From Fierz-Pauli to Einstein-Hilbert

From Fierz-Pauli to Einstein-Hilbert Gravity as a special relativistic field theory Bert Janssen Universidad de Granada & CAFPE A pedagogical review References: R. Feynman et al, The Feynman Lectures on Gravitation, 1995 T. Ortín, Gravity and Strings, 2004 B. Janssen (UGR) Granada, 14 dec 2006 1

Central Idea Orionites with knowlegde of special relativity quantum field theory standard model & QCD NOT general relativity B. Janssen (UGR) Granada, 14 dec 2006 2

Central Idea Orionites with knowlegde of special relativity quantum field theory standard model & QCD NOT general relativity discover gravity F = G N m 1 m 2 r 2 B. Janssen (UGR) Granada, 14 dec 2006 2

Central Idea Orionites with knowlegde of special relativity quantum field theory standard model & QCD NOT general relativity discover gravity F = G N m 1 m 2 r 2 They will try to describe gravity as a special relativistic field theory try to find a bosonic field (graviton) that is responsable for gravity B. Janssen (UGR) Granada, 14 dec 2006 2

Properties of the graviton: Gravity is long range = Massless B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) is universally attractive = even spin (spin 0, 2, 4,...) B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) is universally attractive = even spin (spin 0, 2, 4,...) has deflection of light = not spin 0 spin 0 propagator k 2 only coupling T µ µ k 2 T ν ν electromagnetism (in D = 4): T µ µ = 0 B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) is universally attractive = even spin (spin 0, 2, 4,...) has deflection of light = not spin 0 spin 0 propagator k 2 only coupling T µ µ k 2 T ν ν electromagnetism (in D = 4): T µ µ = 0 theoretical problems for spin 5/2 = spin 2 B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) is universally attractive = even spin (spin 0, 2, 4,...) has deflection of light = not spin 0 spin 0 propagator k 2 only coupling T µ µ k 2 T ν ν electromagnetism (in D = 4): T µ µ = 0 theoretical problems for spin 5/2 = spin 2 = Spin 2 field h µν : Representation theory: 16 = 9 + 6 + 1 with 9 = symmetric & traceless 6 = antisymmetric = spin 1 1 = trace = spin 0 B. Janssen (UGR) Granada, 14 dec 2006 3

Properties of the graviton: Gravity is long range = Massless is static = integer spin (bosonic field) is universally attractive = even spin (spin 0, 2, 4,...) has deflection of light = not spin 0 spin 0 propagator k 2 only coupling T µ µ k 2 T ν ν electromagnetism (in D = 4): T µ µ = 0 theoretical problems for spin 5/2 = spin 2 = Spin 2 field h µν : Representation theory: 16 = 9 + 6 + 1 with 9 = symmetric & traceless 6 = antisymmetric = spin 1 1 = trace = spin 0 = Massless traceless symmetric spin 2 field h µν : B. Janssen (UGR) Granada, 14 dec 2006 3

Outlook 1. Introduction 1: Short review of General Relativity 2. Introduction 2: Electromagnetism as SRFT of spin 1 B. Janssen (UGR) Granada, 14 dec 2006 4

Outlook 1. Introduction 1: Short review of General Relativity 2. Introduction 2: Electromagnetism as SRFT of spin 1 3. Construction of Fierz-Pauli theory 4. Consistency problems with couplings 5. Deser s argument & General Relativity 6. (Philosophical) Conclusions B. Janssen (UGR) Granada, 14 dec 2006 4

1 Short review of General Relativity Einstein & Hilbert (1915): Gravity is curvature of spacetime, caused by matter Interaction between g µν and T µν S = 1 κ d 4 x ] g [R + L matter R µν 1 2 g µνr = κ T µν with R µν = µ Γ λ νλ λ Γ λ µν + Γ λ µσγ σ λν Γ λ µνγ σ λσ ( ) Γ ρ µν = 1 2 gρσ µ g σν + ν g µσ σ g µν T µν = 1 g δl matter δg µν Very geometrical way to understand gravity B. Janssen (UGR) Granada, 14 dec 2006 5

Linear perturbations of Minkowski space Suppose g µν = η µν + εh µν h = h µ µ = η µν h µν B. Janssen (UGR) Granada, 14 dec 2006 6

Linear perturbations of Minkowski space Suppose g µν = η µν + εh µν h = h µ µ = η µν h µν = g µν η µν εh µν R µν 1 2 ε [ 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ ] R µν 1 2 g µνr 1 2 [ ε 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ ( )] η µν 2 h ρ λ h ρλ T µν 0 + ετ µν B. Janssen (UGR) Granada, 14 dec 2006 6

Linear perturbations of Minkowski space Suppose g µν = η µν + εh µν h = h µ µ = η µν h µν = g µν η µν εh µν R µν 1 2 ε [ 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ ] 1 2 R µν 1 2 g µνr [ 1 2 ε 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ ( )] η µν 2 h ρ λ h ρλ T µν 0 + ετ µν Einstein eqn R µν = κ(t µν 1 2 g µνt ) = ] [ 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ κ(τ µν 1 2 η µντ) B. Janssen (UGR) Granada, 14 dec 2006 6

Intermezzo ( Linear perturbations of general solution Suppose g µν = ḡ µν + εh µν h = h µ µ = ḡ µν h µν = R µν R µν + 1 2 ε [ 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ ] T µν T µν + ετ µν Einstein eqn R µν = κ(t µν 1 2 g µνt ) = [ε 0 ] : Rµν = κ( T µν 1 T 2ḡµν ) ] [ε 1 1 ] : 2 [ 2 h µν + µ ν h µ ρ h νρ ν ρ h µρ κ (τ µν 12ḡµντ 12 h T µν + 12ḡµνh ) ρλ Tρλ ) B. Janssen (UGR) Granada, 14 dec 2006 7

Gauge invariance Suppose g µν = η µν + εh µν δx µ = ξ µ, with ξ µ = 0 + εχ µ B. Janssen (UGR) Granada, 14 dec 2006 8

Gauge invariance Suppose g µν = η µν + εh µν δx µ = ξ µ, with ξ µ = 0 + εχ µ = δg µν = ξ ρ ρ g µν + µ ξ ρ g ρν + ν ξ ρ g µρ [ε 0 ] : δη µν = 0 [ε 1 ] : δh µν = µ χ ν + ν χ µ Gauge invariance B. Janssen (UGR) Granada, 14 dec 2006 8

Gauge invariance Suppose g µν = η µν + εh µν δx µ = ξ µ, with ξ µ = 0 + εχ µ = δg µν = ξ ρ ρ g µν + µ ξ ρ g ρν + ν ξ ρ g µρ [ε 0 ] : δη µν = 0 [ε 1 ] : δh µν = µ χ ν + ν χ µ Gauge invariance = δγ ρ µν = ε µ ν χ ρ δr µνρλ = δr µν = 0 B. Janssen (UGR) Granada, 14 dec 2006 8

Gauge invariance Suppose g µν = η µν + εh µν δx µ = ξ µ, with ξ µ = 0 + εχ µ = δg µν = ξ ρ ρ g µν + µ ξ ρ g ρν + ν ξ ρ g µρ [ε 0 ] : δη µν = 0 [ε 1 ] : δh µν = µ χ ν + ν χ µ Gauge invariance = δγ ρ µν = ε µ ν χ ρ δr µνρλ = δr µν = 0 NB: g µν = η µν + εh µν with h µν = µ χ ν + ν χ µ = g µν = η µν since h µν is pure gauge B. Janssen (UGR) Granada, 14 dec 2006 8

Let s forget about all this (for a moment) B. Janssen (UGR) Granada, 14 dec 2006 9

2 Electromagnetism as SRFT of spin 1 Massless spin 1 field A µ : µ µ A ν = j ν B. Janssen (UGR) Granada, 14 dec 2006 10

2 Electromagnetism as SRFT of spin 1 Massless spin 1 field A µ : µ µ A ν = j ν not pos. def. energy density due to non-physical states (A µ : 4 components photon: 2 helicity states) B. Janssen (UGR) Granada, 14 dec 2006 10

2 Electromagnetism as SRFT of spin 1 Massless spin 1 field A µ : µ µ A ν = j ν not pos. def. energy density due to non-physical states (A µ : 4 components photon: 2 helicity states) We can impose Lorenz condition µ A µ = 0 by hand = µ j µ = 0 (charge conservation) B. Janssen (UGR) Granada, 14 dec 2006 10

2 Electromagnetism as SRFT of spin 1 Massless spin 1 field A µ : µ µ A ν = j ν not pos. def. energy density due to non-physical states (A µ : 4 components photon: 2 helicity states) We can impose Lorenz condition µ A µ = 0 by hand = µ j µ = 0 (charge conservation) We would like L such that eqns of motion yield F µ (A) = j µ charge conservation µ F µ (A) = 0 gauge identity implying Lorenz cond. B. Janssen (UGR) Granada, 14 dec 2006 10

Most general Lorentz-invar action, quadratic in A L = α µ A ν µ A ν + β µ A ν ν A µ Eqns of motion: F ν (A) = α 2 A ν + β µ ν A µ = 0 Gauge condition: ν F ν (A) 0 = α = β B. Janssen (UGR) Granada, 14 dec 2006 11

Most general Lorentz-invar action, quadratic in A L = α µ A ν µ A ν + β µ A ν ν A µ Eqns of motion: F ν (A) = α 2 A ν + β µ ν A µ = 0 Gauge condition: ν F ν (A) 0 = α = β The action is [ µ A ν µ A ν µ A ν ν A µ ] L = 1 2 = 1 4 F µνf µν with F µν = µ A ν ν A µ F ν = µ F µν B. Janssen (UGR) Granada, 14 dec 2006 11

Most general Lorentz-invar action, quadratic in A L = α µ A ν µ A ν + β µ A ν ν A µ Eqns of motion: F ν (A) = α 2 A ν + β µ ν A µ = 0 Gauge condition: ν F ν (A) 0 = α = β The action is L = 1 2 [ µ A ν µ A ν µ A ν ν A µ ] = 1 4 F µνf µν with F µν = µ A ν ν A µ F ν = µ F µν Gauge invariance: [ δl δl = µ δ( µ A ν ) δl ] δa ν = µ F µν δa ν µ ν F µν Λ δa ν = δa µ = µ Λ B. Janssen (UGR) Granada, 14 dec 2006 11

Coupling to matter L 0 = 1 2 µφ µ φ 1 4 F µνf µν Eqns of motion: µ µ φ = 0, µ µ φ = 0, µ F µν = 0 Invariant under: δa µ = µ Λ (local U(1)) δφ = ieλφ (global U(1)) B. Janssen (UGR) Granada, 14 dec 2006 12

Coupling to matter L 0 = 1 2 µφ µ φ 1 4 F µνf µν Eqns of motion: µ µ φ = 0, µ µ φ = 0, µ F µν = 0 Invariant under: δa µ = µ Λ (local U(1)) δφ = ieλφ (global U(1)) Noether current for λ: δ λ L 0 ) = ie 2 µλ (φ µ φ φ µ φ = eλ µ j µ (0) j µ (0) preserved on-shell: µj µ (0) ] = [φ i 2 µ µ φ φ µ µ φ B. Janssen (UGR) Granada, 14 dec 2006 12

Coupling to matter L 0 = 1 2 µφ µ φ 1 4 F µνf µν Eqns of motion: µ µ φ = 0, µ µ φ = 0, µ F µν = 0 Invariant under: δa µ = µ Λ (local U(1)) δφ = ieλφ (global U(1)) Noether current for λ: δ λ L 0 ) = ie 2 µλ (φ µ φ φ µ φ = eλ µ j µ (0) j µ (0) preserved on-shell: µj µ (0) ] = [φ i 2 µ µ φ φ µ µ φ j µ (0) is not source term for F µν add coupling term L ea µ j µ (0) B. Janssen (UGR) Granada, 14 dec 2006 12

Coupling term L 1 = 1 2 µφ µ φ 1 4 F µνf µν + ea µ j µ (0) (φ µ φ φ µ φ ) with j µ (0) = ie 2 Eqn of motion: µ F µν = ej ν (0) µ µ φ ie µ A µ φ 2ieA µ µ φ = 0 µ µ φ + ie µ A µ φ + 2ieA µ µ φ = 0 Problem: j µ (0) is not conserved on-shell! 0 = µ ν F µν = µ j µ (0) 0 INCONSISTENCY!!! B. Janssen (UGR) Granada, 14 dec 2006 13

Coupling term L 1 = 1 2 µφ µ φ 1 4 F µνf µν + ea µ j µ (0) (φ µ φ φ µ φ ) with j µ (0) = ie 2 Eqn of motion: µ F µν = ej ν (0) µ µ φ ie µ A µ φ 2ieA µ µ φ = 0 µ µ φ + ie µ A µ φ + 2ieA µ µ φ = 0 Problem: j µ (0) is not conserved on-shell! 0 = µ ν F µν = µ j µ (0) 0 INCONSISTENCY!!! Reason: j µ (0) is Noether current of L 0, not of L 1 Real Noether current: j µ (1) = jµ (0) + eaµ φ φ B. Janssen (UGR) Granada, 14 dec 2006 13

Noether procedure Identify λ and Λ δa µ = µ Λ, δφ = ieλφ B. Janssen (UGR) Granada, 14 dec 2006 14

Noether procedure Identify λ and Λ Then: δl 0 = e Λj µ (0) 0 δa µ = µ Λ, δφ = ieλφ add L ea µ j µ (0) B. Janssen (UGR) Granada, 14 dec 2006 14

Noether procedure Identify λ and Λ Then: δl 0 = e Λj µ (0) 0 δa µ = µ Λ, δφ = ieλφ add L ea µ j µ (0) δl 1 = e Λj µ (0) + e Λjµ (0) e2 µ ΛA µ φ φ 0 add L 1 2 e2 A µ A µ φ φ B. Janssen (UGR) Granada, 14 dec 2006 14

Noether procedure Identify λ and Λ Then: δl 0 = e Λj µ (0) 0 δa µ = µ Λ, δφ = ieλφ add L ea µ j µ (0) δl 1 = e Λj µ (0) + e Λjµ (0) e2 µ ΛA µ φ φ 0 add L 1 2 e2 A µ A µ φ φ L 2 = 1 2 µφ µ φ 1 4 F µνf µν + ea µ j µ (0) + 1 2 e2 A µ A µ φ φ δl 2 = e 2 µ ΛA µ φ φ + e 2 µ ΛA µ φ φ = 0 B. Janssen (UGR) Granada, 14 dec 2006 14

Nota bene: L 2 = 1 2 µφ µ φ 1 4 F µνf µν + ie 2 A µ(φ µ φ φ µ φ ) + 1 2 e2 A µ A µ φ φ = 1 2 ( µφ + iea µ φ )( µ φ iea µ φ) 1 4 F µνf µν = D µ φ D µ φ 1 4 F µνf µν with: D µ φ = µ φ iea µ φ δd µ φ = D µ (δφ) B. Janssen (UGR) Granada, 14 dec 2006 15

Nota bene: L 2 = 1 2 µφ µ φ 1 4 F µνf µν + ie 2 A µ(φ µ φ φ µ φ ) + 1 2 e2 A µ A µ φ φ = 1 2 ( µφ + iea µ φ )( µ φ iea µ φ) 1 4 F µνf µν = D µ φ D µ φ 1 4 F µνf µν with: D µ φ = µ φ iea µ φ δd µ φ = D µ (δφ) Lesson: Lorentz symm & gauge invariance = free theory Gauge invariance & consistent couplings = fully consistent theory coupled to matter B. Janssen (UGR) Granada, 14 dec 2006 15

3 Gravity as a SRFT of spin 2 Massless spin 2 field h µν : ρ ρ h µν = κt µν not pos. def. energy density due to non-physical states (h µν : 4 components graviton: 2 helicity states) B. Janssen (UGR) Granada, 14 dec 2006 16

3 Gravity as a SRFT of spin 2 Massless spin 2 field h µν : ρ ρ h µν = κt µν not pos. def. energy density due to non-physical states (h µν : 4 components graviton: 2 helicity states) Look for action L such that eqns of motion yield H µν (h) = κt µν energy conservation µ T µν = 0 = µ H µν (h) = 0 gauge invariant action B. Janssen (UGR) Granada, 14 dec 2006 16

Most general Lorentz-invar action, quadratic in h L = a µ h νρ µ h νρ + b µ h νρ ν h µρ + c µ h ρ h µρ + d µ h µ h with h = h µ µ = η µν h µν Eqns of motion: H νρ (h) a 2 h νρ + 2b µ (ν h ρ)µ + c (ν ρ) h + c η νρ µ λ h µλ + 2d η νρ 2 h = 0 Gauge condition: ν H νρ (h) 0 = a = 1 2 b = 1 2 c = d B. Janssen (UGR) Granada, 14 dec 2006 17

Most general Lorentz-invar action, quadratic in h L = a µ h νρ µ h νρ + b µ h νρ ν h µρ + c µ h ρ h µρ + d µ h µ h with h = h µ µ = η µν h µν Eqns of motion: H νρ (h) a 2 h νρ + 2b µ (ν h ρ)µ + c (ν ρ) h + c η νρ µ λ h µλ + 2d η νρ 2 h = 0 Gauge condition: ν H νρ (h) 0 = a = 1 2 b = 1 2 c = d The action then is L = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h H µν (h) 2 h µν + µ ν h ρ µ h νρ ρ ν h µρ η µν 2 h + η µν ρ λ h ρλ = 0 (NB: H µν (h) is first order of R µν 1 2 g µνr for g µν = η µν + εh µν ) B. Janssen (UGR) Granada, 14 dec 2006 17

Gauge invariance [ δl δl = µ δ( µ h νρ ) δl ] δh µν = H νρ δh νρ ν H νρ χ δh νρ = δh µν = µ χ ν + ν χ µ (NB: δh µν is first order of δg µν for g µν = η µν + εh µν ) B. Janssen (UGR) Granada, 14 dec 2006 18

Gauge invariance [ δl δl = µ δ( µ h νρ ) δl ] δh µν = H νρ δh νρ ν H νρ χ δh νρ = δh µν = µ χ ν + ν χ µ (NB: δh µν is first order of δg µν for g µν = η µν + εh µν ) Transverse traceless gauge: µ h µν = 0 = h = H µν (h) = ρ ρ h µν = 0 De Donder gauge: µ hµν = 0 with h µν = h µν 1 2 η µνh = H µν (h) = ρ ρ hµν = 0 B. Janssen (UGR) Granada, 14 dec 2006 18

Coupling to matter L = L FP + L φ + κ h µν T µν (φ) Eqns of motion: H µν = κt µν (φ) 2 φ +... = 0 Gauge condition: µ H µν = 0 = µ T µν (φ) = 0 B. Janssen (UGR) Granada, 14 dec 2006 19

Coupling to matter L = L FP + L φ + κ h µν T µν (φ) Eqns of motion: H µν = κt µν (φ) 2 φ +... = 0 Gauge condition: µ H µν = 0 = µ T µν (φ) = 0 Questions: 1. How good is this theory? Experimental verification 2. Is µ T µν (φ) = 0 consistent with 2 φ +... = 0? Consistency problems B. Janssen (UGR) Granada, 14 dec 2006 19

Newtonian limit Point mass: T µν = M δ µ 0 δν 0 δ(x ρ ) = H 00 = κmδ(x ρ ), H ij = 0 = h 00 = Φ, h ii = Φδ ij Φ = M 8π x Action for mass m moving in gravitational potential Φ: 1 } S = m dt{ 1 v2 + [1 + 1 v 2 v2 ]Φ { } 1 dt 2 mv2 mφ 1 4 mv4 + 3 2 mv2 Φ +... (Newtonian potential) = Good agreement for light deflection, time dilatation,... Perihelium Mercury is 0.75 observed value B. Janssen (UGR) Granada, 14 dec 2006 20

4 Consistency problems with couplings L 0 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ Eqns of motion: H µν = 0 µ µ φ = 0 Invariant under: δh µν = µ χ ν + ν χ µ + κ Σ λ λ h µν δφ = κ Σ µ µ φ δx µ = κ Σ µ with Σ µ global symm B. Janssen (UGR) Granada, 14 dec 2006 21

4 Consistency problems with couplings L 0 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ Eqns of motion: H µν = 0 µ µ φ = 0 Invariant under: δh µν = µ χ ν + ν χ µ + κ Σ λ λ h µν δφ = κ Σ µ µ φ δx µ = κ Σ µ Noether current for Σ µ : ] δ Σ L 0 = κ µ Σ [T ν µν (φ) + T µν (h) with Σ µ global symm with T µν (φ) energy-momentum tensor of matter T µν (h) gravitational energy-momentum tensor B. Janssen (UGR) Granada, 14 dec 2006 21

T µν (φ) = µ φ ν φ + 1 2 η µν( φ) 2 T µν (h) = 1 2 µh ρλ ν h ρλ + µ h ρλ ρ h λ ν 1 2 µh ρ h νρ 1 2 ρ h µ h νρ + 1 2 µh µ h + η µν L FP T µν (φ) and T µν (h) preserved on shell: µ T µν (φ) = 2 φ ν φ µ T µν (h) = ν h µρ H µρ B. Janssen (UGR) Granada, 14 dec 2006 22

T µν (φ) = µ φ ν φ + 1 2 η µν( φ) 2 T µν (h) = 1 2 µh ρλ ν h ρλ + µ h ρλ ρ h ν λ 1 2 µh ρ h νρ 1 2 ρ h µ h νρ + 1 2 µh µ h + η µν L FP T µν (φ) and T µν (h) preserved on shell: µ T µν (φ) = 2 φ ν φ µ T µν (h) = ν h µρ H µρ T µν (φ) is not source term for h µν : add coupling term L κh µν T µν (φ) B. Janssen (UGR) Granada, 14 dec 2006 22

Coupling term L 1 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κhµν T µν (φ) Eqns of motion: H µν = κt µν (φ) ] µ µ φ = κ µ [h µν ν φ 1 2 h µφ Problem: T µν (φ) is no longer conserved on shell! 0 = µ H µν = κ µ T µν (φ) = κ 2 φ ν φ 0 (-22) INCONSISTENCY!!! B. Janssen (UGR) Granada, 14 dec 2006 23

Coupling term L 1 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κhµν T µν (φ) Eqns of motion: H µν = κt µν (φ) ] µ µ φ = κ µ [h µν ν φ 1 2 h µφ Problem: T µν (φ) is no longer conserved on shell! Reason: 0 = µ H µν = κ µ T µν (φ) = κ 2 φ ν φ 0 (-22) INCONSISTENCY!!! T µν (φ) is Noether current of L 0, not of L 1 Total energy conserved, not only energy of φ Gravity couples also to own energy non-linear theory B. Janssen (UGR) Granada, 14 dec 2006 23

Gravity self-coupling L 2 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κhµν T µν (φ) + 1 2 κhµν T µν (h) B. Janssen (UGR) Granada, 14 dec 2006 24

Gravity self-coupling L 2 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κhµν T µν (φ) + 1 2 κhµν T µν (h) TOO NAIVE!!! Eqn of motion does not give H µν = κt µν (φ) + κt µν (h) due to contribuition of T µν (h) B. Janssen (UGR) Granada, 14 dec 2006 24

Gravity self-coupling L 2 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κhµν T µν (φ) + 1 2 κhµν T µν (h) TOO NAIVE!!! Eqn of motion does not give H µν = κt µν (φ) + κt µν (h) due to contribuition of T µν (h) T µν (h) too simple for energy-monentum tensor for h µν B. Janssen (UGR) Granada, 14 dec 2006 24

We need L corr = κ h µν L µν κ h( h)( h) such that L 3 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κ hµν T µν (φ) + κ h µν L µν yields B. Janssen (UGR) Granada, 14 dec 2006 25

We need L corr = κ h µν L µν κ h( h)( h) such that L 3 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ yields 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κ hµν T µν (φ) + κ h µν L µν ( ) T µν (h) = L µν σ h ρλ δl ρλ δ σ h µν B. Janssen (UGR) Granada, 14 dec 2006 25

We need L corr = κ h µν L µν κ h( h)( h) such that L 3 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ yields 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κ hµν T µν (φ) + κ h µν L µν ( ) T µν (h) = L µν σ h ρλ δl ρλ δ σ h µν L 3 is invariant upto order κ 2 under δh µν = µ χ ν + ν χ µ + κ χ ρ ρ h µν δφ = κ χ ρ ρ φ B. Janssen (UGR) Granada, 14 dec 2006 25

We need L corr = κ h µν L µν κ h( h)( h) such that L 3 = 1 4 µh νρ µ h νρ 1 2 µh νρ ν h µρ + 1 2 µh ρ h µρ yields 1 4 µh µ h + 1 2 µφ µ φ + 1 2 κ hµν T µν (φ) + κ h µν L µν ( ) T µν (h) = L µν σ h ρλ δl ρλ δ σ h µν L 3 is invariant upto order κ 2 under δh µν = µ χ ν + ν χ µ + κ χ ρ ρ h µν δφ = κ χ ρ ρ φ ) H µν = κt µν (φ) + κt µν (h) (T µ µν (φ) + T µν (h) = 0 B. Janssen (UGR) Granada, 14 dec 2006 25

Most general term for L corr : L corr = 1 2 κ hµν [α µ h ρλ ν h ρλ + β µ h ρλ ρ h λ ν +... ] 20 terms with 16 coefficients determine coefficients by demanding µ T µν (h) = γ ν ρλ Hρλ B. Janssen (UGR) Granada, 14 dec 2006 26

Most general term for L corr : L corr = 1 2 κ hµν [α µ h ρλ ν h ρλ + β µ h ρλ ρ h λ ν +... ] 20 terms with 16 coefficients determine coefficients by demanding µ T µν (h) = γ ν ρλ Hρλ = L corr = 1 2 κ hµν [ 1 2 µh ρλ ν h ρλ ρ h µ λ ρ h λν + λ h ρ (µ ρ h ν)λ +2 λ h ρ (µ ν) h ρλ (µ h ν)λ λ h λ h µν ρ h ρλ λ h λ(µ ν) h + λ h µν λ h + 1 2 µh ν h + η µν L FP ] Eqns of motion: H µν = κt µν (h) + κt µν (φ) ] µ µ φ = κ µ [h µν ν φ 1 2 h m uφ ] Conservation of energy: [T µ µν (h) + T µν (φ) = 0 B. Janssen (UGR) Granada, 14 dec 2006 26

Consistent result upto order κ Right values for perihelium of Mercury previous deficit due to neglecting self-interaction What about order κ 2, κ 3,...? Extremely tedious!! B. Janssen (UGR) Granada, 14 dec 2006 27

Consistent result upto order κ Right values for perihelium of Mercury previous deficit due to neglecting self-interaction What about order κ 2, κ 3,...? Extremely tedious!! WE NEED A MORE GENERAL ARGUMENT! B. Janssen (UGR) Granada, 14 dec 2006 27

5 Deser s argument & General Relativity L 0 = κ h ] ] µν [ µ Γ ρ νρ ρ Γ ρ µν + η [Γ µν λ µργ ρ νλ Γλ µνγ ρ λρ with h µν = h µν 1 2 η µνh Invariant under: δh µν = µ χ ν + ν χ µ δγ ρ µν = µ ν χ ρ Γ ρ µν = Γ ρ νµ Eqns of motion: [ h µν ] : ρ Γ ρ µν 1 2 µγ ρ νρ 1 2 µγ ρ νρ = 0 [Γ ρ µν] : κ ρ hµν + κ λ hλ(µ δρ ν) + 2η λ(µ Γ ν) ρλ η λσ Γ (µ λσ δν) ρ η µν Γ λ ρλ = 0 [Γ ρ µν] is algebraic eqn for Γ ρ µν Constraint on Γ ρ µν B. Janssen (UGR) Granada, 14 dec 2006 28

Constraint on Γ ρ µν: Γ ρ µν = 1 2 κ ηρλ [ µ h λν + ν h µλ λ h µν ] Γ ρ µν is not independent field ( Γ ρ µν is 1st order of Christoffel symbol for g µν = η µν + εh µν ) B. Janssen (UGR) Granada, 14 dec 2006 29

Constraint on Γ ρ µν: Γ ρ µν = 1 2 κ ηρλ [ µ h λν + ν h µλ λ h µν ] Γ ρ µν is not independent field ( Γ ρ µν is 1st order of Christoffel symbol for g µν = η µν + εh µν ) Substitute in [ h µν ]: 2 h µν + µ ν ρ µ h νρ ρ ν h µρ = 0 H µν 1 2 η µνh ρ ρ = 0 ( 1st order of R µν = 0 for g µν = η µν + εh µν ) = L 0 is equivalent to L FP B. Janssen (UGR) Granada, 14 dec 2006 29

We want to find correction to L 0 which takes in account the self-interaction of h µν, such that H µν = κt µν L 1 = κ h ] [ ] µν [ µ Γ ρ νρ ρ Γ ρ µν + (η µν κh µν ) Γ λ µργ ρ νλ Γλ µνγ ρ λρ Eqns of motion: [ h µν ]: R µν (Γ) = 0 [Γ ρ µν]: κ ρ hµν + κ λ hλ(µ δρ ν) + 2(η λ(µ κ h λ(µ )Γ ν) ρλ (η λσ κ h λσ )Γ (µ λσ δν) ρ (η µν κ h µν )Γ λ ρλ = 0 B. Janssen (UGR) Granada, 14 dec 2006 30

We want to find correction to L 0 which takes in account the self-interaction of h µν, such that H µν = κt µν L 1 = κ h ] [ ] µν [ µ Γ ρ νρ ρ Γ ρ µν + (η µν κh µν ) Γ λ µργ ρ νλ Γλ µνγ ρ λρ Eqns of motion: [ h µν ]: R µν (Γ) = 0 [Γ ρ µν]: κ ρ hµν + κ λ hλ(µ δρ ν) + 2(η λ(µ κ h λ(µ )Γ ν) ρλ (η λσ κ h λσ )Γ (µ λσ δν) ρ (η µν κ h µν )Γ λ ρλ = 0 Correction term has no η µν = No contribution to T µν (h) No further correction needed B. Janssen (UGR) Granada, 14 dec 2006 30

We want to find correction to L 0 which takes in account the self-interaction of h µν, such that H µν = κt µν L 1 = κ h ] [ ] µν [ µ Γ ρ νρ ρ Γ ρ µν + (η µν κh µν ) Γ λ µργ ρ νλ Γλ µνγ ρ λρ Eqns of motion: [ h µν ]: R µν (Γ) = 0 [Γ ρ µν]: κ ρ hµν + κ λ hλ(µ δρ ν) + 2(η λ(µ κ h λ(µ )Γ ν) ρλ (η λσ κ h λσ )Γ (µ λσ δν) ρ (η µν κ h µν )Γ λ ρλ = 0 Correction term has no η µν = No contribution to T µν (h) No further correction needed R µν (Γ) = ρ Γ ρ µν µγ ρ νρ + Γλ µν Γρ λρ Γλ µρ Γρ νλ (Surprise!) What is the conecction Γ ρ µν? is Ricci tensor B. Janssen (UGR) Granada, 14 dec 2006 30

Define η µν κ h µν = ĝ µν ĝ µν ĝ νρ = δ ρ µ g µν = ĝ ĝ µν g µν non-linear combination of h µν (infinite series) g µν acts as metric B. Janssen (UGR) Granada, 14 dec 2006 31

Define η µν κ h µν = ĝ µν ĝ µν ĝ νρ = δ ρ µ g µν = ĝ ĝ µν g µν non-linear combination of h µν (infinite series) g µν acts as metric Constraint on Γ ρ µν: Γ ρ µν = 1 2 gρλ [ µ g λν + ν g µλ λ g µν ] Γ ρ µν is Levi-Civita conection R µν (Γ) = R µν (g) [ h µν ] is Einstein eqn R µν = 0 B. Janssen (UGR) Granada, 14 dec 2006 31

Eqn of motion R µν = 0 invariant under general coord transf However L 1 is not! ] add total derivative η µν [ µ Γ ρ νρ ρ Γ ρ µν B. Janssen (UGR) Granada, 14 dec 2006 32

Eqn of motion R µν = 0 invariant under general coord transf However L 1 is not! add total derivative η µν [ µ Γ ρ νρ ρ Γ ρ µν ] = L 2 = (η µν κ h [ ] µν ) µ Γ ρ νρ ρ Γ ρ µν = g µν [ µ Γ ρ νρ ρ Γ ρ µν = g µν R µν (g) [ ] +(η µν κh µν ) Γ λ µργ ρ νλ Γλ µνγ ρ λρ Einstein-Hilbert action! ] [ ] + g µν Γ λ µργ ρ νλ Γλ µνγ ρ λρ B. Janssen (UGR) Granada, 14 dec 2006 32

6 Conclusions Fierz-Pauli L FP is consistent as free theory Couplings to matter = self-coupling of gravity = inconsistent Possible (but difficult) to construct consistent theory upto order κ 2 General Relativity is consistent extension to all orders (unique?) Suprising geometrical interpretation: non-linear function of h µν is metric very elegant & much richer structure B. Janssen (UGR) Granada, 14 dec 2006 33