The Relativistic Motion of a Binary System · 2020-07-29 · The Relativistic Motion of a Binary System Stephan Dinkgreve [email protected] Master Thesis Theoretical Physics Supervised

The Relativistic Motion of a Binary System

Stephan Dinkgreve

[email protected]

Master Thesis Theoretical Physics

Supervised by Jan Pieter van der Schaar

University of Amsterdam

May 28, 2012

Abstract

The Post-Newtonian expansion (PN) is used to find the first relativistic corrections to theequations of motion of a binary system of point particles. The expansion assumes velocities tobe small (v << 1) and the field to be weak (Φ << 1), yet allowing for terms of higher order in Φwith every subsequent term in the expansion. We will derive the orbital equations for the firstcorrection (1PN) following Damour and Deruelle [2]. These results were used in the calculationof the change of the orbital period of the Hulse-Taylor pulsar, the ratio of which comparedto the observed value is an astonishing 1.0013(21) [15]. We also derive effective potentials for1PN and 2PN, which will give us a measure for the accuracy of the expansion. We will thenfollow Maggiore [9] and Poison [11] to show that the equations of motion at 2.5PN containa back-reaction term, which is equivalent to the power of the gravitational radiation emittedfrom the system. Knowing the equations of motion and their corresponding gravitationalwaveforms to high order will be of major importance for the extraction of gravitational wavesfrom the noise in the upcoming gravitational wave detectors.

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Contents

1 Introduction 3

2 The Newtonian Two-Body Problem 42.1 The Radial Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 The Effective Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 The Newtonian Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 The Newtonian Approximation 83.1 The Linearized Einstein Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 The Schwarzschild Solution 114.1 Deriving the Schwarzschild Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 The Schwarzschild Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3 The Precession of Mercury . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.4 Harmonic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Gravitational Waves 175.1 Gravitational Waves in Linear Theory . . . . . . . . . . . . . . . . . . . . . . . . . 175.2 The Energy of Gravitational Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3 The Quadrupole Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245.4 The Order of Magnitude of Gravitational Radiation . . . . . . . . . . . . . . . . . 275.5 Gravitational Waves from a Binary in Circular Orbit . . . . . . . . . . . . . . . . . 275.6 The Inspiraling of a Binary System . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.7 Gravitational Waves from a Binary in Elliptical Orbit . . . . . . . . . . . . . . . . 30

6 The First Order Post-Newtonian Approximation 356.1 Deriving the First Post-Newtonian Metric . . . . . . . . . . . . . . . . . . . . . . . 356.2 The First Order Schwarzschild Radial Equation . . . . . . . . . . . . . . . . . . . . 396.3 The First Order Post-Newtonian Lagrangian . . . . . . . . . . . . . . . . . . . . . 416.4 The Relative Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.5 The Post-Newtonian Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.6 The Pseudo-Effective Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.7 Visualizing the Accuracy of the PN Approximation . . . . . . . . . . . . . . . . . . 486.8 Applying 1PN: The Hulse-Taylor Pulsar . . . . . . . . . . . . . . . . . . . . . . . . 50

7 Radiation Reaction 547.1 The Landau-Lifshitz Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 547.2 The 1PN Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567.3 The Surface Integral Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.4 The Radiation Reaction Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 637.5 The 2.5PN Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

8 Conclusion 67

9 Appendix 689.1 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

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1 Introduction

In 1915 Einstein derived a wonderful set of formulas called the Einstein Field Equations. Inthese equations the Newtonian force of gravity is replaced by the curvature of space-time andrelated to the energy and momentum in the universe. Unfortunately, these equations are verydifficult to solve, even for simple energy-momentum configurations and we shall have to resort toapproximations. In this article we will study the relativistic motion of a two-body system. Tofind the equations of motion we use the post-Newtonian expansion. This expansion is based onthe assumption that the velocities of the particles are low (v << 1 ) and that the fields are weak(Φ << 1), yet allowing for terms of higher order in Φ with every subsequent term in the expansion.

Since our starting point will be low velocities and weak fields, our leading term in the equationof motion will be Newtonian and the other terms will be relativistic corrections. The drawbackof this approach is that our equations will take a deceptively Newtonian form, which wipes outthe conceptual framework of general relativity. Despite this drawback, the obtained results arein strong agreement with observation and constitute strong evidence for general relativity. TheEinstein Equations suggest the existence of gravitational waves, which extract energy and angularmomentum from a binary system, causing a change in the orbital period. This change matches theobserved orbital period change of the Hulse-Taylor pulsar with great precision (the ratio betweenthe theoretical and the observed value is an astonishing 1.0013(21) [15]).

We start in the first chapter by deriving the main results of the Newtonian binary system. Boththe results and the method to obtain them will have close analogies in later chapters. Next wederive the Newtonian equations of motion from Einstein’s theory, assuming low velocities and weakfields (v << 1 and Φ << 1). The result is called the zeroth order term of the Post-Newtonianexpansion (0PN).

In the following chapter we will derive the main results from the Schwarzschild binary system. Inthis case we shall neither assume low velocities nor a weak field. However, we will assume thatone of the particles has negligible mass compared to the other (m2 << m1). In the case of weakfield and low velocities this system will serve as a limiting case to the Post-Newtonian expansion.Following this we study the gravitational radiation that is being generated by the binary system.

The radiation causes a back-reaction on our system and thus influences the equations of motion.In this chapter we will conclude that the radiation will only contribute at 2.5PN.

In the last two chapters we study the first relativistic corrections to the equations of motion,starting with 1PN. This result was first derived by Einstein, Infeld and Hoffmann in 1938 [5]. Wewill then continue to derive the orbital equations, which were derived by Damour and Deruelle in1985 [2]. In the last chapter we shall calculate the equations of motions corresponding to 2.5PN,which describes the largest contribution to the back-reaction on the system caused by the emissionof gravitational waves.

On various occasions we shall derive effective potentials, which give a visual representation ofthe possible orbits associated with different energies and angular momenta. The Newtonian andSchwarzschild effective potentials are discussed in most introductory textbooks on the subject andwill be discussed in chapter two and four. In this article however, we will extend their use to both1PN and 2PN (chapter 6) and to a system emitting gravitational radiation (chapter 5). We shalluse the 1PN and 2PN effective potentials to get an idea of the accuracy of the Post-Newtonianapproximation. In case of gravitational radiation we use it to visualize the circularization of theorbit due to the back reaction.

Let’s start. We have a lot to talk about.

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2 The Newtonian Two-Body Problem

We shall begin studying the two-body problem in the case of Newtonian mechanics. This isnecessary, because a lot of the techniques used in this chapter will have their analogies in laterchapters. Also, the Newtonian theory will serve as a limiting case of the general theory of relativity.More explicitly, the Newtonian theory should reappear when velocities are negligible (v << 1)and the fields are weak (Φ << 1). We’ll start by deriving the radial equation, which describesthe change of the distance between the two bodies. This expression can be written in the formr = E+V , where V is the effective potential. A plot of this potential gives a visual representationof the types of orbits that are possible in our system (circular, elliptical, parabolic and hyperbolicorbits). Finally we shall calculate the paths of the elliptical orbits explicitly. In this chapter we’veoften used [2], [6] and [16].

2.1 The Radial Equation

If we want to describe the location of two par-ticles in three-dimensional space we generallyneed six quantities. We can for instance pickthe two three-component position vectors r1and r2. Alternatively, we can use the centerof mass vector R ≡ 1

m1+m2(m1r1 +m2r2) and

a vector reaching from one particle to the otherr ≡ r2 − r1. This choice is depicted in the firstfigure on the right.

Using Noether’s theorem (appendix) we can find conserved quantities corresponding to invarianceunder resp. spatial translations and Galileo translations:

P = m1v1 +m2v2 K = m1r1 +m2r2 − tP (1)

The first equation expresses conservation of the total momentum. Consequently, we can pick aninertial frame in which the total momentum is zero. This also simplifies the second equation,which now implies the conservation of the center-of-mass vector R. We can therefore pick theframe in which R = K = 0. We end up with:

0 = m1v1 +m2v2 0 = m1r1 +m2r2 (2)

This simplification is depicted on the right side of the previous figure and is called the center-of-mass frame. We rewrite the last two equations to obtain:

r1 = µ/m1r r2 = −µ/m2r (3)

v1 = µ/m1v v2 = −µ/m2v (4)

Here we have use the reduced mass (µ = m1m2/(m1 +m2)) and the difference vector v = v2 − v1.

The Lagrangian of our two body system is:

L =1

2m1v

21 +

1

2m2v

22 +

m1m2

r(5)

In terms of the center-of-mass coordinates this can be rewritten to obtain the so-called relativeLagrangian:

L =1

2µv2 +

m1m2

r(6)

4

Observe that the initial two-body lagrangian now has the simplified shape of a one-body lagrangian.

The lagrangian is invariant under time translation and under rotation. Again using Noether’stheorem, these invariances can be show to correspond to the conservation of energy and momentum(resp. E = v · dL/dv − L and J = r × dL/dv). We find:

E =1

2µv2 − m1m2

rJ = µr × v (7)

It is now convenient to switch to polar coordinates. Because of the conservation of angularmomentum, the motion of the binary system takes place in a plane. This means that the polarangle ϕ is constant. The velocity in polar coordinates therefore becomes:

v2 = r2 + r2θ2 (8)

Also using |r × v| = r2θ, we find:

J = µr2θ J = r2θ (9)

E =1

2µr2 +

J2

2µr2− m1m2

rE =

1

2r2 +

J2

2r2− M

r(10)

Here we’ve used M = m1 +m2, E = E/µ and J = J/µ. From the last equation we find the radialequation of motion:

r2 = 2(E +

M

r− J2

2r2

)(11)

2.2 The Effective Potential

The third term in equation (11) is a centrifugal term. However, since this term is dependent on r,we can pretend it to be a part of the potential energy. We call this fictitious potential the effectivepotential V:

r2 = 2(E − V

)V = −M

r+J2

2r2(12)

The effective potential is plotted in figure to theright. Notice that the first particle is always po-sitioned at the origin of this graph, since thatis where the vector r starts. The second par-ticle is at variable distance r. For large r thegraph starts to look more and more like thegravitational potential, yet for small r the an-gular momentum term starts to dominate, pre-venting the second particle from closing in toomuch.

Since the left hand side of equation (12) is al-ways zero or positive, the energy must alwaysbe higher than the effective potential. Imag-ine the second particle coming in from infinitytoward the particle at the origin. Since thepotential is zero at infinity, this must meanE > 0. At some point the particle reaches apoint where E = V . The radial equation tellsus, that this is the point where the velocity ofthe particle is zero.

5

But what about the acceleration at that point? To answer this we differentiate the radial equationwith respect to time:

d

dtr2 = 2rr = −dV

drr 2r = −dV

dr(13)

The expression dV/dr will only be zero at a minimum or maximum of the last graph, so theacceleration of our particle with E > 0 will be non-zero at E = V . The particle thus closesin to the first particle, until its energy is equal to the effective potential and then goes back toinfinity. In the next paragraph we will show that these trajectories correspond to the hyperbolicand parabolic orbits.

If E < 0 the particle is stuck between two turning points and will thus move back and forthbetween these points. The particle thus oscillates between two fixed radii. These trajectories willturn out to be the elliptical orbits. In the case where the energy is equal to the the minimum ofthe potential, the radius of the particle stays constant. This corresponds to circular motion.

2.3 The Newtonian Orbits

In this paragraph we’ll derive the exact shapes of the orbits by discovering how the angle θ andthe radius r are related. We use:

dθ = dθdt

drdr =

θ

rdr dθ =

J

r2rdr r =

dr

dθ

J

r2(14)

In the second step we have used (9). Inserting this into the radial equation (11) we obtain:

θ(r) = ±∫

J/r2√2(E +M/r − J2

2r2 )dr (15)

The solution to this integral is:

r =J2

M(1 + e cos θ)e =

√1 +

2EJ2

M2(16)

In case of the circular orbits we have e = 0. For elliptical orbitswe have 0 < e < 1. For e = 1 and e > 1 we have the parabolicand hyperbolic orbits respectively. The solutions for different e aregraphically presented in the figure to the right. In case of ellipticalorbits we can rewrite the last equation in terms of the semi-majoraxis a (the longest radius of the ellipse). This is the famous orbitalequation for the ellipse:

r =a(1− e2)

1 + e cos θa =

J2

M(1− e2)(17)

Writing the energy and the angular momentum in terms of the semi-major axis and the eccentricity, we find:

E = −M2a

J2 = −Ma(1− e2) (18)

An ellipse is called closed when after a rotation of θ = θ0 + 2π the particle ends up at the sameradius. If not, it is said to be open. Since cos(θ) = cos(θ+2π) we must conclude from equation (17)

6

that Newtonian ellipses are closed. Newton already realized this and noted that if the ellipticalorbits were found to be open (for which there was no evidence at the time), it would require aforce law different from his inverse-square law [16].

With equation (17) we can find the shape of the orbits, but we do not yet know how the radius isdependent on time. To derive this relation we again start with the radial equation (11):

dr

dt= ±

√2

µ(E +

m1m2

r)− J2

µ2r2(19)

Flipping the fraction on the left and integrating we find:

t =

õ

2

∫ r

0

dr√2µ (E + m1m2

r )− l2

2µr2

(20)

In case of elliptical motion this is most conveniently integrated through an auxiliary variable u(t),called the eccentric anomaly. u(t) is defined by the following relation:

r = a(1− e cosu) (21)

Inserting this into the orbital equation (17) we find:

cos θ =cosu− e

1− e cosu1 + e cos θ =

1− e2

1− e cosu(22)

We can rewrite this, using trigonometric identities, as:

θ = Ae(u) Ae ≡ 2 arctan[√1 + e

1− etan

u

2

](23)

Rewriting the integral in terms of u, a and e we find:

t =

õa3

m1m2

∫ u

0

(1− e cosu)du (24)

Integrating this from 0 to 2π we find Kepler’s law:

ω0 =2π

T=

√m1 +m2

a3(25)

Integrating without fixing u in the integral we arrive at the Kepler equation:

ω0t = u− e sinu (26)

Combining this equation with equation (21) we can relate r to t. Similarly, combining the Keplerequation with equation (22) gives us the relation between θ and t. The solution to the Keplerequation can be expressed in terms of Bessel functions Jn:

u = ω0t+∞∑

n=1

2/nJn(ne) sinω0t (27)

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3 The Newtonian Approximation

In this chapter we will simplify the Einstein equations by assuming the metric to be nearlyminkowskian (gµν = ηµν + hµν with |hµν | ≪ 1). The resulting equation is called the linearizedEinstein equations, since we keep only the terms linear in hµν . The linearized Einstein equationshave the form of a wave function, which suggests the existence of gravitational waves. If we alsoassume low velocities (v << 1) we obtain the Newtonian equations of motion. In this chapter wehave often used [9] and [13].

3.1 The Linearized Einstein Equations

When the metric is nearly flat (nearly Minkowskian), we are able to pick a coordinate system inwhich the metric can be written in this form:

gµν = ηµν + hµν |hµν | ≪ 1 (28)

Since hµν is already very small, quadratic terms in hµν can be neglected. We find the inverse ofthe metric using gµνg

να = δαµ. When guessing:

gµν = ηνα − hνα (29)

we indeed find to first order:

gµνgνα = ηµνη

να + ηµνhνα − hµνη

να = δαµ (30)

In its full form the Einstein equations are invariant under a huge symmetry group, namely thegroup of all possible coordinate transformation xµ → x′µ. Choosing a nearly flat metric meanswe’ve broken this symmetry. We can however still perform the gauge transformations xµ → x′µ =xµ + ξµ(x) which keeps the weak-field metric unchanged. To find how hµν transforms under thisgauge transformation, we use the general metric transformation rule,

gµν(x′) =

∂xρ

∂x′µ∂xσ

∂x′νgρσ(x) (31)

The transformation matrices become:

∂x′µ

∂xρ=∂[xµ + ξµ]

∂xρ= δµρ + ∂µξ

µ ∂xµ

∂x′ρ= δµρ − ∂µξ

µ (32)

In the last step we’ve used that the transformation metric times its inverse gives the identitymatrix. Filling this into the metric transformation rule we find:

hµν(x) → hµν(x′) = hµν(x)− ∂µξν − ∂νξµ (33)

When |∂µξν | is of same order or smaller then hµν , the condition |hµν ≪ 1| is indeed preserved.

In case of the nearly flat metric (28) the Riemann tensor reduces to:

Rµνρσ =1

2(∂ν∂ρhµσ + ∂µ∂σhνρ − ∂µ∂ρhνσ − ∂ν∂σhµρ) (34)

Inserting equation (33) in the last equation we find that the linearized Riemann tensor staysinvariant under gauge transformations.

We now define:

hµν = hµν − 1

2ηµνh h = ηµνhµν (35)

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Observe that h ≡ ηµν hµν = h− 2h = −h, having used ηµνηµν = 4. We can use this to invert the

last equation:

hµν = hµν − 1

2ηµν h (36)

In this notation the Einstein equations become:

hµν + ηµν∂ρ∂σhρσ − ∂ρ∂ν hµρ − ∂ρ∂µhνρ = −16πTµν (37)

This equation would drastically simplify if we could set:

∂ν hµν = 0 (38)

This is called the harmonic gauge condition. Luckily, it is always possible to perform a gauge trans-formation to a system where this condition is satisfied. In terms of hµν the gauge transformation(33) becomes:

hµν(x) → hµν(x′) = hµν(x)− ∂µξν − ∂νξµ + ηµν∂ρξ

ρ (39)

Therefore:

∂ν hµν(x) → ∂ν hµν(x′) = ∂ν hµν(x)−ξµ (40)

So if the initial near-flat field is such that ∂ν hµν(x) = fµ(x), to obtain the harmonic gaugecondition ∂ν hµν(x

′) = 0 in the new frame, we much choose ξµ, such that ξµ = fµ(x). Thisequation always has a solution:

ξµ(x) =

∫d4xG(x− y)fµ(y) xG(x− y) = δ4(x− y) (41)

With the harmonic gauge condition, the resulting linearized Einstein equations take the form of awave equation:

hµν = −16πTµν (42)

Differentiation both sides and using the harmonic gauge condition, we find an expression forthe conservation of energy-momentum ∂µTµν = 0. The physical solutions to linearized Einsteinequations are:

hµν = −16π

∫d3x′G(x− x′)Tµν(x

′) G(x− x′) = − 1

4π|x− x′|δ(tret − t′) (43)

Here tret = t− |x− x′|. This retarded time reflects the fact that gravitational waves from sourcepoint x′ need a certain time to arrive at the observer at position x. From the expression of theretarded time we see the velocity of the gravitational waves is equal to the speed of light. Theother possible solution to the linearized Einstein equations is written in terms of the advancedGreen’s function, but these give unphysical solutions, since this would mean that the observermight notice the effect of gravitational waves before they are emitted. We write the solution as:

hµν(t, x) = 4

∫d3x′

1

|x− x′|Tµν(t− |x− x′|, x′) (44)

In contrast Newton’s theory is governed by Poisson’s equation, which has the solution:

Φ =

∫ρ

|x− x′|d3x′ (45)

Notice there is no time delay in the Newtonian case, which means the potential acts instantly.Einstein showed that in case of gravity there was no mysterious action at a distance.

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When |x| = r is much greater than the source d, we can expand |x− x′| as:

|x− x′| = r − x′ · n+ ... (46)

So far away from the source we arrive at:

hµν(t, x) =4

r

∫d3x′Tµν(t− r + x′ · n, x′) (47)

In the Newtonian limit we will also assume the velocities to be small (v << 1). In this limit theenergy-momentum tensor will be dominated by T 00 = ρ, since the other components are dependenton the velocity. Looking at the linearized Einstein equations (42), we see that the dominance ofT 00 immediately translates to the dominance of h00. Since the velocities are small, the metriconly changes slowly, so we can also neglect the time derivative of the metric. Therefore we areonly left with:

∇h00 = −16πρ (48)

Comparing this to Poisson’s equation of Newtonian mechanics we find h00 = −4Φ.

When h00 is dominant we have h = −h00 = 4Φ and from equation (35) we find h00 = −2Φ. Usingequation (35) for µ = ν = i we find hii = −2Φ. The line element is therefore given by:

ds2 = −(1 + 2Φ)dt2 + (1− 2Φ)(dx2 + dy2 + dz2) (49)

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4 The Schwarzschild Solution

In this chapter we will study the full general relativistic solution of a binary system where oneof the particles has negligible mass compared to the other (m1 >> m2). The metric, called theSchwarzschild metric, is therefore only determined by the large mass. We start by finding themetric for this binary system. Using this we derive the radial equation and find the effectivepotential, which we will compare to the Newtonian versions from chapter 2. We will then studythe shape of the elliptical orbits, which will not turn out to be closed like in the Newtonian case.Finally we will derive the Schwarzschild metric for weak fields in harmonic coordinates, which wewill need in chapter 6. In this chapter we have often used [13] and [18].

4.1 Deriving the Schwarzschild Metric

Since the motion of the massive particle isn’t influenced by the second particle, we can freely picka frame in which the massive particle is at rest. Since the particle will not move, the metric willbe independent of time. For a point particle the metric can also be no directional dependence, soit will be spherically symmetric. Therefore, the metric has this general form:

ds2 = A(r)dt2 +B(r)dr2 + C(r)r2dΩ2 dΩ2 ≡ dθ2 + sin θ2dϕ2 (50)

We can simplify this equation with a change in coordinates. We set C(r)r2 ≡ (r)2. To changefrom dr to dr we write:

dr

dr=d(C(r)r2)

dr=dC(r)

drr2 + 2rC(r) dr =

(C(r)dr

r2 + 2rC(r))dr (51)

We can absorb the extra term that appears before dr in the function B(r). The metric becomes:

ds2 = A(r)dt2 +B(r)dr2 + r2dΩ2 (52)

Since Tµν does not contribute except at the massive particle the Einstein equations simplify toRµν = 0. Writing out these equations in terms of the metric we end up with only three independentequations (the last two equations only differ by a factor sin2 θ):

4BA2 − 2rAAB + rBAA+ rA2B = 0 (53)

rBA+ 2B2A− 2AB − rAB = 0 (54)

−2rAAB + rBAA+ rA2B − 4AAB = 0 (55)

sin2 θ[−2rAAB + rBAA+ rA2B − 4AAB] = 0 (56)

Subtracting equation (53) from equation (55) we find:

AB +AB =d

dt(AB) = 0 → AB = C1 (57)

Here C1 is a constant. Inserting this in equation (54) we end up with:

rA = A(1−A) (58)

The solution to this equation is:

A =(1 +

1

C2r

)−1

(59)

And the metric thus becomes:

ds2 = C1

(1 +

1

C2r

)dt2 +

(1 +

1

C2r

)−1

dr2 + r2dΩ2 (60)

11

To find the constants we compare gtt with the Newtonian metric, which for a point-mass is equalto:

gtt = −(1− 2m1

r

)(61)

We find C1 = −1 and C2 = 1/(2m1). Notice that r turns out to be equal to r. So finally theSchwarzschild metric becomes:

ds2 = −(1− 2m1

r

)dt2 +

(1− 2m1

r

)−1

dr2 + r2dΩ2 (62)

4.2 The Schwarzschild Orbits

The second particle with mass m2 << m1 moves on a geodesic in the metric of particle m1. Thegeodesic equation is:

duρ

dτ+ Γρ

µνuµuν (63)

Multiplying with gρν we find:

duνdτ

= Γρµνu

µuρ = Γρµνu

µρ =

1

2(∂µgσν + ∂νgµσ − ∂σgµν)u

µuσ (64)

If you switch ν and σ in the first and third term in the parentheses you can see this pair isantisymmetric. uσuµ on the other hand is symmetric. A symmetric times an antisymmetrictensor gives zero, so we are only left with:

duνdτ

=1

2∂νgµσu

µuσ (65)

This equation tells us that if all components of gµσ are independent of some xν with a particularν, we have ∂νgµσ = 0. From equation (65) we see this implies that uν is a constant alongthe entire geodesic. Since the Schwarzschild metric is time-independent, we have a constant ofmotion u0. Since u0 = p0/m2, this constant is equal to the energy per unit mass. The metric isalso independent of the angle ϕ, giving us conservation of angular momentum. This implies themotion is confined to a plane, just as in the Newtonian case. For convenience we choose θ = π/2.We find:

p0 = −E pϕ = J (66)

Using pµpν = −m22 we get:

gµνpµpν = g00p20 + grrp2r + gϕϕp2ϕ = −m22 (67)

Filling in the components we arrive at:

−(1− 2m1

r

)−1

E2 +(1− 2m1

r

)−1

m22

(drdτ

)2

+J2

r2= −m2

2 (68)

This equation can be rewritten to find the radial equation and the associated effective potential:(drdτ

)2

= E2 −(1− 2m1

r

)(1 +

J2

r2

)= E2 − V 2

1 (69)

V 21 =

(1− 2m1

r

)(1 +

J2

r2

)(70)

Here we have used E = E/m2.

12

To show that in the limit of weak fields and low velocities we find back the Newtonian radialequation (11), we use:

E2 −m22

m22

= p2 (71)

The radial equation becomes:(drdτ

)2

= 2( p22

+m1

r− J2

2r2+m1J

2

r3) = 2

( p22

− V2

)(72)

V2 = −m1

r+J2

2r2− m1J

2

r3(73)

In the weak field limit, the m1/r3 term can be neglected. In the low velocity limit the term p2/2

becomes the kinetic energy per unit mass. Rewriting the radial equation we get:

˜p2

2=

1

2

(drdτ

)2

− m1

r+J2

2r2− m1J

2

r3(74)

The right hand side of this equation reduces to the total Newtonian energy (10). The curious factthat the kinetic energy seems to incorporate the Newtonian potential reflects the fact that theparticle on a geodesic experiences no potential energy (yet it follows a curved path as though itdoes). So taking p2/2 as the Newtonian equivalent of the total energy, we find back the Newtonianradial equation (11). The effective potential V2 is plotted in the next diagram.

Instead of the effective potential going all theway up for small r, as in the Newtonian case,it goes all the way down. Notice that the leftside of the radial equation is always positive orzero, therefore p2/2 will always be higher thanthe V2. The velocity of the particle is zero whenp/2 = V2. Differentiating the radial equationwith respect to τ we find:

2dr

dτ

d2r

dτ2= −2

dV2dr

dr

dτ

d2r

dτ2= −dV2

dr(75)

From this we find that only at the minima ormaxima both the velocity and the accelerationare zero, describing circular orbits. At a maxi-mum the circular orbit is unstable and thereforeof no physical significance.

For circular orbits the last equation becomes:

0 =d

dr

(− m1

r+J2

2r2− m1J

2

r3

)r =

J2

2m1

(1±

√1− 12m2

1

J2

)(76)

From this equation we see there are two circular orbits for J2 > 12m21. At J2 = 12m2

1 there isonly one orbit left and for J2 < 12m2

1 we have no circular orbits at all.

If a particle is stuck in the potential well (p/2 < 0), we have the Einsteinian equivalent of theNewtonian elliptical orbits. If a particle with 0 < p2/2 < Vmax comes in from a distance it willat some point be equal to the potential and turn around. These trajectories are the Einsteinianequivalents to the Newtonian hyperbolic orbits. If the energy is larger than the maximum of thepotential (p/2 > Vmax), the particle drops right towards r = 0 never to return.

13

4.3 The Precession of Mercury

To find the orbital equations for the quasi-elliptical orbits we start with:

dϕ

dτ≡ uϕ = gϕϕ

pϕm2

=J

r2(77)

Dividing the square of the radial equation (62) by the square of the last equation we find:

(drdτ

)2

/(dϕdτ

)2

=( drdϕ

)2

=E − (1− 2m1

r )(1 + J2

r2 )

J2/r4(78)

Solving this equation, while neglecting terms of order 1/r3 we find back the Newtonian ellipse(16):

1

r=m1

J2+

√E2 +m2

1/J2 − 1

J2cos(ϕ) (79)

To see how this equation is equal to the Newtonian equation we again have to equate p2/2 withthe Newtonian E. As noted before, this formula described a closed ellipse. For a circular orbitwe have 1/r = m1/J

2 (see equation (16)), so we can define y = 1/r −m1/J2 to be the deviation

from circularity. In terms of y the solution becomes:

y =

√E2 +m2

1/J2 − 1

J2cos(ϕ) (80)

In the relativistic case we will not neglect the 1/r3 terms. We will assume the orbit to be nearlycircular, which makes y small enough to be able to neglect terms of order y3. For nearly circularEinsteinian orbits we find this solution:

y =3m3

1

k2J2+

1

k

√E2 +m2

1/J2 − 1

J2+

2m41

J6−( 3m3

1

k2J2

)2

cos(kϕ) (81)

k =

√1− 6m2

1

J2(82)

Since k isn’t equal to 1 the orbit doesn’t return to the same point after a rotation of ϕ = 2π. Thismeans the ellipse precesses. From one periastron to the next need a rotation of ϕ = 2π/k instead.The angle from one periastron to the next therefore given by:

δϕ =2π

k= 2π

(1− 6m2

1

J2

)−1/2

(83)

For nearly circular orbits we have 1/r ≈ m1/J2. For weak fields we have m1/r ≪ 1. Combining

these approximations we get m21/J

2 ≪ 1. Taylor expanding this equation in this approximationwe find:

δϕ ≈ 2π(1 +

3m21

J2

)(84)

The deviation from 2π is called the periastron shift ∆ϕ:

∆ϕ ≈ 6πm21/J

2 ≈ 6πm1

a(1− e2)(85)

In the last equation we used the Newtonian equation (18). The numerical value of this quantityfor Mercury is 43” per century, which is in accord with observation.

14

4.4 Harmonic Coordinates

In chapter 6 we will study the first relativistic effect of a binary system where both masses influencethe metric. To derive these effects it is very convenient to use harmonic coordinates. Since theSchwarzschild solution is a limiting case of this system, we shall need the Schwarzschild solutionin harmonic coordinates as well. In this paragraph we will show how the harmonic coordinatesfollow from the harmonic coordinate conditions, which we introduced in the previous chapter. Wewill start with a generalization of the harmonic coordinate conditions, which will turn out to beequivalent to the expression from the last chapter in case of weak fields. The generalization isgiven by:

gµνΓλµν = 0 (86)

To show that it is always possible to pick a coordinate system in which this is true, we start withthe transformation of the Christoffel symbol:

Γ′λµν =

∂x′λ

∂xρ∂xτ

∂x′µ

∂xσ

∂x′νΓρ

τσ − ∂xρ

∂x′ν

∂xσ

∂x′µ

∂2x′λ

∂xρ∂xσ(87)

Multiplying both sides with g′µν we find:

g′µνΓ′λµν =

∂x′λ

∂xρgτσΓρ

τσ − gρσ∂2x

′λ

∂xρ∂xσ(88)

So if the left hand side of the previous equation doesn’t already vanish, we can define new coor-dinates in which it does by solving the next equation for x′:

gρσ∂2x

′λ

∂xρ∂xσ=∂x

′λ

∂xρgτσΓρ

τσ (89)

The contracted connection can also be written as:

gρσΓνρσ = − 1√

−g∂

∂xν(√ggµν) (90)

Therefore the harmonic coordinate conditions are often written in this form as well:

∂

∂xν(√ggµν) = 0 (91)

For weak fields equation (89) becomes:

(ηρσ + hρσ)∂2x′λ

∂xρ∂xσ=

1

2

∂x′λ

∂xρηµνηρα(∂νhαµ + ∂µhαν − ∂αhµν) (92)

Setting x′j = xj + ξj , the gauge transformation, this takes the form:

(ηρσ + hρσ)∂2(xλ + ξλ)

∂xρ∂xσ=

1

2

∂(xλ + ξλ)

∂xρηµνηρα(∂νhαµ + ∂µhαν − ∂αhµν) (93)

ξλ =1

2ηµνηλα(∂νhαµ + ∂µhαν − ∂αhµν) = ∂µhλµ − 1

2∂λh (94)

Multiplying both sides with ηµλ we find:

ξµ = ∂λhλµ − 1

2ηµλ∂

λh = ∂λhλµ − 1

2ηµλ∂

λh = ∂λhµλ (95)

We saw this equation before in the previous chapter (see the discussion after equation (40)). Thiswas exactly the condition that needed to be satisfied in order for the harmonic gauge condition tobe satisfied. We also saw that this equation always has a solution.

15

To satisfy the harmonic coordinate conditions for the Schwarzschild metric we will need thesecoordinates:

X1 = R(r) sin θ cosϕ X2 = R(r) sin θ sinϕ X3 = R(r) cos θ (96)

For simplicity we will write Schwarzschild metric as ds2 = B(r)dt2 + A(r)dr2 + r2dΩ2. Denotingdifferentiation with respect to r with a prime, we find from equation (89):

gµν∂2Xi

∂xµ∂xµ− ∂Xi

∂xλgµνΓλ

µν =Xi

AR

(( B′

2B+

2

r− A′

2A

)R′ +R′′ − 2A

r2R)= 0 (97)

From this we find that the coordinates are harmonic when R(r) satisfies:

d

dr

(r2√B

A

dR

dr

)− 2

√ABR = 0 (98)

Filling in A and B this equation becomes:

d

dr

(r2(1− 2m1

r

)dRdr

)− 2R = 0 (99)

The solution to the equation is:

R = r −M (100)

In terms of R the Schwarzschild line element takes this form:

ds2 = −1−m1/R

1 +m1/Rdt2 − 1 +m1/R

1−m1/RdR2 − (1 +m1/R)

2R2dΩ2 (101)

Using the following relations:

X · X = R2 dX2 = dR2 +R2dΩ2 RdR = X · dX (102)

We find:

ds2 = −1−m1/R

1 +m1/Rdt2 − (1 +m1/R)

2dX2 − 1 +m1/R

1−m1/R

m21

R4(X · dX)2 (103)

The metric in harmonic coordinates thus becomes:

g00 = −1−m1/R

1 +m1/R(104)

gij =(1 +

m1

R

)2

δij +(m1

R

)2 1 +m1/R

1−m1/R

XiXj

R2

gi0 = 0

Keeping only terms up to the second order in Φ we find:

g00 = −1 + 2m1

r− 2

(m1

r

)2

(105)

gij = δij + 2δijm1

r

g0i = 0

16

5 Gravitational Waves

Unlike the Newtonian equations of motion, the solution to the linearized Einstein equation suggeststhe existence of gravitational waves (as we saw in chapter 3). Since these waves take away energyand angular momentum from the system, they influence the two-body motion and are thus ofinterest to us. In this chapter we will first derive the metric of the gravitational waves in linearizedtheory. To do this it is most convenient to use the transverse-traceless (TT) gauge, which we willintroduce. Using the metric to find the influence of the waves on test-particles, we find that thewaves do work of these particles and thus have energy themselves. If the waves, described by somehµν , have energy, this energy will in turn influences the metric. Therefore we can’t hold on to thelinearized approximation gµν(x) = ηµν +hµν(x). Instead we have to use gµν(x) = gµν(x)+hµν(x),where gµν can be different from the minkowski metric. Finally we will obtain an expression for theenergy, which allows us to find the metric of the gravitational waves for a given energy-momentumconfiguration. This result is called the quadrupole formula, which was first derived by Einstein.In the last three paragraphs we apply this formula to the Newtonian binary system in the case ofcircular and elliptical orbits. Extracting the gravitational energy from these orbits we find thatthe particles start to spiral inward. We also show, using effective potentials, that the inspiralingcircularizes the orbits. In this chapter we have often used [9] and [13].

5.1 Gravitational Waves in Linear Theory

The linearized Einstein equations outside the energy-momentum source are:

h00 = 0 (106)

This is a wave-equation, the solution of which is:

hµν = Aµνeikσx

σ

(107)

Putting this solution in the linearized Einstein equations we get:

h00 = ∂α∂αhµν = kαkαhµν = 0 (108)

This implies kαkα = 0, which means that the wave vector is a null vector (the waves travel with the

speed of light as mentioned in chapter 3). With the solution (107) the harmonic gauge conditionbecomes:

∂µhµν = ikµAµνeikσx

σ

= 0 (109)

We thus find kµAµν = 0, which means that the wave vector is perpendicular to Aµν .

The harmonic gauge condition doesn’t fix the gauge completely. We have still some residual gaugefreedom left, since we haven’t completely specified ξµ yet. Setting ξµ in equation (40) to zero, wefind the solution ξα = Bαe

ikµxµ

. Inserting this solution together with equation (107) into equation(40), we find:

A′µν = Aµν − iBµkν − iBνkµ + iηµνB

ρkρ (110)

This equation can be used to fully specify ξµ. Multiplying with ηµν we obtain:

A′µµ = Aµ

µ + 2iBµkµ (111)

We can take A′µµ = 0 if we demand:

i

2Aµ

µ = Bµkµ (112)

17

This implies that hµµ = h = 0, which in turn gives us hµν = hµν . We secondly impose A′0i = 0.

Under these two restriction the harmonic coordinate condition for µ = 0 simplifies to:

∂0h00 = 0 (113)

Here we see that h00 is independent of time. Therefore h00 is the static part of the metric, theNewtonian potential of the source. The gravitational wave is obviously the time-dependent part,so we can set this term to zero. A′

0i = 0 therefore implies A′0ν = 0. First we investigate the case

ν = 0, which will give us the equation for B0 with which we satisfy A′00 = 0:

A00 − 2ik0B0 − ikµBµ = 0 (114)

A00 − 2ik0B0 −1

2Aµ

µ = 0

B0 =−i2k0

(A00 +

1

2Aµ

µ

)Now for ν = i, finding an expression for Bj to satisfy A′

0i = 0, we obtain:

A0i − ik0Bj − ikjB0 = 0 (115)

A0i − ik0Bj − ikj [−i2k0

(A00 + 1/2Aµµ)] = 0

Bj =i

2k20

[− 2k0A0i + kj

(A00 +

1

2Aµ

µ

)]If we plug these choices for Bµ in equation (112), using kαk

α = 0 and kµAµν = 0, we find it

immediately satisfied. Note that we originally had 10 independent components in the metric. Theharmonic gauge fixed 4 degrees of freedom, leaving only six. Choosing Bµ as we did above fixesall components of the distance vector ξµ, leaving only two degrees of freedom. Note that equation(112) does not require extra degrees of freedom, since it is satisfied by our choice of Bµ withoutextra modifications. So we have now used up all our gauge freedom. Having set all the gaugedependent components to zero, all that is left physical components (as opposed to coordinatedependent contributions). The harmonic gauge with these gauge restrictions now simplifies to∂jhij = 0. Summarizing this, we have:

h0µ = 0 hii = 0 ∂jhij = 0 (116)

From these equations we see that the temporal part of the metric is zero, leaving us with only thespatial part hij . The metric is now written in transverse traceless (TT) form. It is called tracelesssince hii = 0 and transverse since kµA

µν = 0). If we for instance pick a momentum vector pointingin the z-direction, the real of hTT

ij becomes:

hTTij =

h+ h× 0h× −h+ 00 0 0

cos[ω(t− z)] (117)

Notice that we’ve demanded that Aij is traceless and, as usual, symmetric. We’ve also used

|k| = ω. It is useful to invent an operation Lambda that can change a metric just satisfying theharmonic gauge conditions, into a metric in the transverse-traceless form.

hTTij = Λij,klhkl (118)

We do this with the following operator:

Λij,kl = PikPjl −1

2PijPkl Pij = δij − ninj (119)

18

So what frame did we actually pick when we chose the TT gauge? To understand this we haveto look at the motion of test particles. We first have to take the difference between the geodesicequation for a particle on the geodesic xµ and one on the nearby geodesic xµ + ξµ. The lattergeodesic equation takes the form:

d2(xµ + ξµ)

dτ2+ Γµ

νρ(x+ ξ)d(xν + ξν)

dτ

d(xρ + xiρ)

dτ= 0 (120)

The difference between these two equations gives us the geodesic deviation equation:

d2xiµ

dτ2+ 2Γµ

νρ(x)xν

dτ

ξρ

dτ+ ξσ∂σΓ

µνρ(x)

xν

dτ

xρ

dτ= 0 (121)

For a test-particle temporarily at rest at τ = 0 the geodesic equation becomes:

d2xi

dτ2= −Γi

00

(dx0dτ

)2

(at τ = 0) (122)

Expanding the Christoffel symbol we find:

Γi00 =

1

2(2∂0hoi − ∂ih00) (123)

In the TT gauge this immediately vanishes, since h00 and h0i are both zero in this frame. So ifat moment τ = 0 the velocity is zero (by choice), the last equation tells us that the accelerationmust also be zero. This means that if the velocity is zero at time τ = 0 it will be zero always.The coordinates of the TT frame thus seem to stretch themselves in such a way that the positionof free masses at rest do not change as the wave passes by.

This is of course not the frame used by experimentalists. A simple laboratory would by onewhich is freely falling. For a sufficiently small laboratory we have a flat metric, even in thepresence of gravitational waves. This way we can set Γλ

µν to zero. Assuming the detector movesnon-relativistically, the spatial component of the geodesic deviation equation (121) becomes:

d2ξi

dτ2+ ξi∂iΓ

i00 = 0 (124)

Since Ri0j0 = ∂jΓ

i00 in this approximation, we have:

ξi = −Ri0j0ξ

j (125)

We wrote this equation in this form, because the Riemann tensor is invariant in linearized theory.This means we can compute it in any frame we want to with the same result. Picking the TTgauge for convenience we find:

Ri0j0 = Ri0j0 = −1

2hTTij (126)

So the geodesic deviation equation in this frame becomes:

ξi =1

2hTTij ξj (127)

This equation states that the effect of gravitational waves can be described in terms of a Newtonianforce:

F i =m

2hTTij ξj (128)

Let’s consider a ring of test particles in the (x,y) plane centered around the origin and have thegravitational wave come in from the z-direction. For this wave the TT metric components with

19

either i = 3 or j = 3 are zero. Therefore we see from the geodesic deviation equation (127) that aparticle that is initially in the z = 0 plane will stay there. Since hTT

ij ∝ sinωt we can pick t = 0

at a moment when hTTij is zero. If at first we only look at the plus-polarization h+, we have:

hTTij = h+ sinωt

1 0 00 −1 00 0 0

(129)

Let’s consider a particle at the small distance from the origin ξa = (x0 + δx(t), y0 + δy(t)), wherex0 and y0 is the initial position of the particle and δx(t) and δy(t) represent the displacement dueto the gravitational wave. In that case the geodesic deviation equation (127) becomes:

δx = −h+2(x0 + δx)ω2 sinωt δy =

h+2(y0 + δy)ω2 sinωt (130)

Since δx≪ x0 we can neglect it on the right side of both equations. Integrating we find:

δx =h+2x0 sinωt δy = −h+

2y0 sinωt (131)

Similarly for h× we get:

δx =h×2y0 sinωt δy =

h×2x0 sinωt (132)

The deformation of the ring of test-particles is depicted onthe right for different moments in time. From the shape itis clear where the terminology plus and cross polarizationcame from.

5.2 The Energy of Gravitational Waves

We’ve seen that the motion of test particles by a gravitational wave can be described with aNewtonian force. If a gravitational wave momentarily passes two test particle connected by aspring, the spring will start to oscillate. Friction will cause this motion to subside after a while,creating heat. This process suggests that gravitational waves can do work and therefore they musthave energy.

The energy-momentum tensor of for instance electromagnetism is quadratic in the relevant fields(which are described by Fµν). In our weak field limit we only kept linear terms, so we might needto take the expansion a step further to find an expression for this gravitational energy.

The first order vacuum Einstein equations can be used to find hµν :

R(1)µν = 0 (133)

Here R(n)µν contain of terms from the Ricci tensor of n-th order in h. Although hµν is a solution of

the first order, it is generally not a solution of the full Einstein equations and not even the second

order, so R(2)µν = 0. The linearized expansion is therefore not sufficient for higher orders. To be

able to set the Einstein equations of second order to zero again, we need an extra perturbationterm, Rµν of order h2:

Rµν +R(1)µν +R(2)

µν = 0 (134)

To solve this problem we have to incorporate the effect of gravitational wave on the backgroundspace-time. According to general relativity energy influences the metric. So if gravitational waves

20

have energy, they must curve space-time as well. Up until now our background space-time wasassumed flat (or Minkowskian). To have a fixed background excludes the possibility that thegravitational waves can affect it. So instead we have to take:

gµν = gµν(x) + hµν (135)

Far away from any sources gµν(x) turns into the minkowskian metric. If gravitational waves arealso absent we also have hµν = 0. The question becomes how to distinguish between the partsthat belong to gµν(x) and the parts that belong to hµν(x). If we knew the waveform already, thiswould have been an easy task. However, for realistic astrophysical sources these have not beenfound. There are however some conditions in which the distinction is possible. A natural way todo this is by assuming that the typical scale of variation LB of gµν is much larger than the reducedwavelength λ of the gravitational waves:

λ≪ LB (136)

Alternatively, the background can have a maximum frequency that is much smaller than thetypical frequency range of the waves. In this case the background is slowly varying, while thewaves are high-frequency perturbations:

fB,max ≪ fGW (137)

The first term of equation (134) is only dependent on gµν . It therefore contains only low frequency

modes. R(1)µν on the other hand only contains high frequency modes, since it is dependent on terms

linear in hµν . R(2)µν can depend on both high and low frequency modes. Consider for example a

high wave-vector k1 combined with another high wave-vector k2 ≈ −k1. Recalling that hµν ∝ eikx

the quadratic terms might be in the low frequency range. The Einstein equations can thereforebe written as:

Rµν = −[R(2)µν ]

low + 8π(Tµν − 1

2gµνT

)low

(138)

R(1)µν = −[R(2)

µν ]high + 8π

(Tµν − 1

2gµνT

)high

(139)

Low and high denote respectively low and high frequencies or high or low wavelengths. We nowintroduce a length l such that λ ≪ l ≪ L. Averaging over spatial volumes l would leave Lunaffected, since they are basically constant over the volume. However, the quick perturbations λare averaged to zero. Alternatively, if we distinguish by frequency, we use a temporal average overseveral periods of the gravitational wave. This means that we can write the low mode EinsteinEquations as:

Rµν = −⟨R(2)µν ⟩+ 8π⟨Tµν − 1

2gµνT ⟩ (140)

We also write ⟨Tµν⟩ = Tµν . Since Tµν corresponds to the low frequency part, it is a smooth versionof Tµν . Since in typical situations macroscopic matter distributions are already quite smooth wehave Tµν ≈ Tµν .

At large distances from the source covariant derivatives turn into normal derivatives. R(2)µν at large

distances at large distances becomes:

R(2)µν =

1

2

[12∂µhαβ∂νh

αβ + hαβ∂ν∂µhαβ − hαβ∂ν∂βhαµ − hαβ∂µ∂βhαν+ (141)

hαβ∂α∂βhµν + ∂βhαν∂βhαµ − ∂βhαν∂αhβµ − ∂βhαβ∂νhαµ + ∂βh

αβ∂αhµν−

∂βhαβ∂µhαν − 1

2∂αh∂αhµν +

1

2∂αh∂νhαµ +

1

2∂αh∂µhαν

]21

Picking the TT gauge, which includes the harmonic gauge condition ∂µhµν = 0, we can drastically

simplify the average of R(2)µν . To do this we need integration by parts. Note that generally

integration by parts of ∂t is possible only if we have performed an integral over time as well.Similarly, ∂i requires a spatial integral. Since hµν is a function of the combination t − z (seeequation (117)), the spatial integral is equal to minus the temporal integral, so integration byparts can be safely applied. Because we are averaging over all the directions at each event, theaverage of any derivative term vanishes, ⟨∂µX⟩ = 0, we therefore have ⟨A(∂µB)⟩ = ⟨(∂µA)B⟩. We

immediately see that all terms of ⟨R(2)µν ⟩ disappear, accept for the first two which are related by

integration by parts. We find:

⟨R(2)µν ⟩ = −1

4⟨∂µhαβ∂nuhαβ⟩ (142)

Without a matter contribution the low-mode Einstein equations become:

Rµν =1

4⟨∂µhαβ∂nuhαβ⟩ (143)

This expresses the fact that the derivatives of the perturbation affects the curvature of the back-ground metric gµν .

We also define:

tµν = − 1

8π⟨R(2)

µν − 1

2gµνR

(2)⟩ (144)

The trace of tµν is:

t = gµνtµν =1

8π⟨R(2)⟩ (145)

Here we used that gµν⟨R(2)µν ⟩ = gµν⟨R(2)

µν ⟩ = ⟨gµνR(2)µν ⟩ since gµν by definition is a purely low

frequency quantity. Inserting the trace into the equation for tµν we find:

−⟨R(2)µν ⟩ = 8π(tµν − 1

2gµνt) (146)

The low frequency Einstein Equations become:

Rµν = 8π(tµν − 1

2gµνt) + 8π(Tµν − 1

2gµν T ) (147)

Or equivalently:

Rµν − 1

2gµνR = 8π(Tµν + tµν) (148)

This looks like the regular Einstein equations, but now we have an extra term tµν which is notdependent on matter, but only on the gravitational field and is quadratic in hµν . Using the Bianchiidentity we have ∂µ(Rµν − 1

2 gµνR) = 0. This implies:

∂µ(Tµν + tµν) = 0 (149)

The energy of the matter and the gravitational waves are together conserved. This reflects thefact that energy can be exchanged between the two. With Tµν = 0 the last equation simplifies to:

∂µtµν = 0 (150)

The average in the definition of tµν is necessary, since without it the expression would not begauge invariant (and therefore contain some frame dependent terms). Because of the ability to

22

choose local minkowskian coordinates in any space-time, it is impossible to define a local measureof gravitational energy-momentum, we could have expected that some averaging procedure wasnecessary. Averaging over several wavelengths or periods we can pick up the physical curvatureto obtain a gauge invariant measure.

Filling in equation (144) and using ⟨R(2)⟩ = 0, we get:

tµν =1

32π⟨∂µhαβ∂νhαβ⟩ (151)

In particular the energy-density becomes:

t00 =1

32π⟨hTT

ij hTTij ⟩ = 1

16π⟨h2+ + h2×⟩ (152)

The gravitational energy inside a volume is:

EV =

∫V

d3xt00 =

∫V

d3x1

32π⟨hTT

ij hTTij ⟩ (153)

Far from material sources we can use ∂0t00 = −∂it0i to find an expression for the energy flux:

EV =

∫V

d3x∂0t00 = −

∫V

d3x∂iti0 = −

∫S

dAnit0i (154)

In the last step we’ve used the Gauss’s theorem. If S is a spherical surface, we have dA = r2dΩand n = r:

EV = −r2∫S

dΩt0r (155)

As we saw at the end of chapter 3, at large distances we have:

hTTij =

1

rfij(t− r) (156)

From this we find:

∂rhTTij = − 1

r2fij +

1

r∂rfij = − 1

r2fij −

1

r∂0f

′ij ≈ −∂0hTT

ij = ∂0hTTij (157)

So at large distances we have t0r = t00. We find:

EV =r2

32π

∫dΩ⟨ ˙hTT

ij˙hTTij ⟩ (158)

The same equation for tµν can be obtained from Noether’s theorem. We shall apply Noether’stheorem in exactly the way we would electromagnetic theory. In that case we usually use:

LEM = −1

4FµνF

µν (159)

TµνEM = − ∂LEM

∂(∂µAρ)∂νAρ + ηµνLEM (160)

We find:

TµνEM = FµρF ν

ρ − 1

4ηµνF

2 + ∂ρ(FµνAν) (161)

23

The familiar expression doesn’t contain the last term, so we would like to get rid of it. If we take anaverage over a volume large enough that all boundary terms vanish, this term indeed disappears.So we need the average to get the correct answer:

⟨TµνEM ⟩ = ⟨FµρF ν

ρ − 1

4ηµνF

2⟩ (162)

In case of electromagnetism this final averaged expression also happens to be locally true, but thisfact cannot be from Noether’s theorem and requires experimental verification.

Similarly, straight from the full Hilbert-Einstein action to second order we find:

LE = − 1

64π[∂µhαβ∂

µhαβ − ∂µh∂µh+ 2∂µhµν∂νh− 2∂µhµν∂ρh

ρν ] (163)

tµν = ⟨− ∂L∂(∂µhαβ)

∂νhαβ − ηµνL⟩ (164)

We obtain:

tµν =1

32π⟨∂µhαβ∂νhαβ⟩ (165)

Although the electromagnetic expression is also true locally, we have seen that for gravitationalwaves this is the best we can do.

5.3 The Quadrupole Moment

We have seen that the solution to the linearized Einstein equations far away from the mattersources is:

hµν(t, x) =4

r

∫d3x′Tµν(t− r + x′ · n, x′) (166)

Using hTTij = Λij,klhkl = Λij,klhkl (since Λii,kl = Λij,kk = 0), we can write this in the TT gauge:

hTTij (t, x) =

4

rΛij,kl

∫d3x′Tkl(t− r + x′ · n, x′) (167)

Notice that we changed Tµν to Tkl. This is possible since the T00 and T0k are related to each otherby ∂0T

00 = −∂iT i0 and ∂0T0k = −∂iT ik. Applying a Fourier transformation we find:

Tkl(t− r + x′ · n, x′) =∫

d4k

(2π)4Tkl(ω, k)e

iω(t−r+x′·n)+ik·x′(168)

The frequency of the motion inside the source of size d is ωs ∼ v/d. As we shall see later in thischapter, the frequency of the source is of the same order as the frequency of the gravitationalradiation, we thus have ωGW ∼ v/d. For non-relativistic sources this becomes ωsd ≪ 1. Theintegral in the last equation is restricted to the small matter region |x| ≤ d. Therefore we have:

ωx′ · n . ωsd≪ 1 (169)

To see this it is useful to look at the figure at the end of chapter 3. With this we can simplify theexponent in the last integral:

e−iω(t−r+x′·n) = e−iω(t−r)[1− iωx′ini +

1

2(−iω)2x′ix′jninj + ...

](170)

Putting this back in the last integral and inversing the Fourier transformation we get:

hTTij (t, x) =

4

rΛij,kl

∫d3x′(Tkl(t− r, x′) + x′ini∂0Tkl +

1

2x′ix′jninj∂20Tkl+...) (171)

24

We now define:

Sij =

∫d3xT ij Sij,k =

∫d3xT ijxk Sij,kl =

∫d3xT ijxkxl (172)

We find:

hTTµν (t, x) =

1

r4Λij,kl

[Skl + nmS

kl,m +1

2nmnpS

kl,mp + ...]ret

(173)

Where the subscript ’ret’ means that the expression is evaluated at the retarded time t− r. Theformula becomes more physically meaningful if we introduce:

I =

∫d3xT 00 Ii =

∫d3xT 00xi Iij =

∫d3xT 00xixj (174)

P i =

∫d3xT 0i P i,j =

∫d3xT 0ixj P i,jk =

∫d3xT 0ixjxk (175)

Imagine a box with a volume V larger than the source. In that case the energy momentum tensorat the boundary is zero. Using ∂0T

00 = −∂iT 0i we find:

I =

∫V

d3x∂0T00 = −

∫V

d3x∂iT0i = −

∫S

dSiT 0i = 0 (176)

This implies the conservation of energy T 00. At first sight this might seem surprising, sincethe system loses energy by gravitational radiation. The conservation does hold however in thelinearized approximation, where the effect of radiation can still be neglected on the equations ofmotion. A similar calculation for Ii gives:

Ii =

∫V

d3xxi∂0T00 = −

∫V

d3xxi∂iT0j =

∫V

d3x(∂jxi)T 0j =

∫V

d3xδijT0j = P i (177)

Continuing we find:

Iij = P i,j + P j,i Iijk = P i,jk + P j,ki + P k,ij (178)

P i = 0 P i,j = Sij P i,jk = Sij,k + Sik,j (179)

Notice that P i implies conversation of momentum. Finally we derive:

Sij =1

2Iij (180)

...Iijk

= 2(Sij,k + Sik,j + Sjk,i) (181)

Sij,k =1

6

...Iijk

+1

3(P i,jk + P j,ik − P k,ij) (182)

In this notation the leading term in the metric becomes:

hTTµν =

1

r2Λij,klI

kl(t− r) (183)

Notice that the leading term is already a quadrupole term. The monopole term ∝ I and the dipoleterm ∝ Ii = P i are zero because of the conservation of mass and momentum.

When n = z we have:

Pij = δij − ninj =

1 0 00 1 00 0 0

(184)

25

After some algebra, using equation (119), we find:

Λij,klIkl =

(I11 − I22)/2 I12 0

I21 −(I11 − I22)/2 00 0 0

(185)

Comparing to (129) we immediately see:

h+ =1

r(I11 − I22) h× =

2

rI12 (186)

For radiation in the direction of the general angle n can find:

h+ =1

r[I11(cos

2 ϕ− sin2 ϕ cos2 θ) + I22(sin2 ϕ− cos2 ϕ cos2 θ)− (187)

I33 sin2 θ − I12 sin 2ϕ(1 + cos2 θ) + I13 sinϕ sin 2θ + I23 cosϕ sin 2θ]

h+ =1

r[(I11 − I22) sin 2ϕ cos θ + 2I12 cos 2ϕ cos θ − 2I13 cosϕ sin θ + 2I23 sinϕ sin θ] (188)

So if we have Iij we can calculate the angular distribution of the radiation far from a source.

As for any symmetric tensor with two indices, we can decompose Ikl into irreducible representa-tions:

Ikl = (Ikl − 1

3δklIii) +

1

3δklIii (189)

Here Iii is the trace of Iij . The term between the parenthesis is traceless by construction. Whenthe Lambda tensor is contracted with δkl we get zero, so the last term doesn’t contribute. So wefind:

hTTµν =

1

r2Λij,kl(Iij −

1

3δij Ikk) =

1

r2Λij,klQij (190)

The power radiated per unit solid angle, using the property Λij,klΛkl,mn = Λij,mn and equation(158), becomes:

dPrad

dΩ=

r2

32π⟨hTT

ij hTTij ⟩ = 1

8πΛij,kl⟨

...Qij

...Qkl⟩ (191)

It can be shown that:∫dΩΛij,kl =

2π

15(11δikδjl − 4δijδkl + δilδjk) (192)

We therefore find the following expression for the total power, which is Einstein’s famous quadrupoleformula:

Prad =1

5⟨...I ij

...I ij −

1

3(...I kk)

2⟩ (193)

For comparison, the angular momentum carried away per unit time is:

Jrad = −2

5ϵikl⟨

...QkaQla⟩ (194)

26

5.4 The Order of Magnitude of Gravitational Radiation

For Newtonian circular orbits we have v2 = m1/r = Φ (found by equation the centripetal forceand the gravitational force). The virial theorem of classical mechanics states that this equation isapproximately true for all bound orbits. For future reference, from a = m1/r

2 = v2/r we find:

−2Ekin = −m2v2 = −m1m2

r= Epot Etot = Ekin + Epot = −Ekin (195)

The energy density can be expanded as T 00 = O(m/r3) + O(mv2/r3). From the definition ofQij (which is dependent on T 00) therefore find that Qij = O(mr2) + O(mv2r2). Since the timederivatives are of order v/r we find

...Qij = O(mv3/r)+O(mv5/r). From here we can calculate the

order of magnitude of the radiated power:

Prad = ⟨...Qij

...Qij⟩ = 1/T

∫...Qij

...Qijdt = O(m2v6/r2) +O(m2v8/r2) (196)

Using the virial theorem, the leading order term of the radiated power is thus of orderO(m2v6/r2) =O(v10). Using Etot ≈ −Ekin = −1/2mv2 and setting −dEtot/dt = Prad, we find:

−mvv ≈ −m2v6

r2a ≈ mv5

r2= O(v7/r) = O(Φ3.5/r) (197)

So while the Newtonian acceleration is of order Φ/r, the gravitational radiation effects the accel-eration only at the order Φ3.5/r. Counting the Newtonian acceleration as the 0PN contribution,this means the first back reaction affects due to gravitational radiation will appear at 2.5PN.

If we would keep all terms in equation (173), we find this expression for the radiated power:

Prad = O(m2v6/r2) +O(m2v8/r2) +O(m2v10/r2) + ... (198)

The second term in the radiated power gets a contribution from the next-to-leading order termin equation (173), but it will also get a the first relativistic contribution of the first term (sincethis is dependent on T 00 = O(mv0/r3) + O(mv2/r3) + ...). This means that the moment we gobeyond the leading order of equation (173), we go into the realm of relativity. From a Newtonianequations of motion (0PN) we thus find a radiation reaction of order 2.5PN. Similarly, the 1PNcontribution to the equations of motion will only give a radiation reaction effect at 3.5PN.

The post-Newtonian expansion can also be viewed as an expansion of E. Since E = PT = 2πrP/v,the radiated energy per unit mass will be of order:

Erad = O(mv5/r) +O(mv7/r) + ... (199)

The leading radiation term is thus also of order 2.5PN.

5.5 Gravitational Waves from a Binary in Circular Orbit

We will now calculate the energy of gravitational waves from a Newtonian circular orbit. For themoment we will neglect the influence of the gravitational radiation on the orbits. In the centerof mass frame, as usual, the two body problem turns into a one body problem with mass µ andcoordinate x′ = (x′, y′, z′). The circular relative orbit is:

x′ = R cos(ωst+ π/2) y′ = R sin(ωst+ π/2) z′ = 0 (200)

The phase π/2 which just replaces our origin of time will turn out to be a useful choice. In thisframe we have ρ = µδ(x − x′) and therefore we find Iij = µx′ix′j . Explicitly, the non-vanishingcomponents are:

I11 = µR2 1− cos 2ωst

2I22 = µR2 1 + cos 2ωst

2I12 = −1

2µR2 sin 2ωst (201)

27

The second derivatives become:

I11 = 2µR2ω2s cos 2ωst I22 = −I11 I12 = 2µR2ω2

s sin 2ωst (202)

Carefully plugging these equations in equation (187), we find:

h+ =4µω2

sR2

r

(1 + cos2 ϕ

2

)cos(2ωstret + 2ϕ) (203)

h× =4µω2

sR2

rcosϕ sin(2ωstret + 2ϕ) (204)

We also have a ϕ dependence in our sine and cosine. Since the orbit is circular, a rotation of thesource by an angle ∆ϕ around the z-axis is equivalent to a time-translation ∆t with ωs∆t = ∆ϕ.So by shifting the origin of time we can write:

h+ =4µω2

sR2

r

(1 + cos2 ϕ

2

)cos 2ωstret h× =

4µω2sR

2

rcosϕ sin 2ωstret (205)

The power is:

dPrad

dΩ=

r2

16π⟨h2+ + h2×⟩ (206)

Using ⟨cos2 2ωst⟩ = ⟨sin2 2ωst⟩ = 1/2, we find:

dPrad

dΩ=

2µ2R4ω6s

πg(θ) g(θ) =

(1 + cos2 θ

2

)2

+ cos2 θ (207)

Integrating over the solid angle we get:

Prad =32µ2R4ω6

s

5=µ2R4ω6

10(208)

The energy radiated in one period T = 2π/ωs, using v = ωsR, is:

⟨Erad⟩T =64µ2

5Rv5 (209)

5.6 The Inspiraling of a Binary System

For circular orbits we had v2 =M/r, which we can use to find:

ω2s =

v2

r2=m2

r3(210)

This equation is called Kepler’s law. We can use it to eliminate R in favor of ωs in the metric forthe circular orbit (203). We also define the chirp mass:

Mc = µ3/5m2/5 =(m1m2)

3/5

(m1 +m2)1/5(211)

The metric becomes:

h+ =4

rM5/3

c (πfgw)2/3 1 + cos2 ϕ

2cos(2πfgwtret + 2ϕ) (212)

h× =4

rM5/3

c (πfgw)2/3 cosϕ sin(2πfgwtret + 2ϕ) (213)

28

Here we have used ωgw = 2ωs. The radiation per solid angle becomes:

dPrad

dΩ=

2

π

(Mcωgw

2

)10/3

g(θ) (214)

The angular average of g(θ) is:∫dΩ

4πg(θ) =

4

5(215)

Therefore the total power becomes:

Prad =32

5

(Mcωgw

2

)10/3

(216)

The source of the radiated energy is the orbital energy of the system. Using the virial theorem wefind:

Eorbit = −m1m2

2R(217)

From this equation we see that as energy is lost due to gravitational radiation, the radius R mustdecrease in time (remember that the energy is negative for bound orbits). From Kepler’s law(210) we know that as R decreases, ωs increases. From equation (216) we see that the power alsoincreases. Writing Kepler’s law as R3 = m2/ω

2 and taking the time derivative we find:

R = −2

3Rωs

ωs= −2

3(ωsR)

ωs

ω2s

(218)

We see here that |R| is much smaller then the tangential velocity ωsR when ωs ≪ ω2s . This is

called the regime of quasi-circular motion, because the radius is changing very slowly. Again usingKepler’s law to eliminate R in favor of ωs, we find for the energy of the orbit:

Eorbit = −(M5

c ω2gw

32

)1/3

(219)

Equating the decrease of orbital energy per unit time with the radiated power,Prad = −dEorbit/dt,we find:

ωgw =12

521/3M5/2

c ω11/3gw (220)

fgw =96

5π8/3M5/2

c f11/3gw (221)

Integrating this equation we see that fgw formally diverges at a finite moment of time. We call thismoment in time tcoal. Also defining τ ≡ tcoal − t, the time to coalescence, the solution becomes:

fgw =1

π

( 5

256

1

τ

)3/8

M−5/8c (222)

Using Kepler’s law (210) again, noting that d/dτ = −d/dt (where τ is not the proper time, butthe time to coalescence), we find:

R

R= −2

3

ωgw

ωgw= − 1

4τ(223)

29

Integrating we find:

R = R0

( ττ0

)1/4

(224)

This relation is plotted in the graph on theright.

Note that there is a long phase where R decreases smoothly, but then there is a plunge. In thefirst part we can use the quasi-circular approximation, but in the plunge phase we can’t.

5.7 Gravitational Waves from a Binary in Elliptical Orbit

Now we will study the elliptical orbits. Using equation (22) and (21), they are described by:

x(t) = r(t) cos θ(t) = a[cosu(t)− e] y(t) = r(t) sin θ(t) = b[cosu(t)] (225)

In this case Iab becomes:

Iab =

∫d3xT00xixj = µ

(x2 xyyx y2

)= µ

(cos2 θ sin θ cos θ

sin θ cos θ sin2 θ

)(226)

To find the third derivatives it is useful to eliminate r are in favor of θ, using equation (17). Forinstance we find:

M11 = µr2 cos2 θ = µa2(1− e2)2cos2 θ

1(1 + e cos θ)2(227)

Using θ = J/r2 and inserting the equations for e and a (equations (17) and (16)), we find:

θ =

√m

a3(1− e2)−3/2(1 + e cos θ)2 (228)

This can be used to find the third derivatives of Iij :

...I 11 = β(1 + e cos θ)2[2 sin 2θ + 3e sin θ cos2 θ] (229)

...I 22 = β(1 + e cos θ)2[−2 sin 2θ − e sin θ(1 + 3 cos2 θ)]...I 12 = β(1 + e cos θ)2[−2 cos 2θ + e cos θ(1− 3 cos2 θ)]

β2 ≡ 4µ2m3

a5(1− e2)5

The power radiated in the quadrupole approximation becomes:

Prad =1

5[...I211 +

...I222 + 2

...I212 −

1

3(...I211 +

...I222)

2] (230)

=2

15[...I211 +

...I222 + 3

...I212 −

...I211

...I222]

=8

15

µ2m3

a5(1− e2)5(1 + e cos θ)4[12(1 + e cos θ)2 + e2 sin2 θ]

30

Averaging over one period we find:

⟨Prad⟩T =1

T

∫ T

0

dtPrad =ω0

2π

∫ 2π

0

dϕPrad

ϕ(231)

= (1− e2)3/2∫ 2π

0

dϕ

2π

1

(1 + e cosϕ)2Prad

=8µ2m3

15a51

(1− e2)7/2

∫ 2π

0

dϕ

2π[12(1 + e cosϕ)4 + e2(1 + e cosϕ)2 sin2 ϕ]

Performing the integration we find the total radiated power, a result first obtained in 1963 byPeters and Mathews:

⟨Prad⟩T =32µ2m3

5a5f(e) f(e) =

1

(1− e2)7/2

(1 +

73

24e2 +

37

96e4)

(232)

Writing this in terms of a instead of ω0 we find:

⟨Prad⟩T =32

5µ2a4ω6

0f(e) (233)

In case of circular orbits we have e = 0 and f(e) = 1. In case of the famous Hulse-Taylor pulsar,which we’ll discuss in a later chapter, we find e = 0.617 and f(e) = 11.8. So compared to thecircular orbit this is a whole magnitude larger.

Using Kepler’s law (25) and equation (18), we find T = const.× (−E)−3/2 and therefore we have:

T

T= −3

2

E

E(234)

Setting E = −P we find:

T

T= −96

5

µm2

a4f(e) (235)

Writing a in terms of T (using T = 2π/ω0 and Kepler’s law) we finally obtain:

T

T= −96

5µm2/3

( T2π

)−8/3

(236)

This formula is of great importance, since it was used to find the first experimental evidence forthe existence of gravitational waves. We will discus this in a later chapter.

31

Since P = −E we have:

E = −32

5

µ2m3

a51

(1− e2)7/2

(1+

73

24e2 +

37

96e4)

(237)

Similarly, from equation (194) we can find:

J = −32

5

µ2m5/2

a7/21

(1− e2)2

(1 +

7

8e2)

(238)

Using dE/dt = dE/da da/dt and equation (18), we get:

a = −64

5

µm2

a31

(1− e2)7/2

(1 +

73

24e2 +

37

96e4)

(239)

The solution for circular orbits (e = 0), as found by Pe-ters [10], is:

a(t) =[a40 −

256M2µ

5Mt]1/4

(240)

Although in principle we have enough information to findto solution for any e, this integral is hard to evaluate. Us-ing L2 = aM and the formula for the effective potential(12) we find:

V = −Mr

+M [a40 − 256M2µ/(5M)t]1/4

2r2(241)

In the graphs to the right we see the time evolution ofthe effective potential for a0 = 10, M = 1 and µ = 1/4.Notice the long phase where the potential changes slowly,followed by the quick plunge phase.

We can similarly find:

e = −304

15

µm2

a41

(1− e2)5/2

(1 +

121

304e2)

(242)

Notice that if e = 0 we also have e = 0, meaning that a circular orbit will stay circular. Noticealso that for e > 0 we have e < 0, meaning that an elliptical orbit will become more and morecircular in time. So the back reaction of the radiation circularizes the orbit. In terms of effectivepotentials this would mean that the change in the spatial difference between the turning pointsdecreases. At the turning points we have v = 0 and therefore the radial equation (11) becomes:

r =−M ±

√2EJ2 +M2

2E(243)

The difference between these two solutions is called Q:

Q =

√2EJ2 +M2

E(244)

32

The derivative of Q with respect to time is:

Q = −√

2EJ2 +M2

˙E

E2+

J2 ˙E + 2EJ ˙J

E√2EJ2 +M2

(245)

From the equations (237) and (238) we find, using X =M2/J2:

E = JM2

J3f(X) f(X) =

8(148(E/X)2 + 730E/X + 425)

97(14E/X + 15)(246)

The circularization of the orbits implies that Q ≤ 0. Using the last equation we can divide J outof equation (245). Simplifying we find:

Q = −E − M2

J2+

2J2E2

M2f(X)(247)

The energy for bound orbits is between Vmin ≤ E ≤0. We can find Vmin by solving dV/dr = 0. Weeasily obtain the solution r = J2/M . Putting thisr back into the effective potential we find Vmin =−1/2X. The energy for bound orbits is thus be-tween −1/2X ≤ E ≤ 0. With a plot we can verifythat for any X we indeed find that Q ≤ 0 within thisrange. The instant where X = 10 is depicted on theright. Notice that at Q = 0 the energy is exactlyequal to Vmin = −1/2X. This expresses the factthat Q cannot decrease any more when the energyhas reached the minimum of the potential.

In terms of effective potentials we can make an even stronger statement. We can show that intime the energy we always get closer to the bottom of the potential well. Although this statementimplies the circularization of the orbit, the circularization does not imply this statement. This isbecause the decrease of the distance between the turning points has no direct relation to the depthof the potential. To prove this we need to show that the energy drops faster than the minimumof the potential does, so the energy catches up to it. That is, we have to show E ≥ Vmin.

We’ve just seen that Vmin = −1/2X. The derivative with respect to time gives Vmin = −M2J/J3.Writing out E ≥ Vmin and plotting this result for different X indeed proves this statement.

33

Although it is difficult to find e(t) or a(t), we can finda(e) by combining the expression for da/dt and de/dt:

da

de=

12

19a

1 + 73/24 e2 + 37/96 e4

e(1− e2)(1 + 121/304)e2)(248)

The solution to this equation is:

a(e) = a0g(e)

g(e0)(249)

g(e) =e12/19

1− e2

(1 +

121

304e2)870/2299

(250)

We can then use E = −M/(2a) to find the associ-ated energy. To find the angular momentum we useJ2 =M2(e2−1)/(2E). Inserting the angular momentumin the effective potential, we have found the effectivepotential in terms e, e0 and a0. In the next figurewe have plotted the effective potential for different e.We have also plotted the associated energy-level (inblue). The effective potential nicely illustrated thecircularization of the orbit.

34

6 The First Order Post-Newtonian Approximation

In this chapter we will study the binary system where both masses are allowed to influence themetric. We do assume that the velocities are low (v << 1) and the field are weak (Φ << 1),yet we allow for one more order of Φ in the equations of motion than we do in the Newtonianapproximation. This gives us the leading relativistic corrections, called the first Post-Newtoniancorrections (or 1PN). We start this chapter by deriving the 1PN metric from the Einstein equationsand we will use this metric to find the 1PN lagrangian, which was first derived by Lorentz andDroste as early as 1916 [8]. We’ll use this Lagrangian to find the equations of motion, whichwere first found by Einstein, Infeld and Hoffmann in 1938 [5]. Just like in the Newtonian casewe will also use the lagrangian to derive the radial equation and the orbital equations, whichwere derived in quasi-Newtonian form by Damour and Deruelle in 1985 [2]. The resulting orbitsturn out to be slightly different from the precessing ellipses in the Schwarzschild case (they arenow conchoids of precessing ellipses). We will also show that 1PN reduces to the first orderSchwarzschild solution when one of the masses is set to zero. We will then derive the effectivepotential for 1PN. Comparing the effective potentials for 1PN and 2PN with the first and secondorder Schwarzschild solutions we will find an measure for the accuracy of the post-Newtonianexpansion. Finally we will see how the 1PN results were used to obtain the first evidence forthe existence of gravitational radiation from the observation of the Hulse-Taylor pulsar. In thischapter we have often used [2], [3], [9] and [18].

6.1 Deriving the First Post-Newtonian Metric

As mentioned above we assume the fields to be weak, Φ << 1. Since we are studying bound orbitswe know from the virial theorem that v2 ≈ M/r, which is exactly true for circular orbits. Thismeans an expansion in terms of the small parameter Φ is equivalent to an expansion in terms ofthe small parameter v2. For circular orbits we also have J2/r2 = m/r (see equation (17)) and fromthis we can also find J2/r2 = m2/J2. We thus expand in terms of m/r ≈ v2 ≈ J2/r2 ≈ m2/J2.

We will assume that our system fits into a finite region with radius d. We have seen that v/d ≈ωs ∼ ω and therefore λ ∼ d/v. Since we expand in terms of the instantaneous Newtonian potentialsΦ we are neglecting the retardation effect. The metric we will find will thus only be accurate closeto the source, where the retardation effects are negligibly small. This is called the near zone,which is often defined by r ≪ λ.

To know the metric up to the first order we have to solve the Einstein equations. We begin byexpanding the energy-momentum tensor:

T 00 = T 00(0) + T 00(2) + ... (251)

T i0 = T i0(1) + T i0(3) + ...

T ij = T ij(2) + T ij(4) + ...

Here Tµν(N) denotes the term Tµν of order (m/r3)vN . The first term of T 00 is the density ofthe rest-mass (of order m/r3). The second term the non-relativistic part of the energy density (oforder mv2/r3) and so on. Similarly the first term in T i0 gives us the momentum density (of ordermv/r3) and the first term of T ij gives us the flux (of order mv2/r3).

The metric we choose must give us consistent Einstein field equations. In other words, the left andthe right side of the equation must be of the same order. So what will the metric look like? In thelast chapter we saw that gravitational radiation only starts to influence the equation of motion at2.5PN (which is of order Φ3.5). In case of 1PN (which is of order Φ2) we can thus safely neglectit. Without back reaction our system is invariant under time-reversal. Since the velocity changessign under time-reversal, we know g00 and gij can only contain even powers of v, and g0i only odd

35

powers (the dt2 term and the dxidxj term remain invariant under time-reversal, while the crossterms dxidt changes sign). The metric becomes:

g00 = −1 + g(2)00 + g

(4)00 + ... (252)

gij = δij + g(2)ij + g

(4)ij + ...

gi0 = g(3)i0 + ...

The component gi0 does not contain a contribution of order v. This was found a posteriori, sincewithout this the Einstein tensor components where not of same order as the energy-momentumtensor components in the Einstein equations.

As the inverse of the metric we expect:

g00 = −1 + g(2)00 + g(4)00 + ... (253)

gij = δij + g(2)ij + g(4)ij + ...

gi0 = g(3)i0 + ...

Since the inverse components are defined by:

giµg0µ = gi0g00 + gijg0j = 0 (254)

g0µg0µ = g00g00 + g0ig0i = 1

giµgjµ = gi0gj0 + gikgjk = δij

we find:

g(2)00 = −g(2)00 g(2)ij = −g(2)ij g(2)i0 = g(2)i0 g(4)00 = −g(4)00 − (g

(2)00 )

2 (255)

To continue we need to rewrite the geodesic equation as:

d2xµ

dτ2= −Γµ

νλ

dxν

dτ

dxλ

dτ(256)

d2xi

dt2=

( dtdτ

)−1 d

dτ

[( dtdτ

)−1 dxi

dt

]=

( dtdτ

)−2 d2xµ

dτ2−( dtdτ

)−3 d2t

dτ2dxi

dτ

= −Γiνλ

dxν

dt

dxλ

dt+ Γ0

νλ

dxν

dt

dxλ

dt

dxi

dt

This may be written in more detail as:

d2xµ

dτ2= −Γi

00 − 2Γi0j

dxj

dt− Γi

jk

dxj

dt

dxk

dt+

[Γ0

00 + 2Γ00j

dxj

dt+ Γ0

jk

dxj

dt

dxk

dt

]dxidt

(257)

To find the acceleration to order Φ2/r = v4/r, we need Γi00 to order v4/r, Γi

0j to order v3/r,

Γijk to order v2/r, Γ0

00 to order v3/r,Γ00j to order v2/r and Γ0

jk to order v/r.

Since ∂i is of order 1/r and ∂t of order v/r we can insert the metric and the inverse metriccomponents in the explicit expression for the Christoffel symbols:

Γµνλ = Γ

µ(2)νλ + Γ

µ(4)νλ (for Γi

00, Γijk and Γ0

0i) (258)

Γµνλ = Γ

µ(3)νλ + Γ

µ(5)νλ (for Γi

0j , Γ000 and Γ0

ij)

36

In this case Γµ(N)νλ denotes the term in Γµ

νλ of order vN/r. The extra 1/r in the order results fromthe derivative of the metric in the Christoffel symbols. Writing out the terms explicitly we find:

Γi(2)00 = −1

2∂ig

(2)00 (259)

Γi(4)00 = −1

2∂ig

(4)00 + ∂tg

(4)i0 +

1

2g(2)ij ∂ig

(2)00

Γi(3)0j =

1

2

[∂jg

(3)i0 + ∂tg

(2)ij − ∂ig

(3)j0

]Γi(2)jk =

1

2

[∂kg

(2)ij + ∂jg

(2)ik − ∂ig

(2)jk

]Γ0(3)00 = −1

2∂tg

(2)00

Γ0(2)0i = −1

2∂ig

(2)00

Γ0(3)ij = −1

2[∂jg

(3)0i + ∂ig

(3)0j − ∂tg

(2)ij ]

Notice that the last term is of higher order than required. We mention it here anyway, since wewill need it later on when we discuss the harmonic gauge conditions. From these equations we seethat we have to know g00 to order v4, but gij only to order v2 and gi0 to order v3.

We now continue writing the expansion of the Ricci tensor in terms of the Christoffel symbols:

Rµκ = Rλµλκ = ∂κΓ

µµλ − ∂λΓ

λµλ + Γη

µλΓλκη − Γη

µκΓληλ (260)

We find:

R00 = R(2)00 +R

(4)00 + ... (261)

Ri0 = R(3)i0 +R

(5)i0 + ...

Rij = R(2)ij +R

(4)ij + ...

In this caseR(N)µν denotes the term inRµν of order vN/r2. The extra 1/r results from the derivatives

of the Christoffel symbol and the multiplication of two of them. Explicitly we find:

R(2)00 = −∂iΓi(2)

00 (262)

R(4)00 = ∂tΓ

i(3)0i − ∂iΓ

i(4)00 + Γ

0(2)0i Γ

i(2)00 − Γ

i(2)00Γ

j(2)ij

R(3)i0 = ∂tΓ

j(2)ij − ∂jΓ

j(3)0i

R(2)ij = ∂jΓ

0(2)i0 + ∂jΓ

k(2)ik − ∂kΓ

k(2)ij

In terms of the metric these expressions become rather long. But we can simplify this choosing asuitable coordinate system. As mentioned before we take the harmonic coordinate condition:

gµνΓλµν = 0 (263)

The contributing third order terms of gµνΓ0µν = 0 are g00Γ

(3)000 = −Γ

(3)000 and gijΓ

(3)0ij = δijΓ

(3)0ij

(the term with g0i does not contribute, since it’s first term is already of third order and the lowestnon-zero christoffel symbol is of second order). We find:

0 =1

2∂tg

(2)00 − ∂ig

(3)0i +

1

2∂tg

(2)ii (264)

37

The contributing second order terms of gµνΓiµν = 0 give:

0 =1

2∂ig

(2)00 + ∂jg

(2)ij − 1

2∂ig

(2)jj (265)

The next three formulas we find by respectively taking the time derivative of equation (264),taking the spatial derivative of (264) followed by inserting the time derivative of equation (265)and by taking two times the spatial derivative of (265):

1

2∂2t g

(2)ii − ∂i∂tg

(3)i0 +

1

2∂2t g

(2)00 = 0

∂t∂jg(2)ii − ∂i∂jg

(3)i0 − ∂i∂tg

(2)ij = 0

∂k∂jg(2)ij + ∂j∂ig

(2)kj − ∂i∂kg

(2)jj + ∂i∂kg

(2)00 = 0

The components of the Ricci tensor now reduces to:

R(2)00 =

1

2∇2g

(2)00 (266)

R(3)0i =

1

2∇2g

(3)i0

R(2)ij =

1

2∇2g

(2)ij

R(4)00 =

1

2∇2g

(4)00 − 1

2∂2t g

(2)00 − 1

2g(2)ij ∂i∂jg

(2)00 +

1

2(∇2g

(2)00 )

2

Now we can insert this expression in the Einstein Field Equations together with the expansion ofthe energy-momentum tensor. It is convenient to set (Tµν − 1

2gµνTλλ ) = Sµν . From the expansion

of the energy-momentum tensor we find:

S00 = S(0)00 + S

(2)00 + ... (267)

Si0 = S(1)i0 + S

(3)i0 + ...

Sij = S(0)ij + S

(2)ij + ...

In this case S(N)µν denotes the term in Sµν of order Mr3/vN , still the same order as the one used

for the energy-momentum tensor. Explicitly the components become:

S(0)00 =

1

2T 00(0) (268)

S(2)00 =

1

2

[T 00(2) − 2g

(2)00 T

00(0) + T ii(2)]

S(1)i0 = −T 0i(1)

S(0)ij =

1

2δijT

00(0)

The Einstein equations become:

∇2g(2)00 = −8πT 00(0) (269)

∇2g(3)i0 = 16πT (1)i0

∇2g(2)ij = −8πδijT

(0)00

∇2g(4)00 = ∂2t g

(2)00 + g

(2)ij ∂i∂jg

(2)00 −

(∂ig

(2)00

)2

− 8π[T (2)00 − 2g(2)00 T

(0)00 + T (2)ii]

38

The first equation is the Poisson equation, from which we find as expected the solution:

g(0)00 = −2Φ Φ = −

∫d3x′

T (0)00

|x− x′|(270)

The solution of the fourth equation is:

g(2)ij = −2δijΦ (271)

For the third equation we find a new potential ξi:

g(3)i0 ≡ ξi ξi = −4

∫d3x′

T (1)i0

|x− x′|(272)

The solution to the second equation also introduces a new potential:

g(4)00 = −2Φ2 − 2ψ ψ = −4

∫d3x′

|x− x′|

[ 1

4π

∂2Φ

∂t2+ T (2)00 + T (2)ii

](273)

Notice that these are instantaneous potentials, as we expected in the near region. The Post-Newtonian metric becomes:

gµν =

−1− 2Φ− 2Φ2 − 2ψ ξx ξy ξz

ξx 1− 2Φ 0 0ξy 0 1− 2Φ 0ξz 0 0 1− 2Φ

(274)

6.2 The First Order Schwarzschild Radial Equation

The Schwarzschild solution can serve as a limiting case of the Post-Newtonian expansion if we setone of the masses to zero. For this to work we have to write the Schwarzschild metric in harmoniccoordinates and we have to assume low velocities and weak fields as well. We have done this inchapter 4. To first order (that is, to 1PN order) we find:

g00 = −1 + 2m1

r− 2

(m1

r

)2

gij = δij + 2δijm1

rg0i = 0 (275)

In this approximation we do not expect an event horizon (a radius at which g00 = 0). This is tobe expected, since this approximation is only valid for weak fields. Including higher orders, we dofind an event horizon and this horizon can be shown to be moving toward r = m1 the more termswe include.

To find the radial equation of motion we start with the lagrangian for the test particle:

L = −m2

√−gµν

dxµ

dt

dxν

dt= −m2

√−g00 − 2g0i

dxi

dt− gij

dxi

dt

dxj

dt(276)

Inserting the metric components we find:

L = −m2

√1− 2

m1

r+ 2

(m1

r

)2

−(1 + 2

m1

r

)v2 (277)

Expanding the square root and dropping the constant we get:

−L =m1

r− 1

2

(m1

r

)2

+3

2

m1

rv2 +

1

2v2 +

1

8v4

Here we’ve used L = L/m2. Just as in the Newtonian case we use Noether’s theorem to find the

energy and the momentum (E = vdL/dv − L and J = r × dL/dv):

E =1

2v2 +

3

8v4 − m1

r+

1

2

(m1

r

)2

+3

2

m1

rv2 (278)

39

J = r × v(1 + 3

m1

r+

1

2

(m1

r

)2)We can rewrite the energy equation into an equation for the square of the velocity. To do this wehave to find the roots of the equation. We find:

v2 = 2E − 3E2 + 2m1

r− 12E

m1

r− 10

(m1

r

)2

(279)

Again we turn to polar coordinates. Using |r × v| = r2θ we find:

r2θ2 =J2

r2

(1− 6

m1

r− v2

)=J2

r2

(1− 8

m1

r− 2E

)(280)

Using r2 = v2 − r2θ2 we find:

r2 = 2E − 3E2 + 2m1

r− 12

m1

rE − 10

(m1

r

)2

− J2

r2

(1− 8

m1

r− 2E

)(281)

This is the first order Schwarzschild radial equation of motion. We can collect the terms of differentpowers of E. Finding the roots of this equation we obtain:

r2 = a(E − −b2 +√b2 − 4ac

2a)(E − −b2 −

√b2 − 4ac

2a) (282)

a = −3

b = 2 +2J2

r2− 12m1

r

c = −10m21

r2+

2m1

r+

8J2m1

r3− J2

r2

Clearly we cannot find a single effective potential, but we can find two pseudo-effective potentials:

V+ =−b2 +

√b2 − 4ac

2a(283)

V− =−b2 −

√b2 − 4ac

2a

They are called pseudo-effective potentials because although they give the correct turning pointsof the system, we need the combination (E − V+)(E − V−) to find the correct velocities of theparticle. Expanding the square root to the first order we find:

V+ = −m1

r+J2

2r2− 3J2m1

2r3− J4

8r4+m2

1

2r2(284)

V− =2

3− 3m1

r− J2

6r2+

3J2m1

2r3+J4

8r4− m2

1

2r2

40

The first potential closely resembles the full Schwarzschildeffective potential. The second doesn’t. Since r2 is al-ways positive and a is a negative constant, we find thatV− < E < V+ and V+ < E < V−. So the energy is alwaysbetween potentials. Notice that the potentials intersectfor low J. This however happens in the region where thefields are strong and the orbits are unbound and there ourapproximation breaks down. Although V− does influencethe movement of the particle, it does not give rise to turn-ing points in the region where our approximation is valid.Notice for instance that when bound orbits are possible,the second potential is quite far up, while our approxima-tion is only valid for small E.

Zooming in on the lower potential we find a Schwarzschild-like potential:

6.3 The First Order Post-Newtonian Lagrangian

Now we’ll find the 1PN metric for two point particles. Again we start with the free particleLagrangian:

L = −m√g00 + 2g0i

dxi

dt+ gij

dxi

dt

dxj

dt(285)

We can expand the metric in its components and drop the constant term. We get:

L = −m(12

(− xixi+h(2)00 +h

(4)00 +2h

(3)0i x

i+h(2)ij x

ixj)− 1

8

((xixi)2+(h

(2)00 )

2−2h(2)00 x

ixi))

(286)

41

In the case of point particles in curved space we need to use this energy-momentum tensor:

Tµν =1√−g

∑a

γamadxµadt

dxνadt

δ(x− xa) (287)

We can already give 1/√−g and γ to the appropriate order:

−g ≈ 1− g(2)00 +

∑i

g(2)ii = 1− 4Φ

1√−g

≈ 1 + 2Φ γ ≈ 1 +1

2v2 (288)

The metric felt by a particle a is derived from the sum of the energy-momentum tensors of theother particles in the system. This is similar to the Newtonian case where the force acting on oneparticle is the sum of the forces of the other particles acting on it. This will not be true at 2.5PNand higher, where the emission of gravitational waves from the particle causes a back reactionon the particle itself. Using the last equation the energy-momentum components for all but theparticle a become:

T (0)00 =∑b=a

maδ(x− xb) T (2)00 =∑b=a

ma

(12v2b +Φ

)δ(x− xb) (289)

T (1)0i =∑b=a

mbvibδ(x− xb) T (2)ij =

∑b=a

mbvibv

jbδ(x− xb) (290)

Notice how one of the components is dependent on the field Φ. This is an expression of the non-linearity of the Einstein equations. The metric (274) on the 1PN potentials, which in turn aredependent on the energy-momentum components. The metric felt by particle a becomes:

g00 = −1 + 2∑b=a

mb

rab− 2

∑a=b

mbmc

rabrac+ 3

∑b=a

mbv2b

rab− 2

∑b=a

∑c=b,a

mbmc

rabrbc(291)

+2∑b=a

∑c=b,a

mbmc

rabrbc+∑b=a

mb

rab[v2b − (vb · nab)2]

gij = δij + 2δij∑b=a

mb

rabg0i = −4

∑b=a

mb

rabvib

Here we used rab = |xb − xa|. Since we turned to harmonic coordinates, it is quite surprising thatwe exactly find back the Newtonian metric. This is possible since the Taylor expansion of theexpected term −1+2m1/(r−m1) is equal to the usual expression −1+2m1/r. In the Schwarzschildlimit one of the masses will be reduced to a test-particle. This means that the massive particle willbe at rest. Setting the velocity to zero we indeed find back the first order Schwarzschild metric.For the test-particle we set the mass to zero and find the minkowski metric. This is what we wouldexpect, since this particle doesn’t influence space-time.

Inserting the metric components in equation (286), the lagrangian for a particle a becomes:

La =1

2mav

2a +

1

8mav

4a +

∑b=a

mb

rab− 1

2

∑b=a

∑c=a

mbmc

rabrac−∑b=a

∑c=a

∑c=b

mbmc

rabrbc(292)

+3

2v2a

∑b =a

mb

rab+

3

2

∑b =a

mbv2b

rab− 1

2

∑b=a

mb

rab(7va · vb + (va · nab)(vb · nab)) (293)

The total Lagrangian is obtained by adding the single body lagrangians, with a slight modificationof not counting twice the two body interaction terms if they occur in exactly the same form in twoof the single particle lagrangians and not counting three times the three body terms that occur in

42

three different lagrangians. Splitting up the Lagrangian in a Newtonian and a Einsteinian part,L = LN + LE , we find:

LN =∑a

1

2mav

2a +

1

2

∑a=b

mamb

rab(294)

LE =∑a

1

8mav

4a +

3

2

∑a

∑b =a

mambv2a

rab− 1

2

∑a

∑b=a

∑c=a

mambmc

rabrac

+1

4

∑a=b

mamb

rab

[7va · vb + (

rabrab

· va)(rabrab

· vb)]

Notice that some of the factors 1/2 come from the fact that switching a and b gives the sameterm. For the two body case this becomes:

LN =1

2m1v

21 +

1

2m2v

22 +

m1m2

r(295)

LE =1

8m1v

41 +

1

8m2v

42 +

1

2

m1m2

r

[3(v21 + v22)− 7v1 · v2 − (v1 · n12)(v2 · n12)

]−1

2

m1m2(m1 +m2)

r2

In e-mail correspondence professor Maggiore noted that the last term in this equation is a properthree-body term, which is an artifact of the non-linearity of general relativity. It comes fromthe fact that, to this order, we include the gravitational potential to Newtonian order in theenergy-momentum tensor.

6.4 The Relative Lagrangian

In accord with the Newtonian derivation of the orbits we now use Noether’s theorem for spatialtranslation to obtain an expression for the conserved momentum:

P =∂L

∂v1+∂L

∂v2(296)

P = m1v1 +m2v3 +1

2m1v

31 +

1

2m2v

22 +−m1m2

2r(v21 + v22)−

m1m2

2r3((v1 + v2) · r)r (297)

With some persistent computation and using r2r = rr2 and discarding terms of second order itcan be checked that:

P =d

dt

[m1r1

(1 +

v212

− M2

2r

)+m2r2

(1 +

v222

− M1

2r

)](298)

Integrating we find the center of mass expression:

P t+ C = mG1 r1 +mG

2 r2 mGa = ma +

1

2mav

2a −

m1m2

2r(299)

C = mG1 r1 +mG

2 r2 − P t (300)

From the expression of the gravitational mass, mGa , it is more appropriate to call it a center of

energy expression instead. Since the total momentum is constant we are free to choose an inertialframe in which it is zero. Comparing with Newtonian derivation the last equation with P = 0becomes the center of mass (energy) vector. Since this is a constant we can set it to zero. Therelation reduces to:

mG1 r1 +mG

2 r2 = 0 (301)

43

From this we find:

r1 = µ/m2r +µ(m1 −m2)

2M2

(v2 − M

r

)r (302)

r2 = −µ/m1r +µ(m1 −m2)

2M2

(v2 − M

r

)r

Differentiating the center of energy expression we find:

mG1 v1 +mG

2 v2 + mG1 r1 + mG

2 r2 = 0 (303)

We rewrite this as:

mG1 v1 +mG

2 v2 = −mG1 r1 − mG

2 r2 = χ (304)

We find:

v1 =−m2v + χ

m1 +m2v2 =

m1v + χ

m1 +m2(305)

Substituting these equations in our Lagrangian we find the relative Lagrangian. Using ν = µ/M ,we obtain:

L =1

2v2 +

M

r+

1− 3ν

8v4 − 1

2

(Mr

)2

+M

2r

((3 + ν)v2 +

ν

r2(r · v)2

)(306)

This expression is equal to the first order Schwarzschild Lagrangian in the limit where m2 goes tozero (which means ν = µ/M goes to zero, µ goes to m2 and M goes to m1). The 1PN equationof motion in terms of these center of mass coordinates becomes:

a = −Mr2n+

M

r2

(n[Mr(4 + 2ν)− v2(1 + 3ν) +

3

2ν(n · v)2

]+ (4− 2ν)v(n · v)

)(307)

Here n = r/r. The first term is, as expected, equal to the Newtonian acceleration. This equationwas first derived by Einstein, Infeld and Hoffmann [5].

6.5 The Post-Newtonian Orbits

Using Noether’s theorem we can link the invariance under time translation and spatial rotation ofthe relative Lagrangian, to the conservation of energy and angular momentum (E = vdL/dv − L

and J = r × dL/dv):

E =1

2v2 − M

r+

3

8(1− 3ν)v4 +

M

2r

[(3 + ν)v2 +

µ

M(r

r· v)2 + M

r

](308)

J =[1 +

1

2(1− 3ν)v2 + (3 + ν)

M

r

]r × v (309)

Again we turn to plane polar coordinates, using:

v2 = r2 + r2θ2 |r × v| = r2dθ( r · v

r

)= ˙r (310)

We find the following equations of motion:

r2 = A+2B

r+C

r2+D

r3θ = H/r2 + I/r3 (311)

A = 2E − 3(1− 3ν)E2

44

B =M − (6− 7ν)ME

C = −J2 + 2(1− 3ν)J2E)− (10− 5ν)M2

D = (8− 3ν)MJ2

H = J − (1− 3ν)JE

I = −(4− 2ν)MJ

To Newtonian order D and I become zero. The other terms reduce to their Newtonian values. Toobtain the Schwarzschild limit we again set ν = µ/M to zero, µ to m2 and M to m1. To find theorbits Damour and Deruelle [2] switched coordinates:

r = r − D

2J2r = r − I/2H (312)

In these coordinates we find for E < 0 (which corresponds to the bound orbits):

˙r2 = A+2B

r+C +BD/J2

r2(313)

θ = H/r2 (314)

Notice that these equations do not contain a term of order 1/r3 any more. These equations ofmotion can be integrated analytically. We will first consider the first equation. In case of theelliptic-like orbits the solution can be put in a form similar to equation (21) and Kepler’s equation(26):

ω0 (t− t0) = u− et sinu r = aR(1− eR cosu) (315)

aR = −M

2E

[1− 1

2(ν − 7)E

]eR =

√1 +

2E

M2

[1 +

(52ν +

15

2

)E][J2 + (ν − 6)M2

]et =

√1 +

2E

M2

[1 +

(− 7

2ν +

17

2

)E][J2 + (−2ν + 2)M2

]ω0 =

√M

a3R

[1 +

M

2aR(−9 + ν)

]From the zeroth order of the last equation we find Kepler’s law.

From the second equation of motion θ = H/r2 we find:

θk = Aeθ(u) k =

√1− 6M2

J2(316)

Aeθ(u) = 2 arctan[√1 + eθ

1− eθtan

u

2

]eθ =

√1 +

2E

M2

[1 +

(12ν − 15

2

)E][J2 − 6M2

]This equation is similar to equation (23) from the second chapter. We can also derive with sometrigonometry:

cosAeθ =cosu− eθ1− eθ cosu

(317)

Observe that when c → ∞, e2t , e2Rand e

2θ reduce to the Keplerian e2. We can eliminate u using

the expression for Aeθ (316) and r (315) found above. As a first step we rewrite the equation forr as:

r =eReθaR(1− eθ cosu) + aR

(1− eR

eθ

)(318)

45

Using equation (317) we also find:

1− eθ cosu =1− e2θ

1 + eθ cosAeθ(u)=

1− e2θ1 + eθ cos θk

(319)

We also have:

aR

(1− eR

eθ

)=µ

2(320)

So we find:

r =(aR − µ

2

) 1− e2θ1 + eθ cos θk

+µ

2(321)

The first term has the shape of an ellipse, the second term isan additional constant. The constant becomes negligible at theSchwarzschild limit. It is clear from our expression for cos θkthat the perihelion precesses. The relative orbit described iscalled a conchoid of a precessing ellipse. A conchoid of an ellipseis depicted in the first figure on the right. A precessing versionis depicted underneath it.

Expanding 2π/k to first order with respect to the small parameter M2/J2 and subtracting 2π wefind a perihelion shift that is clearly consistent with Schwarzschild:

∆θ = 6πM2

J2=

6πM

aR(1− e2R)=

6πM

a(1− e2)(322)

In the last line we’ve used that the term is already of first order, so eR can be approximated as eand aR as a.

Now we have an expression for the 1PN bound orbits it would make sense to calculate the radiatedpower of this system, as we did for the Newtonian orbits in chapter 5. From the radiated powerwe obtained an expression for the change of the orbital period (236). It would make sense to findthe 1PN correction to this expression as well. This was indeed done by Blanchet and Schafer in1989 [4]. In the chapter on gravitational waves we read that the Newtonian equations of motioncorresponded to the leading term in metric perturbation hTT

ij . For the 1PN equations of motionwe thus must include the next-to-leading term as well:

hTTij =

4

rΛij,kl[S

kl + nmSkl,m]ret (323)

From this expression we can find the radiated power and the change of the orbital period P .The 1PN correction is a long expression which is however far below the present accuracy ofmeasurement.

6.6 The Pseudo-Effective Potentials

Returning to the radial equation of motion (311) and collecting the terms with different powersof E, we find:

r2 = α(E2 +

β

αE +

γ

α

)(324)

46

α = 3(3ν − 1)

β = 2 +2M(−6 + 7ν)

r− 2J2(−1 + 3ν)

r2

γ =2M

r− J2

r2+M2(−10 + 5ν)

r2+MJ2/(8− 3ν)

r3

Finding the roots we obtain:

r2 = α(E − −β +

√β2 − 4γα

2α

)(E − −β −

√β2 − 4γα

2α

)(325)

It is clear from the last expression, that as in the Schwarzschild case, we need two pseudo-effectivepotentials:

r2 = α(E − V+)(E − V−) (326)

V+ =−β +

√β2 − 4γα

2αV− =

−β −√β2 − 4γα

2α

r2 is always positive and α is always negative (since 0 < µ/M ≤ 1/4). This means that V− < E <V1+ and V+ < E < V−. So the energy should always be between the potentials. Expanding thesquare root and keeping only first order terms, the effective potentials simplify to:

V+ = −Mr

+J2 +M2

2r2− J4(−3ν)

8r2−

˜JM2(3 + ν)

2r3(327)

V− =2

3− 9ν+J2 − 3M2

6r2+J4(1− 3ν)

8r2+

˜JM2(3 + ν)

2r3+M(5ν − 9)

3r(1− 3ν)

In the graphs the potential V+ is depicted in red and is compared with the first order Schwarzschildpotential in blue.

The Schwarzschild potential has a more pronounced peak, but it lies in the region where the fieldis strong.

47

Overall, the graph is shifted a little bit to theright. This is in accord with the 1PN orbitalequation (321). Setting θ = 0 in this equation,we find the radius corresponding to the perias-tron. This radius is equivalent to the closestturning point of the effective potential. Usingthe same system as in the right graph above,the orbital equations (321) give the graph onthe right. The graphs (both the 1PN and theSchwarzschild approximation) are in completeagreement with the effective potentials (notethat from the orbital equations we only findsolutions for E < 0, which are the bound or-bits).

V− is even higher as it was in the Schwarzschild limit. Since our approximation isn’t valid for highenergies or small radii, this second potential contributes no new turning points.

6.7 Visualizing the Accuracy of the PN Approximation

The second Post-Newtonian correction gives us terms up to order Φ3 ≈ v6. Following [12] the2PN radial equation of motion is:

r2 = A+2B

r+C

r2+D1

r3+D2

r4+D3

r5(328)

A = 2E − 3(1− 3ν)E2 + (4− 19ν + 16ν2)E3

B =M − (6− 7ν)ME + 3(3− 16ν + 7ν2)ME2

C = −J2 + 2(1− 3ν)EJ2 − 3(1− 5ν + 5ν2)E2J2 − 5(2− ν)M2 + (37− 122ν + 36ν2)EM2

D1 = (8− 3ν)MJ2 + 1/2(53− 83ν + 20ν2)M3 − (16− 73ν + 19ν2)EMJ2

D2 = −3(11− 11ν + 2ν2)J2M2

D3 = −1/4 ν(4 + ν)MJ4

In the Schwarzschild limit this greatly simplifies, because in that case ν = µ/M goes to zero. Wecan rewrite this as:

r2 = C3E3 + C2E

2 + C1E + C0 (329)

C3 = 4

48

C2 = −3 +18M

r− 3J2

r2

C1 = 2− 12M

r+

2J2 + 37M2

r2− 16J2M

r3

C0 =2M

r− J2 + 10M2

r2+

8J2M

r3+

53M3

2r3− 33J2M2

r4

Finding the root of this equation we find an equation of this form:

r2 = C3(E − Va)(E − Vb)(E − Vc) (330)

In this case we have three potentials, but two of them are imaginary and therefore do not contributeextra turning points.

In the graph on the right we compare theNewtonian effective potential (yellow), withthe first, second and full Schwarzschild poten-tials (resp. orange, red and brown). Noticethat with increasing order the potentials alter-nate around the full solution, approximating itcloser and closer. Since we know the full solu-tion in the Schwarzschild case, this plot givesus a good idea of the accuracy of the Post-Newtonian approximation. Notice that for allE < 0 (the bound orbits) the first order poten-tial is indeed more accurate than the Newto-nian term. For E > 0 the potentials starts todiverge, giving very inaccurate results, but inthis region are no bound orbits anyway.

49

In the graph on the right we compare the New-tonian potential (yellow) with the first and sec-ond Post-Newtonian effective potentials (resp.orange and red). Notice that as r goes tozero, the second order potentials goes to +∞just as the Newtonian potential. This isn’t aproblem, since at this radius the approxima-tion breaks down. The third order will, justlike the first one, go to −∞ again. In analogywith Schwarzschild we know that the final re-sult gives rise to a black hole and therefore goesto −∞. We can see from these graphs thatthe full Post-Newtonian approximation mustbe somewhere in the grey area.

This isn’t the full story though, since at 2.5PNwe have to take into account gravitational ra-diation, which will decrease the energy and an-gular momentum of the system. Since the an-gular momentum totally determines the shapeof the potential, the potential will change intime. The three graphs can thus be interpretedas three frames from the evolution of the effec-tive potential in time. At some point in timethe potential barrier vanishes and the systemwill therefore lose it’s ability to harbor boundorbits and the particle will be sucked into theblack hole.

6.8 Applying 1PN: The Hulse-Taylor Pulsar

The binary system named Hulse-Taylor consists of a pulsar (a radiating neutron star) and aregular neutron star (we will call them the pulsar(p) and its companion (c)). The pulsar has beencarefully observed for over thirty years, giving us the first evidence for the existence of gravitationalradiation. Although the gravitational waves aren’t measured directly, the evidence presented isvery convincing, making physicists believe the direct evidence will be found in the near future. Bynow we know this binary system extraordinary well. In the next table we present the measuredvalues;

ap sin ι 2.3417725(8) s (331)

ϵ 0.6171338(4)

Pb 0.322997448930(4) days

ω0 292.54487(8) deg

⟨ω⟩ 4.226595(5) deg

γ 0.0042919(8) s

Pb − 2.4184(9)× 10−12

50

The orbital period of the binary is less than8 hours, which means that the orbital velocityis of order 10−3c (about 300km/s), which isquite relativistic. We take the line of sight tobe in the z-direction. The angle between the z-direction and the normal vector to the plane ofmotion is named ι. The points where the orbitintersects the (x,y)-plane are called nodes andthe line connecting them is called the line ofnodes. In this chapter we will reserve ω for theperiastron angle and not the angular velocity.

For the pulsar the advance of the periastron, ω, is more than four degrees per year. This isconsiderable compared to the precession of Mercury, which is 43 arcsec per century (one arcsec is1/3600th of a degree). This shows that general relativistic effects are important in the binary.

We can find a lot of information about the system from the time of arrival of the pulses. The effectof gravitational redshift and time dilatation on the pulses is encapsulated in the so-called Einsteinparameter. The theoretical expression for the parameter is dependent on the two masses mp andmc. The theoretical expression for the periastron advance ⟨ω⟩ is also dependent on both mp andmc. Combining these equations we can thus find the individual masses of the system. Using the1PN periastron advance (322) and Kepler’s law, we find:

⟨ω⟩ = ∆θ

Pb= 3(mp +mc)

2/3(2πPb

)5/3 1

1− e2(332)

Inserting the measured value for e and P, we find the relation between mc and mp. Combined thisexpression with the expression for γ we find:

mp = 1.4414(2)M⊙ mc = 1.3867(2)M⊙ (333)

With the masses we can then determine the semimajor axis a with the usual Keplerian expression:

a3 =(mc +mp)P

2p

2π(334)

The result is a = 2.2 × 109m. This means semimajor axis is only three times the solar distance.The compactness of the orbit, together with the fact that no eclipse is seen, is a good indicationthat the companion is either a neutron star or a black hole (the value for mc suggests the former).This is fortunate, because this means we can treat them as point-particles, neglecting tidal effects.We can then find the semimajor axis for the two objects individually by using:

ap = amc

mp +mcac = a

mp

mp +mc(335)

Using the measured value ap sin ι from the table we have enough information to find the inclinationangle ι, which is 0.72. Using equation (236) we can find the change of the period because of thegravitational radiation:

Pb = −192π

5mpmc(mp +mc)

−1/3(Pb

2π

)−5/3 1

(1− e2)7/2

(1 +

73

24e2 +

37

96e4)

(336)

The ratio between the experimental and the predicted value was found to be:

Pobserved/Ppredicted = 1.0013(21) (337)

51

This is a wonderful confirmation of general relativity and the existence of gravitation waves.

To find these results we first needed to correct for various distortions in the arrival of the pulses.One of these distortions is caused by the relative motion of both the Earth and the pulsar. Anotherdistortion was caused by the effect of the gravitational fields on the arrival of the pulses. Sinceour solar system is mostly non-relativistic, the Newtonian approximation was accurate enough toaccount for the motion of the Earth. The pulsar system, on the other hand, is a relativistic systemand therefore 1PN was needed. Since this calculation is a direct application of the 1PN orbitalequations, we will describe it in detail.

The distortion where 1PN was used is called the Roemer time delay. It is a small time delay thatis caused by orbital motion of the pulsar around the center of mass of the binary system. Theeffect is given by:

∆R = z · r1(t) (338)

Here r1 is the distance from the pulsar to center of mass of the binary and z is a unit vectorpointing in the direction of Earth. The dot-product with the unit vector takes into account theeffect of the angle of the plane of the binary system as seen from Earth. If we took the Newtonianelliptical orbit we need to use:

r1 = a1[1− e cosu] (339)

Notice that when r1 reaches its minimum at u = 0, we have θ = 0. We therefore know that theangle θ is measured from the periastron. From the previous picture we see that ω+ θ is the anglefrom the line of nodes. We find:

∆R = r1 sin ι sin (ω + θ) (340)

We now use sin(ω + θ) = cos θ sinω + sin θ cosω, where we use the following expressions:

sin θ =√1− e2

sinu

1− e cosucos θ =

cosu− e

1− e cosu(341)

The delay becomes:

∆R =r1

1− e cosusin ι[(cosu− e) sinω + sqrt1− e2 sinu cosω] (342)

= a1 sin ι[(cosu− e) sinω +√1− e2 sinu cosω]

Writing this equation this way, we can directly insert a1 sin ι (the projected semi-major axis),which is a measured value from the table above. Numerically this effect is quite large, making itnecessary to use the 1PN orbits. In this case we use:

r = ar[1− er cosu] θ/k =(1 +

∆θ

2π

)Aeθ (343)

The Roemer time delay becomes:

∆R = r sin ι[sin

(ω0 +

(1 +

∆θ

2π

)Aeθ

)](344)

To continue we need:

cosAeθ =cosu− eθ1− eθ cosu

(345)

52

Notice that if we again use sin(α+β) = cosβ sinα+sinβ cosα, but now with α = ω0+∆θ/(2π)Aeθ

and β = Aeθ , we get a term (1 − er cosu)/(1 − eθ cosu) that we rather avoid. To avoid this weuse the following equation:

∂Ae

∂e=

1

1− e2sinAe (346)

A Taylor expansion around e = 0 to first order becomes:

Aeθ = u+∂Aeθ

∂eθeθ + ... Aer = u+

∂Aer

∂erer + ... (347)

From these expressions we find

Aeθ = Aer + h sinAe + ... h =eθ − er1− e2

(348)

eθ and er both have e as their zeroth order term. eθ − er in h therefore has no zeroth order termand the leading order term in h is thus already of first order, which means we can simply take Ae

in the second term. We now expand the sine with α = ω0 + h sinAe +∆θ/(2π)Ae and β = Aer .Using this we get the pleasant ratio (1 − er cosu)/(1 − er cosu) = 1. The Roemer delay finallybecomes:

∆R = aR sin ι[(cosu− er) sinω +√1− e2θ sinu cosω]/c (349)

Here ω = ω0 + h sinAe + ∆θ/(2π)Ae. Writing er = (1 + δr)e and eθ = (1 + δθ)e, we can alsoexpress the eccentricities in terms of the masses as:

δr =3m2

p + 6mpmc + 2mc

a(mp +mc)(350)

δθ =7/2m2

p + 6mpmc + 2mc

a(mp +mc)(351)

53

7 Radiation Reaction

In chapter 5 we calculated the gravitational radiation for the case of Newtonian orbits. Theassociated acceleration caused by the back reaction of these waves turned out to be of order2.5PN. In this chapter we will derive the 2.5PN equations of motion. To do this we first needto rewrite the Einstein equations in the Landau-Lifshitz form. In this form the full Einsteinequations are written in the form of a wave equation. Will and Wiseman (1996) developed amethod to integrate these equations for the near zone and for the far zone. We will be interestedin the near zone, since this is the region where the Post-Newtonian approximation is valid. Insteadof the metric perturbation h00 we use the potentials Hαβ = ηαβ−

√−ggµν . We will derive the 1PN

expressions for Hαβ using the Will and Wiseman method. We then use these 1PN potentials tofind the 1PN equations of motion using the surface integral method developed by Itoh, Futumaseand Asada in 2002. After this we use the same methods to derive the 2.5PN potentials and the2.5PN acceleration.

When studying the inspiraling system in chapter 5, we started from Newtonian elliptical orbits andmanually drained energy and angular momentum from the system by setting Prad = −dEorbit/dt.This equation is called the energy balance equation. This method is sensible, but also a bitartificial. If the Einstein equations predict the emission of gravitational waves from the system, thisshould automatically turn up in the equations of motion, without having to set Prad = −dEorbit/dtourselves. This indeed is the case at 2.5PN. By calculating the work done by the 2.5PN accelerationand taking the derivative with respect to time we find an expression which is exactly equal to minusthe quadrupole formula (which is the expression for the power of the radiation, Prad). To showthis will be the main purpose of this chapter. In this chapter we have mainly used [11].

7.1 The Landau-Lifshitz Field Equations

Landau and Lifshitz rewrote the Einstein equations in the form of a wave equation. Instead of themetric gµν , they used G =

√−ggµν . Since the determinant of this expression is det(G) = g, we

can always find the usual metric starting from this new variable. The field equations in this newterminology become:

∂µνHαµβν = 16π(−g)(Tαβ + tαβLL) (352)

Hαµβν = GαβGµν − GανGβµ (353)

tαβLL =1

16π

(∂λGαβ∂µGλµ − ∂λGαλ∂µGβµ +

1

2gαβgλµ∂ρGλν∂νGµρ (354)

−gαλgµν∂ρGβν∂λGνρ − gβλgµν∂ρGαν∂λGµρ + gνρgλν∂νGαλ∂ρGβµ (355)

+1

8(2gαλgβµ − gαβgλµ)(2gνρgστ − gρσgντ )∂λGντ∂µGρσ

)(356)

tαβLL is the Landau-Lifshitz pseudo-tensor. The derivation of these formulas is very tedious, al-though quite straightforward and we’ll skip it here. The tensor with four indices can be shown tohave the same symmetries as the Riemann tensor. By virtue of the antisymmetry in the last twoindices we have ∂βµνH

αµβν = −∂βµνHαµβν , having changed the dummy indices. We thus find:

∂βµνHαµβν = 0 ∂β [(−g)(Tαβ + tαβLL)] = 0 (357)

We thus find that tαβLL is conserved. We will also show in a moment that in the TT gauge it reduces

to the familiar expression for the gravitational energy. That is why tαβLL can be loosely interpretedas the gravitational energy in the full Einsteinian theory. To derive the familiar expression for the

54

energy we first define a momentum vector Pα associated with a volume V which encapsulates thesource:

Pα[V ] =

∫V

(−g)(Tα0 + tα0LL)d3x (358)

P 0 will turn out to be the expression for the energy in the TT gauge. From the Landau-LifshitzEinstein equations we find:

Pα[V ] =1

16π

∫V

∂cµHαµ0cd3x (359)

Here we have used that by inverting the last two indices of Hαµ0c we see that Hαµ00 = 0. UsingGauss’ theorem we arrive at:

Pα[V ] =1

16π

∫S

∂µHαµ0cncdS (360)

We shall write ncdS as dSc from now on. The time derivative of this expression is:

∂0Pα[V ] =

1

16π

∫S

∂µ0Hαµ0cdSc (361)

We have ∂µ0Hαµ0c = −∂µ0Hαµc0 = −∂µνHαµcν+∂µdH

αµcd. The first term is equal to 16π(−g)(Tαc+tαcLL), which we can simplify since Tαc vanishes on the surface. The second term is the spatialdivergence of an antisymmetric tensor field and its integral vanishes by virtue of Stoke’s theorem.We find:

∂0Pα[V ] = −

∫S

(−g)tαcLLdSc (362)

We also define the potentials H:

Hαβ = ηαβ − Gαβ (363)

Notice that no approximations have been made yet. Neither H nor G are assumed to be small. Inthis terminology the harmonic coordinate conditions have taken this form:

∂βHαβ = 0 ∂βGαβ = 0 (364)

In the linearized approximation, we’ll find back the usual expressions. Applying the harmonicgauge we find:

∂µνHαµβν = −Hαβ +Hµν∂µνHαβ − ∂µHαν∂νHβµ (365)

We can rewrite this as:

Hαβ = −16πταβ ταβ = (−g)(Tαβ + tαβLL + tαβH ) (366)

(−g)tαβH =1

16π(∂µHαν∂νHβµ −Hµν∂µνHαβ)

ταβ is called the effective energy-momentum tensor. The Einstein equations are now written in awave equation form, yet we still haven’t made any approximations! The method for solving theseequations was developed by Will and Wiseman in 1996.

In the limit of small hαβ we find, using −g = (1 + h):

Hαβ ≈ ηαβ −√1 + h(ηαβ − hαβ) = hαβ − 1

2ηαβh = hαβ (367)

55

From equation (362) for α = 0 and picking the TT gauge and taking V to infinity, we find:

E =1

32π

∫r2hTT

ab habTT dΩ (368)

Here we’ve used that in the TT gauge hTTµν = hTT

µν . Indeed, we have found the usual expressionfor the energy.

Now consider a wave function of the general form:

ψ = −4πµ (369)

The physical solution is dependent on the retarded Green’s function (the other solution, dependenton the advanced Green’s function, gives acausal results):

G(x, x′) =δ(t− t′ − |x− x′|)

|x− x′|(370)

Notice that G(x, x′) vanishes when the integration variable x′ is in the future of x. We there-fore only get contributions from the past. Since gravity travels with the speed of light we getcontributions from the past light cone only. The solution to the Einstein equations becomes:

ψ(x) =

∫G(x, x′)µ(x′)d4x′ =

∫µ(t− |x− x′|, x′)

|x− x′|d3x′ (371)

Notice that x′ also contains the component x′0 = t′. In the last step we’ve integrated over dt′,where the delta functions in the expression of G(x, x′) sets t′ = t − |x − x′| inside the functionµ(x′). We have already remarked that the near zone requires a different treatment from the wavezone. It is therefore reasonable to divide the past light cone of position x into a near zone domainN and a wave zone domain W. The boundary between these zones will be the arbitrarily selectedradius R, which is of the order of the wavelength of gravitational radiation. Our solution can bedivided in a near zone and a wave zone component:

ψ(x) = ψN (x) + ψW(x) (372)

The components will generally be dependent on the boundary R, but since this is an arbitraryboundary, the full solution ψ(x) will not be dependent on this radius. The dependence on R istherefore unimportant and can be ignored.

We are mainly interested in the near zone, since this is the place where our binary system is.Unfortunately, we cannot simply evaluate the near zone only. Even though the integration variablex′ is restricted to the domains, we still have the field point x which is allowed to be in both domains.For a field point in the near zone we thus get a contribution from the near zone and from the farzone. Fortunately, the far zone will only affect the near zone potentials at 3PN, which is outsidethe scope of this chapter. This means we only have to worry about the near zone contributions.

7.2 The 1PN Potentials

The Landau-Lifshitz Einstein equations, with the Newtonian constant G restored, is:

Hαβ = −16πGταβ ταβ = (−g)(Tαβ + tαβLL + tαβH ) (373)

We will take again take the energy-momentum for point masses:

(−g)Tµν =∑A

mAvαAv

βA

√−g√

−gµνvµAvνAδ(x− zA) (374)

56

Here zA is the position of the A-th particle.

For small H we will show we can expand as:

Hαβ = Gkαβ1 +G2kαβ2 +G3kαβ3 + ... (375)

To find the contributions kn we have to solve the Einstein equations iteratively, starting with theMinkowskian metric. In the zeroth-order approximation we have Hαβ

0 = 0, which implies that

Gαβ0 = ηαβ and g0αβ = ηαβ , the desired metric. Since the matter tensor is dependent on the metric,

it is now dependent on the minkowskian metric. The other energy-momentum components do notcontribute when inserting the minkowski metric. Inserting the results into the Einstein equationswe pick up a constant G on the right side. The solution of this first iteration is thus of the formHαβ

1 ≡ Gkαβ1 . The next iteration we insert this new Hαβ1 in on the right side of the Einstein

equations, obtaining Hαβ2 ≡ Gkαβ1 +G2kαβ2 .

Again using√−g =

√1 + h, H = h = −h and the definition of Hαβ we can expand the metric

accordingly:

gαβ ≈ −Hαβ + ηαβ√1−H

≈ ηαβ −Hαβ +1

2Hηαβ (376)

We also have:

gαβ = ηαβ +Hαβ − 1

2Hηαβ (377)

In the first iteration we can keep the matter tensor arbitrary. We define:

T 00 = ρ T 0a = ja T ab = T ab (378)

H00 = 4Φ H0a = 4Aa Hab = 4Bab

Since ταβ has at zeroth order only the matter tensor that contributes, the field equations become:

Φ = −4πGρ Aa = −4πGja Bab = −4πGT ab (379)

The solutions are:

Φ = G

∫ρ(t− |x− x′|, x′)

|x− x′|d3x′ (380)

Aa = G

∫ja(t− |x− x′|, x′)

|x− x′|d3x′ (381)

Φ = G

∫T ab(t− |x− x′|, x′)

|x− x′|d3x′ (382)

The conservation identities become:

∂tρ+ ∂aja = 0 ∂tj

a + ∂bTab = 0 (383)

In the near zone we can expand ρ, since |x− x′| is a small quantity here:

ρ(t− |x− x′|) = ρ− ∂ρ

∂t|x− x′|+ 1

2

∂2ρ

∂t2|x− x′|2 (384)

The first term is the instant source term, the other terms give the retarded corrections. Φ becomes:

Φ = G

∫ρ(t, x′)

|x− x′|d3x′ −G∂t

∫ρ(t, x′)d3x′ +G∂2t

∫ρ(t, x′)|x− x′|d3x′ (385)

57

Using the conservation identities, the second term on the right vanishes by Gauss’ theorem. Wefind:

Φ = U +1

2∂2tX (386)

U(t, x) = G

∫ρ(t, x′)

|x− x′|d3x′ X(t, x) = G

∫ρ(t, x′)|x− x′|d3x′ (387)

Here U is the Newtonian potential. Similarly we find:

Aa = Ua Ua = G

∫ja(t, x′)

|x− x′|d3x′ (388)

Bab = P ab P ab = G

∫T ab(t, x′)

|x− x′|d3x′ (389)

These potentials satisfy the following Poison equations:

∇2U = −4πGρ ∇2X = 2U (390)

∇2Ua = −4πGja ∇2P ab = −4πGT ab (391)

Summarizing, the potentials become:

H00 = 4U + 2∂2tX H0a = 4Ua Hab = 4P ab (392)

Now it’s time for the second iteration. ταβ will now be evaluated to first order in G, so the righthand side of the Einstein equations will be of second order. The metric to first order already given,and we insert this in the matter tensor. From the virial theorem we know that the orders in G areequivalent to orders in v2. Keeping terms to first order in G, only H00 = 4U gives a contribution.Rewriting this in terms of the metric perturbation hµν we find back the Newtonian metric. SinceU = G

∫ρ(t, x′)/|x− x′|d3x′ contains an explicit G already, the ρ-contribution will be of order G0.

For point masses it is ρ =∑

B mBδ(x− zB). U therefore becomes:

U(t, x) =∑B

GmB

|x− zB |(393)

The matter tensor to first order in G becomes:

(−g)Tαβ(t, x) =∑A

mAvαAv

βA

[1 +

1

2v2A + 3[U ]A

]δ(x− zA) (394)

This matter tensor is ill-defined. The delta in the sum makes sure that the potential in the bracketsis evaluated at x = zA. For the particles B = A the potentials gives a finite contribution pointzA, but for the particle A itself we find a singular factor 1/|zA − zA|. We will choose to set thisinfinite contribution to zero. This is called the Hadamard regularization. The infinite self-energyof the particle is not taking into account and we therefore define:

[U ]A =∑B =A

GmB

|zA − zB|(395)

We also have to approximate (−g)tαβLL and (−g)tαβH to first order in G. We will not pursue thiscalculation here. Inserting all the results in the Einstein equations we find:

H00 = −16πG∑A

mA

(1 +

1

2v2A + 3[U ]A

)δ(x− zA) + 14∂cΦ∂

cΦ (396)

58

H0a = −16πG∑A

mAvaAδ(x− zA) (397)

Hab = −16πG∑A

mAvaAv

bAδ(x− zA)− 4

(∂aΦ∂bΦ− 1

2δab∂cΦ∂

cΦ)

(398)

The solutions to these equations are:

H00 = 4U + 4(ψ +

1

2∂2tX − P + 2U2

)H0a = 4Ua Hab = 4P ab (399)

X =∑A

GmA|x− zA| Ua =∑A

GmAvaA

|x− zA|(400)

ψ =∑A

GmA(3/2v2A − [U ]A)

|x− zA|(401)

P =∑A

mAv2A

|x− zA|+

1

4

∑A

m2A

|x− zA|2+

1

2

∑A

∑B>A

mAmB

|x− zA||x− zB |(402)

−1

2

∑A

∑B>A

mAmB

|zA − zB|

( 1

|x− zA|+

1

|x− zB |

)(403)

The explicit expression for P ab is very long, but won’t be necessary in the rest of the chapter.What we found is the same expression in terms of Hµν that is equivalent to the 1PN metric wefound in the previous chapter in terms of hµν .

7.3 The Surface Integral Method

The equations of motions can be found from the potentials Hµν using a method developed by Itoh,Futamase and Asada in 2002. We will use this method to find back the 1PN equations of motion.A similar method was used by Einstein, Infeld and Hoffmann in their classic 1938 paper. Einsteinassumed gµν = ηµν + hµν and separated some of the linear terms in the Ricci tensor componentsand grouped the rest in a function Lµν . He also defined γµν , which is identical to hµν and heimposed the coordinate conditions ∂sh0s − ∂0h00 = 0 and ∂shms = 0. The field equations thantake the simple form:

∂ssh00 = 2Λ00(Lµν) ∂ssh0n = 2Λ0n(Lµν) ∂mnhmn = 22Λmn(Lµν) (404)

From the last two equations, using the coordinate conditions, we can write:

∂s(∂sh0n − ∂nh0s) = 2Λ0n − ∂0nh00 ∂s(∂shmn − ∂nhms) = 2Λmn (405)

Considering a space-time with a number of point particles, a closed surface around each particlegives, via Stokes’ theorem:∫

∂s(∂sh0n − ∂nh0s) · ndS = 0

∫∂s(∂shmn − ∂nhms) · ndS = 0 (406)

∫(∂0nh00 − 2Λ0n) · ndS = 0

∫2Λmn · ndS = 0 (407)

Einstein continued to show that that these surface integrals gave the equations of motion for theparticle. We will now follow Itoh, Futamase and Asada, who picked up on this old idea and usedit for the Einstein equations in the Landau-Lifshitz form. We define a sphere VA, with a surfaceSA centered around every particle. The radius of the sphere sA is described by:

|x− zA| = sA = constant (408)

59

The radius sA is an arbitrary constant. The ball moves rigidly with the body, so its velocity is thesame as the velocity of the particle, vA = dzA/dt. The momentum inside this sphere is given by:

PαA =

∫VA

(−g)(T 0α + t0aLL)d3x (409)

We have to integrate over the moving volume. In a time interval dt, the volume VA(t) moves to a

new volume VA(t+ dt). In the course of this motion the surface element dS sweeps out a volume

dV = (vAdt) · dS. This means that a quantity defined by F (t) =∫VAf(t, x)d3x, will change in

time according to:

∂tF =

∫VA

∂tfd3x+

∫SA

fvA · dS (410)

The first term accounts for the changes intrinsic to the function f, while the second term accountsfor the change in the domain of integration. The momentum changes like:

∂tPαA =

∫VA

∂t[(−g)(T 0α + t0aLL)]d3x+

∫SA

(−g)(T 0α + t0aLL)vA · dS (411)

We use the conservation statement ∂β [(−g)(Tαβ + tαβLL)] = 0 to turn the first term into a surfaceintegral by using Gauss’ theorem. At the surface we have Tαβ = 0, so we are left with:

∂tPαA =

∫SA

(−g)(tαcLL − t0aLLvcA)dSc (412)

We now define:

DαA =

∫VA

(−g)(T 00 + t00LL)(xa − zaA)d

3x (413)

Using the anti-symmetry of Hαµβν we can write the 00 field equations as:

(−g)(T 00 + t00LL) =1

16π∂cdH

0c0d (414)

Using integration by parts we find:

DαA =

1

16π

∫SA

[(∂dH0c0d)(xa − zaA)−H0a0c]dSc (415)

Differentiating the original expression for DαA with respect to time for the moving volume and

realizing that Tµν = 0 at the surface, we find:

∂tDαA =

∫VA

∂t[(−g)(T 00 + t00LL)](xa − zaA)d

3x− vaA

∫VA

(−g)(T 00 + t00LL)d3x (416)

+

∫SA

(−g)(t00LL)(xa − zaA)v

cAdSc (417)

Using the conservation expression and integration by parts we can change the first term into asurface integral:

∂tDαA = −

∫SA

(−g)(t00LL)(xa − zaA)dSc +

∫VA

(−g)(T 0a + t0aLL)d3x (418)

−vaA∫VA

(−g)(T 00 + t00LL)d3x+

∫SA

(−g)(t00LL)(xa − zaA)v

cAdSc (419)

60

The first volume integral is equal to P aA, the second to P 0

A. We therefore find, writing P 0A =MA,

the equation:

P aA =MAv

aA +Qa

A + DaA (420)

QaA =

∫SA

(−g)(t0bLL − t00LLvbA)(x

a − zaA)vcAdSb (421)

Differentiating we find the important law of motion:

MAaA =˙PA − MAvA − ˙

QA − ¨DA (422)

For simplicity we will now focus on particle 1. We now have to decompose the 1PN potentialsfrom the last paragraph in an ’internal potential’ that diverges at x = z1 and ’external potentials’that are smooth at x = z1. Specifically we write:

U =m1

s1+ Uext ψ =

m1µ

s1+ ψext X = m1s1 +Xext (423)

P =1

4U2 + P Ua =

m1va1

s1+ Ua

ext (424)

µ =3

2v21 −

∑A=1

mA

|z1 − zA|P =

m1ν

s1+ Pext ν = v21 −

1

2

∑A=1

mA

|z1 − zA|(425)

Notice that the internal potential of X does not diverge as s1 goes to zero, but its second timederivative that appears in H00 does. The external potentials are given by:

Uext =∑A=1

mA

|x− zA|(426)

ψext = c32∑A=1

mAv2A

|x− zA|−

∑A =1

m1mA

|z1 − zA||x− zA|(427)

−∑A=1

∑B>A

mAmB

|z1 − zA|

( 1

|x− zA|+

1

|x− zB|

)(428)

Xext =∑A=1

mA|x− zA| (429)

Pext =∑A=1

mAv2A

|x− zA|− 1

2

∑A=1

∑B>A

mAmB

|zA − zB |

( 1

|x− zA|+

1

|x− zB |

)(430)

Uaext =

∑A=1

mAvaA

|x− zA|(431)

Notice that by virtue of the non-linearity of the Einstein equations the external part of ψ is stilldependent on m1 and z1. Generally we have s1 = x − z1. For some constant s1 we have oursphere with surface element dSa = s21nadΩ. Let’s say we’re interested in the external potentialsnear s1 = 0. This means we can Taylor expand the potentials around s1 = 0. We get for example:

Uext = Uext(z1) + s1∂aUext(z1)na + ... (432)

Calculating M1, we start by writing ∂bH0b0a in terms of the 1PN potentials:

∂bH0b0a = −4∂aU − 4

(∂aψ +

1

2∂aX − ∂aP + 4U∂aU + Ua

)(433)

61

The expression for M1 therefore becomes:

M1 = −⟨⟨s21na∂aU⟩⟩ − ⟨⟨s21na(∂aψ +1

2∂aX − ∂aP +

7

2U∂aU + Ua)⟩⟩ (434)

Here we have used ⟨⟨...⟩⟩ = 1/(4π)∫...dΩ, which is an average over the spherical surface with

s1 = constant. We now make the distinction between the internal and the external potentials. Weneed the derivatives of the internal potentials and therefore we use the next identities involving sand n = s/s:

s = −n · v ∂as = na na = −1

s(δab − nanb)v

b (435)

∂anb =1

s(δab − nanb) s = −n · a+ 1

s[v2 − (n · v)2] (436)

∂as = −1

s(δab − nanb)a

b − 1

s2(naδbc + 2δabnc − 3nanbnc)v

bvc (437)

As said before, the external contributions will be Taylor expanded in powers of s1. For example,from U = m1/s1 +Uext we find ∂aU = −m1n

a/s21 + ∂aUext, so we find s21na∂aU = −m1 +O(s21),

which leads to ⟨⟨s21na∂aU⟩⟩ = −m+O(s21). In a similar way we can find:

⟨⟨s21na∂aU⟩⟩ = −m1 +O(s21) (438)

⟨⟨s21na∂aψ⟩⟩ = −m1µ+O(s21) (439)

⟨⟨s21na∂aX⟩⟩ = −2

3m1v

21 +O(s1) (440)

⟨⟨s21na∂aP ⟩⟩ = −m1ν +O(s21) (441)

⟨⟨s21na∂aUa⟩⟩ = −1

3m1v

21 +O(s1) (442)

⟨⟨s21na∂aU∂aU⟩⟩ = −m21

s1−m1Uext +O(s1) (443)

The last equation contains as a leading term an external potential, which using the Taylor expan-sion is equal to:

Uext(z1) =∑A=1

mA

|z1 − zA|(444)

This is equal to [U ]1 from before. Notice that in this chapter the infinite part is naturally takenout of the sum, without using an ad-hoc regularization prescription. Finally we arrive at:

M1 = m1

(1 +

1

2v21 + 3Uext +

7

2

m1

s1+O(s1)

)(445)

In a similar way we find:

M1 = m(4va1∂aUext + 3Uext +O(s1)) (446)

P a1 = m1[∂

aUext +O(s1)] +m1

[(32v21 − Uext

)∂aUext + ∂aψext +

1

2∂aXext (447)

+4Uaext − 4(∂aU b

ext − ∂bUaext)v1b +

11

3

m1

s1a1a +O(s1)

](448)

Qa1 =

m21

6s1+O(s1) (449)

62

Da1 = O(s1) (450)

Going back to our law of motion (422) and plugging these equations in we get on each side of theequation terms of order s−1, terms independent of s and terms of higher order in s which have notbeen calculated. Since s is an arbitrarily constant, the equation must be true order-by-order in s(since picking a different constant value, the equations should still be satisfied). The s−1 termscancel out immediately. The terms independent of s give:

aa1 = ∂aUext + (v21 − 4Uext)∂aUext − 4va1v

b1∂bUext − 3va1 Uext + ∂aψext (451)

+1

2∂aXext + 4Ua

ext − 4(∂aU bext − ∂bUa

ext)v1b (452)

aa1 here is the acceleration of particle 1. The external potentials are evaluated at x = z1. The firstterm corresponds to the Newtonian acceleration, while the other terms represent the 1PN correc-tions. Writing out the external potentials and their derivatives explicitly and then generalizingfrom body 1 to an equation valid for any particle A, we find:

aaA = −∑B =A

mB

z2AB

nAB −∑B =A

mA

z2AB

[v2A +2v2B − 4(vA · vB)−

3

2(nAB · vB)2 −

5ma

zAB− 4mA

zAB(453)

−4mA

zAB− 4

∑C =A,B

mC

zAC−

∑C =A,B

mC

zBC+

1

2

∑C =A,B

mCzAB

z2BC

(nAB · nBC)]nAB (454)

+∑B =A

mB

z2AB

nAB · (4vA − 3vB)(vA − vB)−7

2

∑B =A

∑C =A,B

mBmC

zABz2BC

nBC (455)

Here we used zAB = |zA − zB | and n = zAB/zAB . This is the 1PN equation of motion. Thisequation becomes equivalent to the 1PN equation of motion given in the previous chapter whenwe again set the center of gravitational mass to zero and describe the radii in terms of the relativedistance vector (and simplifying to the two body case).

7.4 The Radiation Reaction Potentials

In this paragraph we will derive the potentials for 1.5PN and 2.5PN. We shall skip 2PN, since thesesimply give extra corrections, but does not contribute to the radiation reaction. As expected, the1.5PN terms will not effect the equations of motion and we shall see this in the next paragraph.We start by defining:

H00 = 4V H0a = 4V a Hab = 4W ab (456)

τ00 = ρ τ0a = ja τab = τab (457)

The conservation identities are:

∂tρ+ ∂aja = 0 ∂tj

a + ∂bτab = 0 (458)

The Einstein equations become:

V = −4πρ V a = −4πja W ab = −4πτab (459)

Expanding the solutions to sufficient order, using the Taylor expansion we find:

V =

∫ρ

|x− x′|d3x′ − ∂t

∫ρd3x′ +

1

2∂2t

∫ρ|x− x′|d3x′ − 1

6∂3t

∫ρ|x− x′|2d3x′ (460)

+1

24∂4t

∫ρ|x− x′|3d3x′ − 1

120∂5t

∫ρ|x− x′|4d3x′ (461)

63

V a =

∫ja

|x− x′|d3x′ − ∂t

∫jad3x′ +

1

2∂2t

∫ja|x− x′|d3x′ − 1

6∂3t

∫ja|x− x′|2d3x′ (462)

W ab =

∫τab

|x− x′|d3x′ − ∂t

∫τabd3x′ +

1

2∂2t

∫τab|x− x′|d3x′ − 1

6∂3t

∫τab|x− x′|2d3x′ (463)

We now define:

I =

∫ρd3x Ia =

∫ρxad3x (464)

Pa =

∫jad3x Pab =

∫jaxbd3x (465)

J ab =

∫(jaxb − jbxa)d3x J abc =

∫(jaxb − jbxa)xcd3x (466)

Mab =

∫τabd3x Mabc =

∫τabxcd3x (467)

Each extra upper index d gives an extra component xd in the integral. Using ρ =∑

AmAδ(x− zA)we have to Newtonian order:

I = I =∑A

mA Ia = Ia =∑A

mAzA Iab = Iab =∑A

mAzaAz

bA (468)

Pa = P a =∑A

mAvaA J ab = Jab =

∑A

mA(vaAz

bA − zaAv

aA) (469)

Mab =Mab =∑A

mavaAv

bA (470)

Finding the leading odd terms of ταβ we find after a long computation:

ρ = ρ[0] + ρ[2] + ρ[4] + ρ[5] (471)

ja = ja[0] + ja[2] + ja[4] (472)

τab = τab[0] + τab[2] + τab[3] + τab[4] (473)

Here [n] represents the component of order O(v2vn), or nPN. Inserting the explicit componentsof ταβ into the expressions for the potentials we find:

V [3] = −1

6I(3)cc (474)

V [5] = −Iab(3)∂abX − 1

6I(3)cc [2]− 1

60(r2δab + 2xaxb)I

(5)ab +

1

30xaI(5)acc −

1

120I(5)cdcd (475)

V a[3] =1

6xbI

a(4)b − 1

18Ia(4)cc − 1

9Ja(3)

cc (476)

W ab[1] = −1

2Iab(3) (477)

W ab[3] = −1

2Iab(3)[2]− 1

12r2Iab(5) +

1

18xcIab(5)c +

1

9xc(Ja b(4)

c + Jb a(4)c )− 1

9Mab(3)

cc (478)

Here the number in the brackets represents the number of differentiations with respect to time.An explicit expression for Iab[2] will not be required to find the equations of motion in the nextchapter.

64

7.5 The 2.5PN Equations of Motion

In this paragraph we shall find the 2.5PN radiation reaction acceleration. We again start withthis equation:

MAaA =˙PA − MAvA − ˙

QA − ¨DA (479)

We shall see that 1.5PN indeed does not contribute and we shall derive the contribution at 2.5PN.In the previous paragraph we saw that the components of Hµν contain the potentials U , Ua andX. Like before we divide them into an internal and an external term. We again keep only theterms independent of s1.

We’ve already calculate M1[0] and M1[2] and have found no term for M1[1]. For M1 we have afirst contribution at M1[2]. For D1 we D1[0] = D1[1] = D1[2] = 0. Both M and D depend onthe derivative of H0a0b. Inserting the potentials Hµν from the last paragraph we find: M1[3] =

M1[5] = 0, M1[3] = M1[5] = 0 and D1[3] = D1[5] = 0. For Q1 we found a first contribution atQ1[2]. We now find Q1[3] = Q1[5] = 0. P a

1 is already know to have a contribution at [0] and [2].We now find P a

1 [3] = 0 and:

P a1 [5] = m1

(− I

(3)cd ∂

acdXext + 2Ia(3)c ∂cUext4

3I(3)cc ∂

aUext (480)

+3

5Ia(5)c zc1 −

1

5I(5)cc z

a1 + 2Ia(4)c vc1 −

2

15Ia(5)cc − 2

3Ja(4)

cc

)(481)

We thus find the following expansions:

M1 =M1[0] +M1[2] +M1[4] M1 = M1[2] + M1[4] (482)

˙Q1 =

˙Q1[2] +

˙Q1[4]

˙P1 =

˙P1[0] +

˙P1[2] +

˙P1[4] +

˙P1[5] (483)

Using equation (422) these equations imply:

a1 = a1[0] + a1[2] + a1[4] + a1[5] (484)

a1[5] is the radiation reaction acceleration we are looking for, the first odd term in the expansion.

The term is given by˙P1[5]/m, which is:

a1[5] = −I(3)cd ∂acdXext + 2Ia(3)c ∂cUext

4

3I(3)cc ∂

aUext +3

5Ia(5)c zc1 (485)

−1

5I(5)cc z

a1 + 2Ia(4)c vc1 −

2

15Ia(5)cc − 2

3Ja(4)

cc (486)

Writing out the derivatives of the external potentials explicitly and rewriting the acceleration interms of any particle A, we find:

aaA[5] = −3I(3)bc

∑B =A

mB

z2AB

naABnbABn

cAB − 1

3I(3)cc

∑B =A

mB

z2AB

naAB (487)

+3

5Ia(5)b zbA − 1

5I(5)cc z

aA + 2I

a(4)b vbA − 2

15Ia(5)cc − 2

3Ja(5)

cc (488)

The rate at which this acceleration term does work on the particles is given by:

W =∑A

mAaA[5] · vA (489)

65

The last two terms of the acceleration can be set to zero, because they contain a factor P =∑AmAvA, which is the total Newtonian momentum, which can be set to zero by placing the

origin of the coordinate system at the center of mass. After some further manipulations we find:

W = −1

5I(3)ab I

ab(3) +1

15I(3)cc I

cc(3) (490)

+d

dt

[ 3

10(I

(4)ab I

ab − I(3)ab I

ab)− 1

10(I(4)cc I

cc − I(3)cc Icc) + 2I

(3)ab

∑A

mAvaAv

bA

](491)

Notice that the first two terms are indeed equal to minus Egw, the quadrupole formula. Callingthe other terms d/dt Ebound, we find:

W = −Egw +d

dtEbound (492)

From the chapter about gravitational waves, we know the energy only becomes a valid physicalvariable when we take a time average. Taking this average over several periods we find:

⟨W ⟩ = −⟨Egw⟩+∆Ebound

∆t(493)

In e-mail correspondence with Eric Poisson, he noted that the radiation reaction affects the equa-tion of motion at order v7 (2.5PN), which will affect the energy balance equation at order v15

(remember that Prad = O(v10) for 0PN). We are now considering an energy balance equation oforder v10, so we can assume the orbit to be periodic (similarly, precession is a 1PN effect andeffects the energy balance equation at order v12 and can thus also be neglected) We therefore find:

⟨W ⟩ = −⟨Egw⟩ (494)

This same procedure is necessary in electrodynamics. Here we get the similar equation W =Erad + d/dtEbound. The observed radiation does not contain the last term and it can be made todisappear by taking the average of a period. Since Ebound is not measured far from the source thisenergy must stay bound to the system, hence its name.

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8 Conclusion

We started out with the two-body system in ordinary Newtonian mechanics. This system was im-portant to study carefully, because it would serve as a limit of general relativity when velocities arenegligible (v << 1) and the fields are weak (Φ << 1). We found the Newtonian orbital equationsand also the Newtonian effective potential. The field equations in the Newtonian approximation ofgeneral relativity turned out to be wave equations, which suggested the existence of gravitationalwaves. Another important limiting case of general relativity is the Schwarzschild system wherem1 >> m2. We also found an effective potential for this system, which suggested the existence ofblack holes.

We have shown that gravitational waves can do work on test particles and therefore the wavesmust have energy themselves. We explicitly calculated the power of the gravitational radiation inthe case of Newtonian elliptical orbits. Using effective potentials we have shown that the loss ofenergy and momentum due to this radiation circularizes the orbit.

We then studied the binary system where both masses are allowed to influence the metric. We as-sumed that the velocities were low (v << 1) and the field were weak (Φ << 1), yet we allowed forone more order of Φ in the equations of motion than we did in the Newtonian approximation. Thisgives us the leading relativistic correction, called the first Post-Newtonian correction (or 1PN).Just like in the Newtonian case we found the orbital equations (which turned out to be precessingconchoids of ellipses), but we were not able to find a single effective potential. Instead we neededtwo pseudo-effective potentials, which did give us the correct turning points of the system. Thesepseudo-potentials for 1PN and 2PN gave us a measure of the accuracy of the Post-Newtonianapproximation at different radii. The 1PN results have been used to calculate the first evidenceof gravitational radiation from the observations of the Hulse-Taylor pulsar.

In the final chapter we calculated the 2.5PN equations of motion by rewriting the equations of mo-tion in the Landau-Lifshitz form. Using the Will and Wiseman approach and the Surface integralmethod we found an expression for the 2.5PN acceleration. In the chapter on gravitational waveswe started from Newtonian elliptical orbits and manually drained energy and angular momentumfrom the system by setting Prad = −dEorbit/dt. However, if the Einstein equations predict theemission of gravitational waves from the system, this should turn up automatically at some pointin the Post-Newtonian expansion. This is indeed the case at 2.5PN. The change of the work doneon the system turns out to be equal to minus the quadrupole formula (which is the expression forthe power of the radiation, Prad). The calculation of these higher order terms will be of majorimportance for the extraction of gravitational waves from the noise in the upcoming gravitationalwave detectors.

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9 Appendix

9.1 Noether’s Theorem

Noether’s theorem states that for each symmetry in a system there exist a conserved quantity.Consider a function Qi(s, t) such that Qi(0, t) = qi(t) and let the lagrangian L be dependent onL(Qi(s, t), Qi(s, t); t). Taking the partial derivative of L with respect to s we find:

∂L

∂s=

∂L

∂Qi

∂Qi

∂s+

∂L

∂Qi

∂Qi

∂s(495)

Evaluating this at s = 0 and the using the Euler-Lagrange equation we find:

∂L

∂s

∣∣∣s=0

=∂L

∂qi

∂Qi

∂s

∣∣∣s=0

+∂L

∂qi

∂Qi

∂s

∣∣∣s=0

(496)

=d

dt

( ∂L∂qi

)∂Qi

∂s

∣∣∣s=0

+∂L

∂qi

∂Qi

∂s

∣∣∣s=0

=d

dt

( ∂L∂qi

∂Qi

∂s

∣∣∣s=0

)(497)

We end up with:

∂L

∂s

∣∣∣s=0

=d

dt

( ∂L∂qi

∂Qi

∂s

∣∣∣s=0

)(498)

In the case where ∂L/∂s = 0 (when the Lagrangian is not dependent on some quantity s) wefollowing quantity is conserved at s = 0:

∂L

∂qi

∂Qi

∂s

∣∣∣s=0

(499)

For instance, consider the Lagrangian for a system of N particles:

L =1

2

∑i

miv2i − V (|ri − rj |) (500)

This Lagrangian has the symmetry of translation: qi = ri → Qi = ri+sn. ∂Qi/∂s simply becomesn, so our conserved quantity associated with translation is:∑

i

∂L

∂vi· n =

∑i

pi · n (501)

This is the linear momentum in direction n. Since this holds for all n we find that∑

i pi isconserved. Similarly, the independence of the Lagrangian on time this gives us the conservationof energy and the invariance of the Lagrangian under rotation implies conservation of angularmomentum. These conserved quantities are given by:

p =∑i

∂L

∂vi(502)

E =∑j

xj∂L/∂xj − L (503)

J = r ×∑j

∂L/∂xj (504)

The Lagrangian need not be strictly invariant under a transformation, as long as the equations ofmotion stay unchanged. Look what happens to the action when we add total time derivative ofsome function S times a constant α:

Action =

∫L′dt =

∫Ldt+ α

∫d

dtSdt =

∫Ldt+ αS(t2)− αS(t1) (505)

68

The total time derivative is immediately integrated out of the action, giving two constant termswhich can be taken out of the action. The generalized conserved quantity can be shown to be:∑

i

∂L

∂xi

∂Qi

∂s− S (506)

In the case of, for example, a Galileo transformation (x→ x+ vt) we have:

L(xi + vt, xi + v, t) = L(xi, xi, t) + v∑i

mixi +1

2

∑i

miv2 (507)

The last term is just a constant that can be ignored. We can write the second term as a timederivation of S times a constant alpha. In this case S becomes:

S =∑i

mixi (508)

We also have:

∂Qi

∂s=∂(x+ vt)

∂v= t (509)

With these expression we end up with the following conserved quantity:∑i

mixit−∑i

mixi = tM˙R−MR = tP −MR (510)

R is the center of mass location and P is the center of mass momentum. In this appendix we havemainly used [17].

69

References

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[3] Damour, T. and Deruelle, N. General Relativistic Celestial Mechanics of Binary Systems. II.The Post-Newtonian Timing Formula, Annales de l’I.H.P., section A (1986).

[4] Blanchet, L. and Schafer, G. Higher Order Gravitational Radiation Losses in Binary Systems,Monthly Notices of the Royal Astronomical Society (1989).

[5] Einstein, A. and Infeld, L. and Hoffmann, B. The Gravitational Equations and the Problemof Motion, Annals of Mathematics (1938).

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[9] Maggiore, M. Gravitational Waves, Oxford University Press (2008).

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[11] Poisson, E. Post-Newtonian Theory for the Common Reader, Lecture notes (2007).

[12] Schafer, G. and Wex, N. Second Post-Newtonian Motion of Compact Binaries, Physics LettersA 74 (1993).

[13] Schutz, B. A First Course in General Relativity, Cambridge University Press (1985).

[14] Susskind, L. Einstein’s General Theory of Relativity, www.stanford.edu (2008).

[15] Taylor, J. and Weisberg, M. Relativistic Binary Pulsar B1913+16: Thirty Years of Observa-tion and Analysis, ASP Conference series, Vol. TBD (2004).

[16] Thornton, S. and Marion, J. Classical Dynamics of Particles and Systems, Fifth edition,Brooks/Cole (2008).

[17] Tong, D. Classical Dynamics, www.damtp.cam.ac.uk/user/tong/dynamics.html(2005).

[18] Weinberg, S. Gravitation and Cosmology, John Wiley & Sons (1972).

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