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Faculty of Physics

University of Iceland

2016

Faculty of Physics


2016

Orbits of Particles Around

Compact Objects: From

Newtonian Theory to General

Relativity

Matthías Ásgeir Jónsson

ORBITS OF PARTICLES AROUND COMPACT

OBJECTS: FROM NEWTONIAN THEORY TO

GENERAL RELATIVITY

Matthías Ásgeir Jónsson

16 ECTS thesis submitted in partial ful�llment of aBaccalaureus Scientiæ degree in Physics

Advisor

Gunnlaugur Björnsson

Faculty of Physics

School of Engineering and Natural Sciences


Reykjavik, May 2016

Orbits of Particles Around Compact Objects: From Newtonian Theory to General Rela-tivity16 ECTS thesis submitted in partial ful�llment of a B.Sc. degree in Physics

Copyright c© 2016 Matthías Ásgeir JónssonAll rights reserved

Faculty of PhysicsSchool of Engineering and Natural SciencesUniversity of IcelandHjarðarhagi 2-6107, ReykjavíkIceland

Telephone: 525 4000

Bibliographic information:Matthías Ásgeir Jónsson, 2016, Orbits of Particles Around Compact Objects: FromNewtonian Theory to General Relativity. B.Sc. thesis, Faculty of Physics, University ofIceland.

Abstract

This thesis gives a general overview of orbital trajectories of particles around com-pact objects in Newtonian theory and in general relativity. Only two particle sys-tems are considered, with only gravitational attraction taken into account. Orbitsare calculated from a set of initial conditions, using di�erent methods such as �ndingroots of an e�ective potential, by parameterizing, and solving linear and non-lineardi�erential equations.

v

Contents

List of Figures ix

List of Tables xi

1 Introduction 1

2 The Kepler Problem 3

2.1 Kepler's Laws of Planetary Motion . . . . . . . . . . . . . . . . . . . 32.2 Newtonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Kepler's First Law Derived . . . . . . . . . . . . . . . . . . . . 72.2.2 Kepler's Second Law Derived . . . . . . . . . . . . . . . . . . 92.2.3 Kepler's Third Law Derived . . . . . . . . . . . . . . . . . . . 10

2.3 The Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3.1 Kepler's �rst law derived . . . . . . . . . . . . . . . . . . . . . 122.3.2 The E�ective Potential . . . . . . . . . . . . . . . . . . . . . . 152.3.3 Open and Closed Orbits . . . . . . . . . . . . . . . . . . . . . 172.3.4 Calculated Orbits . . . . . . . . . . . . . . . . . . . . . . . . . 182.3.5 Orbital Velocities . . . . . . . . . . . . . . . . . . . . . . . . . 202.3.6 Other Central Forces . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 A Pseudo-Newtonian Potential . . . . . . . . . . . . . . . . . . . . . . 24

3 Orbits in the Schwarzschild Metric 29

3.1 The Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Time-Like Geodesics: Massive Particles . . . . . . . . . . . . . . . . . 31

3.2.1 Radial Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.2 Bound Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2.3 Unbound Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.3 Null Geodesics: Massless Particles . . . . . . . . . . . . . . . . . . . . 503.3.1 Cone of avoidance and General Orbits . . . . . . . . . . . . . 54

4 Orbits in the Kerr Metric 59

4.1 The Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2 The Ergosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3 Circular Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.3.1 Null Geodesics: Massless particles . . . . . . . . . . . . . . . . 634.3.2 Time-Like Geodesics: Massive Particles . . . . . . . . . . . . . 63

vii

Contents

5 Conclusions 65

Bibliography 67

viii

List of Figures

2.1 A logarithmic plot of the cube of the average distance as a functionof orbital period squared. Data from Nasa [2]. . . . . . . . . . . . . . 5

2.2 The e�ective potential, Veff , and its components, −k/r and L2/2µr2 . 16

2.3 An Earth like system. Initial conditions: m1 = M�, m2 = 3 ×10−6M�, r0 = 0.9832AU , and vθ0 = 6.3853AU/yr. Resulting in anearly circular orbit. . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Initial conditions: m1 = M�, m2 = 0.1M�, r0 = 1AU , and vθ0 =6AU/yr. Resulting in an elliptical orbit with e = 0.17. . . . . . . . . 19

2.5 Initital conditions: m1 = m2 = M�, r0 = 1, vθ0 = 15AU/yr. Result-ing in a hyperbolic orbit with e = 1.848. . . . . . . . . . . . . . . . . 20

2.6 Radial velocity curves due to a particle with mass m2 = 9.54 ×10−4M� = MJupiter placed at r0 = 0.5AU with vθ = 6.35AU/yr,around m1 = M�. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.7 Radial velocity curves due to a particle with massm2 = 3×10−10M� =M⊕ placed at r0 = 0.5AU with vθ = 6.35AU/yr, around m1 = M�. . 21

2.8 Closed generic orbits with C < 0, an attractive force. . . . . . . . . . 23

2.9 Closed generic orbits with C > 0, a repulsive force. . . . . . . . . . . 24

2.10 Orbits in the Psuedo-Newtonian potential. The red circle representsthe event horizon. a) Initial conditions: x0 = 5.06, vθ = 0.427c0. b)Initial conditions: x0 = 2.48, vθ = 0.775c0, e = 0.59. . . . . . . . . . . 26

3.1 The e�ective potential for di�erent values of L. . . . . . . . . . . . . 32

ix

LIST OF FIGURES

3.2 Initial conditions: M = 1, r0 = 12, vθ0 = 0.4c. The red circlerepresents the event horizon. . . . . . . . . . . . . . . . . . . . . . . . 41

3.3 Orbits of the second kind for case B. a) Initial conditions: M = 1and r0 = 5. b) Initial conditions: M = 1 and r0 = 5.6. The red circlerepresents the event horizon. . . . . . . . . . . . . . . . . . . . . . . . 42

3.4 Initial conditions: M = 1 and r0 = 20. a) Orbit of the �rst kindstarting from r0 = 20 approaching r = 4.44. b) Orbit of the secondkind starting from r = 4.44 spiraling towards the center. The redcircle represents the event horizon. . . . . . . . . . . . . . . . . . . . 44

3.5 Capture orbits for case E. a) Initial conditions: M = 1, r0 = 10, andvθ0 = 0.02c. b) Initial conditions: M = 1, r0 = 30, and vtheta0 = 0.05c.The red circle represents the event horizon. . . . . . . . . . . . . . . . 47

3.6 Initial conditions: M = 1, r0 = 8, and vθ0 . The red circle representsthe event horizon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.7 Initial conditions: M = 1 and r0 = 2.8. The red circle represents theevent horizon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.8 Unbound orbits for case C. Parameters from Chandrasekhar (1983).a) Initial conditions: M = 0.3, e = 0.001i, and α = 1. b) Initialconditions: M = 0.3, e = 0.1i, and α = 1. The red circle representsthe event horizon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.9 The photon e�ective potential. . . . . . . . . . . . . . . . . . . . . . . 51

3.10 The cone of avoidance at di�erent distances [9]. . . . . . . . . . . . . 55

3.11 Solutions of equation (3.103), with M = 1, r0 = 12, and varying θ. . . 57

4.1 The positive and negative potentials for a photon with L = 3, arounda rotating black hole of massM = 1, with a = 0.5. The Black dashedline is r+ and the dotted line is r0 at θ = π/2. . . . . . . . . . . . . . 63

4.2 Circular orbits in the equatorial plane (colored) in the Kerr metric. . 64

x

List of Tables

2.1 Orbital energy and eccenetricity . . . . . . . . . . . . . . . . . . . . . 17

xi

1 Introduction

Orbits of particles are an extremely important topic in physics, and are at thefoundation of astrophysics. Understanding of orbits is vital to elucidate the behaviorof astrophysical phenomena such as planetary systems, accretion disks, and blackholes. In this paper we focus on orbits of test particles around stars and compactobjects such as black holes. The goal is to calculate orbital trajectories of particlesfrom a set of initial conditions. Each chapter gets increasingly more complex as wemove towards orbits in general relativity. The aim is then to introduce core conceptsin the earlier chapters for use in the later ones. In all discussions, we consider onlysystems of two particles, i.e the two body problem, which can be reduced to onlyone orbiting particle.

We begin with a discussion on Kepler's laws of planetary motion and then derivethem using Newtonian mechanics. From there we introduce an element from rela-tivity into our Newtonian model before the discussion of orbits in general relativity.A large portion is dedicated to orbits in the Schwarzschild metric, and almost allpossible orbits there are covered. For the Kerr metric we only cover circular orbitsin the equatorial plane.

1

2 The Kepler Problem

In this chapter we focus solely on Kepler orbits (or Keplerian orbits). Kepler orbitsare solutions of the two body problem and describe the motion of two bodies arounda common center of mass due to their gravitational attraction. We treat the orbitingbodies as point particles and neglect other e�ects such as perturbations from externalbodies, radiation pressure, non-uniform gravity, and general relativity. Kepler orbitsare thus solutions of a highly idealized version of the two body problem called theKepler problem. We also assume, in all cases, that our orbits are stable, i.e. willnot decay over time. The orbits, as we shall see, can only take on 4 di�erent shapesdepending on the initial conditions. These shapes (or curves) are known as conicsections. The discussion below follows and is based on Carroll and Ostlie (2014).

2.1 Kepler's Laws of Planetary Motion

The �rst successful description of planetary motion in agreement with observationaldata was put forth by the German mathematician Johannes Kepler. After the deathof Tycho Brahe, the foremost naked-eye-observer at the time, Kepler inherited hismassive amount of data and over a period of about 20 years put forth three lawsof planetary motion famously attributed to him [1]. These three laws, known asKepler's Laws of Planetary Motion, are at the foundation of modern astronomy.They are as follows:

Kepler's First Law

A planet orbits the Sun in an ellipse, with the sun at one focus of the ellipse.

Kepler's Second Law

A line connecting a planet to the Sun sweeps out equal areas in equal timeintervals.

Kepler's Third Law The Harmonic LawThe square of the orbital period of a planet is proportional to the cube of theplanet's average distance from the Sun.

3


The laws are described above in a qualitative manner since they were deduced frompurely empirical data. We can describe them more quantitatively in mathematicalform.

Kepler's First Law

r =a(1− e2)1 + e cos θ

, (2.1)

Kepler's Second Law

∆A

∆t= constant, (2.2)

Kepler's Third Law The Harmonic Law

P 2 = a3. (2.3)

Here r is the planet's distance from the sun (at a focal point), a its average distancefrom the sun (the semi-major axis) and e is the orbit's eccentricity - an indicator ofits shape/structure. The angle θ is the true anomaly, measured from the planet'sclosest distance to the sun, called periapsis (or perihelion), to its current position.The ∆A is the area swept by an orbiting body over time interval ∆t. Finally Pis the orbital period, if P is measured in years and a in astronomical units (AU),then the ratio P 2/a3 ' 1. In order to explain P 2/a3 ' 1 and ∆A/∆t = constant,one requires Newtonian mechanics as these relations are not inherently obvious fromobservations alone.

Using Kepler's third law we can determine a planet's relative distance from the sun ifwe know its orbital period in terms of Earth's. For example, Mars' orbital period isTmars = 1.8808T⊕ and so its average distance from the sun is amars = 1.523AU . Atthe time of Kepler the Astronomical Unit was unknown and was determined manyyears after his death. The orbital parameters of the planets in our solar system havebeen independently measured and agree with Kepler's laws, as shown in �gure (2.1).

4

2.2 Newtonian Mechanics

1.5 1.0 0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0

Square of orbital period P 2 [yr2]

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Cube o

f avera

ge ist

ance

a3[AU

3]

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

Figure 2.1: A logarithmic plot of the cube of the average distance as a function oforbital period squared. Data from Nasa [2].


In order to better understand planetary motion and the forces underlying themwe need Newtonian Mechanics. Isaac Newton was able to deduce and generalizeKepler's laws using his three laws of motion and the law of universal gravitation.

Newton's First Law The Law of InertiaAn object at rest will remain at rest and an object in motion will remain inmotion in a straight line at a constant speed unless acted upon by an externalforce.

Newton's Second Law

The net force acting upon an object of mass m is proportional to the object'smass and its resultant acceleration

Fnet =n∑i=1

Fi = ma, Fnet = mdv

dt=dp

dt. (2.4)

Where v is the velocity vector, a the acceleration vector an p the momentumvector.

Newton's Third Law

For every action there is an equal and opposite reaction,

F12 = −F21. (2.5)

5


Newton's Law of Universal Gravitation

F12 = −Gm1m2|r12|2

r̂12, (2.6)

where G = 6.674−11m3kg−1s−2 is the gravitational constant.

Before deriving Kepler's laws we should choose a reference frame. By far the easiestreference frame to work with is that of the Center-of-Mass. Imagine some arbitraryreference frame with the vectors r1 and r2 pointing from the origin towards theirrespective masses m1 and m2, the displacement vector from r1 to r2 is r = r2 − r1.The system has a center-of-mass vector, R ,which can be thought of as a weightedaverage of the masses and their position.

R ≡ m1r1 +m2r2M

, M = m1 +m2. (2.7)

Rewriting eq. (2.7) and di�erentiating both sides with respect to time we get

MV = m1v1 +m2v2, (2.8)

or

P = P1 +P2. (2.9)

Which means that the momentum of the center-of-mass, P = MV , can describethe system as a whole. If we assume no external forces are acting on the system,then according to Newton's �rst law the total force is zero, i.e. the center-of-massreference frame is an inertial frame. We can simplify things even further by choosinga coordinate system where R = 0, eq. (2.7) then becomes

m1r1 +m2r2M

= 0. (2.10)

Remembering that r = r2 − r1 gives,

r1 = −m2Mr, r2 =

m1Mr. (2.11)

6


We de�ne a useful term, the reduced mass,

µ ≡ m1m2m1 +m2

. (2.12)

The vectors r1 and r2 are then

r1 = −µ

m1r, r2 =

µ

m2r. (2.13)

We can now describe the total energy of the system in terms of the reduced mass,essentially reducing the two-body problem to a one body problem,

E =1

2µv2 −GMµ

r, (2.14)

where T ≡ µv2/2 is the kinetic energy and U ≡ −GMµ/r is the potential energy.

2.2.1 Kepler's First Law Derived

In order to derive Kepler's �rst law we consider the e�ect of gravity on the or-bital angular momentum L. In the center-of-mass coordinates the orbital angularmomentum becomes

L = r× µv = r× p. (2.15)

Taking the time derivative we get,

dL

dt=dr

dt× p+ r× dp

dt= v× p+ r× F. (2.16)

Since v and p are in the same direction their cross product is zero. Also, F isa central force and thus directed inward along r, the cross product of r and F istherefore zero. We then have an important result regarding angular momentum,

dL

dt= 0, (2.17)

7


that is, angular momentum is conserved in a central force �eld. We also note thatthe orbit of the reduced mass µ, lies in a plane perpendicular to L according to eq.(2.15). Since the orbit is planar we conveniently switch to polar coordinates. Nowthe position vector becomes r = rr̂ and the angular momentum can be written

L = µr2r̂× ddtr̂. (2.18)

The force experienced by the reduced mass is given by Newton's law of gravitation,expressed in polar coordinates F = −GMµ

r2r̂, the acceleration is

a =GM

r2r̂. (2.19)

Taking the cross product of a and L gives

a× L = −GMr2

r̂×(µr2r̂× d

dtr̂

)= GMµr̂×

(r̂× d

dtr̂

). (2.20)

Using the triple product expansion A× (B×C) = (A ·C)B− (A ·B)C we �nd

a× L = −GMµ[(r̂ · d

dtr̂

)r̂− (r̂ · r̂) d

dtr̂

], (2.21)

which simpli�es to

a× L = GMµ ddtr̂, (2.22)

where the �rst term in the bracket is zero since 2(r̂ · d

dtr̂)

= r̂· ddtr̂+ d

dtr̂·r̂ = d

dt(r̂·r̂) = 0

and r̂ · r̂ = 1.

Integrating both sides of eq. (2.22) with respect to time and noting that dLdt

= 0 anda = dv

dtwe obtain

v× L = GMµr̂+D, (2.23)

8


where D is a constant vector due to the inde�nite integration. This vector is knownas the Laplace-Runge-Lenz vector (or simply LRL vector) and is a constant of motionin the Kepler problem. It lies in the orbital plane and its magnitude determines theorbit's eccentricity.

Finally taking the dot product

r · (v× L) = GMµr(r̂ · r̂) + r ·D, (2.24)

and using the triple product identity A · (B×C) = (A×B) ·C yields

(r× v) · L = GMµr + r|D| cos θ. (2.25)

Equation (2.15) states that (r× v) = L/µ which gives us

L2

µ= GMµr + r|D| cos θ, (2.26)

solving for r we get Kepler's �rst law

r =L2/µ2

GM(

1 + |D|GMµ

cos(θ)) . (2.27)

We now de�ne the eccentricity e ≡ |D|/GMµ and α ≡ L2/µ2

GMas the semi-latus

rectum to �nally �nd

r =α

1 + e cos(θ). (2.28)

2.2.2 Kepler's Second Law Derived

Kepler's second law describes how an area swept by an orbiting body over a timeinterval is constant, i.e. ∆A

∆t= constant. In order to evaluate that constant we

9


consider an in�nitesimal area element in polar coordinates

dA = dr(rdθ) = rdrdθ. (2.29)

The radius vector sweeps out an area

∆A =1

2r2dθ, (2.30)

dividing by ∆t we obtain an areal velocity

∆A

∆t=

1

2r2dθ

dt. (2.31)

The angular velocity is related to angular momentum |L| = µr2 dθdt

so we can write

∆A

∆t=

1

2

|L|µ, (2.32)

which is Kepler's second law with its constant determined.

2.2.3 Kepler's Third Law Derived

Using Kepler's second law and integrating over a whole orbital period P we get atotal area

A =1

2

|L|µP. (2.33)

The area of an ellipse can also be determined geometrically by its semi-major axis,a, and semi-minor axis, b with the following equation

A = πab. (2.34)

10

2.3 The Lagrangian

Substituting this into equation (2.33) and squaring both sides yields

P 2 =4π2a2b2µ2

L2(2.35)

which leaves us with P 2 ∝ a2. A geometric relationship exists between a, b, and e

b2 = a2(1− e2), (2.36)

substituting this brings us up to P 2 ∝ a4. We also note the relationship betweenthe purely geometrical formula of an ellipse, equation (2.1), and the one derivedfrom Newtonian mechanics, equation (2.27). From which we can derive the angularmomentum

|L| = µ√GMa(1− e2). (2.37)

Inserting this into eq (2.35) we �nally have Kepler's third law,

P 2 =4π2

GMa3. (2.38)

If we express the mass in solar masses, distance in astronomical units, and time inyears then the ratio 4π2/GM = 1. The reason is that for all planets in our solarsystem M� >> Mplanet and so M = M� +Mplanet ∼= M�.

2.3 The Lagrangian

By using Lagrangian mechanics we can more easily derive the equations of motion foran orbiting particle. The discussion in this section is based on Thornton and Marion(2004). To begin we again set up a system of two particles with mutual gravitationalattraction. As shown in the previous section this system can be reduced to a twodimensional orbit of a single particle with a reduced mass µ. The energy given by eq.(2.14) can be expanded to include both radial velocity, vr, and tangential velocity,vθ,

E =1

2µ(ṙ2 + r2θ̇2)−GMµ

r, (2.39)

11


where vr = ṙ and vθ = rθ̇. In polar coordinates the Lagrangian can be written

L = 12µ(ṙ2 + r2θ̇2) +G

Mµ

r. (2.40)

The Euler-Lagrange equations then give us the equations of motion in the coordi-nates r and θ which are,

r̈ − rθ̇2 +GMr2

= 0, (2.41)

and

µr2θ̇ = constant = L. (2.42)

Equation (2.42) shows again the conservation of angular momentum. The positionof the particle as function of time can be found by solving eq. (2.41). This is a nonlinear second-order ODE which can be solved numerically. Another method is tosolve for ṙ in eq. (2.39) which gives

ṙ = ±

√2

µ

(E +G

Mµ

r

)− L

2

µ2r2. (2.43)

Solving for dt and integrating gives t = t(r) which can then be inverted to obtainr = r(t). Equation (2.43) is again a non-linear ODE but this time of the �rst order.We hold o� on numerically solving non-linear ODEs and instead focus our attentionon �nding r as a function of θ, i.e. r = r(θ).

2.3.1 Kepler's �rst law derived

In order to determine r(θ) we make a change of variable u ≡ 1/r, taking the deriva-tive dr/du and dividing by dt we obtain

ṙ = − 1u2du

dt= − 1

u2dθ

dt

du

dθ= − θ̇

u2du

dθ= −L

µ

du

dθ, (2.44)

12

2.3 The Lagrangian

the last step follows from equation (2.42). For r̈ we di�erentiate a second time

r̈ =d

dt

(−Lµ

du

dθ

)= −θ̇L

µ

d2u

dθ2= −u2

(L

µ

)2d2u

dθ2, (2.45)

and writing rθ̇2 in terms of u

rθ̇2 =1

u

(L

µu2)2

=

(L

µ

)2u3, (2.46)

we can now write eq. (2.41) in our new variable

d2u

dθ2+ u−GMµ

2

L2= 0. (2.47)

This is just the equation of a simple harmonic oscillator, its general solution is

u = A cos θ +B sin θ +GMµ2

L2, (2.48)

where A and B depend on initial conditions. For an orbit, be it circular, elliptical,or para- or hyperbolic, there must exist a point where the particle's radial motion iszero in the center-mass frame. We can de�ne such a point at θ = 0 for convenience

du

dθ

∣∣∣∣θ=0

= −B cos(0) = 0, (2.49)

leading to B = 0. Further setting an initial condition u(θ) = 1/r0 at θ = 0

u(θ = 0) = A cos(0) +B sin(0) +GMµ2

L2=

1

r0, (2.50)

and so

A =1

r0−GMµ

2

L2. (2.51)

13


Rearranging terms and solving for r we obtain

r =L2/µ2

GM(1 + AL2

GMµ2cos θ)

, (2.52)

which is Kepler's third law and like before we have α = L2/µ2

GMbut now the ec-

centricity is e = AL2/GMµ2. In the previous section the eccentricity was de�nede = |D|/GMµ. Let's explore how |D| and A relate to one another. As mentionedD is the LRL vector - a conserved quantity in the Kepler problem. It is generallyde�ned

A ≡ p× L−GMµ2r̂, (2.53)

we see from equation (2.23) that |D| = |A|/µ. Multiplying A by L2 and writing Lin terms of initial parameters we �nd

AL2 =L2

r0−GMµ2 = (µr0vθ0)

2

r0= µvθ0L−GMµ2, (2.54)

which means that AL2 = |A|, dropping the cross product since p and L are perpen-dicular. Furthermore we note that |D|µ = AL2.

The LRL vector is conserved along with the energy and angular momentum. Thismeans we have seven conserved quantities, three each (in 3 dimensions) for thevectors A and L and one for a scalar E. These constants are functions of r andp which each have 3 independent components of motion. In all we have sevenconserved quantities and six components of motion. One component of motionmust give information about where a particle is at any given time since A, L, andE are just functions of these components. Therefore we have only �ve independentcomponents of motion describing the orbit and seven conserved quantities, leadingto two relationships between them [4]. These relationships are

A · L = 0, (2.55)

and

|A|2 = µ2k2 + 2µEL2, (2.56)

14

2.3 The Lagrangian

where k = GMµ is the strength of a central force. With these relations ande = |A|/GMµ2 we can write the eccentricity in terms of the energy and angularmomentum

e =

√1 +

2EL2

k2µ. (2.57)

2.3.2 The E�ective Potential

From equation (2.57) we see that the minimum energy allowed is E = −k2µ/2L2which results in e = 0 corresponding to a circular orbit, any less and it does notdescribe a real physical system. To better understand how the energy correspondsto di�erent orbits we introduce an e�ective potential, Veff . The force acting on anorbiting particle is not just the gravitational force, in its own reference frame, thereis also a centrifugal force. The e�ective force is then

Feff = −k

r2+ µ

(rθ̇)2

r= − k

r2+

L2

µr3, (2.58)

which yields an e�ective potential shown on �gure (2.2)

Veff = −k

r+

L2

2µr2. (2.59)

The e�ective potential is very useful and contains information about allowed orbitsincluding their shapes and sizes. For a circular orbit with radius rc the followingcondition must be met

dVeffdr

∣∣∣∣r=rc

= 0, (2.60)

that is

rc =L2

µk, (2.61)

15


r

0Energ

y

Veff

Vmin

−k/r

L 2/2µr2

Figure 2.2: The e�ective potential, Veff , and its components, −k/r and L2/2µr2 .

so the minimum value is

Vmin = −k2µ/2L2, (2.62)

as expected it's just the same we got from (2.57) but it doesn't tell us whether thecircular orbit is a stable one. A stable circular orbit has to ful�ll the additionalcondition

d2Veffdr2

∣∣∣∣r=rc

> 0, (2.63)

that is

d2Veffdr2

∣∣∣∣r=rc

= −2kr3c

+3L2

µr4c, (2.64)

and substituting rc into this equation

d2Veffdr2

∣∣∣∣r=rc

= −2k4µ3

L6+

3k4µ3

L6=k4µ3

L6> 0, (2.65)

16

2.3 The Lagrangian

so circular orbits exist for all r > 0. This makes sense because test particles can inprinciple get arbitrarily close and have no speed limit in classical physics. SolvingVeff − E = 0 for r gives the available turning points, elliptical orbits have twoturning points and the eccentricity can be found with a simple equation

e =rmax − rminrmax + rmin

. (2.66)

Note that the potential depends on initial position and tangential velocity, it doesnot depend on radial velocity. Meaning that an increase in radial velocity cannot change the shape of the potential - only increase the energy and eccentricity.A change in tangential velocity changes the potential and can result in both anincrease and decrease in eccentricity. Table (2.1) shows the allowed orbits and theirrespective energies and eccentricities.

Table 2.1: Orbital energy and eccenetricity

Energy Eccentricity Type Bound/UnboundE = Vmin e = 0 Circle BoundVmin < E < 0 0 < e < 1 Ellipse BoundE = 0 e = 1 Parabola UnboundE > 0 e > 1 Hyperbola Unbound

2.3.3 Open and Closed Orbits

E�ective potentials can not describe how an orbit develops over time. An ellipticalorbit oscillates between its apsides rmax and rmin. These distances are bounded, buttheir position in terms of θ is not. Circular orbits are by de�nition closed and para-and hyperbolic orbits can not be said to be open or closed as they are unbound.If it takes a �nite number of revolutions from the apsides and back to the exactsame position, the orbit is said to be closed. Otherwise the orbit is open and passesthrough every point in the plane between rmax and rmin.

The angle between two consecutive apsides is know as the apsidal angle, ∆θa. Forclosed orbits we would expect the apsidal angle to be a rational fraction of π (dueto symmetry), i.e. ∆θa = π(a/b) to obtain �nite revolutions. Bertrand's theoremstates that the only central force potentials that produce closed orbits are the inverse-square potential and the radial harmonic oscillator potential (Hooke's law) []. Sincethe Kepler problem involves an inverse-square central force, the elliptical orbits are

17


closed and do not change over time. We can con�rm this numerically by writing

dθ =dθ

dt

dt

drdr =

θ̇

ṙdr, (2.67)

then inserting equations (2.42) and (2.43) into the one above, and integrating fromone apside to the next gives the apsidal angle,

∆θa =L√2µ

∫ rmaxrmin

dr

r2√E + k/r − L2

2µr2

. (2.68)

2.3.4 Calculated Orbits

So far we have covered the orbit of the reduced mass µ in a two particle system,if one particle is more massive by many orders of magnitude the reduced massis e�ectively the lighter orbiting particle. The massive one can be thought of asstationary and sits squarely in the origin of the center-mass frame. This is of coursean approximation, in reality both particles orbit around a center of mass known asthe barycenter. For planetary systems the barycenter is usually stationed inside theparent star itself, creating a wobble. This wobble can for example be used to detectexoplanets using Doppler spectroscopy.

Each particle's orbit can easily be determined via the reduced mass orbit's withequations (2.13), their orbital periods and eccentricities must equal those of thereduced mass. What di�ers is their relative distances, as indicated by eq. (2.13),and orbital velocities. The energy and angular momentum is split between them

Ei =µ

miE, Li =

µ

miL, (2.69)

so that

E = E1 + E2 L = L1 + L2. (2.70)

In �gures (2.3) to (2.5) we show calculated orbits of a few systems. In �gures (2.3a)and (2.3b) initial conditions are set to those of Earth's at its perihelion. Sinceµ ≈ m2 the orbit of the reduced mass is essentially just the orbit of m2. The orbit of

18

2.3 The Lagrangian

m1 (the Sun) is also drawn on �gure (2.3b) but due to the huge mass di�erence it isnegligible in comparison to m2. E�ects of reducing the mass di�erence can be seenin �gure (2.4), there we start to see the orbit of the more massive particle comparedto the lighter. In �gure (2.5) we obtain a hyperbolic orbit with two equally sizedmasses at high speeds.

0°

45°

90°

135°

180°

225°

270°

315°

0.20.4

0.60.8

1.01.2

(a) Orbit of µ.

0°

45°

90°

135°

180°

225°

270°

315°

0.20.4

0.60.8

1.01.2

(b) Orbits of both particles.

Figure 2.3: An Earth like system. Initial conditions: m1 = M�, m2 = 3× 10−6M�,r0 = 0.9832AU , and vθ0 = 6.3853AU/yr. Resulting in a nearly circular orbit.

0°

45°

90°

135°

180°

225°

270°

315°

0.20.4

0.60.8

1.0

(a) Orbit of µ.

0°

45°

90°

135°

180°

225°

270°

315°

0.20.4

0.60.8

1.0


Figure 2.4: Initial conditions: m1 = M�, m2 = 0.1M�, r0 = 1AU , and vθ0 =6AU/yr. Resulting in an elliptical orbit with e = 0.17.

19


0°

45°

90°

135°

180°

225°

270°

315°

0.51.0

1.52.0

2.5

(a) Orbit of µ.

0°

45°

90°

135°

180°

225°

270°

315°

0.51.0

1.52.0

2.5


Figure 2.5: Initital conditions: m1 = m2 = M�, r0 = 1, vθ0 = 15AU/yr. Resultingin a hyperbolic orbit with e = 1.848.

2.3.5 Orbital Velocities

As mentioned, particles with di�erent masses will have di�erent orbital speeds, theydi�er from the reduced mass by the same factor as the energy and angular momen-tum

vi =µ

miv, (2.71)

where the speed of the reduced mass, v, is given by the vis-viva equation,

v2 = GM

(1

r− 1a

), (2.72)

as measured in the center-mass frame. An outside observer using Doppler spec-troscopy can only measure the radial velocity component

vr = K sin(i)(cos(θ + ω) + e cos(θ + ω)), (2.73)

where K = (vrmax−vrmin)/2 is the semi-amplitude of the radial curve measurements,i is the orbital inclination, and ω the longitude of periapsis from ascending or re-ceding node [6]. Figures (2.6) and (2.7) show the calculated radial velocity curves

20

2.3 The Lagrangian

for particle with mass m1 = M� in�uenced by an orbiting body. Orbital inclinationis assumed to be i = 90◦ relative to the observer.

0.0 0.2 0.4 0.6 0.8 1.0

Phase

20

10

0

10

20

30

40

50

Radia

l V

elo

city

[m/s]

(a) ω = 0.

0.0 0.2 0.4 0.6 0.8 1.0

Phase

20

10

0

10

20

30

40

Radia

l V

elo

city

[m/s]

(b) ω = 45◦.

Figure 2.6: Radial velocity curves due to a particle with mass m2 = 9.54×10−4M� =MJupiter placed at r0 = 0.5AU with vθ = 6.35AU/yr, around m1 = M�.

0.0 0.2 0.4 0.6 0.8 1.0

Phase

0.06

0.04

0.02

0.00

0.02

0.04

0.06

0.08

Radia

l V

elo

city

[m/s]

(a) ω = 45◦.

0.0 0.2 0.4 0.6 0.8 1.0

Phase

0.08

0.06

0.04

0.02

0.00

0.02

0.04

0.06R

adia

l V

elo

city

[m/s]

(b) ω = 90◦.

Figure 2.7: Radial velocity curves due to a particle with mass m2 = 3× 10−10M� =M⊕ placed at r0 = 0.5AU with vθ = 6.35AU/yr, around m1 = M�.

2.3.6 Other Central Forces

The central force potential U ∼ 1/r has bound orbits that are all closed accordingto Bertrand's theorem. Now lets add a higher order term U ∼ 1/r2 to this potentialand what orbits we �nd. The central force now becomes

F = − kr2

+C

r3, (2.74)

where C is the strength of the 1/r3 central force component. Following the same

21


steps as we did for equation (2.41) we obtain a di�erential equation for u(θ),

d2u

dθ2+ u

(1 +

Cµ

L2

)=kµ

L2, (2.75)

again just an equation for a simple harmonic oscillator. Let's de�ne β2 ≡ 1+Cµ/L2,there are now 3 possibilities:

i) β2 = 0. The solution is then

u =1

r=

kµ

2L2θ2 + Aθ +B (2.76)

so the particle eventually spirals in towards the center.

ii) β2 < 0. The solution is

u =1

r= Aeβθ +Be−βθ − kµ

L2β2, (2.77)

which, depending on the constants A and B, either spirals inwards towards thecenter or outwards to in�nity.

iii) β2 > 0. The solution is

u =1

r= A cos(βθ) +B cos(βθ) +

kµ

L2β2, (2.78)

and using the same initial conditions as before we obtain

r =L2β2/µ2

GM(AL2β2

kµcos(βθ) + 1

) . (2.79)This di�ers slightly from equations (2.52) and (2.27), we now have an additional

22

2.3 The Lagrangian

parameter β. Let's focus on case (iii) where β2 > 0, there we have

β =

√1 +

Cµ

L2> 0. (2.80)

The constant C determines whether the force is attractive or repulsive. For C < 0the force is attractive and 0 < β < 1, and if C > 0 it is repulsive and β > 1. Figures(2.8) and (2.9) show closed generic orbits for varying values of β. Open orbits areobtained when β is an irrational number.

0°

45°

90°

135°

180°

225°

270°

315°

(a) β = 0.3, e = 0.7.

0°

45°

90°

135°

180°

225°

270°

315°

(b) β = 0.9, e = 0.4.

Figure 2.8: Closed generic orbits with C < 0, an attractive force.

23


0°

45°

90°

135°

180°

225°

270°

315°

(a) β = 1.1, e = 0.8.

0°

45°

90°

135°

180°

225°

270°

315°

(b) β = 1.9, e = 0.5.

Figure 2.9: Closed generic orbits with C > 0, a repulsive force.

2.4 A Pseudo-Newtonian Potential

The concept of a black hole, an object whose escape velocity is greater than the speedof light, was �rst put forth by the English natural philosopher John Michell in 1783.This idea was not given much thought for centuries because light in classical physicsis not in�uenced by gravity. An object with such high escape velocity must beextremely massive or at least extremely dense. The escape velocity can be deducedin many ways, for example using the vis-viva equation (2.72). Taking the limita→∞ for an escape trajectory gives

vesc =

√2GM

r, (2.81)

and assuming an escape velocity that of light, c, the radius of a black hole is

rs =2GM

c2, (2.82)

that is called the Schwarzschild radius in general relativity. Even for massive objects,e.g. the sun, the Schwarzschild radius is relatively small, rs ∼ 3km, because G/c2 ∼10−27m2kg−1s−1. For simplicity we set G = c = 1 so that in general rs = 2M anduse a dimensionless parameter x = r/rs. Here we are assuming that one particle

24


is extremely massive and is stationary in the center-mass frame. Introducing thisradius into the Newtonian gravitational potential

U(x) = − krs(x− 1)

= − GmM2M(x− 1)

= − m2(x− 1)

, (2.83)

where M is the mass of the stationary black hole and m the mass of the orbitingparticle. This is called a Pseudo−Newtonian potential because it contains elementsfrom both classical physics and relativity [7]. The e�ective potential now becomes

Vpseudo = −m

2(x− 1)+

L2

2mr2= − m

2(x− 1)+

L2

2mr2sx2, (2.84)

then writing L in terms of initial conditions and dividing by m

2Vpseudom

= − 1x− 1

+(mr0v0,θ)

2

m2r2sx2

= − 1x− 1

+ (x0v0,θ)2 1

x2. (2.85)

De�ning Ṽpseudo ≡ 2Vpseudo/m and l = x0v0,θ

Ṽpseudo = −1/(x− 1) +l2

x2(2.86)

To determine r(θ) for this potential we again make a change of variable u = 1/r.Going through the same steps as before we now get

d2u

dθ2+ u−GMµ

2

L2(u− us)2

u2= 0, (2.87)

a non-linear ODE, but if rmin � rs we can approximate it as eq. (2.47) and obtainequation (2.52) (Kepler's third law), else we must solve it numerically. Insteadof solving this non-linear ODE numerically, we should make use of the e�ectivepotential. We can write Ṽpseudo − E = 0 in the form

Ex3 + (1− E)x2 − lx+ l = 0, x > 1, (2.88)

a third degree polynomial which can easily be solved numerically. For ellipticalorbits there must exist two real roots x1, x2 > 1 and for circular orbits a double root

25


x1 = x2 > 1, otherwise the orbits are unbound. Unbound orbits have an additionalproperty in this potential, they can not only escape but also be captured. Figure(2.10) shows bound orbits for di�erent systems, we consider capture orbits in detailin the next chapter.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(a)

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(b)

Figure 2.10: Orbits in the Psuedo-Newtonian potential. The red circle representsthe event horizon. a) Initial conditions: x0 = 5.06, vθ = 0.427c0. b) Initialconditions: x0 = 2.48, vθ = 0.775c0, e = 0.59.

Let's investigate circular orbits in this potential, where they are possible and theirstability properties. For this we use a di�erent method, the fact that for a stablecircular orbit we have the following condition

dl

dr> 0. (2.89)

For a circular orbit the centrifugal force must equal the gravitational force, which isjust the derivative of the potential, so we can write the angular velocity in terms ofthe potential itself. With a bit of algebra we �nd

θ̇ =

(1

r

dU

dr

) 12

=

(1

xrs· 1rs

dU

dx

) 12

=1√2rs

(1

x(x− 1)2

) 12

. (2.90)

26


The angular momentum, l = x2θ̇, then becomes

l = x2θ̇ =1√2

(x4

x(x− 1)2

) 12

=1√2

(x3

(x− 1)2

) 12

, (2.91)

taking the derivative dl/dx we obtain after some algebra,

dl

dx=

x12

2√

2(x− 1)2(x− 3). (2.92)

So stable circular orbits exist for x > 3. For x < 3 the circular orbits are unstableand a slight inward perturbation causes a particle to spiral into the center. A circularorbit at x = 3 is marginally stable and usually called the innermost stable circularorbit or last stable circular orbit, it is the transition between stable and unstableorbits.

27

3 Orbits in the Schwarzschild

Metric

In classical physics time and space are considered to be independent. Einsteinshowed with his theory of relativity however that space and time are intrinsicallylinked and should not be thought of as space and time but rather as spacetime.Einstein's �eld equations describe how matter and energy curves spacetime andhow spacetime tells matter how to move along geodesics. A geodesic can be thoughtof as the shortest possible distance between two points on a curved surface. On a twodimensional �at surface for example, a geodesic is just a straight line but on a threedimensional curved surface a geodesic is a part of a great circle. Since spacetimeis curved in general relativity the geodesics describes the motion of matter throughit, but we must introduce the time component into our three dimensional space tofully understand the motion.Here we focus our e�orts on understanding the motion of both massive and masslesstest particles around a black hole described by the Schwarzschild metric. We alreadyintroduced an element from relativity, the Schwarzschild radius, into a classicalNewtonian model for the Pseudo-Newtonian potential. This small correction, as weshall demonstrate, is in some aspects a very good approximation of the motion ofmassive particles in the Schwarzschild metric. The discussion below is based largelyon Shapiro and Teukolsky (1983), and Chandrasekhar (1983).

3.1 The Metric

According to Birkho�'s theorem, the Schwarzschild metric is the only static, spher-ically symmetric, vacuum solution to Einstein's �eld equations. It applies every-where up to the surface of a spherical body. It is named after Karl Schwarzschildwho discovered it in 1916. A Schwarzschild black hole carries no charge nor angularmomentum so its only distinguishing property is its mass. It consists of an eventhorizon, which can be thought of as its surface, at the Schwarzschild radius and asingularity at its center. Setting c = G = 1, the Schwarzschild metric in spherical

29

3 Orbits in the Schwarzschild Metric

coordinates is

ds2 = −(1− 2Mr

)dt2 + (1− 2Mr

)−1dr2 + r2dθ2 + r2 sin2 θdφ2. (3.1)

Here r, θ, and φ are the standard spherical coordinates. The Lagrangian in theSchwarzschild metric becomes,

2L = −(

1− 2Mr

)ṫ2 +

(1− 2M

r

)−1ṙ2 + r2θ̇2 + r2 sin2 θφ̇2, (3.2)

where we de�ne ṫ ≡ dt/dλ and an a�ne parameter, λ, so that λ = τ/m where τis the proper time and m a particle'a mass. The Euler-Lagrange equations for thecoordinates θ, φ, and the time component are

θ :d

dλ(r2θ̇) = r2 sin θ cos θφ̇2, (3.3)

φ :d

dλ(r2 sin2 θφ̇) = 0, (3.4)

t :d

dλ

[(1− 2M

r

)ṫ

]= 0. (3.5)

.

Since the Schwarzschild metric is spherically symmetric we can always orient ourcoordinate system in such a way that a particle initially moves in the equatorialplane, i.e. θ = π/2. From equation (3.3) we then see that if θ = π/2 then neces-sarily θ̇ = 0 and the particle remains in the equatorial plane. The Euler-Lagrangeequations with θ = π/2 simplify to

pφ ≡ r2φ̇ = constant = L, (3.6)

pt ≡(

1− 2Mr

)ṫ = E, (3.7)

pθ = 0. (3.8)

30

3.2 Time-Like Geodesics: Massive Particles

By rescaling the a�ne parameter we can write the Lagrangian so that

2L = 2gµνpµpν = m2, (3.9)

where gµν is the Schwarzschild metric tensor and pµ the canonical momenta described

above. In terms of energy and angular momentum the Lagrangian becomes,

2L = E2

1− 2M/r− ṙ

2

1− 2M/r− L

2

r2=

{0, massless particle

m2, massive particle. (3.10)

With this equation in hand we can start to determine geodesics of massive andmassless particles in the Schwarzschild metric. In all of our discussion in this sectionand the next, we assume a static observer at in�nity. We choose this reference framebecause at in�nity the local energy of a particle is equal to the conserved energymeasured at in�nity. The discussion below closely follows Chandrasekhar (1983).


For massive particles moving along time-like geodesics we write equation (3.10) inthe following way

(dr

dτ

)2= E2 −

(1− 2M

r

)(1 +

L2

r2

). (3.11)

Solving this ODE gives the geodesic along which the particle moves, but this ODEis non-linear so we hold o� on solving it. Note however that the above equationresembles equation (2.43), we have a radial velocity term on the left hand side andenergy and angular momentum terms on the right hand side. The radial velocity

is zero when E2 =(1− 2M

r

) (1 + L

2

r2

), this means we have a turning point. We

know from the previous discussion that when we equate energy to radial velocity,and therefore turning points, we have an e�ective potential. The e�ective potentialis then

Veff =

(1− 2M

r

)(1 +

L2

r2

). (3.12)

31


2 4 6 8 10 12 14r

0.85

0.90

0.95

1.00

Veff

L= 2√

3M

L= 3. 7M

L= 4M

rs

Figure 3.1: The e�ective potential for di�erent values of L.

Solving Veff −E = 0 gives us the turning points and after a little algebra we obtainthe equation,

(1− E2)r3 − 2Mr2 + L2r − 2ML2 = 0. (3.13)

The geometry of the geodesic now depends on the nature of the roots of this equation.To actually see how the geodesic develops with time or φ we need to �nd r(φ).Writing equations (3.6) and (3.7) in terms of τ

dφ

dτ=L

r2, (3.14)

dt

dτ=

E

1− 2M/r, (3.15)

then using (3.14) to switch to dr/dφ

(dr

dφ

)2= (E2 − 1) r

4

L2+

2M

L2r3 − r2 + 2Mr, (3.16)

and lastly with the same substitution as in the Newtonian section, u = 1/r, we

32


�nally get

(du

dφ

)2= 2Mu3 − u2 + 2M

L2u− 1− E

2

L2. (3.17)

Note that equations (3.16) and (3.17) are really just reformulations of eq. (3.13). Ifdr/tτ = 0 then we would also expect dr/dφ = du/dφ = 0 at the same time. Thetrick now is that if we �nd the roots of du/dφ (or dr/dτ and dr/dφ), we know theboundaries of r or u. Then we can switch to another variable and parameterize u sothat it takes its boundary values on the boundaries of our new variable. We shallstart our discussion with a radial geodesic of zero anuglar momentum.

3.2.1 Radial Geodesics

For a radial geodesic with zero angular momentum the radial velocity becomes,

(dr

dτ

)2=

2M

r− (1− E2). (3.18)

First we shall consider the case of a particle at in�nity falling from rest and then aparticle at speci�c distance, r0, falling from rest.

Infall from in�nity

A particle at rest at in�nity has an energy E = 1, because spacetime is �at atin�nity (Minkowski spacetime). For an infalling particle we want the radial velocityto be negative and so we take the minus root of eq. (3.18)

dr

dτ= −

√2M

r. (3.19)

Falling from in�nity takes an in�nite amount of time therefore we must integratefrom a starting distance r0 at τ = 0 which the particle has taken an in�nite amountof time to reach and also picked up some speed. Gathering the terms and integrating

33


we �nd

−√

2M

∫ τ0

dτ ′ =

∫ rr0

r12dr′ (3.20)

which yields a simple result

τ =2√

2M

3(r

3/20 − r3/2). (3.21)

This is the time as measured in the frame of the falling particle. What about anoutside observer at in�nity? We can �nd the coordinate time in the following way,

dt

dr=dt

dτ

dτ

dr, (3.22)

equations (3.15) and (3.19) then give,

dt

dr= −

√r

2M

E

1− 2Mr

. (3.23)

Integrating both sides we �nd the coordinate time,

t = − E√2M

∫ rr0

√r

1− 2Mr

. (3.24)

When r → 2M = rs the integral goes to in�nity and we are unable to integrate overit. This means that t→∞ as the particle approaches the horizon, rs.

Infall from a speci�c distance

At a speci�c radial distance r0 a particle's conserved energy is E =√

1− 2GM/r0.The radial velocity is then

dr

dτ= −

√1− 2M

r0−(

1− 2Mr

)= −

√2M

(1

r− 1r0

)=

√2M

r0

r0 − rr

, (3.25)

34


separating the variables and integrating

τ = −√

r02M

∫ rr0

(r

r0 − r

)1/2. (3.26)

Evaluating the integral we obtain a hefty equation,

τ =

√r0

2M

[√r(r − r0) + r0 arctan

(√r(r0 − r)r − r0

)+ r0

π

2

]. (3.27)

The proper time it takes to fall from r0 to the center at r = 0 is therefore

τ0 =π

22M

√r0

2M. (3.28)

The coordinate time is now given by

t = E

√r0

2M

∫ rr0

(r

r0 − r

)1/2(1− 2M

R

)−1, (3.29)

and again we have a singularity at r = 2M = rs so t→∞ as the particle approachesthe horizon.

3.2.2 Bound Orbits

Having considered geodesics of particles with zero angular momentum we now turnto particles with nonzero angular momentum. Looking at the coe�cients of equation(3.17) we see that only one of them can be either positive or negative. The term(1 − E2)/L2 is either positive or negative depending on the energy, E2. If E2 < 1then the term is positive but negative if E2 > 1. The sign determines both thenumber and nature of the roots and thus the number of allowed orbits. A particlewith an energy E2 < 1 is said to be bound and unbound if E2 > 1.

We start by looking at bound orbits, E2 < 1. In this case there are �ve di�erentcon�gurations of roots resulting in a total of eight di�erent types orbits.

35


Case A: Three distinct positive real roots.

Roots u1 and u2 correspond to an elliptical orbit on the interval u1 < u < u2.The third root u3 corresponds to a capture orbit from u

−13 that plunges into

the center at r = 0 (u→∞).

Case B: A double and single root.

The double root u = u1 = u2 corresponds to a stable circular orbit. The singleroot corresponds to a capture orbit from u−13 .

Case C: A Double and single root.

The single root corresponds to an orbit starting at u−11 which asymptoticallyspirals towards a circle with radius u−1 = u−12 = u

−13 . The other orbit starts

from u−1 = u−12 = u−13 and spirals asymptotically inwards away from the circle

in the opposite direction towards the center.

Case D: A Triple root

An unstable circular orbit.

Case E: A single real root and two complex conjugate roots

A capture orbit from u−11 (real). Can be described by an imaginary eccentric-ity. The imaginary eccentricity is obtained from the complex roots.

In cases A, B, C, and E we have, alongside the Kepler orbits (ellipses, parabolasetc.), capture orbits which so far have not been discussed (only brie�y mentionedin the Pseudo-Newtonian case). These capture orbits are called orbits of the secondkind in relativistic theory because they have no Keplerian analogues. The Kepleriantype orbits are then called orbits of the �rst kind.

Circular orbits in cases B and D are a good place to start since they are static inthe equatorial plane and we have already covered the techniques to �nd them. Thecircular orbits are orbits of the �rst kind. We'll cover the orbit of the second kindin case B separately.

Cases B and D: Orbits of the �rst kind

Recalling the condition for a circular orbit dVeff/dr = 0 at r = rc, that is,

dVeffdr

∣∣∣∣r=rc

=2M

r2c− 2L

2

r3c+

6ML2

r4c= 0, (3.30)

36


solving for rc

rc± =L2

2M

(1±

√1− 12M

2

L2

). (3.31)

We see that for L < 2√

3M no circular orbits exist. For L = 2√

3M only onecircular orbit exists at rc = 6M , and two if L > 2

√3M . Let's investigate the

stability of these orbits, recalling the second condition d2Veff/dr2 > 0, if stable and

d2Veff/dr2 < 0 if unstable. For rc = 6M , we �nd after some algebra

d2Veffdr2

∣∣∣∣rc=6M

= 0. (3.32)

So the circular orbit at rc = 6M is neither stable or unstable, it is called marginallystable. For L > 2

√3M the smaller radius rc− must be a local maximum and thus

unstable because we know that Veff → −∞ as r → 0 and Veff → 1 as r → ∞.By the same logic rc+ must be a minimum and thus stable. Therefore rc = 6Mis the limit of stable orbits, i.e. the innermost stable circular orbit. This is thesame result we found from the Pseudo-Newtonian potential, a testament to its goodapproximation.

From the above we have discerned the following

Stable: 6M ≤ rc+


Few examples of these circular orbits are marked as dots on �gure (3.1). Havingcovered circular orbits of massive particles we move on to the next cases.

Case A: Orbits of the �rst and second kind

Here we have tree distinct roots, two of them representing the apsides of an ellipse.The apsides can be expressed in Keplerian form as

u1 =1

α(1− e), u2 =

1

α(1 + e), (3.36)

where as before e is just the eccentricity and α the semi-latus rectum. But whatabout the third root? To determine it use Vieta's formulas. They relate a poly-nomial's coe�cients to the products and sums of its roots, i.e. for a polynomialP = anx

n + an−1xn−1 + · · ·+ a1x1 + a0 with roots xn Vieta's formulas are

x1 + x2 + · · ·+ xn−1 + xn = −an−1an

, x1 · x2 · ... · xn = (−1)na0an. (3.37)

The coe�cients of (3.17) are an−1 = −1, an = 2M , and a0 = (1−E2)/L2 so we �ndthat the third root is

u3 =1

2M− 2α. (3.38)

All the roots in subsequent cases follow the same format but with slight di�erencesdepending on the case. We now de�ne a useful quantity

µ ≡M/α, (3.39)

not to be confused with the reduces mass. Factoring equation (3.17)

(du

dφ

)2= 2M

(u− 1− e

α

)(u− 1 + e

α

)(u− 1

2M+

2

α

), (3.40)

38


we can now write its coe�cients in terms of µ and α

1

L2=

1

αM

(1− µ(3 + e2)

), and

1− E2

L2=

1

α2. (3.41)

Now we introduce the trick of parameterising u in a new variable. This new variable,χ, can be thought of as the relativistic anomaly, an analogue to the true anomaly.It maps the angle φ which in some cases can be unbounded, to a �nite manageableinterval. The substitution we make in this case is

u =1

l(1 + e cosχ), (3.42)

so that u(χ = 0) = (1 + e)/α and u(χ = π) = (1 − e)/α. Equation (3.17) can nowbe written in terms of dχ/dφ with µ and α, by inserting equations (3.39) and (3.40),then obtaining

(dχ

dφ

)2=(

1− 6µ+ 2µe− 4µe cos2 χ2

). (3.43)

Taking the root and factoring (1− 6µ+ 2µe) we get

dχ

dφ= (1− 6µ+ 2µe)1/2

(1− k2 cos2 χ

2

)1/2. (3.44)

Choosing the positive or negative root makes no di�erence since φ is a monotonicfunction. It only changes the direction. Here we have de�ned a constant,

k2 ≡ 4µe1− 6µ+ 2µe

, (3.45)

and note that φ can now be written as an elliptical integral with k2 as its parameter.This particular elliptical integral is incomplete and of the �rst kind. Before wecontinue we should make sure that k2 ≤ 1, else the integral becomes complex.

Recalling the order of the roots u1 < u2 < u3 we require

u3 =1

2M− 2α≥ 1α

(1 + e) = u2 or α ≥ 2M(3 + e), (3.46)

39


writing these inequalities in terms µ we �nd

µ ≤ 12(3 + e)

, and 1− 6µ− 2µe ≥ 0. (3.47)

If 1−6µ−2µe ≥ 0 then 1−6µ+2µe ≥ 4µe, and indeed k2 ≤ 1 is satis�ed. Whereverk appears, the same method can be applied to show k2 ≤ 1, but we'll only show itin this particular case. The integral

F (ψ, k2) =

∫ ψ0

dβ√1− k2 sin2 β

, (3.48)

is an incomplete elliptical integral of the second kind where ψ = (π−χ)/2 is chosenso that φ = 0 at the apoapsis, and is given by

φ =2

(1− 6µ+ 2µe)1/2F (1

2(π − χ), k). (3.49)

For each χ on the interval [0, 2n ·π] where n is the number of revolutions there existsa corresponding φ. Plugging φ into equation (3.40) then gives u = 1/r(φ). To �ndr(τ) we can then use equation (3.14).

This covers the orbit of the �rst kind, let's now determine the orbit of the secondkind. Its root can be written in the form

u3 =1

2M− 2α. (3.50)

Making a substitution to a new variable ξ of the form

u =

(1

2M− 2α

)+

(1

2M− 3 + e

α

)tan2 1

2ξ, (3.51)

ensures that u = u3 at ξ = 0 and u → ∞ (r → 0) as ξ → π. Going through thesame steps as with orbits of the �rst kind, the mapping now becomes

(dξ

dφ

)2= (1− 6µ+ 2µe)(1− k2 sin2 1

2ξ). (3.52)

40


Then integration gives

φ =2

1− 6µ+ 2µeF (1

2ξ, k), (3.53)

which ensures that φ = 0 at the apoapsis (now at ξ = 0). Figure (3.2) showscalculated orbits of both the �rst and second kind for case A.

0°

45°

90°

135°

180°

225°

270°

315°

1020

3040

50

(a) Orbit of the �rst kind.

0°

45°

90°

135°

180°

225°

270°

315°

0.51.0

1.52.0

2.53.0

3.5

(b) Orbit of the second kind.

Figure 3.2: Initial conditions: M = 1, r0 = 12, vθ0 = 0.4c. The red circle representsthe event horizon.

Cases B: Orbit of the second kind

This is an orbit of the second kind starting at u3 = 1/2M−2/α so the same equationsapply from case A. The range of u−13 is 3M ≤ u−13 ≤ 6M . Here however, we havee = 0 and therefore k2 = 0 and equation (3.52) simpli�es to

(dξ

dφ

)2= (1− 6µ)(φ− φ0). (3.54)

Integrating then gives

ξ =√

1− 6µ(φ− φ0), (3.55)

41


where φ0 is a constant of integration. The substitution we make now is

u =1

L+

(1

2M− 3L

)sec2(1

2ξ), (3.56)

and the particle arrives at r = 0 when φ − φ0 = π/(1 − 6µ)1/2. Figure (3.3) showstwo calculated orbits the second kind for case B.

0°

45°

90°

135°

180°

225°

270°

315°

12

34

5

(a) Orbit of the second kind.

0°

45°

90°

135°

180°

225°

270°

315°

12

34

56


Figure 3.3: Orbits of the second kind for case B. a) Initial conditions: M = 1 andr0 = 5. b) Initial conditions: M = 1 and r0 = 5.6. The red circle represents theevent horizon.

Case C: Orbits of the �rst and second kind

In this case the roots are

u1 =1− eα

, u2 = u3 =1 + e

α=

1

2M− 2α. (3.57)

Before we had that α ≥ 2M(3 + e), but now we require that α = 2M(3 + e). Theroots can then we written in the following way,

u1 =1− e

2M(3 + e), u2 = u3 =

1− e2M(3 + e)

. (3.58)

42


We know that 0 < e < 1, and so the ranges for u1 and u2 = u3 are

0 ≤ u1 ≤ 1/6M, (3.59)

and

1/6M ≤ (u2 = u3) ≤ 1/4M or 4M ≤ r ≤ 6M. (3.60)

This orbit has apsides at u1 and u2 = u3 so we can make the same substitution asin case A, but now we de�ne

u =1 + e

2M(3 + e), χ = 0 and u =

1− e2M(3 + e)

, χ = π. (3.61)

Equation (3.17) reduces to

(dχ

dφ

)2= 4µe sin2 1

2χ, (3.62)

and taking the root gives

dχ

dφ= −2√µe sin χ

2. (3.63)

We choose the negative sign to make φ increase as χ goes from π at apoapsis to 0at periapsis. We see that φ = 0 when χ = 0 and φ→∞ as χ→ 0. Integrating bothsides yields

φ = − 1√µe

ln(tanχ

4), (3.64)

Now for the orbit of the second kind. We have the same roots as before but now westart at u2 = u3 = (1 + e)/α and spiral into center in the opposite direction.

43


Making the substitution,

u =1

α(1 + e+ 2e tan2 1

2ξ), (3.65)

ensures that u = u2 = u3 = (1 + e)/α at ξ = 0 and u → ∞ at ξ = π. We have thesame roots as before so φ must change in exactly the same way, that is

φ = − 1√µe

ln(tan 14ξ). (3.66)

Figure (3.4) shows calculated orbits of both the �rst and second kind for case C.

0°

45°

90°

135°

180°

225°

270°

315°

510

1520


0°

45°

90°

135°

180°

225°

270°

315°

12

34


Figure 3.4: Initial conditions: M = 1 and r0 = 20. a) Orbit of the �rst kindstarting from r0 = 20 approaching r = 4.44. b) Orbit of the second kind startingfrom r = 4.44 spiraling towards the center. The red circle represents the eventhorizon.

44


Case E: Orbit of the second kind

Here we shall only cover the orbit of the second kind. The roots can be written ina Keplerian form

u1 =1

2M− 2α, u2 =

1

α(1 + ie) = u∗3, (3.67)

where u∗3 is the complex conjugate. Factoring equation (3.17)

(du

dφ

)2= 2M

(u− 1− ie

α

)(u− 1 + ie

α

)(u− 1

2M+

2

α

), (3.68)

we see that the only di�erence from equation (3.40) is the complex ie. The coe�-cients are therefore almost exactly the same as (3.41) but we must switch the signswherever we �nd e2, that is

1

L2=

1

αM(1− µ(3− e2)), 1− E

2

L2=

1

α2(1− 4µ)(1 + e2). (3.69)

The substitution we now make is,

u =1

α

(1 + e tan 1

2ξ), (3.70)

noting that u→∞ when ξ = π. The other boundary u = u1 is at

tan 12ξ0 =

α

e

(1

2M− 2α

)− 1 = −6µ− 1

2µe. (3.71)

With these substitutions we �nd that (3.17) becomes

(dξ

dφ

)2= 2 [(6µ− 1) + 2µe sin ξ + (6µ− 1) cos ξ] , (3.72)

45


taking the square root and factoring to obtain an elliptical integral; φ is found to be

φ = ± 1√4

∫ ψ0


, (3.73)

where ψ = [−π/2, π/2] and

k2 =1

24(4+ 6µ− 1), (3.74)

also

4 =√

(6µ− 1)2 + 4µ2e2. (3.75)

We want φ = 0 at χ = π but F (π, k) 6= 0 so we introduce a complete integral of the�rst kind,

K(k) =

∫ π/20


(3.76)

so that K(k)− F (ψ = π/2, k) = 0. Now φ given by

φ =1√4

(K(k)− F (ψ, k)). (3.77)

Figure (3.5) shows two calculated orbits the second kind for case E.

46


0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012


0°

45°

90°

135°

180°

225°

270°

315°

510

1520

2530


Figure 3.5: Capture orbits for case E. a) Initial conditions: M = 1, r0 = 10, andvθ0 = 0.02c. b) Initial conditions: M = 1, r0 = 30, and vtheta0 = 0.05c. The redcircle represents the event horizon.

3.2.3 Unbound Orbits

For unbound orbits the energy is E2 > 1 and the term (1−E2)/L2 is negative. Wethen get three di�erent con�gurations of roots resulting in four di�erent orbits. Thecases are the following

Case A: Two positive and one negative

A hyperbolic orbit on the interval 0 < u < u2 and a capture orbit from u3

Case B: Double positive and one negative

An orbit from in�nity asymptotically spiraling towards u2 = u3 and the othera continuation from u2 = u3 spiraling towards the center

Case C: One Negative and a complex-conjugate pair

A capture orbit arriving from in�nity described by an imaginary eccentricity.

Many of these orbits are solved in the same way as the bound orbits but withdi�erent boundary conditions.

47


Case A: Orbit of the �rst and second kind

In this case all the roots are real but one of them is negative. In Keplerian formthey are

u1 = −e− 1α

, u2 =e+ 1

α, u3 =

1

2M− 2α. (3.78)

These roots are very similar to case A for the bound orbits but now we require thate ≥ 1. Factoring equation (3.17) we �nd the coe�cients

1

L2=

1

αM(1− µ(3 + e2)) E

2 − 1L2

=1

α(1− 4µ)(e2 − 1). (3.79)

Making the same substitution as in case A but now with boundary conditions u = u3at χ = π and u = 0 at cosχ = −1/e. Following the same steps as in A we obtain

φ =2√

1− 6µ+ 2µe(K(k)− F (1

2(π − ψ), k)), (3.80)

where K(k) ensures that φ = 0 at χ = 0. The capture orbit for this case is describedin the same manner as the bound orbit in case A. The only di�erence is that e ≥ 1and it starts from u = u2. Figure (3.6) shows calculated orbits of both the �rst andsecond kind for case A.

0°

45°

90°

135°

180°

225°

270°

315°

2 46 8

10121416


0°

45°

90°

135°

180°

225°

270°

315°

0.51.01.52.0

2.53.03.54.0


Figure 3.6: Initial conditions: M = 1, r0 = 8, and vθ0. The red circle represents theevent horizon.

48


Case B: Orbit of the �rst and second kind

Here the roots u2 and u3 coincide so that u2 = (e+1)/α = u3 = 1/2M−2/α. Againthis is described in the same manner as the bound orbits in case C above, but witha new boundary u = 0 at χ = arccos(−1/e). There is also an additional limit of1 ≤ e ≤ 3 and so the double root is limited to 3M ≤ (u−12 = u−13 ) ≤ 4M . Figure(3.7) shows calculated orbits of both the �rst and second kind for case B.

0°

45°

90°

135°

180°

225°

270°

315°

24

68


0°

45°

90°

135°

180°

225°

270°

315°

0.51.01.52.0

2.53.03.5


Figure 3.7: Initial conditions: M = 1 and r0 = 2.8. The red circle represents theevent horizon.

Case C: Orbit of the second kind

Here we have a case a capture orbit described by an imaginary eccentricity. Just likeprevious cases this can be described by the bound orbit in case E but with di�erentboundary conditions. The new boundary condition is tan ξ/2 = −1/e when u = 0.Figure (3.8) shows two calculated orbits of both the �rst and second kind for caseC.

49


0°

45°

90°

135°

180°

225°

270°

315°

0.51.0

1.52.0


0°

45°

90°

135°

180°

225°

270°

315°

0.51.0

1.52.0


Figure 3.8: Unbound orbits for case C. Parameters from Chandrasekhar (1983).a) Initial conditions: M = 0.3, e = 0.001i, and α = 1. b) Initial conditions:M = 0.3, e = 0.1i, and α = 1. The red circle represents the event horizon.

3.3 Null Geodesics: Massless Particles

In this section we will only use the method of parameterizing in one special case.Otherwise we solve a non-linear ODE and also make use of an e�ective potential.

For a massless paricle, the Lagrangian in equation (3.10) is,

L = E2

1− 2M/r− ṙ

2

1− 2M/r− L

2

r2. (3.81)

Solving for ṙ2 and recalling that L = 0 for massless particles, we �nd the radialvelocity is given by

(dr

dλ

)2= E2 − L

2

r2

(1− 2M

r

). (3.82)

50


Here we see that the e�ective potential for a massless particle such as a photon is

Vphoton =L2

r2

(1− 2M

r

). (3.83)

0 5 10 15 20r

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

Vphoton/L

2

rs

Figure 3.9: The photon e�ective potential.

First let's consider a photon with zero angular momentum falling from a distancer0.

Radial Infall

With zero angular momentum the radial velocity becomes

(dr

dλ

)2= E2, (3.84)

and knowing the energy E = ṫ(1− 2M/r) from equation (3.7) we �nd

dr

dt= −

(1− 2M

r

), (3.85)

where we have chosen the negative root. Integrating from a speci�c distance r0 to

51


r, we get the coordinate time,

t = r0 − r + ln(r0 − 2Mr − 2M

). (3.86)

We see clearly that t→∞ when r → 2M = rs as we would expect.

Circular Orbits

Using the e�ective potential we can easily determine circular orbits, recalling thecondition for a circular orbit and evaluating

dVphotondr

∣∣∣∣r=rc

= −2L2

r3c+

6ML2

r4c= 0, (3.87)

we �nd only one solution,

rc = 3M =3

2rs. (3.88)

Circular orbits of massless particles therefore only exist at r = 3M . Let's see whetherthis orbit is stable or not. Taking the second derivative we �nd,

d2Vphotondr2

∣∣∣∣r=rc

=6L2

r4c− 24ML

2

r5c= − 2L

2

81M4< 0, (3.89)

so the circular orbit is unstable, a slight perturbation either sends it �ying o� orspiraling in.

Critical Orbit

In case D for bound time-like geodesics, we had a triple root solution to equation(3.17) which resulted in an unstable circular orbit. We have shown that an unstablecircular orbit exists for massless particles but what about asymptotic spirals as incase C? To answer this question, we should write the radial radial velocity equation

52


analogous to equation (3.17). Switching to u = 1/r and noting that dφ/dλ = L/r2

we obtain

(du

dφ

)2= 2Mu3 − u2 + 1

D2, (3.90)

where D ≡ L/E represents an impact parameter. Here the constant term 1/D2 isalways positive so this is analogous to the unbound time-like geodesics where indeedthere exists a double root. Where might the double roots be in this case? In generala double root must must ful�ll two conditions: �rst f(u = ur) = 0, and secondf ′(u = ur) = 0, where f(u) is a polynomial of degree 2 or higher and ur is a root.Checking the second condition �rst we take the derivative of (3.90)

d

du

(du

dφ

)2= 6Mu2 − 2u = 2u(3u− 1), (3.91)

so u = 1/3M is a double root if

(du

dφ

)2∣∣∣∣∣u=1/3M

= 2M

(1

3M

)3−(

1

3M

)2+

1

D2= 0, (3.92)

that is, u = 1/3M is a double root if Dc = 3√

3M , called the critical impactparameter. If D > Dc the photon escapes but is captured if D < Dc. The singleroot is then u1 = −1/6M (from Vieta's formulas), factoring the equation we get

(du

dφ

)= 2M

(u+

1

6M

)(u− 1

3M

)2. (3.93)

We now make the substitution

u = − 16M

+1

2Mtanh2(φ− φ0), (3.94)

and set tanh2 12φ0 = 1/2 so that u = 0 when φ = 0, and also u = 1/3M as φ→∞.

53


We also have an associated orbit of the second kind starting at u = 1/3M . Thesubstitution here is similar to the ones in the time-like geodesics, namely

u =1

3M+

1

2Mtan2 1

2ξ, (3.95)

so equation (3.90) becomes

(dξ

dφ

)2= sin2

1

2ξ. (3.96)

Integrating then gives

φ = 2 ln(tan1

4ξ), (3.97)

and u→∞ at φ = 0 (ξ = π) and u = 1/3M as φ→∞ (ξ = 2π).

3.3.1 Cone of avoidance and General Orbits

The goal in all previous sections was to set a particle at some initial distance r0,with initial velocity v0, and calculate the resulting orbit. It is clear that the anglebetween a particle's velocity vector and radial vector plays a crucial role in the shapeof its trajectory. If a massive particle, shot at an angle falls to the center, we couldalways increase its velocity (to a point in relativity) and potentially alter its orbit.Massless particles however, always travel at the speed of light in every referenceframe so their trajectories are highly dependent upon the initial angle and distance.The radial and tangential components of massive particles are given by

vφ = cos θ = r0dφ

dtvr = cosφ =

dr

dt, (3.98)

where θ is the initial angle between v and r. Using equations (3.14) and (3.15), wecan write the tangential velocity, as measured from a static observer at in�nity, inthe following way

vφ =1

r· LE

(1− 2M

r

)1/2=D

r

(1− 2M

r

)1/2, (3.99)

54


where D is the impact parameter as before. A massless particle is captured ifD < Dc, so an inward moving particle escapes if

vφ = sin θ >Dcr

(1− 2M

r

), (3.100)

and an outward moving particle escapes if D > Dc, that is if

vφ = sin θ <Dcr

(1− 2M

r

). (3.101)

Here the angle θ can be thought of as the angle of a cone symmetric about the radialdirection. When r > 3M , the angle of the cone is θ > 90◦, meaning its base pointstowards the center. At r = 3M the angle is θ = 90◦ so the cone opens up andessentially forms a vertical barrier. Any forward motion (θ > 90◦) at this distanceresults in a capture. Then at r > 3M , θ < 90◦ so the cone opens in on itself andpoints outward. The cone of avoidance is depicted in �gure (3.10).

Figure 3.10: The cone of avoidance at di�erent distances [9].

Now let us try a di�erent method to determine the null-geodesics. Taking thederivative of both sides of equation (3.90) with respect to du/dφ, we obtain

2du

dφ

d2u

dφ2= 6Mu2

du

dφ− 2udu

dφ. (3.102)

This reduces to

d2u

dφ2− 3Mu2 + u = 0, (3.103)

55


a non-linear second order ODE which we will now solve numerically. But �rst wemust determine the initial conditions. The �rst initial condition is simply u0 = 1/r0but for the derivative du/dφ at u0, the initial conditions are not so obvious. Thetangential and radial velocity components are,

vφ = sin θ = rdφ

dt, (3.104)

and

vr = cos θ =dr

dt, (3.105)

where θ is the initial angle. Noting that u0 = 1/r0 and dr = −du/u2 we can writethe initial velocities in terms of u0

du

dt

∣∣∣∣u=u0

= −u20dr

dt= −u20 cos θ (3.106)

and

dφ

dt

∣∣∣∣u=u0

= u0 sin θ. (3.107)

Now we can write

du

dφ

∣∣∣∣u=u0

=

dudt

∣∣u=u0

dφdt

∣∣u=u0

=−u20 cos θu0 sin θ

= − u0tan θ

, (3.108)

and the second initial condition is determined. Figure (3.11) shows orbits of masslessparticles at an initial distance with varying initial angles.

56


0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(a) θ = 170◦.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(b) θ = 160◦.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(c) θ = 150◦.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(d) θ = 140◦.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(e) θ = 130◦.

0°

45°

90°

135°

180°

225°

270°

315°

24

68

1012

(f) θ = 120◦.

Figure 3.11: Solutions of equation (3.103), with M = 1, r0 = 12, and varying θ.

57

4 Orbits in the Kerr Metric

The Kerr metric, like the Schwarzschild metric, is a solution of Einstein's �eldequations and describes spacetime around a black hole with mass M , but unlikeSchwarzschild metric the black hole now has an angular momentum J . It is a gen-eralization of the Schwarzschild metric in the sense that if we set J = 0 it reducesto the Schwarzshild metric. Also, the Kerr metric is more likely to describe realblack holes, since almost all astronomical objects rotate. The addition of the angu-lar momentum however, severely complicates calculations. The metric is no longerspherically symmetric but axially symmetric around the spin axis. We must take intoconsideration the direction of orbital trajectories relative to the spin direction. Forthis reason we limit our discussion in this section to circular orbits in the equatorialplane.

4.1 The Metric

The Kerr metric, expressed in Boyer-Lindquist coordinates, for a rotating black holewith angular momentum J and mass M is

ds2 = −(

1− 2MΣ

)dt2 − 4aMr sin

2 θ

Σdtdφ+

Σ

∆dr2

+Σdθ2 +

(r2 + a2 +

2Mra2 sin2 θ

Σ

)sin2 θdφ2.

, (4.1)

Here the black hole rotates in the φ direction, the terms a, Σ, and ∆ are de�ned as

a ≡ JM, ∆ ≡ r2 − 2Mr + a2, Σ ≡ r2 + a2 cos2 θ. (4.2)

Note that when a = 0, the Kerr metric reduces to the Schwarzschild metric. Theevent horizon is situated at the point where the sign of the dr term changes, i.e. at

59


∆ = 0. Solving ∆ = 0 for r now gives two horizons,

r± = M ±√M2 − a2. (4.3)

The outer horizon, r+, is the event horizon while the inner one, r−, is called a Cauchyhorizon. We see that a < M , else a black hole can not exist. A black hole witha = M is called a maximally rotating black hole. When a < M the event horizonr+ is smaller than the Schwarzschild radius, and if a = 0 then r+ = rs. The horizonof a rotating black hole is thus smaller than a stationary one.

4.2 The Ergosphere

Consider a photon orbiting in the equatorial plane, so that ds2 = 0, and dθ = dr = 0.The metric is then,

−(

1− 2MrΣ

)dt2 − 4Mar sin

2 θ

Σdtdφ+

(r2 + a2 +

2Mra2 sin2 θ

Σ

)sin2 θdφ2 = 0.

(4.4)

Dividing by dt2 we can write a quadratic equation in terms of φ̇ [11], that is,

Aφ̇2 +Bφ̇+ C, (4.5)

Where the coe�cients are

A = sin2 θ

(r2 + a2 +

2Mra2 sin2 θ

Σ

)(4.6)

B = −4Mar sin2 θ

Σ, (4.7)

and

C = −(

1− 2MrΣ

). (4.8)

60

4.3 Circular Orbits

The solution are then given by

dφ

dt=−B ±

√B2 − 4AC2A

. (4.9)

Note that dφ/dt = 0, when C = 0 for the positive root. This means that masslessparticles such as photons can be stationary at certain places when moving counterto the black hole's rotation. Solving C = 0 for r we �nd the distance to be

r0 ≡M +√M2 − a2 cos2 θ. (4.10)

This forms a region known as the ergosphere. Even light cannot be stationary insideof this boundary, everything is forced to rotate along the direction of rotation. Fromthis we can conclude that no static observers can exist inside the ergospohere.

4.3 Circular Orbits

Before considering circular geodesics we should �rst determine the Lagrangian. Let'sconsider the Lagrangian of particle moving in the equatorial plane, θ = π/2 andθ̇ = 0. In that case we �nd the Lagrangian to be

2L = −(

1− 2Mr

)ṫ2 − 4Ma

rṫφ̇+

r2

∆ṙ2 +

(r2 + a2 +

2Ma2

r

)φ̇2. (4.11)

The Euler-Lagrange equations now give

−pt =(

1− 2Mr

)ṫ+

2Ma

rφ̇ = E, (4.12)

pφ =2Ma

rṫ+

(r2 + a2 +

2Ma2

r

)φ̇ = L, (4.13)

pr =r2

∆ṙ. (4.14)

61


The goal is to �nd an e�ective potential, so we must equate the radial velocity tothe energy. Solving for ṫ and φ̇ we obtain

ṫ =(r3 + a2r + 2Ma2)E − 2MaL

r∆, (4.15)

φ̇ =(r − 2M)L+ 2MaE

r∆. (4.16)

Inserting these results into the the Lagrangian and remembering that 2L = 2gµνpµpν =

m2, we �nd our radial velocity equation

(dr

dλ

)2= E2 +

2M

r2(aE − L)2 + (a

2E2 − L2)r2

− m2∆

r2. (4.17)

Here we could de�ne our e�ective potential as the right-hand side of the equationabove, but in the Schwarzschild metric we de�ned the e�ective potential as the valueof E2 which made dr/dλ = 0, or Veff = E

2. We'll do the same here for consistency,writing equation (4.17) slightly di�erently,

r3(dr

dλ

)2= E2(r3 + a2 + 2Ma2)− 4MaEL− (r − 2M)L2 −m2r∆, (4.18)

and solving for E we obtain

E = Vkerr± ≡−B ±

√B2 − 4AC2A

, (4.19)

which agrees with Pugliese et al. (2011). The coe�cients are

A = r3 + a2 + 2Ma2, (4.20)

B = −4MaL, (4.21)

C = −(r − 2M)L2 −m2r∆. (4.22)

62

4.3 Circular Orbits

Here, and in the subsequent discussion, the plus signs corresponds to a corotatingorbit, and the minus sign to a counterrotating orbit.

4.3.1 Null Geodesics: Massless particles

As before we require Vkerr± = 0 for circular orbits. The circular orbits in theequatorial plane are, from Bardeen et al. (1972),

rphoton± = 2M(1 + cos[

23

arccos(∓a/m)]). (4.23)

We note that for a 6= 0 then r+ < r−. These orbits are also unstable like in theSchwarzschild metric, we can see this by plotting both potentials and overlayingrphoton±, as shown on �gure (4.1).

1 2 3 4 5 6 7 8 9 10r

0.4

0.2

0.0

0.2

0.4

0.6

0.8

Energ

y

rph+ rph−Vkerr+

Vkerr−

Figure 4.1: The positive and negative potentials for a photon with L = 3, around arotating black hole of mass M = 1, with a = 0.5. The Black dashed line is r+ andthe dotted line is r0 at θ = π/2.

4.3.2 Time-Like Geodesics: Massive Particles

For massive particles it is clear that circular orbits can only exist for r > rphoton±.According to Bardeen et al. (1972), stable circular orbits only exist for r > rms±,where rms± is the radius of a marginally stable circular orbit (innermost stable

63


circular orbit) in either direction, given by

rms = M(3 + Z2 ∓ ((3−

orbits of particles around compact objects: from newtonian theory … · 2018. 10. 15. · one...

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