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Motion of gyroscopes around Schwarzschild and Kerr BH exact gravito-electromagnetic analogies Luís Filipe P. O. Costa, Carlos A. R. Herdeiro Centro de Física do Porto, Universidade do Porto - Portugal II International Black Holes Workshop, Lisbon Introduction I It is known, since the works of Mathisson and Papapetrou that spinning particles follow worldlines which are not geodesics; I In linearized theory, the gravitational force exerted on a spinning pole-dipole test particle (hereafter a gyroscope) takes a form: ~ G = ∇(B ~ G .~S ) F similar to the electromagnetic force on a magnetic dipole (Wald 1972). I But only if the gyroscope is at rest in a stationary, weak eld! I This analogy may be cast in an exact form (Natário, 2007) using the Quasi-Maxwell formalism, which holds if the gyroscope's 4-velocity is a Killing vector of a stationary spacetime. Introduction I It is known, since the works of Mathisson and Papapetrou that spinning particles follow worldlines which are not geodesics; I In linearized theory, the gravitational force exerted on a spinning pole-dipole test particle (hereafter a gyroscope) takes a form: ~ G = ∇(B ~ G .~S ) F similar to the electromagnetic force on a magnetic dipole (Wald 1972). I But only if the gyroscope is at rest in a stationary, weak eld! I This analogy may be cast in an exact form (Natário, 2007) using the Quasi-Maxwell formalism, which holds if the gyroscope's 4-velocity is a Killing vector of a stationary spacetime. I There is an exact, covariant and fully general analogy relating the two forces, which is made explicit in the tidal tensor formalism (Costa & Herdeiro 2008). I We will exemplify how this analogy provides new intuition for the understanding of spin curvature coupling. Force on Magnetic Dipole α FEM = I I DP α 1 ;α µν = Fµν Q dτ 2 Fµν ≡ Maxwell tensor Qµν ≡ dipole moment tensor IFor d = 0): a magnetic dipole (~ 0 −d x −d y µz dx 0 Q µν = d y −µz 0 d z µy −µx −d z −µy = q S µν x µ 2m 0 Force on Magnetic Dipole α FEM = I I DP α 1 ;α µν = Fµν Q dτ 2 Fµν ≡ Maxwell tensor Qµν ≡ dipole moment tensor IFor d = 0): a magnetic dipole (~ 0 −d x −d y µz dx 0 Q µν = d y −µz 0 d z µy −µx −d z −µy = q S µν x µ 2m 0 I In Relativity, electric and magnetic dipole moments do not exist as independent entities; I ~d and µ ~ are the time and space components of the dipole moment 2-Form. Force on Magnetic Dipole α FEM = DP α 1 ;α µν = Fµν Q dτ 2 I Fµν ≡ Maxwell tensor Qµν ≡ dipole moment tensor I For a magnetic dipole (~ I d = 0 in its proper frame): 0 0 Q µν = 0 0 0 0 −µz µy 0 µz 0 −µx 0 −µy = σ S µν µx 0 Force on Magnetic Dipole α FEM = DP α 1 ;α µν = Fµν Q dτ 2 I Fµν ≡ Maxwell tensor Qµν ≡ dipole moment tensor I For a magnetic dipole (~ I d = 0 in its proper frame): 0 0 Q µν = 0 0 0 0 0 −µz µy µz 0 −µx 0 −µy = σ S µν µx 0 S µν I ≡ Spin tensor; I σ ≡ gyromagnetic ratio (= q /2m α I FEM = for classical spin) DP α 1 = σ Fµν;α S µν dτ 2 Force on Magnetic Dipole α FEM = DP α 1 = σ Fµν;α S µν dτ 2 I If Pirani supplementary condition S µν Uν = 0 holds, then S µν = µντ λ Sτ Uλ I S α ≡ spin 4-vector; dened as the vector that, in the particle's proper frame, S α = (0, ~S ) Force on Magnetic Dipole α FEM = DP α 1 = σ Fµν;α S µν dτ 2 I If Pirani supplementary condition S µν Uν = 0 holds, then S µν = µντ λ Sτ Uλ I S α ≡ spin 4-vector; dened as the vector that, in the particle's proper frame, S α = (0, ~S ) α FEM = DP α = σµντ λ F µν;α Uλ Sτ = σ Bβα S β dτ Force on Magnetic Dipole α FEM = DP α 1 = σ Fµν;α S µν dτ 2 I If Pirani supplementary condition S µν Uν = 0 holds, then S µν = µντ λ Sτ Uλ I S α ≡ spin 4-vector; dened as the vector that, in the particle's proper frame, S α = (0, ~S ) α FEM = I DP α = σµντ λ F µν;α Uλ Sτ = σ Bβα S β dτ Bαβ ≡ ?Fαγ;β U γ ≡ magnetic tidal tensor I Measures the tidal eects produced by the magnetic eld B α = ?F αγ U γ γ seen by the dipole of 4-velocity U . Force on Magnetic Dipole α FEM = σ Bβα S β I Covariant generalization of the usual 3-D expression (valid in a frame where the dipole is at rest!): ~) F~ EM = ∇(µ. ~B I Yields the force exerted on a magnetic dipole moving with arbitrary velocity. only Force on Gyroscope Papapetrou equation: FGα ≡ 1 DP α = − R αβµν U β S µν Dτ 2 Force on Gyroscope Papapetrou equation: FGα ≡ 1 DP α = − R αβµν U β S µν Dτ 2 I If Pirani supplementary condition S µν Uν S µν = µντ λ Sτ Uλ DP α 1 τ λ µναβ = µν R Uλ Uβ Sτ = Dτ I Hαβ ≡Magnetic 2 =0 holds, then −Hβα S β part of the Riemann tensor Magnetic-type Tidal Tensors The electromagnetic force exerted on a magnetic dipole and the gravitational force causing the non-geodesic motion of a spinning test particle are analogous tidal eects: I Suggests the physical analogy: Bαβ ←→ Hαβ ≡ magnetic tidal tensor; Hαβ ≡ gravito-magnetic tidal I B αβ I tensor. Magnetic-type Tidal Tensors The electromagnetic force exerted on a magnetic dipole and the gravitational force causing the non-geodesic motion of a spinning test particle are analogous tidal eects: I σ = µ/S ≡ ⇒µ ~ ↔ ~S gyromagnetic ratio ⇒ equals 1 for gravity Magnetic-type Tidal Tensors The electromagnetic force exerted on a magnetic dipole and the gravitational force causing the non-geodesic motion of a spinning test particle are analogous tidal eects: I Relative minus sign: mass/charges of the same sign attract/repel one another repel/attract. ⇒ antiparallel charge/mass currents Magnetic Tidal Tensor Gravito-Magnetic Tidal tensor Antisymmetric part: B[αβ] = 1 2 I γ ? Fαβ;γ U − 2παβσγ j Covariant form for ~ ~ = ∂ E + 4π~j ∇×B ∂t Trace: B α α =0 nI Covariant form for ~ =0 ∇·B Antisymmetric part: σ U γ H[αβ] = −4παβσγ J σ U γ of Einstein equations: Rµν = 8π Tµν − 1 α 2 gµν T α Trace: Hαα = 0 ITime-Time projection Magnetic Tidal Tensor Gravito-Magnetic Tidal tensor Antisymmetric part: B[αβ] 1 2 = I γ ? Fαβ;γ U − 2παβσγ j Space projection of Maxwell equations: F αβ ;β = Jβ Trace: α B α I =0 Time projection of Bianchi Identity: ?F αβ;β = 0 Antisymmetric part: σ U γ H[αβ] = −4παβσγ J σ U γ I Time-Space projection of Einstein equations: Rµν = 8π Tµν − 1 α 2 gµν T α Trace: Hαα = 0 ITime-Time projection of Bianchi Identity: ?R γαγβ = 0 Electric-type Tidal Tensors Electric-type tidal forces are described in an invariant way through the wordline deviation equations: which yield the acceleration of the vector particles with the δx α connecting two same (Ciufolini, 1986) 4-velocity U α q /m ratio in the electromagnetic case. (Notation: Fαβ ≡ Maxwell tensor, Rαβγσ ≡ Riemann tensor) same and the Electric-type Tidal Tensors Electric-type tidal forces are described in an invariant way through the wordline deviation equations: I Suggests the physical analogy: I Eαβ ←→ Eαβ Eαβ is the covariant derivative of the electric eld E α = F αµ Uµ measured by the observer with (xed) 4-velocity U α; Electric-type Tidal Tensors Electric-type tidal forces are described in an invariant way through the wordline deviation equations: I Suggests the physical analogy: Eαβ ←→ Eαβ I Hence: I E αβ ≡ electric tidal tensor; Eαβ ≡ gravito-electric tidal tensor. Analogy based on tidal tensors (Costa-Herdeiro 2008) Electromagnetism Gravity Worldline deviation: Geodesic deviation: D Dτ 2 δx α 2 = q α β E δx m β D δx α = −Eαβ δ x β Dτ 2 2 Force on magnetic dipole: Force on gyroscope: Maxwell Equations: Eqs. Grav. Tidal Tensors: DP β q αβ = B Sα Dτ 2m E αα = 4πρc DP β = −Hαβ Sα Dτ Eαα = 4π (2ρm + T αα ) E[αβ] = Fαβ;γ U γ 2 E[αβ] = 0 B αα = 0 Hαα = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 1 2 The Gravitational analogue of Maxwell's Equations Electromagnetism Gravity Maxwell Equations Eqs. Grav. Tidal Tensors E αα = 4πρc Eαα = 4π (2ρm + T αα ) B αα = 0 Hαα = 0 E[αβ] = Fαβ;γ U γ E[αβ] = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 2 1 2 I Strikingly similar when the setups are stationary in the γ γ observer's rest frame (since Fαβ;γ U and ?Fαβ;γ U vanish). The Gravitational analogue of Maxwell's Equations Electromagnetism Gravity Maxwell Equations Eqs. Grav. Tidal Tensors E αα = 4πρc Eαα = 4π (2ρm + T αα ) B αα = 0 Hαα = 0 E[αβ] = Fαβ;γ U γ E[αβ] = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 2 1 2 I Charges: the gravitational analogue of ρc is 2ρm + T αα (ρm + 3p for a perfect uid) ⇒ in gravity, pressure and all material stresses contribute as sources. The Gravitational analogue of Maxwell's Equations Electromagnetism Gravity Maxwell Equations Eqs. Grav. Tidal Tensors E αα = 4πρc Eαα = 4π (2ρm + T αα ) B αα = 0 Hαα = 0 E[αβ] = Fαβ;γ U γ E[αβ] = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 2 1 2 I Ampére law: in stationary (in the observer's rest frame) setups, equations B[αβ] and H[αβ] match up to a factor of 2 ⇒ currents of mass/energy source gravitomagnetism like currents of charge source magnetism. The Gravitational analogue of Maxwell's Equations Electromagnetism Gravity Maxwell Equations Eqs. Grav. Tidal Tensors E αα = 4πρc Eαα = 4π (2ρm + T αα ) B αα = 0 Hαα = 0 E[αβ] = Fαβ;γ U γ E[αβ] = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 2 1 2 I Absence of electromagnetic-like induction eects in gravity: I Eµγ always symmetric ⇒ no gravitational analogue to Faraday's law of induction! The Gravitational analogue of Maxwell's Equations Electromagnetism Gravity Maxwell Equations Eqs. Grav. Tidal Tensors E αα = 4πρc Eαα = 4π (2ρm + T αα ) B αα = 0 Hαα = 0 E[αβ] = Fαβ;γ U γ E[αβ] = 0 B[αβ] = ? Fαβ;γ U γ − 2παβσγ j σ U γ H[αβ] = −4παβσγ J σ U γ 1 2 1 2 I Absence of electromagnetic-like induction eects in gravity: I Induction term ⇒ ?Fαβ;γ U γ in B[αβ] has no counterpart in H[αβ] no gravitational analogue to the magnetic elds induced by time varying electric elds. Magnetic dipole vs Gyroscope Electromagnetic Force Gravitational Force on a Magnetic Dipole on a Spinning Particle β FEM = q αβ B Sα 2m The explicit analogy between FGβ = −Hαβ Sα β FEM and FGβ is ideally suited to: I Compare the two interactions: amounts to compare Hαβ , Magnetic Tidal Tensor B[αβ] = 1 2 Bαβ and which is crystal clear from the equations for tidal tensors: ? Fαβ;γ U γ − 2παβσγ j σ U γ B α α =0 Gravito-Magnetic Tidal tensor H[αβ] = −4παβσγ J σ U γ Hαα = 0 Magnetic dipole vs Gyroscope Electromagnetic Force Gravitational Force on a Magnetic Dipole on a Spinning Particle β FEM = q αβ B Sα 2m The explicit analogy between FGβ = −Hαβ Sα β FEM and FGβ is ideally suited to: I Compare the two interactions: amounts to compare Hαβ , Bαβ and which is crystal clear from the equations for tidal tensors: I Unveils similarities between the two forces which allow us visualize, in analogy with the more familiar electromagnetic ones, gravitational eects which are not transparent in the Papapetrou's original form. I and fundamental dierences which prove especially enlightening to the understanding of spin-curvature coupling. Some Fundamental Dierences Electromagnetic Force Gravitational Force on a Magnetic Dipole on a Spinning Particle β FEM = I I I I FGβ = −Hαβ Sα Bαβ is linear, whereas Hαβ is not In vacuum H[αβ] = 0 (symmetric tensor); B[αβ] = ? Fαβ;γ U γ 6= 0 (even in vacuum) 1 2 Hαβ Uβ force). I q αβ B Sα 2m =0 (spatial tensor) ⇒ FGβ Uβ = 0 (it is a spatial β Bαβ U β 6= 0 ⇒ FEM Uβ 6= 0 (non-vanishing time projection!) Symmetries of Tidal tensors Electromagnetic Force Gravitational Force on a Magnetic Dipole on a Gyroscope β EM = F q 2m B αβ Sα Magnetic Tidal Tensor B[αβ] = 1 2 ? Fαβ;γ U γ − 2παβσγ j σ U γ F β β α G = −Hα S Gravito-magnetic Tidal Tensor H[αβ] = −4παβσγ J σ U γ I If the elds do not vary along the test particle's wordline, ?Fαβ;γ U γ = 0 and the tidal tensors have the same symmetries. I Allows for a similarity between the two interactions. Gravitational Spin-Spin Force i FG ' 3 c " (i j ) (~r · ~J ) ij r J (~r · ~J )r i r j δ + 2 − 5 5 5 7 r r r # S J ↔µ i j = FEM An analogy already known from linearized theory (Wald, 1972), and usually cast in the form: ~ G = −∇(~S · B ~ G )+ ∂ ijk φ,k Si ~ej ∂t F I Holds only if the gyroscope is at rest and the elds are stationary. Gravitational Spin-Spin Force i FG ' 3 c " (i j ) (~r · ~J ) ij r J (~r · ~J )r i r j δ + 2 − 5 5 5 7 r r r # S J ↔µ i j = FEM An analogy already known from linearized theory (Wald, 1972), and usually cast in the form: ~ G = −∇(~S · B ~ G )+ ∂ ijk φ,k Si ~ej ∂t F I Holds only if the gyroscope is at rest and the elds are stationary. Gravitational Spin-Spin Force i FG ' 3 c " (i j ) (~r · ~J ) ij r J (~r · ~J )r i r j δ + 2 − 5 5 5 7 r r r # S J ↔µ i j = FEM An analogy already known from linearized theory (Wald, 1972), and usually cast in the form: ~ G = −∇(~S · B ~ G )+ ∂ ijk φ,k Si ~ej ∂t F I Not suitable to describe motion; accounts only for spin-spin coupling. Gravitational Spin-Spin Force i FG ' 3 c " (i j ) (~r · ~J ) ij r J (~r · ~J )r i r j δ + 2 − 5 5 5 7 r r r # S J ↔µ i j = FEM An analogy already known from linearized theory (Wald, 1972), and usually cast in the form: ~ G = −∇(~S · B ~ G )+ ∂ ijk φ,k Si ~ej ∂t F I Not suitable to describe motion; accounts only for spin-spin coupling. The gyroscope deviates from geodesic motion even in the absence of rotating sources (e.g. Schwarzschild spacetime). An eect readily visualized using the explicit analogy (always valid!!): Force on a Magnetic Dipole F β EM = q 2m B αβ Sα I It the magnetic tidal tensor, Force on a Gyroscope F β αβ G = −H Sα as seen by the test particle, that determines the force exerted upon it; I Hence the gyroscope deviates from geodesic motion by the same reason that a magnetic dipole suers a force even in the coulomb eld of a point charge: in its rest frame, there is a non-vanishing magnetic tidal tensor. Symmetries of Tidal tensors Electromagnetic Force Gravitational Force on a Magnetic Dipole on a Gyroscope β EM = F q 2m B αβ Sα Magnetic Tidal Tensor B[αβ] = 1 2 ? Fαβ;γ U γ − 2παβσγ j σ U γ I If the elds F β β α G = −Hα S Gravito-magnetic Tidal Tensor H[αβ] = −4παβσγ J σ U γ vary along the test particle's wordline, the two interactions dier signicantly. I In vacuum, H[αβ] B[αβ] = ? Fαβ;γ 1 2 is always symmetric, whereas Bαβ Uγ. is not : Radial Motion in Coulomb Field I The dipole sees a time varying electric eld; I Thus, B[αβ] = 1 ? Fαβ;γ U γ 6= 0 2 i = I FEM q qQ B iSα = γ (~v × ~S )i m α 2mr 2 3 Radial Motion in Schwarzschild I No analogous gravitational eect: along a geodesic. FGα = 0 ⇒ gyroscope moves Scalar Invariants The Riemann tensor (20 independent components) splits irreducibly into three spatial tensors (Louis Bel, 1958): Iin vacuum Rαβγδ = Fαβ = −Eαβ n 4 2 o Ũ[α Ũ [γ + g[α[γ Ẽβ]δ] n o +2 αβµν H̃µ[δ Ũ γ] Ũ ν + γδµν H̃µ[β Ũα] Ũν n o +αβµν γδστ Ũ µ Ũσ F̃ντ + Ẽντ − gτν Ẽρρ Fαβ ≡ ?R ?αµβν U µ U ν I Eαβ , Fαβ : spatial, symmetric tensors ⇒ 6 independent components each; I Hαβ : I Fαβ spatial, traceless tensor ⇒ 8 independent components has no electromagnetic analogue. Scalar Invariants The Riemann tensor (20 independent components) splits irreducibly into three spatial tensors (Louis Bel, 1958): I Fαβ = −Eαβ n o [γ δ] = 4 2Ũ[α Ũ [γ + g[α Ẽβ] n o +2 αβµν H̃µ[δ Ũ γ] Ũ ν + γδµν H̃µ[β Ũα] Ũν n o +αβµν γδστ Ũ µ Ũσ F̃ντ + Ẽντ − gτν Ẽρρ in vacuum Rαβγδ Fαβ ≡ ?R ?αµβν U µ U ν I Eαβ , Fαβ : spatial, symmetric tensors ⇒ 6 independent components each; I Hαβ : I Fαβ spatial, traceless tensor ⇒ 8 independent components has no electromagnetic analogue. Scalar Invariants The Riemann tensor (20 independent components) splits irreducibly into three spatial tensors: I Rαβγδ I Fαβ = −Eαβ n o [γ δ] = 4 2Ũ[α Ũ [γ + g[α Ẽβ] n o +2 αβµν H̃µ[δ Ũ γ] Ũ ν + γδµν H̃µ[β Ũα] Ũν n o +αβµν γδστ Ũ µ Ũσ F̃ντ + Ẽντ − gτν Ẽρρ in vacuum Eαβ and Hαβ completely encode the 14 (6+8) independent components of the Riemann tensor in vacuum (Weyl Tensor). Scalar Invariants Although each of these spatial tensors is determined by the 4-velocity Uα of the observer measuring it: Eαβ ≡ Rαµβν U µ U ν Hαβ ≡ ?Rαµβν U µ U ν it can be shown that the following expressions are observer independent (in vacuum):: Eαγ Eαγ − Hαγ Hαγ = Eαγ Hαγ = 1 8 Rαβγδ R αβγδ 1 16 Rαβγδ ? R αβγδ I gravitational tidal tensors form scalar invariants! Scalar Invariants I in vacuum I Eαγ Eαγ − Hαγ Hαγ = Eαγ Hαγ = 1 8 Rαβγδ R αβγδ 1 16 Rαβγδ ? R αβγδ Formally analogous to the electromagnetic scalar invariants: ~ E~ − B 2 2 1 = − Fαβ F αβ 2 ~ = − 1 Fαβ ? F αβ E~ .B 4 Scalar Invariants I in vacuum I Eαγ Eαγ − Hαγ Hαγ = Eαγ Hαγ = 1 8 Rαβγδ R αβγδ 1 16 Rαβγδ ? R αβγδ Formally analogous to the electromagnetic scalar invariants: ~ E~ − B 2 2 1 = − Fαβ F αβ 2 ~ = − 1 Fαβ ? F αβ E~ .B 4 I This is a purely formal analogy, relating electromagnetic elds with gravitational tidal tensors (which are one order higher in dierentiation!) Scalar Invariants Electromagnetism ~ E~ − B 2 2 1 = − Fαβ F αβ 2 ~ = − 1 Fαβ ? F αβ E~ .B 4 I I ~ = 0 and E ~ −B ~ >0 ⇒ E~ .B ~ the magnetic eld B vanishes. 2 2 there are observers for which ~ = 0 and E ~ −B ~ < 0 ⇒ there are observers for which E~ .B ~ vanishes. the electric eld E 2 2 Scalar Invariants Electromagnetism ~ E~ − B 2 2 1 = − Fαβ F αβ 2 ~ = − 1 Fαβ ? F αβ E~ .B 4 I I I ~ = 0 and E ~ −B ~ >0 ⇒ E~ .B ~ the magnetic eld B vanishes. 2 2 there are observers for which ~ = 0 and E ~ −B ~ < 0 ⇒ there are observers for which E~ .B ~ vanishes. the electric eld E 2 ~ E~ − B 2 2 and ~ E~ .B 2 are the only algebraically independent invariants one can dene from the Maxwell tensor F αβ . Scalar Invariants Gravity (Vacuum) In vacuum, one can construct 4 independent scalar invariants from Riemann tensor (would be 14 in general): Eαγ Eαγ − Hαγ Hαγ = Eαγ Hαγ = 1 8 1 16 I R.? R − =0 α β γ 3E β E γ H α and R.R 6= 0 are observers for which I Needs also 16 1 = 1 8 16 . R R Rαβγδ ? R αβγδ ≡ 1 Eαβ Eβγ Eγα − 3Eαβ Hβγ Hγα = Hαβ Hβγ Hγα Rαβγδ R αβγδ ≡ 1 16 R αβλµ R λµρσ R ρσ R ρσαβ ≡ A R αβλµ R λµρσ R ρσ ? R ρσαβ ≡ B is not sucient to ensure that there Hαγ (or Eαγ ) vanishes. M ≡ I /J − 6 to be real positive, where 1 i I ≡ R · R + ?R · R; J ≡ A − iB 3 8 .? R R 2 8 Scalar Invariants Gravity (Vacuum) In vacuum, one can construct 4 independent scalar invariants from Riemann tensor (would be 14 in general): Eαγ Eαγ − Hαγ Hαγ = Eαγ Hαγ = Eαβ Eβγ Eγα − 3Eαβ Hβγ Hγα = Hαβ Hβγ Hγα − 3Eαβ Eβγ Hγα = I 8 Rαβγδ R αβγδ ≡ 1 16 1 8 Rαβγδ ? R αβγδ ≡ 1 16 1 16 . R R 1 16 .? R R R αβλµ R λµρσ R ρσ R ρσαβ ≡ A R αβλµ R λµρσ R ρσ ? R ρσαβ ≡ B M ≥ 0 (real), R.? R = 0 and R.R > 0 ⇒ there are observers for which I 1 Hαγ vanishes (Purely Electric spacetime) . M ≥ 0 (real), R.? R = 0 and R.R < 0 ⇒ there would be observers for which Eαγ vanishes (but there are no known Purely Magnetic vacuum solutions; conjecture: do not exist) αγ αγ E Eαγ − H Hαγ ≈ 6m r Kerr metric: Eαγ Hαγ ≈ 18Jm cos θ r I In the equatorial plane locally Hαγ = 0 7 2 > 0; 6 =0 θ = π/2, I 3 = 6J 2 (Petrov in the plane θ = π/2 there are observers for which : I Observers such that vφ ≡ co-rotating observers!) D) Uφ a = t U a +r 2 2 2 (not the I Impossible in the electromagnetic analog. I A moving dipole sees a time-varying electromagnetic eld; I Thus Bαβ must be non vanishing (consequence of ~ = ∂E ~ /∂ t ): ∇×B B[αβ] = 1 2 ? Fαβ;γ U γ 6= 0 ⇒ Bαβ 6= 0 ~2 ~2 E −B = q r Spinning Charge: 2 4 − µ2 (5 + 3 cos 2θ) 2r 6 ~ .B ~ = 2µq cos θ = 0 E 5 r I In the equatorial plane locally ~ = 0: B I Observers such that observers!) θ = π/2, vφ ≡ in the plane >0 θ = π/2 there are observers for which Uφ J = U t 2mr 2 (not the co-rotating Spinning Charge vs Spinning mass I For r → ∞, the two velocities asymptotically match! factor of 2) (up to a I ~ =0 ⇒ IB I I No precession: d~ S dt ~ =0 I I There is a Force applied: Bαβ α 6= 0 ⇒ FEM = (consequence of Gyroscope precesses: q 2m B βα Sβ 6= 0 ~ = ∂E ~ /∂ t ) ∇×B dS dt 6= 0 No Force on Gyroscope: Hαβ = 0 ⇒ FGα = −Hβα Sβ = 0 Time projection of α FEM in the dipole's proper frame: ˛ ~ ~ ∂B ∂B ∂Φ α ~ FEM Uα = −B i µi = .~ µ= .~nAI = I = −I E~ .ds ∂t ∂t ∂t loop 0 ~ I The magnetic dipole may be thought as a small current loop.E A = 4πa2 ; I ≡ current through the loop, ~n ≡unit (Area of the loop vector normal to the loop) I The magnetic dipole moment is given by µ ~ = IA~n Time projection of α FEM in the dipole's proper frame: ˛ ~ ~ ∂B ∂B ∂Φ α ~ FEM Uα = −B i µi = .~ µ= .~nAI = I = −I E~ .ds ∂t ∂t ∂t loop 0 ~ I The magnetic dipole may be thought as a small current loop.E (Area of the loop A = 4πa2 ; I ≡ current through the loop, ~n ≡unit vector normal to the loop) ~ ~n = Φ ≡ magnetic ux trough the loop I BA Therefore, by Faraday´s law of induction: α FEM Uα = −B i µi = 0 I ˛ ~ ~ ∂B ∂B ∂Φ ~ .~ µ= .~nAI = I = −I E~ .ds ∂t ∂t ∂t loop E~ ≡ Induced electric eld I Hence α U FEM α is minus the power transferred to the dipole by Faraday's law of induction. I α U = DE /d τ FEM α is minus the power transferred to the dipole by Faraday's law of induction; I yields the variation of the energy E as measured in the dipole's center of mass rest frame; I is reected in a variation of the dipole's proper mass m = −P α Uα Time Projection of FGα no gravitational induction Since Hαβ is a spatial tensor, we always have FGα Uα = − I No work is done by induction dm =0 dτ ⇒ the energy of the gyroscope, as measured in its center of mass frame, is constant; I the proper mass m = −P α Uα of the gyroscope is constant. I Spatial character of gravitational tidal tensors precludes induction eects analogous to the electromagnetic ones. Time Projection of FGα no gravitational induction Example: A mass loop subject to the time-varying gravitomagnetic eld of a moving Kerr Black Hole Time Projection of FGα no gravitational induction Example: A mass loop subject to the time-varying gravitomagnetic eld of a moving Kerr Black Hole Time Projection of FGα no gravitational induction Example: A mass loop subject to the time-varying gravitomagnetic eld of a moving Kerr Black Hole Time components on arbitrary frames I Electromagnetism: In an arbitrary frame, in which the dipole has 4-velocity U β = γ(1, ~v ), the time component of the force exerted on a magnetic dipole is: (FEM )0 β Uβ DE FEM i v = − 1 dm + F i v = − FEM ≡ − i EM i dτ γ γ dτ ≡ −(Pmech + Pind ) E ≡ −P is the energy of the dipole and we identify: 1 dm β Pind = = −FEM Uβ /γ ≡ induced power γ dτ i v mechanical power transferred to the dipole Pmech = FEM i i by the 3-force FEM exerted upon it. where I I 0 Static Observers Electromagnetism (FEM )0 = q B S α = ?Fαγ; U γ S α = 0 m α 0 2 0 I no work is done on the magnetic dipole. I Related to a basic principle from electromagnetism: the total amount of work done by a static magnetic eld on an arbitrary system of currents is zero. ~ = q~v × B ~ ⇒ F ~ ⊥ ~v F (Lorentz Force on each individual charge) Static Observers Electromagnetism I When the elds are stationary in the observer's rest frame, (FEM )0 = 0 ⇒ I Pmech and no work is done on the magnetic dipole. Pind exactly cancel out Static Observers Electromagnetism I When the elds are stationary in the observer's rest frame, (FEM )0 = 0 ⇒ I Pmech and no work is done on the magnetic dipole. Pind exactly cancel out Static Observers Electromagnetism I When the elds are stationary in the observer's rest frame, (FEM )0 = 0 ⇒ I Pmech and no work is done on the magnetic dipole. Pind exactly cancel out. Static Observers Gravity I In gravity, since those induction eects are absent, such cancellation does not occur: (FG )0 = − DE = −FGi vi 6= 0 dτ I Therefore, the stationary observer must measure a non-zero work done on the gyroscope. I That is to say, a static gravitomagnetic eld (unlike its electromagnetic counterpart) does work. I And there is a known consequence of this fact: the spin dependent upper bound for the energy released when two black holes collide, obtained by Hawking (1971) from the area law. I For the case with spins aligned, from Hawking's expression one can infer a gravitational spin-spin interaction energy (Wald, 1972). Static Observers Gravitational spin interaction Take the gyroscope to be a small Kerr black hole of spin √ S ≡ S α Sα hole of mass falling along the symmetry axis of a larger Kerr black m and angular momentum J = am. Static Observers Gravitational spin interaction Take the gyroscope to be a small Kerr black hole of spin √ S ≡ S α Sα hole of mass falling along the symmetry axis of a larger Kerr black m and angular momentum J = am. Static Observers Gravitational spin interaction Take the gyroscope to be a small Kerr black hole of spin √ S ≡ S α Sα hole of mass falling along the symmetry axis of a larger Kerr black m and angular momentum J = am. Static Observers Gravitational spin interaction Take the gyroscope to be a small Kerr black hole of spin √ S ≡ S α Sα hole of mass falling along the symmetry axis of a larger Kerr black m and angular momentum J = am. I The time component of the force acting on the small black hole is given by: 2ma 3r − a Ur S dE DP =− =− (FG ) ≡ Dτ dτ (r + a ) 2 2 0 0 2 2 3 Integrating this equation from innity to the horizon one obtains ˆ r+ ∞ (FG )0 ≡ ∆E = aS 2 h √ m m+ m −a 2 2 i , which is precisely Hawking's spin-spin interaction energy for this particular setup. Conclusions I The tidal tensor formalism unveils an exact, fully general analogy between the force on a gyroscope and on a magnetic dipole; I at the same time it makes transparent both the similarities and key dierences between the two interactions; I The non-geodesic motion of a spinning test particle not only can be easily understood, but also exactly described, by a simple application of this analogy. I This analogy sheds light on important aspects of spin-curvature coupling; I the fact that the mass of a gyroscope is constant (as signaling the absence of gravitational eects analogous to electromagnetic induction); I namely, Hawking's spin dependent upper bound for the energy released on black hole collision (as arising from the fact that gravitomagnetic elds do work); I Issues concerning previous approaches in the literature were claried namely, the limit of validity of the usual linear gravito-electromagnetic analogy, and the physical interpretation of the magnetic parts of the Riemann/Weyl tensors. Acknowledgments We thank Rui Quaresma ([email protected]) for the illustrations. L. F. Costa, C. A. R. Herdeiro, Phys. Rev. D 78 024021 (2008) L. F. Costa, C. A. R. Herdeiro, J. Natário, M. Zilhão (work in progress) A. Papapetrou, Proc. R. Soc. London A 209 (1951) 248 Robert M. Wald, Phys. Rev. D 6 (1972) 406 José Natário, Gen. Rel. Grav. 39 (2007) 1477 S. W. Hawking, Phys Rev. 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