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
√g(
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
µ
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 = b
3 / 26
Massive Deformed GR
Add to GR an extra piece such that when gµν = ηµν + hµν
√g(
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
µ
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 = b
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
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
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
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
Taming the zoo of Massive Gravity
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
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
(m2 U
2+
ρ
2 M2pl
)
mGR→ ρeff
M2pl
= m2 U , peff
M2pl
= m2[2U ′ − U + 2a−1
(E ′ − E
2
)]Bianchi → H
(E ′ − 1
2E)
= 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
pl
2
[m2
1s′ik2
k2 + a2m21
s′i −m22 k2 sisi + · · ·
]m2
1 ∝ 2 m2 U ′2
L(s) =a4 Λ4
2 U ′
2 k2 Σ′2 −a4 M2
pl(m22 −m2
3)
2Σ2 + · · · Σ = k2 σ
Dark Energy eq. of state
weff =peff
ρeff= −1 +
2U ′
U
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
mGR Cosmology: perturbations vs DE eq. of state
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 '√
MplH(
HXm
)1/2
(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
mGR Cosmology: perturbations vs DE eq. of state
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 '√
MplH(
HXm
)1/2
(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
mGR Cosmology: perturbations vs DE eq. of state
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 '√
MplH(
HXm
)1/2
(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
mGR Cosmology: perturbations vs DE eq. of state
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 '√
MplH(
HXm
)1/2
(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
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
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
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
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
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
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
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
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
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
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
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)
gab = diag(−1,1,1,1) , gab = diag(−α0, α1, α2, α3)
Index
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:
n =
(−gµν
∂Φ
∂xµ∂Φ
∂xν
)1/2
, 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
2(Πi
i)2)
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
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
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
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
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
mGR
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
mGR
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
mGR
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∑
n=1
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
mGR
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
mGR
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
mGR
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
mGR
{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)
∂i
[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
Index
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(
g ij)
+ N−1 E(γij)
U scalar function of g ijand E scalar function of γij Index
26 / 26