Gravitational duality: a NUT story Francois Dehouck U.L.B. Brussels September 8, 2009 Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 1 / 20
Gravitational duality: a NUT story Francois Dehouck U.L.B. Brussels September 8, 2009 ”Supersymmetry and gravitational duality”: R. Argurio, L. Houart, F.D. [PRD 79:125001, 2009] ”Boosting Taub-NUT to a BPS NUT-wave”: R. Argurio, L. Houart, F.D. [JHEP 0901:045,2009] ”Why not a di-NUT ? ”: R. Argurio, F.D. [hep-th:0909.0542 ] Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 1 / 20
Understanding Quantum Gravity... One of the goals of theoretical physics is to find a quantum theory for gravity... ...through S-duality Duality between weakly and strongly coupled sectors of a theory is a powerful tool to delve into its non-perturbative physics. Supersymmetry helps in providing protected quantities that can be compared in both weakly and strongly coupled (or electric and magnetic) sectors. Goal of this talk presence of dyonic metrics in general relativity and an adapted EM duality in linearized Gravity. Show the presence of this duality in supergravity. Establish the supersymmetry of duality rotated supersymmetric solutions. Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 2 / 20
Outline Electromagnetic duality 1 Duality in Linearized Gravity + Examples 2 N = 2 Supersymmetric solutions with NUT charge 3 N = 1 Supersymmetric solutions with NUT charge 4 Conclusions and future work 5 Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 3 / 20
Duality in EM Duality in electromagnetism states that for every ”electric” field strength, there is a dual ”magnetic” field strength. The duality is a Hodge duality: F µν ≡ ( ∗ F ) µν = 1 F µν ˜ 2 ε µνρσ F ρσ → → Q H Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 4 / 20
Duality in EM Duality in electromagnetism states that for every ”electric” field strength, there is a dual ”magnetic” field strength. The duality is a Hodge duality: F µν ≡ ( ∗ F ) µν = 1 F µν ˜ 2 ε µνρσ F ρσ → → Q H Example: Coulomb charge vs. magnetic monopole Q F = Q A = r dt r 2 dt ∧ dr ˜ ˜ F = H sin θ d θ ∧ d φ A = − H cos θ d φ Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 4 / 20
Duality in EM Duality in electromagnetism states that for every ”electric” field strength, there is a dual ”magnetic” field strength. The duality is a Hodge duality: F µν ≡ ( ∗ F ) µν = 1 F µν ˜ 2 ε µνρσ F ρσ → → Q H Example: Coulomb charge vs. magnetic monopole Q F = Q A = r dt r 2 dt ∧ dr z ˜ ˜ F = H sin θ d θ ∧ d φ A = − H cos θ d φ = − H r ( r 2 − z 2 )( xdy − ydx ) Note: If we look at the gauge potential, the magnetic monopole has a Dirac string singularity along the z-axis. Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 4 / 20
The monopole as a magnetic charge H: Introduce a magnetic current in the Bianchi identity: d ∗ F = 4 π J el dF = 4 π J magn Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 5 / 20
The monopole as a magnetic charge H: Introduce a magnetic current in the Bianchi identity: d ∗ F = 4 π J el dF = 4 π J magn A solution to this set of equations is: ⇒ F = dA R + C J magn = dC Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 5 / 20
The monopole as a magnetic charge H: Introduce a magnetic current in the Bianchi identity: d ∗ F = 4 π J el dF = 4 π J magn A solution to this set of equations is: ⇒ F = dA R + C J magn = dC The charges for the magnetic monopole are: � d ∗ F = 0 � dF = H Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 5 / 20
The monopole as a magnetic charge H: Introduce a magnetic current in the Bianchi identity: d ∗ F = 4 π J el dF = 4 π J magn A solution to this set of equations is: ⇒ F = dA R + C J magn = dC The charges for the magnetic monopole are: � d ∗ F = 0 � dF = H Is there something similar in linearized gravity ? Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 5 / 20
Duality in linearized gravity The Lorentzian Taub-NUT solution found in [Taub ′ 51; Newman, Tamburino, Unti ′ 63] is: ds 2 = − λ 2 R 2 [ dt + 2 Ncos θ d φ ] 2 + R 2 λ 2 dr 2 + R 2 [ d θ 2 + sin 2 θ d φ 2 ] where λ 2 = r 2 − 2 Mr − N 2 and R 2 = r 2 + N 2 Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 6 / 20
Duality in linearized gravity The Lorentzian Taub-NUT solution found in [Taub ′ 51; Newman, Tamburino, Unti ′ 63] is: ds 2 = − λ 2 R 2 [ dt + 2 Ncos θ d φ ] 2 + R 2 λ 2 dr 2 + R 2 [ d θ 2 + sin 2 θ d φ 2 ] where λ 2 = r 2 − 2 Mr − N 2 and R 2 = r 2 + N 2 We will consider linearized theory around flat space. Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 6 / 20
Duality in linearized gravity The Lorentzian Taub-NUT solution found in [Taub ′ 51; Newman, Tamburino, Unti ′ 63] is: ds 2 = − λ 2 R 2 [ dt + 2 Ncos θ d φ ] 2 + R 2 λ 2 dr 2 + R 2 [ d θ 2 + sin 2 θ d φ 2 ] where λ 2 = r 2 − 2 Mr − N 2 and R 2 = r 2 + N 2 We will consider linearized theory around flat space. The duality can be expressed as a Hodge duality on the Riemann tensor: 2 ε µναβ R αβ R µνρσ → ˜ R µνρσ = 1 M → N ρσ [Henneaux, Teitelboim ′ 04] Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 6 / 20
Generalizing the EM idea: [Bunster, Cnockaert, Henneaux, Portugues ′ 06] The duality is taken on the Lorentz indices: ˜ 2 ε µναβ R αβ R µνρσ = 1 ρσ . We introduce a ”magnetic” stress-energy tensor Θ µν : G µν = 8 π T µν R µναβ + R µβνα + R µαβν = 0 ∂ ǫ R γδαβ + ∂ α R γδβǫ + ∂ β R γδǫα = 0 Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 7 / 20
Generalizing the EM idea: [Bunster, Cnockaert, Henneaux, Portugues ′ 06] The duality is taken on the Lorentz indices: ˜ 2 ε µναβ R αβ R µνρσ = 1 ρσ . We introduce a ”magnetic” stress-energy tensor Θ µν : G µν = 8 π T µν − ε ναβγ ˜ G γ − 8 πε ναβγ Θ γ µ ≡ R µναβ + R µβνα + R µαβν = µ ∂ ǫ R γδαβ + ∂ α R γδβǫ + ∂ β R γδǫα = 0 Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 7 / 20
Generalizing the EM idea: [Bunster, Cnockaert, Henneaux, Portugues ′ 06] The duality is taken on the Lorentz indices: ˜ 2 ε µναβ R αβ R µνρσ = 1 ρσ . We introduce a ”magnetic” stress-energy tensor Θ µν : G µν = 8 π T µν − ε ναβγ ˜ G γ − 8 πε ναβγ Θ γ µ ≡ R µναβ + R µβνα + R µαβν = µ ∂ ǫ R γδαβ + ∂ α R γδβǫ + ∂ β R γδǫα = 0 Solution: R αβλµ = r αβλµ + f (Φ) → ∂ α Φ αβ γ = − 16 π Θ β γ Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 7 / 20
Generalizing the EM idea: [Bunster, Cnockaert, Henneaux, Portugues ′ 06] The duality is taken on the Lorentz indices: ˜ 2 ε µναβ R αβ R µνρσ = 1 ρσ . We introduce a ”magnetic” stress-energy tensor Θ µν : G µν = 8 π T µν − ε ναβγ ˜ G γ − 8 πε ναβγ Θ γ µ ≡ R µναβ + R µβνα + R µαβν = µ ∂ ǫ R γδαβ + ∂ α R γδβǫ + ∂ β R γδǫα = 0 Solution: R αβλµ = r αβλµ + f (Φ) → ∂ α Φ αβ γ = − 16 π Θ β γ Charges in general relativity: L µν = ( x µ T 0 ν − x ν T 0 µ ) d 3 x T 0 µ d 3 x � � P µ = L µν = ( x µ Θ 0 ν − x ν Θ 0 µ ) d 3 x ˜ � Θ 0 µ d 3 x � K µ = [Ramaswamy, Sen ′ 81],[Ashtekar, Sen ′ 82],[Mueller, Perry ′ 86], [Bossard, Nicolai, Stelle ′ 09] Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 7 / 20
1. The Kerr-NUT black hole [Carter ′ 68] : − λ 2 ds 2 R 2 [ dt − ( asin 2 θ − 2 Ncos θ ) d φ ] 2 = + sin 2 θ R 2 [( r 2 + a 2 + N 2 ) d φ − adt ] 2 + R 2 λ 2 dr 2 + R 2 d θ 2 , where λ 2 = r 2 − 2 Mr + a 2 − N 2 and R 2 = r 2 + ( N + acos θ ) 2 Gravitational duality on the linearized Schwarzschild (Kerr) solution gives us the linearized NUT (rotating) solution. Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 8 / 20
1. The Kerr-NUT black hole [Carter ′ 68] : − λ 2 ds 2 R 2 [ dt − ( asin 2 θ − 2 Ncos θ ) d φ ] 2 = + sin 2 θ R 2 [( r 2 + a 2 + N 2 ) d φ − adt ] 2 + R 2 λ 2 dr 2 + R 2 d θ 2 , where λ 2 = r 2 − 2 Mr + a 2 − N 2 and R 2 = r 2 + ( N + acos θ ) 2 Gravitational duality on the linearized Schwarzschild (Kerr) solution gives us the linearized NUT (rotating) solution. Linearized Schwarzschild ( a = 0 , N = 0): P 0 = M Francois Dehouck (U.L.B. Brussels) Gravitational duality: a NUT story September 8, 2009 8 / 20
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