The Effect of Non-Static Metrics on Observables

Radboud University

Master thesis The eﬀect of non-static metrics on observables

Renz Bakx Supervisor: Prof. Dr. Wim Beenakker High Energy Physics Contents

1 Theory 5 1.1 The metric and its dynamics ...... 5 1.2 The Einstein equations ...... 6 1.3 Komar integrals ...... 7

2 Dark Matter and MOND 9 2.1 In a nutshell ...... 9 2.2 The Schwarzschild metric ...... 12 2.2.1 Dynamics ...... 13 2.2.2 An extra term ...... 15 2.2.3 The weak ﬁeld limit ...... 17 2.3 Matos et al...... 19

3 Pure General Relativity 23 3.1 Frame Dragging ...... 23 3.2 Cooperstock and Tieu ...... 25 3.2.1 The method ...... 25 3.2.2 Criticism ...... 30 3.3 Rotating Dust ...... 37 3.3.1 Obtaining the metric ...... 37 3.3.2 The Newtonian limit ...... 43 3.3.3 A constant solution ...... 47

4 Conclusion and Outlook 52

A Integrals for the Ernst Potential 54

B Integrals for the Newtonian seed 58

C Documentation 62 C.1 metric ...... 62 C.2 calculation funcs ...... 64 C.3 calculate Ynu ...... 65 C.4 calculate ErnstPot ...... 66

1 Introduction

For a long time, Newton’s theory of mechanics has been an eﬀective tool to describe the motion of all kinds of objects. From small things like apples falling from trees to planets moving around the sun. There are things however, that cannot be explained by this theory. The precession of the orbit of Mercury or the physics of a black hole, for example. In these cases, we need the theory of General Relativity devised by Einstein.

(a) (b)

Figure 1: The rotation curves of the planets in our solar system (a) and of galaxy NGC 6503 (b) [1].

Another problem that Newtonian physics cannot solve is the motion of stars in galaxies. The theory predicts that the velocity should fall of like r−1/2, which works for the planets in our solar system. This agreement can be seen in ﬁgure 1a. However, measurements indicate that the velocity becomes more or less constant at larger radial distance to the center of the galaxy. This can be seen in ﬁgure 1b, where the velocity of the matter in the galaxy NGC 6503 is plotted.

The “extra gravity” mentioned in ﬁgures 1a and 1b refers to the unexplained

2 forces that are needed to correct the theoretical velocity. A common explanation is that there is more matter within galaxies than can be observed: Dark Matter [2]. Our telescopes can only see the stars and luminous gas present in NGC 6503, so perhaps there is more than meets the eye in this galaxy. This Dark Matter may consist of large objects like dead stars or exotic particles that we have yet to discover.

Another explanation is given by Modiﬁed Newtonian Dynamics (MOND), which states that Newtonian physics behaves diﬀerently for low accelerations than for large accelerations, given some reference acceleration [3]. In this case no extra dust or exotic particles are needed, we simply adjust Newtonian physics at low accelerations so that the velocities it predicts match the ones we observe.

It is commonly assumed that the weakness of the gravitational field of galaxies (especially in the outer regions) allows for a Newtonian approach when calculating the angular velocities of the stars within the galaxy. This approach works well enough within our own solar system, where the sun is by far the heaviest object around. In this case the structure of the spacetime is dominated by the curvature generated by the sun. However, in galaxies the spacetime curvature is not dominated by some giant source in the center. It is generated by vast amounts of smaller objects rotating around the center. This combined movement of all the stars within a galactic disk might have a measurable effect on the dynamics of the system. If this effect is big enough, it may help to solve the issue with the rotation curves.

That is precisely the topic of this thesis. Our goal is to find out if the Newtonian approach is valid in this regime, or if relativistic effects do have to be taken into account. This thesis is meant as a comprehensive toolbox, in which all the techniques and literature that can be used to achieve our goal are gathered. We will first show methods to obtain the dynamics of a relativistic space-time and will treat Dark Matter and MOND in this context. After that we will discuss the theory of Cooperstock and Tieu, who claim they can explain rotation curves without resorting to Dark Matter or MOND [4]. Finally, we will attempt to model a galaxy using a flat disk of dust. In order to do this we follow the method described by Ansorg [5] to build our own computer program which can calculate the metric of such a disk.

Conventions We will use a few conventions in this thesis, which will be listed here. Firstly, we will use geometrized units unless mentioned otherwise. These units are char- acterized by setting G = c = 1, meaning all observables will have a dimension of a power of length.

We will also use a metric signature (−, +, +, +), and adopt the Einstein sum- mation convention when using greek indices:

3 µ 0 1 2 3 x xµ = x x0 + x x1 + x x2 + x x3 (1) As for the coordinate systems used in this thesis: (t, r, θ, ϕ) are used when we assume spherical symmetry and (t, ρ, ζ, ϕ) are used in a system with axial symmetry (with θ ∈ [0, π] and ϕ ∈ [0, 2π]). Lastly, partial derivatives will ∂ sometimes be written as ∂x = ∂x, in order to save space.

4 Chapter 1

Theory

In this chapter we will brieﬂy go over some of the aspects of General Relativity that are used in this thesis. The material in this chapter was obtained from books written by Sean Carroll [6] and Robert Wald [7].

1.1 The metric and its dynamics

A metric gµν is usually given by its line element:

2 µ ν ds = gµν dx dx . (1.1) To study the dynamics of a spacetime generated by such a metric we will look at geodesics. These are the paths of shortest distance between two points, the curved spacetime equivalent of a straight line. A deﬁnition of these paths that works in General Relativity is that if a path xµ(λ) is a geodesic, it parallel- d µ transports its own tangent vector dλ x (λ). Parallel transport means that the covariant derivative of the tangent vector along the path vanishes. This is an important requirement, because it ensures that the norm and orthogonality of vectors is preserved along the path. For this demand to be satisﬁed, the path xµ(λ) needs to obey the geodesic equation:

d2xµ dxρ dxσ + Γµ = 0. (1.2) dλ2 ρσ dλ dλ

5 µ In the geodesic equation the Christoﬀel symbols Γρσ are given by:

1 Γµ = gµν (∂ g + ∂ g − ∂ g ) (1.3) ρσ 2 ρ σν σ ρν ν ρσ µρ µ with g gρν = δ ν (1.4)

µ Here δ ν is the Kronecker-delta. In order to satisfy the geodesic equation λ needs to be an aﬃne parameter. This means that λ = aτ +b for some constants a and b, where τ is the proper time. The proper time can be obtained from the line element: dτ 2 = −ds2. Usually we pick λ = τ, so that trajectories are parameterised by the proper time. When using an aﬃne parameter the four-velocity is normalised, meaning:

µ ν ( µ dx dx −1 for massive particles UµU = gµν = (1.5) dτ dτ 0 for massless particles This normalisation condition and the geodesic equations can be combined into a set of differential equations for the path xµ(λ). When there are symmetries present in the metric, conserved quantities can be obtained through Killing vectors that can make solving these differential equations much easier. A Killing vector is the vector in the direction of the symmetry. For example, if a metric has a time-translation symmetry, Kµ = (1, 0, 0, 0) is the appropriate Killing vector. The conserved quantity associated with this symmetry is energy, which can be derived from the following equation. For any Killing vector: dxν dxν K = g Kµ = const. (1.6) ν dλ µν dλ The equations above will aid us in finding the dynamics of any system we might wish to describe, in particular when we look at the angular velocity and the orbits of the metrics we are going to study.

1.2 The Einstein equations

Of course, we cannot just pick any function for the line element. To make sure that the metric describes a system where mass and energy are the source of the curvature of spacetime, it has to obey the Einstein ﬁeld equations. For these equations we ﬁrst need the Riemann tensor:

ρ ρ ρ ρ λ ρ λ Rσµν = ∂µΓνσ − ∂ν Γµσ + ΓµλΓνσ − ΓνλΓµσ. (1.7) We can internally contract indices to obtain the Ricci tensor and Ricci scalar, respectively:

λ µ µν Rµν = Rµλν and R = R µ = g Rµν (1.8) The Einstein equations can then be written as

6 1 R = 8πG T − T g . (1.9) µν µν 2 µν In these equations G is Newton’s constant of gravitation (which will be set to 1), µν Tµν is the energy-momentum tensor and T = g Tµν is its trace. This tensor contains all the information of the matter content of the system we want to describe. We will use it later to describe a ﬂat disk of dust, but it could also be used to represent for example a vacuum. In that case all the tensor components are equal to zero: Tµν = 0, and the Einstein equations reduce to Rµν = 0.

1.3 Komar integrals

If we want to study the rotational dynamics of a given spacetime, it might be very useful to know the total gravitational mass M and the total angular momentum J of the system. In section 1.1 we mentioned a link between conserved quantities and symmetries in the form of the Killing vectors. From a symmetry in time we can derive energy and from a rotation symmetry around an angle ϕ we can derive angular momentum of a test particle. In order to obtain the mass M and the total angular momentum J of the entire spacetime we need the currents

t ν ϕ ν jµ = ξ Rµν and jµ = η Rµν . (1.10) µ µ In these equations Rµν is the Ricci tensor and ξ and η are the Killing vectors related to t and ϕ respectively. For these currents to be conserved we need:

µ µ µ λ ∇µj = ∂µj + Γµλj = 0. (1.11) It can be calculated that for the currents in 1.10, this is indeed the case [6]. In order to obtain the conserved quantities associated with these currents we will need to integrate them: Z 1 µ Q = n jµdV (1.12) 4π S The normalisation is chosen for convenience, the Ricci tensor has a similar factor. This quantity will be independent of the spacelike hypersurface S, and therefore conserved [6]. The hypersurface is a slice of spacetime with t = const. In other words: a spatial volume. In the equation above nµ is the unit normal vector on µ the hypersurface S, normalised to nµn = −1. We can plug the currents into equation 1.12 to obtain the so-called Komar integrals which can be written as [7]:

Z 1 µ ν Komar mass: M = 2 Tµν − T gµν n ξ dV (1.13) S 2 Z µ ν Komar angular momentum: J = − Tµν n η dV (1.14) S

7 t The Komar mass is obtained from jµ and the Komar angular momentum is ϕ obtained from jµ . In section 2.2 we will give an example of a calculation of such an integral, to show that this deﬁnition does indeed make sense.

8 Chapter 2

Dark Matter and MOND

In this chapter we will brieﬂy discuss the theories of Dark Matter and MOND, and how they solve the problem with rotation curves. First we will show how the dynamics of the Schwarzschild metric can be derived. Then we will derive the Newtonian limit of this metric and change it to a Modiﬁed Newtonian limit. Dark Matter and General Relativity are combined in the research of Matos et al., which will be presented at the end of this chapter.

2.1 In a nutshell

The problem of the rotation curves is most commonly explained by the presence of Dark Matter. This is matter that does not radiate or absorb photons. If it did, it would have been detected already. This matter could consist of anything, from dead stars to tiny particles. Examples of these particles are axions or the supersymmetric neutralino. Besides the rotation curves, dark matter evidence is also found in gravitational lensing (e.g. in the Bullet cluster), the Cosmic Microwave Background and Large Scale Structure formation [8].

In Newtonian mechanics, the tangential velocity can be calculated by using the fact that the centripetal force is provided by the gravitational force alone. We can write for an object with mass m rotating circularly around a central mass M:

Fc = Fg mv2 mMG = r r2 r GM ⇒ v = (2.1) r As was said in the introduction, we do not observe this ∼ r−1/2 behaviour. This can be ﬁxed by adding mass to the system, with a certain r-dependence. Then

9 M → M + MDM (r). If the radial behaviour of this Dark Matter is picked just right, a velocity can be obtained that flattens for large r. The dark matter in a galaxy is usually modeled as a halo around the center, with a certain density profile. An example of one such a profile is the one suggested by Navarro, Frenk and White [9]:

−2 rs r ρNFW(r) = ρs 1 + (2.2) r rs

In the equation above r is the distance from the galactic center, rs is a “typical” scale radius and ρs is a “typical” scale density. We can obtain the mass of the Dark Matter added by this profile by integrating the density. If we recalculate the tangential velocity we find that we can obtain flattened rotation curves, as can be seen in figure 2.1. In this figure the contribution from the Dark Matter is also shown.

Figure 2.1: Some rotation curves (solid lines) of three diﬀerent galaxies with the Dark Matter distributed in an NFW proﬁle, compared to observations (dotted lines) [9]. The dashed lines are the contributions from the Dark Matter halo.

Another explanation for the observed rotation curves is given by MOND. This theory is a tweak of regular Newtonian dynamics that entails that forces work diﬀerently for low accelerations. Normally Newton’s laws dictate that force

10 equals mass times acceleration, or F = m a. The modiﬁcation proposed by MOND is that the Newtonian equation gets an extra function f:

a F = m · a · f . (2.3) a0

In this equation the function f(a/a0) determines how MOND deviates from classical mechanics. In the limit of large accelerations a a0 the function f → 1, so that MOND reduces to Newtonian physics. The limit of small accelerations a a0 is the limit where it needs to explain the rotation curves [3]. MOND does not have a speciﬁc function that predicts how the actual acceleration does behave, but many diﬀerent physicists have proposed how f depends on a. Two of these proposals are the simple and standard function given below [10].

1 simple: f(x) = 1 (2.4) 1 + x 1 standard: f(x) = (2.5) q 1 1 + x2

We can solve the equation m·aN = m·a·f(a/a0) to obtain the new acceleration in terms of the Newtonian acceleration aN and the constant a0. This constant is not given, it has to be ﬁtted with experimental data. The MOND acceleration for the simple and standard functions are given below.

1 simple: a(r) = a (r) ± pa (r)2 + 4a (r)a (2.6) 2 N N N 0 1 q 1/2 standard: a(r) = a (r)2 ± a (r)4 + 4a (r)2a2 (2.7) 2 N N N 0

It can be easily seen that if aN a0, the MOND acceleration equals the New- tonian one. For small a however, the expression is changed. For both cases N √ we can then approximate a ≈ a0aN . This in turn affects the gravitational acceleration, leading to a different r-dependence of the acceleration. The rotation curves that result from these adjustments can be seen in figure 2.2.

11 Figure 2.2: The rotation curve for Newtonian physics and the simple and standard MOND accelerations [10].

Now that we have discussed the non-relativistic solutions to the rotation curve problem, let us see what options General Relativity can give us for solving this problem.

2.2 The Schwarzschild metric

The Schwarzschild metric is one of the simplest non-trivial metrics discovered so far. The line element of this metric is given in equation 2.8. Its simplicity does not diminish its physical relevance, as it accurately describes the spacetime around a black hole or star of mass M. This comes from the Birkhoﬀ theorem, which basically states that any spherically symmetric metric can be rewritten as the Schwarzschild metric.

2M 2M −1 ds2 = − 1 − dt2 + 1 − dr2 + r2(dθ2 + sin2(θ)dϕ2) (2.8) r r

Asymptotically this metric gives Newtonian physics, because the gtt and grr components have been chosen this way. However, in close vicinity to the source of the metric, the dynamics are diﬀerent. This can be observed from the precession of Mercury’s orbit, which cannot be explained by Newtonian physics, but comes forth naturally out of the Schwarzschild metric. An interesting property of the Schwarzschild metric is that at r = 2M the tt-component of the metric is zero and the rr-component goes to inﬁnity, this radius is called the Schwarzschild radius. The escape velocity within this radius is higher than the speed of light, meaning that everything that moves inside this radius can never return to the outside.

12 2.2.1 Dynamics In order to study the physical properties of this metric, it is insightful to study the dynamics. To check what kind of orbits are allowed in this spacetime, the geodesic equations will need to be studied. We will focus on circular orbits, so geodesics where the coordinate r does not change. We will also make use of the normalisation of the four-velocity and the conserved quantities E and L. These quantities come from the Killing vectors in the t and ϕ directions: ξµ = (1, 0, 0, 0) and ηµ = (0, 0, 0, 1). We use equation 1.6, with the aﬃne parameter λ = τ, to obtain

dxν dxν −E = g ξµ = const. L = g ηµ = const. µν dτ µν dτ dxν dxν = g = g tν dτ ϕν dτ dt dϕ = g = g tt dτ ϕϕ dτ 2M dt dϕ = 1 − = r2 sin2(θ) (2.9) r dτ dτ For massive particles these are the conserved energy (E) and angular momentum in the ϕ direction (L) per unit mass [6]. The geodesic equations for this metric are listed below:

d2t 2M dr dt + = 0 (2.10) dτ 2 r(r − 2M) dτ dτ d2r M dt 2 M dr 2 + (r − 2M) − dτ 2 r3 dτ r(r − 2M) dτ " # dθ 2 dϕ2 −(r − 2M) + sin2 θ = 0 (2.11) dτ dτ d2θ 2 dθ dr dϕ2 + − sin θ cos θ = 0 (2.12) dτ 2 r dτ dτ dτ d2ϕ 2 dϕ dr cos θ dθ dϕ + + 2 = 0 (2.13) dτ 2 r dτ dτ sin θ dτ dτ This looks like a complicated set of equations, but we can make our lives easier by using the conserved quantities E and L. Equation 2.10 reduces to ∂τ E = 0, and 2.13 to ∂τ L = 0. These do not surprise us, because the fact that these quantities were conserved was already known from the Killing vectors. If we π put θ = 2 , equation 2.12 is automatically satisﬁed. This can be done without loss of generality, because of the spherical symmetry. Lastly, equation 2.11 is connected to the normalisation of the four-velocity. Since the normalisation condition (eq. 1.5) is easier to work with, we will use it instead of the geodesic equation.

13 If we plug the conserved quantities E and L into the normalisation condition for massive particles, we obtain the following:

µ ν gµν U U = −1 dr 2 2M L2 ⇒ + 1 − + 1 = E2 (2.14) dτ r r2 1 dr 2 1 Rewriting gives: + V (r) = E2 (2.15) 2 dτ 2 1 2M L2 with the potential V (r) = 1 − + 1 . (2.16) 2 r r2

Just like in classical mechanics, we will need ∂rV (r) = 0 for circular orbits and 2 ∂r V (r) > 0 for stable orbits. If we do this we ﬁnd the condition

Mr2 − L2r + 3ML2 = 0. (2.17) √ 1 2 4 2 2 This equation can be solved to obtain r± = 2M (L ± L − 12M L ), with 2 2 ∂r V (r−) < 0 and ∂r V (r+) > 0, meaning only the orbit at r+ is stable. We ﬁnd that the smallest possible circular orbit is at r = 6M [6]. To obtain the tangential velocity v(ϕ) of a test particle in this metric we will employ the method used by Matos et al. in their 2002 paper [11]. We again use the normalisation of the four-velocity for particles with mass:

µ ν 1 = −gµν U U " # 2M dt 2 2M −1 dr 2 dθ 2 dϕ2 = 1 − − 1 − − r2 + sin2 θ r dτ r dτ dτ dτ

2M dt 2 2M −2 dr 2 = 1 − 1 − 1 − r dτ r dt " #! r2 dθ 2 dϕ2 − + sin2 θ 2M dt dt 1 − r 2M dt 2 ≡ 1 − 1 − v2 (2.18) r dτ

According to Matos, v is a natural deﬁnition of the spatial velocity and can also 2 2 2 be written as v2 = v(r) + v(θ) + v(ϕ) . We can then write: r dϕ v(ϕ) = sin θ (2.19) q 2M dt 1 − r

14 dϕ To obtain an expression for dt we look at equation 2.11. We assume circular 2 orbits (∂τ r = ∂τ r = 0) and ﬁx θ = π/2, then we can quickly see that r dϕ M = . (2.20) dt r3 Plugging this into equation 2.19 we obtain our ﬁnal result:

r M v(ϕ) = (2.21) r − 2M For large r this expression reduces to the Newtonian case v(ϕ) = pM/r, as we would expect. In the following section we will apply the same method to a Schwarzschild metric with an added term, to see if we can add dark matter or MOND to this solution.

2.2.2 An extra term

As was mentioned before, the terms gtt and grr in the Schwarzschild solution were chosen to give Newtonian physics at large r. In order to see what our options are for extra terms, let us start with the most general form of the spherically symmetric metric:

ds2 = −(1 − h(r))dt2 + (1 − h(r))−1dr2 + r2(dθ2 + sin2(θ)dϕ2). (2.22)

At this point we should mention that the Ricci tensor Rµν 6= 0. This means that this metric does not necessarilly describe a vacuum, what it does describe will depend entirely on h(r). Of course, if we want some dark matter to be present, it is no problem if this metric does not describe a vacuum. However, in order to say anything about the kind of matter (e.g. a dark ﬂuid, some exotic matter...) that is added by the extra term h(r), one should have a look at the Riemann and Ricci tensors to determine the energy-momentum tensor Tµν . Let us start by gathering all the necessary ingredients for the calculation of the dynamics. The non-zero Christoﬀel symbols of this metric are:

−∂ h(r) 1 Γt = Γt = r Γr = − (1 − h(r))∂ h(r) tr rt 2(1 − h(r)) tt 2 r 1 ∂ h(r) Γθ = Γθ = Γr = r rθ θr r rr 2(1 − h(r)) θ r Γϕϕ = − sin θ cos θ Γθθ = −r(1 − h(r)) 1 Γϕ = Γϕ = Γr = −r(1 − h(r)) sin2 θ rϕ ϕr r ϕϕ cos θ Γϕ = Γϕ = θϕ ϕθ sin θ

15 The non-zero components of the Ricci tensor are:

h(r) − 1 ∂2h(r) ∂h(r) 1 ∂2h(r) r ∂h(r) R = r + 2 R = + tt 2r ∂r2 ∂r rr (1 − h(r))r ∂r2 2 ∂r ∂h(r) ∂h(r) R = r + h(r) R = sin2 θ r + h(r) θθ ∂r ϕϕ ∂r

The conserved quantities for this metric can be obtained in the same way as we did for the Schwarzschild metric. The Killing vectors are, again, ξµ = (1, 0, 0, 0) and ηµ = (0, 0, 0, 1). Using this we get:

dxµ dt dxµ dϕ E = −ξ = (1 − h(r)) , L = η = r2 sin2(θ) (2.23) µ dτ dτ µ dτ dτ π Again, we will assume that θ = 2 . From the normalisation of the four-velocity for massive particles we know:

dt 2 1 dr 2 dϕ2 U U µ = −(1 − h(r)) + + r2 = −1 µ dτ (1 − h(r)) dτ dτ dt 2 dr 2 dϕ2 ⇒ −(1 − h(r))2 + + r2(1 − h(r)) = −(1 − h(r)) dτ dτ dτ

We can insert the conserved quantities from equation 2.23 into the expression above. We then obtain:

dr 2 1 − h(r) −E2 + + L2 = −(1 − h(r)) dτ r2 dr 2 L2 ⇒ + (1 − h(r)) + 1 = E2 dτ r2 1 dr 2 1 ⇒ + V (r, L) = E2 (2.24) 2 dτ 2

We have now written the potential V in terms of the function h(r). For stable 2 circular orbits we need ∂rV (r) = 0 and ∂r V (r) > 0, in terms of h(r) this means:

∂V (r) L2 1 ∂h(r) L2 = (h(r) − 1) − + 1 = 0 (2.25) ∂r r3 2 ∂r r2 ∂2V (r) 3L2 ∂h(r) 2L2 ∂2h(r) L2 1 = − (h(r) − 1) + − + > 0 (2.26) ∂r2 r4 ∂r r3 ∂r2 2r2 2

When we have prescribed a function h(r) we will have to check if these conditions are satisﬁed, in order to see if stable circular orbits are allowed. Like before,

16 we will use the geodesic equation for the second derivative of r to derive an expression for the tangential velocity:

d2r 1 dt 2 ∂ h(r) dr 2 − (1 − h(r))∂ h(r) + r dτ 2 2 r dτ 2(1 − h(r)) dτ " # dθ 2 dϕ2 −r(1 − h(r)) + sin2(θ) = 0 (2.27) dτ dτ

π If we assume circular orbits (r constant), and restrict θ to 2 , we get:

1 dt 2 dϕ2 − (1 − h(r))∂ h(r) = r(1 − h(r)) 2 r dτ dτ dϕ2 1 ⇒ = − ∂ h(r) (2.28) dt 2r r Just like in the previous section, we will follow Matos’ procedure to obtain an expression for the tangential velocity [11].

dτ 2 = (1 − h(r))dt2 − (1 − h(r))−1dr2 − r2(dθ2 + sin2(θ)dϕ2) " dt 2 dr 2 ⇒ 1 = (1 − h(r)) 1 − (1 − h(r))−2 dτ dt !# r2 dθ 2 dϕ2 − + sin2(θ) 1 − h(r) dt dt dt 2 ≡ (1 − h(r)) 1 − v2 , where v2 = (v(r))2 + (v(ϕ))2 + (v(θ))2 dτ (2.29)

Using equation 2.28, we obtain for the tangential velocity: s 2 r2 dϕ2 − r ∂ h(r) v(ϕ) = sin2(θ) ⇒ v(ϕ) = 2 r (2.30) 1 − h(r) dt 1 − h(r) Now that we have all the ingredients we need, we only need to ﬁnd some inspi- ration for h(r).

2.2.3 The weak field limit To find a suitable function h(r), let us take a look at MOND. In the weak field limit the Schwarzschild solution corresponds to Newtonian physics. We can pick our function h(r) to make our solution correspond to MOND instead, and see what kind of metric we get. We follow the procedure used by Carroll in section

17 4.1 of his book [6]. For this limit we assume that the test particle is moving slowly and that the weakness of the gravitational ﬁelds allows us to write the metric as Minkowski plus a small pertubation:

dxi dt and g = η + h , |h | 1 (2.31) dτ dτ µν µν µν µν In the equation above, the index i represents a spatial direction (r, θ or ϕ). If we take this weak ﬁeld limit, we ﬁnd that for our metric:

1 d2r ∂ h(r) = = a(r) (2.32) 2 r dt2 2M Normally we would put h(r) = r , the Newtonian potential, but now we need to implement a diﬀerent one. In MOND we can write that the acceleration a M depends on the Newtonian acceleration aN = − r2 like:

1 q simple: a(r) = a ± a2 + 4a a (2.33) 2 N N N 0 1 q 1/2 or standard: a(r) = a2 ± a4 + 4a2 a2 (2.34) 2 N N N 0

We use equation 2.32 to obtain h(r) and will use the simple form of a for now. We get: r ! Z Z M M 2 M h(r) = 2 a(r)dr = − + − 4 a dr (2.35) r2 r4 r2 0 The obtained results are plotted in ﬁgure 2.3 below. As we can see from these plots, v(ϕ) ﬂattens for large r and the potential allows for stable circular orbits, since ∂rV (r) = 0 for r ≈ 8 and we can see that at this radius the second derivative is positive.

18 (ϕ) Figure 2.3: The functional behaviour of v (r) (left) and ∂rV (r) (right) for M = 1, L = 4 and a0 = −0.01 (in arbitrary units).

It seems we have succesfully obtained flat rotation curves and stable orbits with our approach. However, equation 2.32 is derived using |h(r)| 1. But h(r) does p a0 not go to zero for large r. For the simple acceleration, h(r) ∼ arcsinh 2 M r for r 1. Also, when we calculate the Ricci tensor Rµν we find that its components are non-zero. We seem to have added some sort of matter or energy to the metric. Sadly, we have to conclude that this is not a good way to obtain flat rotation curves in a relativistic context without resorting to Dark Matter.

2.3 Matos et al.

We have seen that we cannot add a term to the Schwarzschild metric that makes it asymptotically go to MOND, while still describing a vacuum. The problem is that the Einstein equations are a complicated set of non-linear differential equations. This means that one cannot simply add one solution of these equations to another solution, and still have it obey the Einstein equations. It seems we must take a different approach if we want to study what kind of metric is needed for flat rotation curves.

The method used by Matos et al. is a more fundamental approach [11]. They start with a more general line element, the one of an axisymmetric, stationary spacetime. Axisymmetry is symmetry around an axis, in this case the ζ-axis, meaning that the metric functions are independent of the angle ϕ. Stationary means that the metric functions are independent of the time coordinate t. The metric has terms that “mix” the coordinate t with the coordinate ϕ (dtdϕ and dϕdt). Physically, these can be interpreted as a rotation of the system around the ζ-axis. If it does not have these mixing terms (like the Schwarzschild metric), we call it static. The line element is given by

ds2 = −e2ψ(dt + ωdϕ)2 + e−2ψ[e2γ (dρ2 + dζ2) + µ2dϕ2] (2.36)

19 In this metric ψ, ω, γ and µ are functions of the cylindrical coordinates (ρ, ζ). They claim the slow rotation of galaxies allows for the approximation ω ≈ 0, so that equation 2.36 reduces to a static metric. They make use of the conserved quantities E and L, and the potential V (ρ), both of which can be obtained in the same way as we have done before. The conserved quantities can be obtained µ µ µ µ from the Killing vectors ξ = δt and η = δϕ:

dxν dxν −E = g ξµ L = g ηµ µν dτ µν dτ dxν dxν = g = g tν dτ ϕν dτ dt dϕ = −e2ψ = e−2ψµ2 (2.37) dτ dτ Matos picks −E instead of E as the conserved quantity, but this does not change the physics of the situation. The potential V (ρ) is again obtained from the normalisation of the four-velocity for massive particles:

µ ν gµν U U = −1 ⇒ −e2ψt˙2 + e2(γ−ψ)(ρ ˙2 + ζ˙2) + e−2ψµ2ϕ˙ 2 + 1 = 0 (2.38) ˙ ∂t A dot refers to a derivative w.r.t. proper time (i.e. t = ∂τ ). We are looking for orbits in the plane ζ = 0, so ζ˙ drops out. We can also insert the conserved quantities into the equation above and write:

e2ψ ρ˙2 − e2(ψ−γ) e−2ψE2 − L2 − 1 = 0 µ2 (2.39) 1 1 e2ψ ⇒ ρ˙2 + V (ρ) = 0 with V (ρ) = − e2(ψ−γ) e−2ψE2 − L2 − 1 2 2 µ2 (2.40) Rather than putting E on the right hand side of equation 2.39, Matos includes the energy in the potential. Physically this makes no diﬀerence, as ρ has to satisfy the same equation as before. They are looking for stable circular orbits in the plane ζ = ζ˙ = 0, with a tangential velocity independent of the radius. To obtain these, they pose three conditions that have to be satisﬁed: ρ˙ = 0.

∂ρV (ρ) = 0.

2 ∂ρV (ρ)|extr > 0, in order to have a minimum.

20 dϕ Like before, we also want to obtain an expression for the angular velocity dt . This can be done by evaluating the geodesic for ρ and observing the ﬁrst constraint together with ζ˙ = 0. If we do this, we obtain:

∂ψ dt 2 ∂ψ ∂µ dϕ2 0 = e4ψ−2γ + e−2γ µ2 − µ ∂ρ dτ ∂ρ ∂ρ dτ dϕ e2ψ s ∂ ψ ⇒ = ρ (2.41) dt µ ∂ρµ µ − ∂ρψ

Using their deﬁnition of the spatial velocity v2 = (v(ρ))2 + (v(ζ))2 + (v(ϕ))2, which we have used in the previous sections, they derive an expression for the tangential velocity:

" !# dϕ2 dρ2 dζ 2 −ds2 = dτ 2 = e2ψdt2 1 − e−4ψµ2 − e2γ−4ψ + dt dt dt dt 2 ⇒ 1 = e2ψ [1 − v2] dτ µ dϕ s ∂ ψ with v(ϕ) = = ρ (2.42) e2ψ dt ∂ρµ µ − ∂ρψ

(ϕ) Then they demand that this velocity is independent of ρ: ∂ρv = 0 and (ϕ) (ϕ) v = vc = constant. This leads to a constraint on the metric functions ψ and µ. We can rewrite equation 2.42:

2 (ϕ) 1 + vc ∂ρµ = 2 ∂ρψ (2.43) µ (ϕ) vc It is easily checked that this constraint is satisﬁed if

2 (ϕ) l vc ψ µ e = , with l = 2 . (2.44) µ0 (ϕ) 1 + vc We then end up with a metric that looks like

µ 2l µ −2l ds2 = − dt2 + [e2γ (dρ2 + dζ2) + µ2dϕ2] (2.45) µ0 µ0 They mention that this metric is not asymptotically flat. This means that for large ρ and ζ this metric does not reduce to flat Minkowski space. This means that it at most describes the region of the galaxy with flat rotation curves. The other regions of space will have other metrics that will have to be connected to

21 this one to obtain a complete description of the galaxy and the space around it.

Now Matos et al. want to ﬁnd out what kind of matter could produce a metric like 2.45. They calculate the Einstein tensor and make the assumption that all relations between metric functions obtained in the equatorial plane ζ = 0 hold in a region close to the plane as well. After doing this they are able to derive a relation between the metric functions and an arbitrary energy-momentum tensor. The constraint that any type of matter will have to obey in order to have constant tangential velocities is

1 − (v(ϕ))2 e2ψ − (T + T ) = e−2(ψ−γ) e−2ψT + T . (2.46) 1 + (v(ϕ))2 ρρ ζζ tt µ2 ϕϕ

A vacuum solution Tµν = 0 trivially satisﬁes the condition above. From the Einstein equations Matos et al. derive that the metric in this case has to be

2 ! ρ 2l ρ −2l ρ 2l ds2 = − dt2 + (dρ2 + dζ2) + ρ2dϕ2 . (2.47) ρ0 ρ0 ρ0

The central object that creates this metric is string-like. Although it is an interesting case, we are not looking for matter distributions of this kind, so this is not a solution to our problem with rotation curves.

Another solution to equation 2.46 is the dark perfect ﬂuid. The energy-momentum tensor of this matter is

Tµν = (d + p)UµUν + gµν p (2.48) Here d is the density, p is the pressure and U is the four-velocity. This “ﬂuid” could be composed of planetoids or exotic particles. It could also be composed of baryonic usual matter, but constraints from other observations (concerning the Big Bang nucleosynthesis) make this an unlikely candidate [8, 11].

An interesting remark by Matos states that taking the Newtonian approach to the rotation curve problem fixes the matter to be dust-perfect-fluid-like, while taking the relativistic approach opens up possibilities for other types of matter to fix the problem. They conclude that taking this approach, where more exotic solutions are also considered is the right way to determine what constitutes 90% of the matter content of the universe.

22 Chapter 3

Pure General Relativity

There are physicists that are sceptical about Dark Matter and MOND. It does seem strange that the vast majority of all the matter in the universe is invisible to us, or that we would have to tweak Newtonian theory at large scales, since it has served us so well at smaller scales. In this chapter we will look at models that can solve the problem with rotation curves without resorting to Dark Mat- ter or MOND. We will first briefly discuss a relativistic effect that might impact the motion of stars in a galaxy. Then we will discuss the model of Cooperstock and Tieu [4] and the critisism it received. Finally we will present the model that we are going to use to model a galaxy, using the metric of a rotating disk of dust. We will discuss how our computer program calculates this metric and will present an analytical derivation of the simplest non-trivial case.

At the scales of galaxies, General Relativity is normally invoked, as the masses in these structures are typically large enough. Since the velocities are slow and the gravitational fields are weak, it is often argued that General Relativity is actually not neccesary. In the previous section it was already mentioned that when taking the Newtonian approach, one puts constraints on the types of matter that can fix the problem. Perhaps there are other relativistic effects in galaxies that are overlooked when using Newtonian mechanics?

3.1 Frame Dragging

One such relativistic effect might be frame dragging, an effect that occurs near rotating black holes. The metric describing such a black hole is known as the Kerr metric. It should be noted that for this metric, there is no Birkhoff theorem. This means that the space around a rotating black hole can at most be approximated by the metric below. However, the properties of this metric lead to some interesting physical phenomena.

23 The Kerr metric reads

2Mr 2Mar sin(θ)2 ρ(r, θ)2 ds2 = − 1 − dt2 − (dtdϕ + dϕdt) + dr2 ρ(r, θ)2 ρ(r, θ)2 ∆(r) sin(θ)2 + ρ(r, θ)2dθ2 + (r2 + a2)2 − a2∆(r) sin(θ)2 dϕ2 (3.1) ρ(r, θ)2 where

∆(r) = r2 − 2Mr + a2 (3.2) ρ2(r, θ) = r2 + a2 cos(θ)2 (3.3)

In the Kerr metric a = J/M, where M is the Komar mass and J the Komar angular momentum. The phenomenon known as frame dragging can be shown dϕ when we calculate dt , considering a massless photon emitted in the ϕ direction somewhere in the plane θ = π/2, so that the r and θ components of its four- velocity are zero at the moment of emission. Then we can calculate from the normalisation of the four-velocity:

dt 2 dt dϕ dϕ2 g U µU ν = g + 2g + g = 0 µν tt dτ tϕ dτ dτ ϕϕ dτ dϕ dϕ2 ⇒ g + 2g + g = 0 tt tϕ dt ϕϕ dt s dϕ g g 2 g ⇒ = − tϕ ± tϕ − tt (3.4) dt gϕϕ gϕϕ gϕϕ

Evaluating this expression on the stationary limit surface, for which gtt = 0, gives an interesting set of solutions. The surface is described by the equation

(r − M)2 = M 2 − a2 cos2 θ (3.5) π ⇒ r = 2M at θ = (3.6) 2

If we put gtt = 0 in equation 3.4, we obtain

dϕ dϕ gtϕ a = 0 or = −2 = 2 2 (3.7) dt dt gϕϕ 2M + a The nonzero solution has the same sign as a, we interpret this solution as the photon rotating with the black hole. The other solution does not move in this coordinate system. So the rotation of the black hole “drags” the inertial frame of the photon backwards, such that it appears to stand still. If a particle is at this distance or closer to the black hole, it would have to move faster than the

24 speed of light to counter this eﬀect. Since dϕ/dt ≥ 0, it quite simply has to rotate in the direction of the rotation of the black hole, hence the name stationary limit surface.

Of course this is just an example and not a full solution to the photons geodesic, but it does illustrate the effect frame dragging can have. The magnitude of this effect is most dramatic inside the stationary limit surface, outside this surface it falls off rather quickly. If a rotating black hole can impact its surroundings in this way, perhaps the combined movement of millions of stars could drag the spacetime of a galaxy with it as well, impacting the rotation curves.

3.2 Cooperstock and Tieu

This is not the first time an attempt has been made to obtain flat rotational curves by using General Relativity alone. In their paper [4], Cooperstock and Tieu have attempted to explain galaxy rotation curves with an axisymmetric metric without needing to resort to Dark Matter. In this section we will give a brief description of the Cooperstock-Tieu model and the criticism it received from the scientific community. In this section we will not use geometrized units, as the method of Cooperstock and Tieu relies on an expansion in Newton’s constant G.

3.2.1 The method Cooperstock and Tieu use the general form of the axisymmetric metric:

ds2 = eψ(c dt − Ndϕ)2 − e−ψ eν (dρ2 + dζ2) + ρ2dϕ2 . (3.8) In the metric above ψ, N and ν are functions of the coordinates ρ and ζ, and the metric signature is (+, −, −, −) as opposed to the metrics we have used in the previous sections. Like the Kerr metric (equation 3.1) it has a term, N(ρ, ζ), that mixes the t and ϕ coordinates. This is the term that causes frame dragging. Instead of using the normalisation of the four-velocity and the geodesic equations, they choose a diﬀerent way to obtain the tangential velocity. µ µ They assume a comoving reference frame, meaning U = δ0 . In this frame the observer rotates with the matter, so it appears to stand still. The way they want to derive the tangential velocity is to diagonalise the metric with a local transform (meaning the ρ and ζ coordinates are held ﬁxed). The metric is called diagonal if the terms dtdϕ and dϕdt are gone, so that only the quadratic terms are left. The transform is given by:

ϕ¯ = ϕ + ω(ρ, ζ)t (3.9) Nceψ with ω(ρ, ζ) = (3.10) ρ2e−ψ − N 2eψ

25 They treat ω in the transformation above as the angular velocity and they deﬁne the tangential velocity as follows:

v(ϕ) = ω(ρ, ζ)ρ (3.11) The geodesic equations, along with the co-moving reference frame, can be used to determine the behaviour of ψ. We know that

dxµ U µ = = δµ dτ 0 d2xµ dxλ dxσ and + Γµ = 0. dτ 2 λσ dτ dτ We can combine these two equations to obtain

d2xµ + Γµ = 0 dτ 2 00 1 2 ⇒ Γ00 = 0, Γ00 = 0 (3.12)

We can obtain the Christoﬀel symbols from equation 1.3. Putting them equal to zero gives us constraints on the metric function ψ:

1 1 Γ1 = gρρ (−∂ g ) , Γ2 = gζζ (−∂ g ) 00 2 ρ tt 00 2 ζ tt ψ ψ ⇒ ∂ρgtt = ∂ρ e = 0, −∂ζ gtt = ∂ζ e = 0

⇒ ∂ρψ ≡ ψρ = 0, ∂ζ ψ ≡ ψζ = 0

The geodesic equation, combined with the co-moving reference frame demand that ψ = constant: Cooperstock and Tieu pick ψ = 0. In order to calculate the other metric functions, the Einstein equations need to be calculated. In order to do this Cooperstock and Tieu couple this metric to the energy-momentum that describes a pressure-free ﬂuid:

T µν = σ(ρ, ζ) U µU ν (3.13) In the equation above U is the four-velocity and σ(ρ, ζ) is the mass density. The Einstein equations can then be calculated, they are written below. In these equations subscripts are derivatives w.r.t. the respective variable, e.g. ψρ = ∂ρψ 2 and ψρρ = ∂ρψ.

26 The Einstein equations read:

2 2 2ρνρ + Nρ − Nζ = 0, (3.14)

ρνζ + NρNζ = 0, (3.15) 2 2 2 Nρ + Nζ + 2ρ (νρρ + νζζ ) = 0, (3.16) N N + N − ρ = 0, (3.17) ρρ ζζ ρ h ψ 3 e−ν ψ + ψ + ρ + ρ−2 N 2 + N 2 ρρ ζζ ρ 4 ρ ζ N 1 i 8πGσ(ρ) +Nρ−2 N + N − ρ − (ν + ν ) = (3.18) ρρ ζζ ρ 2 ρρ ζζ c2

Using equations 3.16 and 3.17, equation 3.18 can be rewritten to

N 2 + N 2 8πGσ(ρ) ∇2ψ + ρ ζ = eν (3.19) ρ2 c2 ψ where ∇2ψ ≡ ψ + ψ + ρ (the ﬂat-space Laplacian) (3.20) ρρ ζζ ρ

Since ψ was chosen to be a constant: ∇2ψ = 0, and the ﬁeld equations for N and σ (equations 3.17 and 3.19) are reduced to:

N ∂ N N N + N − ρ = 0 ⇒ ρ = − ζζ (3.21) ρρ ζζ ρ ∂ρ ρ ρ N 2 + N 2 8πGσ(ρ, ζ) ρ ζ = eν (3.22) ρ2 c2

ν 1 2 Now Cooperstock and Tieu perform an expansion e = 1 + ν + 2 ν + ... and ν is treated as being of order G. Then equation 3.22 can be expressed as:

N 2 + N 2 8πGσ(ρ, ζ) ρ ζ = + O(G2) (3.23) ρ2 c2 Equation 3.21 can in turn be written as:

Z ∞ 0 2 N(ρ , ζ) 0 ∇ Φ = 0 where Φ ≡ − 0 dρ (3.24) ρ ρ In the deﬁnition above Φ → 0 for ρ → ∞, in order to ensure asymptotic ﬂatness. An approximation can be made for equation 3.10. We use N 2eψ ρ2e−ψ, since N 2 ∼ O(G), to obtain Nc Nc ω(ρ, ζ) ≈ e2ψ = (3.25) ρ2 ρ2

27 From equation 3.11 follows that: Nc ∂Φ v(ϕ) = = c (3.26) ρ ∂ρ To ﬁt the rotation curve of a certain galaxy, a potential Φ has to be constructed. If we use separation of variables we can obtain the following:

Φ(ρ, ζ) = f(ρ)g(ζ) (3.27) ∂Φ ∂f(ρ) ∂2Φ ∂2f(ρ) ∂2Φ ∂2g(ζ) ⇒ = g(ζ), = g(ζ), = f(ρ) (3.28) ∂ρ ∂ρ ∂2ρ ∂ρ2 ∂2ζ ∂ζ2 Then we can rewrite the diﬀerential equation in equation 3.24:

∂2f(ρ) 1 ∂f(ρ) ∂2g(ζ) ∇2Φ = g(ζ) + + f(ρ) = 0 (3.29) ∂ρ2 ρ ∂ρ ∂ζ2 In order to solve this equation, we can make use of the Bessel functions. Any Bessel function Jα(kρ), with constants α and k, satisﬁes [12]:

d2J (kρ) 1 dJ (kρ) α2 α + α = − k2 J (kρ) (3.30) dρ2 ρ dρ ρ2 α

If we pick f(ρ) = CJ0(kρ) with C some constant, equation 3.29 becomes:

∂2g(ζ) f(ρ) − k2g(ζ) = 0 (3.31) ∂ζ2 In their paper, Cooperstock and Tieu pick g(ζ) = e−k|ζ| in order to solve this equation. This choice receives some critisism (as will be discussed in section 3.2.2), because of the presence of a Dirac delta function in the second derivative. This derivative comes from |ζ|. The absolute value is chosen to provide reﬂectional symmetry in ζ, which is necesarry to correctly model a galaxy. Coop- erstock and Tieu mention the problem of the discontinuity at ζ = 0, but state that it has no impact on the equations of motion and that it is a consequence of the choice of coordinates. They say other coordinates could be found in which this discontinuity would not be present, but these coordinates would only com- plicate the analysis.

We will put the discussion on the discontinuity aside for the moment. The linearity of equation 3.24 allows Φ to be written as a superposition:

X −kn|ζ| Φ(ρ, ζ) = Cne J0(knρ) (3.32) n Using equations 3.10, 3.11 and 3.32 we can write

(ϕ) X −kn|ζ| v = −c knCne J1(knρ) (3.33) n

28 since ∂ρJ0(kρ) = −kJ1(kρ). The constants kn are chosen such that the Bessel R 1 functions are orthogonal, meaning 0 J0(knρ)J0(kmρ)ρdρ = δmn. A velocity curve can be ﬁtted by picking the constants Cn. The function N can then be obtained from the ﬁt and used to calculate the density σ(ρ, ζ) from equation 3.23.

The method has been performed by Cooperstock and Tieu on multiple galaxies. The rotation curve of the Milky Way, together with the ﬁtted curve and the obtained density are displayed in ﬁgure 3.1.

Figure 3.1: The observed rotation curve of the Milky Way and the ﬁtted curve (left) and the obtained density at ζ = 0 (right) (adaptation from [4]). The radius ρ is given in kpc, the velocity is given in m/s and the density in kg/m3.

As mentioned in their paper, General Relativity makes no distinction between Dark and luminous matter. To check if Dark Matter is present in this model, Cooperstock and Tieu calculate the total mass through integration of the mass density, which can be rewritten using equation 3.22 and 3.26:

ZZZ M = σ(ρ, ζ)ρ dρdζdϕ

all space ∞ 2 2! 2 ! 1 Z ∂v(ϕ) ∂v(ϕ) v(ϕ) ∂v(ϕ) = ρ + + + 2v(ϕ) dρdζ 2G ∂ρ ∂ζ ρ ∂ρ 0

29 The value obtained from this integral can be compared to the observed luminous mass in the Milky Way, to determine if Dark Matter is present or not. For the 10 Milky Way an integrated mass of 21 × 10 M is obtained in the Cooperstock- Tieu model. According to Cooperstock and Tieu, this lies at the lower end of the estimated mass range (between 20 × 1010 and 60 × 1010 solar masses) for the Milky Way [4].

From this result, and some other results obtained for other galaxies, Cooperstock and Tieu conclude that relativistic eﬀects can explain the ﬂat rotation curves and that no Dark Matter is needed.

3.2.2 Criticism Their paper(s) have received criticism from others, to which they have provided some response [4]. In this section we will present the concerns expressed by some authors and the reaction of Cooperstock and Tieu.

Disk of exotic matter The main critism was that a singular disk of (exotic) matter was present in their model, according to Korzynski and Vogt & Letelier [13],[14].

The presence of |ζ| in equation 3.32, meant to induce symmetry between above and below the disk, can be thought of as a transform ζ → |ζ|. This induces derivatives in the metric that take the form of the Heaviside step function (H(ζ)) and the Dirac distribution (δ(ζ)), since N = ρ∂ρΦ. If we calculate the derivatives w.r.t. ζ of N we get:

∂2Φ N = ρ ζ ∂ζ∂ρ X ∂ ∂J0(knρ) = ρ C e−kn|ζ| n ∂ζ ∂ρ n X ∂J0(knρ) = ρ C (1 − 2H(ζ)) k e−kn|ζ| n n ∂ρ n ∂3Φ N = ρ ζζ ∂ζ2∂ρ 2 X ∂ ∂J0(knρ) = ρ C e−kn|ζ| n ∂ζ2 ∂ρ n X ∂J0(knρ) = ρ C k2 e−kn|ζ| − 2k δ(ζ) n n n ∂ρ n The Dirac delta distribution and the Heaviside step function enter the Einstein equations through the derivatives of N, and aﬀect the energy-momentum tensor

30 in this way. According to Korzynski and Vogt & Letelier this leads to the presence of a disk at ζ = 0. Calculating T µν reveals that this disk is made out of exotic material, namely either cosmic strings or material with a negative energy density.

Cooperstock and Tieu claim that this disk is not a physical property of the metric but rather a mathematical construct, meaning that the disk has no eﬀect on the dynamics of test particles in the system. In order to check wether the singular disk was purely mathematical or physical they study the ζ-geodesic equation:

d2ζ N N dζ 2 = ρ ζ (3.34) dτ 2 2ρ dτ If the disk is indeed physical, the negative mass of the disk would mean that a test particle would experience a violent repulsion, according to Cooperstock and Tieu. They calculated the right hand side of equation 3.34 for ρ ∈ [0.1, 30] and ζ ∈ [0.001, 1] for galaxy NGC7331 and found that all points gave a negative value. They conclude that since the acceleration is always negative, a test particle would never feel a repulsion. Therefore, the disk is not physical.

As a second check, instead of picking g(ζ) = e−kn|ζ| in equation 3.27, they pick: ( cosh (κ ζ) |ζ| < ζ g(ζ) = n 0 (3.35) −kn|ζ| e |ζ| > ζ0 where ζ0 > 0. This choice is continuous in ζ = 0, so would not lead to a singular disk and for large ζ it still asymptotically goes to ﬂat space. The “interior” solution |ζ| < ζ0 needs to be matched to the “exterior” solution |ζ| > ζ0 at ζ = ±ζ0. In other words, constants Cn, kn and κn will have to be chosen carefully s.t.:

Nint|ζ=±ζ0 = Next|ζ=±ζ0 (3.36)

∂N ∂N int = ext (3.37) ∂ζ ∂ζ ζ=±ζ0 ζ=±ζ0 These conditions require that

X X −knζ0 Cnκn cosh (κnζ0) J1(κnρ) = Cnkne J1(knρ) (3.38) n n

X 2 X 2 −knζ0 Cnκn sinh (κnζ0) J1(κnρ) = − Cnkne J1(knρ) (3.39) n n This is not easy to do, but with additional terms the rotation curve of a galaxy can still be ﬁtted while ensuring that the conditions above are satisﬁed.

31 Having shown that the disk has no repulsive effect in the ζ-direction on test particles, and that the issue can be evaded by choosing a different ζ-dependence for Φ, Cooperstock and Tieu feel confident that their model is still valid.

Ill-deﬁned expansion Another concern was expressed by Cross [15]. He commented that the expansion done in [4] in G (equation 3.23) was not strictly well deﬁned, as G has a dimension. Instead, he performs an expansion in the dimensionless quantity q GM λ = Lc2 , where M and L are some characteristic mass and length of the system respectively. Cross expands the metric as follows:

2 4 gtt = −1 − ψ + O(λ ) (3.40) 1 2 1 3 5 gtϕ = gϕt = N + ψN + N + O(λ ) (3.41) 1 2 2 2 2 4 gϕϕ = ρ − N − ρ ψ + O(λ ) (3.42) 2 2 4 gρρ = gζζ = 1 − ψ + ν + O(λ ) (3.43)

2 The overscript indicates the order of the term in λ, so ψ is the part of the metric function ψ that is of the order λ2. We can see that the zeroth-order terms are simply ﬂat space in cylindrical coordinates. Cross chooses the reference frame U ϕ dϕ 1 ω where the ﬂuid rotates with an angular velocity ω, meaning U t = dt c = c . Knowing this, and using the normalisation of the four-velocity, we can write:

µ ν 2 gµν U U = −c (3.44) dt 2 dt dϕ dϕ2 = g c2 + 2g c + g (3.45) tt dτ tϕ dτ dτ ϕϕ dτ dt 2 ω ω2 = c2 g + 2g + g (3.46) dτ tt tϕ c ϕϕ c2 dt ω ω2 −1/2 ⇒ = − g + 2g + g (3.47) dτ tt tϕ c ϕϕ c2

Then the energy-momentum tensor components T µν = σ(ρ, ζ) U µU ν have the expansion:

2 ω 1 ρ2ω2 T tt = σ(ρ, ζ)c2 1 + ψ − 2 N − + O(λ4) (3.48) c c2 T tϕ = T ϕt = σ(ρ, ζ)cω + O(λ3) (3.49) T ϕϕ = σ(ρ, ζ)ω2 + O(λ4) (3.50)

32 Since 8πG/c2 ∼ λ2, the Einstein equations are of order λ2 and higher. There- fore, the ﬁrst order Einstein equation for T tφ is

1 1 1 1 N − N + N = 0. (3.51) ρρ ρ ρ ζζ

1 Because of this, Cross chooses N = 0. With this choice, through third order the Einstein equations become:

2 1 2 2 8πGσ(ρ, ζ) ∇2ψ − ν + ν = (3.52) 2 ρρ ζζ c2 2 νρ = 0 (3.53) 2 2 νρρ + νζζ = 0 (3.54) 2 νζ = 0 (3.55) 1 3 1 3 3 8πGσ(ρ, ζ) N − N + N = ω (3.56) 2ρ2 ρρ ρ ρ ζζ c3 We can immediately see that ν is a constant. The remaining equations can be rewritten in order to get the following.

2 8πGσ(ρ, ζ) ∇2ψ = (3.57) c2 3 1 3 3 16πGσ(ρ, ζ) N − N + N = ρ2ω (3.58) ρρ ρ ρ ζζ c3 The coupling of N to the mass density is of third-order, and all the metric terms depend linearly on the mass density. This contradicts the expansion in G that was done by Cooperstock and Tieu, which led to a square root dependency: N ∼ σ(ρ, ζ)1/2. Now we can write

2 2Φ ψ = − N (3.59) c2

Where ΦN is the Newtonian gravitational potential. To study the eﬀects of this approximation on the tangential velocity, we take a look at the geodesic equation for the four-velocity in ρ:

d2ρ dxµ dxν + Γ1 = 0 dτ 2 µν dτ dτ d2ρ dt 2 dt dϕ dϕ2 ⇒ + Γ1 c2 + 2Γ1 c + Γ1 = 0 dτ 2 00 dτ 03 dτ dτ 33 dτ d2ρ dt 2 ω ω2 ⇒ + c2 Γ1 + 2 Γ1 + Γ1 = 0 (3.60) dτ 2 dτ 00 c 03 c2 33

33 For circular orbitsρ ˙ =ρ ¨ = 0, so then the geodesic equation requires:

ω ω2 Γ1 + 2 Γ1 + Γ1 = 0 (3.61) 00 c 03 c2 33 From the geodesic equation (equation 1.2) we can calculate:

1 Γ1 = − gρρ∂ g 00 2 ρ tt 1 Γ1 = − gρρ∂ g 03 2 ρ tϕ 1 Γ1 = − gρρ∂ g 33 2 ρ ϕϕ Then equation 3.61 can be rewritten to:

1 ω ω2 − gρρ ∂ g + 2 ∂ g + ∂ g = 0 2 ρ tt c ρ tϕ c2 ρ ϕϕ ω ω2 ⇒ ∂ g + 2 ∂ g + ∂ g = 0 ρ tt c ρ tϕ c2 ρ ϕϕ 3 2Φ ω2 ⇒ (Φ ) + ωcN + ρω2 1 + N + ρ2 (Φ ) = 0 N ρ ρ c2 c2 N ρ

Using v(φ) = ω ρ this can be rewritten to:

2 2 2 v(φ) v(φ) v(φ)c 3 v(φ) = − (Φ ) − (Φ ) − N − 2 Φ (3.62) ρ N ρ c2 N ρ ρ ρ ρc2 N Cross interprets this result as the ﬁrst term being Newtonian physics, and the following terms as being higher order contributions. He states that corrections of these orders cannot possibly account for the observed rotation curves.

1 Cooperstock and Tieu react to his critisism saying that putting N = 0 is precisely what they want to avoid with their calculation. Putting this term to zero is indeed the standard approach that comes to the conclusion that Newto- nian gravity is the lowest order contribution and that any further eﬀects cannot account for the observed rotation curves. The novel aspect of the Cooperstock- Tieu model is that the lowest order contribution now comes from the metric function N, and this contribution neatly ﬁxes the discrepancy between theory and observation.

Velocity calculation A third critisism was given by Menzies and Mathews [16]. They claim that the way the angular velocity was calculated is wrong. The transform used by Cooperstock and Tieu:

34 ϕ → ϕ¯ = ϕ + ω(ρ, ζ)t (3.63) diagonalises the metric for ω(ρ, ζ) = gtϕ . But when we calculated the frame gϕϕ dragging earlier on, we found that this had a similar form (equation 3.7). Ac- cording to Menzies and Mathews, this means that the quantity calculated by Cooperstock and Tieu is not the actual angular velocity of the dust particles, but rather the frame dragging induced by the metric.

The actual angular velocity comes from the geodesic equations, and needs to be calculated in the way that we have done before in chapter 2. This calculation would involve derivatives of the metric, since they appear in the Christoﬀel symbols. The ω used by Cooperstock and Tieu does not contain these derivatives. Cooperstock and Tieu have not reacted to this criticism.

Test of model While Cooperstock and Tieu come to the conclusion that their model explains galaxy rotation curves and comes up with the right mass density distribution without resorting to Dark Matter, de Almeida et al. disagree [17]. According to them, there is no reliable astrophysical data on the density distributions of galaxies. They propose a test that takes the uncertainties in these measurements into account and conclude that a signiﬁcant amount of Dark Matter is needed to ﬁx the Cooperstock-Tieu model.

The claim of de Almeida et al., is that non-Newtonian models like the Cooperstock- Tieu model lack a detailed investigation with respect to the baryonic matter inferred from their observations. Since there is no obvious way to compare the density proﬁle obtained from a model with one obtained from observation, they propose a diﬀerent method. They have devised a test for any model that receives a rotation curve as input and comes up with a mass density distribution. An essential part of this test is that it uses the error bars present in the observed rotation curve. The biggest contribution to these error bars comes from changes in redshift data at the same radius, which corresponds to violation of axial symmetry. Therefore these errors are a good way to test the reliability of axisymmetric models. The test can be summarised as follows:

(φ) The model’s circular velocities at ζ = 0: v (ρ, pi)(pi are model parameters) are fitted to the observed rotation curve of a galaxy. The fit determines the best fit parametersp ¯i. The 1σ-error bars from the rotation curve are translated into error bars on these parameters.

(φ) From these v (ρ, p¯i) the density proﬁle σ(ρ, ζ, p¯i ± δpi) can be constructed. This density can be plugged into the convential Newtonian equations to (φ) get an eﬀective Newtonian rotation curve: veN (ρ, p¯i ± δpi). The errors

35 on the parameters now translate into an error in the eﬀective Newtonian rotation curve. From the assumed matter content of the studied galaxy, a Newtonian (φ) rotation curve (vN ) can be constructed in the usual way. Say we assume the galaxy consists of a bulge, a disk, some dust and some Dark Matter, then we can add each contribution to the velocity separately. We then get 2 2 2 2 2 (φ) (φ) (φ) (φ) (φ) for the velocity: vN = vbulge + vdisk + vgas + vDM .

If the theory considered is compatible with both the observed rotation (φ) (φ) curve and the assumed matter content of the galaxy, veN and vN should be mutually compatible.

After performing this test, de Almeida concludes that a large amount of Dark Matter is needed to ﬁx the Cooperstock-Tieu model.

At first glace it seemed like the Cooperstock-Tieu model was exactly what we were looking for: a relativistic model that predicts flat rotation curves without the need for Dark Matter. However, the critiques given by different authors are valid and the results of Cooperstock and Tieu could not be reproduced with corrected models. We should remember the flaws in the Cooperstock-Tieu model for when we want to build our own model. In the next section we will discuss an approach that we think is more rigorous, as the metric proposed there very accurately describes an infinitesimally thin rotating disk of dust without resorting to a weak field limit, or an expansion in G.

36 3.3 Rotating Dust

Neugebauer, Kleinwachter and Meinel have derived a metric that describes a inﬁnitesimally thin, rotating disk of dust [18]. We could use this metric to simulate a galaxy, the dust representing the stars and gas that it contains. This metric is not easy to obtain analytically, this is why we have chosen to write a computer program instead, which can calculate the metric functions numerically. The way the program calculates the metric follows the procedure described by Ansorg [5]. In the section below we will explain the diﬀerent aspects of this calculation. Then we will show how we calculated the simplest non-trivial case analytically. The results of this calculation can be used to check if the program works.

3.3.1 Obtaining the metric Staying as close as possible to the notation used by Ansorg, we use an axisymmetric metric:

ds2 = −e2ψ (dt + a dϕ)2 + e−2ψ e2k dρ2 + dζ2 + ρ2dϕ2 (3.64) We make use of an axisymmetric, stationary metric so the metric functions depend on ρ and ζ alone: ψ(ρ, ζ)1,a(ρ, ζ) and k(ρ, ζ). This metric is equivalent to the one used by Cooperstock and Tieu (equation 3.8). We want to describe a rotating disk of dust, so the energy-momentum tensor looks like:

T µν = σ(ρ)eψ−kδ (ζ) U µU ν , (3.65) where σ(ρ) is the surface density of the inﬁnitesimally thin disk and U is the four-velocity of the dust material. We assume the inertial frame of an observer that sees the disk rotate with an angular velocity Ω(ρ), without moving in the ρ or ζ direction. Then our four-velocity is

 1  µ −A(ρ) µ µ −A(ρ)  0  U = e (ξ + Ω(ρ)η ) = e   . (3.66)  0  Ω(ρ) Like before, ξµ and ηµ are the Killing vectors corresponding to stationarity and axisymmetry.

1Note that in our computer program the metric function ψ is named U, since this is the notation used by Ansorg in his paper. However, since U is associated with the four-velocity in this thesis we have chosen ψ to represent this metric function instead.

37 The function A(ρ) is there to ensure that the four-velocity is normalised, it can be calculated by checking the normalisation condition:

µ −2A(ρ) i i U Uµ = e (ξ + Ωη )(ξi + Ωηi) = −1 2A(ρ) i i ⇒ e = −(ξ + Ωη )(ξi + Ωηi) 2 = −(gtt + 2Ωgtϕ + Ω gϕϕ) = e2ψ(1 + aΩ)2 − ρ2Ω2e−2ψ (3.67)

Now that we know the components of the energy-momentum tensor, we need to calculate the Einstein equations. For a metric like this one, these are equivalent to a single complex equation - the Ernst equation [5]:

(<(f)) ∆f = (∇f)2 (3.68)

In this equation <(f) is the real part of f, and

∂2 1 ∂ ∂2 ∂ ∂ ∆ = + + , ∇ = , (3.69) ∂ρ2 ρ ∂ρ ∂ζ2 ∂ρ ∂ζ Solutions to the Ernst equation are given by the Ernst potential f, from which the metric functions can be obtained:

e4ψ e4ψ f = e2ψ + ib with b = a , b = − a (3.70) ζ ρ ρ ρ ρ ζ The metric function k can be calculated as follows:

k 1 h i ρ = (ψ )2 − (ψ )2 + e−4ψ (b )2 − (b )2 (3.71) ρ ρ ζ 4 ρ ζ k 1 ζ = 2ψ ψ + e−4ψb b (3.72) ρ ρ ζ 2 ρ ζ So, if the metric functions depend on each other in the way described above, the metric satisﬁes the Einstein equations for a vacuum.

38 We also need a way to connect the density proﬁle and the angular velocity (σ and Ω) to the metric functions if we want our metric to describe an observed galaxy. Ansorg has integrated the Einstein equations from the lower to the upper side of the disk (so that he integrates over the Dirac delta distribution), to obtain the following conditions on the metric functions [5]:

1 2πσ = eψ−k ψ + Q (3.73) ζ 2 4ψ 2 4ψ 2 e Q + Q e ζ + (bρ) = 0 (3.74) Q Ω = (3.75) aζ − aQ −4ψ h 2ψ 2ψ i where Q = −ρe bρbζ + e ρ e ζ (3.76) These equations are valid for ζ = 0+ (meaning ζ approaches zero while remaining positive) and 0 ≤ ρ ≤ ρ0, where ρ0 is the radius of the disk. In these equations Ω is the angular velocity Ω = Ω(ρ) = U ϕ/U t of the disk. The second equation does not feature σ or Ω because this condition refers to the nature of the material that the disk is made of, namely dust. Note that only one of the quantities σ or Ω can be prescribed, as they describe diﬀerent boundary conditions on the same Ernst potential. Physically, this makes sense as the motion of the matter in the disk is determined completely by the matter distribution since the only force considered is gravity.

When we have obtained such a metric, having fitted the angular velocity Ω to experimental data, we want to check if the Komar mass and the Komar angular momentum are similar to the observed quantities. In order to calculate the Komar integrals we will use their definitions (eqns. 1.13 and 1.14) together with the four velocity defined in 3.66. Explicit expressions for M and J have been derived for a ρ-independent rotation curve by Neugebauer, Kleinwachter and Meinel [18]. We can generalize these expressions when we want to calculate our case, where Ω does depend on ρ. We get:

Z ρ0 1 −A µ M = ψζ + Q (1 + 2Ωe U ηµ)ρdρ (3.77) 0 2 Z ρ0 1 −A µ J = ψζ + Q e U ηµρdρ (3.78) 0 2 with A coming from our inertial frame (equation 3.66), so that

−A µ −2A e U ηµ = e (gtϕ + Ωgϕϕ) Ωρ2 − ae4ψ(1 + aΩ) = e4ψ(1 + aΩ)2 − ρ2Ω2 We end up with expressions for M and J in terms of the metric functions.

39 Obtaining the potential In order to find the metric functions we need to know the Ernst potential f(ρ, ζ). For a given set of complex numbers {Yν }q = {Y1, ..., Yq} with integer q ≥ 1, and an even, real analytic function g(x) defined on the interval x ∈ [0, 1], the expression below ensures that f satisfies the Ernst equation (3.68) [5]. A bar denotes complex conjugation and an asterisk is defined by c∗ = (¯c)−1.

1 1 1 1 1 ··· 1 1 ∗ ∗ ∗ ∗ ∗ ∗ −1 α1λ1 α1λ1 α2λ2 α2λ2 ··· αqλq αq λq 2 ∗ 2 2 ∗ 2 2 ∗ 2 1 λ1 (λ1) λ2 (λ2) ··· λq (λq ) 3 ∗ ∗ 3 3 ∗ ∗ 3 3 ∗ ∗ 3 −1 α1λ1 α1(λ1) α2λ2 α2(λ2) ··· αqλq αq (λq )

...... 2q ∗ 2q 2q ∗ 2q 2q ∗ 2q 1 λ1 (λ1) λ2 (λ2) ··· λq (λq ) f = f0 , (3.79) 1 1 1 1 1 ··· 1 1 ∗ ∗ ∗ ∗ ∗ ∗ 1 α1λ1 α1λ1 α2λ2 α2λ2 ··· αqλq αq λq 2 ∗ 2 2 ∗ 2 2 ∗ 2 1 λ1 (λ1) λ2 (λ2) ··· λq (λq ) 3 ∗ ∗ 3 3 ∗ ∗ 3 3 ∗ ∗ 3 1 α1λ1 α1(λ1) α2λ2 α2(λ2) ··· αqλq αq (λq )

...... 2q ∗ 2q 2q ∗ 2q 2q ∗ 2q 1 λ1 (λ1) λ2 (λ2) ··· λq (λq ) with

Z 1 (−1)qg(x)dx f0 = exp − (3.80) −1 ZD p 2 2 with ZD = (ix − ζ/ρ0) + (ρ/ρ0) , <(ZD) < 0 (3.81) r Yν − iz¯ 1 ∗ ¯ λν = with z = (ρ + iζ), λν λν = 1 (3.82) Yν + iz ρ0 Z 1 q 1 − γν (−1) g(x)dx αν = with γν = exp λν (Yν + iz) , (3.83) 1 + γν −1 (ix − Yν )ZD ∗ αν α¯ν = 1 (3.84)

But which set {Yν }q and which function g(x) do we take? This all depends on the Ω and σ that we want our disk of dust to have. Ansorg prescribes a way to get from an input rotation curve to its corresponding g and {Yν }q.

40 Ansorg writes the function g and the input rotation curve or density proﬁle as Chebyshev polynomial expansions [19]. These functions satisfy the recurrence relation Tn+1(x) = 2xTn(x) − Tn−1(x) (3.85)

If x ∈ [−1, 1], Tn(x) = cos(n arccos(x)) is the n-th Chebyshev polynomial of the ﬁrst kind. For a given function f(x) and some integer N, if

N N 2 X X 1 c ≡ f(x )T (x ) then f(x) ≈ c T (x) − c . (3.86) j N k j k n n 2 0 k=0 n=0

The points xk at which f is evaluated are the roots of the Chebyshev poly- 1 nomials: xk = cos(π k + 2 /N). The function f(x) can then completely be described by an array of the coefficients cn. Another advantage of Chebyshev polynomials is that the derivative or primitive of a Chebyshev polynomial can be written as different Chebyshev polynomials. This means that df/dx and R f(x)dx can be calculated through a few operations on the array of coefficients cn. In a computer program, this is much more accurate and efficient than performing numerical differentiation or integration. Ansorg uses these polynomials to write

N X1 1 g(x) ≈ g T (x) − g (3.87) n 2n 2 0 n=0 N X2 1 Ω(y) ≈ Ω T (y) − Ω (3.88) n 2n 2 0 n=0 N X2 1 or σ(y) ≈ σ T (y) − σ (3.89) n 2n 2 0 n=0

The higher N1 and N2, the better approximations these expansions will be. The fitting parameters are the numbers gi and Ωi or σi. The arguments of Ω and σ have been substituted by y = ρ/ρ0, so that y represents the relative radius within the disk. Then they are defined on y ∈ [0, 1], and the Chebyshev polynomials are equal to the definition given above. Only even terms will be taken into account, to preserve the symmetry ρ → −ρ. The reason that we choose a Chebyshev polynomial expansion is that it is easy to work with such polynomials in a program. Instead of defining some complicated function we just need an array of numbers. We want to do the same for the set {Yν }q, so we describe the set of numbers with a function ξ(x):

" q # 1 Y iYν − x ξ(x) = ln , x ∈ [−1, 1] (3.90) x iY + x ν=1 ν with ξ(−x) = ξ(x). The inverse procedure; extracting the numbers from a given function ξ, is also prescribed by Ansorg.

41 We start by writing

q q Y iYν − x Pq(−x) X exp [xξ(x)] = = with P (x) = c xa (3.91) iY + x P (x) q a ν=1 ν q a=0

Here Pq(x) is simply a polynomial. The polynomial coefficients ca can be found by evaluating the function at q +1 different points xi ∈ [0, 1], with i = 0, 1, ..., q. Then we get a system of equations from which we can obtain the coefficients:

q q X a X a exp [xiξ(xi)] caxi = ca(−xi) . (3.92) a=0 a=0 Having obtained the coeﬃcients, the entire polynomial is known. We can then easily calculate its roots xk and write

q Y Pq(x) = (x − xk). (3.93) k=1

Comparing this expression with equation 3.91, we can identify: Yk = ixk. Like g, Ω and σ we can expand ξ into Chebyshev polynomials:

N X2 1 ξ(x) = ξ T (x) − g with T (x) = cos(n arccos(x)) (3.94) n 2n 2 0 n n=0 So given two real, analytic functions g(x) and ξ(x), deﬁned on x ∈ [0, 1], we can obtain the Ernst potential f(ρ, ζ). The set {Yν }q can be calculated from ξ(x), and then f can be obtained from equation 3.79. The integrals in equations 3.80 and 3.83 have rather complicated solutions, which will be discussed in appendix A. The question that is left is: how are the expansions of the input rotation curve and proﬁle density connected to g and ξ?

In order to find the g and ξ belonging to a specific input σ or Ω, Ansorg pro- poses the Newton-Raphson method. This is a method for finding the roots of a function. For a given function h(x), we guess x0 is a root of h. The method of Newton-Raphson states that

h(x0) x1 = x0 − 0 (3.95) h (x0) is an improved guess for the root. Here h0(x) is the derivative of h w.r.t. x. This can be seen when we perform a Taylor expansion around the actual rootx ˜ (for which h(˜x) = 0) and compare it to h(x0 + δ), where δ is the diﬀerence between x˜ and x0:

T aylor 0 2 0 = h(˜x) = h(x0 + δ) = h(x0) + δh (x0) + O(δ ) (3.96)

42 0 Then to second order: δ = −h(x0)/h (x0). The more orders we take into account, the closer our guess will come to the actual root. We can repeat the process:

h(xn) xn+1 = xn − 0 (3.97) h (xn) until we get a satisfactory approximation. The method only works when our initial guess x0 is close enough, however. If it is not, the Taylor approximation will not work and the method fails.

This method can also be applied to functions with multiple variables. We can write the rotation curve as a function of the input parameters gi and ξj where i = 0, 1, ..., N1 and j = 0, 1, ..., N2. We can then deﬁne a function:

h (g0, ..., gN1 , ξ0, ..., ξN2 , x) = Ω (g0, ..., gN1 , ξ0, ..., ξN2 , x) − Ωobs(x) (3.98)

. Here Ωobs is the observed angular velocity and Ω is the angular velocity from a metric that was obtained from guessed functions g and ξ. The root of this function corresponds to the parameters of the unique metric that has the prescribed rotation curve. Instead of the rotation curve, the density proﬁle could also be used in the same way. We can use the Newton-Raphson method to ﬁnd a good approximation of the root of this function, and in that way obtain a metric with the correct rotation curve. All we need now is a good initial guess. For this guess we will look at the Newtonian limit, as will be discussed in the next section.

3.3.2 The Newtonian limit The initial guess required by the Newton-Raphson method can be obtained from the Newtonian limit, where the dependency of the metric functions on the rotation curve can be simpliﬁed. To calculate the metric in this limit, Ansorg deﬁnes the parameter

M 2 ε = . (3.99) J where M is the Komar mass and J is the Komar angular momentum. If this parameter is small (i.e. ε 1) we ﬁnd ourselves in the Newtonian regime. When ε → 1 we go to the ultrarelativistic regime and our metric reduces to the Kerr metric. In the limit of small ε Ansorg expands the functions g and ξ as

2 4 g (x) = ε g0 (x) + O(ε ) (3.100) 2 ξ (x) = εξ0 (x) + O(ε ) (3.101)

43 In this expansion, the Ernst potential becomes [5]:

2 3 4 f = 1 + e2(ρ, ζ)ε + ib3(ρ, ζ)ε + O(ε ). (3.102)

In equation 3.102 the functions e2 and b3 are given by:

1 1 Z Z g0(x)dx ix g0(x)ξ0(x)dx e2(ρ, ζ) = − , b3(ρ, ζ) = − (3.103) ZD ZD −1 −1

The boundary conditions are given either by the rotation curve or the density proﬁle, which are also expanded into powers of ε:

p 2 2 4 σ(ρ) = σ0σ2(ρ) 1 − (ρ/ρ0) ε + O(ε ) with σ2(0) = 1 or (3.104) 3 Ω(ρ) = Ω0Ω1(ρ)ε + O(ε ) with Ω1(0) = 1. (3.105)

So, to prescribe a certain density proﬁle, the constant σ0 and the function σ2(ρ) should be provided as input. If one prefers to prescribe a rotation curve, the constant Ω0 and the function Ω1(ρ) are the input parameters. If we plug these expansions into equations 3.70 to 3.76, we can obtain simpliﬁed relations between the metric functions and the input parameters. In the following equations + 0 < ρ < ρ0 and ζ = 0 , meaning ζ approaches zero while remaining positive. This last demand is very important, as it means many terms will drop out of calculations. It also means we will have to be careful to avoid the singularities that many of our equations possess at ζ = 0. First, using equation 3.102, we can write:

1 e2ψ ≈ 1 + 2ψ ⇒ ψ = e ε2 + O(ε4) (3.106) 2 2 3 b = b3ε (3.107)

Then we can immediately see from equations 3.71 and 3.72 that

k k ρ = ζ = O(ε4) (3.108) ρ ρ This means that in the regime of small ε, k is a constant. We pick k = 0. From the deﬁnition of Q (equation 3.76) we can obtain

4 O(ε6) O(ε ) z }| { −4ψh z}|{ 2ψ 2ψ i 6 Q = −ρe bρbζ + e ρ e ζ = −4ρψρψζ + O(ε ) (3.109)

44 We can plug this into equation 3.73 to get:

1 2πσ(ρ) = eψ−k ψ + Q ζ 2 ψ 4 = ψζ e + O(ε ) 4 = ψζ + O(ε ) 2 4 ⇒ (e2)ζ ε = 4πσ(ρ) + O(ε ) p 2 ⇒ (e2)ζ = 4πσ0σ2(ρ) 1 − (ρ/ρ0) (3.110) So the input density proﬁle can be linked directly to the Ernst potential. We can also plug equation 3.109 into equation 3.74 to obtain:

O(ε8) O(ε6) z }| { z }| { 4ψ 2 4ψ 2 e Q + Q e ζ + (bρ) = 0 2 ⇒ (bρ) = −4Qψζ 2 = 16ρψρψζ 2 2 ⇒ (b3)ρ = 2ρ (e2)ρ (e2)ζ (3.111) We can link the rotation curve to the metric functions, through equation 3.75. 4 3 Since Q ∼ O(ε ) and aζ ∼ O(ε ) (cf. equation 3.70), aQ aζ and we can approximate:

Q Q Ω = ≈ aζ − aQ aζ This equation can in turn be rewritten to:

Q Ω = −4ψ −ρbρe Q ≈ −ρbρ ψ ψ = 4 ρ ζ + O(ε3) bρ

Since we expanded the angular velocity to ﬁrst order only (equation 3.105), we will cut oﬀ the expansion at the third order term, in order to obtain:

(e2)ρ (e2)ζ Ω0Ω1 = (3.112) (b3)ρ

45 Now we can use equation 3.111 to get the following expressions:

2 2 (e2)ρ = 2ρΩ0Ω1 (3.113)

(b3)ρ = 2ρΩ0Ω1(e2)ζ (3.114)

It will also be useful to calculate the Komar integrals (equations 3.77 and 3.78) in the regime for small ε. The Q-term in these integrals will drop out right away, since it is of order ε4. We will also need the following:

Ωρ2e−4ψ − a(1 + aΩ) e−AU µη = µ (1 + aΩ)2 − ρ2Ω2e−4ψ ≈ Ωρ2e−4ψ − a(1 + aΩ)

The approximation that we made in the previous equation is that the leading term in the denominator is 1, the rest is order ε or higher. Since ε 1, the fraction is approximately equal to its numerator. We can rewrite the expression even further:

−A µ 2 4 e U ηµ ≈ Ωρ (1 − 4ψ) − a(1 + aΩ) + O(ε ) = Ωρ2 − a(1 + aΩ) + O(ε4) (3.115)

The ﬁrst term is of order ε. Then we can write M to ﬁrst order:

Z ρ0 1 −A µ M = ψζ + Q 1 + 2Ωe U ηµ ρ dρ 0 2 Z ρ0 3 = ψζ ρ dρ + O(ε ) 0 Z ρ0 = 2π σ(ρ)ρ dρ + O(ε3) (3.116) 0

−A µ For the Komar angular momentum we will use the ﬁrst order of e U ηµ to obtain:

Z ρ0 1 −A µ J = ψζ + Q e U ηµρ dρ 0 2 Z ρ0 3 4 = ψζ Ω(ρ)ρ dρ + O(ε ) 0 Z ρ0 = 2π σ(ρ)Ω(ρ)ρ3 dρ + O(ε4) (3.117) 0 These ﬁrst order approximations are precisely the classical expressions for the total mass and total angular momentum in cylindrical coordinates, so we can feel conﬁdent that this is indeed the Newtonian limit.

46 Now that we have obtained simpliﬁed relations between the metric and the input parameters, we need to ﬁnd out which functions g0 and ξ0 generate such a potential. The functions g0 and ξ0 that generate the metric can be obtained by inverting the expressions in equation 3.103. Ansorg has done this in his paper [5], and came up with the expressions below.

Z π/2 2 2 2 2 2 g0(x) = −4σ0(1 − x ) sin (ϕ) σ2 cos ϕ + x sin ϕ dϕ (3.118) 0 Z π/2 2 2 2 2 2 g0(x)ξ0(x) = 8σ0Ω0(1 − x ) sin (ϕ) Ω˜ 1 cos ϕ + x sin ϕ dϕ (3.119) 0

with Ω˜ 1(ρ) = Ω1(ρ)σ2(ρ) (3.120)

Using these equations, one can express the metric functions ψ, a, k as functions of the given σ(ρ) or Ω(ρ) and a prescribed value for ε.

3.3.3 A constant solution In this section we will look at a speciﬁc solution in the regime ε 1. To illustrate the complexity of the method developed by Ansorg, let us start with the simplest non-trivial solution; we prescribe σ2 = 1 so that we obtain a density proﬁle

p 2 2 σ(ρ) = σ0 1 − (ρ/ρ0) ε . (3.121) This density proﬁle is not meant to represent any known galaxy, it is simply meant as an example. From this density it is straightforward to obtain g0:

π/2 Z 2 2 g0(x) = −4σ0(1 − x ) sin (φ) dφ 0 2 = −πσ0(1 − x ) (3.122)

We mentioned in the previous section that in the limit of small ε the functions e2 and b3 can be written as integrals of g0 and ξ0 (equation 3.103). For these q 2 2 calculations we take ZD = − (ix − ζ/ρ0) + (ρ/ρ0) , s.t. <(ZD) < 0.

47 If we take the derivatives with respect to ρ and ζ we get the following:

ζ Z 1 g0(x) ix − ρ0 (e2)ζ = dx (3.123) 2 23/2 −1 ζ ρ ix − + ρ0 ρ0 ρ0

1 ρ Z g0(x) ρ0 (e2)ρ = − dx (3.124) 2 23/2 −1 ζ ρ ix − + ρ0 ρ0 ρ0

1 ρ Z (ix)g0(x)ξ0(x) ρ0 (b3)ρ = − dx (3.125) 2 23/2 −1 ζ ρ ix − + ρ0 ρ0 ρ0

Since we have g0, we can calculate (e2)ρ and (e2)ζ . The complete calculation is rather long, and can be found in appendix B. The result of all that work is:

∂e2 2 = π ρσ0 (3.126) ∂ρ ζ→0 If we compare this expression to equation 3.114 we obtain

π2σ Ω2Ω2 = 0 = constant. (3.127) 0 1 2

Since it is required that Ω1(0) = 1, it must be true that Ω1(ρ) = 1. We can also p 2 identify Ω0 = π σ0/2. Using the expansions from equations 3.116 and 3.117, the Komar integrals can now be expressed in terms of the constants σ0,Ω0, ρ0 and ε:

ρ ρ Z 0 Z 0 M = 2π σ(ρ)ρ dρ J = 2π σ(ρ)Ω(ρ)ρ3 dρ (3.128)

0 0 ρ0 s ρ0 s Z 2 Z 2 2 ρ 3 ρ 3 = 2πσ0ε 1 − 2 ρ dρ = 2πσ0Ω0ε 1 − 2 ρ dρ (3.129) ρ0 ρ0 0 0 2 4 = πρ2σ ε2 = πρ4σ Ω ε3 (3.130) 3 0 0 15 0 0 0 2 We can use the relation ε = M /J to obtain the constants σ0 and Ω0. We simply solve the system of equations:

2 p 2 M 5πσ0 Ω0 = π σ0/2, ε = = ε (3.131) J 3Ω0 3π 9 ⇒ Ω = , σ = (3.132) 0 10 0 50

48 4 We will take the Milky Way as a typical disk of dust, meaning ρ0 ≈ 1.5 · 10 −5 11 pc. If we put ε = 1.14 · 10 , we get that M ≈ 0.011 pc ∼ 10 M , which is approximately the mass of the Milky Way. However, we then also obtain J ≈ 10.6 pc2 ∼ 1069Js, while the actual value of the angular momentum of the Milky Way is more in the region of 0.026 pc2 ∼ 1067 Js. This discrepancy is to be expected, since Ω(ρ) = constant implies v(φ) ∼ ρ. In ﬁgure 3.2 we have plotted the mass density proﬁle and the angular velocity for this disk.

Figure 3.2: The surface mass density σ(ρ) (left) and the angular velocity Ω(ρ) (right) for a disk with ψ2 = Ω1 = 1.

Now that we have obtained the observables, we can compare them to the output generated by our computer program as a check. We have already obtain the function g0(x), but we will also need ξ0(x). Now that we know Ω˜ 1, this function is easy to obtain from equation 3.119:

Z π/2 2 2 g0(x)ξ0(x) = 8σ0Ω0(1 − x ) sin (φ)dφ 0 2 = 2πσ0Ω0(1 − x )

⇒ ξ0(x) = −2Ω0

49 When we plug these functions into our program, it returns the following output:

Figure 3.3: The surface mass density σ(ρ) (upper) and the angular velocity Ω(ρ) (lower) for a disk with ψ2 = Ω1 = 1, as returned by our program.

The scaling is a little bit different, but we can see that the density profile σ(ρ) matches the shape of the one in figure 3.2 very well. The scaling issue comes from a different value of ε used in the program, this was done because using the value ε = 1.14 · 10−5 leads to issues with precision and ruins the calculation. The angular velocity Ω(ρ) does not match its analytically derived counterpart very well, it starts out close to zero but quickly starts behaving eratically with the function blowing up to ±∞ for certain values of ρ.

50 If we take a look at the Ernst Potential derived by the program we can see why:

Figure 3.4: The real (left) and imaginary (right) parts of the Ernst potential derived by the program.

The real part is calculated correctly, but the imaginary part (originating from the ratio of the matrix determinants in equation 3.79) cannot be calculated for small values of ρ/ρ0 and approaches zero for larger values of ρ/ρ0, while this should not be the case. The reason for this error is that the terms in the integral for γν (equation 3.83) are very large, but need to combine to a small number. The precision of the method used by our program (as explained in appendix A) is not high enough to correctly calculate these contributions, and the program rounds them off to zero. In order for the program to work, this will need to be fixed first.

51 Chapter 4

Conclusion and Outlook

We have tried many different ways to find relativistic effects on the dynamics of different kinds of systems. Although we have not been able to explain flat rotation curves with relativity alone, we have built up a nice toolbox for studies to come.

First, we constructed a way to obtain the tangential velocity from a given metric. We used this method in chapter 2 to see how the combination of MOND and General Relativity would aﬀect the rotation of matter in a spherically symmetric spacetime. We found that it was not so easy to ensure circular orbits with a ﬂattened rotation curve while still keeping a vacuum solution.

After that we looked at the approach used by Matos et al.[11]. He first constructed an axisymmetric spacetime and then looked at the constraints on the energy-momentum tensor imposed by demanding circular orbits with ρ- independent tangential velocities. He found that some type of exotic matter was needed, but that General Relativity allows for more different kinds of exotic matter than the Newtonian approach. This caused him to stress the importance of researching these effects further.

In chapter 3 we first took a look at the paper written by Cooperstock and Tieu [4]. Their model explained the flat rotation curves without resorting to any type of exotic matter, as the density profiles matched observed ones quite well. Their approach was not without flaws however, as we have seen in the critical response given by other scientists. Since we are trying to do the same as Cooperstock and Tieu, it would serve us well to study their model and the response of the crit- ics. Firstly, Cooperstock and Tieu chose to implement the ζ-symmetry present in galaxies as a term with a factor |ζ|. The discontinuity in the derivatives of this function resulted in a disk of exotic matter in the energy-momentum tensor [13][14]. Secondly, Cooperstock and Tieu’s expansion parameter was not dimensionless, which makes the expansion ill-defined [15]. Lastly, there are subtle differences between rotations, velocities and frame dragging, that Cooperstock

52 and Tieu did not take into account [16].

When we build our own model, we will want to avoid the mistakes Cooperstock and Tieu made. We base our model on the metric derived by Neugebauer, Kleinwachter and Meinel [18]. This metric describes a rotating disk of dust, which seems to us is a perfect way to simulate a rotating galaxy. A fundamental difference between this method and the others is that a rotation curve is given as an input parameter in this model and the density profile is returned as output. In previous attempts the rotation curve always followed from a given metric. Now we look for the right metric to describe a given rotation curve. The metric is rather complicated and has to be computed numerically, but could give us insight in how the stars in a galaxy move. The computer program written can so far calculate the metric function from a given Ernst potential. Obtaining the Ernst potential from given gi and ξi is still a problem. Something seems to go wrong with the calculation of the imaginary part of the Ernst potential. If this is fixed, the program could be implemented in a Newton-Raphson scheme. Then not only Newtonian solutions, but also solutions where the relativistic effects are stronger could be calculated [5].

Although we have not yet found a deﬁnitive answer to the question wether General Relativity has a measurable eﬀect on the rotation curves of galaxies, we do feel that we have taken a step in the right direction. This thesis will provide a nice starting point for any research continuing in this direction.

53 Appendix A

Integrals for the Ernst Potential

The calculation of the Ernst potential 3.70 involves two integrals that depend on the expansion coeﬃcients gn and the complex numbers {Yν }q, as described in [5]. The way our program calculates these integrals will be described in this appendix. For the sake of brevity the coordinates ρ and ζ used in these calculations are the relative radius and height, so to obtain the results in the original notation perform ρ → ρ/ρ0 and ζ → ζ/ζ0. The integrals are

Z 1 (−1)qg (x) dx f0(ρ, ζ) = exp − (A.1) −1 ZD(ρ, ζ) Z 1 (−1)qg(x)dx and γν = exp λν (Yν + iz) (A.2) −1 (ix − Yν )ZD(ρ, ζ) p 2 2 with ZD(ρ, ζ) = (ix − ζ) + ρ (A.3)

If we write g(x) as its Chebyshev polynomial expansion, we get1

Z 1 N1 Z 1 g(x)dx X T2n(x)dx = gn p 2 2 p 2 2 −1 (ix − ζ) + ρ n=0 −1 (ix − ζ) + ρ Z 1 N1 Z 1 g(x)dx X T2n(x)dx and = gn p 2 2 p 2 2 −1 (ix − Yν ) (ix − ζ) + ρ n=0 −1 (ix − Yν ) (ix − ζ) + ρ

1 1 One might have noticed that a term 2 g0 is missing in the integral. Since T0(x) = 1, we 1 can redeﬁne g0 → 2 g0 and then the extra term is not neccesary anymore. We have done this for the sake of brevity.

54 We can rewrite the ﬁrst integral to obtain:

Z 1 T (x)dx Z 1 T (x)dx n = n p 2 2 p 2 2 −1 (ix − ζ) + ρ −1 (ix − ζ) + ρ Z 1 T (x)dx = n p 2 2 −1 ρ − (x + iζ) Z 1 Tn(x)dx = q −1 x ζ 2 ρ 1 − ( ρ + i ρ ) x ζ dy 1 Now substitute: y = + i , = : ρ ρ dx ρ Z a+ib T ( 1 (y − ib))dy 1 ζ = n a with a = , b = p 2 −a+ib 1 − y ρ ρ

The idea is to split the Chebyshev term into its polynomial coeﬃcients. Then we get, for the even terms:

Z x2n √ dx = arcsin(x) for n = 0 1 − x2 1 p = arcsin(x) − x 1 − x2 for n = 1 2 3 p 2 = arcsin(x) − x 1 − x2 x2 + 1 for n = 2 8 3 5 p 8 2 = arcsin(x) − x 1 − x2 x4 + x2 + 1 for n = 3 16 15 3 . . . .

For the odd terms we get:

Z x2n+1 p √ dx = − 1 − x2 for n = 0 1 − x2 2p 1 = − 1 − x2 1 + x2 for n = 1 3 2 8 p 1 3 = − 1 − x2 1 + x2 + x4 for n = 2 15 2 8 16p 1 3 5 = − 1 − x2 1 + x2 + x4 + x6 for n = 2 35 2 8 16 . . . .

55 Looking at the ﬁrst few values of n, we can come up with the following equivalent expressions:

" n # Z x2n X p √ dx = f(n) arcsin(x) − g(m)x2m−1 1 − x2 2 1 − x m=1 " n # Z x2n+1 p X √ dx = −g(n + 1) 1 − x2 f(m)x2m 2 1 − x m=0 (2n − 1)!! (2n − 2)!! where f(n) = and g(n) = (2n)!! (2n − 1)!!

These expressions were checked with Maple, and are valid for all integers n. P2n m Then we can write T2n(ρy − iζ) = m=0 αmy , collect the powers of y and come to the following expression:

a+ib Z a+ib " n # T2n(ρy − iζ)dy X = f(k)α2k arcsin(y) p 2 −a+ib 1 − y k=0 −a+ib a+ib  n  n   X X 2k−1p 2 −   f(j)g(k)α2j y 1 − y  k=1 j=k −a+ib a+ib n−1 n−1   X X 2kp 2 −   f(k)g(j + 1)α2j+1 y 1 − y  k=0 j=k −a+ib Now the program can easily calculate the value of the integral for all (ρ, ζ). To solve the second integral we can follow a similar procedure. We start again by substituting:

Z 1 T (x)dx 1 Z a+ib T (ρy − iζ)dy 2n = 2n p 2 2 −1 (ix − Yν ) (ix − ζ) + ρ ρ −a+ib 1 p 2 iy − ρ (Yν − ζ) 1 − y

56 Then, like before, we split the integral into powers of y and gather similar terms for the full integral:

Z xn i √ dx = √ ln(h(x, c)) for n = 0 (ix − c) 1 − x2 1 + c2 c ih(x, c) = √ ln( ) − i arcsin(x) for n = 1 1 + c2 c ic2 h(x, c) p = −√ ln(− ) − c arcsin(x) + i 1 − x2 for n = 2 1 + c2 c2 c3 h(x, c) 1 = −√ ln( ) + i c2 − arcsin(x) 1 + c2 c3 2 ix p + c + 1 − x2 for n = 3 2 . . . .

In the expressions above we have deﬁned √ √ 2 1 + c2 1 − x2 + icx + 1 h(x, c) = √ 1 + c2(c − ix) We can again combine these terms to get an expression for the complete integral. We end up with:

Z a+ib T (ρy − iζ)dy 2n = p 2 −a+ib (iy − c) 1 − y

" n  k  X X 2j−2 k−j 2k−1 2k  c f(k − j)(−1)  (i) α2k−1 + (i) cα2k arcsin(y) k=1 j=1

2n k X k+1 c k h(x, c) + αk(−i) √ ln (i) 2 ck k=0 1 + c n k−1 2n X X k+1 2j X m−1 m−2k + (−1) f(j)g(k)x αm(i) (−c) k=1 j=0 m=2k n−1 k 2n ! #a+ib X X k 2j−2 X m m−(2k+1) p 2 + (−1) yf(k)g(j)x αm(i) (−c) 1 − y k=1 j=1 m=2k+1 −a+ib

1 1 If we then take c = ρ (Yν − ζ) and multiply the result by a factor ρ , we get the answer to the integral we wanted to calculate.

57 Appendix B

Integrals for the Newtonian seed

In order to obtain the function Ω1 belonging to a prescribed σ2 in the Newtonian regime mentioned in section 3.3.2, we need to solve the integrals for e2 and b3. The solutions for these integrals will be derived in this section for the case σ2 = 1. For the sake of brevity the coordinates ρ and ζ used in these calculations are the relative radius and height, so to obtain the results in the original notation perform ρ → ρ/ρ0 and ζ → ζ/ζ0. For the constant solution, we had

2 g0(x) = −σ0π(1 − x ) (B.1)

Then we want to calculate e2. We get:

Z 1 g0(x) e2(ρ, ζ) = − dx −1 ZD Z 1 1 − x2 = −σ0π dx (B.2) p 2 2 −1 ρ + (ix − ζ)

Now we can see that a problem may occur. We need to evaluate the derivative of this function at ζ → 0+. That means that we are integrating over a line that gets earily close to the pole at pρ2 − x2 = 0, since the relative radius satisfies −6 0 ≤ ρ ≤ 1. When we plot the integrand g0(x)/(ZDσ0) for (ρ, ζ) = (0.5, 1·10 ), we get figure B.1. We can see that at x = ρ, the value of the function turns from purely imaginary (|x| > ρ) to purely real (|x| < ρ). This does not have to be a problem, if we are careful at these points. It also looks like the imaginary part is antisymmetric, meaning that the result of the integral will be purely real. This sounds right, since the function is coupled to the observables σ and Ω. If we continue our integration (before letting ζ → 0+) we find

58 Figure B.1: Here the integrand of equation B.2 is plotted for (ρ, ζ) = (0.5, 1 · 10−6). The blue line is the real part and the red line is the imaginary part.

3π 1 1 e (ρ, ζ) = σ (iζ − )p(i − ζ)2 + ρ2 − (iζ + )p(i + ζ)2 + ρ2 2 0 2 3 3 ρ2 iζ − 1 iζ + 1 + πσ ζ2 − + 1 arcsin − arcsin (B.3) 0 2 ρ ρ

To check if our expression for e2 is correct, we can derive e2 w.r.t. ζ and then let ζ → 0+. Let us ﬁrst perform the derivation w.r.t. ζ:

∂e 3π 2 = σ ip(i − ζ)2 + ρ2 − ip(i + ζ)2 + ρ2 ∂ζ 0 2 ! (iζ − 1 )(ζ − i) (iζ + 1 )(ζ + i) + 3 − 3 p(i − ζ)2 + ρ2 p(i + ζ)2 + ρ2 iζ − 1 iζ + 1 + 2πζσ arcsin − arcsin 0 ρ ρ 2 ! 2 ρ i i + σ0π ζ − + 1 − (B.4) 2 p(i + ζ)2 + ρ2 p(i − ζ)2 + ρ2

59 If we then let ζ → 0+, we ﬁnd that some terms we can ignore right away while some others we will have to carefully add up. The ζ terms in the numerators on the second line, the arcsin terms on the third line and the quadratic ζ term on the last line will drop out right away. Our intermediate function is then

∂e 3π 2 = σ ip(i − ζ)2 + ρ2 − ip(i + ζ)2 + ρ2 ∂ζ 0 2

i i ! + 3 − 3 p(i − ζ)2 + ρ2 p(i + ζ)2 + ρ2 ! ρ2 i i + σ0π − + 1 − (B.5) 2 p(i + ζ)2 + ρ2 p(i − ζ)2 + ρ2

We can put these terms together using that p(i − ζ)2 + ρ2 − p(i + ζ)2 + ρ2 ≈ −2ip1 − ρ2 if ζ is suﬃciently small. We then end up with

∂e2 p 2 = 4πσ0 1 − ρ . (B.6) ∂ζ ζ→0 Which is precisely what we prescribed. Now we can calculate the derivative w.r.t ρ to obtain the function Ω1. We obtain

1 1 ! ∂e2 3π (iζ − 3 )ρ (iζ + 3 )ρ = σ0 − ∂ρ 2 p(i − ζ)2 + ρ2 p(i + ζ)2 + ρ2 iζ − 1 iζ + 1 − πρσ arcsin − arcsin 0 ρ ρ 2 ! 2 ρ iζ − 1 iζ + 1 1 + σ0π ζ − + 1 − + (B.7) 2 p(i + ζ)2 + ρ2 p(i − ζ)2 + ρ2 ρ Again, we let ζ → 0. Now all but the arcsin terms vanish, leaving us with an intermediate form

∂e iζ + 1 iζ − 1 2 = πρσ arcsin − arcsin (B.8) ∂ρ 0 ρ ρ √ Using the formula arcsin(a) = −i ln 1 − a2 + ia we can rewrite the expression above. In figure B.2 we have plotted the arcsin terms for different values of ζ. We see that the closer ζ approaches zero, the closer the terms approach π. We see a drop-off just before the value ρ = 1. This drop-off gets sharper the lower the value for ζ. This has to do with the fact that the arcsin terms are constant because of the domain of ρ. As we said before, ρ is the relative radius, meaning 0 ≤ ρ ≤ 1. This in turn means that the term inside the arcsin is larger than 1, while the regular domain for this function is [−1, 1]. When we plot ρ above 1, the argument of the arcsin changes to the regular domain, resulting in the drop-off that we can see.

60 Figure B.2: A plot of the arcsin terms for diﬀerent values of ζ. The green line has ζ = 0.01, the blue line ζ = 0.001 and the red line ζ = 0.00001.

So, if we rewrite the arcsin terms to the constant π, we end up with:

∂e2 2 = π ρσ0 (B.9) ∂ρ ζ→0 This is the result that will use in our derivations in section 3.3.2.

61 Appendix C

Documentation

In this section I will write a documentation for the metric generator program. I will shortly discuss the important functions so that anyone using or adapting my program will be able to understand their workings.

C.1 metric

The ﬁles metric.cpp and metric.hh contain the implementation of the metric class. An instance of this class is meant to have all the properties assigned to it by Ansorg. These properties include the metric functions, observables like the rotation curve and surface density and the dust condition. metric::metric(double g[],int N1, double xi[], int N2, double precision) The constructor takes the functions g and ξ as arrays of numbers, with g[] and xi[] having length N1 and N2 respectively. It also needs the desired precision, which is a double number between 0 and 1. The program gives an error when requesting a precision which is too small, as this would take too much time. This function initialises the variables and arrays of the metric and calls the functions that calculate the metric functions. metric::∼metric() The destructor function cleans up the pointers called by the initialisation function initFunc(). void metric::initFunc() This function allocates the pointers needed to store the metric functions and the arrays for g[] and xi[]. It also determines the resolution based on the precision entered into the constructor. void metric::print(string x) This is a printing function that can print a variety of metric properties, depend-

62 ing on what is entered in string x. double metric::gExpansion(double x squared) double metric::xiExpansion(double x squared) These functions calculate the value of the g and ξ expansions. void metric::obtainSetY() This function calls the necessary functions to obtian the set {Yν }q needed for the matrices in equation (7) from Ansorg. It also makes sure the parameters satisfy the requirements set by Ansorg in section 1.2. More information on these functions is given in section C.3. void metric::obtainErnstPot() This function calls the necessary functions to obtain the Ernst potential for given function g and {Yν }q, following the procedure of section 1.2 in Ansorg. More information on these functions is given in sectionC.4. void metric::obtain a() void metric::obtain k() These functions perform a Riemann sum integration to obtain the metric functions a(ρ, 0) and k(ρ, 0). The integrals are obtained from the expressions:

ρ a = b (C.1) ρ e4U ζ 1 h i k = ρ (U )2 − (U )2 + e−4U (b )2 − (b )2 (C.2) ρ ρ ζ 4 ρ ζ double metric::exp2U(int rho index, int zeta index) double metric::exp2U rho(int rho index, int zeta index) double metric::exp2U zeta(int rho index, int zeta index) 2U 2U 2U These functions return e ,(e )ρ and (e )ζ respectively. The last two perform numerical differentiation to obtain the relevant quantity. The first one simply extracts the real part of the Ernst potential at the requested coordinates. double metric::b(int rho index, int zeta index) double metric::b rho(int rho index, int zeta index) double metric::b zeta(int rho index, int zeta index) These functions return b,(b)ρ and (b)ζ respectively. The last two perform numerical differentiation to obtain the relevant quantity. The first one simply extracts the imaginary part of the Ernst potential at the requested coordinates. double metric::Q(int rho index, int zeta index) This function returns the function Q as defined by Ansorg in equation (5). double metric::D(int rho index) This function returns the so-called dust condition, defined as:

63 4U 2 4U 2 D = e Q + Q e ,ζ + (b,ρ) (C.3) This quantity should be equal to zero always, and comes from the nature of the material the disk is made out of. double metric::Omega(int rho index) double metric::sigma(int rho index) These function return the rotation curve Ω and the surface density σ respectively. They are obtained from equations (3) and (6) from Ansorg. double metric::Komar M() double metric::Komar J() These functions calculate and return the Komar integrals for mass M and total angular momentum J. They are calculated from the expressions below.

Z ρ0 1 −V i M = U,ζ + Q 1 + 2e u ηi (C.4) 0 2 Z ρ0 1 −V i J = U,ζ + Q e u ηi (C.5) 0 2 In these equations ηµ is the Killing vector w.r.t. axisymmetry.

C.2 calculation funcs

The files calculation funcs.cpp and calculation funcs.hh contain the implementation of some general mathmatical functions that are used troughout the full metric calculation. Most of them are simple definitions or basic calculus functions that require no further explanation, the rest will be briefly explained in this section. vector gauss(vector< vector > A) This piece of code was obtained from [20]. It uses Gaussian elimination to solve a system of equations. This routine should only be used if the system does not have to be solved more than once, for example solving:

q q 2 X ν X ν exp xµξ xµ; {Yν }q bν xµ = bν (−xµ) (C.6) ν=0 ν=0 to obtain the set {Yν }q. When solving equation 7 in [5] it is better to use the LUP functions below, as these are more eﬃcient. int LUPDecompose(complex **A, int N, double Tol, int *P) int LUPDecompose(double **A, int N, double Tol, int *P)

64 void LUPSolve(double **A, int *P, double *b, int N, double *x) complex LUPDeterminant(complex **A, int *P, int N)

These functions were obtained from [21]. The first two decompose a complex or real N × N matrix A into the lower-upper form required by the last two functions to either solve a system of equations or calculate a determinant. The vector P is also required to store the permutations made by the decomposition functions in order to get the lower-upper form. void ChebyshevExp(double (*func)(double), int order N, double output[]) void ChebyshevExp Even(double (*func)(double), int number of coeff, double output[]) void ChebFit(double coeff[], int number of terms, double (*func)(double x)) void ChebFit discr(double coeff[], int number of terms, double func array[], double precision) These functions calculcate a given number of Chebyshev coefficients for a given function. The different allow the function to be entered as an array of points, or a function in a program. The second one only calculates the even coefficients, for even functions like g, ξ or the other functions of the axisymmetric metric. The last two calculate functions on the domain [0, 1] instead of the standard [−1, 1]. They all follow a similar procedure for the calculation of the coefficients cj:

N 2 X c = f (x ) T (x ) . (C.7) j N k j k k=0

In this formula the xk are the N zeroes of Tj(x), the Chebyshev polynomial, and f(x) is the function to be approximated. For N → ∞ the approximation

N−1 X 1 f(x) ≈ c T (x) − c (C.8) k k 2 0 k=0 becomes exact.

C.3 calculate Ynu

The ﬁles calculate Ynu.cpp and calculate Ynu.hh contain the implementa- 2 tion of the functions needed to obtain the set {Yν }q from the function ξ x . The calculation follows the procedure described in section 3 from [5]. void obtainPolynom(int q, double xi input[], int N2, double output[]) This function obtains the polynomial Pq in the following equation:

65 q q Y iYν − x Pq(−x) X exp xξ x2; {Y } = = with P = b xν (C.9) ν q iY + x P (x) q ν ν=1 ν q ν=1

As Ansorg describes in [5], evaluating this function at q arbitrary points leads to a system of equations that can be solved using the functions in calculation funcs.cpp. void DurandKerner(int degree n, double polynom coeff[],list< complex > &roots output) The parameters {Yν }q are determined by the roots of the polynomial Pq. The method to find these roots is called the Durand-Kerner method. The found roots are written in the roots output vector. void cleanSetY(list< complex >& setY) Lastly, the parameters Yν need to have certain properties. As mentioned in section 1.2 of [5], for each parameter Yν there must also be a parameter Yµ with Yν = −Y µ. This is to ensure reflectional symmetry in ζ. Also, the parameters are assumed to lie outside the imaginary interval [−i, i]. The cleaning function makes sure that the obtained set meets these demands. The parameters that do not fit the demands are thrown away.

C.4 calculate ErnstPot

In the process of calculating the metric, calculating the Ernst potential from 2 g(x ) and {Yν }q is the most complicated and time-consuming step. This is because of some nasty integrals that need to be numerically calculated. The routines that perfom these calculations are implemented in calculate ErnstPot.cpp and calculate ErnstPot.hh. void obtain coeff(long double a, long double b, int n, vector< complex >& coeff) This function obtains the polynomial coefficients of a Chebyshev polynomial Tn(ay − ib). This is done so that the relevant integrals can be written as a sum of powers of x, rather than a Chebyshev function. After this function there are a few routines that calculate the integrals in the way that was described in the previous section. void prep integrateT2n(int number of terms, double precision) 2 The integral for f0 is only dependent on g(x ) through its coefficients gk. This means that this integral can be calculated once as a preparation and then the k-terms only need to be multiplied with their respective coefficients gk to get R f0. This function performs those integrals T2n(x)/ZDdx in advance. These integrals are calculated by calling complex integral T2n(int n, long double rho rel, long double zeta rel).

66 complex integral factorY(int n, long double rho rel, long double zeta rel, complex Ynu) This function performs the integral needed for γν . This integral cannot be calculated in advance, as it depends on the parameter Yν . These parameters are different for every metric. void obtain f0(int number of terms, int q, double g[]) The first part of the Ernst potential, f0, is obtained by calling this function. For it to work properly it needs the integrals calculated by prep integrateT2n. It will give an error if this function has not been called yet. If it has these integrals, the rest of the calculation is easy. The individual terms simply need to be multiplied with the coefficients of g(x2). void constructMatrices(double rho rel, double zeta rel, list< complex > Ynu, int q, double g[], complex** output upper, complex** output lower) This function constructs the matrices needed for the second part of the Ernst potential; the ratio of the determinants. It uses the helper functions declared above itself to calculate the matrix elements, and gives an error if some element is not equal to itself, for instance when it is not a number or infinite. void obtain ErnstPot(int number of terms g, double g[], int number of terms Ynu, list< complex > Ynu, double precision, complex** ErnstPot output) This function calls all the functions defined above in order to perform the complete calculation for the Ernst potential, and writes the result to the output vector. void cleanup Ernst() The cleanup function takes care of the pointers that have been declared in other functions, to avoid memory leaks.

67 Bibliography

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