Orbits of Particles Around Compact Objects: From Newtonian Theory to General Relativity
Matthías Ásgeir Jónsson
FacultyFaculty of of Physics Physics UniversityUniversity of of Iceland Iceland 20162016
ORBITS OF PARTICLES AROUND COMPACT OBJECTS: FROM NEWTONIAN THEORY TO GENERAL RELATIVITY
Matthías Ásgeir Jónsson
16 ECTS thesis submitted in partial ful llment of a Baccalaureus Scientiæ degree in Physics
Advisor Gunnlaugur Björnsson
Faculty of Physics School of Engineering and Natural Sciences University of Iceland Reykjavik, May 2016 Orbits of Particles Around Compact Objects: From Newtonian Theory to General Rela- tivity 16 ECTS thesis submitted in partial ful llment of a B.Sc. degree in Physics
Copyright c 2016 Matthías Ásgeir Jónsson All rights reserved
Faculty of Physics School of Engineering and Natural Sciences University of Iceland Hjarðarhagi 2-6 107, Reykjavík Iceland
Telephone: 525 4000
Bibliographic information: Matthías Ásgeir Jónsson, 2016, Orbits of Particles Around Compact Objects: From Newtonian Theory to General Relativity. B.Sc. thesis, Faculty of Physics, University of Iceland. 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. Orbits are calculated from a set of initial conditions, using dierent methods such as nding roots of an eective potential, by parameterizing, and solving linear and non-linear dierential 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 ...... 3 2.2 Newtonian Mechanics ...... 5 2.2.1 Kepler's First Law Derived ...... 7 2.2.2 Kepler's Second Law Derived ...... 9 2.2.3 Kepler's Third Law Derived ...... 10 2.3 The Lagrangian ...... 11 2.3.1 Kepler's rst law derived ...... 12 2.3.2 The Eective Potential ...... 15 2.3.3 Open and Closed Orbits ...... 17 2.3.4 Calculated Orbits ...... 18 2.3.5 Orbital Velocities ...... 20 2.3.6 Other Central Forces ...... 21 2.4 A Pseudo-Newtonian Potential ...... 24
3 Orbits in the Schwarzschild Metric 29 3.1 The Metric ...... 29 3.2 Time-Like Geodesics: Massive Particles ...... 31 3.2.1 Radial Geodesics ...... 33 3.2.2 Bound Orbits ...... 35 3.2.3 Unbound Orbits ...... 47 3.3 Null Geodesics: Massless Particles ...... 50 3.3.1 Cone of avoidance and General Orbits ...... 54
4 Orbits in the Kerr Metric 59 4.1 The Metric ...... 59 4.2 The Ergosphere ...... 60 4.3 Circular Orbits ...... 61 4.3.1 Null Geodesics: Massless particles ...... 63 4.3.2 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 function of orbital period squared. Data from Nasa [2]...... 5
2 2 2.2 The eective potential, Veff , and its components, −k/r and L /2µr . 16
2.3 An Earth like system. Initial conditions: m1 = M , m2 = 3 × −6 , , and . Resulting in a 10 M r0 = 0.9832AU vθ0 = 6.3853AU/yr nearly circular orbit...... 19
2.4 Initial conditions: , , , and m1 = M m2 = 0.1M r0 = 1AU vθ0 = 6AU/yr. Resulting in an elliptical orbit with e = 0.17...... 19
2.5 Initital conditions: , , . Result- m1 = m2 = M r0 = 1 vθ0 = 15AU/yr ing in a hyperbolic orbit with e = 1.848...... 20
2.6 Radial velocity curves due to a particle with mass m2 = 9.54 × −4 10 M = MJupiter placed at r0 = 0.5AU with vθ = 6.35AU/yr, around m1 = M ...... 21
−10 2.7 Radial velocity curves due to a particle with mass m2 = 3×10 M = 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 represents
the 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 eective potential for dierent values of L...... 32
ix LIST OF FIGURES
3.2 Initial conditions: , , . The red circle M = 1 r0 = 12 vθ0 = 0.4c represents the event horizon...... 41
3.3 Orbits of the second kind for case B. a) Initial conditions: M = 1 and r0 = 5. b) Initial conditions: M = 1 and r0 = 5.6. The red circle represents the event horizon...... 42
3.4 Initial conditions: M = 1 and r0 = 20. a) Orbit of the rst kind starting from r0 = 20 approaching r = 4.44. b) Orbit of the second kind starting from r = 4.44 spiraling towards the center. The red circle represents the event horizon...... 44
3.5 Capture orbits for case E. a) Initial conditions: M = 1, r0 = 10, and . b) Initial conditions: , , and . vθ0 = 0.02c M = 1 r0 = 30 vtheta0 = 0.05c The red circle represents the event horizon...... 47
3.6 Initial conditions: , , and . The red circle represents M = 1 r0 = 8 vθ0 the event horizon...... 48
3.7 Initial conditions: M = 1 and r0 = 2.8. The red circle represents the event 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) Initial conditions: M = 0.3, e = 0.1i, and α = 1. The red circle represents the event horizon...... 50
3.9 The photon eective potential...... 51
3.10 The cone of avoidance at dierent 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, around a rotating black hole of mass M = 1, with a = 0.5. The Black dashed line 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 the foundation of astrophysics. Understanding of orbits is vital to elucidate the behavior of astrophysical phenomena such as planetary systems, accretion disks, and black holes. In this paper we focus on orbits of test particles around stars and compact objects such as black holes. The goal is to calculate orbital trajectories of particles from a set of initial conditions. Each chapter gets increasingly more complex as we move towards orbits in general relativity. The aim is then to introduce core concepts in the earlier chapters for use in the later ones. In all discussions, we consider only systems of two particles, i.e the two body problem, which can be reduced to only one orbiting particle.
We begin with a discussion on Kepler's laws of planetary motion and then derive them 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 all possible orbits there are covered. For the Kerr metric we only cover circular orbits in the equatorial plane.
1
2 The Kepler Problem
In this chapter we focus solely on Kepler orbits (or Keplerian orbits). Kepler orbits are solutions of the two body problem and describe the motion of two bodies around a common center of mass due to their gravitational attraction. We treat the orbiting bodies as point particles and neglect other eects such as perturbations from external bodies, radiation pressure, non-uniform gravity, and general relativity. Kepler orbits are thus solutions of a highly idealized version of the two body problem called the Kepler problem. We also assume, in all cases, that our orbits are stable, i.e. will not decay over time. The orbits, as we shall see, can only take on 4 dierent shapes depending on the initial conditions. These shapes (or curves) are known as conic sections. 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 observational data was put forth by the German mathematician Johannes Kepler. After the death of Tycho Brahe, the foremost naked-eye-observer at the time, Kepler inherited his massive amount of data and over a period of about 20 years put forth three laws of planetary motion famously attributed to him [1]. These three laws, known as Kepler'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 time intervals.
Kepler's Third Law The Harmonic Law The square of the orbital period of a planet is proportional to the cube of the planet's average distance from the Sun.
3 2 The Kepler Problem
The laws are described above in a qualitative manner since they were deduced from purely empirical data. We can describe them more quantitatively in mathematical form.
Kepler's First Law
a(1 − e2) r = , (2.1) 1 + e cos θ
Kepler's Second Law
∆A = constant, (2.2) ∆t
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 distance from the sun (the semi-major axis) and e is the orbit's eccentricity - an indicator of its shape/structure. The angle θ is the true anomaly, measured from the planet's closest 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 P is 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 from observations alone.
Using Kepler's third law we can determine a planet's relative distance from the sun if we know its orbital period in terms of Earth's. For example, Mars' orbital period is
Tmars = 1.8808TL and so its average distance from the sun is amars = 1.523AU. At the time of Kepler the Astronomical Unit was unknown and was determined many years after his death. The orbital parameters of the planets in our solar system have been independently measured and agree with Kepler's laws, as shown in gure (2.1).
4 2.2 Newtonian Mechanics
2.0 ] 3 1.5 Neptune U A
[ Uranus 3 a
e 1.0 Saturn c n
a Jupiter t s i
e 0.5 g a
r Mars e
v Earth a 0.0
f
o Venus
e
b Mercury u 0.5 C
1.0 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]
Figure 2.1: A logarithmic plot of the cube of the average distance as a function of orbital period squared. Data from Nasa [2].
2.2 Newtonian Mechanics
In order to better understand planetary motion and the forces underlying them we need Newtonian Mechanics. Isaac Newton was able to deduce and generalize Kepler's laws using his three laws of motion and the law of universal gravitation.
Newton's First Law The Law of Inertia An object at rest will remain at rest and an object in motion will remain in motion in a straight line at a constant speed unless acted upon by an external force.
Newton's Second Law The net force acting upon an object of mass m is proportional to the object's mass and its resultant acceleration
n X dv dp F = F = ma, F = m = . (2.4) net i net dt dt i=1 Where v is the velocity vector, a the acceleration vector an p the momentum vector.
Newton's Third Law For every action there is an equal and opposite reaction,
F12 = −F21. (2.5)
5 2 The Kepler Problem
Newton's Law of Universal Gravitation m m F 1 2 ˆr (2.6) 12 = −G 2 12, |r12| 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 easiest reference frame to work with is that of the Center-of-Mass. Imagine some arbitrary reference frame with the vectors r1 and r2 pointing from the origin towards their respective 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 weighted average of the masses and their position.
m r + m r R ≡ 1 1 2 2 ,M = m + m . (2.7) M 1 2
Rewriting eq. (2.7) and dierentiating 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 describe the 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-mass reference frame is an inertial frame. We can simplify things even further by choosing a coordinate system where R = 0, eq. (2.7) then becomes
m r + m r 1 1 2 2 = 0. (2.10) M
Remembering that r = r2 − r1 gives,
m m r = − 2 r, r = 1 r. (2.11) 1 M 2 M
6 2.2 Newtonian Mechanics
We de ne a useful term, the reduced mass,
m m µ ≡ 1 2 . (2.12) m1 + m2
The vectors r1 and r2 are then
µ µ r1 = − r, r2 = r. (2.13) m1 m2
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,
1 Mµ E = µv2 − G , (2.14) 2 r 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 eect of gravity on the or- bital angular momentum L. In the center-of-mass coordinates the orbital angular momentum becomes
L = r × µv = r × p. (2.15)
Taking the time derivative we get,
dL dr dp = × p + r × = v × p + r × F. (2.16) dt dt dt
Since v and p are in the same direction their cross product is zero. Also, F is a central force and thus directed inward along r, the cross product of r and F is therefore zero. We then have an important result regarding angular momentum,
dL = 0, (2.17) dt
7 2 The Kepler Problem that is, angular momentum is conserved in a central force eld. We also note that the 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. Now the position vector becomes r = rˆr and the angular momentum can be written
d L = µr2ˆr × ˆr. (2.18) dt
The force experienced by the reduced mass is given by Newton's law of gravitation, expressed in polar coordinates F GMµ r, the acceleration is = − r2 ˆ
GM a = ˆr. (2.19) r2
Taking the cross product of a and L gives
GM d d a × L = − ˆr × µr2ˆr × ˆr = GMµˆr × ˆr × ˆr . (2.20) r2 dt dt
Using the triple product expansion A × (B × C) = (A · C)B − (A · B)C we nd
d d a × L = −GMµ ˆr · ˆr ˆr − (ˆr · ˆr) ˆr , (2.21) dt dt which simpli es to
d a × L = GMµ ˆr, (2.22) dt where the rst term in the bracket is zero since r d r r d r d r r d r r 2 ˆ · dtˆ = ˆ· dtˆ+ dtˆ·ˆ = dt (ˆ·ˆ) = 0 and ˆr · ˆr = 1.
Integrating both sides of eq. (2.22) with respect to time and noting that dL and dt = 0 a dv we obtain = dt
v × L = GMµˆr + D, (2.23)
8 2.2 Newtonian Mechanics where D is a constant vector due to the inde nite integration. This vector is known as the Laplace-Runge-Lenz vector (or simply LRL vector) and is a constant of motion in the Kepler problem. It lies in the orbital plane and its magnitude determines the orbit's eccentricity.
Finally taking the dot product
r · (v × L) = GMµr(ˆr · ˆr) + 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 = GMµr + r|D| cos θ, (2.26) µ solving for r we get Kepler's rst law
L2/µ2 r = . (2.27) |D| GM 1 + GMµ cos(θ)
We now de ne the eccentricity and L2/µ2 as the semi-latus e ≡ |D|/GMµ α ≡ GM rectum to nally nd
α r = . (2.28) 1 + e cos(θ)
2.2.2 Kepler's Second Law Derived
Kepler's second law describes how an area swept by an orbiting body over a time interval is constant, i.e. ∆A . In order to evaluate that constant we ∆t = constant
9 2 The Kepler Problem consider an in nitesimal area element in polar coordinates
dA = dr(rdθ) = rdrdθ. (2.29)
The radius vector sweeps out an area
1 ∆A = r2dθ, (2.30) 2 dividing by ∆t we obtain an areal velocity
∆A 1 dθ = r2 . (2.31) ∆t 2 dt
The angular velocity is related to angular momentum 2 dθ so we can write |L| = µr dt
∆A 1 |L| = , (2.32) ∆t 2 µ 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 a total area
1 |L| A = P. (2.33) 2 µ
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
4π2a2b2µ2 P 2 = (2.35) L2 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 between the purely geometrical formula of an ellipse, equation (2.1), and the one derived from Newtonian mechanics, equation (2.27). From which we can derive the angular momentum
|L| = µpGMa(1 − e2). (2.37)
Inserting this into eq (2.35) we nally have Kepler's third law,
4π2 P 2 = a3. (2.38) GM
If we express the mass in solar masses, distance in astronomical units, and time in years then the ratio 4π2/GM = 1. The reason is that for all planets in our solar ∼ system 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 for an 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 gravitational attraction. As shown in the previous section this system can be reduced to a two dimensional 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,
1 Mµ E = µ(r ˙2 + r2θ˙2) − G , (2.39) 2 r
11 2 The Kepler Problem
˙ where vr =r ˙ and vθ = rθ. In polar coordinates the Lagrangian can be written
1 Mµ L = µ(r ˙2 + r2θ˙2) + G . (2.40) 2 r
The Euler-Lagrange equations then give us the equations of motion in the coordi- nates r and θ which are,
M r¨ − rθ˙2 + G = 0, (2.41) r2 and
µr2θ˙ = constant = L. (2.42)
Equation (2.42) shows again the conservation of angular momentum. The position of the particle as function of time can be found by solving eq. (2.41). This is a non linear second-order ODE which can be solved numerically. Another method is to solve for r˙ in eq. (2.39) which gives
s 2 Mµ L2 r˙ = ± E + G − . (2.43) µ r µ2r2
Solving for dt and integrating gives t = t(r) which can then be inverted to obtain r = 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 attention on 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
1 du 1 dθ du θ˙ du L du r˙ = − = − = − = − , (2.44) u2 dt u2 dt dθ u2 dθ µ dθ
12 2.3 The Lagrangian the last step follows from equation (2.42). For r¨ we dierentiate a second time
d L du L d2u L2 d2u r¨ = − = −θ˙ = −u2 , (2.45) dt µ dθ µ dθ2 µ dθ2 and writing rθ˙2 in terms of u
1 L 2 L2 rθ˙2 = u2 = u3, (2.46) u µ µ we can now write eq. (2.41) in our new variable
d2u Mµ2 + u − G = 0. (2.47) dθ2 L2
This is just the equation of a simple harmonic oscillator, its general solution is
Mµ2 u = A cos θ + B sin θ + G , (2.48) L2 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 is zero in the center-mass frame. We can de ne such a point at θ = 0 for convenience
du = −B cos(0) = 0, (2.49) dθ θ=0
leading to B = 0. Further setting an initial condition u(θ) = 1/r0 at θ = 0
2 Mµ 1 (2.50) u(θ = 0) = A cos(0) + B sin(0) + G 2 = , L r0 and so
2 1 Mµ (2.51) A = − G 2 . r0 L
13 2 The Kepler Problem
Rearranging terms and solving for r we obtain
L2/µ2 r = , (2.52) AL2 GM(1 + GMµ2 cos θ) which is Kepler's third law and like before we have L2/µ2 but now the ec- α = GM centricity is e = AL2/GMµ2. In the previous section the eccentricity was de ned e = |D|/GMµ. Let's explore how |D| and A relate to one another. As mentioned D is the LRL vector - a conserved quantity in the Kepler problem. It is generally de ned
A ≡ p × L − GMµ2ˆr, (2.53) we see from equation (2.23) that |D| = |A|/µ. Multiplying A by L2 and writing L in terms of initial parameters we nd
L2 (µr v )2 2 2 0 θ0 2 (2.54) AL = − GMµ = = µvθ0 L − GMµ , r0 r0 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. This means we have seven conserved quantities, three each (in 3 dimensions) for the vectors A and L and one for a scalar E. These constants are functions of r and p which each have 3 independent components of motion. In all we have seven conserved quantities and six components of motion. One component of motion must give information about where a particle is at any given time since A, L, and E are just functions of these components. Therefore we have only ve independent components of motion describing the orbit and seven conserved quantities, leading to 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 and e = |A|/GMµ2 we can write the eccentricity in terms of the energy and angular momentum
s 2EL2 e = 1 + . (2.57) k2µ
2.3.2 The Eective Potential
From equation (2.57) we see that the minimum energy allowed is E = −k2µ/2L2 which results in e = 0 corresponding to a circular orbit, any less and it does not describe a real physical system. To better understand how the energy corresponds to dierent orbits we introduce an eective potential, Veff . The force acting on an orbiting particle is not just the gravitational force, in its own reference frame, there is also a centrifugal force. The eective force is then
k (rθ˙)2 k L2 F = − + µ = − + , (2.58) eff r2 r r2 µr3 which yields an eective potential shown on gure (2.2)
k L2 V = − + . (2.59) eff r 2µr2
The eective potential is very useful and contains information about allowed orbits including their shapes and sizes. For a circular orbit with radius rc the following condition must be met
dVeff = 0, (2.60) dr r=rc that is
L2 r = , (2.61) c µk
15 2 The Kepler Problem
L 2/2µr2
0 Energy
Veff
Vmin
k/r −
r
2 2 Figure 2.2: The eective potential, Veff , and its components, −k/r and L /2µr . so the minimum value is
2 2 Vmin = −k µ/2L , (2.62) as expected it's just the same we got from (2.57) but it doesn't tell us whether the circular orbit is a stable one. A stable circular orbit has to ful ll the additional condition
2 d Veff > 0, (2.63) dr2 r=rc that is
2 2 d Veff 2k 3L = − + , (2.64) dr2 r3 µr4 r=rc c c
and substituting rc into this equation
2 4 3 4 3 4 3 d Veff 2k µ 3k µ k µ = − + = > 0, (2.65) dr2 L6 L6 L6 r=rc
16 2.3 The Lagrangian so circular orbits exist for all r > 0. This makes sense because test particles can in principle get arbitrarily close and have no speed limit in classical physics. Solving
Veff − E = 0 for r gives the available turning points, elliptical orbits have two turning points and the eccentricity can be found with a simple equation
r − r e = max min . (2.66) rmax + rmin
Note that the potential depends on initial position and tangential velocity, it does not depend on radial velocity. Meaning that an increase in radial velocity can not 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 an increase and decrease in eccentricity. Table (2.1) shows the allowed orbits and their respective energies and eccentricities.
Table 2.1: Orbital energy and eccenetricity Energy Eccentricity Type Bound/Unbound
E = Vmin e = 0 Circle Bound Vmin < E < 0 0 < e < 1 Ellipse Bound E = 0 e = 1 Parabola Unbound E > 0 e > 1 Hyperbola Unbound
2.3.3 Open and Closed Orbits
Eective potentials can not describe how an orbit develops over time. An elliptical orbit oscillates between its apsides rmax and rmin. These distances are bounded, but their 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 exact same position, the orbit is said to be closed. Otherwise the orbit is open and passes through every point in the plane between rmax and rmin.
The angle between two consecutive apsides is know as the apsidal angle, ∆θa. For closed orbits we would expect the apsidal angle to be a rational fraction of π (due to symmetry), i.e. ∆θa = π(a/b) to obtain nite revolutions. Bertrand's theorem states that the only central force potentials that produce closed orbits are the inverse- square potential and the radial harmonic oscillator potential (Hooke's law) []. Since the Kepler problem involves an inverse-square central force, the elliptical orbits are
17 2 The Kepler Problem closed and do not change over time. We can con rm this numerically by writing
dθ dt θ˙ dθ = dr = dr, (2.67) dt dr r˙ then inserting equations (2.42) and (2.43) into the one above, and integrating from one apside to the next gives the apsidal angle,
L Z rmax dr ∆θa = √ . (2.68) q L2 2µ rmin 2 r E + k/r − 2µr2
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 mass is eectively the lighter orbiting particle. The massive one can be thought of as stationary and sits squarely in the origin of the center-mass frame. This is of course an approximation, in reality both particles orbit around a center of mass known as the barycenter. For planetary systems the barycenter is usually stationed inside the parent star itself, creating a wobble. This wobble can for example be used to detect exoplanets using Doppler spectroscopy.
Each particle's orbit can easily be determined via the reduced mass orbit's with equations (2.13), their orbital periods and eccentricities must equal those of the reduced mass. What diers is their relative distances, as indicated by eq. (2.13), and orbital velocities. The energy and angular momentum is split between them
µ µ Ei = E,Li = L, (2.69) mi mi 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. Since
µ ≈ 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 dierence it is negligible in comparison to m2. Eects of reducing the mass dierence can be seen in gure (2.4), there we start to see the orbit of the more massive particle compared to the lighter. In gure (2.5) we obtain a hyperbolic orbit with two equally sized masses at high speeds.
90° 90°
135° 45° 135° 45°
1.2 1.2 1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of µ. (b) Orbits of both particles.
−6 Figure 2.3: An Earth like system. Initial conditions: m1 = M , m2 = 3 × 10 M , , and . Resulting in a nearly circular orbit. r0 = 0.9832AU vθ0 = 6.3853AU/yr
90° 90°
135° 45° 135° 45°
1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of µ. (b) Orbits of both particles.
Figure 2.4: Initial conditions: , , , and m1 = M m2 = 0.1M r0 = 1AU vθ0 = 6AU/yr. Resulting in an elliptical orbit with e = 0.17.
19 2 The Kepler Problem
90° 90°
135° 45° 135° 45°
2.5 2.5 2.0 2.0 1.5 1.5 1.0 1.0 0.5 0.5 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of µ. (b) Orbits of both particles.
Figure 2.5: Initital conditions: , , . Resulting m1 = m2 = M r0 = 1 vθ0 = 15AU/yr in a hyperbolic orbit with e = 1.848.
2.3.5 Orbital Velocities
As mentioned, particles with dierent masses will have dierent orbital speeds, they dier from the reduced mass by the same factor as the energy and angular momen- tum
µ vi = v, (2.71) mi where the speed of the reduced mass, v, is given by the vis-viva equation,
1 1 v2 = GM − , (2.72) r a 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 is the semi-amplitude of the radial curve measurements, K = (vrmax −vrmin )/2 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 inclination is assumed to be i = 90◦ relative to the observer.
50 40
40 30
] 30 ] s s
/ / 20 m m [ [ y y t t
i 20 i c c o o
l l 10 e e V V
10 l l a a i i
d d 0 a a
R 0 R
10 10
20 20 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Phase Phase (a) ω = 0. (b) ω = 45◦.
−4 Figure 2.6: Radial velocity curves due to a particle with mass m2 = 9.54×10 M = MJupiter placed at r0 = 0.5AU with vθ = 6.35AU/yr, around m1 = M .
0.08 0.06
0.06 0.04
] 0.04 ] 0.02 s s / / m m [ [ y y t t
i 0.02 i 0.00 c c o o l l e e V V
0.00 0.02 l l a a i i d d a a
R 0.02 R 0.04
0.04 0.06
0.06 0.08 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Phase Phase (a) ω = 45◦. (b) ω = 90◦.
−10 Figure 2.7: Radial velocity curves due to a particle with mass m2 = 3 × 10 M = 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 according to Bertrand's theorem. Now lets add a higher order term U ∼ 1/r2 to this potential and what orbits we nd. The central force now becomes
k C F = − + , (2.74) r2 r3 where C is the strength of the 1/r3 central force component. Following the same
21 2 The Kepler Problem steps as we did for equation (2.41) we obtain a dierential equation for u(θ),
d2u Cµ kµ + u 1 + = , (2.75) dθ2 L2 L2 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
1 kµ u = = θ2 + Aθ + B (2.76) r 2L2 so the particle eventually spirals in towards the center. ii) β2 < 0. The solution is
1 kµ u = = Aeβθ + Be−βθ − , (2.77) r L2β2 which, depending on the constants A and B, either spirals inwards towards the center or outwards to in nity. iii) β2 > 0. The solution is
1 kµ u = = A cos(βθ) + B cos(βθ) + , (2.78) r L2β2 and using the same initial conditions as before we obtain
L2β2/µ2 r = . (2.79) AL2β2 GM kµ cos(βθ) + 1
This diers 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
r Cµ β = 1 + > 0. (2.80) L2
The constant C determines whether the force is attractive or repulsive. For C < 0 the 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 are obtained when β is an irrational number.
90° 90°
135° 45° 135° 45°
180° 0° 180° 0°
225° 315° 225° 315°
270° 270°
(a) β = 0.3, e = 0.7. (b) β = 0.9, e = 0.4.
Figure 2.8: Closed generic orbits with C < 0, an attractive force.
23 2 The Kepler Problem
90° 90°
135° 45° 135° 45°
180° 0° 180° 0°
225° 315° 225° 315°
270° 270°
(a) β = 1.1, e = 0.8. (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 speed of 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 physics is not in uenced by gravity. An object with such high escape velocity must be extremely massive or at least extremely dense. The escape velocity can be deduced in many ways, for example using the vis-viva equation (2.72). Taking the limit a → ∞ for an escape trajectory gives
r 2GM v = , (2.81) esc r and assuming an escape velocity that of light, c, the radius of a black hole is
2GM r = , (2.82) s c2 that is called the Schwarzschild radius in general relativity. Even for massive objects, 2 e.g. the sun, the Schwarzschild radius is relatively small, rs ∼ 3km, because G/c ∼ −27 2 −1 −1 10 m kg s . For simplicity we set G = c = 1 so that in general rs = 2M and use a dimensionless parameter x = r/rs. Here we are assuming that one particle
24 2.4 A Pseudo-Newtonian Potential is extremely massive and is stationary in the center-mass frame. Introducing this radius into the Newtonian gravitational potential
k GmM m U(x) = − = − = − , (2.83) rs(x − 1) 2M(x − 1) 2(x − 1) where M is the mass of the stationary black hole and m the mass of the orbiting particle. This is called a P seudo−Newtonian potential because it contains elements from both classical physics and relativity [7]. The eective potential now becomes
2 2 m L m L (2.84) Vpseudo = − + 2 = − + 2 2 , 2(x − 1) 2mr 2(x − 1) 2mrs x then writing L in terms of initial conditions and dividing by m
V 1 (mr v )2 1 1 pseudo 0 0,θ 2 (2.85) 2 = − + 2 2 2 = − + (x0v0,θ) 2 . m x − 1 m rs x x − 1 x
˜ De ning Vpseudo ≡ 2Vpseudo/m and l = x0v0,θ
l2 V˜ = −1/(x − 1) + (2.86) pseudo x2
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 Mµ2 (u − u )2 + u − G s = 0, (2.87) dθ2 L2 u2 a non-linear ODE, but if rmin rs we can approximate it as eq. (2.47) and obtain equation (2.52) (Kepler's third law), else we must solve it numerically. Instead of solving this non-linear ODE numerically, we should make use of the eective ˜ potential. We can write Vpseudo − 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 elliptical orbits there must exist two real roots x1, x2 > 1 and for circular orbits a double root
25 2 The Kepler Problem
x1 = x2 > 1, otherwise the orbits are unbound. Unbound orbits have an additional property in this potential, they can not only escape but also be captured. Figure (2.10) shows bound orbits for dierent systems, we consider capture orbits in detail in the next chapter.
90° 90°
135° 45° 135° 45°
12 12 10 10 8 8 6 6 4 4 2 2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) (b)
Figure 2.10: Orbits in the Psuedo-Newtonian potential. The red circle represents
the event horizon. a) Initial conditions: x0 = 5.06, vθ = 0.427c0. b) Initial conditions: x0 = 2.48, vθ = 0.775c0, e = 0.59.
Let's investigate circular orbits in this potential, where they are possible and their stability properties. For this we use a dierent method, the fact that for a stable circular orbit we have the following condition
dl > 0. (2.89) dr
For a circular orbit the centrifugal force must equal the gravitational force, which is just the derivative of the potential, so we can write the angular velocity in terms of the potential itself. With a bit of algebra we nd
1 1 1 1 dU 2 1 1 dU 2 1 1 2 ˙ √ (2.90) θ = = · = 2 . r dr xrs rs dx 2rs x(x − 1)
26 2.4 A Pseudo-Newtonian Potential
The angular momentum, l = x2θ˙, then becomes
1 1 1 x4 2 1 x3 2 l = x2θ˙ = √ = √ , (2.91) 2 x(x − 1)2 2 (x − 1)2 taking the derivative dl/dx we obtain after some algebra,
1 dl x 2 = √ (x − 3). (2.92) dx 2 2(x − 1)2
So stable circular orbits exist for x > 3. For x < 3 the circular orbits are unstable and a slight inward perturbation causes a particle to spiral into the center. A circular orbit at x = 3 is marginally stable and usually called the innermost stable circular orbit or last stable circular orbit, it is the transition between stable and unstable orbits.
27
3 Orbits in the Schwarzschild Metric
In classical physics time and space are considered to be independent. Einstein showed with his theory of relativity however that space and time are intrinsically linked 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 and how spacetime tells matter how to move along geodesics. A geodesic can be thought of as the shortest possible distance between two points on a curved surface. On a two dimensional at surface for example, a geodesic is just a straight line but on a three dimensional curved surface a geodesic is a part of a great circle. Since spacetime is curved in general relativity the geodesics describes the motion of matter through it, but we must introduce the time component into our three dimensional space to fully understand the motion. Here we focus our eorts on understanding the motion of both massive and massless test particles around a black hole described by the Schwarzschild metric. We already introduced an element from relativity, the Schwarzschild radius, into a classical Newtonian model for the Pseudo-Newtonian potential. This small correction, as we shall demonstrate, is in some aspects a very good approximation of the motion of massive particles in the Schwarzschild metric. The discussion below is based largely on 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 Schwarzschild who discovered it in 1916. A Schwarzschild black hole carries no charge nor angular momentum so its only distinguishing property is its mass. It consists of an event horizon, which can be thought of as its surface, at the Schwarzschild radius and a singularity at its center. Setting c = G = 1, the Schwarzschild metric in spherical
29 3 Orbits in the Schwarzschild Metric coordinates is
2M 2M ds2 = −(1 − )dt2 + (1 − )−1dr2 + r2dθ2 + r2 sin2 θdφ2. (3.1) r r
Here r, θ, and φ are the standard spherical coordinates. The Lagrangian in the Schwarzschild metric becomes,
2M 2M −1 2L = − 1 − t˙2 + 1 − r˙2 + r2θ˙2 + r2 sin2 θφ˙2, (3.2) r r where we de ne t˙ ≡ dt/dλ and an a ne parameter, λ, so that λ = τ/m where τ is the proper time and m a particle'a mass. The Euler-Lagrange equations for the coordinates θ, φ, and the time component are
d θ : (r2θ˙) = r2 sin θ cos θφ˙2, (3.3) dλ d φ : (r2 sin2 θφ˙) = 0, (3.4) dλ d 2M t : 1 − t˙ = 0. (3.5) dλ r .
Since the Schwarzschild metric is spherically symmetric we can always orient our coordinate system in such a way that a particle initially moves in the equatorial plane, i.e. θ = π/2. From equation (3.3) we then see that if θ = π/2 then neces- sarily θ˙ = 0 and the particle remains in the equatorial plane. The Euler-Lagrange equations with θ = π/2 simplify to
2 ˙ pφ ≡ r φ = constant = L, (3.6)
2M p ≡ 1 − t˙ = E, (3.7) t r
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
µ ν 2 2L = 2gµνp p = m , (3.9)
µ where gµν is the Schwarzschild metric tensor and p the canonical momenta described above. In terms of energy and angular momentum the Lagrangian becomes,
( E2 r˙2 L2 0, massless particle 2L = − − = . (3.10) 1 − 2M/r 1 − 2M/r r2 m2, massive particle
With this equation in hand we can start to determine geodesics of massive and massless particles in the Schwarzschild metric. In all of our discussion in this section and the next, we assume a static observer at in nity. We choose this reference frame because at in nity the local energy of a particle is equal to the conserved energy measured at in nity. The discussion below closely follows Chandrasekhar (1983).
3.2 Time-Like Geodesics: Massive Particles
For massive particles moving along time-like geodesics we write equation (3.10) in the following way
dr 2 2M L2 = E2 − 1 − 1 + . (3.11) dτ r r2
Solving this ODE gives the geodesic along which the particle moves, but this ODE is non-linear so we hold o on solving it. Note however that the above equation resembles equation (2.43), we have a radial velocity term on the left hand side and energy and angular momentum terms on the right hand side. The radial velocity 2 is zero when 2 2M L , this means we have a turning point. We E = 1 − r 1 + r2 know from the previous discussion that when we equate energy to radial velocity, and therefore turning points, we have an eective potential. The eective potential is then
2M L2 V = 1 − 1 + . (3.12) eff r r2
31 3 Orbits in the Schwarzschild Metric
L = 4M 1.00
rs 0.95 f f e L = 3. 7M V
0.90 L = 2 3M p
0.85 2 4 6 8 10 12 14 r
Figure 3.1: The eective potential for dierent values of L.
Solving Veff − E = 0 gives us the turning points and after a little algebra we obtain the 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 L = , (3.14) dτ r2
dt E = , (3.15) dτ 1 − 2M/r then using (3.14) to switch to dr/dφ
dr 2 r4 2M = (E2 − 1) + r3 − r2 + 2Mr, (3.16) dφ L2 L2 and lastly with the same substitution as in the Newtonian section, u = 1/r, we
32 3.2 Time-Like Geodesics: Massive Particles
nally get
du2 2M 1 − E2 = 2Mu3 − u2 + u − . (3.17) dφ L2 L2
Note that equations (3.16) and (3.17) are really just reformulations of eq. (3.13). If dr/tτ = 0 then we would also expect dr/dφ = du/dφ = 0 at the same time. The trick now is that if we nd the roots of du/dφ (or dr/dτ and dr/dφ), we know the boundaries of r or u. Then we can switch to another variable and parameterize u so that it takes its boundary values on the boundaries of our new variable. We shall start 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 2 2M = − (1 − E2). (3.18) dτ r
First we shall consider the case of a particle at in nity falling from rest and then a particle 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 at in nity (Minkowski spacetime). For an infalling particle we want the radial velocity to be negative and so we take the minus root of eq. (3.18)
r dr 2M = − . (3.19) dτ r
Falling from in nity takes an in nite amount of time therefore we must integrate from a starting distance r0 at τ = 0 which the particle has taken an in nite amount of time to reach and also picked up some speed. Gathering the terms and integrating
33 3 Orbits in the Schwarzschild Metric we nd
√ Z τ Z r 0 1 0 − 2M dτ = r 2 dr (3.20) 0 r0 which yields a simple result
√ 2 2M τ = (r3/2 − r3/2). (3.21) 3 0
This is the time as measured in the frame of the falling particle. What about an outside observer at in nity? We can nd the coordinate time in the following way,
dt dt dτ = , (3.22) dr dτ dr equations (3.15) and (3.19) then give,
dt r r E = − . (3.23) dr 2M 2M 1 − r
Integrating both sides we nd the coordinate time,
√ Z r E r (3.24) t = −√ 2M . 2M r0 1 − r
When r → 2M = rs the integral goes to in nity and we are unable to integrate over it. This means that t → ∞ as the particle approaches the horizon, rs.
Infall from a speci c distance
p At a speci c radial distance r0 a particle's conserved energy is E = 1 − 2GM/r0. The radial velocity is then
s s dr 2M 2M 1 1 r2M r − r = − 1 − − 1 − = − 2M − = 0 , (3.25) dτ r0 r r r0 r0 r
34 3.2 Time-Like Geodesics: Massive Particles separating the variables and integrating
r r Z r r 1/2 τ = − 0 . (3.26) 2M r0 r0 − r
Evaluating the integral we obtain a hefty equation,
r " p ! # r0 p r(r0 − r) π τ = r(r − r0) + r0 arctan + r0 . (3.27) 2M r − r0 2
The proper time it takes to fall from r0 to the center at r = 0 is therefore
π r r τ = 2M 0 . (3.28) 0 2 2M
The coordinate time is now given by
r r Z r r 1/2 2M −1 t = E 0 1 − , (3.29) 2M r0 r0 − r R
and again we have a singularity at r = 2M = rs so t → ∞ as the particle approaches the horizon.
3.2.2 Bound Orbits
Having considered geodesics of particles with zero angular momentum we now turn to 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 < 1 then the term is positive but negative if E2 > 1. The sign determines both the number and nature of the roots and thus the number of allowed orbits. A particle with 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 dierent con gurations of roots resulting in a total of eight dierent types orbits.
35 3 Orbits in the Schwarzschild Metric
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 corresponds to a capture orbit from −1 that plunges into u3 u3 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 single root corresponds to a capture orbit from −1. u3
Case C: A Double and single root. The single root corresponds to an orbit starting at −1 which asymptotically u1 spirals towards a circle with radius −1 −1 −1. The other orbit starts u = u2 = u3 from −1 −1 −1 and spirals asymptotically inwards away from the circle u = u2 = u3 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 −1 (real). Can be described by an imaginary eccentric- u1 ity. The imaginary eccentricity is obtained from the complex roots.
In cases A, B, C, and E we have, alongside the Kepler orbits (ellipses, parabolas etc.), capture orbits which so far have not been discussed (only brie y mentioned in the Pseudo-Newtonian case). These capture orbits are called orbits of the second kind in relativistic theory because they have no Keplerian analogues. The Keplerian type 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 in the equatorial plane and we have already covered the techniques to nd them. The circular orbits are orbits of the rst kind. We'll cover the orbit of the second kind in 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,
2 2 dVeff 2M 2L 6ML = − + = 0, (3.30) dr r2 r3 r4 r=rc c c c
36 3.2 Time-Like Geodesics: Massive Particles
solving for rc
r ! L2 12M 2 r = 1 ± 1 − . (3.31) c± 2M L2
√ √ We see that for no circular orbits exist. For only one L < 2 3M √ L = 2 3M circular orbit exists at rc = 6M, and two if L > 2 3M. Let's investigate the 2 2 stability of these orbits, recalling the second condition d Veff /dr > 0, if stable and 2 2 d Veff /dr < 0 if unstable. For rc = 6M, we nd after some algebra
2 d Veff = 0. (3.32) dr2 rc=6M
So the circular orbit at is neither stable or unstable, it is called marginally √ rc = 6M stable. 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 = 6M is the limit of stable orbits, i.e. the innermost stable circular orbit. This is the same result we found from the Pseudo-Newtonian potential, a testament to its good approximation.
From the above we have discerned the following
Stable: 6M ≤ rc+ < ∞, (3.33) but what of the boundaries for unstable circular orbits? Clearly they cannot exist inside the Schwarzschild radius. The minimum value of rc− must be when the square root term is equal to zero, but that only happens when L2 → ∞. This is a problem since there is already a L2 term outside the square root in eq. (3.31) making it 2 di cult to nd the limit. We can get around this by instead solving rc− for L , we then nd,
Mr2 L2 = . (3.34) (r − 3M)
Clearly L2 → ∞ when r → 3M so the boundaries of unstable orbits are
Unstable: 3M ≤ rc− < 6M. (3.35)
37 3 Orbits in the Schwarzschild Metric
Few examples of these circular orbits are marked as dots on gure (3.1). Having covered 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
1 1 u = (1 − e), u = (1 + e), (3.36) 1 α 2 α where as before e is just the eccentricity and α the semi-latus rectum. But what about 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 polynomial n n−1 1 P = anx + an−1x + ··· + a1x + a0 with roots xn Vieta's formulas are
an−1 n a0 x1 + x2 + ··· + xn−1 + xn = − , x1 · x2 · ... · xn = (−1) . (3.37) an an
2 2 The coe cients of (3.17) are an−1 = −1, an = 2M, and a0 = (1 − E )/L so we nd that the third root is
1 2 u = − . (3.38) 3 2M α
All the roots in subsequent cases follow the same format but with slight dierences depending 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)
du2 1 − e 1 + e 1 2 = 2M u − u − u − + , (3.40) dφ α α 2M α
38 3.2 Time-Like Geodesics: Massive Particles we can now write its coe cients in terms of µ and α
1 1 1 − E2 1 = 1 − µ(3 + e2) , and = . (3.41) L2 αM L2 α2
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 manageable interval. The substitution we make in this case is
1 u = (1 + e cos χ), (3.42) l so that u(χ = 0) = (1 + e)/α and u(χ = π) = (1 − e)/α. Equation (3.17) can now be written in terms of dχ/dφ with µ and α, by inserting equations (3.39) and (3.40), then obtaining
dχ2 χ = 1 − 6µ + 2µe − 4µe cos2 . (3.43) dφ 2
Taking the root and factoring (1 − 6µ + 2µe) we get
dχ χ1/2 = (1 − 6µ + 2µe)1/2 1 − k2 cos2 . (3.44) dφ 2
Choosing the positive or negative root makes no dierence since φ is a monotonic function. It only changes the direction. Here we have de ned a constant,
4µe k2 ≡ , (3.45) 1 − 6µ + 2µe 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 we continue we should make sure that k2 ≤ 1, else the integral becomes complex.
Recalling the order of the roots u1 < u2 < u3 we require
1 2 1 u = − ≥ (1 + e) = u or α ≥ 2M(3 + e), (3.46) 3 2M α α 2
39 3 Orbits in the Schwarzschild Metric writing these inequalities in terms µ we nd
1 µ ≤ , and 1 − 6µ − 2µe ≥ 0. (3.47) 2(3 + e)
If 1−6µ−2µe ≥ 0 then 1−6µ+2µe ≥ 4µe, and indeed k2 ≤ 1 is satis ed. Wherever k appears, the same method can be applied to show k2 ≤ 1, but we'll only show it in this particular case. The integral
Z ψ dβ F (ψ, k2) = , (3.48) p 2 0 1 − k2 sin β is an incomplete elliptical integral of the second kind where ψ = (π − χ)/2 is chosen so that φ = 0 at the apoapsis, and is given by
2 φ = F ( 1 (π − χ), k). (3.49) (1 − 6µ + 2µe)1/2 2
For each χ on the interval [0, 2n·π] where n is the number of revolutions there exists a corresponding φ. Plugging φ into equation (3.40) then gives u = 1/r(φ). To nd r(τ) we can then use equation (3.14).
This covers the orbit of the rst kind, let's now determine the orbit of the second kind. Its root can be written in the form
1 2 u = − . (3.50) 3 2M α
Making a substitution to a new variable ξ of the form
1 2 1 3 + e u = − + − tan2 1 ξ, (3.51) 2M α 2M α 2
ensures that u = u3 at ξ = 0 and u → ∞ (r → 0) as ξ → π. Going through the same steps as with orbits of the rst kind, the mapping now becomes
dξ 2 = (1 − 6µ + 2µe)(1 − k2 sin2 1 ξ). (3.52) dφ 2
40 3.2 Time-Like Geodesics: Massive Particles
Then integration gives
2 φ = F ( 1 ξ, k), (3.53) 1 − 6µ + 2µe 2 which ensures that φ = 0 at the apoapsis (now at ξ = 0). Figure (3.2) shows calculated orbits of both the rst and second kind for case A.
90° 90°
135° 45° 135° 45°
50 3.5 3.0 40 2.5 30 2.0 1.5 20 1.0 10 0.5 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the rst kind. (b) Orbit of the second kind.
Figure 3.2: Initial conditions: , , . The red circle represents M = 1 r0 = 12 vθ0 = 0.4c the 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 equations apply from case A. The range of −1 is −1 . Here however, we have u3 3M ≤ u3 ≤ 6M e = 0 and therefore k2 = 0 and equation (3.52) simpli es to
dξ 2 = (1 − 6µ)(φ − φ ). (3.54) dφ 0
Integrating then gives
p ξ = 1 − 6µ(φ − φ0), (3.55)
41 3 Orbits in the Schwarzschild Metric
where φ0 is a constant of integration. The substitution we make now is
1 1 3 u = + − sec2( 1 ξ), (3.56) L 2M L 2
1/2 and the particle arrives at r = 0 when φ − φ0 = π/(1 − 6µ) . Figure (3.3) shows two calculated orbits the second kind for case B.
90° 90°
135° 45° 135° 45°
6 5 5 4 4 3 3 2 2 1 1 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the second kind. (b) Orbit of the second kind.
Figure 3.3: Orbits of the second kind for case B. a) Initial conditions: M = 1 and r0 = 5. b) Initial conditions: M = 1 and r0 = 5.6. The red circle represents the event horizon.
Case C: Orbits of the rst and second kind
In this case the roots are
1 − e 1 + e 1 2 u = , u = u = = − . (3.57) 1 α 2 3 α 2M α
Before we had that α ≥ 2M(3 + e), but now we require that α = 2M(3 + e). The roots can then we written in the following way,
1 − e 1 − e u = , u = u = . (3.58) 1 2M(3 + e) 2 3 2M(3 + e)
42 3.2 Time-Like Geodesics: Massive Particles
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 as in case A, but now we de ne
1 + e 1 − e u = , χ = 0 and u = , χ = π. (3.61) 2M(3 + e) 2M(3 + e)
Equation (3.17) reduces to
dχ2 = 4µe sin2 1 χ, (3.62) dφ 2 and taking the root gives
dχ √ χ = −2 µe sin . (3.63) dφ 2
We choose the negative sign to make φ increase as χ goes from π at apoapsis to 0 at periapsis. We see that φ = 0 when χ = 0 and φ → ∞ as χ → 0. Integrating both sides yields
1 χ φ = −√ ln(tan ), (3.64) µe 4
Now for the orbit of the second kind. We have the same roots as before but now we start at u2 = u3 = (1 + e)/α and spiral into center in the opposite direction.
43 3 Orbits in the Schwarzschild Metric
Making the substitution,
1 u = (1 + e + 2e tan2 1 ξ), (3.65) α 2
ensures that u = u2 = u3 = (1 + e)/α at ξ = 0 and u → ∞ at ξ = π. We have the same roots as before so φ must change in exactly the same way, that is
1 φ = −√ ln(tan 1 ξ). (3.66) µe 4
Figure (3.4) shows calculated orbits of both the rst and second kind for case C.
90° 90°
135° 45° 135° 45°
20 15 4 3 10 2 5 1 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the rst kind. (b) Orbit of the second kind.
Figure 3.4: Initial conditions: M = 1 and r0 = 20. a) Orbit of the rst kind starting from r0 = 20 approaching r = 4.44. b) Orbit of the second kind starting from r = 4.44 spiraling towards the center. The red circle represents the event horizon.
44 3.2 Time-Like Geodesics: Massive Particles
Case E: Orbit of the second kind
Here we shall only cover the orbit of the second kind. The roots can be written in a Keplerian form
1 2 1 u = − , u = (1 + ie) = u∗, (3.67) 1 2M α 2 α 3 where ∗ is the complex conjugate. Factoring equation (3.17) u3
du2 1 − ie 1 + ie 1 2 = 2M u − u − u − + , (3.68) dφ α α 2M α we see that the only dierence 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 signs wherever we nd e2, that is
1 1 1 − E2 1 = (1 − µ(3 − e2)), = (1 − 4µ)(1 + e2). (3.69) L2 αM L2 α2
The substitution we now make is,
1 u = 1 + e tan 1 ξ , (3.70) α 2 noting that u → ∞ when ξ = π. The other boundary u = u1 is at
α 1 2 6µ − 1 tan 1 ξ = − − 1 = − . (3.71) 2 0 e 2M α 2µe
With these substitutions we nd that (3.17) becomes
dξ 2 = 2 [(6µ − 1) + 2µe sin ξ + (6µ − 1) cos ξ] , (3.72) dφ
45 3 Orbits in the Schwarzschild Metric taking the square root and factoring to obtain an elliptical integral; φ is found to be
1 Z ψ dβ φ = ±√ , (3.73) p 2 4 0 1 − k2 sin β where ψ = [−π/2, π/2] and
1 k2 = (4 + 6µ − 1), (3.74) 24 also
4 = p(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,
Z π/2 dβ K(k) = (3.76) p 2 0 1 − k2 sin β so that K(k) − F (ψ = π/2, k) = 0. Now φ given by
1 φ = √ (K(k) − F (ψ, k)). (3.77) 4
Figure (3.5) shows two calculated orbits the second kind for case E.
46 3.2 Time-Like Geodesics: Massive Particles
90° 90°
135° 45° 135° 45°
12 30 10 25 8 20 6 15 4 10 2 5 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the second kind. (b) Orbit of the second kind.
Figure 3.5: Capture orbits for case E. a) Initial conditions: M = 1, r0 = 10, and . b) Initial conditions: , , and . The red vθ0 = 0.02c M = 1 r0 = 30 vtheta0 = 0.05c circle 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. We then get three dierent con gurations of roots resulting in four dierent orbits. The cases 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 other a 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 with dierent boundary conditions.
47 3 Orbits in the Schwarzschild Metric
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 form they are
e − 1 e + 1 1 2 u = − , u = , u = − . (3.78) 1 α 2 α 3 2M α
These roots are very similar to case A for the bound orbits but now we require that e ≥ 1. Factoring equation (3.17) we nd the coe cients
1 1 E2 − 1 1 = (1 − µ(3 + e2)) = (1 − 4µ)(e2 − 1). (3.79) L2 αM L2 α
Making the same substitution as in case A but now with boundary conditions u = u3 at χ = π and u = 0 at cos χ = −1/e. Following the same steps as in A we obtain
2 φ = √ (K(k) − F ( 1 (π − ψ), k)), (3.80) 1 − 6µ + 2µe 2 where K(k) ensures that φ = 0 at χ = 0. The capture orbit for this case is described in the same manner as the bound orbit in case A. The only dierence is that e ≥ 1 and it starts from u = u2. Figure (3.6) shows calculated orbits of both the rst and second kind for case A.
90° 90°
135° 45° 135° 45°
16 14 4.0 12 3.5 10 3.0 8 2.5 6 2.0 4 1.5 2 0.51.0 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the rst kind. (b) Orbit of the second kind.
Figure 3.6: Initial conditions: , , and . The red circle represents the M = 1 r0 = 8 vθ0 event horizon.
48 3.2 Time-Like Geodesics: Massive Particles
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. Again this is described in the same manner as the bound orbits in case C above, but with a new boundary u = 0 at χ = arccos(−1/e). There is also an additional limit of and so the double root is limited to −1 −1 . Figure 1 ≤ e ≤ 3 3M ≤ (u2 = u3 ) ≤ 4M (3.7) shows calculated orbits of both the rst and second kind for case B.
90° 90°
135° 45° 135° 45°
3.5 8 3.0 2.5 6 2.0 4 1.5 1.0 2 0.5 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the rst kind. (b) Orbit of the second kind.
Figure 3.7: Initial conditions: M = 1 and r0 = 2.8. The red circle represents the event horizon.
Case C: Orbit of the second kind
Here we have a case a capture orbit described by an imaginary eccentricity. Just like previous cases this can be described by the bound orbit in case E but with dierent boundary 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 case C.
49 3 Orbits in the Schwarzschild Metric
90° 90°
135° 45° 135° 45°
2.0 2.0 1.5 1.5 1.0 1.0 0.5 0.5 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) Orbit of the second kind. (b) Orbit of the second kind.
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 eective potential.
For a massless paricle, the Lagrangian in equation (3.10) is,
E2 r˙2 L2 L = − − . (3.81) 1 − 2M/r 1 − 2M/r r2
Solving for r˙2 and recalling that L = 0 for massless particles, we nd the radial velocity is given by
dr 2 L2 2M = E2 − 1 − . (3.82) dλ r2 r
50 3.3 Null Geodesics: Massless Particles
Here we see that the eective potential for a massless particle such as a photon is
L2 2M V = 1 − . (3.83) photon r2 r
0.035
0.030 rs
0.025 2 L /
n 0.020 o t o h p V 0.015
0.010
0.005
0.000 0 5 10 15 20 r
Figure 3.9: The photon eective potential.
First let's consider a photon with zero angular momentum falling from a distance r0.
Radial Infall
With zero angular momentum the radial velocity becomes
dr 2 = E2, (3.84) dλ and knowing the energy E = t˙(1 − 2M/r) from equation (3.7) we nd
dr 2M = − 1 − , (3.85) dt r
where we have chosen the negative root. Integrating from a speci c distance r0 to
51 3 Orbits in the Schwarzschild Metric r, we get the coordinate time,
r − 2M t = r − r + ln 0 . (3.86) 0 r − 2M
We see clearly that t → ∞ when r → 2M = rs as we would expect.
Circular Orbits
Using the eective potential we can easily determine circular orbits, recalling the condition for a circular orbit and evaluating
2 2 dVphoton 2L 6ML = − + = 0, (3.87) dr r3 r4 r=rc c c we nd only one solution,
3 r = 3M = r . (3.88) c 2 s
Circular orbits of massless particles therefore only exist at r = 3M. Let's see whether this orbit is stable or not. Taking the second derivative we nd,
2 2 2 2 d Vphoton 6L 24ML 2L = − = − < 0, (3.89) dr2 r4 r5 81M 4 r=rc c c so the circular orbit is unstable, a slight perturbation either sends it ying o or spiraling 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 unstable circular orbit exists for massless particles but what about asymptotic spirals as in case C? To answer this question, we should write the radial radial velocity equation
52 3.3 Null Geodesics: Massless Particles analogous to equation (3.17). Switching to u = 1/r and noting that dφ/dλ = L/r2 we obtain
du2 1 = 2Mu3 − u2 + , (3.90) dφ D2 where D ≡ L/E represents an impact parameter. Here the constant term 1/D2 is always positive so this is analogous to the unbound time-like geodesics where indeed there exists a double root. Where might the double roots be in this case? In general a double root must must ful ll two conditions: rst f(u = ur) = 0, and second 0 f (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 du2 = 6Mu2 − 2u = 2u(3u − 1), (3.91) du dφ so u = 1/3M is a double root if
du2 1 3 1 2 1 = 2M − + = 0, (3.92) dφ 3M 3M D2 u=1/3M
√ that is, u = 1/3M is a double root if Dc = 3 3M, called the critical impact parameter. If D > Dc the photon escapes but is captured if D < Dc. The single root is then u1 = −1/6M (from Vieta's formulas), factoring the equation we get
du 1 1 2 = 2M u + u − . (3.93) dφ 6M 3M
We now make the substitution
1 1 u = − + tanh2(φ − φ ), (3.94) 6M 2M 0 and set 2 1 so that when , and also as . tanh 2 φ0 = 1/2 u = 0 φ = 0 u = 1/3M φ → ∞
53 3 Orbits in the Schwarzschild Metric
We also have an associated orbit of the second kind starting at u = 1/3M. The substitution here is similar to the ones in the time-like geodesics, namely
1 1 u = + tan2 1 ξ, (3.95) 3M 2M 2 so equation (3.90) becomes
dξ 2 1 = sin2 ξ. (3.96) dφ 2
Integrating then gives
1 φ = 2 ln(tan ξ), (3.97) 4 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 angle between a particle's velocity vector and radial vector plays a crucial role in the shape of its trajectory. If a massive particle, shot at an angle falls to the center, we could always 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 reference frame so their trajectories are highly dependent upon the initial angle and distance. The radial and tangential components of massive particles are given by
dφ dr v = cos θ = r v = cos φ = , (3.98) φ 0 dt r dt where θ is the initial angle between v and r. Using equations (3.14) and (3.15), we can write the tangential velocity, as measured from a static observer at in nity, in the following way
1 L 2M 1/2 D 2M 1/2 v = · 1 − = 1 − , (3.99) φ r E r r r
54 3.3 Null Geodesics: Massless Particles where D is the impact parameter as before. A massless particle is captured if D < Dc, so an inward moving particle escapes if
D 2M v = sin θ > c 1 − , (3.100) φ r r
and an outward moving particle escapes if D > Dc, that is if
D 2M v = sin θ < c 1 − . (3.101) φ r r
Here the angle θ can be thought of as the angle of a cone symmetric about the radial direction. When r > 3M, the angle of the cone is θ > 90◦, meaning its base points towards the center. At r = 3M the angle is θ = 90◦ so the cone opens up and essentially forms a vertical barrier. Any forward motion (θ > 90◦) at this distance results in a capture. Then at r > 3M, θ < 90◦ so the cone opens in on itself and points outward. The cone of avoidance is depicted in gure (3.10).
Figure 3.10: The cone of avoidance at dierent distances [9].
Now let us try a dierent method to determine the null-geodesics. Taking the derivative of both sides of equation (3.90) with respect to du/dφ, we obtain
du d2u du du 2 = 6Mu2 − 2u . (3.102) dφ dφ2 dφ dφ
This reduces to
d2u − 3Mu2 + u = 0, (3.103) dφ2
55 3 Orbits in the Schwarzschild Metric a non-linear second order ODE which we will now solve numerically. But rst we must determine the initial conditions. The rst initial condition is simply u0 = 1/r0 but for the derivative du/dφ at u0, the initial conditions are not so obvious. The tangential and radial velocity components are,
dφ v = sin θ = r , (3.104) φ dt and
dr v = cos θ = , (3.105) r dt
2 where θ is the initial angle. Noting that u0 = 1/r0 and dr = −du/u we can write the initial velocities in terms of u0
du 2 dr 2 = −u = −u cos θ (3.106) dt 0 dt 0 u=u0 and
dφ = u0 sin θ. (3.107) dt u=u0
Now we can write
du du −u2 cos θ u dt u=u0 0 0 (3.108) = dφ = = − , dφ u=u u0 sin θ tan θ 0 dt u=u0 and the second initial condition is determined. Figure (3.11) shows orbits of massless particles at an initial distance with varying initial angles.
56 3.3 Null Geodesics: Massless Particles
90° 90°
135° 45° 135° 45°
12 12 10 10 8 8 6 6 4 4 2 2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (a) θ = 170◦. (b) θ = 160◦.
90° 90°
135° 45° 135° 45°
12 12 10 10 8 8 6 6 4 4 2 2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (c) θ = 150◦. (d) θ = 140◦.
90° 90°
135° 45° 135° 45°
12 12 10 10 8 8 6 6 4 4 2 2 180° 0° 180° 0°
225° 315° 225° 315°
270° 270° (e) θ = 130◦. (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 eld equations and describes spacetime around a black hole with mass M, but unlike Schwarzschild 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 reduces to the Schwarzshild metric. Also, the Kerr metric is more likely to describe real black holes, since almost all astronomical objects rotate. The addition of the angu- lar momentum however, severely complicates calculations. The metric is no longer spherically symmetric but axially symmetric around the spin axis. We must take into consideration the direction of orbital trajectories relative to the spin direction. For this reason we limit our discussion in this section to circular orbits in the equatorial plane.
4.1 The Metric
The Kerr metric, expressed in Boyer-Lindquist coordinates, for a rotating black hole with angular momentum J and mass M is
2M 4aMr sin2 θ Σ ds2 = − 1 − dt2 − dtdφ + dr2 Σ Σ ∆ , (4.1) 2Mra2 sin2 θ +Σdθ2 + r2 + a2 + sin2 θdφ2. Σ
Here the black hole rotates in the φ direction, the terms a, Σ, and ∆ are de ned as
J a ≡ , ∆ ≡ r2 − 2Mr + a2, Σ ≡ r2 + a2 cos2 θ. (4.2) M
Note that when a = 0, the Kerr metric reduces to the Schwarzschild metric. The event horizon is situated at the point where the sign of the dr term changes, i.e. at
59 4 Orbits in the Kerr Metric
∆ = 0. Solving ∆ = 0 for r now gives two horizons,
√ 2 2 r± = M ± M − a . (4.3)
The outer horizon, r+, is the event horizon while the inner one, r−, is called a Cauchy horizon. We see that a < M, else a black hole can not exist. A black hole with a = M is called a maximally rotating black hole. When a < M the event horizon r+ is smaller than the Schwarzschild radius, and if a = 0 then r+ = rs. The horizon of 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,
2Mr 4Mar sin2 θ 2Mra2 sin2 θ − 1 − dt2 − dtdφ + r2 + a2 + sin2 θdφ2 = 0. Σ Σ Σ (4.4)
Dividing by dt2 we can write a quadratic equation in terms of φ˙ [11], that is,
Aφ˙2 + Bφ˙ + C, (4.5)
Where the coe cients are
2Mra2 sin2 θ A = sin2 θ r2 + a2 + (4.6) Σ
4Mar sin2 θ B = − , (4.7) Σ and
2Mr C = − 1 − . (4.8) Σ
60 4.3 Circular Orbits
The solution are then given by
√ dφ −B ± B2 − 4AC = . (4.9) dt 2A
Note that dφ/dt = 0, when C = 0 for the positive root. This means that massless particles such as photons can be stationary at certain places when moving counter to the black hole's rotation. Solving C = 0 for r we nd the distance to be
√ 2 2 2 r0 ≡ M + M − a cos θ. (4.10)
This forms a region known as the ergosphere. Even light cannot be stationary inside of this boundary, everything is forced to rotate along the direction of rotation. From this 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's consider the Lagrangian of particle moving in the equatorial plane, θ = π/2 and θ˙ = 0. In that case we nd the Lagrangian to be
2M 4Ma r2 2Ma2 2L = − 1 − t˙2 − t˙φ˙ + r˙2 + r2 + a2 + φ˙2. (4.11) r r ∆ r
The Euler-Lagrange equations now give
2M 2Ma −p = 1 − t˙ + φ˙ = E, (4.12) t r r
2Ma 2Ma2 p = t˙ + r2 + a2 + φ˙ = L, (4.13) φ r r
r2 p = r.˙ (4.14) r ∆
61 4 Orbits in the Kerr Metric
The goal is to nd an eective potential, so we must equate the radial velocity to the energy. Solving for t˙ and φ˙ we obtain
(r3 + a2r + 2Ma2)E − 2MaL t˙ = , (4.15) r∆
(r − 2M)L + 2MaE φ˙ = . (4.16) r∆
µ ν Inserting these results into the the Lagrangian and remembering that 2L = 2gµνp p = m2, we nd our radial velocity equation
dr 2 2M (a2E2 − L2) m2∆ = E2 + (aE − L)2 + − . (4.17) dλ r2 r2 r2
Here we could de ne our eective potential as the right-hand side of the equation above, but in the Schwarzschild metric we de ned the eective potential as the value 2 2 of E which made dr/dλ = 0, or Veff = E . We'll do the same here for consistency, writing equation (4.17) slightly dierently,
dr 2 r3 = E2(r3 + a2 + 2Ma2) − 4MaEL − (r − 2M)L2 − m2r∆, (4.18) dλ and solving for E we obtain
√ −B ± B2 − 4AC E = V ≡ , (4.19) kerr± 2A 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 corotating orbit, 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 the equatorial plane are, from Bardeen et al. (1972),
2 (4.23) rphoton± = 2M(1 + cos 3 arccos(∓a/m) ).
We note that for a 6= 0 then r+ < r−. These orbits are also unstable like in the Schwarzschild metric, we can see this by plotting both potentials and overlaying rphoton±, as shown on gure (4.1).
0.8
0.6
Vkerr + rph + rph 0.4 −
0.2
Energy 0.0
0.2
Vkerr − 0.4
1 2 3 4 5 6 7 8 9 10 r
Figure 4.1: The positive and negative potentials for a photon with L = 3, around a rotating black hole of mass M = 1, with a = 0.5. The Black dashed line is r+ and the 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 4 Orbits in the Kerr Metric circular orbit) in either direction, given by
1/2 rms = M(3 + Z2 ∓ ((3 − Z1)(3 + Z1 + 2Z2)) ), (4.24) where
a2 1/3 a 1/3 a 1/3 Z = 1 + 1 − 1 + + 1 − , (4.25) 1 M 2 M M
a2 Z = 3 + Z2 . (4.26) 2 M 2 1
It is not clear however, whether these orbits are bound or not. A marginally bound particle with E = 1 has a circular orbit, given by Bardeen et al., to be at
p rmb± = 2M ∓ a + 2 M(M ∓ a). (4.27)
These results, along with the null-geodesics, are summarized in gure (4.2). There we can see that the marginally stable orbit is indeed bound, as rms± > rmb± for all a. Note that if a = 0 we get the same allowed ranges as in the Schwarzschild metric, namely: 3M < r < 4M for unbound orbits (case C), 3M < r < 6M for bound orbits (case B), and r = 3M for photons.
9
8
rms 7 −
6 rms + 5
M rmb
/ − r 4 rmb + rph 3 − rph + rs 2 r + 1 r − 0 0.0 0.2 0.4 0.6 0.8 1.0 a
Figure 4.2: Circular orbits in the equatorial plane (colored) in the Kerr metric.
64 5 Conclusions
Orbits of two particle systems in a central force eld, can be reduced to a two dimensional system of one particle with a reduced mass. Dierent methods can be used to explicitly determine the orbit of the reduced mass, such as through vector algebra or by using the Lagrangian and the corresponding equations of motion. Both methods reduce to Kepler's third law of planetary motion, more speci cally to conic sections, i.e. ellipses, para- and hyperbolas, and circles. Another powerful method is the use of an eective potential. The eective potential yields information about possible orbits and their apsides. In addition the eective potential can be applied in general relativity.
Introducing an element of general relativity with a Pseudo-Newtonian potential allows for capture orbits, which have no Newtonian analogues. An eective potential allows us to determine the Keplerian orbits (conic sections) in a relatively simple fashion, where otherwise we would have to solve a non-linear ODE.
The Schwarzschild metric is spherically symmetric, so any orbit can always be de- scribed in two dimensions. Again using an eective potential we determine the boundaries (apsides) of the allowed orbits. Expressing the boundaries in Keplerian form, we parameterize the orbits in a new variable analogous to the true anomaly by making a substitution that ful lls the boundary conditions. For massless particles we mainly solve a non-linear ODE to see the eect the initial angle.
In the Kerr metric, the orbital trajectories relative to the black hole's spin direction must be taken into account. Circular orbits are determined via an eective potential, and now exist for both corotating and counterrotating orbits. We also see how the circular orbits in Kerr metric converge to a corresponding circular orbit in the Schwarzschild metric the slower a black hole spins. Finally the Kerr metric reduces to the Schwarzschild metric for a non-rotating black hole.
65
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