Dipole anisotropy in radio surveys

Bachelorarbeit

zur Erlangung des Grades eines Bachelor of Science der Fakultät für Physik der Universität Bielefeld

vorgelegt von Matthias Rubart

Universität Bielefeld Fakultät für Physik Oktober 2009 Betreuer & 1. Gutachter: Prof. Dr. Dominik Schwarz 2. Gutachter: PD Dr. York Schröder

1 Contents

1 Motivation 3

2 A brief introduction to cosmology 3 2.1 Cosmological principle ...... 3 2.2 Expanding Universe ...... 3 2.3 Friedmann equations ...... 4

3 Cosmic Microwave Background 5 3.1 The origin of the CMB ...... 5 3.2 Observations of the CMB ...... 6

4 Radio galaxies 7 4.1 Synchrotron radiation ...... 7 4.2 Properties of radio galaxies ...... 8

5 Survey data 9

6 Anisotropy of radio surveys 10 6.1 Introduction ...... 10 6.2 The dipole signal ...... 10 6.3 Local clustering ...... 12

7 Literature on the velocity dipole in radio surveys 13 7.1 Baleisis et al. 1998 ...... 13 7.2 Blake and Wall 2002 ...... 16 7.3 Crawford 2009 ...... 17 7.4 Conclusion ...... 20

8 One alternative method 21 8.1 Two-point correlation function ...... 21 8.2 Angular two-point correlation function ...... 22 8.3 Magnitude of the signal ...... 23 8.4 Shot noise problem ...... 24

9 Outlook and Summary 25 9.1 Outlook ...... 25 9.2 Summary ...... 26

10 Erklärung 26

2 1 Motivation

In this thesis I consider the dipole anisotropy in radio surveys. Our motion through the Universe should lead to a dipole signal not only in the Cosmic Microwave Background but also in the distribution of radio galaxies. The physical reasons for this are the redshift and stellar aberration. I will calculate the theoretical magnitude of the dipole signal and discuss some publications on this issue. The question if the dipole can be observed with today’s radio surveys, seems not to be fully resolved. Blake and Wall claimed in 2002 (BW02) to have found the dipole signal. But Crawford made a publication in 2009 (FC09) which says that the detection of this signal is not possible yet. This contradiction is the reason I engage the issue of the dipole anisotropy in radio surveys.

2 A brief introduction to cosmology

2.1 Cosmological principle

Cosmology deals with the Universe as a whole. The most important foundation of cosmology is the Copernican principle. This states that we (as humans on the Earth) are typical and not special. So we are not in the centre of the Universe, nor is our Solar System (or our Galaxy) something special. Of course, there is no proof of this assumption yet. From the Cosmic Mircowave Background (see below) we know that the Universe is isotropic around our point of view. This, combined with the Copernican principle, leads to an isotropic Universe around every point of view. If the Universe is isotropic around every possible point of view, it also has to be homogeneous. Altogether we have the Cosmological principle which claims: “ The Universe is isotropic and homogeneous.“ The Cosmological principle can only hold for big scales and in a statistical sense. One should always bear in mind that this principle is only an assumption.

2.2 Expanding Universe

We do not live in a static Universe. The stars and galaxies are moving with certain velocities.

Because of the Doppler effect, the emitted frequency (νE) of a photon is not necessarily the same as the observed (νO) one. If the object is moving away from us, the frequency is decreased. In account of this fact, the redshift z is defined by ν − ν e o =: z. (1) νo The observation of distant galaxies shows that there is a linear dependence of redshift and distance d, called the Hubble law:

c z = H0 d. (2)

3 Here c is the velocity of light and H0 is called the Hubble constant. Measurements (FW01) showed that km H ≈ 72 ± 8 Mpc−1. (3) 0 s So we can conclude that distant galaxies are moving away from us. Because of the Cosmo- logical principle this leads to an expanding Universe. The unit of the Hubble constant is 1 time . If the Universe had always been expanding with the same velocity, the inverse of the Hubble constant would equal the age of the Universe, 1 tH = ≈ 13.9 Gyr. (4) Ho 2.3 Friedmann equations

In this subsection the evolution of the Universe will be discussed more detailed. According to the general theory of relativity we have to consider the space-time of the Universe. The line element of this space-time is defined by

2 µ ν d s = gµν dx dx , (5) where gµν denotes the metric. If we assume the Cosmological principle, the line element can be written as the Robertson-Walker line element  dr2  ds2 = −c2dt2 + a2(t) + r2dΩ2 . (6) 1 − kr2 This line element describes a homogeneous and isotropic Universe with a certain curvature k. The value of k is either 1 for a spherical space, 0 for flat space or -1 for hyperbolic space. Important to notice is that all space-like scales are taken care of by the scale factor a(t). The expression dΩ2 stands for dθ2 + sin2(θ)dφ2. It can be shown that the singularity at r = 1 for k = 1 is just an effect of the coordinate system and has no physical meaning. If a(t) is increasing with time, typical distances between galaxies will also increase with time. This will lead to the observed redshift, because those galaxies would be moving away from us. So the Hubble parameter, or the expansion rate, is defined by

a˙(t) H(t) := . (7) a(t)

To describe the evolution of the space-time one needs to use the Einstein equation 8πG G + Λg = T . (8) µν µν c4 µν

On the left hand side Gµν denotes the Einstein tensor and Λ is the cosmological constant.

The right hand side is given by the energy-momentum tensor Tµν . If we describe the content of the Universe as an isotropic fluid with pressure p and energy density ε, we obtain u u T = (ε + p) µ ν + pg . (9) µν c2 µν

4 Including the cosmological principle we arrive at the first

 kc2  8πGε 3 H2 + − Λc2 = (10) a2 c2 and second  kc2  8πGp − 2H˙ + 3H2 + + Λc2 = (11) a2 c2

Friedmann equations. With the help of these formulas we can calculate the evolution of the Universe depending on its content. For example, the Einstein-de Sitter model assumes 2 Λ = k = p = 0. The solution for the scale factor of this model is a(t) = Ct 3 , where C is some constant. So, in this model, there has been a time when the scale factor vanished. This moment is called the Big Bang. Einstein found a solution which describes a static Universe with k = 1, p = 0 and a¨(t) = 0, by choosing Λ = 4πGε/c2. Today we believe to live in a Universe with accelerated expansion (a¨(t) > 0). In our current model of the Universe, there has also been a Big Bang. The best evidence for the Big Bang model will be discussed in the next section.

3 Cosmic Microwave Background

3.1 The origin of the CMB

Because of the finite speed of light, we do not observe the Universe like it is today. Every photon we observe needed a certain time to travel the huge distances of outer space. For example the light from the next galaxy (Andromeda) took 2.5 × 106 years to reach us. Be- cause of the Hubble law such a time can also be expressed by the redshift of the galaxy

d z tγ = = . (12) c H0 If we observe the temperature of something in the Universe, we actually observe a tem- perature that has been emitted a certain time ago. For photons the relation T T = e (13) o 1 + z can be shown. The observed temperature is To, while the emitted temperature is Te. This relation is needed below. When the scale factor a(t) has been very small (compared to today), the content of the Universe had less space. This leads to high mass densities and temperatures. Soon after the Big Bang the Universe was very hot and it has been cooling down since then (accept at a time called ”reheating“, but this was well before any processes leading to the CMB).

5 As long as the temperature of the Universe is bigger then T ≈ 3000 K the reaction

e + p ↔ H + γ (14) is in thermal equilibrium. If the temperature falls beneath this value, the reaction goes only from the left to the right side. So protons and electrons combine to hydrogen emitting photons. This process is called recombination. After recombination the Universe is trans- parent to photons, because hydrogen is neutral, and the photons will no longer scatter. So all photons are moving freely through the Universe after the recombination. Because they have not interacted since then, they still have the same temperature distribution. These photons are what we call the Cosmic Microwave Background.

3.2 Observations of the CMB

In the year 1965 two radio astronomers have published the first publication about the CMB. Arno Penzias and Robert Wilson found a background noise in their observations at 4080 MHz. Further analysis showed that this background noise is a perfect black body spectrum of T = 3.5 ± 1.0 K. The COBE (1989) and WMAP(2001) satellites measured this signal more detailed. A temperature of T = 2.725 ± 0.002 K was observed (JM99). With (13) we can calculate, that the recombination happened at z = 1100. The most significant anisotropy in this signal is a dipole. In one direction the CMB is a bit warmer (10−3K above the mean 2.73K of the CMB), while in the opposite direction it is about the same amount colder. Magnitude d and galactic coordinates l, b of the dipole are observed to be (GH09)

(d, l, b) = (3.355 ± 0.008 mK, 263.99◦ ± 0.14◦, 48.26◦ ± 0.03◦). (15)

If we subtract this dipole (and local anisotropies like the Milky Way) we see an isotropic background with very small temperature fluctuations of

∆T  ≈ 2 · 10−5. (16) T rms Thus the dipole anisotropy is interpreted as a Doppler effect, due to our motion in the Universe. If it was an effect of cosmic structures, we would expect it to be of the same order as the quadrupole and higher multipole moments.

6 Figure 1: WMAP 2008 (NASA), local contamination and dipole removed

4 Radio galaxies

4.1 Synchrotron radiation

To understand the synchrotron radiation we have to consider relativistic electrodynamics. That means, besides the Maxwell equations, we also have to take into account the finite speed of light. If we look at the electro-magnetic field of a moving (velocity β~) point-like particle, we see photons that have been emitted a certain time ago. Both, the E~ and B~ field, can be expressed in the following way (the exact expressions can be found in the Landau Lifschitz 2):

~µ ~ν + . (17) R2 R

~ν ~˙ The second term R is proportional to the acceleration β. We are interested in the power that is emitted in an infinitesimal solid angle dΩ and a certain direction ~n:

dP = R2S~ · ~n dΩ. (18)

~ c ~ ~ −1 Here S = 4π E × B is the Poynting-vector. So only terms proportional to R lead to a contribution for large distances R. This leads to

dP q2 ˙ = |~n × [(~n − β~) × β~]|2. (19) dΩ 4πc(1 − β~ · ~n)5 The denominator of the prefactor is

4πc(1 − β~ · ~n)5 = 4πc(1 − β n cos(θ))5, (20) where θ is the angle between the direction of motion and the direction of the emitted power. If β ≈ 1 (meaning we have relativistic particles) and θ  1 (but not θ = 0), then this term

7 will be very close to zero. In this case the prefactor of dP/dΩ will increase. ˙ One can show, that the radiation for circular motions (β~ ⊥ β~) is bigger than for linear ˙ motions (β~ k β~). Because an electron in a homogeneous magnetic field is moving on a circle, it emits the so called synchrotron radiation. This kind of radiation is typically polarized.

4.2 Properties of radio galaxies

Radio galaxies belong to the class of active galaxies. The difference between normal and active galaxies is the AGN, which by definition only active galaxies have. AGN stands for Active Galactic Nucleus. This Nucleus is a region in the galaxy which is very small (about the size of our Solar System) compared to the mother galaxy. In this region a huge amount of energy is released. The radiation from the Nucleus is often much bigger than that of the rest of the galaxy. It is assumed that every AGN is powered by a super-massive black hole. So this active galaxies can be seen from larger distances than normal galaxies. If the emission of the AGN is observed mainly in the radio band, we call this a radio galaxy. Because the emission from radio galaxies is polarized, we can conclude that this is due to synchrotron radiation. In those galaxies ultra relativistic electrons are propelled from the centre to the outer region of the galaxy in typically two jets. In the outer regions they are captured by the magnetic field of the galaxy and forced to move on circles. This is were most of the synchrotron radiation is created. The brightest radio galaxy is Cygnus A. In the picture of this galaxy one can see the AGN as well as the two radio lobes, which emit most of the radio waves. It is noteworthy that the galaxy itself lies perpendicular to the two lobes.

Figure 2: The radio galaxy Cygnus A (NRAO)

Of course, if a radio galaxy is very far away, you can no longer see the jets. Typically we can only see a point like radio source.

8 5 Survey data

In this section I will briefly summarize the important facts about the radio surveys, used in search for the velocity dipole. The Green Bank Telescope, located in Green Bank, West Virginia, USA, is the world’s biggest radio telescope with the ability to change the direction of view. At 4.85 GHz this telescope has been used to create the 87 Green Bank survey between 1986 and 1988. The data covers most of the northern hemisphere with declinations of 0◦ < δ < +75◦. Griffith and Wright conducted the Parkes-MIT- NRAO survey in 1993 at the same frequency as the 87GB. They used the Parkes 64m telescope in New South Wales, Australia. This survey cov- ers the declinations of −87.5◦ < δ < +10◦. The survey with the most data points available today is the NVSS survey.

• NVSS stands for NRAO VLA Sky Survey

• NRAO stands for National Radio Astron- omy Observatory Figure 3: The isotropic NVSS survey • VLA stands for Very Large Array (NRAO)

Each of the 27 antennas of the VLA has a diameter of 25m. They are combined electroni- cally and work as a single huge radio telescope. The VLA is located in New Mexico, USA. Be- tween 1993 and 1997 the VLA has produced the data for the NVSS at 1.4GHz. This catalogue covers Ω ≈ 10.3 sr of the sky. Excluded from the survey due to a limited visual field are areas with declinations of δ < −40◦. The data of the NVSS is available in the internet for free use. In this catalogue one can find about 1.8×106 radio sources. Later we will see that it is not conve- nient to use all this data points in search for the dipole. As one can see in Figure 3 and 4, the Figure 4: A detail of the NVSS survey NVSS survey shows a very isotropic distribution (NRAO) of radio sources.

9 6 Anisotropy of radio surveys

6.1 Introduction

Very often the Cosmic Microwave Background (CMB) is taken as evidence for an isotropic Universe (at least around our point of view). Our motion in the Universe is supposed to be responsible for the dipole anisotropy in the CMB. If this interpretation of the dipole is correct, it must also occur in observations at different wavelengths. I consider the dipole anisotropy at radio frequencies. To find the dipole signal in radio surveys is more difficult than to find it in the CMB because:

• Radio surveys provide discrete sources.

• The mean red-shift of radio galaxies is about z = 1 and so local anisotropies (large scale structure) are much more important.

• We do not have a catalogue, made by one observatory, that covers the whole sky, like the COBE or WMAP satellites do.

• Combining two catalogues leads to serious calibration problems.

6.2 The dipole signal

It is not possible to analyse the radio survey data in the the same way as the CMB, because we only have discrete sources. If we look in the direction of our motion (where we assume an increase of the flux) we will find more sources, because very dim radio galaxies would get bright enough to count as an extra source. So, not the mean flux will change, but the number density of sources. How big this dipole should be has been shown by Ellis and Baldwin (EB84) in the following way: The flux S of an radio galaxy depends on the frequency ν as

S ∝ ν−α, (21) where α is determined (by observations) to be about 0.75. The number source count per unit solid angle (with minimum flux S) is assumed to behave like dN (> S) = k S−x, (22) dΩ with x ≈ 1 (empirical fact). Now, we have to take two independent effects into account. First the Doppler effect, which enhances the energy and therefore the frequency, in the direction of our motion

νobs = νrest δ. (23)

10 It can be shown that δ is given by 1 + v cos(θ) δ = c , (24) p v 2 1 − ( c ) where θ is the angle between our direction of motion and the direction of observation. The W 2 unit of the flux S is m2 Hz , where m corresponds to the unit antenna area of the observatory. So we do not have to transform the m2 factor due to relativistic motion, because the motion 1 is normal to the area of the observatory. So the unit of the flux S transforms like Hz an therefore gets a factor δ:

−α −α α 1+α −α 1+α Sobs(νobs) = a δ νrest = a δ νobs δ = a δ νobs = Srest δ . (25) A second effect, the stellar aberration, has to be taken into account. If we are moving in a certain direction, all angles that we see are bend towards this direction. This effect is just due to the fact that the velocity of light is finite. So the angle θ changes like

−1 θobs = θrest δ . (26)

Because

dΩ = dθ sin(θ) dφ ≈ dθ θ dφ, (27) we get

−2 dΩobs = dΩrest δ . (28)

If we now combine those two effects we get dN dN = δ2+x(1+α). (29) dΩ obs dΩ rest Now we make an approximation for the case v  c, then v δ ≈ 1 + cos(θ) (30) c and h v  i2+x(1+α) v  v 2 1 + cos(θ) = 1 + [2 + x(1 + α)] cos(θ) + O . (31) c c c Together the equations (29), (30) and (31) lead to dN dN h v  i = 1 + [2 + x(1 + α)] cos(θ) . (32) dΩ obs dΩ rest c After defining the dipole amplitude A of a signal σ by

σ(θ) = σo(1 + A cos θ), (33) we finally identify the velocity dipole amplitude Avel with v  A = [2 + x(1 + α)] . (34) vel c km −3 Inserting x = 1, α = 1 and v ≈ 370 s we get Avel ≈ 4.6 × 10 .

11 6.3 Local clustering

The Cosmological principle holds for the largest scales only. Up to distances of about 300 Mpc the Universe is neither isotropic nor homogeneous. Our Galaxy lies in the Local Group, consisting of about 30 galaxies. The biggest galaxy in the Local Group is Andromeda, followed by the Milky Way. This group has a diameter of about 2 Mpc and is gravitationally bound. If a group has much more than 30 galaxies, one talks about a cluster. But even clusters with thousands of galaxies have typically about the same size as the Local Group. Yet, this is not the biggest observed structure in the Universe. It has been shown (by Abell’s survey), that the distribution of galaxy clusters is not uni- formly. Our Local Group belongs to the Local Supercluster. The diameter of the Local Supercluster is about 30 Mpc. A super cluster is just a cluster of galaxy clusters. Those super clusters are not homogeneously distributed in space. There are regions with relatively high densities of super clusters and region with very view super clusters. The second type of region is called a void. At this scales (about 250 Mpc) one talks about large scale structure. One evidence for large scale structure is the SDSS map:

Figure 5: SDSS map (SDSS)

12 This means, if we look at the distribution of galaxies at those distances, we do not neces- sarily see an isotropic universe. In 2000, Rowan-Robinson M. et al. (RR00) found a dipole signal in the distribution of local galaxies with distances < 425 Mpc. For their search, they used the PSCz (Point Source Catalogue Redshift Survey) IRAS (the Infra-Red Astronom- ical Satellite) galaxy redshift survey. This infra-red survey includes for each found galaxy not only the observed luminosity, but also the redshift. In this survey 16% of the sky are masked, because they were biased due to emissions from the Mily Way. In order to find a Dipole signal anyway, they filled the masked areas with artificial data. They found a dipole signal with convergence at about 285 Mpc and has a direction (RR00) of

(l, b) = (249.1◦, 34.3◦). (35)

Now we assume that this dipole in the density of galaxies leads also to a dipole in the local distribution of mass. If this is true, the increased density in mass could be the reason for the movement of our galaxy with respect to the CMB. Because the angle between those two dipoles is only 14◦ (with a not negligible uncertainty because of masked areas), this argumentation seems very likely.

7 Literature on the velocity dipole in radio surveys

The first publication about the dipole anisotropy in radio source counts has been published by Ellis and Baldwin in 1984. They showed (with the same derivation I used in section 5) that the expected dipole should have an amplitude of (2 + x(1 + α) β). In the year 1984 no sufficient radio survey was available, so Ellis and Baldwin were not able to find the dipole in observations. In the following subsections I will outline the most important publications on this issue. I have modified some of the original formulas in order to keep an uniform notation.

7.1 Baleisis et al. 1998

The authors of this publication considered the dipole anisotropy as an phenomenon of two effects. Not only our motion in the Universe should lead to a dipole, but also the large scale structure of matter. In this publication the Green Bank 1987 (87GB) and the Parkes-MIT- NRAO (PMN) catalogue are used, which leads to some calibration problems. How exactly those two catalogues are combined is described in the publication. In the combined 87GB- N PMN catalogue the mean density of radio sources with a flux S > 70 mJy is σ0 := Ω ≈ 2878. To predict the dipole, spherical harmonic analysis is used. This means, that the distribution of radio sources is expanded in spherical harmonics like

X m m σ(~x) = al Yl (~x) (36) l,m

13 with

N m X m∗ al = Yl (~xi) . (37) i=1 It is not convenient to use the full sky data in order to calculate the dipole, because certain anisotropies (e.g. our galaxy) will bias the dipole signal. So it is useful to exclude certain areas of the sky from the analysis, like the Galactic and the super Galactic Planes.

Figure 6: The masked distribution of radio sources (S > 70 mJy) taken from (AB98).

Of course such masked areas could lead to an artificial dipole signal, so one needs to compensate this signal in some way. One way (realised in Baleisis et al.) is to use

Ngal Z m X m∗ m∗ cl,obs = Yl (~xi) − N Yl (Ω), (38) i=1 Ωobs where the second term simulates the masked areas.

For this kind of problem (fluctuations δm with respect to a mean value) the spherical har- monic coefficients can be written as Z m m∗ al = dV (1 + δm(~r)) Φ(r) Yl (~r). (39)

Here Φ(r) is the probability that a galaxy at distance r and luminosity above the flux limit will be detected by the survey. If we change to a spherical coordinate system and only look at coefficients with l > 0 we get Z m 2 m∗ al = dΩ dr r δm(~r) Φ(r) Yl (~r). (40)

14 Here Baleisis et al. had assumed a model which fits the IRAS as well as the optically observed dipole. The exact derivation can be found in their publication from 1997. We are interested in the dipole amplitude A from: dN (θ) = σ (1 + A cos θ). (41) dΩ 0 They found, for the local clustering term, a value of

−3 Alcd ≈ 2.7 × 10 . (42)

The shot noise contribution to the dipole is 3 1 Asnd = √ √ . (43) 4π σo

So we have for a survey with a mean radio source density of σo = 2878 a shot noise value of

−3 Asnd ≈ 15.8 × 10 . (44)

The velocity Dipole term is calculated by: v  A = [2 + x(1 + α)] . (45) vel c

km Using α = 0.75, x = 1 and v ≈ 370 s (like the CMB Dipole shows) this leads to

−3 Avel ≈ 4.6 × 10 . (46)

So it is expected that the observed dipole signal is strongly biased due to shot noise. This problem would decrease if more sources were used, like in the NVSS (see below). In order to find the dipole signal in the 87GB-PMN catalogue they used a different ansatz than the one above. Their method to detect a dipole signal is to calculate

N X D~ = wi ~xi, (47) i=1 where ~xi is the direction of one source. With wi it is possible to weight the different sources, for example by the flux. Because it would introduce an extra error source to weight the signals with the flux, Baleisis et al. choose wi = 1 ∀ i. Using this method they find for

Smin > 70 mJy a dipole signal with (AB98)

(D, l, b) = (307, 286◦, 15◦) (48)

◦ 1 and an angle to the CMB dipole of about 38 . In section (7.3) the relation D = 3 AN −3 will be shown. This leads to Aobs ≈ 25.5 × 10 . Because this dipole is much bigger than expected, it is likely not to be dominated by the velocity dipole. So Baleisis et al. have not been able to detect the velocity dipole. In their publication an error estimation of the calculated dipole was not included.

15 7.2 Blake and Wall 2002

Blake and Wall used the NVSS described in section 4. They did not want to find the whole Dipole signal, but the contribution coming from our velocity. So they have to remove the local clustering dipole in some way. On the one hand this will reduce the dipole signal, but on the other hand it will proof the existence of the velocity dipole. To suppress contamination of local anisotropies they removed almost all local (z<0.03) radio sources. To identify those nearby sources they used the PSCz and RCBG3 catalogues. Additionally they removed all data within 15◦ of the galactic plane. In order to find the dipole Blake and Wall made use of the spherical harmonics, just like Baleisis et al. (1998) did. Using N radio sources they computed

N m X ∗ al = Yl,m(~ri) . (49) i=1 0 If the radio survey was perfectly isotropic,we would only get a signal in the a0 coefficient. A dipole signal will lead to signals in all the l = 1 spherical harmonic coefficients. Coefficients with l > 1 should be zero. But because we can not use data from the whole sky, this is not the case. A dipole will bias the higher harmonic coefficients in this case. So Blake and Wall measured all spherical harmonic coefficients with 0 < l < 4. The dipole model has three degrees of freedom only. One is the amplitude and the other two are for the observed direction. After Blake and Wall calculated the spherical harmonics, they searched for the best fitting parameters for the dipole model with different flux cutoffs.

2 Flux N Best A Best φ Best θ χred / mJy ×10−3 / o / o > 40 125,603 7 ± 5 149 ± 49 135 ± 38 1.02 > 35 143,524 9 ± 4 161 ± 44 117 ± 39 0.74 > 30 166,694 11 ± 4 156 ± 32 88 ± 33 1.01 > 25 197,998 11 ± 3 158 ± 30 94 ± 34 1.01 > 20 242,710 11 ± 3 153 ± 27 93 ± 29 1.32 > 15 311,037 8 ± 3 148 ± 29 59 ± 31 1.81 > 10 431,990 5 ± 2 132 ± 29 25 ± 19 4.96

Table 1: The best fits fot the three dipole parameters (BW02)

The notation differs from the one above. Blake and Wall used a ordinary spherical coordinate system. In this coordinate system the direction of the CMB dipole is θ = 97.2 ± 0.1◦, φ = 168.0 ± 0.1◦. For the flux cutoffs S < 15 mJy the NVSS data is strongly biased, due to technical reasons.

16 The observed value of the dipole amplitude is about 1.5σ higher than the prediction made by Ellis and Baldwin, i.e. it is in agreement. The direction of the dipole observed is consistent with the CMB dipole as can be seen in the following graph:

Figure 7: Taken from (BW02). The great circles represent 1σ and 2σ errors. The dot represents the CMB Dipole.

7.3 Crawford 2009

Crawford claims that it is not possible to find the dipole signal with statistical significance using today’s radio survey catalogues. This publication is just theoretical, meaning Crawford did not try to detect the dipole using observations. He considered the case of a perfect radio survey, meaning a full sky coverage and no calibration problems. This will make the following derivations easier. From a radio catalogue one can calculate

N X D~ = ~xi. (50) i=1

Here ~xi is the normalized position of a radio source. The coordinate system can be rotated, so that our direction of motion (and so the direction of the dipole) is the z-Axis. For sufficiently

17 high number of sources the sum can be changed into an integral. The z-component of this integral is

Z π Z 2π Dz = dθ dφ σ(θ) cos θ sin θ, (51) 0 0 where σ(θ) = (1 + A cos θ)σo is the number density of radio sources with angle θ to the direction of motion. This integral is solved by 4π D = A σ . (52) z 3 0

N v With σ0 = 4π and A = (2+x(1+α))( c ) the value of the dipole measured in the ideal case is

1 D = (2 + x (1 + α))β N. (53) 3 Because of the finite number of sources available we will also measure a shot noise. The question arises how significant a measured dipole signal is. Even if we assume that our motion in the Universe is zero, a dipole signal can be measured due to shot noise. So if we measure a dipole signal, we will not necessarily have found the velocity dipole. Crawford represents this uncertainty by a probability cloud with characteristic size ρ. The value of ρ is,by a random walk in three dimensions, computed to be

r N ρ = . (54) 3

The probability that the vector D~ is displaced by a ~r is given by a three dimensional Gaussian probability distribution

4πr2 −r2  p(r) = √ 3 exp 2 . (55) ρ3 2π 2ρ Those two formulas lead to

r 54 −3r2  p(r) = r2 exp . (56) πN 3 2N

With this equation we can calculate the probability to measure a dipole D with magni- tude r in the zero motion case. By integrating of r from D to N (which is the maximum possible dipole value (FC09)) the probability p(r > D) can be computed. So if we measure a dipole signal D, the confidence level at which the zero motion case is excluded is given by 1 − p(r > D).

At a certain confidence level the measured dipole can be expressed as D ± rc. This rc can also be calculated by the above formula in the following way. We choose rc so that the

18 integral over p(r) from rc to N gives the wanted confidence level. This can be translated to the uncertainty angle of the dipole direction as seen by an observer. For rc < D we get r δθ = arcsin ( c ) . (57) c D

If rc > D, no conclusion about the direction of the dipole can be made. This formula was used to make the following graph.

Figure 8: Uncertainty in the measured diople direction (δθ) vs. the number of survey source counts (N). The data used by Blake and Wall is represented by the NVSSselect line. Taken from (FC09).

19 7.4 Conclusion

Crawford claims, that it is not possible to find the dipole anisotropy in today’s radio sur- veys. Even the zero motion case cannot be excluded at confidence level of 1σ. This result is surprising, because Blake and Wall claims to have found the dipole signal. They even have a direction, with a uncertainty of about 35◦, which contradicts the results of Crawfords publication. The magnitude of the expected dipole amplitudes can be calculated for any catalogue by the same method Baleisis et. al. used (AB98). For 200 000 sources (like the NVSS) and full sky coverage (Ω = 4π) the expected velocity and shot noise dipoles are:

3 1 −3 Asnd = √ √ ≈ 6.7 × 10 (58) 4π σo and

v  A = [2 + x(1 + α)] ≈ 4.6 × 10−3. (59) vel c It is assumed, that in the 87GB-PMN catalogue almost every radio galaxy with d < 400 Mpc is included. So the calculated value for the local clustering dipole will not change with the use of new catalogues. This values can be compared with the dipole signal calculated by Blake and Wall:

−3 ABW02 ≈ 11 × 10 . (60)

This value is significant greater, than the expected shot noise dipole amplitude. So the zero motion case can be exlcuded from the analysis of Blake and Wall. If the direction of the shot noise dipole would be opposite to the velocity dipole direction , the resulting amplitude −3 would be clearly smaller than ABW02 ≈ 11 × 10 . So it seems that the direction of the detected dipole cannot be biased more than 90◦, like Crawford claims in his publication. Crawford used in his analysis a different way to derive the dipole signal, as Blake and Wall did in 2002. It is quite possible, that the dipole can be detected by one method, while the other model does not show any clear signal. I do not believe, that Crawford analysis can be applied to the work of Blake and Wall. This would also mean, that the method Blake and Wall used is more powerful than the naive ansatz of just adding all galaxy positions up. So it makes sense to look for different methods to find the dipole. Maybe there is a method even more powerful that the one used by Blake and Wall. This could lead to more tight constraints on the dipole direction in radio surveys.

20 8 One alternative method

8.1 Two-point correlation function

In this section I want to introduce the concept of a two-point correlation function (like in (PP80)). This formalism is useful in many physical problems. I will use the two-point correlation function for galaxy distributions, but in principle it would also work for e.g. the distibution of plankton in oceans. Let n be the mean number of galaxies per volume. Then the probability to find a galaxy in an infinitesimal volume δV is given by

δP = n δV . (61)

This only holds if galaxies are distributed randomly. Because of the gravitational force it is possible that galaxies cluster. Then it is likely to find in the proximity of a galaxy more galaxies and equation (61) has to be modified. So the probability of finding a galaxy in δV1 and another one in δV2 with separation r12 is given by

2 δP = n δV1 δV2(1 + ξ(r12)). (62)

This is the definition of the two-point correlation function ξ. The value of ξ must always be bigger or equal to minus one. Assuming that galaxies would always come in pairs at separation r12, the correlation function would have a peak at this distance. If the positions of two galaxies are not correlated, ξ would just vanish. In this case we would just have the probability of (61) multiplied for the two galaxies:

2 δP = δP1 δP2 = n δV1 δV2. (63)

We can also ask for the conditional probability to find a galaxy in δV , assuming there is at galaxy a distance r. For finding this probability we just can divide (62) by δP1 = n δV1. After dropping the indices we arrive at

δP = n δV (1 + ξ(r)). (64)

This is the probability to find a galaxy in distance r of another galaxy. To find the mean number of galaxies around a random galaxy, we have to integrate this from zero to the desired distance R:

4 Z R hNi = πR3n + n ξ(r) 4π r2 dr. (65) 3 0 Clustering of galaxies means that the probability to find a galaxy near another one is in- creased. So for R ≤ 8 Mpc the value of hNi would be increased due to galaxy clustering. R Rcatalogue 2 For any finite catalogue of diameter Rcatalogue we have 0 ξ(r) 4π r dr = 0, as n := Ncatalogue . Vcatalogue

21 8.2 Angular two-point correlation function

In today’s radio surveys we only have the position, luminosity and polarisation of galaxies listed. Not included (for most of the galaxies) is information about the redshift or the distance. So it would be hard to find the two-point correlation function, which depends on the distances between galaxies. The solution for this problem is the angular two-point correlation function ω(θ). Here we do not look for correlations at certain distances r but at certain angles θ. For a mean number density per solid angle N N = (66) Ω the probability to find a galaxy in the solid angle δΩ is

δP = N δΩ. (67)

The definition of the angular two-point correlation function is

2 δP = N δΩ1 Ω2(1 + ω(θ12)), (68) where δP is the probability to find a galaxy in δΩ1 and another one in δΩ2 with angular separation θ12. In a perfectly isotropic universe a comoving observer would measure ω(θ) = 0. The next two steps are analogue to those for the spatial two-point correlation function.

We divide (68) by δP1 = N δΩ1 and drop all indices:

δP = N δΩ(1 + ω(θ)). (69)

Having a dipole signal means that δP is in one direction above the mean and in the opposite below the mean. So this would lead to a negative ω at θ ≈ 180◦ and a positive ω at small angles. Again we get the mean number of galaxies around a random galaxy by

Z θ hNi = N Ω + N 2π dθ ω(θ). (70) 0

R π N Again we have N 0 dΩ ω(θ) = 0 for any catalogue if N = Ω . We know that the dipole signal would leave an imprint on the angular two-point correlation function. In order to use this, we need to find a way to calculate ω(θ) from radio surveys. From a 0 survey we can count all pairs of galaxies np with a separation by a certain angle θ . But we have to give a certain tolerance level δθ in order to find some pairs. So if we take a random galaxy, every galaxy located in the ring between θ and θ + δθ will count as one pair. Of course, in this way every pair will be counted twice, so we have to multiply by one half at the end. Now we have a way to compute the number of pairs np from a survey. With equation (69) we can figure out how many pairs there should be, depending on the angular two-point correlation function. Let hδΩi be the solid angle of the ring between θ N−1 and θ + δθ. So for each galaxy we find in this ring about Ω hδΩi(1 + ω(θ)) galaxies with

22 would be counted as a partner. After doing this for all N galaxies and dividing by two (again we count every pair twice) we obtain (for large N)

1 N − 1 1 N n = N hδΩi(1 + ω(θ)) ≈ N hδΩi(1 + ω(θ)). (71) p 2 Ω 2 Ω In this formula the only unknown quantity is ω(θ). After we transform this equation we arrive at 2 n Ω ω(θ) = p − 1. (72) N 2 hδΩi

Now we can compute ω(θ) for a given survey and look for an imprint of the dipole signal. If the dipole signal can be identified with this method, then ω would be negative at 180◦. It is not clear which value of hδΩi is the most useful. A very small hδΩi would lead to no galaxy pairs at all. On the other hand, a big value of hδΩi could smear out the signal. For the test of correlation at 180◦ it is convenient to make δΩ as big as possible, as long as there is only one galaxy found in this region.

8.3 Magnitude of the signal

In this section I want to compare the two methods metioned above to find the dipole signal. Only the magnitude of the signal, and not the statistical significances will be considered here. The first method would be the vector addition ansatz (vaa) used by Baleisis et al. :

N X D~ = ~xi. (73) i=1

The second method would be the one of the angular two-point correlation function (acf) described above. First of all I will look at the case of a perfect isotropic distribution. In the vaa this would lead to D~ = 0 because two opposite vectors cancel each other. Also ω(θ) has to vanish in this case, but this is not easy to see from the formula (72). For a distribution of one thousand ◦ points (N = 1000) it would look like this: At 180 we have five hundred pairs, so np = 500. If the whole sky is observed we have Ω = 4π. As hδΩi we can choose a value of 4π × 10−3, so that there is always only one data point in the region. All in all we obtain 2 n Ω 2 · 500 · 4π ω(180◦) = p − 1 = − 1 = 0. (74) N 2 hδΩi 106 · 4π · 10−3

If we choose hδΩi to be smaller 4π×10−3, then ω(180◦) will become a larger positive number. This would just be due to our model of opposite galaxy direction. In reality there would just be a galaxy anywhere in the solid angle hδΩi = 4π × 10−3. This means we have to choose hδΩi as large as possible, with the constraint that in each hδΩi is no more than one galaxy. Now I add a dipole signal in the following way: In the direction of motion we will see ten extra galaxies, while in the opposite direction ten galaxies will no longer be seen. So we

23 only have np = 490 (for the galaxies still isotropic distributed) now:

2 n Ω 2 · 490 · 4π ω(180◦) = p − 1 = − 1 = −0.02 . (75) N 2 hδΩi 106 · 4π · 10−3

D 20 For vaa such a distribution would lead to N = 1000 = 0.02 with the direction of motion. So the signal would have the same magnitude. I will now consider the case of more background galaxies, meaning those which do not contribute to the dipole signal. The number of sources can be split into the number of sources distributed isotropicly NI (background) and the number of sources accountable for the dipole signal NS. The number of all sources will then be N = NI + NS and the number of pairs will be np = NI /2. Choosing hδΩi = Ω/N we get N Ω N N ω(180◦) = I − 1 = I − 1 ≈ − s . (76) 2 Ω (NI + NS) NI + NS NI NI +NS Now we see, that for the acf a great number of isotropic sources will reduce the dipole signal. This is not helpful for the purpose of finding a small signal in a great number of sources. Now I want to analyse what happens with both methods when the number of sources is doubled while the relation between NS and NI stays the same. The magnitude of the dipole per number of sources in the vaa can be expressed as

NI NS ! D 1 X X 0 + NS NS = | ~x | + | ~x | ≈ ≈ . (77) N N i i N + N N i=1 i=1 S I I Here I used the approximation that all vectors contributing to the dipole signal have the same direction. So the magnitude of the signal should be nearly the same for both methods. But the magnitude is not the most important value to compare. We have to ask how significant an observed signal is. In principle a large signal could be dominated by shot noise. This problem will be discussed in the next section. Even if the magnitude of the dipole signal in the acf is quite small, this ansatz is of great use in the search of clustering of galaxies. This is because galaxies cluster at small angles and so the number of signals corresponding to a signal NS would be much greater than the isotropic background NI .

8.4 Shot noise problem

The statistical uncertainty in the vaa has already been discussed by Crawford (FC09). So I will look at the acf in this section. Equation (72) can be converted into

N 2hδΩi N 2 hδΩi ω = 2n − . (78) Ω p Ω

24 After a longer analysis (see (PP80)) of this term one finds that the absolut error is 1 δω = √ . (79) np So the percentage error is given by δω N 2hδΩi N = 2 √ = √ , (80) ω (2npΩ − N hδΩi) np (2np − N) np where the error decreases, if np increases. But there is a serious problem if the number of pairs is about half of the absolute number of sources. This means the estimate of ω is bad as long as the data is dominated by an isotropic background. Because this is the case for the radio surveys, we can conclude that the signal ω(180◦) would be very small and strongly biased. I conclude that it is not a good idea to search the dipole signal via the acf ansatz.

9 Outlook and Summary

9.1 Outlook

In this thesis I only considered three methods to find the dipole signal: the acf, vaa and the method used by Blake and Wall. In principle there are a lot more possible ways to compute this signal. For example one can separate the sky into NP equal parts. Then in each part the mean flux of radio galaxies F¯ must be determined. After this we can look, if F¯ is greater in one direction. This would correspond to a dipole signal and the method would be very similar to the one used for CMB analysis. Another possible way to find the dipole would be, to calculate the Maximum likelihood:

  2  2  X X  min  ~xi −  alm Ylm(~xi)  . (81) l,m

The Problem here is that with the new surveys it could take a lot of time to find this minimum.

The vaa can be improved by weighting each signal with wi like

N X D~ = wi ~xi . (82) i=1

Here one has to be careful to the choose wi. The first idea would be just to weight the signals by the observed flux. But then nearby (and very bright) galaxies could bias the signal.

In the publication of Crawford ((FC09)) ” the characteristic size ρ of the probability cloud is determined by a random walk of N steps in three dimensions ”. In a random walk process, by definition, a grid is considered. Each step can only go along one of the axis of

25 this grid. But the direction of radio galaxies is not restricted to such a grid. I suppose the value of ρ should be changed into √ N ρ = , (83) 2 which holds for a Wiener process. This change would alter the results of Crawfords analysis. Further investigations are necessary to make any final conclusions.

9.2 Summary

Our motion through the universe should lead to a dipole anisotropy in the distribution of radio galaxies. I considered the question if this dipole can be detected in today’s radio surveys. Blake and Wall claim to have found this signal using spherical harmonic analysis. The work of Baleisis et al. would support this statement. But Crawford computed that the dipole can not yet be detected by the vaa. This is not necessarily a contradiction to Blake and Walls publication, because they used different methods. I found out that using the acf would not lead to a result with bigger statistical significance. In the near future there will be a new radio survey available. The Low Frequency Array will be used for the creation of a survey with N = 3.5 × 108 sources. With such a survey the dipole anisotropy in radio galaxies will be detected without a doubt.

10 Erklärung

Hiermit erkläre ich die vorliegende Arbeit selbstständig und ohne unerlaubte Hilfe verfasst und keine weiteren Hilfsmittel als die im Literaturverzeichnis angegebenen Quellen benutzt zu haben.

Matthias Rubart

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