general massive gravity and cosmology · general massive gravity and cosmology l. pilo1 1department...

General Massive Gravity and Cosmology

L. Pilo1

1Department of Physical and Chemical SciencesUniversity of L’Aquila

Tales of Lambda, Nottingham , July 2013

Based onComelli-Crisostomi-Nesti-LP Phys.Rev. D86 (2012) 101502

Comelli-Nesti-LP arXiv:1302.4447, PRD in pressComelli-Nesti-LP arXiv:1305.0236 , JHEP in press

Comelli-Nesti-LP to appear

1 / 26

Einstein’s GR

A 95 year-long successful theorya single free parameter and it works great

Weak Equivalence principle (10−13 )Solar system tests (weak field) (10−3 − 10−5 )Binary pulsar (nonlinear) (10−3)Newton’s Law tested between 10−2mm and 1016mmAs an effective field theory the cutoff is hugeΛ ∼

(10−33 cm

)−1 ∼ M2pl ∼ 1019 GeV

Modify GR at large distance to explain the present acceleration ?

2 / 26

Einstein’s GR

A 95 year-long successful theorya single free parameter and it works great

Weak Equivalence principle (10−13 )Solar system tests (weak field) (10−3 − 10−5 )Binary pulsar (nonlinear) (10−3)Newton’s Law tested between 10−2mm and 1016mmAs an effective field theory the cutoff is hugeΛ ∼

(10−33 cm

)−1 ∼ M2pl ∼ 1019 GeV

Modify GR at large distance to explain the present acceleration ?

2 / 26

Massive Deformed GR

Add to GR an extra piece such that when gµν = ηµν + hµν

R − m2 V)

= Lspin 2 +14(m2

0 h200 + 2m2

1 h0ih0i −m22 hijhij

+ m23 h2

ii − 2m24 h00hii

)+ · · ·

Beyond lin. approximation building V requires extra stuffOne is possibility is to introduce gµν = ∂µφ

A∂νφB ηAB

Scalars made out of Xµν = gµαgαν , τn = Tr(X n)

Unitary gauge: ∂µφA = δA

a (τ1 − 4)2 + b (τ2 − 2τ1 + 4) =(

a hµνhµν + b h2)

+ · · ·

Lorentz invariant mass term:m2

0 = a + b , m21 = −b , m2

2 = −a , m24 = b , m2

3 / 26

Massive Deformed GR

Add to GR an extra piece such that when gµν = ηµν + hµν

R − m2 V)

= Lspin 2 +14(m2

0 h200 + 2m2

1 h0ih0i −m22 hijhij

+ m23 h2

ii − 2m24 h00hii

)+ · · ·

Beyond lin. approximation building V requires extra stuffOne is possibility is to introduce gµν = ∂µφ

A∂νφB ηAB

Scalars made out of Xµν = gµαgαν , τn = Tr(X n)

Unitary gauge: ∂µφA = δA

a (τ1 − 4)2 + b (τ2 − 2τ1 + 4) =(

a hµνhµν + b h2)

+ · · ·

Lorentz invariant mass term:m2

0 = a + b , m21 = −b , m2

2 = −a , m24 = b , m2

3 / 26

Massive Deformed GR

Nondynamical g breaks inevitably local Lorentz (grav. sector)

gab = diag(−1,1,1,1) , gab = diag(−α0, α1, α2, α3)

Accidental Lorentz symm. of V (unitary gauge) unrelated withLorentz symmetry in the equivalence principleAlternatively only rotational invariance can be retained in theunitary gauge. Helps with phenomenology Rubakov ’04

fµν = ∂µΦm ∂νΦnδmn , n = (−gµν∂µΦ∂νΦ)1/2 , nµ = n ∂µΦ

Γµν = gµα fαν + nµ nαfαν , gµαgαµ = Xµν = Γµµ − n2

Γµµ + c1 n2 , τ1 = Tr(X ) = Γµµ − n2

Matter always couples with gµν , weak equivalence principle is OK

4 / 26

Massive Deformed GR

Γµµ + c1 n2 , τ1 = Tr(X ) = Γµµ − n2

4 / 26

Massive Deformed GR

Γµµ + c1 n2 , τ1 = Tr(X ) = Γµµ − n2

4 / 26

Taming the zoo of Massive Gravity

How many degrees of freedom propagate ?

Canonical analysis is best suitednonperturbativebackground independent

Variable in the ADM form : lapse N, shift vector N i and spatial metric γij

gµν =

(−N2 + N iN jγij γijN j

γijN j γij

)NA = (N,N i)

V = m2√

(g) V = m2 N γ1/2 V (NA, γij)

5 / 26

Results from Canonical Analysis

1 No condition on V ⇒ 6 DoF propagatearound flat space : 2 tensors + 2 vectors + 1+ 1 scalarsOne of the scalars is a the Bolware-Deser ghost. No good

2 5 DoFs propagate if and only if NB V = det(γij)1/2 V

det(∂V

∂NA∂NB ) = 0 and rank(VAB) = 3 Monge-Ampere eq.

Vi + 2 ξAξj ∂VA

∂γ ij = 0 , ξA = (1, ξi) VAB ξB = 0

6 / 26

Results from Canonical Analysis

1 No condition on V ⇒ 6 DoF propagatearound flat space : 2 tensors + 2 vectors + 1+ 1 scalarsOne of the scalars is a the Bolware-Deser ghost. No good

2 5 DoFs propagate if and only if NB V = det(γij)1/2 V

det(∂V

∂NA∂NB ) = 0 and rank(VAB) = 3 Monge-Ampere eq.

Vi + 2 ξAξj ∂VA

∂γ ij = 0 , ξA = (1, ξi) VAB ξB = 0

6 / 26

Taming the zoo of Massive Gravity: bottom line

All massive gravity theories with residual rotational invariance (unitarygauge) with five DOF are of the form (for any U and E )

V (N,N i , γij) = U + N−1(E +Qi Uξi

)U(Kij) , E(γij , ξ

i) Kij = γ ij − ξi ξj

ξi is defined by N i − N ξi =

(∂2U∂ξi∂ξj

)−1∂E∂ξj ≡ Q

i(γ ij , ξi)

Simplest case: E(γij) ⇒ ξi = N i/N

Kij = γ ij − N−2 N i N j ≡ g ij and V = U(g ij) + N−1 E(γij)

Z+dRGT Lorentz inv. is recovered. The price is a very low cutoffΛ3 = (m2 Mpl)

1/3 ∼ 103 Km−1 in conflict with Newton’s law testsgenerically (no Lorentz inv. only rotation inv.) the cutoff isΛ2 = (m Mpl)

1/2 >> Λ3. OK with Newton’s law tests

7 / 26

Massive Gravity Cosmology

CMB isotropy frame ≡ massive GR preferred frame (unitarygauge) with Stuckelbergs are trivialThe only reasonable option if mGR is relevant for cosmologyFRW metric in conformal time (zero spatial curvature for simplicity)

ds2 = a2 ηµν dxµdxν , 3H2 = a2

2 M2pl

mGR→ ρeff

= m2 U , peff

= m2[2U ′ − U + 2a−1

(E ′ − E

)]Bianchi → H

(E ′ − 1

= 0∂U∂γ ij = U ′γij

Either there is no cosmology or E must be tuned: for instance Ehomogeneous of degree -3/2 in γij or simply E = 0dRGT choice for V ∼ Tr(X 1/2) does not satisfy Bianchi:ELI = (1− ξiγijξ

j)−1/2 no FRW solution8 / 26

mGR Cosmology: perturbations vs DE eq. of state

Perturbationsgµν = a2(ηµν + hµν) hij = χij + ∂isj + ∂jsi + δij τ + ∂i∂jσ

L(v) =M2

1s′ik2

k2 + a2m21

s′i −m22 k2 sisi + · · ·

1 ∝ 2 m2 U ′2

L(s) =a4 Λ4

2 U ′

2 k2 Σ′2 −a4 M2

pl(m22 −m2

2Σ2 + · · · Σ = k2 σ

Dark Energy eq. of state

weff =peff

ρeff= −1 +

2U ′

dS and/or Minkowski perturbations are strongly coupled if FRWsolutions exist: perturbatively instead of 5 DOF only 2 DOF (tensors)General conclusion for any mGR theory with 5 DOF !

9 / 26

How much DE dominated Universe can differ from dS ?Away from dS the strong coupling scale for vectors and scalar is

Λscal '√

MplH(weff + 1)1/4

Λvect '√

(weff + 1)1/2 X =2U ′

2U ′ + E ′′v

(Mpl H)1/2 ∼ 10−1 mm−1 and with the present uncertainty on weffwe can still have a sub-millimiter cutoff scaleFurther progress on small scale test of 1/r would give a predictionfor weff to stay away from strong coupling

10 / 26

Λscal '√

MplH(weff + 1)1/4

Λvect '√

(weff + 1)1/2 X =2U ′

2U ′ + E ′′v

10 / 26

Λscal '√

MplH(weff + 1)1/4

Λvect '√

(weff + 1)1/2 X =2U ′

2U ′ + E ′′v

10 / 26

Λscal '√

MplH(weff + 1)1/4

Λvect '√

(weff + 1)1/2 X =2U ′

2U ′ + E ′′v

10 / 26

Conclusions

A randomly picked deforming potential propagates 5+1 DOF;one is a ghost around Minkowski spaceThe condition for having 5 DoF in encoded in a set differentialequationsBesides dGRT, there are a lot of new interesting solutions(infinitely many)Generically, the new solutions has no vDVZ discontinuity and nostrong coupling is needed in the solar systemAs an effective field theory the cutoff Λ2 = (m Mpl)

1/2 seems to bethe highest possible in the absence of a Higgs mechanism forgravityIf FRW solution exist then perturbations around dS and/orMinkowski are strongly coupleddeviations from Minkowski are needed

11 / 26

Conclusions

11 / 26

Conclusions

11 / 26

Conclusions

11 / 26

Conclusions

11 / 26

Conclusions

11 / 26

Directions

Canonical Analysis

Stuckelberg Formulation

Derivation of the new solutions

Conclusions

12 / 26

The Stuckelberg Trick in Massive GR I

The extra metric is non-dynamical flat given metric

To recover diff (gauge) invariance introduce 4 (Stuckelberg)scalars to recast the fixed metric as

gµν =∂φA

∂xµ∂φB

∂xνηAB

Minimal set of DOF to recover diff invariancegµν , Xµ

ν = gµαgαν are tensors and τn = Tr(X n) scalarsGeometrically φA are coordinates of some fictitious flat spaceMpoint-wise identified with the physical spacetime with a tetradbasis EA = dφA

One can chose coordinates such that (Unitary gauge)

xµ = φAδµA ,∂φA

∂xµ= δA

µ ⇒ gµν = ηµν

13 / 26

gµν =∂φA

∂xµ∂φB

∂xνηAB

∂xµ= δA

13 / 26

gµν =∂φA

∂xµ∂φB

∂xνηAB

∂xµ= δA

13 / 26

gµν =∂φA

∂xµ∂φB

∂xνηAB

∂xµ= δA

13 / 26

The Stuckelberg Formulation II

Weak equivalence principle is OK, matter still couples with gResidual Lorentz symm. in the unitary gauge is not related withEinstein equivalence principleThe extra metric g breaks inevitably local Lorentz invariance in thegravitation sectorLocally (tetrad basis)

14 / 26

The Stuckelberg Formulation II

Galileian structure inMuniversal time field Φ defines Φ = const . slices, on each of them a flatmetric f is given

fµν = ∂µΦm ∂νΦnδmn .

Galileian Unitary gauge (defined up to rotations)

x0 = Φ, x i = Φi fµν = δiµδ

jνδij .

Basic objects:

(−gµν

∂xµ∂Φ

∂xν

, nµ = n ∂µΦ , Yµν = gµα fαν

Γµν = (gµα + nµ nα) fαν = Yµν + nµ nαfαν

Dictionary

N → n , γ ikδkj → Γµν N i → n nµ = n gµνnν

15 / 26

The Stuckelberg Formulation III

The deforming potential is made of SO(3) scalars. For instance

Γµν , nµnµ

The two approaches are related but do not coincide. For instance,

gµαgαµ = Xµν = Γµµ − n2

but a general scalar function of n,nµ, Γµν is not function of Tr(X n)The residual symmetry in the unitary gauge is differentSO(3,1) vs SO(3)

Action

SmGR =

∫d4x√

g M2pl

[R(g) + Lmatt (ψ,g)−m2 V (g, g)

]Index

16 / 26

Hamiltonian Analysis

ADM decompositions

gµν =

(−N2 + NiNjhij Ni

Ni γij

)Hamiltonian of GR and mGR in the unitary gauge

H = M2pl

∫d3x

[NAHA + m2 N

√γ V]

HA = (H, Hi)

Πij → Conj. momenta of γij

PA = (P0,P i) Conjugate momenta of NA = (N,N i)

Hi = −2γijDk Πjk , H = −γ1/2 R(3) + γ−1/2(

ΠijΠij − 1

No time derivatives of NA → PA = 0 Constrained theory !

17 / 26

Constrained Theory: Dirac treatment in a nutshell

1 Momenta are not all independent→ introduce Lagrangemultipliers (LMs) to enforce the constraints

2 Time evolution us generated by the the total Hamiltonian:canonical + constraints + LMs

HT = H +

∫d3x λAΠA ,

EoMs: dynamical + time evolution of primary (PA = 0) constraints3 enforcing the consistency of constrs. with time evolution produces

new constraints or determine some of the LMs4 Full set of conserved constraints {Cs , i = 1,2, · · · c} reduces the

number of DoF from 10 down to (10 + 10− c)/2 = 10− c/2If some of the LMs are not determined→ gauge invariance

18 / 26

HT = H +

∫d3x λAΠA ,

18 / 26

HT = H +

∫d3x λAΠA ,

18 / 26

HT = H +

∫d3x λAΠA ,

18 / 26

Example: GR

Time evolution of PA = 0 via Poisson brackets are just the Eqs. ofNA, being H linear in NA

{PA(t , x), HT (t)} = {PA(t , x), H} = HA = 0

Thanks to the GR algebra the four secondary constraints areconserved and no LM is determined (Diff invariance)

{H(x), H(y) = Hi (x) ∂(x)i δ

(3)(x − y)−Hi (y) ∂(y)i δ

(3)(x − y)

{H(x), Hj (y)} = H(y) ∂(x)j δ

(3)(x − y)

{Hi (x), Hj (y)} = Hj (x) ∂(x)i δ

(3)(x − y)−Hi (y) ∂(x)j δ

(3)(x − y)

In GR four diffs have to be gauge fixed adding 4× 2 additionalconstraints

DoF = (10 + 10− 4− 4− 8)/2 = 2

The analysis is nonpertutbative and background independent

19 / 26

Example: GR

Time evolution of PA = 0 via Poisson brackets are just the Eqs. ofNA, being H linear in NA

{PA(t , x), HT (t)} = {PA(t , x), H} = HA = 0

Thanks to the GR algebra the four secondary constraints areconserved and no LM is determined (Diff invariance)

{H(x), H(y) = Hi (x) ∂(x)i δ

(3)(x − y)−Hi (y) ∂(y)i δ

(3)(x − y)

{H(x), Hj (y)} = H(y) ∂(x)j δ

(3)(x − y)

{Hi (x), Hj (y)} = Hj (x) ∂(x)i δ

(3)(x − y)−Hi (y) ∂(x)j δ

(3)(x − y)

In GR four diffs have to be gauge fixed adding 4× 2 additionalconstraints

DoF = (10 + 10− 4− 4− 8)/2 = 2

The analysis is nonpertutbative and background independent

19 / 26

When V deforming potential is turned on, the time evolution ofPA = 0 still gives NA Eqs

{PA(t , x), HT (t)} = SA = HA + VA 4 new secondary constraints

V = m2 N γ1/2 V∂V∂NA = VA

Is time evolution consistent with SA ?VAB ≡ V,AB =

∂2V/∂NA∂NB

TA ≡ {SA, HT} = {SA, H} − VAB λB = 0

If the r = Rank(VAB) = 4: deforming pot. has non degenerateHessianall LMs λA are determined and we are doneDoF=10 - (4 + 4)/2 = 6 = 5 + 1Around Minkowski: massive spin 2 (5) plus a ghost scalar (1)Boulware-Deser mode

20 / 26

When V deforming potential is turned on, the time evolution ofPA = 0 still gives NA Eqs

{PA(t , x), HT (t)} = SA = HA + VA 4 new secondary constraints

V = m2 N γ1/2 V∂V∂NA = VA

Is time evolution consistent with SA ?VAB ≡ V,AB =

∂2V/∂NA∂NB

TA ≡ {SA, HT} = {SA, H} − VAB λB = 0

If the r = Rank(VAB) = 4: deforming pot. has non degenerateHessianall LMs λA are determined and we are doneDoF=10 - (4 + 4)/2 = 6 = 5 + 1Around Minkowski: massive spin 2 (5) plus a ghost scalar (1)Boulware-Deser mode

20 / 26

The Hessian matrix of V has a single zero mode χA, r = 3

VAB χB = 0 , VAB EB

n = κn EAn

λA = z χA +3∑

dn EAn

def≡ z χA + λA .

If det(Vij) 6= 0, then χA = (1,−V−1ij V0j) ≡ (1, ξi)

Projection of TA = 0 along χA is a single new constraintProjection on the remaining eigenvectors gives Three (λA) out of thefour LMs

χA{SA, H} = TAχA = T = 0

EAn {SA, H} − dn κn = 0 No sum in n

21 / 26

Time evolution of T

Q(x) = {T (x),HT}

= {T (x),H}+

∫d3y {T (x), λA(y)ΠA(y)} = 0

1 If Q does not depend on z, the last LM, we have a new constraint

z is determinate by the time evolution of Q. OK

Total # of constraints 4 (PA) + 4 (SA) + 1 (T ) + 1 (Q) = 10

DoF: 10− 10/2 = 5

2 If Q = 0 determines z, There is no additional constraintsTotal # of constraints 4 (PA) + 4 (SA) + 1 (T ) = 8 + 1

DoF: 10− 9/2 = 5 + 1/222 / 26

Time evolution of T

Q(x) = {T (x),HT}

= {T (x),H}+

∫d3y {T (x), λA(y)ΠA(y)} = 0

DoF: 10− 10/2 = 5

DoF: 10− 9/2 = 5 + 1/222 / 26

Time evolution of T

Q(x) = {T (x),HT}

= {T (x),H}+

∫d3y {T (x), λA(y)ΠA(y)} = 0

DoF: 10− 10/2 = 5

DoF: 10− 9/2 = 5 + 1/222 / 26

{T , λA · ΠA} = terms z indep.−∫

d3y Θ(x , y) z(y) = · · · − I[z]

Θ(x , y) = χA(x) {SA(x),SB(y)}χA(y) = Ai(x , y)∂iδ(3)(x − y)

A(x , y) = A(y , x)

Only in field theory Θ can be non zero !

I[z] = − 12z(x)

[z(x)2Ai(x , x)

]Q is free from z if Ai(x , x) = 0, which consists in the following condition

Vi + 2χAχj ∂VA

∂γ ij = 0 , V = γ1/2V

23 / 26

mGR: Summary of the Canonical Analysis

Necessary and sufficient conditions for having 5 DoF in mGR

Rank(VAB) = 3⇒ V00 − V0i(Vij)−1Vj0 = 0 (1)

Vi + 2χAχj ∂VA

∂γ ij = 0 χA = (1,−V−1ij V0j) (2)

Notice: If only (1) holds 5+1/2 DoF propagate

5+1/2 DoF found also in a class of Horava-Lifshitz modified gravitytheory(1) is a homogeneous Monge-Ampere equation(2) is much more restrictive: it involves of the the spatial metric

24 / 26

New solutions: sketch of the derivation I

Change variable from N i to ξ ≡ χi and then Monge-Ampere eq. ⇒V is a degree zero homogeneous function of (N, χi)

Introduce U = χA VA of ξi and γij , sort of Legendre transformation

V0 = U − Ui ξi , Vi = Ui . (3)

Any V that satisfies (3) solves the Monge-Ampere eq.Eqs. (3) can be integrated using the integtability conditionN i − N ξi = − U ij Ej ≡ Qi

where U ij ≡ (Uij)−1 hessian inverse of U and Ei ≡ ∂ξiE gradient of

a generic function E(ξi , γ ij). The general solution is

V = N U + E + UiQi

25 / 26

New solutions: sketch of the derivation II

The remaining equation for having 5 Dofs reads

∂U∂ξi + 2 ξj ∂U

∂γ ij = 0

The spatial metric γij enters here. The Monge-Ampere involvedonly the lapse and the shiftsKey observation: any function of K ij = γ ij − ξi ξj is a solution.SO(3) invariant solutions ara made from the invariants of K inδnj

Pick up a function E of ξi and γij and express ξi in terms of N i

Take ∂E/ξi = 0⇒ ξi = N−1 N i and K ij = γ ij − ξiξj ≡ g ij

The potential is given by

V = U(

+ N−1 E(γij)

U scalar function of g ijand E scalar function of γij Index

26 / 26

general massive gravity and cosmology · general massive gravity and cosmology l. pilo1 1department...

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