Download Motion of gyroscopes around Schwarzschild and Kerr BH

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.
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Robert M. Wald, Phys. Rev. D 6 (1972) 406
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Thank you