n y g (Gµν + κt )g µν µν 1 2κ d n y g R m dt g µν ẋ µ ẋ ν + where: T µν (y) = m 2 g dt ẋ µ ẋ ν δ(y x) g αβ ẋ α ẋ β Notice that T µν is t-parametrization invariant. Call κ 2 = κ 2 κ. It is dimensionless. 7

8 The new action has a term: S p = m dt ẋ µ ẋ ν (g µν + κ 2 g αβ ẋ α ẋ β 2 g µν )=m dt ḡ µν ẋ µ ẋ ν (6) g αβ ẋ α ẋ β Since far from the sources, we must have free particles in Minkowski space, it follows that we are describing the motion of a particle of mass m : m = m(1 + κ 2 ) 2 Since outside the sources g µν = g µν, the geodesic equation outside the sources is the same as Einstein s. Equation of motion for massive particles: d(ẋ µ ẋ ν ḡ µν ẋ β g αβ + 2ẋ β ḡ αβ ) dτ 1 2 ẋµ ẋ ν ḡ µν ẋ β ẋ γ g βγ,α ẋ µ ẋ ν ḡ µν,α = 0, ẋ µ ẋ ν g µν = 1 (7) Massless particle moves in a null geodesic of g µν =g µν + κ 2 g µν : K = ẋ µ ẋ ν g µν = 0, L 0 = dt 1 4 (ẋµ ẋ ν g µν ) 8

9 Distances and time intervals Remark 1. Eq.(7) satisfies the important property of preserving the form of the proper time in a particle in free fall. Notice that in our case the quantity that is constant using (7) is ẋ µ ẋ ν g µν. This single out this definition of proper time and not other. Proper time: dτ = g 00 dx 0 (8) Following Landau, the change in x 0 for a roundtrip of a light ray from a point with coordinates x α to a neighboring point with coordinates x α + dx α is: x 0 = 2 (g 0α g 0β g αβ g 00 ) g 00 dl = g 00 g αβ g 0αg 0β g 00 g 00 dl 2 = γ ij dx i dx j, γ ij = g 00 g 00 (g ij g 0ig 0j g 00 ) (9) 9

10 Energy momentum tensor for a perfect fluid Then: T µν = pg µν + (p + ρ)u µ U ν, g µν U µ U ν = 1 (10) δt µν δg γσ g µν = pg γσ (p + ρ)(u γ U ν g σν + U σ U ν g γν ) (11) Friedman-Robertson-Walker metric In this case, assuming flat three dimensional metric: ds 2 = dt 2 R(t) 2 {dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 } d s 2=Ã(t)dt 2 B (t){dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 } Equations (3,11) give: R B 1 2 prb R 1 R 2B 1 6 ρr3 Ã RR 2Ã = 0 (12) p B 2B R 2 R 2B + 2R 1 R B + 2R 1 R B + ρr 2 Ã + R 2Ã + 2RR Ã + 2R ÃR = 0 (13) 10

11 Einstein s equations are: ( ) d 3 R 2 ( ) ( ) dt d 2 d 2 R 2 = κρ, 2R dt 2 R + dt R = κr 2 p We use the state equation: to get, for w 1 : p = wρ R = R 0 t 2 3(1+w), Ã = 3wl 1 (w w+1 2 t ), (14) l 2 :arbitrary constant B = R 2 0 l 2 t b, b= 4 3w w 1 w

12 Red Shift Photons moves on a null geodesic of g: 0 = (1 + κ 2 Ã)dt 2 + (R 2 + κ 2 B )(dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 ) (15) Introduce: Then: R (t)= R2 + κ 2 B (t) 1 + κ 2 Ã z = R (t 0 ) R (t 1 ) 1 (16) To make the usual connection between redshift and the scale factor, we consider light waves traveling to r = 0, from r = r 1, along the r direction with fixed θ, φ. Photons moves on a null geodesic of g: 0 = (1 + κ 2 Ã) d t 2 + (R 2 + κ 2 B )(d r 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2 ) (17) So, t 1 t 0 d t 1 + κ 2 t A R 2 + κ 2 t B = r 1 (18) 12

13 A typical galaxy will have fixed r 1, θ 1, φ 1. If a second wave crest is emitted at t = t 1 + δ t 1 from r = r 1, it will reach r = 0 at t 0 + δ t 0, where t 0 +δt κ d t 2 t A R 2 + κ = r 1 2 t B t 1 +δt 1 Therefore, for δt 1, δt 0 small, which is appropiate for light waves, we have: Introduce: 1 + κ δ t 2 t A 1 + κ 0 R 2 + κ (t 0 ) = δt 2 ta 1 2 t B R 2 + κ (t 1 ) (19) 2 t B R (t)= R2 + κ 2 t B 1 + κ (t) 2 t A We get: δ t 0 δ t 1 = R (t 0 ) R (t 1 ). A crucial point is that, according to equation (8), δ t measure the change in proper time. That is: ν 1 ν 0 = R (t 0 ) R (t 1 ), where ν 0 is the light frequency detected at r = 0 corresponding to a source emission at frequency ν 1. Or in terms of the redshift parameter z, defined as the fractional increase of the wavelength λ: z = R (t 0 ) R (t 1 ) 1 = λ 0 λ 1 λ 1 (20) 13

14 We see that R replaces the usual scale factor R in the computation of z. Distances Luminosity distance: d L = R (t z 0 ) R (t 1 ) 0 z dz H(z) 0 = (1 + z) dz H (z ), H R = R (21) Let us consider a mirror of radius b that is receiving light from a distant source. The photons that reach the mirror are inside a cone of half-angle ε with origin at the source. Let us compute ε.the light path of rays coming from a far away source at xq1 is xq given by (ρ) = ρnˆ + xq1, ρ > 0 is a parameter and nˆ is the direction of the light ray.the path reaches xq us at = 0 for ρ = xq1 = r 1. So nˆ εq = xˆ1 +. Since nˆ, xˆ1 have modulus 1, εq ε = < <1 is precisely the angle between xq1 and nˆ at the source.the impact parameter is the proper distance of the path from the origin, when ρ = xq1. The proper distance is determined by the 3-dimensional metric (9). That is b = R (t 0 ) r 1 θ = R (t 0 ) r 1 ε, i.e. ε = b. R (t 0 ) r 1 A Then the solid angle of the cone is π ε 2 = r 2 1 R (t 0 ) 2, where A = π b2 is the proper area of the mirror.the A fraction of all isotropically emitted photons that reach the mirror is f = 4 π r 2 1 R (t 0 ) 2. Each photon carries an energy h ν 1 at the source and h ν 0 at the mirror. Photons emitted at intervals δ t 1 will arrive at inter- 14

15 vals δ t 0. We have ν 1 = R (t 0 ), δ t 0 = R (t 0 ) ν 0 R (t 1 ) δ t 1 the luminosity of the source. The apparent luminosity is l = P 0 = LR (t 1 ) 2 1 A R (t 0 ) 2 the luminosity decreases with distance d according to l = L.This permits to define the luminosity dis- 4 π d 2. Using (18) we can write this in terms of the red shift: tance:d L = L 4 π l = R (t 0 ) 2 r 1 R (t 1 ) R (t 1 ). Therefore the power at the mirror is P 0 = L R (t 1 ) 2 4 π r 1 2 R (t 0 ) f, where L is R (t 0 ) 2 2. In Euclidean space, z d L = (1 + z) 0 d z,h R = H (z ) R (22) 15

16 Supernova Ia data The supernova Ia data gives, m (apparent or effective magnitude) as a function of z. This is related to distance d L by: m = M + 5log( d L 10pc ) Here M is common to all supernova and m changes with d L alone. We compare δ gravity to General Relativity(GR) with a cosmological constant: H 2 = H 0 2 (Ω m (1 + z) 3 + (1 Ω m )), Ω Λ = 1 Ω m Notice that à = 0 for w = 0 in (14). So, it seems that we cannot fit the supernova data. However w = 0 is not the only component of the Universe. The massless particles that decoupled earlier still remain. It means that true 0 w < 1, but very close to w = 0. So, we will fit the data with w = 0.1, 0.01, and 3 see how sensitive the predictions are to the value of w. 16

17 Using data from Essence, we notice that R 2 test does not change much with w. Moreover l 2 scales with w, l 2 a. As an approximation to the limit w = 0, we get: 3w 1 3w parameter. a R (t)=r(t) (23) a t renormalizes the derivative of R at t = 0. It is not divergent, because for t 0, w 1. a is a free 3 Of course, the complete model must include the contribution of normal matter(w = 0) plus relativistic matter (w = 1 ). But, at later times, the data should tend to (23). 3 Let us fit the data to the simple scaling model (23). We get: Ω m = 0.22 ± 0.03, M = ± 0.03, χ 2 (p e r p oint)=1.0328, General Relativity a = 2.21 ± 0.12, M = ± 0.06,χ 2 (p e r p oint)=1.0327, Delta Gravity δ-gravity with non-relativistic(nr) matter alone give a fit to the data as good as GR with NR matter plus a cosmological constant. 17

18 Below, we have the graphs for general relativity and delta gravity versus the data in Essence Figure 1. General Relativity with NR matter plus a cosmological constant and Essence data.on the vertical axis, we have m. On the horizontal axis we have z. Figure 2. Delta Gravity with NR matter without a cosmological constant and Essence data.on the vertical axis, we have m. On the horizontal axis we have z. 18

19 According to the fit to data, a Big Rip will happen at t = 2.21 in unities of t 0 (today). It is a similar scenario as in R. R. Caldwell, M. Kamionkowski, and N. N.Weinberg, Phys. Rev. Lett.91(2003) Finally, we want to point out that since for t 0, we have w 1 3, then R (t) = R(t). Therefore the accelerated expansion is slower than (23) when we include both matter and radiation in the model. 19

20 The Newtonian limit The motion of a non relativistic particle in a weak static gravitational field is obtained using: g µν = which solves Einstein equations to first order in ǫ if: 1 2 Uǫ Uǫ Uǫ Uǫ 2 U = 1 2 κρ The solution for g µν is: g µν = ǫ Ũ ǫ ( Ũ 2U ) ǫ ( Ũ 2 U ) ǫ ( Ũ 2U ) Solving (3),to first order in ǫ we get: 2 Ũ = 1 2 κρ 20

21 To recover the Minkowsky metric far from the sources, ρ 0, we must require there: (8) implies : U 0,Ũ ǫ 1 d 2 x i d t 2 = φ,i φ = U κ 2 (2U + Ũ ) That is: 2 φ = κ 2 (1 3κ 2 )ρ, κ 2 1 The whole effect is a small redefinition of Newton constant. Gravitational red shift experiments can be used to put bounds on κ 2. According to (8), the shift in frequency of a source located at x 1, compared to the same source located at x 2 due to the change in gravitational potential is: ν 2 ν 1 ν 1 = φ N (x 2 ) φ N (x 1 ) 21

22 where φ N is the usual Newtonian potential, computed with κ as Newton constant. We get; ν ν = ( ± )(ϕ S ϕ E +.) where ϕ S is the gravitational potential at the spacecraft position and ϕ E is the gravitational potential on Earth. accounts for additional effects not related to the gravitational potential. We can ascribe the uncertainty of the experiment to κ 2, to get the bound: κ 2 < This bound is conservative because the Newton constant itself has a larger error (CODATA): G = ± m 3 kgs 2 22

23 Dark Matter Far from a source the gravitational field correspond to the Schwarzschild solution: pointlike source, spherically symmetric. We get: The exact solution is: g µν = (1 a r ) g µν = 0 1 a 0 0 r 0 0 r r 2 s i n(θ) 2 (1 a r + ba r ) ab 0 1 a r(1 a 0 0 r r )2 0 0 r r 2 sin(θ) 2 Boundary condition:g µν η µν g µν η µν for r.notice that still there are 2 arbitrary constants. 23

24 Massive Particles g µν = η µν + h µν, h 00= g µν = η µν + h µν,h 00 = a r,a=2m a(1 b),a(1 b) = 2M r The Newtonian potential is: 1 φ = (( 1 + κ )h κ κ 2 2 h 00)= M T r So the total mass of the source is: So, the dark matter mass is: M T = M κ 2 bm 1 + κ 2 2 M DM = κ 2 bm 1 + κ 2 (24) 2 M is the mass coming from the fluid density in Einstein equations. b is a new constant to accomodate DM. 24

25 Photons: The photon trajectory is given by: So, according to photons: [ (1 ar ) κ 2 (1 ar + bar ] [ ) dt 2 + [ r (a κ 2 ba 1 + κ 2 ) ] 1 1 a + κ 1 ab 2 ( 1 a r(1 a )2) dr 2 = 0 r] [ r r dt r (a κ ] 2 ba 1 + κ ) dr 2 = 0 2 M T = M κ 2 bm 1 + κ 2 Notice that photons and massive particles see different M T, but since κ 2 hard to detect. is very small, this difference is 25

26 Conclusions and open problems Delta Gravity agrees with General Relativity when T µν = 0. In a homogeneous and isotropic universe, we get accelerated expansion without a cosmological constant or additional scalar fields. Some solutions of Delta Gravity can accomodate Dark Matter. Further research is needed. Growth of Density perturbations? Anisotropies in the CMBR? THANK YOU! 26