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\title{Generation of cosmological large lepton asymmetry from a
  rolling scalar field}
%
\author{Masahide Yamaguchi}
%
\affiliation{Research Center for the Early Universe, University of Tokyo,
Tokyo 113-0033, Japan \\ and \\
Physics Department, Brown University, Providence, RI 02912, USA}
%
\date{\today}
%%
\begin{abstract}
  We propose a new scenario to simultaneously explain a large lepton
  asymmetry and a small baryon asymmetry. We consider a rolling scalar
  field and its derivative coupling to the lepton number current. The
  presence of an effective non-zero time derivative of the scalar
  field leads to CPT violation so that the lepton asymmetry can be
  generated even in thermal equilibrium as pointed out by Cohen and
  Kaplan. In this model, the lepton asymmetry varies with time.  In
  particular, we consider the case where it grows with time. The final
  lepton asymmetry is determined by the decoupling of the lepton
  number violating interaction and can be as large as order unity. On
  the other hand, if the decoupling takes place after the electroweak
  phase transition, a small baryon asymmetry is obtained from the
  small lepton asymmetry at that time through sphaleron effects.  We
  construct a model, in which a rolling scalar field is identified
  with a quintessence field.
\end{abstract}

\pacs{98.80.Cq \hspace{8cm} BROWN-HET-1333}
\maketitle

%]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\label{sec:introduction}

The baryon number density dominates over the anti-baryon number
density in our universe. The magnitude of the baryon asymmetry is
mainly estimated by the two different methods. One is big bang
nucleosynthesis (BBN). By comparing predicted primordial abundances of
light elements (D, $^3$He, $^4$He and $^7$Li) with those inferred from
observations, the baryon-to-entropy ratio $\eta$ is estimated as $\eta
\sim 10^{-10}$ \cite{BBN,BBN2}. The other is observations of small
scale anisotropies of the cosmic microwave background radiation (CMB)
\cite{BOOMERANG,MAXIMA,DASI,CBI,Archeops,BM}. However, the
baryon-to-entropy ratios $\eta$ inferred from these two methods do not
necessarily coincide. Furthermore, the baryon-to-entropy ratios $\eta$
inferred from the abundances of different light elements may not
necessarily agree \cite{BBN2,BBN3}.  As observations are improved and
their errors are reduced, such discrepancies may be completely
removed, however it is probable that such small discrepancies are
genuine and suggest additional physics in BBN and the CMB.

An interesting possibility to eliminate such discrepancies is the
presence of a large and positive lepton asymmetry of electron type
\cite{KS}. Roughly speaking, such an asymmetry causes two effects on
the predicted primordial abundance of $^{4}$He. The excess of electron
neutrinos shifts the chemical equilibrium between protons and neutrons
toward protons, which reduces the predicted primordial abundance of
$^{4}$He. On the other hand, the excess of electron neutrinos also
causes increase of the Hubble expansion, which makes the predicted
primordial abundance of $^{4}$He increase. In practice, the former
effect overwhelms the latter so that the presence of a large and
positive electron type asymmetry decreases the predicted primordial
abundance of $^{4}$He, which often solves the discrepancies as
mentioned above. However, it is very difficult for such a large lepton
asymmetry to be compatible with a small baryon asymmetry. This is
mainly because, if we take the sphaleron effects into account, lepton
asymmetry is converted into baryon asymmetry of the same order with
the opposite sign \cite{sphaleron}.

There are several ways to overcome this difficulty. One possibility is
to generate the large lepton asymmetry after electroweak phase
transition but before BBN, which may be realized through oscillations
between active neutrinos and sterile neutrinos \cite{oscillation}.
Another is to disable the sphaleron effects. It was pointed out that
the presence of a large chemical potential prevents restoration of the
electroweak symmetry \cite{nonrestoration}. Based on this
non-restoration mechanism, the generation of a large lepton asymmetry
compatible with the small baryon asymmetry was discussed
\cite{second}.  March-Russell {\it et al.} discussed another
possibility \cite{MRM}.  In their model, a positive electron type
asymmetry but no total lepton asymmetry, that is, $L_{e} = - L_{\mu} >
0$ and $L_{\tau} = 0$, is generated by the Affleck-Dine mechanism for
some flat direction \cite{AD}. Then, the small positive baryon
asymmetry is produced through thermal mass effects of sphaleron
processes.  Recently, Kawasaki, Takahashi, and the present author
discussed another possibility \cite{KTY}, in which a positive electron
type asymmetry but negative total lepton asymmetry is produced by the
Affleck-Dine mechanism and almost all the produced lepton numbers are
absorbed into L-balls. A small amount of negative lepton charges are
evaporated from the L-balls due to thermal effects. These are
converted into the observed small baryon asymmetry by virtue of
sphaleron effects. However, the rest of lepton charges are protected
from sphaleron effects and released into thermal plasmas by the decay
of L-balls before BBN.

In this paper, we discuss another possibility, in which a spontaneous
leptogenesis mechanism proposed by Cohen and Kaplan is used \cite{CK}.
We consider a scalar field, which slow rolls like a quintessence field
and couples derivatively to the lepton number current. The presence of
an effective non-zero time derivative of the scalar field leads to CPT
violation so that the lepton asymmetry can be generated even in
thermal equilibrium. The produced lepton asymmetry varies with time
according to the dynamics of the rolling scalar field. Then, the final
lepton asymmetry is determined by the decoupling of the lepton number
violating interaction and it can be as large as order
unity.\footnote{Note that the decoupling of the lepton number
  violating interaction need not take place before BBN. We have only
  to obtain the large lepton asymmetry at BBN.} On the other hand, in
the case that the lepton asymmetry grows with time, it can be small at
the electroweak phase transition, which is converted into the observed
small baryon asymmetry. In a similar way to \cite{KTY}, we can
generate a positive electron type asymmetry but a negative total
lepton asymmetry according to the coupling constants. Thus, a large
positive electron type asymmetry and the small positive baryon
asymmetry are realized simultaneously.

In the next section, after reviewing the spontaneous
baryo/leptogenesis mechanism proposed by Cohen and Kaplan, we explain
our scenario to produce the desired asymmetries. In Sec III, we give a
model as an example, in which a rolling scalar field is identified
with a quintessence field. Several authors have already discussed
similar scenarios, in which only the present small baryon asymmetry is
explained by a Nambu-Goldstone boson \cite{DF} or a quintessence field
\cite{LFZ,FNT}, and its derivative coupling to a $B$ current or a
$B-L$ current.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Spontaneous leptogenesis and large lepton asymmetry}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\label{sec:spontaneous}

First of all, we introduce the derivative coupling of a rolling scalar
field $\phi$ with the lepton number current $J_{i}^{\mu}$,
%
\beq
  \CL_{\rm eff} = \sum_{i} \frac{c_{i}}{M}\,f(\phi)\, 
                    \del_{\mu}\phi\,J_{i}^{\mu},
\eeq
%
where $i$ denotes the generation, $c_{i}$ are coupling constants,
$f(\phi)$ is a function of $\phi$, and $M$ is the cutoff scale, which
is set to be the reduced Plank mass $M_{G} \sim 10^{18}$ GeV in this
paper. Assuming that $\phi$ is homogeneous,
%
\beq
  \CL_{\rm eff} = \sum_{i} \frac{c_{i}}{M}\,f(\phi)\, 
                    \dot\phi\,n_{L}^{i}
                = \sum_{i} \frac{c_{i}}{M}\,f(\phi)\, 
                    \dot\phi\,(n_{l}^{i} - n_{\bar{l}}^{i}) 
                = \sum_{i} \mu_{i}(t)\,n_{L}^{i},
\eeq
%
where the effective time-dependent chemical potential $\mu_{i}(t)$ is
given by
%
\beq
  \mu_{i}(t) \equiv \frac{c_{i}}{M} f(\phi)\, \dot\phi.
\eeq
%
As pointed out by Cohen and Kaplan, this interaction induces CPT
violation if the time derivative $\dot\phi$ is effectively nonzero,
which generates the lepton asymmetry even in a state of thermal
equilibrium. Then, in thermal equilibrium, the lepton asymmetry
$n_{L}^{i}$, for $\mu_{i} < T$, is given by
%
\beq
  n_{L}^{i} = \frac{g}{6} T^3 \lhk 
                  \frac{\mu_{i}}{T} 
                     + \CO\lkk\lmk\frac{\mu_{i}}{T}\rmk^3\rkk
                                \rhk,
\eeq
%
where $g$ represents the number of degrees of freedom of the fields
corresponding to $n_{L}$. Since the entropy density $s$ is given by $s
= \frac{2\pi}{45}g_{\ast}T^3$ with $g_{\ast}$ the total number of
degrees of freedom for the relativistic particles, the ratio between
the lepton number density and the entropy density is given by
%
\beq
  \frac{n_{L}^{i}}{s} \simeq \frac{15}{4\pi^2}\frac{g}{g_{\ast}}
                                \frac{\mu_{i}}{T}
                        = \frac{15}{4\pi^2}\frac{g}{g_{\ast}}
                           \frac{c_{i}}{MT} f(\phi)\, \dot\phi.
  \label{eq:lepton}
\eeq

The above generation mechanism is effective as long as the lepton
number violating interaction is in thermal equilibrium. Thus, the
final lepton asymmetry is determined by the decoupling temperature
$T_{D}$ of the lepton number violating interaction. Such a lepton
number violating interaction is, for example, given by a dimension
five operator,
%
\beq
  \CL_{\not L} = \frac{2}{v}\,l\,l\,HH + {\rm h.c.},
\eeq
%
where $v$ is the scale characterizing the lepton number violation, $l$
and $H$ represent the left handed lepton doublet and the Higgs
doublet. For $T < m_{H}$, the interaction rate $\Gamma_{\not L}$ is
given by $\Gamma_{\not L} \sim m_{H}^3 / v^2$, which yields the
decoupling temperature $T_{D} \sim 0.1 \sqrt{\Gamma_{\not L} M_{G}}
\sim 10^7 m_{\nu}$. Here, $m_{H}$ is the Higgs mass, $m_{\nu}$ is the
neutrino mass given by $m_{\nu} \sim \la H \ra^2 / v$, and $\la H \ra
\sim m_{H} \sim 100$ GeV. For $m_{\nu} \sim 10^{-1}$ eV, the
decoupling temperature $T_{D}$ becomes $T_{D} \sim 1$ MeV. However, we
do not specify the lepton number violating interaction in this paper.
Instead, for our purpose, we have only to demand that the decoupling
of the lepton number violating interaction takes place after the
electroweak phase transition.

If $c_1$ is positive but the total sum of $c_{i}$ is negative,
positive electron type asymmetry but negative total lepton asymmetry
is realized. Depending on the values of the scalar field $\phi$ and
its time derivative $\dot\phi$, the absolute magnitude of the lepton
asymmetry can be as large as order unity. On the other hand, the
lepton asymmetry at the electroweak phase transition can be as small
as the order of $10^{-10}$ according to the dynamics of the scalar
field. Then, a part of the lepton asymmetry at that time is converted
into the baryon asymmetry through the sphaleron processes, which can
be estimated as \cite{sphaleron}
%
\beq
  \eta = \frac{n_B}{s} \simeq \left. 
                  - \frac{8}{23} \sum_{i} \frac{n_{L}^{i}}{s}\,
                  \right|_{T=T_{\rm EW}}     
            = - c \left.  
                    \frac{30}{23\pi^2}\frac{g}{g_{\ast}}
                           \frac{1}{MT} f(\phi)\, \dot\phi\,
                  \right|_{T=T_{\rm EW}},      
\eeq
%
where $c = \sum_{i} c_{i}$, $T_{\rm EW}$ represents the temperature at
the electroweak phase transition, and we have assumed the standard
model with two Higgs doublets and three generations. Thus, the large
positive electron type asymmetry and the small positive baryon
asymmetry are obtained.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Quintessential baryo/leptogenesis model}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\label{sec:quintessential}

In this section, as an example, we consider the case in which the
scalar field is identified with a quintessential field. Though a lot
of quintessence models are proposed
\cite{RP,FHW,CSN,TW,CDS,BS,AS,COY,AMS}, we take a simple model, whose
kinetic term is canonical and whose potential is given by the inverse
power law \cite{RP,ZWS,LS,Binetruy}.

We consider the quintessence field $\phi$ with the following potential
of inverse power law,
%
\beq
  V(\phi) = \frac{5}{72} \frac{M_s^8}{\phi^4}.
\eeq
%
Here $M_s$ is a new scale, which is determined by requiring that this
potential energy is responsible for the present acceleration of the
universe. In this case, there is a tracker solution given by
\cite{RP,ZWS,LS},
%
\beq
 \left\{
 \begin{array}{l}
 \phi = M_s^{\frac43}\,t^{\frac13}\, \\
 \dot\phi = \frac13\,M_s^{\frac43}\,t^{-\frac23}, \qquad ({\rm RD})
 \end{array}\right.
\eeq
%
\beq
 \left\{
 \begin{array}{l}
 \phi = \lmk \frac58 \rmk^{\frac16}\,M_s^{\frac43}\,t^{\frac13}\, \\
 \dot\phi = \frac13 \lmk \frac58 \rmk^{\frac16}\,
              M_s^{\frac43}\,t^{-\frac23}, \qquad ({\rm MD})
 \end{array}\right.
\eeq
%
where RD applies in the radiation dominated universe, and MD applies
in the matter dominated universe. The present value of the
quintessence field $\phi_0$ is determined by requiring that $V''\sim
H^2$, and is roughly given by $\phi_0 \sim M_{G}$, which yields $M_s
\sim 10^3$ GeV. Here the derivative of the potential are taken with
respect to $\phi$. On the other hand, the value of the quintessence
field at BBN is given by
%
\beq 
  \phi(T=T_{\rm BBN}) 
       \sim M_s^{\frac43}\,t_{\rm BBN}^{\frac13}
       \sim \lmk \frac{M_s^4 M_{G}}{T_{\rm BBN}^2} \rmk^{\frac13}
       \sim 10^{12}\,{\rm GeV}
\eeq
%
with $T_{\rm BBN} \sim 1$ MeV.

Now, we consider the derivative coupling of the quintessence field
$\phi$ with the lepton number current $J_{i}^{\mu}$,
%
\beq
  \CL_{\rm eff} = \sum_{i} \frac{c_{i}}{M}\,f(\phi)\, 
                    \del_{\mu}\phi\,J_{i}^{\mu}.
  \label{eq:coupling}
\eeq
%
Here we assume that $f(\phi)$ is given by
%
\beq
  f(\phi) = \lmk \frac{\phi(\phi + M)^2}{\phi^3+M_s M^2} \rmk^3.
\eeq
%
Such a coupling may be obtained by the extension of the shift symmetry
introduced in \cite{KYY,YY}. This function $f(\phi)$ is roughly
classified into three regions,
\beq
  f(\phi) \simeq \left\{
             \begin{array}{ll}
              1, & \quad {\rm for} \quad M \lesssim \phi \\
              \lmk \frac{M}{\phi} \rmk^6, 
                 & \quad {\rm for} \quad 
                   M_s^{\frac13}M^{\frac23} \lesssim \phi \lesssim M \\   
              \lmk \frac{\phi}{M_s} \rmk^3, 
                 & \quad {\rm for} \quad
                   \phi \lesssim M_s^{\frac13}M^{\frac23}
             \end{array}\right.
\eeq 
%
with $M_s^{\frac13}M^{\frac23} \simeq 10^{13}$ GeV for $M = M_{G}
\simeq 10^{18}$ GeV. Thus, we can use the third region for the
function $f(\phi)$ at least until BBN.

Using Eq. (\ref{eq:lepton}), the lepton asymmetry is estimated as
%
\beq
  \frac{n_{L}^{i}}{s} 
    \sim \frac{c_{i}}{MT} f(\phi)\, \dot\phi
    \sim c_{i} \lmk \frac{M_s^7 M_{G}}{M^3 T^5} \rmk^{\frac13}
    \propto T^{-\frac53}.
\eeq                             
%
By inserting $T_{\rm EW} \sim 1$ TeV and $M = M_{G}$ into the above
equation, the total lepton asymmetry at the electroweak phase
transition is given by
%
\beq
  \left. \frac{n_{L}}{s} \right|_{T = T_{\rm EW}} 
    \sim c \,10^{-10}.
\eeq     
%
Thus, taking $c = \sum_{i} c_{i} = - \CO(1)$, the present baryon
asymmetry is given by
%
\beq
  \eta = \frac{n_B}{s} \sim 10^{-10}.
\eeq 

On the other hand, if the decoupling temperature $T_{D}$ of the lepton
number violating interaction is nearly equal or lower than $T_{\rm
  BBN}$, the lepton asymmetry at BBN is given by
%
\beq
  \left. \frac{n_{L}^{i}}{s} \right|_{T = T_{\rm BBN}} 
    \sim c_{i}.
\eeq
%
Taking $c_{1} = \CO(1)$, the electron type asymmetry becomes positive
and of order unity at BBN.

The equation of motion for the quintessence field $\phi$ is modified
due to the presence of the derivative coupling in Eq.
(\ref{eq:coupling}) becoming
%
\beq
  \ddot{\phi} + 3H\dot\phi +V'(\phi) = 
     - c \frac{f(\phi)}{M} (\dot{n} + 3Hn)
     \sim \frac{gc^2}{6} \frac{T^2}{M} f(\phi) 
             (\ddot\phi + 3H\dot\phi). 
\eeq
%
We can easily show that the correction to the equation of motion for
$\phi$ is negligible for $T \ll M = M_{G}$.

Finally, we must check the present constraint on the derivative
coupling given in Eq. (\ref{eq:coupling}) with those from laboratory
experiments. For the time component, the coefficient $\mu(t_0)$ is
constrained as $|\mu(t_0)| \lesssim 10^{-25}$ GeV \cite{CKMP}. Here
$t_0$ is the present age of the universe. In our model, $\mu(t_0)$ is
given by
%
\beq
  |\mu(t_0)| \sim \frac{|c|}{M} 10^{-25}\,{\rm GeV}^2
           \sim 10^{-43} {\rm GeV}
           \ll 10^{-25} {\rm GeV}
\eeq
%
for $f(\phi) \sim 1$, $|c|=\CO(1)$, and $M=M_{G}$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Discussion and conclusions}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\label{sec:con}

In this paper, we considered the large lepton asymmetry from a rolling
scalar field. Considering the derivative coupling of the scalar field
to the lepton number current, the presence of an effective non-zero
time derivative of the scalar field leads to CPT violation, which
generates the lepton asymmetry even in thermal equilibrium. This
lepton asymmetry changes with time. So, depending on the dynamics of
the scalar field, it is possible that the lepton asymmetry is small at
the electroweak phase transition but large at BBN. A part of the
lepton asymmetry is converted into the baryon asymmetry with the
opposite sign through sphaleron effects. We pointed out that by
choosing the sign of the coupling constants properly, a large positive
lepton asymmetry of electron type and a small positive baryon
asymmetry can be realized simultaneously.

As an example, we considered the rolling scalar field, which is
identified with a quintessence field. By considering the potential of
the inverse power law, we showed that this model manifests just such
asymmetries. However, the coupling to the lepton number current is a
bit complicated, though such a coupling may be realized via some
symmetry argument. This is mainly because the time derivative of the
tracker solution of the quintessence field, generally speaking,
decreases with time. For a simple shape of the Lagrangian, in the best
case, it is a constant in k-essence \cite{COY}. If we abandon the
tracking behavior, there may be evolution of the quintessence field,
in which the magnitude of the time derivative of the quintessence
field at BBN is much larger than that at the electroweak phase
transition. For such evolution, the model may work, in which the
coupling to the lepton number current is simple, that is, $f(\phi)
\equiv 1$. Such a possibility will be considered in a further
publication.

Finally, we comment on the recent discussion of neutrino oscillations
around BBN. It was pointed out that complete or partial equilibrium
between all active neutrinos may be accomplished through neutrino
oscillations in the presence of neutrino chemical potentials,
depending on neutrino oscillation parameters \cite{equilibrium}. In
case of partial equilibrium, we need not change our scenario. Complete
equilibrium may spoil our scenario. In that case, if we choose the
coupling constants $c_{i}$ as $c_{1} = - c_{2} = - c_{3}$ for some
symmetry reason, our scenario still works. In such a case, no mixing
takes place because of the cancellation, as pointed out in
\cite{equilibrium}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{ACKNOWLEDGMENTS}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

M.Y. is much grateful to W. Kelly for useful comments and correcting
the English. M.Y. is also grateful to T. Chiba and F. Takahashi for
useful comments. M.Y. is partially supported by the Japanese
Grant-in-Aid for Scientific Research from the Ministry of Education,
Culture, Sports, Science, and Technology.

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\end{document}



