The effect of Minimal Dark Matter annihilation
on the hydrogen recombination process
Pieter Cornelis van der Wijk
University of Groningen
Centre for Theoretical Physics
Supervisor: Prof. dr. Mees de Roo
May 9, 2010
Abstract
The Minimal Dark Matter model states that the dark matter is a quintuplet
of WIMPs added to the Standard Model. No other new physics beyond the
Standard Model is added. There is one neutral component in the multiplet
(without electrical and weak charge) has a mass of 9.6 TeV/c2 . The other
components are charged and more massive due to a mass splitting ∆M = 166
MeV/c2 . The neutral component can annihilate into two W -bosons. The bare
annihilation cross section of this annihilation is hσvi = 4.1 · 10−33 cm3 s−1 .
The annihilation cross section is enhanced by the Sommerfeld enhancement:
a summation over ladder-diagrams corresponding to gauge boson exchanges
prior to the annihilation. The Sommerfeld enhancement factor is 1 · 103 . Using
Pythia we analyze the decay of the two W -bosons produced in the annihilation.
Applying the so-called on-the-spot approximation we conclude that 40% of the
energy involved in the annihilation process, is transferred on to the electrons
and photons of the intergalactic medium during recombination. We modify
RecFast to incorporate the extra energy from the dark matter annihilations
and calculate the evolution of the free electron fraction during the hydrogen
recombination. We obtain a result in which the free electron fraction with
dark matter annihilation is not different from the free electron fraction without
−5
e
). Hence, the annihilation of
the annihilation taken into account ( ∆x
xe ∼ 10
Minimal Dark Matter does not change the recombination process. The energy
release from the annihilations is too small because of a very small annihilation
cross section. Adding physics beyond the Standard Model could increase the
energy released from dark matter annihilations.
Contents
1 Introduction
4
2 Minimal Dark Matter model
2.1 Introduction to the Minimal Dark Matter model
2.2 Properties of MDM . . . . . . . . . . . . . . . . .
2.2.1 Multiplet size . . . . . . . . . . . . . . . .
2.2.2 Interactions . . . . . . . . . . . . . . . . .
2.2.3 Mass splitting . . . . . . . . . . . . . . . .
2.3 Annihilation cross section χ0 χ0 → W + W − . . .
2.3.1 Amplitude 1 . . . . . . . . . . . . . . . .
2.3.2 Amplitude 2 . . . . . . . . . . . . . . . .
2.3.3 Mixed amplitudes terms . . . . . . . . . .
2.3.4 Kinematics and total cross section . . . .
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9
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3 Sommerfeld Enhancement
3.1 Introduction to the Sommerfeld Enhancement . .
3.2 Ladder diagrams . . . . . . . . . . . . . . . . . .
3.3 Yukawa and Coulomb potential (1-state system)
3.4 Excited dark matter . . . . . . . . . . . . . . . .
3.4.1 Two states . . . . . . . . . . . . . . . . .
3.5 Sommerfeld enhancement for MDM . . . . . . . .
3.5.1 The Sommerfeld enhancement factor . . .
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25
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4 W + W − Decay
4.1 Branching ratios . . . . . .
4.2 Pythia event generator . .
4.2.1 String fragmentation
4.3 The simulation . . . . . . .
4.4 Results . . . . . . . . . . . .
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41
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48
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. . . . . . . . . . .
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and hadronization
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5 Recombination
5.1 Introduction to cosmology . . . . . . . . . . .
5.2 Standard Recombination Process . . . . . . .
5.2.1 Recombination with DM annihilation
5.3 RecFast Results . . . . . . . . . . . . . . . . .
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6 Conclusion
62
7 Acknowledgements
64
2
A Complete χ0 χ0 → W + W − cross section calculation
65
B Mathematica notebook for Sommerfeld enhancement
81
3
Chapter 1
Introduction
Observations tell us that the energy and matter we are familiar with constitutes
only 4% procent of the total energy content of the universe. The remaining 96%
consist of an unkown form of nonbaryonic dark matter (22%) and dark energy
(74%) [1]. What is this dark matter and dark energy? Numerous articles have
been published on many theories about this dark matter and dark energy [2–4].
It is through observations that we hope to rule out or confirm the different
models, in the hope to find the correct theory describing the universe.
Dark matter observations
How do we know there ought to be extra dark matter, and why do we know
so little about it? There are numerous indication for dark matter. Three very
important indications are: random motion of galaxies in clusters, X-ray emission
from clusters, gravitational lensing, and rotation curves of galaxies. All four
indicators tell us that there should be more mass in the universe than we observe
from counting the mass that is luminous. Dark matter is simply some source of
mass which does emit light.
There are many clusters (a cluster is a group of galaxies) which have large
velocity dispersions. Assuming the cluster would be in a dynamical equilibrium
we can apply the virial theorem to the motion of the galaxies. With the application of the virial theorem we can infere a gravitational mass of the cluster. We
can also count the mass of the stars in the galaxies of the cluster and obtain an
expression of the mass of the luminous matter. Comparing the luminous mass
with the gravitational mass shows that the gravitational mass is, for the case of
for example the Coma cluster, 400 times larger than the luminous mass. [5]
Another way of determining the gravitational mass of a cluster is by looking
at the X-ray emission of the cluster. Between the galaxies in the cluster there
is intracluster gas. This gas is trapped and heated to around 105 K by the
gravitational potential of the cluster. This hot gas produces X-ray emission.
Comparing the mass inferred from this X-ray emission with the mass from the
luminous matter shows us again that the gravitational mass is much larger. [5]
The third indication for dark matter is given by gravitational lensing. Dark
matter not only influences the dynamics of galaxies and intracluster gas, but
also the trajectories of photons. Large concentrations of mass like clusters or
galaxies can alter the direction of the photon trajectory, as if it were a lense. This
4
gravitational lensing is a third way of determining the gravitational mass of the
object that produces the lensing effect. The mass derived from the gravitational
lensing is also in agreement with the presence of dark matter. [6]
Hints at dark matter are also present at the smaller scale of galaxies. In
spiral disk galaxies, the stars follow circular orbits around the galaxy center.
Assuming the circular motion is a consequence of the gravitational pull of the
mass M (R) inside the orbit, we would expect that at a large radii R where the
luminous mass has converged to some constant value M , the velocity would fall
as v ∝ √1R . However, observations show us that the velocity of stars at a large
distance from the galaxy center remains constant (see Figure 1.1). This suggest
that there should be a form of non-luminous mass in the outer region of the
galaxy: a dark matter halo. [6]
Figure 1.1:
The rotation curve of disk galaxy NGC 3198 [7].
Dark matter models
What is this dark matter? We already know some general properties from the
observations described above. For example, dark matter should have mass and
interacts very weakly. Furthermore we know that dark matter must be stable,
otherwise it would have decayed by now. And dark matter must be ‘cold’,
meaning that it decoupled while it was nonrelativistic. If dark matter would
have been relativistic while it decoupled, it would have erased the large scale
structure in the universe. [3, 5]
But more specific properties differ from dark matter model. There are two
classes of dark matter candidates: Massive Astrophysical Compact Halo Objects
(MACHO) and Weakly Interacting Massive Particles (WIMP) [6]. MACHOs are
a baryonic form of dark matter. It is a collective term for brown dwarfs, primordial black holes and steller remnants like black holes and neutron stars. But
5
observations put strong limits on the abundance of these MACHOs, preventing
it to account for the full 22% of dark matter. [8]
WIMPs are elementary particles with a mass between 10 GeV/c2 and a
few TeV/c2 . It is assumed they interact via the weak interaction. They were
produced via a thermal freeze-out. Before the moment of thermal freeze-out,
the creation and annihilation of the dark matter particles χ was in thermal
equilibrium χχ *
) ψψ with ψ being some Standard Model particle. As the
universe expanded and cooled, the Standard Model particles ψ no longer had
enough energy to create dark matter particles, and the interaction went in one
direction: the annihilation of dark matter particles. This lead to a decrease in
the dark matter abundance. The annihilation would eventually stop when the
expansion rate of the universe was larger than the annihilation rate: the dark
matter abundance would ‘freeze-out’. The moment of this thermal freeze-out
depends on for example the annihilation cross section and dark matter particle
density. The cross section that gives WIMPs the proper abundance during the
freeze-out is of the same order as the cross section for a generic weak-scale
interaction. This is called the ‘WIMP Miracle’. [2, 9]
However, there are a lot of theories on what this WIMP could be. Two
important candidates for a WIMP are the neutralino from supersymmetry and
the B boson from theories with universal extra dimensions. Both theories not
only try to solve the dark matter problem, but have the dark matter particle
embedded in a larger theory which tries to solve other (high-energy) physics
problems. [2]
In this report we will focus on a new WIMP model: Minimal Dark Matter [10–12]. It is an attempt to solve the dark matter problem with no free
parameters and a minimal amount of new physics. It only adds a new multiplet
to the Standard Model. Because the theory does not have free parameters, it
can immediately be put to the test.
Recombination as a probe for dark matter
In order to test the dark matter theories to find the correct one, the theories must
be tested using various probes. The dark matter model must explain or predict
certain phenomena and be in accordance with observations. An example of a
dark matter probe is the electron positron excess coming from the center of the
galaxy [13]. Attemps are made to explain this excess in terms of dark matter
annihilations. [14] Other probes which form possible constraints on the dark
matter models are structure formation and the Cosmic Microwave Background.
We will focus on the question whether the Minimal Dark Matter model is
in accordance with the hydrogen recombination process. We will calculate how
annihilations of Minimal Dark Matter particles influences the hydrogen recombination process. This probe for dark matter theories was already considered
by several authors, see [15–18], and can be applied to any theory of dark matter
that can annihilate.
Recombination is the process of the formation of hydrogen (and helium)
atoms in the early universe. Before the recombination, the universe consisted
of a hot plasma of protons, electrons and photons. Electrons and protons tried
to form hydrogen, but the hydrogen was ionized due to an energetic photon
very quickly. The free electrons were constantly scattering photons, enabling
the photons to travel only for short distances in a straight line: the direction
6
was quickly changed due to scattering. Before recombination, the universe was
opaque.
As the universe expanded, the energy of the photons decreased below the
ionization energy of hydrogen and eventually the photons did not have enough
energy to prevent the production of hydrogen: the protons and electrons could
sucesfully combine. The number of free electrons and protons decreased sharply.
At the end of recombination, there was only a very small fraction of free electrons left over. Therefore, the photons were not scattered anymore and they
were able to move in straight lines: the universe became transparent. These photons coming form this recombination form the Cosmic Microwave Background
(CMB). Measurements of the CMB performed by satellites like the Wilkinson
Microwave Anisotropy Probe (WMAP) showed that there are small anisotropies
in the angular photon energy distribution [1]. The angular power spectrum of
these anistropies is shown below in Figure 1.2.
Figure 1.2:
The TT-powerspectrum measured by WMAP [1].
The CMB powerspectra can be explained using a physical picture with baryonic acoustic oscillations [19]. With this understanding of the CMB it is possible
to determine some cosmological parameters such as the density parameters of
baryonic matter, dark matter and dark energy. The parameter values derived
from CMB were in accordance with values derived from other observations.
The CMB powerspectra is a probe for dark matter theories because any dark
matter model should be able to explain this powerspectra. If a dark matter
model predicts (using the physics of baryonic acoustic oscillations) the CMB
powerspectra should be different from the powerspectra we observe, then there
are two possible conclusions: 1.) the dark matter model is not correct; or 2.)
the physics explaining the CMB is not correctly understood. Since the current
physical picture of the CMB allows us to determine values for the cosmological
parameters which correspond to values obtained by other observations and we
have no other reasons to question our understanding of the CMB, we are not
inclined to draw conclusion nr 2. Moreover, we will see it as an indication that
7
the dark matter model is not correct.
In [15–18] already several dark matter models with decaying or annihilating
dark matter were tested using the CMB as a probe. In those articles it was shown
that if at the time of recombination, energy from dark matter annihilations
or decays was able to heat the hot plasma of protons, electrons and photons
or if this energy caused extra ionizations or excitations of hydrogen atoms, it
could change the recombination process. And if the amount of energy released
was large enough, it could alter the CMB powerspectra so significant that the
dark matter model can be ruled out by observations of the CMB performed by
WMAP.
Minimal Dark Matter and recombination
In this report we will test the Minimal Dark Matter model by looking at how
Minimal Dark Matter particle annihilations affects the recombination process.
If it would change the recombination process significantly, this may lead to a
prediction of the CMB that differs from observations.
First we need to calculate the chance of an annihilation of two dark matter
particles, this we will perform in chapters 2 and 3. After a short description of
the Minimal Dark Matter model we will calculate the annihilation cross section
in chapter 2. There we will only consider the tree-level annihilation into two
W -bosons.
In chapter 3 we will discuss the Sommerfeld enhancement. The Sommerfeld
enhancement leads to an increase in the annihilation cross section [20, 21] This
is due to multiple exchanges of a gauge boson that corresponds to an attractive
force between the two dark matter particles. The Sommerfeld enhancement
increases with decreasing velocity. This effect enables dark matter particles to
have a larger annihilation cross section at recombination than at freeze-out. In
chapter 3 we will numerically calculate the increase in the annihilation cross
section.
After having calculated the total annihilation cross section, we discuss in
chapter 4 how much energy is released during an annihilation, and in what
form. In that chapter we will study the decay of the W -boson produced in the
annihilation, using a computer simulation called Pythia.
Finally, we will calculate the effect of dark matter annihilation on the hydrogen recombination process in chapter 5.
8
Chapter 2
Minimal Dark Matter
model
Before we can calculate the cross section for the dark matter annihilation, we
first have to discuss some basics of the dark matter theory we will use: Minimal
Dark Matter model. First we will discuss the motivations for this model and
tell how it was designed. After that we will say some things about important
properties like mass and interactions of the dark matter particles in this model.
And finally, we will calculate the annihilation cross section of two neutral dark
matter particles into two W -bosons at the end of this chapter.
2.1
Introduction to the Minimal Dark Matter
model
The Minimal Dark Matter (MDM) model was first published by M. Cirelli,
N. Fornengo and A. Strumia in 2007 [11]. They wanted to introduce a very
minimalistic dark matter model.
With the MDM model they wanted to solve the dark matter problem with a
minimal amount of new physics introduced [11,12]. Other models like supersymmetry or the Kaluza-Klein model do not only introduce new particles, but also
new physics like R-parity and extra dimensions. This new physics beyond the
Standard Model is not yet observed in experiments. M. Cirelli and co-authors
point out three drawbacks of the current dark matter models, which motivated
them to introduce this minimalistic model. The drawbacks are:
1. the parameters bounded by collider experiments need a high degree of
finetuning;
2. the presence of a large undetermined parameter space obscures the dark
matter particle phenomenology;
3. the stability of the dark matter particle needs extra ad hoc features which
are introduced by hand (such as R-parity).
The MDM only introduces a new dark matter SU(2) multiplet. Its quantum
numbers are chosen such that it is a viable dark matter candidate. In this way,
9
the only remaining parameter is the dark matter mass which can be determined
by calculations of the cosmic dark matter abundance from the thermal freezeout. [11, 12] As a result, there are no free parameters and therefore the theory
can be used to do univocal predictions that can be tested.
To choose the right quantum numbers, there are four demands put on the
model: [11]
1. the lightest component must be stable on a cosmological time-scale;
2. the only renormalizable interactions are of the gauge-type;
3. quantum corrections generate mass splitting;
4. the MDM is still allowed by current dark matter observations (thus Y = 0
and no strong and electromagnetic interactions).
The fourth demand requires the lightest dark matter component to have no
strong interactions [22], no electrical charge (no electromagnetic interactions)
and no hypercharge Y (no neutral weak interactions). [23–25]
In MDM the assumption is made that dark matter is a WIMP, and is thus
a n-tuplet of the SU(2) gauge group. We can write down the most minimalistic
Lagrangian density for the MDM extension of the Standard Model (LSM )
1
/ +M χ
L = LSM + χ̄ iD
2
/ the gaugewith χ being the dark matter multiplet with tree-level mass M and D
covariant derivative of SU(2). The factor 1/2 is placed in front because χ is a
Majorana fermion (Cirelli and co-authors also discussed the case of scalar dark
matter but they found it not to be a viable dark matter particle [11]).
2.2
Properties of MDM
The main properties of the MDM model we use in this report will be discussed in
this section. We use will first discuss which size the multiplet must have in order
to have a non-decaying dark matter particle. It turns out it must be quintuplet
under SU(2). Secondly we will look into the possible interactions. And thirdly
we will explain the small mass splittings between the different components of
the multiplet. For information about other properties we refer to [11, 12].
2.2.1
Multiplet size
In order to have a stable component in the SU(2) dark matter multiplet χ without resorting to new physics, one has to look for a multiplet which is only able
to have renormalizable (decay) couplings to Standard Model particles such that
the dark matter lifetime is longer than the age of the universe. For different
multiplet sizes, we can write down different decay interactions. Thus the possibility of writing down a decay interaction, depends on the multiplet size. If we
can find an interaction term which would give the dark matter the possibility of
already being decayed, then, because we do not want to introduce new physics
to forbid that interaction, we do not have reasons as to why the dark matter
should not have this possible coupling.
10
A typical decay coupling would contain one dark matter multiplet, a coupling
constant, and several Standard Model fields. We assume that if the coupling
constant is dimensionless (it has mass dimension 0), it is of the same order
as the Standard Model couplings. If we can write down a dark matter decay
interaction with such a dimensionless coupling constant, the dark matter has a
short lifetime and has probably already decayed into Standard Model particles.
However, if the coupling constant in the dark matter interaction term is very
small due to a suppression, then the dark matter would have a very long lifetime
and it has perhaps not yet decayed.
For example, if the dark matter multiplet χ is a scalar doublet, then we can
write down an interaction term where χ can couple to a left-handed electron
doublet and right-handed electron neutrino singlet with a dimensionless coupling
constant. The coupling constant is dimensionless because the dimension is the
sum of the different components of the term and should be equal to 4: the
two fermionic fields each have mass dimension 3/2, the scalar field has mass
dimension 1 and together with the mass dimension 0 of the coupling constant
this makes 4. Because we can write down a decay interaction for this scalar
dark matter doublet, the neutral dark matter component within the doublet is
allowed to decay into for example two neutrino’s.
In order to be certain the dark matter does not decay, one needs to find a dark
matter multiplet for which it is impossible to write a proper decay interaction
terms or which has decay interactions with a very small coupling constant.
Any proper interaction term should conserve charge and be renormalizable.
Furthermore, since we consider a dark matter particle which couples to the
weak gauge sector, it should be invariant under global SU(2) transformations.
Any interaction term in the Lagrangian should conserve charge and therefore
not contain any net charge (electric and or hypercharge). Take for example the
interaction term
λχij Hk Hl ik jl
(2.1)
where χ is a dark matter triplet under SU(2), λ is the coupling constant and H is
a Higgs doublet. The are 2-dimensional anti-symmetric Levi-Cita tensors that
are included to make the term an invariant tensor under SU(2) transformations.
The two Higgs doublets H both have hypercharge Y = 1, thus in order for this
interaction to be invariant under SU(2) transformations and hence conserve
hypercharge, the χ triplet should have Y = −2.
A renormalizable term in the Lagrangian should have a total mass dimension
of 4. From the Dirac equation it follows that fermions have mass dimension 3/2
and from the Klein-Gordon equation it follows that scalars have mass dimension
1.1
Renormalizable interaction terms are of the form [26]
λφ4
gφψ † ψ
µφ3
eψ̄γ µ ψAµ
c2 |φ|2 A2
(2.2)
where φ represents some scalar field, ψ a fermion field, Aµ a (electromagnetic)
vector-field and λ, g, µ, e and c are the coupling constants.
The interaction term between the dark matter multiplet χ and Standard
Model particles must be invariant under global SU(2) transformations. Therefore the multiplet χ must be of the same size as the Standard Model particles
1 Unless
explicitly stated otherwise, we use h̄ = c = 1 throughout this report.
11
part in total. This is displayed in the number of indices the multiplet has:
the number of indices is equal to the multiplet size minus one. For example,
a triplet has two indices. Because χ must be of the same size as the sum of
the Standard Model particles (ignoring singlets), the number of indices must be
equal. For equation (2.1) this means that χ is a triplet since it must have two
indices because the two Higgs doublets together carry two indices. Acting with
the two anti-symmetric Levi-Cita tensors on the term gives the invariant tensor
λχkl Hk Hl . [27]
Because the interaction term in equation (2.1) contains besides the multiplet
χ only two scalar doublets, χ cannot be a fermion because of the spin a fermionic
particle has. Hence, the multiplet χ is a scalar triplet. The three scalar fields
together have a mass dimension 3. Therefore, the coupling constant λ should
have a mass dimension of 1 in order for the total interaction term to have mass
dimension 4.
For different interaction terms, for example if the terms in (2.2) would contain more scalar or fermion fields, it is possible that the coupling constant λ has
negative mass dimension. Then λ can be written in terms of a dimensionless
coupling constant divided by a cut-off mass: λ = λ0 /Λ. Λ is a large cut-off scale
which is initially free to choose, for example the Planck-scale 1028 eV. The cutoff scale describes at what energy the theory is no longer valid. In the case of a
coupling constant with mass dimension −n, one could say that the interaction
is suppressed by order Λn .
Example: Fermion triplet
Suppose the dark matter multiplet χ is a fermion triplet under SU(2) with
Y = 0. An invariant decay interaction term of the MDM multiplet χ with some
Standard Model particles such as an electron doublet and a Higgs doublet is
given by
λχij Lk Hl ik jl
(2.3)
with L the left-handed electron doublet eνLe with Y = −1 which contains a
left-handed
electron eL and a left-handen neutrino νe . And H the Higgs doublet:
+
φ
with
Y = 1.
φ0
Using the relation Q = Y /2 + T3 which follows from weak interaction symmetries, the relation gives the electric charge Q of the three weak isospin components T3 in the triplet χ. Every component of the multiplet has a different
weak isospin T3 quantum number: 1, 0, −1. The weak isospin component with
for example T3 = 1 has electric charge Q = 1 (because Y = 0).
Working out the indices in equation (2.3) gives:
+
+
0
+
λχ11 νe φ0 + λχ−
12 νe φ + λχ21 eL φ + λχ22 eL φ
From this equation we see that for example the weak isospin component χ11
can decay into an electron anti-neutrino (the νe wavefunction corresponds to
an ingoing fermion field or an outgoing anti-fermion field) and a Higgs boson.
Applying conservation of electric charge tells us that χ11 has Q = 0. (We do
not mind that this interaction violates conservation of lepton number, because
that also occurs in the Standard Model with neutrino oscillations.) Using the
same reasoning we see that χ22 is also neutral and corresponds to the same
12
component in the triplet as χ11 . χ12 and χ21 have electric charge −1 and +1
respectively.2
Because of these decay-channels, this triplet is unstable and therefore not a
suitable choice for the Minimal Dark Matter.
Example: Fermion quadruplet
Suppose we add an extra component to multiplet in the previous example. In
this case the four different weak isospin components are T3 = − 32 , − 21 , 12 , 32 . As
stated in the previous section, we want a multiplet with hypercharge Y = 0 and
one component of the multiplet to have electric charge Q = 0 because these
are constraints given by experiments. With Q = Y /2 + T3 we can choose any
weak isospin component and then calculate what hypercharge is needed for that
specific weak isospin to be electrical neutral. For example, demanding T3 = 12
to have charge Q = 0 gives a hypercharge Y = 1. It seems that for the case
of a quadruplet, we can not have Q = Y = 0 for any single component in the
multiplet.
If we would forget about the Y = 0 constraint, we can analyse some possible
decay-channels to illustrate the line of reasoning. We choose Y = 1 and thus
have QT3 =1/2 = 0. A possible interaction term is
λχijk Ll Hm H̃n il jm kn
φ0 where H̃ = −φ
is the conjugated Higgs doublet with hypercharge Y = −1.
−
Calculating the mass dimensions tells us that the coupling constant has mass
dimension −1. χ and L contain fermions which have mass dimension 32 , and
both Higgs doublets contain scalars with mass dimension 1. In order for this
interaction term to have an overall mass dimension of 4, the coupling constant
must have a mass dimension of −1. Hence, the coupling constant is suppressed
by Λ. Although the effective coupling constant is small, it would give a lifetime
for the dark matter particles which is smaller than the age of the universe.
This means that the neutral component of the multiplet can have for example
already decayed into an electron anti-neutrino and two Higgs bosons [12]. Even
though we losened the constraint of the hypercharge, this multiplet is not a
viable extension of the Standard Model. We will continu our search for the
proper multiplet size.
Fermion quintuplet
Using the line of reasoning demonstrated in the examples above, we can eliminate all fermionic multiplets with size smaller than five.
Any fermion quintuplet must interact with at least four doublets. To write
an interaction which is invariant under SU(2) transformations, the quintuplet
must interact with multiplets which have an equivalent size. Since we need four
doublets to compose a quintuplet, we need four doublets to write an interaction
term with the quintuplet which is invariant under SU(2). And at least one
2 These interactions are for the case without the Higgs symmetry-breaking. After symmetry
breaking φ+ becomes zero and φ0 = v + η, where v is given by the vacuum expectation value
and η is the Higgs boson. But still, after symmetry-breaking there are two interaction terms,
each a possible way for the dark matter particle to decay.
13
fermion doublet is required to get the proper spin relation. Performing a mass
dimension analysis we notice that the coupling constant is at least suppressed by
Λ2 (counting the mass dimensions of the fields in the interaction term: 4−2 23 −
3 (1) = −2). Because of this large suppression the quintuplet is considered to
be stable. Thus the fermion quintuplet is a proper choice for a dark matter
multiplet. Hence we will use this quintuplet as our dark matter.
Our dark matter multiplet is the fermion quintuplet with Y = 0 (because of
observational constraints)
 ++ 
χ
 χ+ 


0 
(2.4)
χ=
 χ− 
 χ 
χ−−
in which the neutral component χ0 has Y = T3 = Q = 0. It turns out that
the neutral component is also the lightest component compared to the charged
components in the multiplet. In the section below we discuss the interactions of
this quintuplet and it turns out that the charged components are able to decay
into the neutral component via the weak interaction. Therefore we expect χ0 is
the dark matter that is present in our universe.
2.2.2
Interactions
Because the dark matter is a quintuplet under SU(2)L it couples to the electroweak gauge bosons: W , Z and γ. The quintuplet hass mass and should
therefore also couple to the Higgs boson. The lightest component is neutral
(Q = Y /2 + T3 = 0) and therefore only couples to the W and Higgs boson.
Below we will work out the coupling to the weak gauge bosons in more detail
and obtain the correct coupling coefficients for the quintuplet.
The coupling to the electroweak gauge bosons is given by the standard covariant derivative in the Lagrangian
~ µ − Y ig 0 Bµ χ
Dµ χ = ∂µ − ig~τ · A
(2.5)
2
with Y = 0. The τ -matrices are the 2×2 Pauli matrices and form the generators
of SU(2) in a 2-dimensional representation.
For the case of the quintuplet we first have to derive the generators τ for this
5-dimensional representation. Since these generators τ have the same structure
as the angular momentum operators Lx,y,z , we can use the properties of the
raising and lowering angular momentum operators J+ and J− .
The three Pauli matrices of the 2-dimensional SU(2) representation correspond to the Lx , Ly and Lz matrices for the case l = 1 with m = ±1, 0. For
the five-dimensional representation we have to find the Lx , Ly and Lz matrices
for the case l = 2 with m = ±2, ±1, 0. To do this, we make use of the relations
τ1 = 21 (J+ + J− ), τ2 = − 2i (J+ + J− ) and of the fact that acting with τ3 on a
state should give its eigenvalue m.
Acting with J− on the state |l = 2, m = 1i changes the direction of the an-
14
gular momentum m

0
1
0
0
0


J− 






 √ 
 = 6




0
0
1
0
0






(2.6)
√
where the coefficient 6 is the Clebsch-Gordon coefficient. This equation tells
us that the second column in the J− matrix should look like the right-handside of equation (2.6). Performing this computation for all m values gives the
complete J− matrix (e.g. the analogue computation of J− acting on the state
|l = 2, m = 2i gives the first column). J+ can be determined in the same way.
The J+ and J− matrices together give the first two generators τ1 and τ2 . The
third generator is obtaind by performing the Lz |l = 2, mi = m |l = 2, mi calculation for every m-state. The result of these calculations are


0
2
√0
0
0
2
0
2 √0
6 √0
0 
1


6 √0
6 0 
τ1 =  0
2 0 0

6
0
2
0

τ2 =
i


2
0
2
0
0
0
−2
√0
6
0
0
0
√
− 6
√0
6
0
0
0
√
− 6
0
2
0
0
0
0
−2
0
0








τ3 = 

2
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
−1
0
0
0
0
0
−2





Using Wµ± = √12 A1µ ∓ iA2µ , where Aiµ are the vector-fields of SU(2) symmetry,
we can rewrite equation (2.5). Also applying the explicit expressions for the
generators τ obtained above and the expression for the dark matter quintuplet
χ we can write the weak interaction term
 ++ 
χ
 χ+ 


~ µ χ = ig χ++ χ+ χ0 χ− χ−− S  χ0 
ig χ̄~τ · A
(2.7)


2
 χ− 
χ−−
with



S≡


3
√4Aµ −
2 2W
0
0
0
√
2 2W +
3
√2Aµ −
2 3W
0
0
√0
2 3W +
√0
2 3W −
0
0
0
√
2 3W +
3
√2Aµ −
2 2W
0
0
√0
2 2W +
4A3µ






From this follows that the quintuplet component can be changed by exchanging
a W -boson. For example χ+ can decay into a χ0 under the emission of a W +
gauge boson.
The third component of the vector gauge boson A3µ correspond to a superposition of the Zµ gauge boson with the photon field Aµ :
A3µ = Zµ cos θW + Aµ sin θW
15
with the Weinberg angle θW . Only the charged components of the quintuplet
couple to the Z-boson and the photon.
Using the reasoning applied in the previous Section 2.2.1, one can write down
several interaction terms of the multiplet with Higgs bosons. An example is
λ
¯
χ̄χH̃H
Λ
Counting the mass-dimensions (4 −
coupling is suppressed by Λ.
2.2.3
3
2
−
3
2
− 1 − 1 = −1) it follows that the
Mass splitting
All components of the multiplet χ have the same mass at tree-level: M =
9.6 TeV/c2 . Assuming all multiplet components have the same mass minimizes
the number of free parameters. The mass M is determined by computing the
required dark matter abundance in the universe using the assumption that the
WIMPs are produced during a thermal freeze-out [11].
Loop contributions due to the Standard Model gauge bosons interactions
give small corrections to the masses of the charged components of χ. The
charged components have, compared to the neutral component χ0 , additional
couplings to the photon and Z-boson. These extra loop contributions produce
heavier components. These contributions correspond to the classical effect in
which Coulomb energy is stored in electroweak fields around the particle.
The mass splitting is
MQ − M0 ≈ Q2 ∆M
with ∆M the electroweak energy stored
θW
≈ 166 MeV/c2
∆M = α2 MZ sin2
2
M0 is the (tree-level) mass of the neutral component, MQ and Q are respectively
the mass and the charge of the charged dark matter particles and α2 is the
coupling constant of the weak interaction. [11, 12]
The mass splitting is in most calculations negligible small compared to the
tree-level mass.
2.3
Annihilation cross section χ0 χ0 → W + W −
In this section we will calculate the annihilation cross section corresponding to
the process χ0 χ0 → W + W − . (See Appendix A for the calculation in more
detail.) The two Feynman graphs that describes this process at tree-level are
given in Figure 2.1. The t-channel graph has an intermediate χ+ particle, the
u-channel graph has an intermediate χ− particle. Other possible annihilations
contain loops, for example the annihilation into two photons or one photon
and one Z-boson. These annihilations have smaller cross sections than the
annihilation into two W -bosons and therefore we will not consider them. [11]
The neutral component χ0 is a Majorana particle; it has no charge. It is said
to be in a superposition of particle and anti-particle [28]. To calculate Feynman
diagrams with Majorana particles we use the Feynman rules as described in [29].
16
An ingoing Majorana fermion is represented by u(p, s) and an outgoing one by
v̄(p, s). To determine whether a Majorana fermion is ingoing or outgoing, one
uses the direction of the (curved) arrow which corresponds to the direction in
which the fermion charge flows. [29]
Figure 2.1:
The two Feynman diagrams corresponding to the annihilation of the neutral
component χ0 into two W -bosons at tree-level.
The annihilation cross section σ depends on the matrix element M with
Z
Z
dσ
1
|~k1 |
σ = dΩ
= dΩ
|M|2
dΩ
2E1 2E2 |v1 − v2 | 16π 2 ECM
Thus in order to find the cross section σ, we first have to find an expression for
|M|2 . |M|2 depends on the amplitudes of the two Feynman diagrams
|M|2 = A1 A†1 + A1 A†2 + A2 A†1 + A2 A†2
with A1 the amplitude of the Feynman diagram on the left, and A2 is the
amplitude of the Feynman diagram on the right in Figure 2.1.
First we will find expressions for A1 A†1 and A2 A†2 in Sections 2.3.1 and
2.3.2. Secondly we will find expressions for the mixed amplitude terms A2 A†1
and A1 A†2 . And finally, using the expression for the matrix element |M|2 we
will use kinematics in the center-of-mass frame to find an explicit value for the
annihilation cross section in section 2.3.4.
2.3.1
Amplitude 1
Using the Feynman rules for Majorana interactions we obtain the amplitude of
the left diagram:
A1 = v̄(p2 , s2 )iΓiS(q)iΓu(p2 , s2 )∗ν (k1 )∗µ (k2 )
where the ∗µ and ∗ν are the two polarization vectors of the outgoing W -bosons
[30]. The spinors are functions of momentum pi and spin si , with index i = 1, 2.
The W -bosons only couple to the left-handed fields and the coupling strength
is given by equation (2.7). The term for the interaction vertices is
√
1
Γ = −i 3gW γ µ (1 − γ 5 )
2
17
(2.8)
√
with gW = 4πα2 .
The propagator S(q) is a charged dark matter fermion and is given by
q +M
/
q 2 −M 2 +i with M the mass of the dark matter particle. The internal momentum
q can alternatively be expressed as q = p1 − k1 = k2 − p2 .
Using equation (2.8) and re-arranging the projection operators 21 (1 − γ 5 )
(combining the two (1 − γ 5 ) terms leads to eliminating the mass term in the
center due to the left and right-projection operators acting on it) gives
A1 =
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 ) ∗ν (k1 )∗µ (k2 )
2
2
2(q − M + i)
Secondly we calculate the Hermitian conjugate A†1 and then sum the amplitude squared A†1 A1 over the dark matter spins and W -boson polarizations. For
the summation over spins we use the standard relations
X
ua,r (p)ūb,r (p) = (p
(2.9)
/ + M )ab
r
X
va,r (p)v̄b,r (p)
=
(p
/ − M )ab
(2.10)
r
To sum over polarizations of the W -boson we use the relation for massive gauge
bosons in unitary gauge:
X
ν,λ1 ∗σ,λ1
= −gνσ +
k1ν k1σ
2
MW
(2.11)
µ,λ2 ∗ρ,λ2
= −gµρ +
k2µ k2ρ
2
MW
(2.12)
λ1
X
λ2
where λi are the indices for the polarizations.
The second term on the right hand-side of equations (2.11, 2.12) is often
neglected because the W -boson is considered to be very heavy compared to the
momenta of the interaction. However, since we have dark matter particles with
a rest-mass of 9.6 TeV which is converted into a W -boson (MW = 80 GeV),
the W -bosons produced are relativistic. Hence the momenta of the W -boson
are very large compared to their rest-mass and this may lead to considerable
contributions of the right hand-side in equations (2.11, 2.12).
Performing the summation of spins r, s and polarizations λ1 , λ2 yields
X
2
|A1 |
r,s,λ1 ,λ2
4
9gW
4(q 2 − M 2 + i)2
h
i
µ
ν
σ
× (p/1 + M )ja γab
(1 − γ 5 )bc /qcd γde
(p/2 − M )ef γfρg (1 − γ 5 )gh /qhi γij
k2µ k2ρ
k1ν k1σ
k1ν k1σ k2µ k2ρ
×
gνσ gµρ − gνσ
− gµρ
+
(2.13)
2
2
4
MW
MW
MW
=
Expanding the last term between brackets gives four seperate traces proportional to different orders of W -boson mass MW . Calculating these traces is
quite some work. It requires carefull bookkeeping of all the γ-matrices.3
3 See
Appendix A for the complete calculation.
18
With the help of both projection operators 12 (1 − γ 5 ) we can talk one term
proportional to the mass M in (p
/ ± M ) to zero. The trick for this is to move
one of the projection operators to the other one, in order to absorb one projection operator into the other. By doing this, it will encounter several other
γ-matrices, and change a left-handed into a right-handed projection operator
and vice versa because of the anticommutation of a γ 5 -matrix with another γmatrix. The terms proportional to mass M contain one γ-matrix less than the
p
/ term. Eventually we have a right and left-handed projection operator acting
on M which gives zero: 12 (1 − γ 5 ) 21 (1 + γ 5 )M = 0.
Furthermore, we can get rid of the second mass-term M as well, just by
noticing that it is contained within a trace of an odd number of γ-matrices.
And all traces containing a γ 5 are zero as well.
Finally we obtain the expression for the first term in |M|2
X
2
|A1 | =
r,s,λ1 ,λ2
4
9gW
(I1 + II1 + III1 + IV1 )
(q 2 − M 2 + i)2
(2.14)
with
I1
II1
III1
IV1
(16(p1 · q)(p2 · q) − 8(q · q)(p1 · p2 ))
4 2 2
=
(q )(k2 )(p1 · p2 ) − 2(q 2 )(p1 · k2 )(p2 · k2 )
2
MW
−2(k22 )(p1 · q)(p2 · q) + 4(k2 · p2 )(k2 · q)(q · p1 )
4 2 2
=
(q )(k1 )(p1 · p2 ) − 2(q 2 )(p1 · k1 )(p2 · k1 )
2
MW
−2(k12 )(p1 · q)(p2 · q) + 4(k1 · q)(k1 · p1 )(p2 · q)
2 =
−(k12 )(k22 )(q 2 )(p1 · p2 ) + 2(k22 )(q 2 )(k1 · p2 )(k1 · p1 )
4
MW
=
+2(k12 )(q 2 )(k2 · p2 )(k2 · p1 ) − 4(q 2 )(k1 · p1 )(k1 · k2 )(k2 · p2 )
+2(k12 )(k22 )(p1 · q)(p2 · q) + 8(k1 · q)(k1 · p1 )(k2 · q)(k2 · p2 )
−4(k12 )(k2 · p2 )(k2 · q)(p1 · q) − 4(k22 )(k1 · p1 )(k1 · q)(p2 · q)
2.3.2
Amplitude 2
A2 is the amplitude corresponding to the diagram on the right in Figure 2.1. The
two differences between this diagram and the left diagram are 1.) the internal
momentum is b = p1 − k2 = k1 − p2 and 2.) the vertices of the W -bosons are
interchanged, interchanging the indices in their epsilon terms. The mass of the
χ− particle is equal to the mass of the χ+ particle. Therefore the amplitude
looks much like A1 and is given by
A2
=
×
4(b2
2
3igW
− M 2 + i)
v̄(p2 , s2 )γ µ (1 − γ 5 ) /b + M γ ν (1 − γ 5 )u(p1 , s1 ) ∗µ (k1 )∗ν (k2 )
After performing the same operations done on A1 on this amplitude, calculating the Hermitian conjugate, re-arranging the projection operators and then
19
sum over spins and polarizations using the same relations (2.9, 2.10, 2.11, 2.12),
we obtain
4
X
9gW
2
|A2 | =
2
4(b − M 2 + i)2
r,s,λ1 ,λ2
i
h
ν
5
µ
ρ
5
σ
× (p
/1 + M )γ (1 − γ )/bγ (p
/2 − M )γ (1 − γ )/bγ
k1µ k1ρ
k2ν k2σ
k2ν k2σ k1µ k1ρ
×
gνσ gµρ − gνσ
− gµρ
+
(2.15)
2
2
4
MW
MW
MW
Similar as to the first amplitude, we get an expression of four terms, each containing a trace. These traces can be calculated in an analogue way and we
get
4
X
9gW
2
(I2 + II2 + III2 + IV2 )
(2.16)
|A2 | = 2
(b − M 2 + i)2
r,s,λ1 ,λ2
with
IV2
=
I2
=
II2
=
III2
=
(16(p1 · b)(p2 · b) − 8(b · b)(p1 · p2 ))
4 2 2
(b )(k1 )(p1 · p2 ) − 2(b2 )(p1 · k1 )(p2 · k1 )
2
MW
−2(k12 )(p1 · b)(p2 · b) + 4(k1 · p2 )(k1 · b)(b · p1 )
4 2 2
(b )(k2 )(p1 · p2 ) − 2(b2 )(p1 · k2 )(p2 · k2 )
2
MW
−2(k22 )(p1 · b)(p2 · b) + 4(k2 · b)(k2 · p1 )(p2 · b)
2 −(k22 )(k12 )(b2 )(p1 · p2 ) + 2(k12 )(b2 )(k2 · p2 )(k2 · p1 )
4
MW
+2(k22 )(b2 )(k1 · p2 )(k1 · p1 ) + 2(k22 )(k12 )(p1 · b)(p2 · b)
−4(b2 )(k2 · p1 )(k2 · k1 )(k1 · p2 ) + 8(k2 · b)(k2 · p1 )(k1 · b)(k1 · p2 )
−4(k22 )(k1 · p2 )(k1 · b)(p1 · b) − 4(k1 )2 (b · k2 )(b · p2 )(k2 · p1 )
The subscript 2 indicates that these are the terms of the second amplitude.
2.3.3
Mixed amplitudes terms
To calculate the total amplitude at tree-level one also has to evaluate the mixed
amplitude terms A1 A†2 and A2 A†1 . The two mixed terms are given by
A1 A†2
A2 A†1
=
4(q 2
M2
4
9gW
+ i)(b2 − M 2 + i)
−
× v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 )ū(p1 , s1 )γ σ (1 − γ 5 )/bγ ρ v(p2 , s2 )
× ∗ν (k1 )∗µ (k2 )σ (k2 )ρ (k1 )
4
9gW
−
+ i)(q 2 − M 2 + i)
× v̄(p2 , s2 )γ ρ (1 − γ 5 )/bγ σ u(p1 , s1 )ū(p1 , s1 )γ ν (1 − γ 5 )/qγ µ v(p2 , s2 )
× ∗ρ (k1 )∗σ (k2 )ν (k1 )µ (k2 )
=
4(b2
M2
20
Again we sum over spins and polarizations, again we absorb one projection
operator into the other, and again we calculate the traces. Both mixed terms
give the same contribution. The final result is
X
r,s,λ1 ,λ2
A1 A†2 + A2 A†1
=
(b2
−
M2
4
9gW
+ i)(q 2 − M 2 + i)
× [Imix + IImix + IIImix + IVmix ]
(2.17)
with
Imix
IImix
IIImix
IVmix
=
= −32(b · q)(p2 · p1 )
8 =
(k2 )2 [(b · q)(p1 · p2 ) + (b · p1 )(p2 · q) − (b · p2 )(p1 · q)]
2
MW
+2(k2 · p1 ) [(b · p2 )(k2 · q) − (b · k2 )(p2 · q)]
+2(k2 · p2 ) [(b · k2 )(p1 · q) − (b · p1 )(k2 · q)]}
8 =
(k1 )2 [(b · q)(p1 · p2 ) + (b · p2 )(p1 · q) − (b · p1 )(p2 · q)]
2
MW
+2(k1 · p1 ) [(b · k1 )(p2 · q) − (b · p2 )(k1 · q)]
+2(k1 · p2 ) [(b · p1 )(k1 · q) − (b · k1 )(p1 · q)]}
4 (k1 )2 (k2 )2 [(b · q)(p1 · p2 ) − (b · p1 )(p2 · q) − (b · p2 )(p1 · q)]
4
MW
+2(k1 )2 [(b · p1 )(p2 · k2 )(k2 · q) + (b · k2 )(k2 · p1 )(p2 · q)
−(b · q)(k2 · p2 )(k2 · p1 ) − (b · k2 )(k2 · q)(p1 · p2 )]
+2(k2 )2 [(b · p2 )(k1 · q)(k1 · p1 ) + (b · k1 )(p2 · k1 )(p1 · q)
−(b · q)(p2 · k1 )(p1 · k1 ) − (b · k1 )(k1 · q)(p1 · p2 )]
+4 [(b · k1 )(k1 · q)(k2 · p1 )(k2 · p2 ) + (b · k2 )(k1 · p2 )(k1 · p1 )(k2 · q)]
+2(b · q)(k1 · k2 ) [(k1 · p2 )(p1 · k2 ) + (k1 · p1 )(k2 · p2 ) − (k1 · k2 )(p1 · p2 )]
+2(b · p1 )(k1 · k2 ) [(k1 · k2 )(p2 · q) − (k1 · q)(k2 · p2 ) − (k1 · p2 )(k2 · q)]
+2(b · p2 )(k1 · k2 ) [(k1 · k2 )(p1 · q) − (k1 · q)(k2 · p1 ) − (k1 · p1 )(k2 · q)]
+2(b · k1 )(k1 · k2 ) [(k2 · q)(p1 · p2 ) − (k2 · p1 )(p2 · q) − (k2 · p2 )(p1 · q)]
+2(b · k2 )(k1 · k2 ) [(k1 · q)(p1 · p2 ) − (k1 · p2 )(p1 · q) − (k1 · p1 )(p2 · q)]}
2.3.4
Kinematics and total cross section
In order to get a more explicit expression of how the matrix element |M|2
depends on the energy and momenta of the particles, we have to choose a frame
of reference and perform explicit calculations of the kinemetics. We will use the
center-of-mass frame as our frame of reference. We will use the values for the
dark matter mass M = 9.6 TeV/c2 , and the W -boson mass MW = 80.4 GeV/c2 .
We also incorporate the mass difference due to the splitting between the masses
of the dark matter particles in the quintuplet (see Section 2.2.3). The charged
dark matter particles are heavier than the neutral dark matter particle: Mχ+ =
M + ∆M = Mχ− with ∆M = 166 MeV/c2 . The kinematics are expressed in
the Mandelstam variable s = (p1 + p2 )2 .
21
χ01 : p1
χ02 : p2
=
=
W − : k1
=
+
W : k2
p
|~
p| = E 2 − M 2
|~k| =
q
=
(E, p~z)
(E, −p~z)
(ω, ~k)
(ω, −~k)
r
M2
1− 2 =E
E
= E
r
E2 −
2
MW
= E
1−
r
2
MW
=E
E2
1−
r
4M 2
≡ Eβ1
s
1−
2
4MW
≡ Eβ2
s
The general equation of the differential cross section for the interaction process p1 , p2 → k1 , k2 is [26]:
dσ
1
|~k1 |
=
|M|2
dΩ CM
2E1 2E2 |v1 − v2 | 16π 2 ECM
with |v1 − v2 | = Ep~11 − Ep~22 = 2p
E as the relative velocity of the two particles
as viewed from the laboratory-frame, E1 and E2 follow from the normalization
of the two ingoing wavepackets χ01 and χ02 and are the energies of the ingoing
particles. Putting in the appropriate numbers for the dark matter annihilation
into two W -bosons gives
dσ
(h̄c)2
=
|M|2
(2.18)
dΩ CM
256π 2 Ep
where we inserted the factor (h̄c)2 to get the corrects units of barn. And |M|2
is given by:
X
r,s,λ1 ,λ2
|M|2
=
4
9gW
[I1 + II1 + III1 + IV1 ]
(q 2 − M12 + i)2
4
9gW
[I2 + II2 + III2 + IV2 ]
(b2 − M12 + i)2
4
9gW
+ 2
2
(q − M1 + i)(b2 − M12 + i)
× [Imix + IImix + IIImix + IVmix ]
+
(2.19)
where M1 = M + ∆M is the mass of a singly charged dark matter particle χ± .
Integrating the differential cross section over the solid angle gives us the
total cross section
Z
Z
2π
σ=
π
dφ
0
dσ sin θdθ
0
We perform this intergration for each term in equation (2.19) seperately, after
which we take the limit of → 0 and finally perform a series expansion in
momentum p. The explicit expression for the lowest order contribution to the
cross section is
4
2
4
M 2M 4 − 3M 2 MW
+ 4MW
(h̄c)2 36gW
2 (2M 2 + 2M ∆M − M 2 + ∆M 2 )2 p
128π MW
W
22
With the appropriate values for the specific parameters we can obtain a numerical result for the cross section. We use the expression that for small momentum p = M v with v being in units of c (thus v ranges from 0 to 1) and
we fill in all the numbers of the
√ masses and the coupling constants: 1 barn =
1
, gW = 4πα2 = 0.66, (h̄c)2 = 0.389379 GeV2 mbarn,
10−24 cm2 , α2 = 29
MW = 80.398 GeV/c2 , ∆M = 166 MeV/c2 , M = 9.6 TeV/c2 . Then the expression for the cross section becomes (writing the largest contribution at the
next-to-lowest order)
σ=
4.1 · 10−33
+ 4.0 · 10−29 v cm2
v
(2.20)
We can compare the cross section we obtained with the one published by
Cirelli and co-authors [11]. They calculated a cross section of
hσvi =
18π(h̄c)2 α22
≈ 2.8 · 10−37 cm3 s−1
M2
(2.21)
with v being the velocity as a fraction of the speed of light c. Comparing this
with the first-order result we obtained (hσvi = 4.1 · 10−33 cm3 s−1 in equation
(2.20)) shows the cross section of Cirelli is independent of the mass splitting ∆M ,
is independent of the W -boson mass and is much smaller. It is independent of
the mass splitting ∆M because they simply state that they neglected the mass
splitting.
The reason for equation (2.21) to be independent of W -boson mass MW is
that they used a different summation over the W -boson polarizations (equations
(2.11, 2.12))
X
ν,λ1 ∗σ,λ1
= −gνσ +
k1ν k1σ
2
MW
µ,λ2 ∗ρ,λ2
= −gµρ +
k2µ k2ρ
2
MW
λ1
X
λ2
If one only uses the metric term and sets the second term on the right hand-side
containing the W -boson momenta and mass to zero (let us define this term as
the ‘kinetic term’), then one obtains an expression for the cross section which is
at first order independent of the W -boson mass in the nominator. Performing
the cross section calculation without
kinetic terms in the summation over
P the
polarizations, effectively reduces
|M|2 (equation (2.19)) to the sum of
X
I1
I2
4
|M|2 = 9gW
+ 2
+
(q 2 − M12 + i)2
(b − M12 + i)2
r,s,λ1 ,λ2
Imix
+ 2
(q − M12 + i)(b2 − M12 + i)
And in this case one is able to obtain equation (2.21).
Although it is often allowed to ignore the kinetic term in the sum over polarizations because the mass of the W -boson is in most Standard Model and
QED calculations very large (the W -boson is very heavy compared to the momenta involved), we argue that in this case it is not permitted. The two neutral
23
Minimal Dark Matter particles χ0 χ0 which annihilate into the two W -bosons
W + W − are much more massive: the total energy involved in the annihilation
(ECM > 19.2 TeV) is much larger than the rest-mass energy of the two W bosons produced. Hence, the two W -bosons are (highly) relativistic and have
large momenta. Consequently, the kinetic terms are large compared to the
metric terms.
Taking the kinetic terms into account leads to a larger cross section: the
summation over polarizations is increased and therefore one calculates a larger
amplitude compared to the calculation without the kinetic terms. In this report,
we will use the value for the cross section as calculated in this section:
hσvi = 4.1 · 10−33 cm3 s−1
24
(2.22)
Chapter 3
Sommerfeld Enhancement
After having calculated the annihilation cross section, we have to take the Sommerfeld enhancement of the annihilation cross section into account. The Sommerfeld enhancement of the cross section is an additional multiplicative factor
due to extra interactions which are not taken into account in the annihilation
cross section calculation we have performed. In some cases, this Sommerfeld
effect can lead to a significant boost of the annihilation cross section. In this
chapter we will first discuss the Sommerfeld enhancement in general and sketch
the outline of the derivation of the Sommerfeld enhancement in Section 3.2.
In Section 3.3 we will discuss the main features for the Coulomb and Yukawa
potential for single state dark matter. And in Section 3.4 we discuss the sommerfeld enhancement for a 2-state dark matter system. Finally, in Section 3.5
we numerically calculate the Sommerfeld enhancement for the Minimal Dark
Matter model.
3.1
Introduction to the Sommerfeld Enhancement
Normally one calculates the annihilation cross section using the perturbation
theory given by the Feynman diagrams. In this calculation one often only considers the lower order diagrams, the higher order diagrams are ignored because
they would give negligible contributions. The incoming and outgoing particles
are described by plane waves and the interactions are point-like. For relativistic
particles this theory gives reasonably correct results for the cross section.
For nonrelativistic particles which interact via an attractive long range force
the perturbation theory is not always valid: in specific cases the probability
is increased dramatically due to the attractive force. Consider two particles
(e.g. an dark matter and its anti-particle) are approaching each other and for
the moment we ignore the attractive force, then the chance of an interaction
is given by the cross section σ0 , calculated using the perturbative theory of
the Feynman diagrams. Because of the attractive force, the cross section σenhc
of the annihilation process which takes the attractive interaction priot to the
annihilation into account can be larger than the cross section σ0 .
For a classical analogue, consider the example given in [20] in which a point
particle scatters on a star. If gravity is switched off, the cross section for the
25
particle scattering on the star with radius R is given by the geometrical size of
the star: σ0 = πR2 . If the incoming particle has an impact parameter equal to
or smaller than R it will scatter on the star. Gravity is switched back on. Due
to the attractive nature of gravity, the point particle is attracted in the direction
towards the star. Because of this long range force, it is possible for the particle
to scatter with the star even though the impact parameter is larger than R: the
scattering cross section is given by σ = πb2max where bmax is the largest possible
impact parameter such that the particle barely hits the star. bmax depends
on the velocity of the particle with respect to the star. For exmaple, if the
particle is relativistic (bmax ≈ R) and has an impact parameter larger than R,
its direction can be slightly altered by the gravitational pull of the star, but it
will continue to move more or less in a straight line (a plane wave). However,
if the particle would travel at a low velocity, the effect of the gravity is larger
with possibly a scattering event as a result (bmax > R). The lower the velocity,
the larger the effect of the potential: the larger the enhancement of the cross
section.
The interaction of the two particles due to this attractive force, prior to
the moment of annihilation, is mediated by gauge bosons corresponding to that
force. The more gauge bosons are exchanged, the larger the attraction. We
can visualize this in so-called ladder diagrams. See for example Figure 3.1
where two hypothetical dark matter particles χ exchange a massive gauge boson
φ corresponding to an attractive interaction, before they annihilate into two
Standard Model particles ψSM .
Figure 3.1:
An example of a ladder diagram.
If the relative velocity of the two dark matter particles is very small, they
are able to exchange more gauge bosons than when the relative velocity is large.
Each gauge boson φ exchange in Figure 3.1 corresponds to a factor of αM
mφ in the
amplitude of this Feynman diagram [21], with α being the coupling constant of
the attractive interaction, M the mass of the dark matter particle χ and mφ the
mass of the gauge boson φ. If αM
mφ > 1, the amplitude increases as we considers
more gauge boson exchanges, hence: the amplitude of a higher order diagram
is larger than the amplitude of a lower order diagram. Therefore, the lower the
relative velocity, the larger the amplitude, the larger the enhanced annihilation
cross section.
Hence, for the annihilation of relative slow particles with an attractive interaction and have αM
mφ > 1, one must take the higher order diagrams into account.
In the next Section we argue that calculating the amplitudes of these higher-
26
order diagrams is equivalent to solving the (radial) Schrödinger equation for
a wavefunction ψ(r) in the presence of the potential V (r) from the attractive
interaction:
p2
− ψ(r) + V (r)ψ(r) = Eψ(r)
(3.1)
M
with M the reduced mass of the annihilating 2-particle system and E the kinetic
energy in the center-of-mass frame. In the case we would ignore the potential
(as we would do in the calculation of the lowest order Feynman diagrams) the
solution is given by a plane wave ψ0 (r). However, if we take into account this
potential V (r), the wavefunction ψenhc (r) is different from a plane wave: the
incoming particle is distorted due to the presence of the potential. [20, 31]
The probability of an annihilation at point r = 0 is proportional to amplitude squared of the time-independent 2-particle wavefunction ψ(r): |ψ(0)|2 .
If there would be no attractive interaction, the probability is proportional to
|ψ0 (0)|2 . However, if do consider this attractive interaction, the probability is
proportional to |ψenhc (0)|2 . We define the Sommerfeld enhancement as the relative increase of the annihilation cross section σenhc with interaction potential,
compared to the cross section σ0 calculated without potential [20]:
Sk ≡
|ψenhc (r = 0)|
σenhc
=
2
σ0
|ψ0 (r = 0)|
2
→
σenhc ≡ Sk σ0
(3.2)
The Sommerfeld enhancement can lead to significant enhancement of the annihilation cross section.
An equivalent expression often used can be derived via the optical theorem.
ikr
Using a different boundary condition (ψ = f (θ)e
as r → ∞), the optical
r
theorem states that σ = 4π
k =f (0) [32]. Instead of depending on the wavefunction
at r = 0, the Sommerfeld enhancement is given by the wavefunction ψ(r) as
r→∞
2
|ψenhc (r = ∞)|
Sk =
(3.3)
2
|ψenhc (r = 0)|
The Sommerfeld enhancement is only relevant for slow particles. In the case
of very slow particles, we only have to consider the s-wave annihilations. The
enhancement Sk can be obtained from solving the radial Schrödinger equation.
From equation (3.2) we see that the total cross section is obtained from multiplying the Sommerfeld enhancement factor Sk with the bare annihilation cross
section σ0 , which is calculated using the Feynman diagrams.
3.2
Ladder diagrams
The derivation of the Sommerfeld enhancement from field theory was done by
R. Iengo [33] and S. Cassel [34] using the Bethe-Salpeter. The Bethe-Salpeter
equation is an integral equation that describes a bound state which can have
multiple vector bosons exchanges [35]. They showed that the radial Schrödinger
equation is the nonrelativistic version of the Bethe-Salpeter integral equation.
The (annihilation or scattering) amplitude is a multiplication of the amplitude
without Sommerfeld enhancement with the amplitude of the solution from the
Schrödinger equation (3.1).
27
Figure 3.2:
Irreducible diagrams of first and (some) of second order. Time flows in the
vertical direction.
Consider two particles in a bound state (the total energy is smaller than
the total kinetic energy of the two particles) as a result of an attractive force
between them. For bound states, one can generally assume there will be multiple
exchanges of vector bosons mediating the force. There are several irreducible
diagrams which give contributions to this bound state (see Figure 3.2). In case
of small coupling constants, we only have to consider the first diagram with one
vector boson exchange at a time. If the two particles would be free waves for
r → ∞, then these vector boson exchanges would destroy this free state and
create the bound system.
Figure 3.3:
Ladder diagrams: higher orders of the first diagram in Figure 3.2. Time flows
in the vertical direction.
The irreducible diagram corresponding to one vector boson exchange can
be expanded in powers of the coupling constant α2 . The higher order terms
correspond to the succesive exchange of multiple vector bosons (Figure 3.3). The
resulting Bethe-Salpeter integral equation for the wavefunction corresponding
to 2-particle state with these multiple exchanges is [35]:
ZZ
ψ(3, 4; 1, 2) = ψa (3; 1)ψb (4; 2) − i
Kb (4; 6)Ka (3; 5)G(6, 5)ψ(5, 6; 1, 2)dx5 dx6
(3.4)
where the first term corresponds to no vector boson exchange: the 2-particle
state remains consisting of two free wavefunctions. Kb (4; 6) is the amplitude
function for particle b traveling from point 6 to point 4 (K is the Green’s function
satisfying i ∂ψ
∂t = Hψ [36]). G(6, 5) is the function describing the exchange of
the vector boson from point 5 to 6.
For nonrelativistic bound states, Bethe and Salpeter argue that the two
particles are able to interact over a long time, exchanging vector bosons several times. This would destroy the free particle state. Therefore they take the
homogenous Bethe-Salpeter equation with the boundary value that the wavefunction ψ is given by two free particle wavefunctions as r → ∞ . [35]
Going to the center-of-mass frame in the momentum space, they define the
total and the relative momentum: P = pa + pb , p = pa − pb . For the nonrelativistic case we can make the approximation of an instantaneous interaction;
removing the relative time component p20 = 2 dependence in the denominator
28
of the G(k) term, putting the vector boson on-shell. One takes only into account
the ‘instant’ Coulomb interaction, and neglects possible retarded magnetic interactions. This can be justified as the retarded magnetic interactions are a relative
factor of order (mφ /M )2 smaller compared to the Coulomb interaction. [33, 35]
Taking the nonrelativistic limit of the Bethe-Salpeter equation, Bethe and
Salpeter introduce the function
Z ∞
φ(~
p) =
d0 ψ(~
p, 0 )
(3.5)
∞
with ψ(~
p, 0 ) being the wavefunction ψ(3, 4; 1, 2) in the center-of-mass frame in
the coordinate space. After inserting this function and a Yukawa potential into
the nonrelativistic Bethe-Salpeter equation, they perform the integration over
relative energy . With this integration, they obtain the Schrödinger equation
in momentum space for two particles interacting by the Yukawa potential: [35]
2
Z
α2
p
− E φ(~
p) = − d3 kφ(~
p + ~k) 2 2
(3.6)
M
2π (k + m2φ )
R
2
with the Fourier transform − d3 k 2π2 (kα2 +m2 ) =
φ
−e−mφ r
.
r
[37]
The interactions described by the ladder diagrams change the two incoming
wavefunctions of free particles into one wavefunction φ. The annihilation amplitude A of these two bound particles is different from the annihilation amplitude
A0 that contains no summation over ladder diagrams. For the s-wave annihilation channel this new amplitude A is given in terms of the old annihilation
amplitude A0 multiplied by the solution to the Schrödinger equation φ(~
p) [33]
Z
d3 p
A=
p)A0
(3.7)
3 φ(~
(2π)
For an s-wave A0 is a constant c and thus A = cφ (r = 0). [33, 34] The Sommer|φ(r=0)|2
feld enhancement factor is given by |φplane
2.
wave (r=0)|
Another way to look at the Sommerfeld enhancement in field theory, is to
view each vector boson exchange in the ladder diagrams corresponding to an
extra factor of αM/mφ in φ. The more exchanges, the larger the enhancement
(provided αM > mφ ). This corresponds to a threshold singularity. [21]
3.3
Yukawa and Coulomb potential (1-state system)
Coulomb potential
α
In the case of an attractive Coulomb potential V (r) = − 2r
between the two
annihilating particles, the Schrödinger equation can be written in dimensionless
parameters
1
−ψ 00 (x) − ψ(x) = v ψ(x)
x
where we have performed the coordinate transformation x ≡ αM r and divided
all terms by M α2 /2. The only free parameter on which the differential equation
29
depends, is v = αv with v in units of the light speed c and α as the coupling
constant of the Coulomb potential. v is dimensionless.
The solution can be obtained analytically using hypergeometric functions
(see [38]) as is shown in [20] and [39]. The corresponding Sommerfeld enhancement is given
π
v
(3.8)
Sk = − πv 1−e
For v → 1 the enhancement goes to 1, and for v → 0 the enhancement goes to
infinity. The Sommerfeld enhancement in the Coulomb case is independent of
the mass M of the particles involved.
Note that the Sommerfeld enhancement only depends on the dimensionless
parameter v . In the analysis that follows next, we will always make use of the
coordinate transformation x ≡ αM r and write the Schrödinger equation and the
Sommerfeld enhancement in terms of dimensionless parameters. For example, in
the next subsection where we discuss the Yukawa interaction, the enhancement is
also a function of the ratio between the mass of the force mediating gauge boson
mφ and the mass of the two annihilating particles M which is represented by
m
the dimensionless parameter φ = αMφ . By using these dimensionless parameter,
one can make a more general analysis of the Sommerfeld enhancement.
Yukawa potential
The Coulomb case can be seen as a special case of the Yukawa potential in
which the mass mφ of the gauge boson is set to zero. The general form of the
−mφ r
(attractive) Yukawa potential is V (r) = − αe r .
The solution to the radial Schrödinger equation containing a Yukawa potential can not be obtained analytically. Therefore the equation has to be solved
numerically. This is done with Mathematica (see appendix B for the notebook
used).
m
The solution depends on two quantities: v and φ = αMφ , with M the mass
of the annihilating particle. See Figure 3.4 for the Sommerfeld factor dependence
on these two parameters.
There are three important features for the enhancement by the Yukawa
potential:
1. it has resonances;
2. it saturates;
3. it contains the Coulomb limit for φ < v (the lower right part in Figure
3.4).
In the region of the Coulomb limit, the deBroglie wavelength of the annihilating
particles is smaller than the range of interaction (the energy of the gauge boson
is very small compared to the energy of the annihilating particles, hence it seems
the gauge boson is massless). As the velocity decreases, the wavelength increases
and becomes larger than the range of interaction (φ > v ) and the Sommerfeld
enhancement saturates. [20] When we look at the Sommerfeld enhancement as
a function of the velocity (v ) and take φ constant, then we see in Figure 3.4
that as v decreases from large to small, the Sommerfeld enhancement is at first
30
Figure 3.4:
Sommerfeld enhancement contourplot for a Yukawa interaction. Figure is obtained from [20].
increasing, but for v < φ the Sommerfeld enhancement saturates: it becomes
constant.
In the saturation region there are resonances: regular sharp increases in the
Sommerfeld enhancement. The position of the resonances is a function of φ :
when we look at Figure 3.4 we see that as φ changes (and take v to be constant
and v < φ ) there are several sharp increases in the Sommerfeld enhancement,
they are the
q resonance peaks. The position of the resonances are approximately
(2n−1)π
αM
given by
with n = 1, 2, . . .. These resonances correspond to
mφ =
2
zero-energy bound states.
The resonance behavior can be explicitly studied by approximating the
Yukawa potential with a potential well (for Sommerfeld enhancement in a potential well see: [20, 21, 39]). The Yukawa potential is approximated by V (r) =
−αmφ θ(R − r) with R = 1/mφ as the range of the Yukawa potential. Solving
the Schrödinger equation for this potential is done in [40],
it is shown
where
2 αM
that the zero-energy bound states are obtained for cos mφ = 0. Thus the
q
(2n−1)π
resonance positions correspond to the equation αM
.
mφ =
2
3.4
Excited dark matter
Excited dark matter is dark matter which has the ability that a dark matter particle can be ’excited’ to a more massive dark matter particle by the absorption
or emission of a gauge boson. The excited state is associated with a dark matter particle in the multiplet that has a larger mass than the stable and neutral
dark matter particle in the multiplet. In the case of Minimal Dark Matter, the
charged dark matter particles are the excited states of the neutral component.
31
Consider a 2-particle state, for example χ0 χ0 , then the (first) excited state corresponds to χ+ χ− and the second excited state to χ++ χ−− . A 2-particle state
can be excited by the exchange of a W -boson. Considering the exchange of the
W -boson in the ladder diagrams of the Sommerfeld enhancement, then there is
a constant process of being excited to the first or second excited state and a
form of radiative decay to a lower state (see Figure 3.5).
Figure 3.5: An example of a ladder diagram for the case of Minimal Dark Matter. With the
exchange of a W -boson, a 2-particle dark matter state can be excited to an other 2-particle
dark matter state.
Before we discuss the Sommerfeld enhancemet of Minimal Dark Matter in
Section 3.5, we will first discuss the Sommerfeld enhancement of a 2-particle
state with one excited state. The qualitative effects of adding a first excited state
are comparable of having a two excited states, which is the case for Minimal
Dark Matter.
3.4.1
Two states
Suppose a state ψ1 consists of two χ0 both with mass M . And suppose state
ψ2 contains one χ− and one χ+ , both with mass M + ∆M . ∆M is the mass
splitting between χ0 and χ± . This mass splitting is small compared to the mass
M of χ0 (hence M + ∆M ≈ M ). However, the mass splitting does produce a
difference in rest-mass energy between the two 2-particle states χ0 χ0 and χ+ χ−
which need to be taken into account. This mass splitting results in the 2∆M
term in the form of a potential energy. The radial Schrödinger equation is given
by
ψ100
+ V (r)ψ2
M
= Eψ1
ψ200
+ V (r)ψ1 + 2∆M ψ2
M
= Eψ2
−
−
where E is the kinetic energy and with boundary conditions
ψ1 (0) = ψ2 (0) = 1
ψ10 (∞) = iM vψ1 (∞)
p
ψ20 (∞) = i 2M (E − 2∆M )ψ2 (∞)
Note that we have not yet performed the subsitution x ≡ αM r.
The Sommerfeld enhancement for dark matter with an excited state does
m
not only depend on the dimensionless parameters v = αv and φ = αMφ . It also
32
q
1
depends on ratio of the mass splitting over the dark matter mass δ = 2∆M
M α
with α being the coupling constant for the potential.
The Sommerfeld enhancement for a two-state dark matter with a Yukawa
potential has been extensively studied by T. Slatyer [31]. The main features for
a simple 1-state Yukawa interaction are slightly altered as a result of adding a
first excited state: [31]
1. the enhancement in the non-resonant saturated region is increased by a
factor 2 (see left graph in Figure 3.6 where the upper solid-line, which
corresponds to a 2-state Sommerfeld enhancement, is twice as large as the
lower dashed-line which corresponds to a single state);
2. the resonance positions are slightly shifted to lower values of mφ if v < δ ,
and the heights are increased by a factor 4 (see the graph on the right in
Figure 3.6);
3. the Sommerfeld enhancement near threshold is a factor two larger than
its saturated value (see Figure 3.7).
Figure 3.6: Sommerfeld enhancement as a function of φ , calculated numerically with Mathematica. The solid-line corresponds to a 2-state (δ = 0.1) and the dashed-line to a 1-state.
The graph on the left corresponds to v = 0.1 and the graph on the right to v = 0.01.
3.5
Sommerfeld enhancement for MDM
The Sommerfeld enhancement for the annihilation χ0 χ0 → W + W − is calculated
with three coupled Schrödinger equations: the χ0 χ0 system can be excited to
χ+ χ− by the exchange of a W -boson, and the χ+ χ− system can be excited to
χ++ χ−− by another exchange of a W -boson. The Schrödinger equations for
systems of charged dark matter particles have extra terms because they couple
to photons (Coulomb potential), neutral Z-bosons (Yukawa potential) and have
a mass splitting.
We will introduce these extra terms seperately as to facilitate the explanations of the effect the features have on the Sommerfeld enhancement. First we
will discuss the third excited state, then we will say some things about the different coupling constant, consequently we will discuss the addition of an extra
Yukawa potential due to the Z-boson exchange. After that we will discuss what
happens when we incorporate a Coulomb potential in the sytem. Finally we will
33
Figure 3.7:
Sommerfeld enhancement as a function of v . The solid line corresponds to
the enhancement of a 2-state, the dashed-line to a 1-state with the same φ . Figure obtained
from [31].
apply the correct values for the different parameters and determine the Sommerfeld enhancement for the Minimal Dark Matter annihilation χ0 χ0 → W + W −
cross section.1
A third excited state
Adding a third excited state to the system:
−ψ100 −
−ψ200 −
e−φ r
ψ2
r
e−φ r
e−φ r
ψ1 −
ψ3 + 2δ ψ2
r
r
e−φ r
ψ2 + 42δ ψ3
−ψ300 −
r
= 2v ψ1
= 2v ψ2
= 2v ψ3
(3.9)
with E as the kinetic energy and boundary conditions
ψ1 (0) = ψ2 (0) = ψ3 (0) = 1
ψ10 (∞) = iM vψ1 (∞) → iv ψ1 (∞)
q
p
ψ20 (∞) = i 2M (E − 2∆M )ψ2 (∞) → i 2v − 2δ ψ2 (∞)
q
p
ψ30 (∞) = i 2M (E − 8∆M )ψ3 (∞) → i 2v − 42δ ψ3 (∞)
Note that we have performed the subsitution αM r → r in equation (3.9) and
in the boundary conditions as expressed on the right-hand side of the arrow.
2
The Sommerfeld enhancement is given by Sk = |ψ1 (r = ∞)| . Our choice
of boundary conditions gives ψ1 (r = 0) = 1. (Remember that equation (3.3)
2
i (r=∞)|
states that Sk = |ψ
.) Trying to calculate the Sommerfeld enhancement
|ψi (r=0)|2
for the ψ2 and ψ3 wavefunctions gives zero. Because as r → ∞ they behave as
1 For small values of and there are problems with numerical calculations: they become
v
φ
unstable.
34
p
normal plane waves with for example ψ2 ∝ exp i 2M (E − 2∆M ) . For low
velocities, hence low kinetic energy E, and sufficiently large mass splitting ∆M ,
the square-root is imaginary and the exponent becomes negative and real: ψ2
is a rapid decreasing function towards zero. The same holds for ψ3 . Hence both
ψ2 and ψ3 are zero as r → ∞.
As the radial distance r decreases, the potential energy of the 2-particle state
in the ground state ψ1 increases. Eventually, the state ψ1 has enough energy
to overcome the mass difference and become excited to the first excited state
ψ2 . The potential energy has to increase more in order to go the second excited
state ψ3 . The smaller the velocities of the particles, the smaller r has to become
to be excited to ψ2 or ψ3 . At r is almost zero, the 2-particle state can be in any
of all three states and all three states can produce the annihilation. As an order
of magnitude approximation, we assume all three systems have the same (bare)
annihilation cross sections hσ0 vi = 4.1 · 10−33 cm3 s−1 , see equation (2.22). The
small mass splittings are negligible in the calculation of σ0 .
Figure 3.8:
Sommerfeld enhancement as a function of φ (v = 1/100) above and as a
function of v (φ = 1/10) below (δ = 5/100), for a 2-state (dashed-line) and a 3-state (solidline).
The effect of adding a second excited state to a 2-state is illustrated in Figure
3.8. The resonant positions are shifted to higher values of φ . The reason for
35
the shift is that the W -boson interaction between the first and second excited
state increases the potential energy available in the first excited state ψ2 , this
potential energy is effectively decreasing the mass splitting between the ground
−φ r
and first excited state: Vef f = δ − α2 er ψ3 .
The enhancement saturates at the same value of v but the enhancement is
approximately doubled. This doubling is due to the shift of the resonances to
larger values of φ . Furthermore, there are two irregularities in the graph on
the right at v = δ and v = 2δ . These are the same irregularities as v = δ
in the 2-state case (see Figure 3.7).
Coupling coefficients
The interactions are proportional to some coupling
constant √
multiplied by a
√
coupling coefficient. For example: an A = g2 2 3 with g = 4πα2 coupling
gives a constant of proportionality of A2 /4π = 3α2 . Adding the proper coupling
coefficients for the Yukawa potentials gives
−ψ100 −
−ψ200 −
3e−φ r
ψ2
r
2e−φ r
3e−φ r
ψ1 −
ψ3 + 2δ ψ2
r
r
2e−φ r
ψ2 + 42δ ψ3
−ψ300 −
r
= 2v ψ1
= 2v ψ2
= 2v ψ3
Because the coupling coefficients 3 and 2 are larger than one, they effectively
increase the coupling strength. From Figure 3.9 we see the effects of the larger
coupling coefficients on the Sommerfeld enhancement. We observe that the
resonances are shifted to larger values of φ and they are more densely packed.
And we see that the Sommerfeld enhancement in the non-resonant region of v
is increased by a factor ∼ 2.
Additional Yukawa potential
Apart from the W -boson exchange, there is also a neutral Z-boson exchange
for the excited state (e.g. a 2-particle state consisting of (double) charged dark
matter particles). A Z-boson exchange between the two particles does not
induce a change of the state as a charged W -boson exchange does. This Zboson exchange leads to resonant suppresions of the χ0 χ0 annihilation as it
leads to sharp decreases in the Sommerfeld enhancement for specific values of
φ . Because the Z-boson exchange does not excite or de-excite the charged
particle state, it leads to suppression of the ground-state ψ1 which does not
have Z-boson exchange: the system remains in the excited state rather than
being de-excited to the ground state χ0 χ0 (see Figure 3.10).
However, for the annihilation process we are interested in, an annihilation of
χ+ χ− has the same effect as an annihilation of two χ0 ’s. If the Z-boson exchange
suppresses the occupation of the ground state, the number with which this
occupation decreases, should be equal to the number increase of the occupation
of the excited states. Neglecting the mass splitting, these excited states have
almost the same annihilation cross section for the annihilation into two W bosons. Thus instead of a χ0 χ0 → W + W − annihilation, one would have for
example χ+ χ− → W + W − which would have the same net effect. To conclude:
36
Figure 3.9: Sommerfeld enhancement as a function of φ (v = 1/100) on the top and as
a function of v (φ = 1/10) below (δ = 5/100), for a 3-state with the correct coupling
constants (solid purple-line) and a 3-state from Section 3.5 (dashed blue-line).
the suppression of the χ0 χ0 state in favor of the excited states does not affect
the annihilation process, and hence we can ignore these suppressions.
Including the potential of the Z-boson in the coupled system of equations
gives
−ψ100 −
−ψ200 −
3e−φ r
ψ2
r
3e−φ r
2e−φ r
cos2 θW e−ωφ r
ψ1 −
ψ3 −
ψ2 + 2δ ψ2
r
r
2r
2e−φ r
4 cos2 θW e−ωφ r
−ψ300 −
ψ2 −
ψ3 + 42δ ψ3
r
r
= 2v ψ1
= 2v ψ2
= 2v ψ3
with ω = MZ /MW the ratio of the Z and W -boson mass: 91.4/80.4, and
cos2 θW = 0.774.
Apart from the resonant suppressions, we conclude from Figure 3.11 that
adding the extra potential shift the positions of the resonances: they are shifted
to larger values of φ . And we infere that there is a slight change in the behavior
of the Sommerfeld enhancement for large values of v : it is slightly decreased.
37
Figure 3.10:
An example of a ladder diagram with a Z-boson exchange. With the Z-boson
exchange, the 2-particle state does not change.
Figure 3.11:
Sommerfeld enhancement as a function of φ (v = 1/100) above and as a
function of v (φ = 1/10) below (δ = 5/100), for a 3-state with the extra Yukawa potential
of the Z-boson (solid purple-line) and a 3-state with only the correct coupling constants from
Section 3.5 (dashed blue-line).
Coulomb potential
The addition of a Coulomb term leads to a large instability in the numerical
calculation in the region 0.05 < v < 0.50, as is shown in the Figure 3.12. The
figure shows the Sommerfeld enhancement for a 3-state with and without an
extra Coulomb potential in the excited states. Note that in the calculation
38
for this figure, we decreased the coupling constants by a factor of four because
otherwise the computation would be too unstable to perform. Performing the
same calculation with only slightly different parameters, drastically changes the
behavior in the same region of v . Ignoring the instabilities we notice that for
smaller velocities, both lines for the Sommerfeld enhancement (with and without
Coulomb potential) converge to the same value.
Figure 3.12 shows that the potential energy of the Yukawa interaction increases earlier (at larger r) than the Coulomb potential energy. This, and that
with Coulomb term we obtain the same Sommerfeld enhancement motivates
us to state that the Coulomb energy does not play a role in obtaining enough
potential energy to compensate for the mass splitting. Combining this with the
fact of the numerical instabilities of the computation with the Coulomb term,
we choose to neglect the Coulomb term in the calculation of the Sommerfeld
enhancement in the Minimal Dark Matter annihilation.
Figure 3.12:
Left diagram: potential energies of W -boson Yukawa (large dashed), Z-boson
Yukawa (solid), Coulomb (small dashed) and the first mass splitting (horizontal solid) potential terms as a function of distance r. Right diagram: Sommerfeld enhancement for a 3-state
as a function of v with and without Coulomb term in the excited states (respectively dashed
and solid line)(φ = 0.24, δ = 0.17).
3.5.1
The Sommerfeld enhancement factor
W
Filling in the accurate values for the parameters (φ = αm
= 24/100, δ =
2 Mχ
q
2∆M
= 17/100) gives the Sommerfeld enhancement as a function of v which
α2 M
2
saturates at a certain value. The relevant velocities of dark matter at the time
of recombination is wel below v/c = 10−4 and hence we can use the saturated
value of the Sommerfeld enhancement.
From Figure 3.13 we see that as the velocity decreases, the Sommerfeld
enhancement saturates at around
Sk,MDM ≈ 1 · 103
In the same figure we can also see that the parameter φ = 0.24 (at the position
of the vertical solid gold-line) correspond to a point between a minimum and a
maximum of a resonance. A slightly different value of φ would correspond to a
large change in the Sommerfeld enhancement. The positions of the resonances
is an interplay of all the different features of the Minimal Dark Matter model:
39
Figure 3.13: Sommerfeld enhancement as a function of φ and v for the MDM case, without
the Coulomb potential.
the extra third state compared to a 2-state, the correct coupling coefficients,
the extra Yukawa potential of the Z-boson and finally the correct value of the
parameter δ . And the Sommerfeld enhancement factor in the non-resonant
regions is larger compared to a generic 2-state Yukawa interaction, due to the
larger coupling coefficients.
Now we have obtained the total cross section. Combining the bare cross
section σ0 calculated in Section 2.3.4 with the Sommerfeld enhancement given
above, gives us the total cross section
hσenhc vi = Sk hσvi = 4.1 · 10−30 cm3 s−1
40
(3.10)
Chapter 4
W +W − Decay
The W -bosons created in the annihilation process χ0 χ0 → W W are unstable:
they have a lifetime of about 10−24 seconds. With the cross section obtained
in the previous sections, we are able calculate the number of dark matter annihilations, and hence the number of W -bosons produced. But considering the
short lifetime of the W -boson compared, they have decayed before they have
any significant amount of interactions with the electron-proton-photon plasma
in the universe. Therefore we will discuss the decay of the W -boson in this
section.
First we will discuss how we how we analyze the decay process of the W boson. We will use a computer simulation program Pythia to calculate the
products of the W -decay. And then we will present our results of the analysis
of the decay, for example which particles are produced and how many energy
they have.
4.1
Branching ratios
The branching ratios of the W -boson decay are (first column: [41], second column: [42]):
ΓW + →e+ ν
ΓW + →µ+ ν
ΓW + →τ + ν
ΓW + →hadrons
= 10.80 ± 0.09%
= 10.75 ± 0.13%
= 10.57 ± 0.15%
= 67.60 ± 0.27%
ΓW + →ud¯
ΓW + →us̄
ΓW + →cs̄
ΓW + →cd¯
ΓW + →ub̄
ΓW + →cb̄
=
=
=
=
=
=
32.04065%
1.74396%
31.96649%
1.74162%
0.00053%
0.05985%
as fractions of the total width Γ.
A large fraction of the W -bosons will decay into high energetic quarks. Since
quarks couple to the strong interaction they are color confined. This leads
to hadronization. This is the production of hadrons out of the quark gluon
plasma that confined the two initial quarks. This hadronization event cannot
41
be calculated analytically; a simulation program is needed. Here we will make
use of Pythia.
4.2
Pythia event generator
Pythia [42,43] is a Monte-Carlo based event generator to simulate high-energy
events like electron-positron collisions. It contains a lot of physics (possible
interactions, decays, fragmentation, initial- and final state radiation, etc.) and
uses Monte Carlo techniques to choose the right variables to simulate the amplitude character of the involved particles. We have to perform the same simulation
numerous times in order to get the right averages and distributions of the resulting quantities, such that these results correspond to the result one would have
obtained analytically. Pythia originates from Jetset which is a simulation
program for hadronization processes using the Lund stringmodel.
To simulate the whole process of a high-energy collision of two particle
beams, Pythia decomposes the entire event into small sub-events and solves
these sub-events one by one. For example, given some user-specified beam of
particles that collides, we first have the creation of a initial-state shower from a
particle in each beam (e → eγ). After that, one particle from each beam enters
a specific hard-process (the process the user is interested in) which produces
outgoing particles (for example e+ e− → W + W − ). Some particles produced in
the hard-process are unstable and decay (W → q q̄). Unstable outgoing particles can decay further. At the same time, Pythia calculates how a quark-gluon
plasma fragmentates and hadronizes. The particles formed in the hadronization
process can decay into (eventually) stable particles. The final particles produced
are the stable particles with possibly some final state radiation. [42]
The output generated by Pythia is a full description of the evolution of the
particles involved in the interaction. It tells which particles are involved, what
their properties are (mass, momentum and energy) and with which particles
they interact to form other particles.
4.2.1
String fragmentation and hadronization
A system consisting of a quark (e.g. down quark d) and an antiquark (anti-down
¯ with color (red r) and anti-color (anti-red r̄) respectively, behaves
quark d)
differently than a dipole of a positive and a negative electric charged particle.
The fieldlines of the electric dipole are spread out in space, while the field lines
of the quark ‘dipole’ are directed along a tube. The potential for the quark
antiquark pair is linear with their separation. As the separation between the
quarks increases, the (attractive) force increases and the field energy increases.
When the field energy reaches a certain threshold, a new quark antiquark pair
(for example a uū with colors rr̄) is formed inside the tube and the tube splits.
As a result there are now two color-neutral quark anti-quark pairs (dū and ūd)
connected by a tube. [28]
In the Lund stringmodel, the tube is seen as a one-dimensional string, and
the above process is therefore called string fragmentation. When the fragmentation process is finished (there is not enough energy for further fragmentation),
the quarks and anti-quark pairs combine to form mesons (quark-antiquark pair)
or baryons (three quarks): hadronization.
42
Figure 4.1: The decay of a W − boson into two quarks. These quarks exchange gluons (of
which one creates another quark antiquark pair) which will enter the string at moment of
hadronization. Some of the gluons emitted were left out for simplification.
Baryons can be formed if the quark-antiquark pair (q2 q̄2 ) formed inside the
tube in the example above would not have had the color red and anti-red, but
for example green and anti-green. These quarks do not gain energy from the
red anti-red field, but are able to move to the other red quarks, producing dq2
¯ 2 . This would create an anti-blue (r + g = b̄) and blue field. A second
and dq̄
string fragmentation could form a q3 q̄3 quark pair with color bb̄ that combines
¯ 2 q3 . For multiparton-states (e.g. multiple quarks
with the other quark pairs: dq
and gluons) the process of string fragmentation and subsequent hadronization
is more complicated. [42]
For an example of a hadronization process, see Figure 4.1. It shows how
a W -boson decays into a strange and charm quark. Both quarks exchange
¯
multiple gluons (one of which creates a dd-pair)
as they seperate: more energy
is stored in the string between the strange and charm quark. Eventually, the
strings hadronize into a set of mesons and some baryons.
4.3
The simulation
To study the decay of the W -boson, we first have to produce them. Since
Pythia does not know the Minimal Dark Matter model, we simulate the production of the W -bosons in the dark matter annihilation by electron-positron
annihilation. An electron beam and a positron beam are chosen as the beams
that collide, and we choose the hard-process ffbar→WW with a center-of-mass
43
energy 19200 GeV, the (rest) energy of two neutral Minimal Dark Matter particles χ0 .
Initial state radiation
Usually, the beam particles like electrons or protons are able to have initial state
photon or gluon radiation before the particles collide. As a result, part of the
initial energy (and momentum) is radiated away. However, since the neutral
dark matter does not couple to photons or gluons, it is not able to have initial
state radiation. Therefore, to simulate the dark matter annihilation through
colliding an electron with a positron, this initial state radiation is not allowed
and switched off using the PDF:lepton switch.
Maximum W -boson mass
In the calculation of the hard process, Pythia uses a finite number of diagrams.
Only the lowest orders are used such that it gives correct behavior close to the
on-shell energy of the W -boson. In this specific case of ee → W W there is
a delicate cancellation of the contributions of the s (Z, γ) and t-channel (νe )
diagrams (see Figure 4.2). However, the center-of-mass energy used is much
larger than the resonance energy of 160 GeV. Resumming the finite number
of diagrams at this energy gives an imperfect cancellation between the s and
t-channel [42].
Figure 4.2:
The two s-channel and one t-channel Feynman diagrams corresponding to the
production of two W -bosons from lepton annihilation.
The imperfect cancellation leads to ‘strange’ behavior at high energies. The
mass-distribution of the W -boson has in addition to the expected Breit-Wigner
peak an additional lower but much broader second peak around the total energy
of center-of-mass frame. The consequence of these heavy W -bosons is that the
number of particles produced in the W -boson decay into a quark-antiquark pair
has increased by a factor of two to four. In order to prevent the production of
heavy W -bosons, a limit is set on the maximum mass of the W -boson produced.
This mass limit does not have any effects on the branching ratios.
Semi-stable particles
The hadronization process produces semi-stable particles. For the purpose of
simulating the signatures of a process observed in a collider like the LHC, these
particles are considered stable. In Pythia their lifetime is long enough to let
them decay outside all detectors (lifetime between brackets are in seconds):
µ+/− (2.2 · 10−6 ), π +/− (2.6 · 10−8 ), K+/− (1.2 · 10−8 ), K0L (5.1 · 10−8 ) [42].
44
However, on the timescale of the recombination (years) the only stable particles are protons and electrons (and perhaps neutrons, provided it combines
sufficiently fast with a proton) and their antiparticles. Therefore, these semistable particles should decay in our simulation.
4.4
Results
W → lνl decay
This decay gives leptons with an energy of the order TeV. Possible final state
radiation l → γl produces energetic photons and can reduce the energy of the
lepton dramatically. In case of a W -boson decay into a muon or a tau, the lepton
produced will subsequently decay into an electron, an electron antineutrino and
a muon or tau neutrino. This type of W -boson decay produces a spectrum of
higly energetic particles.
W → q q̄ decay
The quarks produced in this way emit numerous gluons which can also emit
gluons themselves. The energies of the gluons vary from order TeV to GeV.
Energetic gluons are able to produce new quark antiquark pairs. The sum of
these quarks and gluons will hadronize to form mesons and baryons.
The mesons and unstable baryons decay eventually into leptons and photons.
Most of the neutral mesons decay (via neutral pions) into photons, and most
of the charged mesons will decay via charged pions into muons and muon antineutrinos [41].
The energy spectrum of the resulting leptons and photons is much softer
than from the W → lνl decay. The energy is divided among a larger number of
particles. The typical energies are in the range of up to 500 Gev.
Particle multiplicities
Figure 4.3 shows the average number of particle-species produced in one annihilation process. The neutrinos are by far the most abundant1 : every production
of an electron involves the creation of one or more neutrinos. Also note the
ratio muon neutrinos over electron neutrinos is almost 2: almost all electrons
and positrons come from the decay of a muon into an electron. The production
of the muon produces one muon (anti-)neutrino, and the decay of the muon into
an electron produces another muon neutrino and an electron neutrino.
Energy distribution
Figure 4.3 also shows how the energy is distributed among the different particles.
51.6% of the energy is put into the neutrinos. Because the neutrinos are very
weakly interacting particles, they do not transfer their energy to photons or
electrons and hence have no influence on the recombination process.
21.1% of the energy released during the W -boson decay is contained in photons. Figure 4.4 shows a histogram of the photon energy. The bars have a
width of around the 3 GeV. As the photon energy increases, the number of
1 Note
that we use the term ‘neutrino’ for both neutrino and antineutrino.
45
Figure 4.3:
The top graph contains the average number of particles produced in one annihilation process. The bottom graph shows the total amount of energy released during the
annihilation distributed among the particle types.
photons with that energy decreases. The high-energetic photons produced in
the W → lνl decay ‘drown’ in the large amount of photons produced in the
W → q q̄ decay. The same can be said about the electrons (and positrons) energy histogram in Figure 4.4. The electrons and positrons both contain around
10% of the energy of the interaction.
46
Figure 4.4:
Two histograms of the photon and electron energies released in one annihilation
process. They are the averages taken over 1,000,000 annihilation interactions.
47
Chapter 5
Recombination
Before we will derive the equations that describe the recombination process,
there will be a small introduction to cosmology in general that will introduce
some concepts used in the derivation of the equations for the recombination. After that, we will discuss the recombination process without dark matter annihilation. Subsequently, we will talk about how the energy freed in the annihilation
can influence the recombination process and then calculate the recombination
process including dark matter annihilation.
5.1
Introduction to cosmology
The early universe
The ΛCDM Cosmology Model describes the universe that has its origin in a ‘Big
Bang’ and expands ever after. As the universe expanded, it underwent several
phase transitions.
Shortly after the Big Bang, the universe was very hot and dense. It consisted
of quarks, electrons, photons, neutrinos (and dark matter and dark energy). All
Standard Model particles were strongly interacting. One second after the Big
Bang, the universe had expanded and cooled enough (the photons are redshifted
to longer wavelength due to the expansion), such that the quarks were able to
form protons and neutrons in a period that is called baryogenesis. Some 400
seconds later the photons decreased that much in energy that they could not
prevent or destroy the nuclei being formed by the protons and neutrons. As a
result the light elements were produced (nucleosynthesis). The abundances of
the elements formed was 75% hydrogen, 24% helium and some smaller traces of
lithium and berillium [6].
As the universe expanded for another 400,000 years, the photon energy
dropped below the average ionization energy of hydrogen atoms, and the electrons and protons could succesfully combine to atoms without being immediately ionized by a photon. This period of combining protons and electrons is
called recombination. Heavier elements like helium and lithium had already
combined with electrons to form atoms before recombination.
At the moment of recombination, the microwave background radiation is
formed. Before recombination, the photons were constantly scattered by the
free electrons (and protons) and they had a very short mean free path. However,
48
after recombination, there are very few free electrons and the photons became
able to move freely in a straight line: the mean free path had increased from
very short to infinity. These free travelling photons form the cosmic microwave
background and are still observed today by satellites and other detectors.
Scale factor and redshift
On large scales, the universe is homogeneous and isotropic. Furthermore, recent observations indicate that it is expanding [44]. The solution to the Einstein equations of general relativity which satisfies these three conditions is the
Friedmann-Robertson-Walker-Lemaı̂tre metric
ds2 = −c2 dt2 + a(t)2 dr2 + Sk (r)2 dθ2 + sin2 θdφ2
with comoving coordinates r, θ and φ, cosmic time t (t = 0 at the moment of
Big Bang), the scale factor a(t) that describes how the distances grow with time
(normalized at the present (t = t0 ): a(t0 ) = 1). Sk (r) depends on the geometry
of the universe, for the flat universe that we are probably living in Sk (r) = r.
The expansion rate of the universe can be expressed in terms of the Hubble
constant H(t)
ȧ
H=
a
Note that the scale factor is not necessarily a function linear in time.
For nearby objects we can calculate the speed at which objects recede from
us, due to the expansion of the universe, using a linear relation between the
distance d to the object and its receding velocity v: v = Hd. A galaxy close to
us recedes less fast than a galaxy far away. This tells us that the light emitted
by the galaxy far away is more redshifted than the light coming from a nearby
galaxy. If we observe two galaxies at a different (radial) distance at the same
time, the light of the galaxies farthest away is emitted earlier in time than the
light from the nearby galaxy; hence that light was emitted when the universe
had a smaller scale factor a(t). The relation between redshift z and the scale
factor a(t) is
a(t0 )
1
1+z =
=
a(te )
a(te )
with a(te ) the scale factor at the moment of emission and a(t0 ) at the moment
of observation. An increasing z means looking at a smaller a(te ), thus further
back in time, thus at a larger distance (as it takes time for light to travel a
certain distance).
5.2
Standard Recombination Process
The early universe contained so many energetic photons that every time a free
electron combined with a proton to form hydrogen, the atom was quickly ionized
again through the absorption of a photon: they were in thermal equilibrium
e+p *
) H +γ. And as long they are in this equilibrium, they obey the MaxwellBoltzman distribution. This distribution tells us that the number density of the
1s-state (ground state of hydrogen) is, at the start of the recombination process,
larger by a factor of exp(−∆E/kB T ) ∼ 10+13 (at T =4,000 K) compared to the
49
number densities of the excited states. Thus the amount of hydrogen atoms is
almost only determined by the amount of 1s-state atoms.1
As the universe expands, the radiation cools: the photons are not only diluted but also redshifted as the space expands. The temperature is inversely
proportional to the expansion T ∝ a1 . At a temperature lower than 50,000 K,
the mean energy of the blackbody spectrum of the photons is below the 13.6
eV. Because the baryon to photon ratio is 0.22 to 3.7 · 108 there are still a lot of
photons in the tail of the spectrum that have energy above the 13.6 eV and can
easily ionize the hydrogen atoms. The universe has to cool to a temperature
around the 3.500 K for the hydrogen atoms to form.
To form hydrogen in the ground state, one only has to consider the radiative
transitions 2s → 1s and 2p → 1s. The first transitions emits two photons, both
photons do not have enough energy to excite another hydrogen atom from the
ground state and are thus absorbed in the CMB spectrum. Although this is a
forbidden transition, it occurs sufficiently rapid because the other transition is
so inefficient. The photon emitted in the 2p → 1s transition has just enough
energy to excite another 1s hydrogen atom. If the interval between emission
and absorption is too long (the mean free path of the photon is larger than the
Hubble distance c/H), the photon is redshifted too much due to the expansion
of the universe. As a result, the photon does not have enough energy to excite
a hydrogen atom from the ground state.
Recombinations or radiative cascades from an excited state n ≥ 3 directly
into the ground state emit photons with a large energy and resonant cross
section. The resonant cross section enables the photon to find and excite or
even ionize another hydrogen atom from the ground state. The absorption of
the emitted photon cancels the effect of the recombination or cascade into the
1s-state: no net result in the amount of hydrogen atoms in the ground state. So
we are only interested in recombinations or cascades into the n = 2 level that
subsequenlty falls into the 1s-state.
For the relevant temperature range we can consider the excited states to
be in a thermal equilibrium: collisions between atoms and the radiative decays
between the excited states occur very rapidly compared to the Hubble expansion. At the start of the recombination (T ∼ 4, 000 K) there are 105 photons
with more energy than the energy spread of the excited levels of 3.4 eV for each
hydrogen. At the end of recombination (T ∼ 2, 000 K), there are 20 times as
much photons.
The thermal equilibrium enables us to express the number density of one
excited state in terms of an other excited state using the Maxwell-Boltzmann
relation
nnl = (2l + 1) n2s exp (− (B2 − Bn ) /kB T )
(5.1)
with Bn being the binding energy of the n-th state.
The 1s-state is not in thermal equilibrium and is only reached through the
slow and inefficient processes described above.
As said, the net change in the amount of hydrogen atoms in the 1s-state
is given by the transition from the 2p and 2s-state into the 1s-state minus the
number of ionizations and excitations from the 1s-state to a n = 2 state. (All
other processes in which the ground state is reached are cancelled by either
1 The treatment of the recombination presented in this section largely follows the derivation
presented in the book ‘Cosmology’ written by S. Weinberg [19].
50
ionization or excitation by the photon emitted.) The transition rates 2p → 1s
and 2s → 1s are 4.699 · 108 s−1 and 8.225 s−1 respectively [45].
The recombination happens through collisions between free electrons and
protons and is proportional to the number densities of the free electrons ne and
protons np in a comoving volume a3 . Every succesful recombination decreases
the free electron fraction ne a3 in a comoving volume. The rate of recombinations
is given by α(T )np ne a3 with α(T ) as the effective recombination rate of all
recombinations into an excited hydrogen state.
The rate of ionization increases the number density of free electrons. Just as
for the recombination, we only consider ionizations of the excited states. The
ionization rate is proportional with the population of the excited states. The
effective recombination coefficient only depends on the temperature and can be
defined as
"
#
X
X
n2s β(T ) ≡
βn (T )nnl = n2s
βn (T ) (2l + 1) exp (− (B2 − Bn ) /kB T )
n>1
n>1
with βi as the ionization rate for excited state i. The ionization rate is given by
β(T )n2s a3 .
Thus the amount of hydrogen formed is given by the decrease in the number
density of the free protons
d
np a3 = −α(T )np ne a3 + β(T )n2s a3
dt
(5.2)
As said, the effective recombination and ionization coefficients only
depend
d
on the temperature T . In case of a thermal equilibrium ( dt
np a3 = 0) they
are related via the density of states
β
=
α
ne np
n2s
=
eq
me kB T
2πh̄2
3/2
exp(−B2 /kB T )
(5.3)
However, if there is a difference between the recombination rate and the
d
np a3 6= 0, this difference is equal to the change in number
ionization rate, dt
of hydrogen in the 1s-state. The net difference is given by the transition rates
Λ2s and Λ2p of the radiative decays from the 2s- and 2p-states into the ground
state and the excitation rate from the ground state to the 2s- and 2p-state.
However, the radiative decay of the 2p- into the 1s-state has only a net effect
if the emitted photon is unable to quickly excite another hydrogen atom from
the ground state, in other words: if the photon is redshifted enough. This is
incorporated into the P factor: the probability the photon will escape to infinity.
(Below we will discuss this P factor more extensively.) The net change in free
protons is given by
α(T )ne np − β(T )n2s = (Λ2s + 3P Λ2p ) n2s − n1s
(5.4)
The change in number of protons is so slow that we can take the average rates
for the transitions. The factor 3 is included because of the threefold degeneracy
of the p-states, and because the energy difference between the 2s and 2p-state
is negligible small we have: n2p = 3n2s .
51
Using equation (5.1) and considering that the temperature T is much smaller
than (B2 − B3 )/kB = 21, 900 K one can assume that all states nnl with n > 2
are much less occupied than n2s . Making it reasonable to state
nH = n1s + n2s + n2p = n1s + 4n2s
→
n1s = nH − 4n2s
(5.5)
This approximation enables us to express n1s in terms of n2s and nH . Rewriting
equation (5.4) gives:
αne np + nH
n2s =
(5.6)
Λ2s + 3P Λ2p + β + 4
In case of a thermal equilibrium, we can consider the number density of
hydrogen atoms in the 1s-state to be constant. The left-hand side of equation
(5.4) would become zero and in combination with equation (5.1) this would give:
n2s
=
= exp (−(B1 − B2 )/kB T ) ≡ D
(5.7)
Λ2s + 3P Λ2p
n1s eq
Using the expression for the number density n2s we rewrite the differential
equation of the proton number density (5.2)
d np =
dt n
Λ2s + 3P Λ2p
[Λ2s + 3P Λ2p ] [1 + 4D] + β
αne np
nH β
× −
(1 + 4D) +
n
n (Λ2s + 3P Λ2p )
(5.8)
with n as the total number of protons: free and hydrogen nuclei n = np + nH .
Because of the low temperature during recombination, the factor 1 + 4D can
be considered one: the exponent −(B1 − B2 )/kB T is decreasing from around
−10 as the temperature drops from 4,000 K to 1,000 K (B1 − B2 ∼ 3, 4 eV).
Thus:
Λ2s + 3P Λ2p
d np =
dt n
(Λ2s + 3P Λ2p ) + β
βnH exp (−(B1 − B2 )/kB T )
αne np
+
(5.9)
× −
n
n
where the first factor is the suppression of the recombination if the radiative
decay rates are much smaller than the ionization of the n = 2 state.
Photon survival probability P
All that is left to do is to find an expression for the photon survival probability
P . It gives the probability that the photon emitted in the 2p → 1s is not able to
excite another hydrogen atom from the ground state. For a Lyman-α photon P
is between 0 and 1, for any other transition photon P can be considered equal
to 1: these photons escape to infinity. P depends on the energy of the photon
emitted during the transition 2p → 1s, the number density of hydrogen atoms
in the ground state, the cross section of the excitation 1s → 2p and the amount
of redshift of the photon (i.e. the expansion rate given by the Hubble constant
H). (For detailed derivation see [19] or [46].)
52
The photon survival probability is given by:
P =
8πH
3λ3α Λ2p n1s
(5.10)
, with λα being the wavelength of a Lyman-α
The quantity 3P Λ2p equals λ8πH
3n
α 1s
photon. We define a new (often used) parameter K:
K≡
λ3α
8πH
Fudge Factor F
As pointed out by Seager et al. [45], not all of the assumptions (e.g. the thermal
equilibrium between the states, a simple recombination coefficient, collisional
processes are negligible, as is stimulated de-excitation of Lyman-α lines, ignoring
helium recombination) made above are that well justified. They made a full
calculation of the recombination of H, He I and He II using multilevel atoms
consisting of 300 energy levels (instead of a three level atom: n = 1, 2, +)
and calculated the recombination to and photoionization from each level and
also included all bound-bound transitions. They did not assume a thermal
equilibrium and a simple recombination coefficient but instead calculated it
level-by-level of the 300 level atom. Seager et al. also made a more careful
treatment of the matter temperature, which we will also use.
The result of the more exact calculation was that the recombination went
faster, a smaller ionization fraction at lower redshift and a delay in the He I recombination. Collisions were of negligible influence. They found that one could
multiply the recombination coefficient α in the effective three-level calculation
with a so-called fudge factor F = 1.14 to speed up the recombination and reproduce the result of the 300 level atom calculation. [45, 46]
The final result is a differential equation for the fraction of free protons:
dxp
dz
=
×
α(T )
[1 + KH Λ2s n(1 − xp )]
[(1 + z) H(z)]
αH xe xp n − βH (1 − xp )e−(B1,H −B2,H )/kB TM
[1 + KH (Λ2s,H + βH )n(1 − xp )]
= F
atb
· 10−19
1 + ctd
m3 s−1
(5.11)
(5.12)
with xp = np /n, n1s = n − np = n(1 − xp ); β is given by equation (5.3); the
values for the parameters a, b, c, d in the expression for α(T ) are respectively:
4.309, -0.6166, 0.6703, and 0.5300; and t = T /104 [47]. And we made use of the
d
d dz
d
relation dt
= dz
dt = dz (−(1 + z)H(z)) with H as the Hubble constant. Note
that since the universe is on large scale electrically neutral, we can replace xp
by the free electron fraction xe .
Solving this differential equation will tell us how the recombination of electrons and protons occurred. It will tell us how the free proton fraction evolves
with time (redshift).
53
Temperature
Before decoupling, matter and radiation are coupled and have the same temperature. Via Coulomb scattering they exchange energy: when a high frequency
photon scatters onto a stationary electron, it transfers energy onto the electron (and vice versa). However, as the universe expands, matter and radiation
cool adiabaticaly. Radiation cools as T ∝ (1 + z)−1 while matter cools with
T ∝ (1 + z)−2 , thus after decoupling, the temperature of matter is different
from the radiation temperature, and this difference affects the end of recombination. As a consequence we should distinguish between these two temperatures
as some processes depend on the radiation temperature (photoionization and
stimulated recombination) and other on the matter temperature (spontaneous
recombination).
The equation for the matter temperature is [46]
8σT aR TR4 ne
dTM
=
(TR − TM ) − 2TM
dt
3me cntot
(5.13)
where the first term is due to the Coulomb cooling (σT Thomson scattering cross
section, aR TR4 radiation field density, ntot total matter number density (which
is a sum of free protons, free electrons, hydrogen atoms, helium atoms, helium
ions)): it is the energy transfer between matter and radiation. The second term
is due to the adiabatic expansion of the universe.
Other cooling processes which were taken into account in the calculation
made by Seager et al. [46] like bremsstrahlung, photoionization heat, photorecombination cooling, line cooling, collisional ionization and recombination heating were found not to be relevant according to their results.
5.2.1
Recombination with DM annihilation
The dark matter annihilations enable the energy stored in the dark matter to
influence the recombination process. The annihilation products in the form
of photons, electrons and positrons affect the recombination process in three
ways: extra ionizations, excitations and heating. The neutrinos produced in
the annihilation are very weakly interacting and not able to convey their energy
to other particles: their energy is lost. Neutrons and protons are also poor at
transfering their energies to the intergalactic medium (IGM) which consists of
electrons, protons and photons. [48]
We will use the so-called ‘on-the-spot approximation’: the energy of the
annihilation is instantaneously deposited in the IGM with some efficiency factor
f . We will discuss this approximation below in this section. But first we will
discuss the possible ways in which energy from the dark matter annihilation
influences the IGM.
Energy Transfer
First we will give a short discussion of how the electrons, positrons and photons
are able to deposit their energy in the IGM.
Electrons and positrons Low energetic electrons have Coulomb collisions
with thermal electrons that cause extra heating, or they have collisions with
54
atoms that excite or ionize the atoms. How the electron energy is distributed
among heating, excitation and ionization depends on the fraction of free electrons [49]. A crude approximation for the fraction used in previous articles
[15, 16, 18] is that the fraction (1 + 2xe )/3 of the electron energy is used for
heating and (1 − xe )/3 for both excitation and ionization. In the case of a large
free electron fraction, most of the energy goes into the heating. The smaller the
free electron fraction, the more atoms that can be excited or ionized and hence
more energy goes into these processes.
Relativistic electrons have inverse Compton scattering with CMB photons
in which they upscatter the CMB photon to higher energies and thus effectively
transfer the electron energy spectrum into a photon energy spectrum. The
electrons need several scattering events (producing many low energy photons)
before they cool down and deposit their remaining energy via heating, excitation
or ionization.
Positrons behave identically to electrons, except for low energies. Thus relativistic positrons have inverse Compton scattering. But as they cool down, they
form positronium with a free electron. Eventually, the positron and the electron
will annihilate, producing an energy spectrum which is continuous down to an
energy of 511 keV.
All electron and positron processes happen fast compared to the Hubble
time, so they can be regarded as depositing the energy locally and no redshifting
needs to be taken into account.
Photons The relevant processes for the photons to deposit their energy are
discussed in [50]. They are (in order of increasing energy) photoionization,
Compton scattering, pair-production off nuclei and atoms, photon-photon scattering, pair-production off CMB photons.
Photons with an energy below the 1 keV (UV and soft X-ray photons) have
a large photoionization cross section and are quickly absorbed locally [48]. Photons with energy around the 108 − 1010 eV mainly pair-produce off nuclei and
atoms. But this is a slo process for the photons to deposit their energy. High
energy photons with an energy of the TeV order scatter with CMB photons or
produce pairs, producing softer X-ray photons which have a larger cross section
for absorption, or relativistic electron-positron pairs. This process also occurs
sufficiently rapid to be local. [17]
However, in the energy range of 103 − 105 and 108 − 1010 eV at redshift z ∼
1000 (the redshift at which recombination occurs) is a so-called ‘transparency
window’ [17, 48] (see Figure 5.1). For photons within this energy range the
universe almost looks transparent: it takes some time for them before they
interact and deposit (the cooling time) their energy into the IGM. They deposit
their energy on a timescale comparable the Hubble time. The Hubble time is
the inverse of the Hubble constant and is an order of magnitude estimate for
the age of the universe.
A cooling time that is of the same order of magnitude as the Hubble time
means that a part of the photon energy is lost due to redshifting, and that said
photon does not deposit its energy fast enough to affect the recombination. Only
a small part of the photon energy is transferred into the IGM by for example
Compton scattering. At energies around 106 eV the Compton scattering is able
to effectively deposit part of the energy into the IGM, but at lower energies of
55
Figure 5.1:
A photon cooling time (including all processes) compared to the Hubble time
at the redshift around recombination z ∼ 1000. (Figure obtained from [17].)
103 to 105 eV these Compton scatterings become elastic. Due to the redshifting,
the photon energy eventually decreases to a value lower than 103 eV, when it
becomes able to deposit its energy via photoionization.
At high redshift, the cooling time of the photons is smaller than the cooling
time at low redshift. This is a result of for example higher CMB photon and
nuclei densities. And hence the effect of redshifting is smaller at high redshift.
Thus photons with energies outside the transparency window are able to
deposit their energy fast and locally. Photons inside the transparancy window lose large part of their energy to redshifting until they redshift out of the
transparency window.
On-the-spot approximation
As stated above, we will incorporate the dark matter annihilation via the onthe-spot approximation (as published in [15,16,18]). This method assumes that
the relevant part of the annihilation energy that is transferred onto the IGM, is
transferred locally and instantaneously.
Looking at the photon energy spectrum (Figure 4.4) we conclude that a
substantial part (60% of the photons, 30% of the photon energy) is in the region
of the Compton scattering, between the two transparent domains. Should they,
via Compton scattering, enter the transparency window, they will only redshift
slightly before they reach the 103 eV exit and deposit their energy locally via
photoionization. However, a large part of the photon energy is in the highenergy part of the transparency window and will have trouble to deposit their
energy fast compared to the Hubble time.
56
The same reasoning can be applied to the electrons and positrons, which are
almost all relativistic and need inverse Compton scattering to cool down and
deposit their energy. This creates a photon energy spectrum which, although
it has lower energies (multiple scatterings are needed for one electron to cool
down), will have a substantial part in the transparency window.
Although the on-the-spot approximation can be considered to be crude, it
will give us a qualitative estimate of the effect of Minimal Dark Matter annihilation on the recombination process. We will assume all photons, electrons and
positrons deposit their energy into the IGM locally and instantaneously.
The dark matter annihilation influences the standard recombination in three
ways:
1. extra excitations;
2. extra ionizations;
3. heating.
First we will discuss the excitations and ionizations. The excitations and ionizations free electrons from the atoms and thus change the ionization fraction
xp . Hence they are incorporated via an extra term IX into equation (5.11):
1
dxp
=
(Rs (z) − Is (z) − IX (z))
dz
[(1 + z) H(z)]
(5.14)
with Rs (z) and Is (z) being the standard recombination and ionization terms:
they are given by the right-hand-side in equation (5.11).
IX (z) is split into two terms:
IX (z) = IXi (z) + IXe
an ionization (IXi ) and an excitation part (IXe ).
The ionization term corresponds to collisions of electrons from the annihilation with atoms and from possible photoionization from the photons produced
in the annihilation. These photoionizations from the ground state have a net
effect of reducing the number of atoms because the photon is produced in the
W -decay and then absorbed. Using the energy partition between heating, ionizations and excitations as proposed in [49]:
IXi (z) =
1 − xe dE/dt
3
nEi
with Ei = 13.6 eV the ionization energy of hydrogen, and dE
dt the rate at
which energy from the annihilation is deposited into the IGM (note dE
dz =
1
dE 2
).
(1+z)H(z) dt
The excitation term corresponds to excitations from the ground state to the
first excited state n = 2. This does not necessarily lead to the ionization. If the
2 In
[15] S. Galli et al. give an almost similar expression for the extra ionization-term IXi ,
Λ2s +3P Λ2p
the only difference is that they include an extra factor C =
. However, the
(Λ2s +3P Λ2p )+β
dependence of this term on the radiative decay and ionization rates only seems logical if the
ground state would be barely populated and all the atoms would be in a n = 2 state, but that
is not the case and this factor only decreases IXi since C ≤ 1.
57
ionization rate β(T ) for the excited level is large compared to the radiative decay
rates Λ2s and Λ2p , this excitation will lead to an ionization. The excitation term
IXe is given by
1 − xe
β
dE/dt
IXe =
3 Λ2s + 3P Λ2p + β nEα
with Eα = 10.4 eV, which is the energy corresponding to a Lyman-α photon.
Besides ionization and excitation, we have heating of the IGM due to the
dark matter annihilations. A fraction of the annihilation energy is put into
heating of the IGM, effectively adding a term to the equation for the temperature
(using E = 32 kB T ):
dTM
1
8σT aR TR4 ne
=
(TR − TM ) − 2TM
dz
(1 + z)H(z)
3me cntot
2
(1 + 2xe )(dE/dt)
−
(5.15)
3kB ntot
3n
Rate of energy release
To calculate dE
dt we first have to find an expression for the total annihilation
rate Γ. The rate of annihilation of one dark matter particle Γ0 is proportional
to the chance of finding another dark matter particle multiplied by the chance
of an interaction: Γ0 = nDM Sk hσvi including the Sommerfeld enhancement Sk .
Then the total rate of annihilation is:
Γ = nDM Γ0 = n2DM Sk hσvi
(5.16)
The particle number density nDM is a function of redshift z (or expansion
paramater a = 1/(1 + z)). The particle number in a co-moving volume a3 is
constant: nDM (z) = nDM,0 (1 + z)3 with nDM,0 being the current dark matter
number density. Given the relative dark matter energy density ΩDM = 0.222 [1]
and using the critical density ρc,0 = 5200 MeV m−3 [6]:
nDM,0 =
ρDM,0
ΩDM ρc,o
=
= 1.2 · 10−4 m−3
MDM
MDM
(5.17)
(For reference: proton number density np,0 = 0.22 m−3 and CMB photon number density nγ,0 = 3.7 · 108 m−3 [51]). This gives a dark matter particle density
at recombination (z ≈ 1000): nDM (z = 1000) = 1.2 · 105 m−3 .
The typical velocities of dark matter at the time of recombination are very
small: the kinetic energy is negligible compared to their rest mass energy of
9.6 TeV. After the dark matter decoupled from ordinary matter and radiation
during their thermal freeze-out at a temperature of T ≈ MDM /26 [12], the
temperature of the dark matter was proportional with T ∝ a−2 while the normal
matter (still coupled to the radiation) cooled with T ∝ a−1 . At the moment
of recombination, the scale factor a increases with a factor 1012 compared to a
at dark matter decoupling. And normal matter has an average kinetic energy
of order eV at the moment of recombination. Thus average kinetic energy of
dark matter is around the 10−12 eV, not including possible velocity boosts due to
dark matter clustering. Although it is very likely that dark matter has clustered
during the recombination producing gravitational potentials, it would still have
only small velocities as it would otherwise erase this structure.
58
The cross section at recombination is dominated by the lowest order term
and given by (equation 3.10):
Sk hσvi = 4.1 · 10−30 cm3 s−1
(5.18)
The energy release-rate of the annihilations is given by:
dE
= f MDM c2 n2DM,0 (1 + z)6 Sk hσvi
dt
(5.19)
with f being an efficiency factor since not all energy released in the annihilation
will be transferred to the IGM.
Two small sample calculations will serve as an illustration: energy release
rate in one cubic meter nowadays and during recombination. The (constant)
parameters are: MDM c2 = 9.6 · 1012 eV, nDM,0 = 1.4 · 10−4 m−3 , Sk hσvi =
4.1 · 10−30 cm3 s−1 . Then we get (assuming f = 1)
dE
= 8 · 10−31 eV
dt z=0
dE
= 8 · 10−13 eV
(5.20)
dt z=1000
The only thing left do is to determine the efficiency f . In Figure 4.3 we
see that the electrons and positrons both contain 9.9% and photons 21% of
the energy released in the annihilation process. The other part of the energy
contained by neutrinos, protons and neutrons can be considered lost. Making
the naı̈ve assumption that all of the energy of the electrons and photons is
deposited into the IGM we choose
f = 0.40
(5.21)
We say ‘naı̈ve’ because this is a simplified and over-estimate of the efficiency.
We assume here that all of the energy contained within the electrons, positrons
and photons is transfered into the IGM while we know that for example a large
part of the photons are within the transparency window and will not deposit
their energy fast enough to affect the recombination.
5.3
RecFast Results
To calculate the evolution of the free electron fraction during the recombination
period, we modified the publicly available code RecFast [45,46] to solve equation
(5.14) which is the modified differential equation for the free electron fraction.
We performed the calculation using the cosmological parameters as published by
the WMAP 7-year results [1]: H0 = 71 km/s/Mpc, ΩΛ = 0.734, ΩDM = 0.222,
Ωb = 0.0449, YHE = 0.28.
The result is shown in Figure 5.2. We also plotted two lines with a boosted
energy-release rate to show how the recombination process changes with extra
release of energy due to annihilations.
The first thing to observe is that the Minimal Dark Matter annihilations do
not produce significant changes to the recombination history: it traces the line
59
Figure 5.2:
The free electron fraction during recombination as a function of redshift z. The
red line corresponds to a recombination process without dark matter annihilation. The blue
dots correspond to the recombination process with minimal dark matter annihilations. The
green and yellow line correspond to energy release of the minimal dark matter annihilation
boosted by a factor of 104 and 105 respectively.
corresponding to the recombination without dark matter annihilations. The
difference in the final free electron fraction is
∆xe
∼ 10−5
(5.22)
xe
The difference is smaller than the accuracy of some of the parameter values used.
Hence, the result is that Minimal Dark Matter annihilations have no effect on
the recombination process.
The energy release is too small to affect the recombination process. Because
the recombination with Minimal Dark Matter annihilations traces the normal
recombination line, we can infere that the dark matter annihilations do not cause
extra ionizations of the ground state or ionizations of the first excited state.
And we can also conclude that there is no significant change in temperature:
an increase in temperature would for example increase the number of photons
that can ionize a hydrogen atom. The number of extra ionizations, excitations
and heating caused by dark matter annihilations depends only on the energy
release-rate dE
dt . Apparently, the energy release-rate is just too low, which could
be expected considering the result of the sample calculation in equation (5.20).
If we would want to observe changes in the recombination, the energy release
should increase drastically: an increase of 104 gives only slight changes. If the
60
energy release-rate would have been larger by a factor of at least 105 (the yellow
line in Figure 5.2), it would predict a CMB powerspectrum that is different from
the observed powerspectrum by WMAP [15]. Dark matter annihilations that
have less effect on the recombination process than the Minimal Dark Matter
boosted by 105 would predict a powerspectrum similar or within the error of
the powerspectrum observations.
If the energy release-rate would have been larger by this factor of 105 , then
it would only change the residual free electron fraction and not so much the
process of recombination. In the beginning the recombination rate is fast and
the universe still dense. As a result, the ionized electron is able to find another
proton soon, cancelling the ionization. At the end, the free proton and electron
density in a co-moving volume drops significantly, and the size of the co-moving
volume increases, making it harder for the electron to recombine.
61
Chapter 6
Conclusion
Although we made the naı̈ve assumption that the electrons, positrons and photons are able to deposit all their energy into the IGM, leading to extra ionizations, excitations and heating, the recombination process is not (significantly)
changed. Neither is the residual free electron fraction altered. The small change
∆xe
−5
is smaller than the accuracy of some of the parameters used.
xe ∼ 10
The annihilation of the neutral dark matter component of the Minimal Dark
Matter model does not affect the recombination. Since this dark matter model
does not predict a different recombination process, it predicts a CMB powerspectrum which is in accordance with current observations. Minimal Dark
Matter is allowed by the CMB observations.
There is no change in the recombination process because the energy released
due to Minimal Dark Matter annihilations is too small. The energy releaserate depends on the the annihilation cross section, the dark matter mass, the
deposition efficiency and the Sommerfeld enhancement factor. The large dark
matter mass and the small annihilation cross section are the direct reasons
for the small energy release-rate. And they are subsequently determined by
the assumptions of a minimal amount of new physics and the weak interaction
allowed by thermal freeze-out physics.
The cross section for the annihilation of two neutral Minimal Dark Matter
particles into two W -bosons is very small: hσvi = 4.1 · 10−33 cm3 s−1 . Even
though we obtained a larger result than published earlier [11] because we performed a different summation over W -boson polarizations, the cross section is
smaller than the typical weak cross section allowed by thermal freeze-out. The
typical weak interaction cross section is of the order 10−26 cm3 s−1 .
The cross section is small because it is a weak interaction (hence small
coupling strength) and because the dark matter particles are very heavy. A
smaller dark matter mass would increase the cross section, but since the 9.6
TeV/c2 stems from the thermal freeze-out physics calculation, we can only have
a different dark matter mass if we add physics that changes the production of
dark matter (e.g. an alternative for the thermal freeze-out) or if we add a new
type of interaction that changes the calculation with the thermal freeze-out.
If the dark matter has a different interaction type than the weak interaction,
it can also have a larger cross section and circumvent the bounds on the cross
section from thermal freeze-out. However, all changes would imply abandoning
the assumption of minimal amount of new physics. Because decreasing the mass
62
would introduce new physics and a different interaction would also ask for new
physics beyond the Standard Model.
The Sommerfeld enhancement factor of 1 · 103 we calculated also depends
on the mass and the interaction type. A different coupling constant, a different
mass of the dark matter particle or even a different mass of the gauge boson
of a new interaction can significantly change the Sommerfeld enhancement. So
changing the mass or type of interaction by adding new physics changes the
Sommerfeld enhancement. But due to the resonance behaviour it is complicated
to predict whether the enhancement will increase or decrease.
Perhaps there is some other way of boosting the annihilation cross section.
As an example we can assume the dark matter is not distributed uniformly
but concentrated into small clumps of dark matter. This substructure would
increase the chance an encounter of two dark matter particles and boost the
annihilation cross section.
With Pythia we analysed the decay products of the W -bosons and enabled
us to infere an efficiency factor of 0.40, stemming from the energy contained by
the photons, electrons and positrons. This parameter in the energy release-rate
function will not change drastically if we introduce a new type of interaction:
it will remain at the same order of magnitude.
The energy released and the dark matter particle number density are simple functions of the dark matter mass. The energy released is the dark matter
mass and hence linear proportional to the mass. The number density is inversely proportional to the mass. Since the enery release-rate is linear with the
energy released per annihilation and quadratic in the particle number density,
it is (ignoring the cross section and the Sommerfeld enhancement) inversely
proportional with the dark matter mass.
Hence, because we wanted only to introduce a new multiplet and no new
interactions, we have dark matter that annihilates via the weak interaction and
is very massive. Adding new physics to this model can lead to an increase in
the effect on the recombination and can predict a different CMB powerspectrum
than we observe.
Because the Minimal Dark Matter model is in accordance with observations
of the recombination, we have to look for other probes, other observations in
order to possibly exclude this model. For example, another possible tests closely
related to the CMB powerspectrum, would be to look for the influence on dark
matter annihilations on the black-body spectrum of the CMB.
63
Chapter 7
Acknowledgements
First of all I would like to thank Prof. dr. Mees de Roo for his supervising. I
am especially grateful for his patience and tips in the long calculations and for
his flexibility, even in marvellous winter wheather. Furthermore I would like to
thank Joost, Hans and Geert-Wiebe for their tips and advice while writing this
report. Without them, this report would not have been accessible for somebody
else.
And finally I would like to thank Dieuwertje for her mental support and her
enthusiasm about the topic of dark matter.
64
Appendix A
Complete χ0χ0 → W +W −
cross section calculation
In this appendix we present the calculation of the annihilation cross section in
more detail than presented in Section 2.3.
To calculate Feynman diagrams with majorana particles we use the Feynman
rules as described in [29]. An ingoing Majorana fermion is represented by u(p, s)
and an outgoing one by v̄(p, s).
The two Feynman diagrams contributing at tree-level to the annihilation
cross section are given in Figure A.1. The t-channel graph has an intermediate
χ+ particle, the u-channel graph has an intermediate χ− particle.
Amplitude 1
Using the Feynman rules for Majorana interactions we obtain the amplitude for
the left diagram
A1
= v̄(p2 , s2 )iΓiS(q)iΓu(p2 , s2 )∗ν (k1 )∗µ (k2 )
= (−i)v̄(p2 , s2 )ΓS(p1 − k1 )Γ∗ν (k1 )∗µ (k2 )
Figure A.1:
The two Feynman diagrams corresponding to the annihilation of the neutral
component into two W -bosons at tree-level. The small arrow corresponds to the chosen
direction of the fermion charge flow (see [29]).
65
where the ∗µ and ∗ν are the two polarization states of the outgoing W -bosons
[30].
The W -bosons only couple to the left-handed fields and the coupling strength
is given by equation (2.7). The terms for the interaction vertices is
√
−i 3gW µ
Γ=
γ (1 − γ 5 )
(A.1)
2
√
with gW = 4πα2 .
The propagator S(q) is a charged dark matter fermion and is given by
q +M
/
q 2 −M 2 +i with M the mass of the dark matter particle. The internal momentum
q can alternatively be expressed as q = p1 − k1 = k2 − p2 .
Using equation (A.1) we can re-express amplitude A1 into
A1
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 ) /q + M γ ν (1 − γ 5 )u(p1 , s1 )
4(q 2 − M 2 + i)
×∗ν (k1 )∗µ (k2 )
=
It is possible to bring the second projection operator to the front and absorb
it by the first projection operator: 12 (1−γ 5 ) 12 (1−γ 5 ) = 14 (1−2γ 5 +1) = 12 (1−γ 5 ).
However, because the mass-term misses an extra γ matrix compared to /q the
projection operator acting on M changes
(1 − γ 5 )(/q + M )γ ν (1 − γ 5 ) = (1 − γ 5 )(1 − γ 5 )/qγ ν + (1 − γ 5 )(1 + γ 5 )M γ ν (A.2)
Notice a left and right projection operator acting on M , hence the M -term
becomes zero. This gives:
A1 =
2(q 2
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 ) ∗ν (k1 )∗µ (k2 )
2
− M + i)
For the cross section we need to calculate the square amplitude . Therefore
we first have to find an expression for the Hermitian conjugate of the amplitude
A†1 . First we calculate the Hermitian conjugate of the center-term:
v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 )
†
= u† (p1 , s1 )γ ν† /q† (1 − γ 5 )† γ µ† v̄ † (p2 , s2 )
= ū(p1 , s1 )γ ν /q(1 + γ 5 )γ µ v(p2 , s2 ) (A.3)
The Hermitian conjugate term is therefore
A†1
=
2
−3igW
ū(p1 , s1 )γ ν (1 − γ 5 )/qγ µ v(p2 , s2 )ν (k1 )µ (k2 )(A.4)
2(q 2 − M 2 + i)
and the amplitude squared A†1 A1 becomes
2
|A1 |
=
4(q 2
4
9gW
− M 2 + i)2
ū(p1 , s1 )γ ν (1 − γ 5 )/qγ µ v(p2 , s2 )
× v̄(p2 , s2 )γ ρ (1 − γ 5 )/qγ σ u(p1 , s1 )
× ν (k1 )µ (k2 )∗σ (k1 )∗ρ (k2 )
×
66
We are interested in the cross section averaged over the dark matter spins
and W -bosons polarizations. A summation over the spin and polarizations of
the external particles gives us (dropping the explicit functions of spin (si ) and
introducing them with subscripts s and r):
X
2
|A1 |
=
r,s,λ1 ,λ2
X
r,s,λ1 ,λ2
4(q 2
4
9gW
− M 2 + i)2
ūs (p1 )γ ν (1 − γ 5 )/qγ µ vr (p2 )
× v̄r (p2 )γ ρ (1 − γ 5 )/qγ σ us (p1 )
× ν,λ1 (k1 )µ,λ2 (k2 )∗σ,λ1 (k1 )∗ρ,λ2 (k2 )
×
We use the standard relation to sum over the polarizations of the massive
gauge boson in unitary gauge
X
ν,λ1 ∗σ,λ1
= −gνσ +
k1ν k1σ
2
MW
(A.5)
µ,λ2 ∗ρ,λ2
= −gµρ +
k2µ k2ρ
2
MW
(A.6)
λ1
X
λ2
We have to include the second term on the right handside because the W -bosons
have large momenta. The Minimal Dark Matter particles that annihilate into
the two gauge bosons are very heavy (9.6 TeV/c2 ) and the W -bosons produced
are relativistic particles.
Performing the summation over spins r and s using
X
ua,r (p)ūb,r (p) = (p
/ + M )ab
r
X
va,r (p)v̄b,r (p) = (p
/ − M )ab
r
and using equations (A.5,A.6) gives
X
2
|A1 |
=
X
r,s
r,s,λ1 ,λ2
4(q 2
4
9gW
− M 2 + i)2
i
µ
ν
ūs,a (p1 )γab
(1 − γ 5 )bc /qcd γde
vr,e (p2 )
h
i
σ
× v̄r,f (p2 )γfρg (1 − γ 5 )gh /qhi γij
us,j (p1 )
k1ν k1σ
k2µ k2ρ
×
−gνσ +
−g
+
µρ
2
2
MW
MW
×
h
4
9gW
4(q 2 − M 2 + i)2
h
µ
ν
5
× (p
/1 + M )ja γab (1 − γ )bc /qcd γde (p
/2 − M )ef
i
σ
× γfρg (1 − γ 5 )gh /qhi γij
k2µ k2ρ
k1ν k1σ
k1ν k1σ k2µ k2ρ
×
gνσ gµρ − gνσ
− gµρ
+
2
2
4
MW
MW
MW
=
67
Expanding the term between brackets gives four terms with different orders
in W -boson mass MW . After acting with the two metric tensors on the γmatrices, changing their indices, the first term becomes
I1
=
×
h
4
9gW
µ
ν
5
(p
/1 + M )ja γab (1 − γ )bc /qcd γde (p
/2 − M )ef
2
2
− M + i)
i
γµ,f g (1 − γ 5 )gh /qhi γν,ij
(A.7)
4(q 2
The other three terms are:
II1
=
×
III1
=
×
IV1
=
×
h
4
9gW
µ
ν
5
(p
/1 + M )ja γab (1 − γ )bc /qcd γde (p
/2 − M )ef
4(q 2 − M 2 + i)2
i
k2µ k2ρ
σ
γfρg (1 − γ 5 )gh /qhi γij
−gνσ
(A.8)
2
MW
h
4
9gW
µ
ν
5
(p
/1 + M )ja γab (1 − γ )bc /qcd γde (p
/2 − M )ef
4(q 2 − M 2 + i)2
i
k1ν k1σ
σ
(A.9)
γfρg (1 − γ 5 )gh /qhi γij
−gµρ
2
MW
h
4
9gW
µ
ν
5
(p
/1 + M )ja γab (1 − γ )bc /qcd γde (p
/2 − M )ef
4(q 2 − M 2 + i)2
i k k k k 1ν 1σ 2µ 2ρ
σ
(A.10)
γfρg (1 − γ 5 )gh /qhi γij
4
MW
The subscript 1 indicates these are the terms of the first amplitude.
We will calculate the trace of the first term I1 explicitly, the traces of the
other terms are just given. First we rewrite the first term by moving the projection operator (1 − γ 5 ) on the right towards the left:
nh
i
o
ν
5
µ
ρ
5
σ
Tr (p
/1 + M )γ (1 − γ )/qγ (p
/2 − M )γ (1 − γ )/qγ gνσ gµρ
nh
i
o
ν
5
µ
5
5
ρ
σ
= Tr (p
/1 + M )γ (1 − γ )/qγ ((1 − γ )p
/2 − (1 + γ )M )γ /qγ gνσ gµρ
and see how a left handed and right-handed projection operator act on the
mass-term: the projection operators cancel the mass-term to zero. Then one
can bring the 1 − γ 5 on the right to the left to absorb it into the first 1 − γ 5
term, giving effectively a factor 2: (1 − γ 5 )(1 − γ 5 ) = 2(1 − γ 5 ). Because it
permutes with an even number of gamma matrices while bringing it to the left,
there is no change in sign.
nh
i
o
ν
5
µ
ρ
σ
2Tr (p
(A.11)
/1 + M )γ (1 − γ )/qγ p
/2 γ /qγ gνσ gµρ
We can get rid of the remaining mass-term as well since it is the trace of an
odd number of gamma-matrices, hence it is zero. This way we have to calculate
nh
i
o
ν
5
µ
ρ
σ
2Tr p
/1 γ (1 − γ )/qγ p
/2 γ /qγ gνσ gµρ =
nh
i
o
nh
i
o
ν
µ
ρ
σ
ν 5
µ
ρ
σ
2Tr p
/1 γ /qγ p
/2 γ /qγ gνσ gµρ − 2Tr p
/1 γ γ /qγ p
/2 γ /qγ gνσ gµρ
68
The first term is evaluated as follows using the cyclic property of the trace;
γµ γ ν γ µ = −2γ ν (in the third step) and γµ γ ν γ λ γ σ γ µ = −2γ σ γ λ γ ν (in the fifth
step):
nh
i
o
ν
µ
ρ
σ
2Tr p
/2 γ /qγ gνσ gµρ =
/1 γ /qγ p
o
n
ν
µ
= 2Tr p
/2 γµ /qγν
/1 γ /qγ p
= 2p1σ qλ p2ρ qς Tr γ σ γ ν γ λ γ µ γ ρ γµ γ ς γν
= 2p1σ qλ p2ρ qς Tr γν γ σ γ ν γ λ γ µ γ ρ γµ γ ς
= −4p1σ qλ p2ρ qς Tr γ σ γ λ γ µ γ ρ γµ γ ς
= −4p1σ qλ p2ρ qς Tr γµ γ ς γ σ γ λ γ µ γ ρ
= 8p1σ qλ p2ρ qς Tr γ λ γ σ γ ς γ ρ
= 8p1σ qλ p2ρ qς 4 g λσ g ςρ − g λς g σρ + gλρg σς
= 32 [(p1 · q)(p2 · q) − (q · q)(p1 · p2 ) + (q · p2 )(q · p1 )]
= −32(q · q)(p1 · p2 ) + 64(p1 · q)(p2 · q)
The second trace goes almost analogous to the first term. First we let the
two metric tensors act on the γ-matrices, lowering the ν and µ indices
nh
i
o
n
o
ν 5
µ
ρ
σ
ν 5
µ
2Tr p
(A.12)
/1 γ γ /qγ p
/2 γ /qγ gνσ gµρ = 2Tr p
/1 γ γ /qγ p
/2 γµ /qγν
After that we cyclically permute the γ 5 matrix to the left and then we use
γµ γ ν γ λ γ µ = 4g νλ and γµ γ ν γ λ γ σ γ µ = −2γ σ γ λ γ ν again to rewrite the expression:
n
o
n
o
ν 5
µ
ν
µ 5
2Tr p
= 2Tr p
(A.13)
/1 γ γ /qγ p
/2 γµ /qγν
/1 γ /qγ γ p
/2 γµ /qγν
σ λ µ ρ
= −4p1σ qλ p2ρ qς Tr γ γ γ γ γµ γ 5 γ ς (A.14)
= 8p1σ qλ p2ρ qς Tr γ λ γ σ γ ς γ ρ γ 5
(A.15)
=
32ip1σ qλ p2ρ qς λσςρ
(A.16)
Using the fact that the momentum q is symmetric under interchange of indices
and λσςρ is anti-symmetric, this expression becomes zero:
p1σ qλ p2ρ qς λσςρ = p1σ qς p2ρ qλ ςσλρ = −p1σ qς p2ρ qλ λσςρ = 0
(A.17)
Thus we get:
nh
i
o
ν
5
µ
ρ
5
σ
(p
/1 + M )γ (1 − γ )/qγ (p
/2 − M )γ (1 − γ )/qγ gνσ gµρ =
nh
i
o
ν
µ
ρ
σ
2Tr p
/1 γ /qγ p
/2 γ /qγ gνσ gµρ
Tr
=
= −32(q · q)(p1 · p2 ) + 64(p1 · q)(p2 · q)
(A.18)
Putting this back into the equation for the first term of amplitude (A.7) gives:
I1 =
(q 2
4
9gW
(16(p1 · q)(p2 · q) − 8(q · q)(p1 · p2 ))
− M 2 + i)2
69
The traces of the terms 2 to 4 are respectively:
h
i
4
9gW
k2µ k2ρ
ν
µ
ρ
σ
II1 =
(−2)Tr
g
p
γ
q
γ
p
γ
q
γ
=
/1 / /2 /
νσ
2
4(q 2 − M 2 + i)2
MW
4
4 2 2
9gW
=
(q )(k2 )(p1 · p2 ) − 2(q 2 )(p1 · k2 )(p2 · k2 )
2
(q 2 − M 2 + i)2 MW
−2(k22 )(p1 · q)(p2 · q) + 4(k2 · p2 )(k2 · q)(q · p1 )
h
i
4
k1ν k1σ
9gW
ν
µ
ρ
σ
(−2)Tr p
=
III1 =
/1 γ /qγ p
/2 γ /qγ gµρ
2
4(q 2 − M 2 + i)2
MW
4
9gW
4 2 2
=
(q )(k1 )(p1 · p2 ) − 2(q 2 )(p1 · k1 )(p2 · k1 )
2
2
2
2
(q − M + i) MW
−2(k12 )(p1 · q)(p2 · q) + 4(k1 · q)(k1 · p1 )(p2 · q)
h
i
4
9gW
ν
µ
ρ
σ k1ν k1σ k2µ k2ρ
2Tr
p
γ
q
γ
p
γ
q
γ
IV1 =
=
/
/
/
/
4
1
2
4(q 2 − M 2 + i)2
MW
=
(q 2
4
2
9gW
4
2
2
− M + i) MW
−(k12 )(k22 )(q 2 )(p1 · p2 ) + 2(k22 )(q 2 )(k1 · p2 )(k1 · p1 )
+2(k12 )(q 2 )(k2 · p2 )(k2 · p1 ) − 4(q 2 )(k1 · p1 )(k1 · k2 )(k2 · p2 )
+2(k12 )(k22 )(p1 · q)(p2 · q) + 8(k1 · q)(k1 · p1 )(k2 · q)(k2 · p2 )
−4(k12 )(k2 · p2 )(k2 · q)(p1 · q) − 4(k22 )(k1 · p1 )(k1 · q)(p2 · q)
The final result for the first diagram is:
X
2
|A1 | = I1 + II1 + III1 + IV1
r,s,λ1 ,λ2
Amplitude 2
A2 is the amplitude corresponding to the diagram on the right in Figure A.1.
The two differences between this diagram and the left diagram are: the internal
momentum is b = p1 − k2 = k1 − p2 and the vertices of the W -bosons are
interchanged. This results in the interchange of indices in their epsilon terms.
The mass of the χ− particle is equal to the mass of the χ+ particle. Therefore
the amplitude looks much like A1 and is given by:
A2
2
3igW
4(b2 − M 2 + i)
× v̄(p2 , s2 )γ µ (1 − γ 5 ) /b + M γ ν (1 − γ 5 )u(p1 , s1 ) ∗µ (k1 )∗ν (k2 )
=
After performing the same operations on this amplitude as done on A1 (combining the two 1 − γ 5 terms and eliminating the mass term in the middle due
to the left and right-projection operators acting on it) we obtain:
A2 =
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 )/bγ ν u(p1 , s1 ) ∗µ (k1 )∗ν (k2 )
2
2
2(b − M + i)
70
The part between the rectangular brackets is of similar shape as in A1 so we
can write the Hermitian conjugate term
A†2 =
2(b2
2
−3igW
ū(p1 , s1 )γ ν (1 − γ 5 )/bγ µ v(p2 , s2 )ν (k2 )µ (k1 )
2
− M + i)
Thus the square of the amplitude becomes
2
|A2 |
=
4(b2
4
9gW
− M 2 + i)2
ū(p1 , s1 )γ ν (1 − γ 5 )/bγ µ v(p2 , s2 )
× v̄(p2 , s2 )γ ρ (1 − γ 5 )/bγ σ u(p1 , s1 )
× ν (k2 )µ (k1 )∗σ (k2 )∗ρ (k1 )
×
The summation over spins and polarizations gives:
X
2
|A2 |
r,s,λ1 ,λ2
4
9gW
4(b2 − M 2 + i)2
h
ν
5
µ
× (p
/1 + M )γ (1 − γ )/bγ
i
ρ
5
σ
× (p
/2 − M )γ (1 − γ )/bγ
k2ν k2σ k1µ k1ρ
k1µ k1ρ
k2ν k2σ
×
gνσ gµρ − gνσ
− gµρ
+
2
2
4
MW
MW
MW
=
The trace of the first term is the same as for the first diagram with the
momenta k1 and k2 interchanged. This term is already calculated (see equation
(A.18)). Uniting the two projection operators gives an extra factor of 2 which
is put in front of the traces.
h
i
ν
5
µ
5
Tr (p
/1 + M )γ (1 − γ )/bγ (p
/2 − M )γµ (1 − γ )/bγν =
= −32(b · b)(p1 · p2 ) + 64(p1 · b)(p2 · b)
The traces of the terms I2 to IV2 are respectively:
I2
=
4
9gW
(16(p1 · b)(p2 · b) − 8(b · b)(p1 · p2 ))
(b2 − M 2 + i)2
The factor (−2) from the projection operators in front of the first two traces
is multiplied by the factor (−2) coming from applying γ σ p
/γ ν gνσ = γν p
/γ ν = −2p
/
and the total factor of 4 cancels the 4 in the nominator. The resulting equations
are
h
i
4
9gW
k1µ k1ρ
ν/ µ
ρ/ σ
II2 =
(−2)Tr
p
γ
b
γ
p
γ
b
γ
g
=
/
/
νσ
2
1
2
4(b2 − M 2 + i)2
MW
4
9gW
4 2 2
=
(b )(k1 )(p1 · p2 ) − 2(b2 )(p1 · k1 )(p2 · k1 )
2
2
2
2
(b − M + i) MW
−2(k12 )(p1 · b)(p2 · b) + 4(k1 · p2 )(k1 · b)(b · p1 )
71
III2
=
=
IV2
=
=
h
i
4
k2ν k2σ
9gW
ν/ µ
ρ/ σ
(−2)Tr p
=
/1 γ b γ p
/2 γ b γ gµρ
2
4(b2 − M 2 + i)2
MW
4
9gW
4 2 2
(b )(k2 )(p1 · p2 ) − 2(b2 )(p1 · k2 )(p2 · k2 )
2
2
2
2
(b − M + i) MW
−2(k22 )(p1 · b)(p2 · b) + 4(k2 · b)(k2 · p1 )(p2 · b)
h
i
4
9gW
ν/ µ
ρ / σ k2ν k2σ k1µ k1ρ
2Tr p
=
/1 γ b γ p
/2 γ b γ
4
4(b2 − M 2 + i)2
MW
4
9gW
2
4
(b2 − M 2 + i)2 MW
2
2
2
−(k2 )(k1 )(b )(p1 · p2 ) + 2(k12 )(b2 )(k2 · p2 )(k2 · p1 )
+2(k22 )(b2 )(k1 · p2 )(k1 · p1 ) + 2(k22 )(k12 )(p1 · b)(p2 · b)
−4(b2 )(k2 · p1 )(k2 · k1 )(k1 · p2 ) + 8(k2 · b)(k2 · p1 )(k1 · b)(k1 · p2 )
−4(k22 )(k1 · p2 )(k1 · b)(p1 · b) − 4(k1 )2 (b · k2 )(b · p2 )(k2 · p1 )
And we obtain the total matrix element
X
2
|A2 | = I2 + II2 + III2 + IV2
Mixed amplitudes terms
To calculate the total amplitude at tree-level we also have to evaluate the mixed
2
amplitude terms: |M| = A1 A†1 + A1 A†2 + A2 A†1 + A2 A†2 .
To summarize the previous results:
A1
=
A†1
=
A2
=
A†2
=
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 ) ∗ν (k1 )∗µ (k2 )
2
− M + i)
2
−3igW
ū(p1 , s1 )γ ν (1 − γ 5 )/qγ µ v(p2 , s2 ) ν (k1 )µ (k2 )
2
2
2(q − M + i)
2
3igW
v̄(p2 , s2 )γ µ (1 − γ 5 )/bγ ν u(p1 , s1 ) ∗µ (k1 )∗ν (k2 )
2
2
2(b − M + i)
2
−3igW
ū(p1 , s1 )γ ν (1 − γ 5 )/bγ µ v(p2 , s2 ) ν (k2 )µ (k1 )
2
2
2(b − M + i)
2(q 2
The first mixed amplitude squared is
A1 A†2
=
4(q 2
M2
4
9gW
+ i)(b2 − M 2 + i)
−
× v̄(p2 , s2 )γ µ (1 − γ 5 )/qγ ν u(p1 , s1 )ū(p1 , s1 )γ σ (1 − γ 5 )/bγ ρ v(p2 , s2 )
× ∗ν (k1 )∗µ (k2 )σ (k2 )ρ (k1 )
and the other term is
A2 A†1
4
9gW
−
+ i)(q 2 − M 2 + i)
× v̄(p2 , s2 )γ ρ (1 − γ 5 )/bγ σ u(p1 , s1 )ū(p1 , s1 )γ ν (1 − γ 5 )/qγ µ v(p2 , s2 )
× ∗ρ (k1 )∗σ (k2 )ν (k1 )µ (k2 )
=
4(b2
M2
72
Performing the summation over spins and polarizations while using the appropriate relations, we get:
X
A1 A†2
s,r,λ1 ,λ2
4
9gW
4(q 2 − M 2 + i)(b2 − M 2 + i)
o
n
µ
5
ν
σ
5
ρ
×
(p
/2 − M )γ (1 − γ )/qγ (p
/1 + M )γ (1 − γ )/bγ
k1ν k1ρ
k2µ k2σ
×
−gρν +
−gµσ +
(A.19)
2
2
MW
MW
=
4
9gW
2(q 2 − M 2 + i)(b2 − M 2 + i)
n
o
µ
5
ν
σ
ρ
× Tr (p
/2 − M )γ (1 − γ )/qγ p
/1 γ /bγ
k2µ k2σ
k1ν k1ρ
k1ν k1ρ k2µ k2σ
×
gρν gµσ − gρν
−
g
+
(A.20)
µσ
2
2
4
MW
MW
MW
=
where we moved the right 1 − γ 5 to the left and eliminated the mass-term again
the same way as before. We do the same for the other mixed term:
X
s,r,λ1 ,λ2
A2 A†1
=
2(b2
M2
4
9gW
+ i)(q 2 − M 2 + i)
−
n
o
ρ
5
σ
ν
µ
× Tr (p
/2 − M )γ (1 − γ )/bγ p
/1 γ /qγ
k1ρ k1ν
k1ρ k1ν k2σ k2µ
k2σ k2µ
− gσµ
+
(A.21)
×
gρν gσµ − gρν
2
2
4
MW
MW
MW
Traces of mix-term A1 A†2
9g 4
W
The term A1 A†2 consist of four trace terms, all proportional to 2(q2 −M 2 +i)(b
2 −M 2 +i) .
In all these terms we can leave out the mass term because it is contained in a
trace of an odd number of gamma matrices. The four terms are defined as
follows with the subscript ‘12’ indicating they are the terms of A1 A†2
n
o
0
µ
5
ν
σ
ρ
I12
= Tr p
/2 γ (1 − γ )/qγ p
/1 γ /bγ gρν gµσ
n
o
k2µ k2σ
µ
5
ν
σ
ρ
0
II12
= −Tr p
/2 γ (1 − γ )/qγ p
/1 γ /bγ gρν
2
MW
n
o
k1ν k1ρ
0
µ
5
ν
σ
ρ
III12
= −Tr p
/2 γ (1 − γ )/qγ p
/1 γ /bγ gµσ
2
MW
n
ok k k k
1ν 1ρ 2µ 2σ
0
µ
5
ν
σ
ρ
IV12
= Tr p
/2 γ (1 − γ )/qγ p
/1 γ /bγ
4
MW
0
First we calculate the traces of the term I12
containing the double metric
tensors explicitly. First one can split the term due to the 1 − γ 5 term:
n
o
n
o
µ
5
ν
µ
ν
Tr p
= Tr p
/2 γ (1 − γ )/qγ p
/1 γµ /bγν
/2 γ /qγ p
/1 γµ /bγν
n
o
µ 5
ν
−Tr p
/2 γ γ /qγ p
/1 γµ /bγν
1
5
≡ I12
− I12
73
with the superscript 1 as an indication for the first term with the unit matrix,
and the superscript 5 for the term with the γ 5 -matrix.
With the use of relations γµ γ ν γ µ = −2γ ν , γµ γ ν γ λ γ σ γ µ = −2γ σ γ λ γ ν and
Tr(γ µ γ ν ) = 4g µν one can calculate the trace of the first term:
1
I12
= p2λ qδ p1φ bρ Tr γ λ γ µ γ δ γ ν γ φ γµ γ ρ γν
= p2λ qδ p1φ bρ Tr γν γ λ γ µ γ δ γ ν γ φ γµ γ ρ
= −2p2λ qδ p1φ bρ Tr γ δ γ µ γ λ γ φ γµ γ ρ
= −2p2λ qδ p1φ bρ Tr γµ γ ρ γ δ γ µ γ λ γ φ
= −8p2λ qδ p1φ bρ Tr g ρδ γ λ γ φ
= −8p2λ qδ p1φ bρ g ρδ g λφ
= −32(p2 · p1 )(q · b)
5
The calculation of I12
goes along the same line as the previous calculation
5
for the part with the γ -matrix and gives zero:
5
I12
= p2λ qδ p1φ bρ Tr γ λ γ µ γ 5 γ δ γ ν γ φ γµ γ ρ γν
= −p2λ qδ p1φ bρ Tr γν γ λ γ µ γ δ γ ν γ φ γ 5 γµ γ ρ
= 2p2λ qδ p1φ bρ Tr γ δ γ µ γ λ γ φ γ 5 γµ γ ρ
= 2p2λ qδ p1φ bρ Tr γµ γ ρ γ δ γ µ γ λ γ φ γ 5
= 8p2λ qδ p1φ bρ Tr g ρδ γ λ γ φ γ 5
= 0
For the calculation of the other traces, we split these terms analogous to I12
into a unit matrix part and a γ 5 -part. The γ 5 -part always gives zero. First we
1
calculate II12
. To do this, we need the trace-rule for six gamma matrices:
λ δ ν φ σ θ
Tr γ γ γ γ γ γ
= 4 g λδ g νφ g σθ + g λδ g νθ g φσ − g λδ g νσ g φθ − g νλ g φδ g σθ
−g νλ g δθ g φσ + g νλ g σδ g φθ + g φλ g νδ g σθ + g φλ g δθ g νσ
−g φλ g σδ g νθ − g λσ g νδ g φθ − g λσ g δθ g νφ + g λσ g φδ g νθ
+ g λθ g νδ g φσ + g λθ g σδ g νφ − g λθ g φδ g νσ
(A.22)
We use the equation above to calculate the trace of the six γ-matrices. Notice
that some terms appear twice due to the two k2 ’s and some cancel:
1
II12
n
ok k
2µ 2σ
µ
ν
σ
5
= −Tr p
/2 γ /qγ p
/1 γ (1 − γ )/bγν
2
MW
8 =
(k2 )2 [(b · q)(p1 · p2 ) + (b · p1 )(p2 · q) − (b · p2 )(p1 · q)]
2
MW
+2(k2 · p1 ) [(b · p2 )(k2 · q) − (b · k2 )(p2 · q)]
+2(k2 · p2 ) [−(b · p1 )(k2 · q) + (b · k2 )(p1 · q)]}
Using the Dirac notation we see that the trace of the third term III12
1
is analogous to the one of the second term hII12
with the
i p1 and p2 inter1
1
2
and III12
=
changed and k1 substituted by k2 : II12 = Tr p
/2 k/2 /qp
/1 k/2 /b /MW
74
h
i
2
Tr p
. thus:
/2 k/1 /b/qk/1 p
/1 /MW
1
III12
=
8 (k1 )2 [(b · q)(p1 · p2 ) + (b · p2 )(p1 · q) − (b · p1 )(p2 · q)]
2
MW
+2(p1 · k1 ) [(b · k1 )(p2 · q) − (b · p2 )(k1 · q)]
+2 (p2 · k1 ) [(b · p1 )(k1 · q) − (b · k1 )(p1 · q)]}
And finally
1
IV12
o
n
µ
ν
σ
ρ k1ν k1ρ k2µ k2σ
Tr p
/1 γ /bγ
/2 γ /qγ p
4
MW
p2λ k2µ qδ k1ν p1φ k2σ bθ k1ρ λ µ δ ν φ σ θ ρ =
Tr γ γ γ γ γ γ γ γ
4
MW
4 =
(k1 )2 (k2 )2 {(b · q)(p1 · p2 ) − (b · p2 )(q · p1 ) − (b · p1 )(q · p2 )}
MW
−2(k1 )2 [(b · q)(k2 · p1 )(k2 · p2 ) − (b · p1 )(q · k2 )(p2 · k2 )
−(b · k2 )(k2 · p1 )(p2 · q) + (b · k2 )(k2 · q)(p1 · p2 )]
−2(k2 )2 [(b · q)(k1 · p1 )(k1 · p2 ) − (b · p2 )(q · k1 )(p1 · k1 )
−(b · k1 )(k1 · p2 )(p1 · q) + (b · k1 )(q · k1 )(p1 · p2 )]
+4 [(b · k1 )(k1 · q)(k2 · p1 )(k2 · p2 ) + (b · k2 )(q · k2 )(k1 · p2 )(k1 · p1 )]
+2(b · q)(k1 · k2 ) [(k1 · p1 )(k2 · p2 ) + (k1 · p2 )(k2 · p1 ) − (k1 · k2 )(p1 · p2 )]
+2(b · p1 )(k1 · k2 ) [(k1 · k2 )(p2 · q) − (k1 · p2 )(k2 · q) − (k1 · q)(k2 · p2 )]
+2(b · p2 )(k1 · k2 ) [(k1 · k2 )(p1 · q) − (k1 · p1 )(k2 · q) − (k1 · q)(k2 · p1 )]
+2(b · k1 )(k1 · k2 ) [(p2 · p1 )(k2 · q) − (k2 · p1 )(p2 · q) − (k2 · p2 )(p1 · q)]
+2(b · k2 )(k1 · k2 ) [(p1 · p2 )(k1 · q) − (k1 · p1 )(p2 · q) − (k1 · p2 )(p1 · q)]
=
The final result for this mixed term is:
X
A1 A†2 =
s,r,λ1 ,λ2
4
1
9gW
1
1
1
I12 + II12
+ III12
+ IV12
2
2
2
2
2(q − M + i)(b − M + i)
Traces of mix-term A2 A†1
The second mixed term is:
X
s,r,λ1 ,λ2
A2 A†1
4
9gW
2(b2 − M 2 + i)(q 2 − M 2 + i)
n
o
ρ
5 / σ
ν
µ
× Tr (p
−
M
)γ
(1
−
γ
)
b
γ
p
γ
q
γ
/2
/1 /
k2σ k2µ
k1ρ k1ν
k1ρ k1ν k2σ k2µ
×
gρν gσµ − gρν
−
g
+
σµ
2
2
4
MW
MW
MW
=
It consists like the other mixed term, of four trace terms proportional to
75
4
9gW
2(b2 −M 2 +i)(q 2 −M 2 +i) :
I21
II21
III21
IV21
n
o
ρ
5
σ
ν
µ
Tr (p
/2 − M )γ (1 − γ )/bγ p
/1 γ /qγ gρν gσµ
o
n
k2σ k2µ
ρ
5
σ
ν
µ
= −Tr (p
/1 γ /qγ gρν
/2 − M )γ (1 − γ )/bγ p
2
MW
o
n
k1ρ k1ν
ρ
5
σ
ν
µ
= −Tr (p
/1 γ /qγ gσµ
/2 − M )γ (1 − γ )/bγ p
2
MW
n
o
ρ
5
σ
ν
µ k1ρ k1ν k2σ k2µ
= Tr (p
/2 − M )γ (1 − γ )/bγ p
/1 γ /qγ
4
MW
=
with the subscript ‘21’ indicating these are the terms in the second mixed term
A2 A†1 .
We can drop the part with the mass-term again because it is the trace of an
odd number of γ-matrices. And we will also drop the terms with the γ 5 -matrix
becuase they will give zero as well. We are only interested in the terms with
1
the unitary matrix (hence the superscript 1). I21
will be calculated explicitly,
the others will just be given.
n
o
ν
µ
1
I21
= Tr p
/2 γν /bγµ p
/1 γ /qγ
= p2λ bθ p1φ qδ Tr γ λ γν γ θ γµ γ φ γ ν γ δ γ µ
= −2p2λ bθ p1φ qδ Tr γ λ γν γ θ γ δ γ ν γ φ
= −8p2λ bθ p1φ qδ g θδ Tr γ λ γ φ
= −32p2λ bθ p1φ qδ g θδ g λφ
= −32(b · q)(p2 · p1 )
1
where we used the same relations as for the other mixed term I12
. For the
1
1
calculation of II21 and III21 we make use of the trace rule for six γ-matrices
(A.22):
1
II21
n
o
k2σ k2µ
ρ
5
σ
ν
µ
= −Tr (p
/2 − M )γ (1 − γ )/bγ p
/1 γ /qγ gρν
2
MW
n
o
σ
ν
µ k2σ k2µ
= −Tr p
/2 γν /bγ p
/1 γ /qγ
2
MW
p2λ bθ k2σ p1φ qδ k2µ λ
Tr γ γν γ θ γ σ γ φ γ ν γ δ γ µ
= −
2
MW
2p2λ bθ k2σ p1φ qδ k2µ λ φ σ θ δ µ =
Tr γ γ γ γ γ γ
2
MW
8 =
(k2 )2 [(b · q)(p1 · p2 ) + (b · p1 )(p2 · q) − (b · p2 )(p1 · q)]
2
MW
+2(k2 · p1 ) [(b · p2 )(k2 · q) − (b · k2 )(p2 · q)]
+2(k2 · p2 ) [(b · k2 )(p1 · q) − (b · p1 )(k2 · q)]}
76
and
1
III21
o
n
k1ρ k1ν
ρ
5
σ
ν
µ
= −Tr (p
/1 γ /qγ gσµ
/2 − M )γ (1 − γ )/bγ p
2
MW
n
o
ρ
ν
µ k1ρ k1ν
= −Tr p
/2 γ /bγµ p
/1 γ /qγ
2
MW
p2λ k1ρ bθ p1φ k1ν qδ λ ρ θ
= −
Tr γ γ γ γµ γ φ γ ν γ δ γ µ
2
MW
2p2λ k1ρ bθ p1φ k1ν qδ λ ρ θ δ ν φ =
Tr γ γ γ γ γ γ
2
MW
8 =
(k1 )2 [(b · q)(p1 · p2 ) + (b · p2 )(p1 · q) − (b · p1 )(p2 · q)]
2
MW
+2(k1 · p1 ) [(b · k1 )(p2 · q) − (b · p2 )(k1 · q)]
+2(k1 · p2 ) [(b · p1 )(k1 · q) − (b · k1 )(p1 · q)]}
The fourth term includes the trace of eight γ-matrices:
1
IV21
=
=
=
n
o
ρ
5
σ
ν
µ k1ρ k1ν k2σ k2µ
Tr (p
/2 − M )γ (1 − γ )/bγ p
/1 γ /qγ
4
MW
p2λ k1ρ bθ k2σ p1φ k1ν qδ k2µ λ ρ θ σ φ ν δ µ Tr γ γ γ γ γ γ γ γ
4
MW
4 (k1 )2 (k2 )2 [(b · q)(p1 · p2 ) − (b · p1 )(p2 · q) − (b · p2 )(p1 · q)]
4
MW
+2(k1 )2 [(b · p1 )(p2 · k2 )(k2 · q) + (b · k2 )(k2 · p1 )(p2 · q)
−(b · q)(k2 · p2 )(k2 · p1 ) − (b · k2 )(k2 · q)(p1 · p2 )]
+2(k2 )2 [(b · p2 )(k1 · q)(k1 · p1 ) + (b · k1 )(p2 · k1 )(p1 · q)
−(b · q)(p2 · k1 )(p1 · k1 ) − (b · k1 )(k1 · q)(p1 · p2 )]
+4 [(b · k1 )(k1 · q)(k2 · p1 )(k2 · p2 ) + (b · k2 )(k1 · p2 )(k1 · p1 )(k2 · q)]
+2(b · q)(k1 · k2 ) [(k1 · p2 )(p1 · k2 ) + (k1 · p1 )(k2 · p2 ) − (k1 · k2 )(p1 · p2 )]
+2(b · p1 )(k1 · k2 ) [(k1 · k2 )(p2 · q) − (k1 · q)(k2 · p2 ) − (k1 · p2 )(k2 · q)]
+2(b · p2 )(k1 · k2 ) [(k1 · k2 )(p1 · q) − (k1 · q)(k2 · p1 ) − (k1 · p1 )(k2 · q)]
+2(b · k1 )(k1 · k2 ) [(k2 · q)(p1 · p2 ) − (k2 · p1 )(p2 · q) − (k2 · p2 )(p1 · q)]
+2(b · k2 )(k1 · k2 ) [(k1 · q)(p1 · p2 ) − (k1 · p2 )(p1 · q) − (k1 · p1 )(p2 · q)]}
The final result for the second mixed term is:
X
s,r,λ1 ,λ2
A2 A†1 =
4
1
9gW
1
1
1
I + II21
+ III21
+ IV21
2(q 2 − M 2 + i)(b2 − M 2 + i) 21
To obtain one expression for the mixed matrix elements we add the two
mixed amplitude terms. Adding the two mixed terms together is easy since the
77
two mixed terms are equal: I12 = I21 , . . . , IV12 = IV21 . Thus:
Imix
≡
IImix
≡
IIImix
≡
IVmix
≡
1
(I12 + I21 )
2
1
(II12 + II21 )
2
1
(III12 + III21 )
2
1
(IV12 + IV21 )
2
So the contribution of the mixed terms is given by:
X
A1 A†2 + A2 A†1
4
9gW
(b2 − M 2 + i)(q 2 − M 2 + i)
=
s,r,λ1 ,λ2
[Imix + IImix + IIImix + IVmix ]
Kinematics
2
Now that we have obtained the expressions for the matrix element |M| in terms
of the momenta, we need to use explicit expressions for these momenta in order
to obtain an expression for the cross section. We choose the kinematics to be in
the center-of-mass frame. The dark matter has mass M = 9.6 TeV, and the W boson has mass MW = 80.4 GeV. We also incorporate the mass difference due to
the splitting between the masses of the dark matter particles in the multiplet:
the charged dark matter particles are heavier than the neutral dark matter
particle. M = Mχ+ − ∆M = Mχ− − ∆M = Mχ++ − 2∆M = Mχ−− − ∆M .
And ∆M = 166 MeV. The kinematics are expressed in the Mandelstam variable
s (using c = h̄ = 1).
DM1 : p1
DM2 : p2
=
=
W − : k1
=
+
W : k2
p
|~
p| = E 2 − M 2
|~k| =
q
=
(E, p~z)
(E, −p~z)
(ω, ~k)
(ω, −~k)
r
M2
1− 2 =E
E
= E
r
E2
−
2
MW
= E
1−
r
2
MW
=E
E2
1−
r
4M 2
≡ Eβ1
s
1−
2
4MW
≡ Eβ2
s
q
q
4M 2
4M 2
with β1 = 1 − s , β2 = 1 − sW where M is the tree-level mass of the
dark matter.
Some equations about the kinematics in the CM-frame (using p2 = E 2 −M 2 )
78
are:
= M2
= M2
s
p1 · p2 =
− M2
2
s s
− β1 β2 cos θ
p1 · k1 =
4 4
s s
p1 · k2 =
+ β1 β2 cos θ
4 4
s s
p2 · k1 =
+ β1 β2 cos θ
4 4
s s
− β1 β2 cos θ
p2 · k2 =
4 4
2
k1 · k1 = MW
2
k2 · k2 = MW
s
2
k1 · k2 =
− MW
2
q = p1 − k1 = k2 − p2
b = p1 − k2 = k1 − p2
p1 · p1
p2 · p2
Cross section
The general equation of the differential cross section for the interaction process
p1 , p2 → k1 , k2 is [26]:
dσ
dΩ
=
CM
1
|~k1 |
|M|
2E1 2E2 |v1 − v2 | 16π 2 ECM
p
~2 with |v1 − v2 | = Ep~11 − E
= 2p
E putting in the relevant values. Applying the
2
equations above to the differential equation gives:
dσ
(h̄c)2 1
(h̄c)2
=
|M|
=
|M|
(A.23)
dΩ CM
8Ep 32π 2
256π 2 Ep
where we inserted the factor (h̄c)2 to get the right units of barn. |M| is given
by:
X
r,s,λ1 ,λ2
|M| =
(q 2
4
9gW
[I1 + II1 + III1 + IV1 ]
− M12 + i)2
4
9gW
[I2 + II2 + III2 + IV2 ]
(b2 − M12 + i)2
4
9gW
+ 2
2
(q − M1 + i)(b2 − M12 + i)
× [Imix + IImix + IIImix + IVmix ]
+
where M1 = M + ∆M = M + 166 MeV/c2 and M is the mass of the neutral
dark matter particle χ0 .
79
To obtain the total cross section, we integrate the differential cross section
over the solid angle
Z 2π
Z π
σ=
dφ
dσ sin θdθ
0
0
We will perform the integration over θ for each term seperately, including
the fraction of the corresponding internal propagator, then take the limit of
→ 0 and perform a series expansion in p (the dark matter momentum). So
(h̄c)2
for term I1 we make the expansion of (dropping the term 256π
2 ) with respect
to momentum p:
Z π
4
9gW
1
lim
2
Ep →0 0 (q − (M + ∆M )2 + i)2
[−8(q · q)(p1 · p2 ) + 16(p1 · q)(p2 · q)] sin θdθ
(A.24)
with E 2 = M 2 + p2 .
From this calculation, we notice that for this choice of reference frame, the
contribution from the A1 A†1 and A2 A†2 terms are equal.
After performing the expansion for every term, we add these terms. Adding
all terms of order 1/p (the lowest order), we see that for this order only the
terms II1 , III1 , II2 and III2 give an (equal) contribution of
4
4
M 2M 4 − 3M 2 MW + 4MW
(h̄c)2 36gW
(A.25)
2 (2M 2 + 2M ∆M − M 2 + ∆M 2 )2 p
128π MW
W
All the other terms, giving a contribution which is proportional to the frac2
M (M 2 −MW
)
cancel.
tion
2
2
(2M 2 +2M ∆M −MW +∆M 2 ) p
The next order in the expansion which is nonzero are the terms of order p.
This time there is no large cancellation and the result is given by a very large
expression. For simplicity we will only present the largest contribution at this
order:
4
(h̄c)2
1536gW
M 10 p
(A.26)
4 M (2M 2 + 2M ∆M − M 2 + ∆M 2 )4
128π MW
W
Using the expression that for small momentum p = M v with v being in units
of c (thus v ranges from 0 to 1) and putting in all the numbers of√the masses and
1
the coupling constants: 1 barn = 10−24 cm2 , α2 = 29
, gW = 4παW = 0.66,
2
2
(h̄c) = 0.389379 GeV mbarn, MW = 80.398 GeV/c2 , ∆M = 166 MeV/c2 ,
M = 9.6 TeV/c2 ; one gets a numerical expression for the cross section:
σ=
4.13 · 10−33
+ 3.97 · 10−29 v cm2
v
(A.27)
The second terms starts to dominate at velocities of order 10−3 . Even for large
velocities (v ≈ 0.1), the first two terms are dominating the cross section.
80
Appendix B
Mathematica notebook for
Sommerfeld enhancement
The Mathematica code for the numerical calculation of the Sommerfeld enhancement for a generic three-state coupled via Yukawa interactions. We use
the NDSolve-code adapted from [52].
a = 10^(-2);
b = 10^3;
epsv = 1/100;
epsphi = 1/10;
epsd = 5/100;
sol3 = NDSolve[{-P1’’[r] -(Exp[-epsphi r]/r)P2[r] == epsv^2 P1[r],
-P2’’[r] - (Exp[-epsphi r]/r)P1[r] - (Exp[-epsphi r]/r)P3[r]
+ epsd^2 P2[r] == epsv^2 P2[r],
-P3’’[r] - (Exp[-epsphi r]/r)P2[r]
- 4 epsd^2 P3[r] == epsv^2 P3[r],
P1[a]==1, P2[a]==1, P3[a]==1, P1’[b]==I*epsv P1[b],
P2’[b]==I*(epsv^2-epsd^2)^(1/2) P2[b],
P3’[b]==I*(epsv^2-4*epsd^2)^(1/2) P3[b]},{P1, P2, P3},{r,a,b},
Method -> {Chasing, ExtraPrecision -> 30}, AccuracyGoal -> 10,
PrecisionGoal -> 10, MaxSteps -> Infinity, MaxStepSize -> 1];
value = Max[Table[Re[P1[r] /. sol3],
{r, IntegerPart[b/2], b - 1}]];
81
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