1 Astrophysical and Cosmological Applications of General Relativity ETH Zurich and University of Zurich FS 011 Prof. Dr. Philippe Jetzer typeset and r...

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FS 2011

Prof. Dr. Philippe Jetzer ¨ hl typeset and revision: Felix Ha

June 2, 2011

CONTENTS

Contents 1 Time Delay of Radar Echoes

2

2 Geodetic Precession 2.1 Geodetic Precession . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 De Sitter Precession of the Moon . . . . . . . . . . . . . .

5 6 10

3 Linearized Field Equations 3.1 The Energy Momentum Tensor of the Grav. Field . . . . . . . . . 3.2 Linearized Field Equations . . . . . . . . . . . . . . . . . . . . . .

12 12 14

4 Lense-Thirring Effect 4.1 Metric of the Rotating Earth . . . . . . . . . . . . . . . . . . . . . 4.1.1 Rotation of the Local IS . . . . . . . . . . . . . . . . . . . 4.2 Gravitomagnetic Forces . . . . . . . . . . . . . . . . . . . . . . . .

16 16 18 21

5 Gravitational Waves 5.1 Electromagnetic Waves . . . . . . . . . . . . . . 5.2 The Gravity Case . . . . . . . . . . . . . . . . . 5.2.1 Helicity . . . . . . . . . . . . . . . . . . 5.3 Particles in the Field of a Wave . . . . . . . . . 5.4 Energy and Momentum of a Gravitational Wave 5.5 Quadrupole Radiation . . . . . . . . . . . . . . 5.5.1 Dipole Radiation in Electromagnetism . 5.5.2 Gravitational Quadrupole Radiation . .

. . . . . . . .

22 22 23 24 25 27 30 30 30

6 Sources of Gravitational Waves 6.1 Rigid Rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Example: The Double Star System . . . . . . . . . . . . . 6.2 The PSR 1913+16 System . . . . . . . . . . . . . . . . . . . . . .

37 37 38 40

7 The post-Newtonian Approximation 7.1 The 1PN Approximation . . . . . . . . . . . . . . . . . . . . . . . 7.2 Motion of Test Particles in the 1PN Metric . . . . . . . . . . . . . 7.2.1 Two Body Problem in the 1PN Approximation . . . . . .

43 45 47 50

8 The 8.1 8.2 8.3

56 56 58 59

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Kerr Solution Interpretation of the Parameters a and m . . . . . . . . . . . . . . Kerr-Newman Solution . . . . . . . . . . . . . . . . . . . . . . . . Equations of Motion for Test Particles . . . . . . . . . . . . . . .

1

1. TIME DELAY OF RADAR ECHOES

1

Time Delay of Radar Echoes

In 1964 Shapiro proposed a new test of GR consisting in a measurement of the time delay of radar signals which are transmitted from Earth through a region near the Sun to another planet or spacecraft and reflected back to Earth. Since the radar signal is affected by the gravitational field of the Sun, it will return to Earth with a certain time delay. The radar signal is emitted from Earth and then sent back by a reflector as sketched in Fig. 1. In calculations we set θ = π2 . The trajectory of the signal is curved. This is not shown in the figure because the curvature is, of course, only a tiny effect and the light ray seems almost straight. Let’s compute the time that the signal needs to get from Earth to the reflector. To this end we use Eq. (23.15) for the orbit in a static isotropic gravitational field with ε = 0 (since m = 0): A

dr dλ

2 +

F2 l2 − = 0. r2 B

(1.1)

If we divide this by l2 and use x0 = ct, we find 1 dr dx0 dr = dλ c dt dλ

(23.12)

=

1 dr F . c dt B

(1.2)

This yields AF 2 c2 B 2 l 2

dr dt

2 +

1 F2 − = 0. r2 Bl2

(1.3)

For the minimal distance r0 (from the Sun) it holds dr =0 dt

F2 B(r0 ) = . 2 l r02

⇒

(1.4)

Sun ϕR

ϕE

rE

rR

r0

Earth

Reflector

Figure 1: Sketch of the situation as described in the text. The angles ϕE and ϕR and the distances rE , rR describe the position of the Earth and the reflector relative to the Sun. The whole setup is assumed to lie in the θ = π2 -plane. 2

1. TIME DELAY OF RADAR ECHOES We insert this into Eq. (1.3) to obtain A 2 cB

dr dt

2 +

r02 B(r) − 1 = 0. r2 B(r0 )

This differential equation is solved by the following integral: r −1/2 Z r02 B(r0 ) 1 r 0 A dr 1 − 02 t(r, r0 ) = c r0 B r B(r0 )

(1.5)

(1.6)

where t(r, r0 ) is the time that the radar signal needs to travel from r0 to r. Note that this is the time which would be shown by a clock at rest at infinity (as space is asymptotically Minkowskian at infinity). This actually forces us to introduce a correction since our clock rests at Earth, not at infinity. However, the correction which is needed to compensate this effect is much smaller than the time delay and can thus be neglected. Using the Robertson expansion from Eq. (22.3), A(r) = 1 + γ

2a + ..., r

B(r) = 1 −

2a + ..., r

(1.7)

we get r02 B(r) r02 1 1 1− 2 = 1 − 2 1 + 2a − r B(r0 ) r r0 r r02 2ar0 = 1− 2 1− . r r(r + r0 )

(1.8)

Inserting Eqs. (1.7), (1.8) into (1.6) and expanding, we get −1/2 Z a 1 r 0 ar0 r02 + (1 + γ) 0 t(r, r0 ) ' dr 1 − 02 1+ 0 0 c r0 r r (r r0 ) r ! p p r r2 − r02 a r − r0 r + r2 − r02 a = + + (1 + γ) log . c2 r r + r0 c r0 √

r2 −r2

(1.9)

0 corresponds to the travelling time assuming a straight The first term c2 trajectory in Euclidean space as can easily be seen from the figure and Pythagoras’ theorem. The other terms account for the general relativistic time delay due to the gravitational field of the Sun. For the situation drawn in the figure, the total delay is " # p p rE2 − r02 rR2 − r02 δt = 2 t(rE , r0 ) + t(rR , r0 ) − − . (1.10) c c

3

1. TIME DELAY OF RADAR ECHOES where the factor of 2 accounts for the fact that the signal travels from Earth to the reflector and back again. Significant delays occur if the radar signal passes nearby the Sun, i.e. if r0 is of the order of some Sun radii. In this case we have rE , rR R and also rE , rR r0 . We thus can perform the following approximations in Eq. (1.10):1 1+γ 4rE rR 4a 1+ log . (1.11) δt ' c 2 r02 We see that δt is maximal if the signal just grazes the surface of the Sun, i.e. r0 = R . In order to see the orders of magnitude, we use rE ∼ rR ∼ 108 km, R ' 7·105 = 2GM km. This yields 2a ' 10−5 s and thus c rc2 4a 1+γ 4rE2 rR2 δtmax = 1+ log 2 c 2 R ' 2 · 10−4 s. (1.12) Performing the measurements has been a very difficult task since the precise distances rE and rR were not known with sufficient precision. Nevertheless, in the seventies these measurements were performed using Venus and Mercury and afterwards also using spacecrafts as reflectors (e.g. the Vikings which landed on Mars or, more recently, the Cassini spacecraft). The results are2 Vikings: γ = 1.000 ± 0.001, Cassini: γ = 1 + (2.1 ± 2.3) · 10−5 .

(1.13)

Another experimental verification of the Shapiro delay is the measurement of the PSR J1614-2230 system. It consists of a pulsar which emitts signals in very regular time intervals, and a white dwarf that orbits the pulsar. When the white dwarf is in front of the pulsar and the light of the pulsar arrives at Earth only by passing the white dwarf, then the signal arrives with a certain delay. Measuring the Shapiro delay3 , one can infer the mass of the white dwarf to be 0.500 ± 0.006M . With an orbital period of 8.7 days, this yields a neutron star mass of 1.97 ± 0.04M . This reslt is important for the modelling of neutron stars since the largest neutron stars (so far) had masses of about 1.4M .

1

I. I. Shapiro, Phys. Rev. Lett. 13, 789 (1964) B. Bertotti et al., Nature 425, 374 (2003) 3 P. Demorest et al., Nature 467, 1081 (2010) 2

4

2. GEODETIC PRECESSION

2

Geodetic Precession

Consider a particle with a “classical” angular momentum (for instance the intrinsic angular momentum of a rigid body like a gyroscope). In the local inertial system in which the body is at rest, the spin (i.e. angular momentum) is given by S = S i ei . To the three-vector S i we assign a Lorentz vector S α . Consider now a locally inertial coordinate frame IS’ which is momentarily at rest with respect to the rigid body (or particle): S 0α = (0, S 0i ).

(2.1)

We can transform this to some arbitrary inertial system IS by means of a Lorentz transformation. In the rest frame IS’, the velocity of the body is u0α = (c, 0).

(2.2)

u0α S 0α = 0.

(2.3)

Therefore we have in IS’

Since this quantity is a Lorentz scalar, we have uα S α = 0 in any arbitrary IS. Consider first the case without any forces acting on the particle and no torque acting on its spin. In an arbitrary IS we have dS α = 0. dτ

(2.4)

∂xµ α S ≡ αS ∂ξ

(2.5)

We define the Riemann vector µ

which describes the transition from the coordinate system IS with coordinates (ξ α ) to a general system (xµ ). According to the covariance principle, the generalization of Eq. (2.4) reads DS µ =0 dτ

dS µ = −Γµ νλ uν S λ . dτ

⇔

(2.6)

This equation describes the spin precession in a gravitational field. The condition uα S α = 0 reads uµ S µ = 0 in the general coordinate system. Note that Sµ S µ =const., which implies that (2.6) describes the rotation or precession of the spin vector. Since we assumed no external forces, Eq. (2.6) contains only gravitational effects. We conclude that Eq. (2.6) describes the precession of the spin of a particle which is freely falling in a gravitational field as, for example, the precession of a rotating satellite (or gyroscope).

5

2.1 GEODETIC PRECESSION If there are other external forces f µ besides gravity, then one finds instead of Eq. (2.6) 1 Duµ DS ν =− 2 Sµ uν dτ c dτ

(2.7)

which is also called Fermi transport. It describes the spin precession of an acµ µ celerated particle on which a gravitational field acts (c.f. Eq. (19.2), Du = fm ). dτ The special case f µ = 0 is just parallel transport. Based on the above considerations, we shall study the following effects (in the gravitational field of the Earth): 1. Geodetic precession: the precession of a freely falling gyroscope. In order to simplify the analysis we will assume the gravitational field to be isotropic and static. 2. Lense-Thirring effect: the precession of a gyroscope in the gravitational field of the Earth which is due to the rotation of the Earth.

2.1

Geodetic Precession

Gyroscopes are rigid bodies which can also perform rotations described by S µ besides the movement of its center of mass. To compute the geodetic precession of a gyroscope we use Eq. (2.6). In the local rest frame of the satellite (orbiting the Earth) we have for the spin vector: S 0α = (0, l)

(2.8)

where l describes the angular momentum of the gyroscope. We use the standard form of the static and isotropic metric in spherical coordinates (cf. chapter (22.1)): xµ = (ct, r, θ, ϕ) g = diag B(r), −A(r), −r2 , −r2 sin2 θ . Assume that the satellite is on a circular orbit, i.e. r = const.,

θ=

π , 2

ϕ = ω0 τ.

(2.9)

Therefore the velocity of the satellite reads uµ =

dxµ = u0 = const., 0, 0, u3 = ω0 = const. . dτ

6

(2.10)

2.1 GEODETIC PRECESSION If we insert θ = π2 into the Christoffel symbols from Eq. (22.6), we obtain for the non-zero components Γ1 00 =

B0 , 2A

Γ0 01 = Γ0 10 =

Γ1 11 =

B0 , 2B

A0 , 2A

Γ2 12

r r Γ1 22 = − , Γ1 33 = − A A 1 1 = Γ2 21 = , Γ3 13 = Γ3 31 = . r r

(2.11)

Eq. (2.6) reads then dS 0 dτ dS 1 dτ dS 2 dτ dS 3 dτ

= −Γ0 01 u0 S 1

(2.12)

= −Γ1 00 u0 S 0 − Γ1 33 u3 S 3

(2.13)

=0

(2.14)

= −Γ3 31 u3 S 1 .

(2.15)

We can immediately solve the third of these equations: S 2 (τ ) = const.

(2.16)

that is the component of the spin (or angular momentum) of the gyroscope which is perpendicular to the satellite’s orbit (θ-direction) is constant. Because of r =const. all coefficients of the system of linear differential equations (2.12)-(2.15) are constants. We differentiate Eq. (2.13) with respect to τ and insert (2.12) and (2.15) on the right-hand side. This yields 1 d2 S 1 1 0 0 2 1 3 3 2 = Γ Γ (u ) + Γ Γ (u ) S 00 01 33 31 dτ 2 ≡ −ω 2 S 1 .

(2.17)

With (u3 )2 = ω02 and inserting the Christoffel symbols from (2.11), we get " # 0 2 02 B u 1 ω 2 = ω02 − + . (2.18) 4AB u3 A 0

In order to understand the ratio uu3 better, we look at the equation for the trajectory of the satellite (geodetic equation): duµ = −Γµ νλ uν uλ . dτ

(2.19)

The µ = 1 component of this equation is (u1 = 0) 0=

du1 = −Γ1 00 (u0 )2 − Γ1 33 (u3 )2 dτ 7

(2.20)

2.1 GEODETIC PRECESSION from which we infer

u0 u3

2

s

1 A

=

2r . B0

(2.21)

Inserting this into (2.18), we find ω = ω0

rB 0 1− . 2B

Using the Schwarzschild solution (B = A−1 = 1 − r 1−

ω = ω0

2a ), r

(2.22) this yields

3a r

(2.23)

or in terms of the Robertson expansion r a ω = ω0 1 − (1 + 2γ) . r

(2.24)

Since r =const., Eq. (2.17) is of type S¨1 + ω 2 S 1 = 0

(2.25)

which describes a harmonic oscillator. With initial conditions S 1 (0) = S and S˙ 1 (0) = 0 the solution reads S 1 (τ ) = S cos(ωτ ),

S 2 = const..

(2.26)

Inserting this into Eq. (2.15), we obtain by integration S 3 (τ ) = −

Sω0 sin(ωτ ). rω

(2.27)

We proceed by studying the time dependence of the projection (S 1 , S 3 ) (or (r, ϕ)-components) of the spin vector onto the orbital plane (θ = π2 ), c.f. Fig. 2. Consider the constant vector ex along the orbit of the satellite: ex = cos( ω0 τ )er − sin( ω0 τ )eϕ . |{z} |{z} =ϕ

(2.28)

=ϕ

The orbital period of the satellite is τ0 = ω2π0 . After each orbit, the argument ϕ in (2.28) increases by τ0 ω0 = 2π whereas the argument in (2.26) or (2.27) increases

8

2.1 GEODETIC PRECESSION ~ey

~eϕ

~er

ωτ ~ex

Figure 2: Projection of the motion orbit of precession of the angular momentum vector onto the orbital plane of the satellite. The precession takes place in the (r, ϕ)-plane since the θ-component of the spin vector (i.e. the component perpendicular to the orbital plane) is constant. by τ0 ω which differs slightly from 2π. The phase difference after one orbit is given by ∆α = τ0 (ω0 − ω) r 1−

= 2π − 2π

(1 + 2γ)a r

(1 + 2γ)a . (2.29) r Consider the vector S which is the projection of S µ onto the orbital plane: =π

S = Sr er + Sϕ eϕ .

(2.30)

The components of this vector are given by (Sr )2 = −g11 S 1 S 1

and

(Sϕ )2 = −g33 S 3 S 3

(2.31)

with −g11 = A(r)

and

− g33 = r2 sin2 θ θ= π = r2 .

(2.32)

2

Therefore Sr ∝ cos(ωτ ) and Sϕ ∝ sin(ωτ ). For τ = 0, S is thus parallel to ex . However, after an orbit, ωτ differs slightly from 2π as we have calculated in Eq. (2.29). The geodetic precession after one orbit is given by ∆α =

3πa (1 + 2γ) . r 3 9

(2.33)

2.1 GEODETIC PRECESSION Consider the concrete example of a satellite in a circular orbit around the Earth. We have 1/2 GME RE 2 ω0 RE = =g and τ0 = 2π . (2.34) 2 RE g After one year (i.e. after

t τ0

orbits with t = 1 year), we find (assuming r ' RE )

∆α(t) = ∆α

t 3πgRE t q ' 800 .4 yr−1 . = τ0 c2 2π RE g

(2.35)

For a general radius r, one finds ∆α(t) ' 800 .4 (RE /r)5/2 yr−1 . On April 20, 2004 the satellite Gravity Probe B has been launched to measure the geodetic precession. At an altitude of 642 km, general relativity predicts a geodetic precession of 6606 mas yr−1 . The measured result was 6673 mas yr−1 (1 mas = 1 milliarcsecond). This result therefore matches the predictions reasonably well.4 2.1.1

De Sitter Precession of the Moon

The Earth-Moon system can be considered as a “gyroscope” with an axis “perpendicular” to the orbital plane in an orbit around the Sun. We denote by l the angular momentum of the Earth-Moon system with respect to the common center of mass (l is essentially the orbital angular momentum of the Moon because the common center of mass almost coincides with the center of the Earth). The angular momentum l can be decomposed in a component l⊥ perpendicular to the Earth’s orbital plane around the Sun and a parallel component l|| . The parallel component does not vanish because the orbital plane of the Moon around the Earth is tilted by 5 with respect to the orbital plane of the Earth around the Sun. In terms of the previously defined quantities, l⊥ corresponds to S 2 and stays constant. The component l|| corresponds to S and lies in the orbital plane of the Earth around the Sun. It is this parallel component which is affected by geodetic precession. Therefore the orbital plane of the Moon rotates slightly (this was first noticed by De Sitter in 1916). We can calculate this precession per century: ∆αDe Sitter = 100

3πa rEarth-Sun

' 200 per century

(2.36)

Additionally we have a Newtonian precession (as in any three body system) with a period of 18.6 yr. This Newtonian effect is 107 times larger than the De Sitter precession. Nevertheless the De Sitter precession has been measured. A 4

C. Everitt et al., Classical and Quantum Gravity 25, (2008) 114002

10

2.1 GEODETIC PRECESSION laser beam has been sent to the the Moon where it was reflected back to Earth by mirrors previously brought to the Moon by the Apollo mission (1969 and following years). The measurements that have been performed from 1970 till 1986 confirmed the De Sitter precession with a precision of about 1%.5

5

Shapiro et al., Phys. Rev. Lett. 61, 2643 (1988) and M¨ uller et al., Astrophys. J. 382, L101 (1991)

11

3. LINEARIZED FIELD EQUATIONS

3

Linearized Field Equations

In order to find solutions to the Einstein field equations in the weak field regime, one can linearize the equations. The results of this section will, for example, be used to calculate the Lense-Thirring effect and to describe gravitational waves. Since the field equations are non-linear, there is no standard procedure to solve the equations given a source for the fields Tµν (for example, in electrodynamics we didn’t have to struggle with such difficulties). Besides numerical methods, there are essentially three possibilities as for which type of solutions one can find: • Exact solutions assuming simplifying conditions (as, for example, staticity, isotropy, ...). An example for this case is the Schwarzschild solution. • Solutions of the linearized field equations for weak gravitational fields. • Systematic expansion of the field equations and the equations of motion for weak fields and small velocities. This method is also called postNewtonian approximation. For example, in planetary systems we have v2 ∼ cφ2 which is very small. Such results should reproduce the Newtonian c2 limit in lowest order (linearized in φ). We shall now elaborate the second possibility.

3.1

The Energy Momentum Tensor of the Gravitational Field

The field itself is a form of energy and thus also a source of the field. This effect is purely due to the non-linearities, of course. Considering a weak field, we can work with small deviations from the Minkowski metric: with |hµν | 1.

gµν = ηµν + hµν

(3.1)

One proceeds as follows. First, Gµν has to be expanded in powers of hµν . The first order terms will lead to a linear wave equation. Neglecting terms of third order, the second order equations give the energy-momentum tensor of the gravitational field. The expansion of the Ricci tensor can be written as (1) (2) Rµν = Rµν + Rµν + ...

(3.2)

(0)

with Rµν = 0. In order to compute the first order term of (3.2), we write down the expansion of the curvature tensor: Rρµσν =

1 (gρσ,µ,ν + gµν,ρ,σ − gµσ,ν,ρ − gρν,σ,µ ) + O(h2 ) 2 12

(3.3)

3.1 THE ENERGY MOMENTUM TENSOR OF THE GRAV. FIELD where the derivatives are non-covariant (the additional terms due to covariant derivatives are of higher order). We can thus write the first order of the Ricci tensor in terms of hµν : (1) Rµν =

1 (hµν + hρ ρ,µ,ν − hρ µ,ρ,ν − hρ ν,ρ,µ ) . 2

(3.4)

The d’Alembert operator can be used instead of ∂µ ∂ µ because in the approximation (3.1) the coordinates are “almost” Minkowskian, so ∂µ ∂ µ = + O(h). The first order Ricci scalar is given by (1)

R(1) = η λρ Rλρ .

(3.5)

We proceed by considering the second order equations. The left-hand side of the field equations can be written in terms of the quantity tµν which is defined by the equation (2) Rµν

−

Rgµν 2

(2) =:

8πG tµν . c4

(3.6)

We take these terms to the right-hand side of Einstein’s equations and find at second order in hµν : (1) Rµν −

R(1) 8πG ηµν = − 4 (Tµν + tµν ). 2 c

(3.7)

This can be interpreted as a linear wave equation in hµν with the source terms τµν = Tµν + tµν .

(3.8)

We have to think of τµν as being the energy-momentum tensor which also includes the contribution of the gravitational field itself. We interpret (3.7) as follows: since Gµν ;ν = 0 (Bianchi identity), we find for the left-hand side of (3.7): R(1) ∂ (1) Rµν − ηµν = 0. (3.9) ∂xν 2 Therefore the right-hand side satisfies ∂τµν = 0. ∂xν

(3.10)

This gives the momentum Z Pµ =

d3 r τµ0 = const. 13

(3.11)

3.2 LINEARIZED FIELD EQUATIONS which is conserved (in time). We can thus interpret τµ0 as the momentum density and τµν as an energy-momentum tensor (indeed we know that T µν ;ν = 0 but so far we did not necessarily conclude τ µν ;ν = 0). Since Tµν includes all nongravitational sources and τµν is interpreted as the “complete” energy-momentum tensor, tµν clearly describes energy-momentum which is purely due to the gravitational field: (2) ! 4 c Rg µν (2) tgrav. = Rµν − (|hµν | 1). (3.12) µν 8πG 2

3.2

Linearized Field Equations

With Eq. (3.3) we find for the field equations at first order in h T 16πG ρ ρ ρ Tµν − ηµν . hµν + h ρ,µ,ν − h µ,ρ,ν − h ν,ρ,µ = − 4 c 2

(3.13)

We use ηµν instead of gµν in this equation because both sides are already of order h. Since the field equations are covariant, we are free to perform a coordinate transformation. But note that due to |hµν | 1 we can only perform coordinate transformations which deviate only slightly from Minkowski coordinates: xµ −→ x0µ = xµ + εµ (x) 0µ

with ε 1.

0ν

∂x From g 0µν = ∂x g λρ we find the transformation of hµν . With ∂xλ ∂xρ inserted into g 0µν we get

g 0µν = η µν − h0µν ∂εµ ∂εν µ ν = δλ + λ δρ + ρ η λρ − hλρ ∂x ∂x

(3.14) ∂x0µ ∂xλ

= δλµ +

∂εµ ∂xλ

(3.15)

where we used that from gµν = ηµν + hµν it follows g µν = η µν − hµν . From Eq. (3.15) we infer h0µν = hµν −

∂εν ∂εµ − . ∂xν ∂xµ

(3.16)

Since this is already a first order equation (in h), we can raise and lower indices with gµν ' ηµν and g µν ' η µν . Thus ∂εµ ∂εν − µ. (3.17) ν ∂x ∂x In analogy to electrodynamics this transformation of the “potentials” gµν is called gauge transformation. We can choose four functions εµ (x) which gives four constraints on the “potentials” hµν . For instance, h0µν = hµν −

2hµ ν,µ = hµ µ,ν . 14

(3.18)

3.2 LINEARIZED FIELD EQUATIONS We insert the gauge condition (3.18) into (3.13) and obtain the decoupled linearized field equations: hµν

16πG =− 4 c

Tµν

T − ηµν 2

.

(3.19)

This can easily be seen if we differentiate (3.18) (i.e. hρ ρ,µ = 2hρ µ,ρ ) with respect to xν : hρ ρ,µ,ν = 2hρ µ,ρ,ν = hρ µ,ρ,ν + hρ ν,ρ,µ

(3.20)

(we used hµν = hνµ ). This implies −hρ µ,ρ,ν − hρ ν,ρ,µ + hρ ρ,µ,ν = 0

(3.21)

which is just another form of our gauge condition from which it can be seen that (3.13) indeed reduces to (3.19). Furthermore, it can be shown that from (3.17) it follows that if hµν does not satisfy (3.18), then we can find a transformed h0µν that does so. This can be done using the coordinate transformation (3.14) with εν = hµ ν,µ − 12 hµ µ,ν . The linearized field equation Eq. (3.19) has the same structure as the field equations in electrodynamics. We can therefore immediately write down the well-known solution for the retarded potentials: |r−r 0 | 0 Z S r , t − µν c 4G d3 r 0 (3.22) hµν (r, t) = − 4 c |r − r 0 | T with Sµν = Tµν − ηµν . 2 The interpretation is the same as in electrodynamics: a change in Sµν at place r 0| does not affect the place r 0 before some time |r−r has passed. c

15

4. LENSE-THIRRING EFFECT

4

Lense-Thirring Effect

The Lense-Thirring effect is the precession of a gyroscope in the gravitational field of the Earth due to the fact that the Earth is rotating. To set up an analogy with electrodynamics, we note that the gravitational field of the Schwarzschild metric corresponds to the Coulomb field outside of a static, spherical charge distribution. If a charge distribution rotates with constant angular velocity, this results in the presence of a static, non-isotropic magnetic field. Similarly, the rotation of the Earth will cause a “gravitomagnetic” field. We will treat this problem by using the linearized field equations. Another approach would be to start from the exact Kerr solution (i.e. the metric outside of a rotating black hole) and apply the weak field limit.

4.1

Metric of the Rotating Earth

We assume a weak field caused by a slowly rotating planet, so |hµν | 1. The linearized field equations read T 16πG Tµν − ηµν . (4.1) hµν = − 4 c 2 The coordinates xµ = (x0 , ..., x3 ) are Minkowski coordinates up to corrections of O(h). In the energy-momentum tensor (19.9) we can neglect the pressure since p ρc2 . Since the velocities (rotation of the Earth) are small compared to c, we 2 neglect terms of order vc . The energy-momentum tensor thus reads 1 vci 2 Tµν ' ρc vi . (4.2) 0 c The terms proportional to vi generate the gravitomagnetic field. This has to be compared to electrodynamics where magnetic fields are generated by currents. The mass distribution of the Earth can be approximated as follows: ( ρ0 (r ≤ RE ) ρ(r) = (4.3) 0 (r > RE ) The angular velocity of the Earth is ω = ωe3

with ω =

2π . 1 day

(4.4)

We consider the Earth as a rigid body, so we can write the velocity field as v(r) = ω ∧ r

or vi = εijk ω j rk . 16

(4.5)

4.1 METRIC OF THE ROTATING EARTH Because this velocity is constant, Tµν does not depend on time and therefore the field equations (4.1) have stationary solutions. We can thus replace by −∆ and Eq. (4.1) becomes 8πg ρ(r), c2 16πG ∆h0i = ρ(r)εijk ω j rk c3

∆hµµ =

where we used ηµν = diag(1, −1, −1, −1), Using ∆

T 2

=

ρc2 2

(4.6) (4.7)

and T00 = ρc2 , Tii = 0.

1 = −4πδ (3) (r − r 0 ), |r − r 0 |

(4.8)

we can immediately solve the equations (4.6) and (4.7) inside the planet: Z 2G ρ(r 0 ) hµµ (r) = − 2 , (4.9) d3 r 0 c |r − r 0 | Z 0 0n 4G k 3 0 ρ(r )x dr (4.10) h0i (r) = − 3 εikn ω c |r − r 0 | where x0n denotes the n-th component of r 0 . For the region outside of the mass distribution (r > RE ), we can use the expansion X 4π r0l ∗ 0 1 = Y (ˆ r )Ylm (ˆ r) l+1 lm |r − r 0 | (2l + 1) r l,m =

1 xj x0j − 3 + ... r r

(r > r0 ) (4.11)

where Ylm are the spherical harmonics and rˆ, rˆ0 denote unit vectors in directions of r and r 0 , respectively. Note that for the Cartesian components we have xi = gik xk = −xi + O(h). The corrections of O(h) can thus be neglected because the right-hand sides of (4.9) and (4.10) are already of first order in h. Since ρ(r) is spherically symmetric, only the first term of (4.11) contributes in Eq. (4.9): Z 2G 2GME hµµ (r) = − 2 d3 r0 ρ(r 0 ) = − 2 (r ≥ RE ) (4.12) cr cr Since ρx0n is proportional to Y1m , only terms with l = 1 contribute to (4.10): Z 4G ω j xk h0i (r) = 3 εijn 3 d3 r0 ρ(r 0 )x0n x0k c r 4GME RE2 ω j xn =− ε (r ≥ RE ). (4.13) ijn 5c3 r3 17

4.1 METRIC OF THE ROTATING EARTH 02

Since ρ(r) is spherically symmetric, we integrate using x0n x0k = −δkn r3 . Considering h0i as a vector, i.e. h0i → h = h0i ei , we can write (4.7) as ∆h(r) =

16πG ρω ∧ r c3

(4.14)

and (4.13) becomes 4GME RE2 ω ∧ r 2GI ω ∧ r h(r) = − =− 3 3 3 5c r c r3

(4.15)

where I = 52 ME RE2 is the moment of inertia of a homogeneous sphere. We want to consider the analogy with electrodynamics again. In magnetostatics, the vector potential A of a homogeneously charged rotating sphere with radius R and total charge q satisfies 4π ρe ω ∧ r c qR2 ω ∧ r A= 5c r3

∆A = − ⇒

(4.16) (4.17)

which has the same form as Eq. (4.15). Eqs. (4.12) and (4.13) determine the metric of the rotating Earth (valid for r ≥ RE ): 2GME 2 2 2GME ds = 1 − 2 c dt − 1 + 2 dr 2 + 2ch0i dxi dt cr cr 2

(4.18)

E where dr 2 = −dxi dxi . Note that this metric at O GM and for ω = 0 does c2 r not reduce to the Schwarzschild metric since Eq. (4.1) implies that we chose other coordinates as compared to the standard form. However, the metric (4.18) asymptotically (r → ∞) becomes the Minkowski metric. One can perform a coordinate transformation such that dr 2 has the usual angular dependence, i.e. dr 2 → r2 (dθ2 + sin2 θdϕ2 ). Distant “fixed” stars (which live in the asymptotic Minkowski spacetime) have constant values for the angles (θ, ϕ). Therefore changes in the angles due to spin precession refer to rotations with respect to the distant fixed stars. 4.1.1

Rotation of the Local IS

The metric (4.18) implies a rotation of the local insertial system IS. In order to see this, consider the axis of a gyroscope which is described by dS µ = −Γµ κν S κ uν . dτ 18

(4.19)

4.1 METRIC OF THE ROTATING EARTH We solve the equations for a gyroscope at rest with S µ = (0, S i ),

uµ = (c, 0).

(4.20)

Note that this gyroscope is freely falling, without external forces acting on it besides gravity. The forces act only on the center of mass. i Because we are taking into account only terms up to O(h), we can replace dS dτ i . From Eqs. (4.19) and (4.20) it follows that by dS dt dS i = −cΓi 0j S j . dτ For time independent hµν we find at first order 1 η ik ∂h0k ∂h0j i i i = ∂ h − ∂ h . Γ 0j = − j 0 0j 2 ∂xj ∂xk 2

(4.21)

(4.22)

Inserting this into (4.21) and lowering the index i, we find dSi c = − (∂j h0i − ∂i h0j ) S j dt 2 c = − εikl εkmn ∂m h0n S l 2 = εikl Ωk S l

(4.23)

with εikl εlmn = δlm δin − δln δim and the gravitomagnetic field c Ωk = − εkmn ∂m h0n 2

or

c Ω(r) = − ∇ ∧ h(r) 2

(4.24)

The indices i, j, m run over 1, 2, 3. We call h the gravitomagnetic potential. Using h from Eq. (4.15) we obtain the following expression for the angular velocity in the local IS: Ω(r) =

2GM RE2 3(ω · r)r − ωr2 . 5c2 r5

(4.25)

Note that Ω has the same form as B = ∇ ∧ A in electrodynamics. In vector form, Eq. (4.23) reads dS =Ω∧S dt

or

dS = (Ωdt) ∧ S .

(4.26)

This implies a precession of the spin of the gyroscope’s axis with angular velocity Ω. This precession is due to the rotation of the Earth with angular velocity ω. The precession of the gyroscope’s axis is equivalent to the rotation of the local IS because in the local IS we have S =const.. Therefore the local IS rotates with 19

4.1 METRIC OF THE ROTATING EARTH angular velocity Ω as compared to the global coordinate system described by (4.18). For r → ∞ Eq. (4.18) becomes Minkowskian, i.e. it describes a system which doesn’t rotate with respect to the fixed star system. To summarize, the physical meaning of the angular velocity Ω is that the local IS rotates with Ω with respect to the fixed star system. The rotation of the Earth drags the local IS (“frame dragging”). Now that we derived geodetic precession and the Lense-Thirring effect, we note that in fact both effects take place at the same time and thus sum up. Inserting r = RE in Eq. (4.25) we get at the north pole and at the equator, respectively: ( 2 (North pole, θ = 0) 2GME (4.27) Ω= 2 ω· 5c RE (−1) (equator, θ = π2 ) where we used |ω · r| = ωr cos θ. Numerically, one finds 2GME ω ∼ 10−9 ω. 5c2 RE

(4.28)

We conclude that the Lense-Thirring precession affects a Foucault pendulum located at the north pole. The rotation with respect to the distant stars amounts to ∆φ = ΩLT · 1yr =

4GME 2π · 365 = 0.200 per year. 3c2 RE

(4.29)

This effect was first computed by Lense and Thirring in 1918.6 The NASA satellite Gravity Probe B aims to measure both geodetic precession and Lense-Thirring precession. It was launched in 2004 and orbits the Earth on a polar orbit. If one chooses the spin perpendicular to the orbital plane, the LenseThirring effect vanishes for equatorial orbits since in this case we have S||w||Ω and Eq. (4.26) gives dS = 0. The expected total frame dragging is 000 .05 per year dt and it is yet unclear whether the instrumental precession suffices to detect the effect. However there are claims that the effect has been measured on the orbit of the LAGEOS satellites.7 Update: in May 2011 the final results of Gravity Probe B have been released and they indeed confirm the GR predictions to a very high accuracy.8 6

Lense and Thirring, Phys. Zeitschr. 19, 156 (1918) I. Ciufuloni and Pavli, Nature 431, (2004) 958 8 C. W. F. Everitt et al., Phys. Rev. Lett. 106, 221101 (2011)

7

20

4.2 GRAVITOMAGNETIC FORCES

4.2

Gravitomagnetic Forces

In the discussion of the Lense-Thirring precession, we saw a formal equivalence to electromagnetism: h Ω

↔ ↔

A B.

This analogy persists even for the equation of motion of a particle in the metric (4.18), duµ = −Γµ γν uγ uν . dτ Neglecting terms of O reads

2 v c2

(4.30)

we have dτ ≈ dt and uµ = (c, v i ). Therefore Eq. (4.30)

dv i = −Γi 00 c2 − 2cΓi 0j v j + O dt

v2 c2

.

(4.31)

The first term on the right-hand side corresponds to the gradient of the Newtonian potential. In analogy to to (4.21)-(4.23), the second term can be shown to give −Γi 0j v j = (Ω ∧ v)i .

(4.32)

This implies that Eq. (4.31) can be written as dv = −gradφ + 2Ω ∧ v. dt

(4.33)

This is the equation of motion in the presence of gravitomagnetic forces and it has the same structure as the equation for the Lorentz force 1 K =q E+ v∧B . (4.34) c This analogy is the origin of the notion of gravitomagnetism. Note, however, that this analogy is only true if we consider the linearized field equations of general relativity. In the above identification the electromagnetic fields are full solutions to Maxwell’s equations whereas the gravitomagnetic potential and field are approximations. Furthermore, the analogy is quite formal and certainly not complete due to the absence of negative “gravitational charges”.

21

5. GRAVITATIONAL WAVES

5

Gravitational Waves

For weak gravitational fields (i.e. |hµν | = |gµν − ηµν | 1) the Einstein field equations read 16πG T hµν = − 4 Tµν − ηµν . (5.1) c 2 In the vacuum (Tµν = 0) the equation reduces to hµν = 0

(5.2)

which has plane waves as its simplest solution. The above equation is quite similar to the wave equation in electromagnetism, Aµ = 0 with the electromagnetic vector potential Aµ . As we will see, the solutions are similar, as well. Note that the wave equation in electromagnetism is exact whereas the general relativistic wave equation arises from the approximate linearized field equations.

5.1

Electromagnetic Waves

Physical fields are invariant under gauge transformations Aµ → A0µ = Aµ + ∂ µ χ

(5.3)

so that we can choose ∂µ Aµ = 0 (Lorenz gauge) and get Aµ =

4π µ j . c

(5.4)

Due to the gauge condition, only three out of four components of Aµ are independent. In vacuum we have that j µ = 0 which allows us to perform another gauge transformation satisfying χ = 0. We use this condition to set A0 = 0. Finally we are left with two degrees of freedom (polarizations). The conditions read then Aµ = 0,

A0 = 0,

∂i Ai = 0.

(5.5)

This is solved by the ansatz Aµ = eµ exp[−ikν xν ] + c.c.

(5.6)

where kµ k µ = 0 and ei k i = 0 (polarizations are transverse to propagation direction.

22

5.2 THE GRAVITY CASE

5.2

The Gravity Case

Due to the symmetry hµν = hνµ , 10 out of 16 components of hµν are independent. With a gauge transformation of the form (3.18) we can impose four additional conditions. This leaves us with 6 degrees of freedom that are really independent. If we consider the vacuum case hµν = 0,

(5.7)

in addition to (3.17) we can perform a further transformation of the form hµν → h0µν = hµν − ∂µ εν − ∂ν εµ

(5.8)

provided that εµ satisfies εµ = 0.

(5.9)

Such a transformation leaves Eq. (5.7) and the gauge condition (3.18) invariant (this is in complete analogy to electromagnetism, of course). With these four additional conditions we are left with two independent components of hµν . The solution to (5.7) can be written in terms of plane waves hµν = eµν exp[−ikν xν ] + c.c.

(5.10)

where ω2 η kλ kν = k kν = 0 ⇔ = 2 = k2 = k 2 . (5.11) c The amplitude of the wave eµν is called polarization tensor. Inserting (5.10) into the gauge condition (3.18) (2hµ ν,µ = hµ µ,ν ) leads to λν

ν

k02

2kµ η µρ eρν = kν η µρ eρµ .

(5.12)

Clearly eµν inherits the symmetry of hµν , thus eµν = eνµ . Let us assume a wave travelling along the x3 -axis. This yields the wave solution hµν = eµν exp ik(x3 − ct) (5.13) where we used Eq. (5.11). The components of the wave vector are then ω k1 = k2 = 0, k0 = −k3 = k = . c In this case the gauge condition (5.12) reads 1 e00 + e30 = (e00 − e11 − e22 − e33 ), 2 e01 + e31 = 0, e02 + e32 = 0, 1 e03 + e33 = − (e00 − e11 − e22 − e33 ). 2 23

(5.14)

(5.15) (5.16) (5.17) (5.18)

5.2 THE GRAVITY CASE With eµν = eνµ and these four conditions the polarization tensor can be described by six components. All the other components can be expressed in terms of the six independent components e00 , e11 , e33 , e12 , e13 and e23 .

(5.19)

The other components are givevn by e01 = −e31 = −e13 ,

e02 = −e32 ,

e22 = −e11 ,

1 e03 = (e00 + e33 ). 2

(5.20)

We can perform yet another transformation (3.14) (x0µ = xµ + εµ ) with functions εµ satisfying εµ = 0. The functions are solutions of the wave equation, therefore we can write them as εµ (x) = δ µ exp [−ikµ xµ ] + c.c..

(5.21)

As noted before, such a transformation with arbitrary δ µ is not violating the gauge condition (3.18). We choose k µ in (5.21) equal to the wave vector of a given gravitational wave. Using (5.21) in (3.16) we obtain a new solution h0µν in which all the terms have the same exponential dependence given by exp[−ikµ xµ ]. Thus only the amplitudes transform as e011 e012 e013 e023 e033 e000

= e11 , = e12 , = e13 − iδ1 k, = e23 − iδ2 k, = e33 − 2iδ3 k, = e00 + 2ikδ0 .

(5.22) (5.23) (5.24) (5.25) (5.26) (5.27)

We can choose δµ such that e000 = e013 = e033 = e023 = 0. This new solution is equivalent to the old one. From the physical point of view, only polarizations corresponding to e011 and e012 are relevant. Neglecting primes in our notation from now on, we get for the gravitational wave propagating in x3 -direction after gauging away all redundancies 0 0 0 0 0 e11 e12 0 3 hµν = (5.28) 0 e12 −e11 0 · exp ik(x − ct) + c.c. 0 0 0 0 5.2.1

Helicity

The direction of k is the x3 -axis. We ask now the question how (5.28) transforms under a rotation around this axis. Since we are in an almost Minkowskian metric 24

5.3 PARTICLES IN THE FIELD OF A WAVE we can realize this transformation as a Lorentz transformation described by the matrix 1 0 0 0 ¯ µ ν = 0 cos ϕ sin ϕ 0 . Λ (5.29) 0 − sin ϕ cos ϕ 0 0 0 0 1 Therefore the polarization tensor transforms as ¯ ρµΛ ¯ σ ν eρσ . e0µν = Λ

(5.30)

e011 = e11 cos(2ϕ) + e12 sin(2ϕ), e012 = −e11 sin(2ϕ) + e12 cos(2ϕ).

(5.31) (5.32)

This yields

If we consider e± ≡ e11 ± ie12 instead, we thus have e0± = e±2iϕ e± .

(5.33)

The vectors e± have helicity ±2, whereas the wave solutions in electrodynamics have polarization ±1. This just means that gravitons have spin 2 whereas photons are spin 1 bosons.

5.3

Particles in the Field of a Wave

Similarly to electromagnetic waves also gravitational waves exert forces on massive particles. We want to explore how the positions of particles are affacted in the field of a gravitational wave. For this analyis consider a plane wave described by 0 0 0 0 0 e11 e12 0 3 hµν (x3 , t) = (5.34) 0 e12 −e11 0 exp ik(x − ct) + c.c 0 0 0 0 The corresponding metric has the form ds2 = ηµν + hµν (x3 , t) dxµ dxν .

(5.35)

The trajectory xσ (τ ) of a particle in the gravitational field satisfies the equation of motion dxµ dxν d2 x σ = −Γ µν dτ 2 dτ dτ 25

(5.36)

5.3 PARTICLES IN THE FIELD OF A WAVE where Γσ µν can be taken from Eq. (5.28). We assume that there are no other forces but gravity which act on the particles. Inserting (5.34) into 1 σλ ∂hνλ ∂hµλ ∂hµν σ Γ µν = η (5.37) + − 2 ∂xµ ∂xν ∂xλ it follows i

Γ 00

1 =− 2

∂h0i ∂h0i ∂h00 + − ∂x0 ∂x0 ∂xi

.

(5.38)

As initial conditions we choose dxi x˙ (0) = = 0. dτ τ =0

(5.39)

d2 xi = −Γiµν x˙ µ (0)x˙ ν (0) = 0. dτ τ =0

(5.40)

i

This implies

Therefore the acceleration vanishes. This means that the velocities of the particles don’t change. The solution of the equations of motion (5.36) thus reads dxi =0 dτ

⇒

xi (τ ) = const.

(5.41)

In the chosen coordinates the particles in the field of the gravitational wave can thus be described by constant spatial coordinates. However, this does not mean that the particles are at rest. In fact their distances vary due to the time dependence of the metric tensor gµν as in Eq. (5.35). Consider now particles which are arranged on a circle in the x1 -x2 -plane. The particles are initially at rest on a circle ((x1 )2 + (x2 )2 = R2 ). We want to examine the effect of an incident gravitational wave along the x3 -axis. To do so, we write (5.35) in the form ds2 = c2 dt2 − dl2 − (dx3 )2 with dl2 = (δmn − hmn (t)) dxm dxn

(m, n = 1, 2).

(5.42)

In the x1 -x2 -plane we have e11 e12 hmn (t) = hmn (x = 0, t) = exp(−iωt) + c.c. e12 −e11 3

(5.43)

where ω 2 = c2 k 2 . With (5.42) we can compute the physical distance ρ of a particle p from the center of the circle. According to Eq. (5.41) the coordinates x1p and x2p of the particle are constant. We insert in (5.42) the finite values of the 26

5.4 ENERGY AND MOMENTUM OF A GRAVITATIONAL WAVE m coordinates xm p of p instead of dx (this is allowed because the metric coefficients do not depend on x1 and x2 ): n ρ2 = (δmn − hmn (t)) xm p xp

with

x1p

= R cos ϕ,

x2p

(m, n = 1, 2) = R sin ϕ.

Using Eqs. (5.43)-(5.45) we find the solution ( 1 − 2h cos(2ϕ) cos(ωt) if e11 = h, e12 = 0, ρ2 = R2 1 − 2h sin(2ϕ) cos(ωt) if e11 = 0, e12 = h.

(5.44) (5.45)

(5.46)

The cos(ωt) term comes from e−iωt + c.c., whereas the cos(2ϕ) term, for example, comes from cos2 ϕ − sin2 ϕ = cos(2ϕ). The distinction that we made in (5.46) concerns the two possible linear polarization states. Contrary to the coordinates x1 , x2 which are constant, the physical variables x = ρ cos ϕ and y = ρ sin ϕ describe the distance relative to the center. The physical oscillations lead to a particle configuration which is an ellipse with very tiny eccentricities (h 1). From the type of oscillation one can infer the polarization of the incoming wave. The two independent polarization states form an angle of π4 . Therefore the oscillations correspond to a quadrupole moment of the mass distribution: the gravitational waves induce a quadrupole oscillation of the mass distribution. Conversely we expect that mass distributions with oscillating quadrupole moment should emit gravitational waves. In order to study this phenomenon further, we have to learn more about the energy and momentum of a gravitational wave.

5.4

Energy and Momentum of a Gravitational Wave

We now want to determine the energy-momentum tensor of a gravitational wave which is a solution of the first order equation in h (|hµν | 1) (1) Rµν = 0.

(5.47)

The solution to this equation is the weak wave solution that we derived before: hµν = eµν exp −ikλ xλ + c.c. (5.48) The energy-momentum tensor of a gravitational field is known from Eq. (3.19): ! (2) 4 (g R) c µν (2) Rµν − (5.49) tgrav. = µν 8πG 2

27

5.4 ENERGY AND MOMENTUM OF A GRAVITATIONAL WAVE (0)

(1)

with gµν = ηµν + hµν . Using Rµν = Rµν = 0 and R = g ρσ Rρσ we get c4 (2) (1) (1) (2) − hµν η ρσ Rσρ + ηµν hρσ Rσρ 2Rµν − ηµν η ρσ Rσρ 16πG c4 (2) (2) = 2Rµν − ηµν η ρσ Rσρ 16πG

= tgrav. µν

(5.50)

where we used g µν = η µν − hµν which follows from g µν gνλ = δλµ . In order to (2) consider this expression further, we need the Ricci tensor Rµν . In order to be able to calculate the Ricci tensor (2) (2) (1) (2) Rµκ = g λν Rλµνκ = η λν Rλµνκ − hλν Rλµνκ (5.51) we need the Riemann tensor which is given by Eq. (17.4): 2 ∂ gλν ∂ 2 gµκ ∂ 2 gµν ∂ 2 gλκ 1 + − λ κ− µ ν + Rλµνκ = 2 ∂xµ ∂xκ ∂xλ ∂xν ∂x ∂x ∂x ∂x η σ η σ + gησ (Γ νλ Γ µκ − Γ κλ Γ µν ) .

(5.52)

The first line of this expression (which contains only first order terms in h) gives rise to the second term in (5.51) with gij = hij . The second line (which is of second order in h) gives rise to the first term in (5.51) if we insert the Christoffel symbols 1 ∂hσ µ ∂hσ ν ∂hµν σ(1) Γ µν = + − . (5.53) 2 ∂xν ∂xµ ∂xσ This yields the first term in (5.51) if we multiply with ηµν . In total, we get the following expression for the Ricci tensor that we searched for in order to evaluate 5.50: ∂ 2 hµκ ∂ 2 hµν ∂ 2 hλκ hλν ∂ 2 hλν (2) Rµκ = − + − λ κ− µ ν + 2 ∂xµ ∂xκ ∂xλ ∂xν ∂x ∂x ∂x ∂x ν σ ν ν ∂h µ ∂hσ κ ∂hµκ 1 ∂h σ ∂h σ ∂h ν + − + − + − 4 ∂xν ∂xν ∂xσ ∂xκ ∂xµ ∂xσ 1 ∂hσκ ∂hσλ ∂hλκ ∂hσ µ ∂hσλ ∂hλ µ . (5.54) − + − + − 4 ∂xλ ∂xκ ∂xσ ∂xλ ∂xµ ∂xσ The second line of this expression vanishes because of the gauge condition (3.18). The remaining terms are quadratic in h and of the form hµν hσκ = eµν exp(−ikλ xλ ) + c.c. eσκ exp(−ikλ xλ ) + c.c. = eµν eσκ exp(2ixλ k λ ) + 2 + exp(−2ixλ k λ ) . (5.55) We can see that on the one hand there appear oscillating terms of the form exp(±2ikλ xλ ) and on the other hand there are also terms which do not depend 28

5.4 ENERGY AND MOMENTUM OF A GRAVITATIONAL WAVE on the coordinates xµ at all. If we average over time, this implies that the oscillating terms drop out (their average over time being zero) such that we are left with terms of the form λ λ (5.56) eµν exp(−ikλ x ) + c.c. eσκ exp(−ikλ x ) + c.c. = 2< e∗µν eσκ where h· · · i denotes time-average. For plane wave solutions derivatives correspond to multiplication with k, so ∂hµν = −ikλ hµν . ∂xλ

(5.57)

We can now plug all these terms which are quadratic in h into Eq. (5.54) replacing all partial derivatives by factors of k: (2) hRµκ i = < (eλν )∗ (kµ kκ eλν + kλ kν eµκ − kλ kκ eµν − kµ kν eλκ ) 1 ∗ λ σ σλ σ λ (5.58) − (kλ eσκ + kκ eσλ − kσ eκλ ) · k e µ + kµ e − k e µ 2 where we used that the gauge condition 2∂µ hµ ν = ∂ν hµ µ sets the second line of (5.54) to zero. In order to simplify this expression further, we use the gauge condition again, essentially replacing the h-terms in it according to Eq. (5.57) and find 1 (5.59) (eλµ )∗ kκ kλ eµν = (eλ λ )∗ k µ kκ eµν . 2 Imposing the null condition kµ k µ = 0 we obtain 1 1 λ 2 (2) λν ∗ hRµκ i = kµ kκ (e ) eλν − |e λ | . 2 2

(5.60)

Thus the energy-momentum tensor (5.50) of the gravitational wave reads 1 λ 2 c4 λκ ∗ grav. kµ kν (e ) eλκ − |e λ | tµν = (5.61) 16πG 2 where we used that (2) ηµν η ρσ hRρσ i ∝ ηµν η ρσ kρ kσ = ηµν k σ kσ = 0.

(5.62)

We can further simplify the energy-momentum tensor by specializing to the case of linearly polarized waves with either e11 = −e22 6= 0, e12 = e21 = 0 or e11 = −e22 = 0, e12 = e21 6= 0: tgrav. = µν

c4 kµ kν h2 . 8πG 29

(5.63)

5.5 QUADRUPOLE RADIATION Energy in this formula being proportional to frequency squared is exactly the type of relation that we would intuitively expect. Furthermore it is clear that tµν ∝ kµ kν because t0i is the current of momentum which should be proportional to ki . We see immediately that measuring such energies will be extremely difficult because h2 is very small. A wave propagating in the x3 -direction has the ω ω wavevector kµ = c , 0, 0, c . The energy current density Φgrav. = ct03 grav. =

c5 0 3 2 k k h 8πG

(5.64)

for such a wave is then given by Φgrav.

5.5

energy c3 2 2 = = ω h. time · surface 8πG

(5.65)

Quadrupole Radiation

Oscillating charge distributions emit electromagnetic waves. In analogy we expect oscillating mass distributions to emit gravitational waves. We quickly repeat the case of electromagnetic dipole radiation before turning to the case of oscillating mass distributions. 5.5.1

Dipole Radiation in Electromagnetism

In electromagnetism one finds that an oscillating dipole moment p(t) = p0 exp(−iωt) + c.c.

(5.66)

emits electromagnetic waves whose power per solid angle is given by ω4 dP = |p|2 sin2 θ 2 dΩ 8πc

(5.67)

where θ is the angle between p and k where k is the direction of propagation. This is sketched in Fig. (3). The total emitted power can be obtained by integration over θ: P =

5.5.2

ω4 2 |p| . 3c3

(5.68)

Gravitational Quadrupole Radiation

The computation of the emitted gravitational radiation is similar to electromagnetism but also more involved since the source terms are rank 2 tensors. We will proceed in the following steps: 30

5.5 QUADRUPOLE RADIATION

θ

Figure 3: Sketch of the waves that are emitted by a dipole in electromagnetism. 1. Calculate the asymptotic fields emitted by a source Tµν . 2. Reduce the result to spatial components. 3. Apply the long wavelength approximation. The setup is sketched in Fig. 4. In contrast to the electromagnetic case there is no gravitational dipole radiation. The density is given by Z ρ(r, t) = ρ(r) exp(−iωt) + c.c. ⇒ p = d3 r rρ(r) = M Rc.m. (5.69) where M is the total mass and Rc.m. is the center of mass. If we choose the center of mass system as the inertial system then p = 0. Consequently p = 0 in

~r

~r ′ r0

Figure 4: Sketch of the setup for gravitational wave emission. The source has spatial extent r0 . The observer is at position r. 31

5.5 QUADRUPOLE RADIATION all inertial systems. We shall now assume an oscillatory mass distribution of the form ( 6= 0 if r ≤ r0 Tµν (r, t) = Tµν (r) exp(−iωt) + c.c. (5.70) = 0 ifr > 0. This is only one Fourier component. Thus a generalization is possible by integrating over ω. According to (3.22) the retarded potentials are given by Z exp(ik|r − r 0 |) 4G + c.c. (5.71) hµν (r, t) = − 4 exp(−iωt) d3 r0 Sµν (r 0 ) c |r − r 0 | where we used |r − r 0 | −iωtr = −iω t − = −iωt + ik|r − r 0 | c

(5.72)

to obtain the phase factors. Furthermore we have introduced 1 Sµν (r) = Tµν (r) − ηµν T (r). 2 We now assume r0 r with k = and thus

ω c

=

2π . λ

(5.73)

For large distances we have |r| r0

h r i r · r0 0 |r − r | = r − =r 1+O r r

(5.74)

exp [ik|r − r 0 |] = exp[ikr] exp[−ikr 0 ]

(5.75)

0

and

where we defined k = k rr = ker . This way we obtain for (5.71) Z 4G 1 λ d3 r0 Sµν (r 0 ) exp [−ikr 0 ] +c.c. hµν (r, t) = − 4 exp −ikλ x c r | {z }

(5.76)

=:Sµν (k)

where Sµν (k) is the spatial Fourier transform of Sµν (r). This yields hµν (r, t) = eµν (r, ω) exp −ikλ xλ + c.c..

(5.77)

The amplitudes are defined as eµν (r, ω) = −

4G 1 Sµν (k) c4 r

32

(5.78)

5.5 QUADRUPOLE RADIATION and are proportional to 1r . They depend on rr = er and ω via k = ker . The energy current passing through a surface element r2 dΩ is given by grav. i dP = ctgrav. 0i df = ct0i

xi 2 r dΩ. r

Plugging in Eq. (5.61) for tgrav. we obtain µν dP 1 λ 2 c4 k0 ki xi 2 λν ∗ =c r (e ) eλν − |e λ | . dΩ 16πG r 2

(5.79)

(5.80)

In the derivation of (5.61) we assumed eµν = const., whereas here we have eµν ∝ 1r . The energy-momengum tensor tgrav. contains partial derivatives of the hµν which µν 1 would lead to additional term ∝ r . With eµν = const., the derivatives just lead to factors of kµ ∝ λ1 . In the far field and distant observer approximation we have r λ and can thus neglect the additional terms since 1r λ1 . Using k·r ω ki xi = =k= r r c

(5.81)

dP Gω 2 1 2 µν ∗ = T (k) Tµν (k) − |T (k)| dΩ πc5 2

(5.82)

in Eq. (5.80) we get

where Tµν (k) is the Fourier transform of the source distribution. We proceed with the second step as outlined in the beginning: we want to reduce our results to spatial components. The source distribution can be written as Z 1 Tµν (r, t) = d3 k T µν (k) exp −ikλ xλ + c.c.. (5.83) 3 (2π) For weak fields the covariant derivative in the energy-momentum conservation simplifies to an ordinary derivative and the continuity equation reads kµ T µν (k) = 0.

(5.84)

In particular we find for ν = 0 and ν = i k0 T 00 = −kj T 0j

and k0 T i0 = −kj T ij .

We can define a three dimensional unit vector kˆi = T 0i = T i0 = −kˆj T ij , T 00 = kˆi kˆj T ij . 33

ki k

(5.85)

and obtain for (5.85) (5.86) (5.87)

5.5 QUADRUPOLE RADIATION All non-spatial components in (5.82) can thus be eliminated and we calculate T µν∗ Tµν = ηµρ ηνσ T µν∗ T ρσ X X = T 00∗ T 00 − 2 T 0i∗ T 0i + T ij∗ T ij i

i,j

= kˆi kˆj kˆl kˆm T ij∗ T lm − 2kˆj kˆm δil T ij∗ T lm + δil δjm T ij∗ T lm ,

(5.88)

T λ λ = ηλρ T ρλ X = T 00 − T ii i

= kˆi kˆj T ij − δij T ij ,

(5.89)

|T λ λ |2 = kˆi kˆj kˆl kˆm T ij∗ T lm − δij kˆl kˆm T ij∗ T lm − δlm kˆi kˆj T ij∗ T lm .

(5.90)

Inserting these expressions into (5.82) we get Gω 2 dP = Λij,lm T ij (k)∗ T lm (k) dΩ πc5

(5.91)

where we introduced 1 1 1 1 Λij,lm (θ, ϕ) = δil δjm − δij δlm − 2δil kˆj kˆm + δij kˆl kˆm + δlm kˆi kˆj + kˆi kˆj kˆl kˆm . 2 2 2 2 (5.92) Having reduced the formula for the radiated power to spatial components, we now turn to the last step that we outlined in the beginning: we apply the long wavelength approximation, i.e. we assume λ r0 which simplifies the energy-momentum tensor as follows: Z ij T (k) = d3 r T ij (r) exp(−ikr) Z = d3 r T ij (r)(1 − ikr + ...) Z ω2 (5.93) ≈ d3 r T ij (r) =: − Qij . 2 The object Qij will turn out to be a quadrupole tensor. From covariant conservation of energy-momentum, T µν ,ν = 0, we get ∂j T ij (r, t) = −∂0 T i0 (r, t)

∂i T 0i (r, t) = −∂0 T 00 (r, t).

and

(5.94)

Using Eq. (5.70) we obtain ω 2 00 ∂i ∂j T (r, t) = = − 2 T (r, t) c 2 ω ⇒ ∂i ∂j T ij (r) = − 2 T 00 (r). c ij

∂02 T 00 (r, t)

34

(5.95) (5.96)

5.5 QUADRUPOLE RADIATION Since we are in the non-relativistic regime (λ r0 , v c) we have T 00 ' ρc2 . Therefore Eq. (5.93) yields Z Z Z ω2 3 ij 3 i j lk 2 d r T (r) = d r x x ∂k ∂l T (r) = − 2 d3 r xi xj T 00 (r) (5.97) c where we integrated by parts twice in the first step and used Eq. (5.96) in the second step. According to the definition in Eq. (5.93), we find Z Z 1 ij 3 i j Q = d r x x ρ(r) = 2 d3 r xi xj T 00 (r) (5.98) c which we can obviously interpret as the quadrupole tensor of the mass distribution9 . Because we are in almost Minkowskian spacetime, we can compute Qij in three-dimensional Cartesian coordinate. Inserting (5.93) into (5.91) we get dP Gω 6 = Λij,lm Qij∗ Qlm . dΩ 4πc5

(5.99)

We observe that the corresponding formula in electrodynamics looks very similar but it depends on ω 4 rather than ω 6 . This is just a reflection of the fact that the electromagnetic radiation is dipole radiation whereas gravitational dipole radiation does not exist (the dipole moment of any mass distribution vanishes in the center of mass system). Furthermore we note that Qij are constants (they do not depend on θ or ϕ). The complete angular dependence is encoded in Λij,lm in which the vector kˆ appears. This is the unit vector which indicates the direction from the mass distribution to the observer and therefore clearly depends on θ and ϕ: (kˆi ) = (kˆx , kˆy , kˆz ) = (sin θ cos ϕ, sin θ sin ϕ, cos θ).

(5.100)

This simplifies calculations, of course, because the quadrupole moment can be calculated once and forever and the specific angular dependence is only to be considered in the form of Λij,lm . As an example we consider a quadrupole mass distribution in the principal axis system: the only non-vanishing elements are on the diagonal, Qij = diag(Q11 , Q22 , Q33 ). In this case the only non-vanishing terms in (5.99) are those with i = j and l = m. This means that there appear only even powers of kˆx , kˆy , kˆz in Eq. (5.92). The emitted power has an angular dependence of the form dP ∼ a1 cos4 θ + a2 cos2 θ sin2 θ + a3 sin4 θ. dΩ 9

(5.101)

In the literature one also finds different definitions of this tensor. For example, one can define a traceless version where xi xj is replaced by xi xj − 13 r2 δ ij .

35

5.5 QUADRUPOLE RADIATION To get the total emitted power, we have to integrate Eq. (5.99) over dΩ: Z 2π (11δil δjm − 4δij δlm + δim δjl ) (5.102) dΩ Λij,lm = 15 where we used 4π dΩ kˆi kˆj = δij , 3 Z 4π dΩ kˆi kˆj kˆl kˆm = (δij δlm + δil δjm + δim δjl ) . 15 Z

Inserting this result into (5.99) we obtain for the total emitted power 2 ! 3 Z 3 6 X X dP 2Gω 1 ii ij 2 Q P = dΩ = |Q | − . dΩ 5c5 3 i=1 i,j=1

(5.103) (5.104)

(5.105)

Note that Qij can be defined traceless such that the second sum in the brackets vanishes. Furthermore one can assume a more general time dependence than just e−iωt . If we had defined the quadrupole moments in a traceless form, Z 1 2 ij ij 3 i j Q (t) = d x x x − r δ ρ(t, x), (5.106) 3 then instead of Eq. (5.105) we would have found P =

G ... ... hQ Q i 5c5 kl kl

(5.107)

where h· · · i denotes a time average.10

... ... G In the literature, one finds also the formula P = 45c 5 h Q kl Q kl i. The different prefactor arises if one defines the quadrupole tensor with an additional factor of 3 as compared to Eq. (5.106). 10

36

6. SOURCES OF GRAVITATIONAL WAVES

6

Sources of Gravitational Waves

We want to consider different physical systems whose dynamics leads to the emission of gravitational waves.

6.1

Rigid Rotator

As a first example we consider the emission of gravitational waves by a rigid rotating body. Consider a coordinate system KS’ with coordinates x0m in which the body is fixed. In KS’ the mass density ρ0 (r 0 ) is time independent. We choose KS’ such that the quadrupole tensor Θ0ij is diagonal:

Θ0ij =

Z

I1 0 d3 r0 x0i x0j ρ0 (r 0 ) = 0 I2 0 . 0 0 I3

(6.1)

We assuume that the body rotates with angular velocity Ω around the x03 -axis. The orthogonal transformation to an inertial system IS with coordinates xn can be written as cos Ωt − sin Ωt 0 xn = αm n (t)x0m with αm n (t) = sin Ωt cos Ωt 0 . (6.2) 0 0 1 The tensor Θij in IS reads Z Θij (t) = Z =

d3 r xi xj ρ(r, t) d3 r0 (αn i x0n ) (αm j x0m ) ρ0 (r 0 )

= (α(t)Θ0 αT (t))ij

(6.3)

where we used d3 r = d3 r0 and ρ0 (r 0 ) = ρ(r) since the density transforms as a scalar quantity. With Eqs. (6.1) and (6.2) we can compute (6.3): I1 + I2 I1 − I2 + cos(2Ωt) 2 2 I1 − I2 sin(2Ωt) Θ12 (t) = 2 I1 + I2 I1 − I2 − cos(2Ωt) Θ22 (t) = 2 2 Θ33 (t) = I3 , Θ13 (t) = Θ31 (t) = 0.

Θ11 (t) =

37

(6.4)

6.1 RIGID ROTATOR This is of the form Θij = const. + [Qij exp(−2iΩt) + c.c.] 1 i 0 I1 − I2 i −1 0 . with Qij = 4 0 0 0

(6.5) (6.6)

Comparing Eqs. (6.3) and (6.5) we conclude that a rotating rigid body can be interpreted as a mass distribution whose rotational frequency Ω leads to gravitational waves of frequency ω = 2Ω. We introduce the moment of inertia I with respect to the rotation axis and the ellipticity of the body ε, I = I1 + I2 ,

ε=

I1 − I2 . I1 + I2

(6.7)

We can then write P =

32GΩ6 2 2 εI 5c5

(6.8)

which is clearly the type of formula that one expects for quadrupole radiation (Q ∼ eI). 6.1.1

Example: The Double Star System

As an example we consider a binary star system (masses M1 and M2 ) rotating on a Keplerian ellipse. Assuming a circle with constant radius r, we can consider the system as a rigid rotator: I ' I1 =

M1 M2 2 r , M1 + M2

I2 ' 0,

ε ' 1.

(6.9)

The circular orbit is characterized by GM1 M2 M1 M2 Ω2 r = M +M r2 | 1 {z 2}

⇒

Ω2 =

GM r3

(6.10)

=µ

where µ is the reduced mass and M = M1 + M2 . Inserting this into Eq. (6.8) we find P =

32 G4 3 2 32G4 M12 M22 (M1 + M2 ) = M µ. 5c5 r5 5 c5 r 5

(6.11)

It is convenient to express the emitted power in terms of L =

c5 = 3.63 × 1059 erg s−1 G = 2.03 × 105 M c2 s−1 38

(6.12)

6.1 RIGID ROTATOR which yields 32 P = 5 GM c2 r

use

GM c2 r

5

µ2 L . M2

(6.13)

For example, coalescing neutron stars in the final stage have r ≈ RS and thus ∼ O(1) and P ≈ L . In general, for order of magnitude estimates one can P = L

GM c2 R

5 (6.14)

for systems of typical scale R. Due to the emission of gravitational waves the system loses energy and thus its distance R shrinks, till the two bodies coalesce after a time tspir. (inspiral time). In the Kepler problem the total energy is E=−

Gm1 m2 . 2r

(6.15)

During the inspiral process, the system loses potential energy which is emitted in the form of gravitational waves. Thus dE = −P dt and P =−

dE Gm1 m2 dr 32 G4 m21 m22 (m1 + m2 ) =− = . dt 2r2 dt 5 c5 r5

(6.16)

Upon substituting x(t) = [r(t)/r(0)]4 we can rewrite the last equality in (6.16) as 256G3 m1 m2 (m1 + m2 ) 1 dx . =− ≡− 5 4 dt 5c r (0) tspir.

(6.17)

This is solved by x = 1 − t/tspir. such that we find r(t) = r(0) 1 −

t

1/4

tspir.

.

(6.18)

Next, we want to calculate the strain of such a system. We evaluate the expression (5.78) for eµν in analogy to what we did with Eq. (5.79). We can express Tµν in terms of its spatial components (k0 T 0i = kj T ij and k0 T 00 = ki T 0i ) which in turn can be expressed in terms of Qij : Tij (k) = −

ω2 Qij . 2

(6.19)

With the definitions in (6.6) and (6.8) we obtain e11 =

1 GIε 1 (2Ω)2 4 2 c D 39

(6.20)

6.2 THE PSR 1913+16 SYSTEM where D is the distance between source and observer. If there are two polarizations and e11 = e12 then we have for the dimensionless strain q √ GIε 1 (6.21) h = e211 + e212 = 2 2 4 Ω2 . c D For a binary system characterized by masses m = m1 = m2 on an orbit of radius r, this yields r2 Ω2 =

Gm , r

I=

mr2 2

⇒

h≈

RS2 Dr

(6.22)

where RS = 2Gm/c2 is the Schwarzschild radius of the system. For two neutron stars with m = 1.4M , r = 100 km (i.e. T = 10−2 s, Ω = 102 Hz) at a distance D = 30000 ly we get h ≈ 10−18 .

(6.23)

This corresponds to the relative amplitude of the oscillation of a ruler’s length when the gravitational wave passes by.

6.2

The PSR 1913+16 System

The Hulse-Taylor binary system is a famous example for the confirmation of the gravitational wave theory. Russell Alan Hulse and Joseph Taylor got the 1993 Nobel Prize for the discovery and measurement of this system. The system consists of two neutron stars, one of which is a pulsar whose radio signals have been observed for many years. From the phase shift of the pulsar one can infer the orbital parameters τ T m1 m2

= 0.06 s, = 27906.980894 ± 0.000002 s, = (1.442 ± 0.003)M , = (1.386 ± 0.003)M

where T is the orbital time of the stars. Because of these very accurate pulse times, very precise astrophysical measurements can be done with this system. Let us ignore ellipticity of the orbit for an order of magnitude estimate. We obtain for the spiral time tspir. ∼ 109 yr.

(6.24)

From Kepler’s law we have T 2 ∝ r3 such that 2

dT dr =3 . T r 40

(6.25)

6.2 THE PSR 1913+16 SYSTEM Furthermore Eq. (6.18) implies that dr 1 r(0) =− . dt 0 4 tspir.

(6.26)

Therefore dT 3 dr T 3 T = =− ∼ 10−12 . dt 2 dt r 8 tspir.

(6.27)

Through the phase shift of the pulses one can measure the variation in the orbital time and finds experimentally dT = −(2.43 ± 0.03) × 10−12 . dt

(6.28)

This is the same order of magnitude as the above simplified GR estimate. So far we approximated the actual orbit by a circular one. We want to refine the analysis by taknig into account that the real orbit is actually elliptical in order to make more precise predictions. As we will see, this more detailed analysis yields very good agreement with the experimental data (6.28). Accounting for elliptical orbits (see the textbooks of Maggiore or Straumann for derivations), we have for the semi-major axis (G = c = 1) a=−

m1 m2 2E

(6.29)

where E is the total energy. Denoting by L the angular momentum, one finds for the squared eccentricity of the orbit 2EL2 (m1 + m2 ) . m31 m32

(6.30)

a(1 − e2 ) . 1 + e cos ϑ

(6.31)

8m21 m22 2 2 2 12(1 + cos ϑ) + e sin ϑ ϑ˙ 15a2 (1 − e2 )2

(6.32)

e2 = 1 + The orbit can be described as r= The emitted power reads P =

where h· · · i denotes the time average over a period T , Z Z 1 T 1 2π dϑ h· · · i = (· · · )dt = (· · · ) . T 0 T 0 ϑ 41

(6.33)

6.2 THE PSR 1913+16 SYSTEM Using the third Kepler law T =

2πa3/2 , (m1 + m2 )1/2

(6.34)

one finds after averaging 32G4 m21 m22 (m1 + m2 ) P = 5c5 a5 (1 − e2 )7/2

73 2 37 4 1+ e + e . 24 96

(6.35)

Then averaging da m1 m2 dE =− dt 2E 2 dt

(6.36)

over time, one finds da 2a2 dE 64G4 m1 m2 (m1 + m2 ) 73 2 37 4 = =− 5 1 + e + e . (6.37) dt m1 m2 dt 5c a3 (1 − e2 )7/2 24 96 Using Kepler’s third law ˙ T = T

again, this yields 3 a˙ 96 G4 = − 5 4 m1 m2 (m1 + m2 )f (e) 2 a 5c a 1 73 2 37 4 . where f (e) := 1 + e + e 24 96 (1 − e2 )7/2

(6.38) (6.39)

If we replace a by T in the right-hand side of Eq. (6.38), we conclude ˙ T m1 m2 96 =− 5 f (e). 8/5 T 5c (T /2π) (m1 + m2 )1/3

(6.40)

Plugging in the measured ellipticity of e ≈ 0.617, one gets T˙theoretical = (−2.40243 ± 0.00005) × 10−12

(6.41)

which is in perfect agreement with the measured data. This is indirect evidence for the existence of gravitational waves.

42

7. THE POST-NEWTONIAN APPROXIMATION

7

The post-Newtonian Approximation

Astrophysical sources of gravitational radiation are held together by gravitational forces. For a self-gravitating system of mass m we have v c

∼

RS d

(7.1)

where RS is the Schwarzschild radius and d is a typical distance. This relation follows immediately from r2 ≈ Gm/r ≈ Gm/d. We note that RS /d is a measure of the strength of the gravitational field close to the source. As soon as we consider v/c corrections to the orbital motion, for consistency we have to consider also (RS /d)1/2 corrections to the metric (corrections to the flat background). Moderately relativistic systems require a post-Newtonian treatment. The assumptions that we will use are • The systems under consideration are slowly moving, weakly self-gravitating systems such that an expansion in v/c or (RS /d)1/2 is possible. • The energy-momentum tensor Tµν has a spatially compact support (Tµν (r) = 0 for r ≥ d). If ωS is a typical frequency of the system, then typical velocities are v ≈ ωS d. As we saw before, the frequency of the radiation of the emitted gravitational radiation is ω = 2ωS ≈ 2v/d. In non-relativistic systems we have v c, thus c/v 1 and the wavelength of the emitted radiation satisfies λ = c/ω ∼ cd/v d. In analogy to the electromagnetic case, for non-relativisic sources it is convenient to distinguish between: • near field regime (r, d λ) where retardation is negligible and potentials are static, • far field regime (r λ) where retardation is crucial and we have waves. The small parameter in powers of which we will perform an expansion, is11 ε∼

RS d

1/2

v ∼ . c

(7.2)

We demand that |T ij | ∼ O(ε2 ). T 00 11

Note that some references also use the convention ε ∼ (v/c)2 .

43

(7.3)

7. THE POST-NEWTONIAN APPROXIMATION For a fluid with pressure p and energy density ρ, we thus have p ∼ ε2 . (7.4) ρ We expand the metric and the energy-momentum tensor in the near field regime in powers of ε. As long as the emission of radiation is neglected, a classical system subject to conservative forces is invariant under time reversal. Under time reversal, g00 and gij are even (i.e. there appear even powers of v and thus of ε, as well) while g0i is odd (i.e. only odd powers of v and thus of ε appear). By inspection of Einstein’s field equations, one finds that in order to work consistently to a given order ε, if we expand g00 up to O(εn ), then we have to expand g0i up to O(εn−1 ) and gij up to O(εn−2 ). The metric is expanded as follows: g00 = −1 + (2) g00 + (4) g00 + ... (3)

g0i =

gij = δij + where

(n)

(2)

g0i + (5) g0i + ...

gij +

(4)

(7.5)

gij + ...

gµν denotes terms of O(ε). Similarly, for the energy-momentum tensor: T 00 = (0) T 00 + (2) T 00 + ... T 0i =

(1)

ij

(2)

T

=

T 0i + (3) T 0i + ... ij

T +

(4)

(7.6)

ij

T + ...

We now want to insert these expansions into Einstein’s field equations and equate terms of the same order in ε. Considering v c, the time derivatives of the metric are smaller than the spatial derivatives by O(ε):

∂ ∂ = O(v) or ∂0 ∼ O(ε)∂i , (7.7) ∂t ∂x where we used that ∂0 = 1c ∂t . The d’Alembert operator applied to the metric, to lowest order becomes the Laplacian: 1 ∂2 − 2 2 + ∆ = (O(ε) + 1) ∆. (7.8) c ∂t This means that retardation effects are small corrections. Consequently, we also have to expand the geodesic equation

µ ν d 2 xi i dx dx = −Γ . (7.9) µν dτ 2 dτ dτ In chapter 9.3 of GR I, we considered the Newtonian limit where we just had to use g00 = −1 + (2) g00 , g0i = 0 and gij = δij . It thus follows that the terms (4) g00 , (3) g00 and (2) gij give the first post-Newtonian order for which we use the notation 1PN. The terms (6) g00 , (5) g0i , (4) gij give the 2PN approximation (which is already highly complicated), and so on.

44

7.1 THE 1PN APPROXIMATION

7.1

The 1PN Approximation

It is useful to choose a simplifying gauge condition right from the beginning. A convenient choice is the de Donder gauge condition √ ∂µ ( −gg µν ) = 0. (7.10) (this is a harmonic gauge condition, i.e. the coordinate functions satisfy the d’Alembert equation). The next step is to insert Eqs. (7.5), (7.6) into Einstein’s equations (together with (7.7) and (7.10)). We skip the explicit computations but state only the results12 . One finds for (2) g00 the Newtonian equation ∆

(2)

8πG g00 = − 4 (0) T 00 c

(7.11)

while the 1PN correction to the metric yields 8πG gij = − 4 δij (0) T 00 , c (3) 16πG (1) 0i g0i = ∆ T , 4 c ∆ (4) g00 = ∂02 (2) g00 + (2) gij ∂i ∂j (2) g00 − ∂i (2) g00 ∂i (2) g00 8πG − 4 (2) T 00 + (2) T ii − 2(2) g00 (0) T 00 , c ∆

(2)

(7.12) (7.13)

(7.14)

where ∆ = δ ij ∂i ∂j and the sum over repreated spatial indices is performed with δ ij . In order to write down a solution of (7.11) with the boundary condition that the metric is asymptotically flat, is Z (0) G T (x0 , t) (2) (7.15) g00 = −2φ with φ(t, x) = − 4 d3 x0 c |x − x0 | where the Newtonian potential is U = −c2 φ. Similarly Eqs. (7.12) and (7.13) are solved by (2) (3)

gij = −2φδij

g0i = ξi

(7.16) (7.17)

where we defined 4G ξi (x, t) = − 4 c 12

Z

d 3 x0

T 0i (x0 , t) . |x − x0 |

(1)

For details, see section 9.1. in Weinberg’s book or section 5.2 in Straumann’s.

45

(7.18)

7.1 THE 1PN APPROXIMATION In order to solve (7.14), we replace Furthermore we use the identity

(2)

g00 on the right-hand side by −2φ.

1 (∇φ)2 = ∂i φ∂i φ = ∆(φ2 ) − φ∆φ. 2

(7.19)

and we introduce a new potential ψ such that (4)

g00 = −2(φ2 + ψ).

(7.20)

Eq. (7.14) then reads ∆ψ = ∂02 φ +

4πG c4

(2)

T 00 + (2) T ii .

(7.21)

Using the boundary condition that ψ vanishes at spatial infinity, ψ can be written as Z G (2) 00 0 d 3 x0 1 2 (2) ii 0 ψ(x, t) = − ∂ φ+ 4 T (x , t) + T (x , t) . (7.22) |x − x0 | 4π 0 c Notice that φ and ξi are not independent due to the gauge condition (7.10) which imposes the contraint 4∂0 φ + ∇ξ = 0.

(7.23)

From Eqs. (7.15) and (7.18) one can see that this condition is indeed satisfied due to energy-momentum conservation at the 1PN order (since T µν is covariantly conserved in the exact solution, it has to be conserved at all post-Newtonian orders independently). We observe that φ, ψ, ξi are instantaneous potentials. Our order of approximation is thus insensitive to retardation effects. Note also that g00 can be expressed very simply as 2

g00 = −e2V /c + O(ε6 )

(7.24)

where V = −c2 (φ + ψ). This follows immediately if we expand the exponential and write (7.24) as 2V 2 1 2V g00 = −1 + 2 − 4 + O 6 c c c = −1 − 2(φ + ψ) − 2(φ + ψ)2 + O(ε6 ) = −1 − 2φ − 2(φ2 + ψ) − 2(ψ 2 + 2φψ) + O(ε6 ). (7.25) Using that φ = O(ε2 ) and ψ = O(ε4 ), we see that this is just g00 = −1 + (2) g00 + (4) g00 + O(ε6 ). 46

(7.26)

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC Putting together (7.11) (with

(2)

g00 = −2φ) and (7.21), we have

∆(φ + ψ) = ∂02 φ +

4πG (0) 00 (2) 00 (2) ii T + T + T . c4

(7.27)

To this order we can set ∂02 φ = ∂02 (φ + ψ) and replace ∆ by . We then obtain V = −

4πG 00 4πG T + T ii ≡ − 2 σ 2 c c

(7.28)

where we replaced (0) T 00 + (2) T 00 → T 00 and (2) T ii → T ii . The solution of Eq. (7.28) is given by a retarded integral Z |x − x0 | 0 1 00 ii 3 0 T +T ,x (7.29) t− V (x, t) = G d x |x − x0 | | c {z } =σ

and similarly for ξi defined as Vi , given by Z 1 |x − x0 | 0 3 0 Vi (x, t) = G d x σi t − ,x |x − x0 | c

(7.30)

where σi ≡ T 0i . To summarize, in harmonic coordinates the 1PN solution can be written in terms of two functions V and Vi in the following way: 2 2 2 1 g00 = −1 + 2 V − 4 V + O 4 , c c c 1 4 (7.31) g0i = − 3 Vi + O 5 , c c 1 2 gij = δij 1 + 2 V + O 4 . c c

7.2

Motion of Test Particles in the 1PN Metric

To get the equations of motion of a particle of mass m in the near zone, we have to solve the geodesic equation which is encoded in the following action which reads in curved background 1/2 Z dxµ dxν S = −mc dt −gµν dt dt 1/2 Z vi vivj 2 dt −g00 − 2g0i − gij 2 + . . . . (7.32) = −mc c c We are interested in the equations of motion for a binary system. If we restrict ourselves to the lowest PN corrections, it is possible to treat the two masses as point-like. 47

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC In curved space the energy-momentum tensor of a set of point-like particles is given by T

µν

N dxµ dxν 1 X γa ma a a δ (3) (x − xa (t)) . =√ −g a=1 dt dt

(7.33)

where the masses are denoted by ma and xµa are the coordinates (a = 1, ..., N ). Furthermore we used the definitions 1 γa = q 1−

and

va2 c2

pµa = γa ma

dxµa . dt

(7.34)

In an N -body system (a > 2) the metric felt by a particle b is obtained by taking the energy-momentum tensor of all other particles as a source. This amounts to P P replacing a by a(6=b) in Eq. (7.33). We expand the determinant of the metric to second order and using (2) g00 = −2φ, we get X (2) gii = 1 − 4φ. (7.35) −g = 1 − (2) g00 + i

Therefore the expansion of (7.33) gives (0)

T 00 (x, t) =

X a(6=b)

(2)

00

T (x, t) =

X

ma c2 δ (3) (x − xa (t)) ma

a(6=b) (1)

0i

T (x, t) = c

X

1 2 2 v + 2φc δ (3) (x − xa (t)) 2 a

ma vai δ (3) (x

a(6=b) (2)

ij

T (x, t) =

X a(6=b)

(7.36)

− xa (t))

ma vai vaj δ (3) (x − xa (t))

Inserting these expressions into Eqs. (7.15), (7.18) and (7.22), one can obtain the metric in which the particle b propagates. Inserting this metric into (7.32), one can calculate its action P Sb . The total action can then be obtained by summing over all particles, S = b Sb . Expanding the square root that appears in the integral of the action and demanding consistency of the expansion, only terms up to O ((v/c)4 ) give the 1PN correction. In terms of the Lagrangian, one can verify the following results for the two-

48

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC body system: 1 L1 (7.37) c2 1 Gm1 m2 1 , (7.38) with L0 = m1 v12 + m2 v22 + 2 2 r 1 1 L1 = m1 v14 + m2 v24 + 8 8 Gm1 m2 G(m1 + m2 ) 2 2 + 3(v1 + v2 ) − 7v1 v2 − (ˆ r v1 )(ˆ r v2 ) − 2r r (7.39)

L = L0 +

where r denotes the separation vector between the two particles, r = |r| and rˆ = rr . The Lagrangians L0 and L1 describe the Newtonian part and the first post-Newtonian correction, respectively. The same can be obtained for N particles (Einstein-Infeld-Hoffmann Lagrangian) : L0 = L1 =

X1 a

2

X1 a

8

ma va2 +

X Gma mb , 2rab

(7.40)

a(6=b)

ma va4 −

X Gma mb [7va vb + (ˆ rab va )(ˆ rab vb )] + 4rab

a(6=b)

3G X X ma mb va2 G2 X X X ma mb mc + − , 2 a b6=a rab 2 a b6=a c6=a rab rac

(7.41)

where a = 1, ..., N labels the particles, rab is the distance between particles a and b, and rˆab is the corresponding unit vector. From this Lagrangian one can derive the Euler-Lagrange equations of the N particle system including 1PN corrections. These equations are also called the Einstein-Infeld-Hoffmann equations. Denoting rab = |xa − xb | and xab = xa − xb , one finds after lengthy calculations X mc X X xab xab xbc 1 +G + − v˙a = −G mb 3 1 − 4G mc − 3 rab rac rbc 2rbc c6=a c6=a,b b6=a 2 ma 3 vb xab 2 2 − 5G + va − 4va vb + 2vb − − rab 2 rab X X 7 X mb mc xbc xab − G G +G mb (4va − 3vb ) · (va − vb ). 3 3 2 b6=a rab rbc rab c6=b,a b6=a (7.42)

49

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC

7.2.1

Two Body Problem in the 1PN Approximation

The Einstein-Infeld-Hoffmann equations for the two body problem imply that the center of mass X=

m∗1 x1 + m∗2 x2 m∗1 + m∗2

(7.43)

with v 2 1 m m G 1 a a b m∗a := ma + ma − 2 c 2 rab c2

(a, b = 1, 2)

(7.44)

is not accelerated, i.e. d2 X = 0. dt2 We can choose X = 0 such that m2 µδm 2 m v − x1 = + x m1 2m2 r m1 µδm 2 m + v − x2 = − x m 2m2 r

(7.45)

(7.46) (7.47)

where x = x1 − x2 , v = v1 − v2 , m = m1 + m2 , δm = m1 − m2 , µ = m1mm2 (reduced mass). For the relative motion we obtain from Eqs. (7.38), (7.39) with (7.46), (7.47) after dividing by µ: L = L0 + L1

(7.48)

with the Newtonian part L0 and the post-Newtonian perturbation L1 : Gm 1 L0 = v 2 − 2 r 3 3µ v 4 Gm µ 2 µ vx 2 G2 m2 2 L1 = 1− + 3v + v + − . 8 m c2 2rc2 m m r 2r2 c2

(7.49) (7.50)

The corresponding Euler-Lagrange equation is (c = 1) Gm 2µ 3µ 3µ vx 2 Gm 2 v˙ = − 3 x 1 − 4+ + 1+ v − + r r m m 2m r Gm 2µ + 3 v(vx) 4 − . (7.51) r m

50

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC

y

a

ϕ

~r

S

x

Figure 5: Elliptical Kepler orbit. Consider the Kepler problem with motion in the plane z = 0 and the periastron lying on the x-axis. Without 1PN corrections one has an elliptic orbit (c.f. fig. 5) where e is the eccentricity and r=

a(1 − e2 ) . 1 + e cos ϕ

(7.52)

One finds r = a(1 − e cos u)

where u − e sin u = ω0 t

(7.53)

(Kepler’s equation). Here ω0 = 2π with orbital period T (t = 0 for pasT sage at perihelion) and ω is the so called eccentric anomaly (ω0 t is the mean anomaly). We define f = ϕ, the true anomaly, such that (c.f. fig. 6) cos f = cos ϕ =

cos u − e . 1 − e cos u

(7.54)

Taking L1 from Eq. (7.50), we define ε=

E = L0 + L1 µ

(7.55)

and J = j(r ∧ v) µ 1 3µ v 2 µ Gm = 1+ 1− + 3+ (r ∧ v). 2 m c2 m rc2

j=

51

(7.56)

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC

y

~r

u

ϕ

S

x

Figure 6: The variable u takes the role of the angle as measured from the center of the ellipse. After some lengthy calculations (including an integration over time), one can get an equation which is analogous to the Kepler equation but with different coefficient: 2π t = u − et sin u (7.57) Tb r = ar (1 − er cos u) (7.58) where µ ε i Gm h 1− −7 2ε M 2c2 h ε i µ G2 m2 2ε µ 2 = 1 + 2 2 1 + 5 − 15 j + −6 Gm m 2c2 m c2 2ε h µ ε i 2 2µ G2 m2 = 1 + 2 2 1 + 17 − 7 j + 2− Gm m 2c2 m c2 µ ε i (−2ε)3/2 h = 1− − 15 Gm m 4c2

ar = −

(7.59)

e2r

(7.60)

e2t 2π Tb

(7.61) (7.62)

(in a bound orbit, ε is negative). The eccentricity e of the Keplerian orbit is now split into a “radial eccentricity” er and a “time eccentricity” et . The Newtonian limit is found by considering c → ∞:

e2r c→∞

Gmµ ar c→∞ = −2E 2EL2 = e2t c→∞ = 1 + 2 2 3 , Gmµ

52

(7.63) L2 ≡

(r ∧ v)2 µ3

etc.

(7.64)

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC One finds for the true anomaly cos u − eθ ϕ(u) = 1 − eθ cos u

" or

ϕ(u) = Aeθ (u) = 2 arctan

1 + eθ 1 − eθ

1/2

# u +g 2 (7.65)

where eθ = e for Kepler orbits. This yields ϕ(u) = ω0 + (1 + k)Aeθ (u)

with k =

3Gm − e2 )

c2 a(1

(7.66)

with ω02 = (2π/T )2 = Gm/a3 . The quantity eθ is called “angular eccentricity” and it satisfies µ ε i 6G2 m2 2ε h 2 2 eθ = 1 + 2 2 1 + − 15 j − . (7.67) Gm m 2c2 c2 For c → ∞, we have to make the replacement eθ → e. One can also find the perihelion precession (or periastron precession) per orbit (c.f. section 25 of GR). After some calculations one finds: δϕ =

6πG(m1 + m2 ) . c2 a(1 − e2 )

(7.68)

Note that a in this formula is the semi-major axis (a(1 − e2 ) = p); in section 25 it was half of the Schwarzschild radius, a = Gm . The relevant difference as c2 compared to section 25 is the fact that we had only one mass (the Sun) in section 25. In (7.68) m = m1 + m2 is the sum of the two masses. In view of the nonlinearities that are involved in the describption of the system, this simple result is far from being obvious. For the binary pulsar PST 1913+16 (c.f. section 6.2), the measured periastron shift is ω˙ obs. = 4.226607 ± 0.00007 deg yr−1

(7.69)

The GR prediction which follows from Eq. (7.68) and the known orbital element (given a period of about 7.75 hours, i.e. ω˙ ∼ δϕ × 1130 deg yr−1 ) is 2/3 m1 + m2 deg yr−1 (7.70) ω˙ GR = 2.11 M where we used

G a= (m1 + m2 ) ω02

1/3 (7.71)

due to Kepler’s thir law. If we set ω˙ obs. = ω˙ GR , it follows m1 + m2 = 2.83M . In order to characterize the pulsar system, several parameters are relevant: 53

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC • The parameters which characterize the pulsar itself: right ascension α, declination δ, proper motion, the ineratial pulse φ0 , the frequency ν, and the spindown parameter. • The five Keplerian parameters (c.f. fig. 7): Tb (period), t0 (time of passage at periastron), x = ar sin i (i denotes the inclination angle of the orbital plane with respect to the observer), e (eccentricity), ω (angular position of periastron as measured from the ascending node).

~r ω i

Figure 7: Sketch of spatial arrangement of post-Keplerian orbit relative to observer. • There are eight independent measurable post-Keplerian parameters (we state only the five main parameters): – ω˙ (periastron shift), – γ (not to be confused with γ from Robertson expansion), – ∆E (Einstein time delay) which is related to the transformation from the pulsar proper time to the coordinate time of the pulsar-companion barycenter system. One finds ∆E = γ sin u (7.72) 1/3 Tb G2/3 m2 (m1 + 2m2 ) where γ = e 2 2π c a(m1 + m2 ) −4/3 m2 m1 + 2m2 m1 + m2 = 2.93696 ms · M M M (7.73) – r=

Gm2 c3

(Shapiro time delay) which corresponds to 54

a c

= ν in Eq. (1.12),

7.2 MOTION OF TEST PARTICLES IN THE 1PN METRIC Tb −2/3 m2/3 m−1 2 2π G 7 2 m + 6m1 m2 + c2 am 2 1

– s = sin i = cG−1/3 t – δθ = eθe−e = t and (7.61).

2m22 where we used Eqs. (7.67)

Seven parameters are needed to fully specify the dynamics of the two-body system (up to uninteresting rotation about the line of sight). Therefore, the measurement of any two post-Keplerian parameters (besides five Keplerian parameters) allows to predict the remaining ones. These parameters thus constitute a consistency check for GR.

55

8. THE KERR SOLUTION

8

The Kerr Solution

The Kerr solution is one of the most important solutions of Einstein’s (vacuum) equations. It describes stationary rotating blak holes and it was found in 1963 by R. Kerr13 . Later on, it was generalized to the Kerr-Newman solution which describes rotating, electrically charged black holes. In this chapter we will give an overview over the Kerr solution and the KerrNewman solution without detailed calculations. The Kerr solution is axisymmetric and stationary. We use the so-called Boyer-Lindquist coordinates (t, r, θ, ϕ) and we use the following abbreviations: ∆ = r2 − 2mr + a2 ,

ρ2 = r2 + a2 cos2 θ

(8.1)

where we set G = c = 1. The Kerr metric reads 1 2 ds = 2 −(∆ − a2 sin2 θ)c2 dt2 + 2a sin2 θ(∆ − r2 − a2 )c dtdϕ+ ρ 2 2 2 2 2 2 2 2 2 2 dr + dθ . + sin θ (r + a ) − ∆a sin θ dϕ + ρ ∆

8.1

Interpretation of the Parameters a and m

In order to interpret a and m we look at the asymptotic form of the metric (8.2) for large “radial coordinate” r: 2m 1 1 4am 2 2 2 ds = − 1 − +O 2 sin θ + O 2 dt − dtdϕ+ r r r r 1 dr2 + r2 (dθ2 + sin2 θ dϕ2 ). (8.2) + 1+O r The examination is easier by transforming to asymptotically Lorentzian coordinates (Cartesian coordinates): x = r sin θ sin ϕ,

y = r sin θ cos ϕ,

z = r cos θ.

(8.3)

This yields the following form of the asymptotic metric: 4am 1 2am 1 2 2 ds = − 1 − +O 2 dt − +O 4 (x dy − y dx)dt+ 3 r r r r 1 + 1+O (dx2 + dy 2 + dz 2 ) (8.4) r 13

Phys. Rev. Lett. 11, 237 (1963)

56

8.1 INTERPRETATION OF THE PARAMETERS A AND M where we used sin2 θ dϕ = x dy − y dx. In section 4 we computed the metric of the rotating Earth assuming slow rotation. In Eq. (4.18) we found 2GME 2GME 2 2 2 c dt − 1 + 2 dr 2 + 2ch0i dxi dt. (8.5) ds = 1 − 2 cr cr E Identifying GM → m, dr 2 = dx2 + dy 2 + dz 2 and including an overall sign (since c2 we used another sign convention in section 4), this metric should coincide with (8.4). Using Eq. (4.1), Z 4G ω k xj d3 re x en ρ(e r )e xj , (8.6) h0i = 3 εikm 3 c r

we infer that T 0i is proportional to ρ vci where vi = εikn ω k xn with ω k being the angular velocity of the rotating body. We define Z Sk = εklm d3 x xl T m0 (8.7) which is the intrinsic angular momentum of the rotating body. Therefore, h0i =

4G xm S k ε . ikm c3 r3

(8.8)

With the sign convention of (8.4), Eqs. (8.4) and (8.5) indeed take the common form 1 2m 1 2m 2 2 +O 2 dt + 1 + +O 2 δij dxi dxj − ds = − 1 − r r r r k l S x 1 − 4εikl 3 + O 4 dtdxi . (8.9) r r Clearly S k is proportional to the body’s mass M and its angular momentum (cf. Eq. (4.13)): Sk =

aGM ∂ GM =a 2 × 2 c ∂z c

unit vector along polar axis of Boyer-Lindquist coordinates

.

(8.10)

Therefore, m is just the mass, and a can be interpreted as the angular momentum (0 ≤ a ≤ 1).

57

8.2 KERR-NEWMAN SOLUTION

8.2

Kerr-Newman Solution

The Kerr-Newman solution is the extension of the Kerr solution which also describes electrically charged black holes (c = G = 1). We use the three parameters ∆ = r2 − 2M r + a2 + Q2 ρ2 = r2 + a2 cos2 ϕ Σ2 = (r2 + a2 )2 − a2 ∆ sin2 θ

(8.11) (8.12) (8.13)

where M is the total mass, a the intrinsic angular momentum and Q the total charge. The metric coefficients of the Kerr-Newman metric are grr =

ρ2 , ∆

gtt = −1 +

gθθ = ρ2 ,

Σ2 sin θ, ρ2 2M r − Q2 = −a sin2 θ ρ2

gϕϕ =

2M r − Q2 , ρ2

gtϕ

where we assumed a > 0 without loss of generality. This metric contains the following special cases: • Q = a = 0: Schwarzschild solution, • a = 0: Reissner-Nordstrøm solution, • Q = 0: Kerr solution.

The electromagnetic field of the Kerr-Newman solution is F = Qρ−4 (r2 − a2 cos2 θ) dr ∧ (dt − a sin2 θ dϕ)+

+ 2Qρ−4 ar cos θ sin θ dθ ∧ (r2 + a2 )dϕ − adt

(8.14)

where ∧ denotes the exterior product defined in section 15 of GR. From (8.15) one can deduce the asymptotic expressions for electric and magnetic fields (in r, θ, ϕ directions): Q 1 Er = Frt = 2 + O 3 r r Fθt 1 Eθ = =O 4 r r r Eϕ Fϕt = =0 r sin θ r sin θ (8.15) Fθϕ 2Qa 1 Br = 2 = 3 cos θ + O 4 r sin θ r r Fϕr Qa 1 Bθ = = 3 sin θ + O 4 r sin θ r r Frθ Bϕ = = 0. r We see immediately that asymptotically the electric field is a Coulomb field. 58

8.3 EQUATIONS OF MOTION FOR TEST PARTICLES

8.3

Equations of Motion for Test Particles

Let a test particle with electric charge e and rest mass µ move in the external fields of a Kerr-Newman black hole. The geodesic equations are β d 2 xα dxβ dxγ α dx α = eF + Γ β βγ dλ2 dλ dλ dλ

(8.16)

The best way to solve this equation turns out to be the Hamiltonian formalism.14 We can simplify the analysis by assuming that the metric is that of a Kerr black hole (Q = 0) and that the motion is confined to the equatorial plane (θ = 0).15 In this case, the metric has the following non-vanishing components: 2GM gtt = − 1 − 2 , (8.17) cr a 2GM gtϕ = − , (8.18) 2 r c a2 2GM gϕϕ = r2 + a2 + . (8.19) r c2 Denoting by K = particle, one finds

E µc

the total energy and by l the angular momentum of the 1 2 µr˙ + µVeff. = const. 2

(8.20)

with the effective potential 2GM (l − aK)2 l2 − a2 (K 2 − c2 ) c2 − 2r 2µ2 r2 2µ2 c2 r3 GM l2 − a2 (K 2 − c2 ) GM (l − aK)2 =− + − . r 2µ2 r2 µ 2 c2 r 3

Veff. = −

2GM 2 c c2

+

(8.21)

For a = 0 this reduces to the Schwarzschild case. For a black hole which is spinning extremely fast, it can be shown that for a particle which spirals in towards the black hole in an accretion disc from very far 1 away to the innermost circular stable orbit, the fraction 1 − √3 of its rest energy is set free. The innermost stable circular orbit can easily be determined from Eq. (8.21). Thus a rotating black hole allows a gravitational energy conversion with an efficiency up to ≈ 42.3%! These considerations are of astrophysical importance since quasars in the center of galaxies are supermassive rotating black holes. 14

Details can be found in Straumann’s book or the book by Misner, Thorne, Wheeler. Note that the equatorial plane of a rotating black hole is distinguished. When we solved the geodesic equations in the Schwarzschild background, any arbitrary plane which includes the origin was equivalent. 15

59

Literature • T. Fliessbach, Allgemeine Relativit¨atstheorie, Spektrum Verlag, 1995 • S. Weinberg, Gravitation and Cosmology, Wiley, 1972 • M. Maggiore, Gravitational Waves: Volume 1: Theory and Experiments, Oxford University Press, 2007 • N. Straumann, General Relativity with Applications to Astrophysics, Springer Verlag, 2004 • C. Misner, K. Thorne and J. Wheeler, Gravitation, Freeman, 1973 • R. Wald, General Relativity, Chicago University Press, 1984 • B. Schutz, A first course in General Relativity, Cambridge, 1985 • R. Sachs and H. Wu, General Relativity for mathematicians, Springer Verlag, 1977 • J. Hartle, Gravity, An introduction to Einstein’s General Relativity, Addison Wesley, 2002 • H. Stephani, General Relativity, Cambridge University Press, 1990

Index Anomaly Eccentric, 51 Mean, 51 True, 51 Boyer-Lindquist coordinates, 56

post-Newtonian, 43 Post-Newtonian approximation, 12 Quadrupole Radiation, 30 Quadrupole Tensor, 34 Retarded Potential, 15

Dipole Radiation, 30

Shapiro delay, 4 Einstein-Infeld-Hoffmann Lagrangian, 49 Strain, 39 Ellipticity, 38 Time reversal, 44 Exact solution, 12 Far field regime, 43 Fermi transport, 6 Frame dragging, 20 Gauge transformation, 14 de Donder gauge, 45 harmonic, 45 Geodetic Equation, 7 Geodetic precession, 6, 9 Gravitomagnetic, 16, 21 field, 19 potential, 19 Helicity, 25 Inspiral Time, 39 Kepler’s equation, 51 Kerr solution, 16, 56 Kerr-Newman solution, 56, 58 Lense-Thirring effect, 6, 16 Linearized field equations, 12 decoupled, 15 Long Wavelength Approximation, 31 Near field regime, 43 Polarization tensor, 23 61

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