Abstract \\ \vspace{1cm} %PACS number: 98.80.Cq
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\begin{document}
%\twocolumn[
\centerline{\bf July 2001 \hfill hep-ph/0109080
} \vspace{3.5cm}

\centerline{\Large\bf Large Gauged Q Balls} \vspace{2.cm}

\centerline{ {\large \bf K.N. Anagnostopoulos$^{a}$}\footnote {e-mail
address : konstant@physics.uoc.gr}{\large ,} {\large \bf
M. Axenides$^{b}$}\footnote {e-mail address :
axenides@mail.demokritos.gr }{\large ,} {\large \bf E.G.
Floratos$^{b,c}$}\footnote {e-mail address :
manolis@mail.demokritos.gr }{\large ,} {\large and} {\large\bf N.
Tetradis$^{a,c}$}\footnote {e-mail address : tetradis@physics.uoc.gr}
} \vspace{0.3cm} \centerline{\em a) Department of Physics, University
of Crete, GR-71003 Heraklion, Greece}
\vspace{0.2cm} \centerline{\em b) Institure of Nuclear Physics,
N.C.R.P.S. Demokritos, GR-15310 Athens, Greece} \vspace{0.2cm}
\centerline{\em c) Nuclear and Particle Physics Sector, University of
Athens, GR-15771 Athens, Greece} \vspace{3.cm}

\centerline{\large\bf Abstract}
\begin{quote}\large\indent
We study Q-balls associated with local $U(1)$ symmetries.  Such
Q-balls are expected to become unstable for large values of their
charge because of the repulsion mediated by the gauge force.  We
consider the possibility that the repulsion is eliminated through the
presence in the interior of the Q-ball of fermions with charge
opposite to that of the scalar condensate. Another possibility is that
two scalar condensates of opposite charge form in the interior.  We
demonstrate that both these scenaria can lead to the existence of
classically stable, large, gauged Q-balls.  We present numerical
solutions, as well as an analytical treatment of the ``thin-wall''
limit.



\end{quote}
\vspace{1cm}
%]

\newpage

\section{Introduction}
Non-topological solitons named Q-balls can appear in scalar field theories
with $U(1)$ symmetries \cite{coleman}.
(For a review of the early literature see ref. \cite{leepang}.)
These objects can be viewed as coherent states of the scalar field
with fixed total $U(1)$ charge. The case of global $U(1)$
symmetries has attracted much attention. The reason is the presence
of such symmetries in the Standard Model, related to baryonic or
leptonic charge. In supersymmetric extensions of the Standard Model,
the scalar superpartners of baryons or leptons can form coherent
states with fixed baryon or lepton number, making the existence
of Q-balls possible. Their properties \cite{existence,iiro},
cosmological origin
\cite{formation} and experimental implications \cite{experiment}
have been the subject of several recent studies.
Non-abelian global symmetries can also lead to the existence of
Q-balls \cite{nonabelian}.

We are interested in the less popular case of
Q-balls resulting from local $U(1)$ symmetries \cite{axen,local}.
Such Q-balls become unstable for large values of their charge
because of the repulsion mediated by the gauge force.
However, small Q-balls can still exist.
A possibility that has not been considered before is that
the repulsion is eliminated through the presence in the interior
of the Q-ball of fermions with charge opposite to that of
the scalar condensate.
The fermions must carry
an additional conserved quantum number that prevents their
annihilation against the condensate. This scenario can
lead to the existence of large Q-balls. The fermion gas may also be replaced
by another scalar condensate, of opposite charge to the first, such that
the interior of the Q-ball remains neutral.

In the following we discuss in detail the above scenaria in the context
of a toy model. We show that arbitrarily large Q-balls can exist
and examine the constraints imposed on the parameters by the requirement
of classical stability.
%We also discuss charge evaporation from the surface and
%demonstrate that it induces only small modifications to the basic picture.

\section{Small Gauged Q-Balls}
For completeness we summarize briefly the basic properties of gauged Q-balls
in a toy model (see ref. \cite{local} for the details).
We consider a complex scalar field
$\phi(\rv,t)=f(\rv,t)\,\exp(-i\,\theta(\rv,t))/
\sqrt{2}$, coupled to an Abelian gauge field
$A^\mu$. The Lagrangian density is
\be
{\cal L} = \frac{1}{2} \partial_\mu f\partial^\mu f+ \frac{1}{2} f^2
\left( \partial_\mu\theta-e A_\mu \right)^2 - U(f)
-\frac{1}{4} F_{\mu\nu} F^{\mu\nu}.
\label{toylag} \ee
The total $U(1)$ charge of a particular field configuration is
\be
Q_\phi=\int d^3\rv f^2 \left( \dot\theta -e A_0 \right).
\label{charge0} \ee
Without loss of generality we assume $e,Q_\phi \geq 0$ in the following.
The value $e=0$ leads
to decoupling of the scalar from the gauge field.

We consider a spherically symmetric ansatz
that neglects the spatial components of the
gauge field $A_i=0,~i=1,2,3$
and assumes $\theta = \omega t$ \cite{coleman}. The component
$A_0$ of the gauge field corresponds to the
electrostatic potential that is responsible for the repulsive force
destabilizing the Q-ball.
The equations of motion for the fields are
\beq
f''+\frac{2}{r}f'+fg^2-\frac{dU(f)}{df}&=&0
\label{eom1} \\
g''+\frac{2}{r}g'-e^2f^2g&=&0,
\label{eom2} \eeq
with $r=|\rv|$, $g(r)=\omega-eA_0(r)$, and primes denoting derivatives
with respect to $r$.
The total charge and energy are
\beq
Q_\phi &=& \int dV \, \rho_\phi=4 \pi \int  r^2\, dr \, f^2 g
\label{charge} \\
E&=& \int dV \, \ex=4 \pi \int r^2\, dr \left[
\frac{1}{2} f'^2 + \frac{1}{2e^2}g'^2 +\frac{1}{2} f^2 g^2
+U(f)\right].
%= \frac{1}{2} \omega Q_\phi + \frac{4\pi}{3} R^3 U(F).
\label{energy}
\eeq

The Q-ball solution of the equations of motion
involves
an almost constant non-zero scalar field $f(r)=F$
in the interior of the Q-ball,
which moves quickly to zero (the vacuum value) at the surface.
We are interested in the limit in which
the radius $R$ of the Q-ball is much larger than the thickness of its
surface and the total charge can be big.
In this limit and for
$eFR \ll 1$,
the energy can be expressed as \cite{local}
\be
E= Q_\phi \left[ \frac{2\,U(F)}{F^2} \right]^{1/2}
\left[ 1+ \frac{C^{2/3}}{5} \right]=
 Q_\phi \left[ \frac{2\,U(F)}{F^2} \right]^{1/2}
+\frac{3e^2Q_\phi^2}{20\pi R},
\label{approx2} \ee
with
\beq
R&=& \left[
\frac{3Q_\phi}{4\pi F\sqrt{2\, U(F)}} \right]^{1/3}
\left[ 1+ \frac{C^{2/3}}{45} \right]
\label{approx1a} \\
C&=&\frac{3e^3Q_\phi}{4\pi}
 \sqrt{\frac{F^4}{2U(F)}}.
\label{approx1b} \eeq


The ratio $E/Q_\phi$ increases with $Q_\phi$ because of the
presence of the electrostatic term
in eq. (\ref{approx2}).
This means that large Q-balls are unstable and
tend to evaporate
scalar particles from their surface
in order to increase their binding energy.
For small $Q_\phi$ the above expressions are not applicable. Numerical
solutions indicate that
the ratio $E/Q_\phi$ becomes large again because of the contribution
from the field derivative terms that we neglected.
Therefore, there is a value
$\left( Q_\phi \right)_{min}$ for which $E/Q_\phi$ is minimized.
Classical stability requires that
$\left( E/Q_\phi \right)_{min} < d^2U(0)/df^2$ (assuming that the
absolute minimum of the potential is at $f=0$), so that
the Q-ball does not disintegrate into scalar particles of unit charge.



\section{Large Gauged Q-Balls with fermions}
It is clear from the above discussion that gauged
Q-balls with very large $Q_\phi$ become unstable because of electrostatic
repulsion. One possibility that could remedy this problem is that
fermions with charge
opposite to that of the scalar background neutralize the
electrostatic field and eliminate the repulsion.
%These fermions carry
%an additional conserved quantum number that forbids their annihilation
%against the background.
A model that realizes this scenario has a Lagrangian density
\be
{\cal L} = \frac{1}{2} \partial_\mu f\partial^\mu f+ \frac{1}{2}
f^2 \left( \partial_\mu\theta-e A_\mu \right)^2 - U(f) +i
\bar{\psi}_\alpha \gamma^\mu \left( \partial_\mu +ie' A_\mu
\right) \psi^\alpha -\frac{1}{4} F_{\mu\nu} F^{\mu\nu}
\label{toyflag} \ee
In the absence of Yukawa couplings, the scalar
and fermionic fields carry independent conserved $U(1)$ charges. A
linear combination of these charges is gauged while the orthogonal
one remains global. We assume that there are $N$ fermionic degrees
of freedom, labelled by $\alpha=1...N$, with gauge coupling $e'$
and negligible mass. Realistic scenaria could involve condensates
of electrically charged mesonic fields, with characteristic scales
for their potentials ${\cal O}$(100 MeV -- 1 GeV), and Higgs or
squark fields, with characteristic scales ${\cal O}$(100 GeV -- 1
TeV). In all these cases, neutralizing fermions, such as
electrons, can be considered effectively massless.

The equation of motion (\ref{eom1}) is not altered by the presence
of fermions, but eq. (\ref{eom2}) becomes
\be
g''+\frac{2}{r}g'-e^2f^2g- e e' \psi^\dagger_\alpha \psi^\alpha=0.
\label{eom3} \ee
Instead of solving the
Dirac equation,  we approximate the fermions as
a non-interacting Fermi gas with position dependent density. This is
the Thomas-Fermi approximation \cite{schiff}.
The fermionic $U(1)$ charge and energy density are
\be
\left\langle \psi^\dagger_\alpha  \psi^\alpha \right\rangle
= N \rho_\psi = N \frac{k_F^3}{3 \pi^2},
~~~~~~~~~~~~~~
\left\langle \psi^\dagger_\alpha
\left( -i \vec{\alpha}\cdot \vec{\nabla} \right) \psi^\alpha \right\rangle
=N \ex_\psi = N \frac{k_F^4}{4 \pi^2},
\label{erhopsi} \ee
in terms of the Fermi momentum $k_F$.
The Dirac equation for a fermion near the Fermi surface results in the
expression
\be
\mu_\psi  = k_F(r) + e' A_0(r)
= k_F(r) + \frac{e'}{e} \left (\omega -g(r) \right).
\label{mupsi} \ee
We see that $\mu_\psi$ can be interpreted as the chemical potential,
i.e. the energy cost in
order to add an extra fermion on the top of the Fermi sea. The
fermions rearrange themselves so that $\mu_\psi$ is position
independent. It is convenient to define the gauge-invariant
chemical potential
\be
{\tilde{\mu}} = \mu - \frac{e'}{e} \omega
= k_F(r) - \frac{e'}{e} g(r).
\label{mutpsi} \ee

The total
energy is now given by
\be
E= 4 \pi \int r^2\, dr \left[
\frac{1}{2} f'^2 + \frac{1}{2e^2}g'^2 +\frac{1}{2} f^2 g^2
+U(f) + N \ex_\psi +
\left(
\vec{E} \cdot \vec{\nabla} A_0 + N e' \rho_\psi A_0 + e \rho_\phi A_0
\right)
\right],
\label{energyf} \ee
where $\rho_\phi$ is given by eq. (\ref{charge}), $\rho_\psi$ by the first of
eqs. (\ref{erhopsi})  and $\vec{E}$
is the electric field.
The equations of motion can be obtained by minimizing the energy under
constant scalar and fermionic charge. This can be achieved through the
use of Lagrange multipliers $\omega$ and $\mu_\psi$. Minimization of
$E - \omega \int \rho_\phi
 \,dV - \mu_\psi N \int \rho_\psi\, dV$ with respect to
$A_0$, $f$ and $k_F$ results in eqs. (\ref{eom3}), (\ref{eom1}) and
(\ref{mupsi}), respectively.
Finally, the quantity in parentheses in the rhs of eq. (\ref{energy})
vanishes through the application of Gauss' law, eq. (\ref{eom3}).

\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=pplot01.eps,width=12cm}}
\caption{The magnitude of the
scalar field $f$ as a function of the radial distance $r$ for
Q-balls of increasing size.
}
\label{fig1}
\end{figure}


The above considerations provide a simple method for the
determination of
the properties of large Q-balls. We are interested in the ``thin-wall''
limit, in which the effects of the surface of the Q-ball may be neglected.
(A more careful discussion of the validity of this approximation is given
in the next section.) In this limit, the total energy is given by
\be
E = \left( \frac{1}{2}f^2 g^2 + U(f) + N \frac{k_F^4}{4 \pi^2} \right) V,
\label{enthin} \ee
with $V$ the volume of the Q-ball.
The charges of the scalar condensate and the fermions are
\be
Q_\phi = f^2 g V, ~~~~~~~~~~~~~~~~~~~ Q_\psi = N \frac{k_F^3}{3
\pi^2} V. \label{charges} \ee In terms of the constant scalar and
fermion charges $Q_\phi$ and $Q_\psi$ the total energy to be
minimized is given by:
\be
E= \frac{1}{2} \frac{Q_\phi^2}{f^2 V} + U(f) V +
\frac{(3 \pi^2)^{4/3}}{4 \pi^2 N^{1/3}} \,\,\frac{Q_\psi^{4/3}}{V^{1/3}}.
\label{enthin2} \ee

Minimization with respect to $f$ and use of the first of eqs. (\ref{charges})
gives
\be
fg^2 = \frac{dU(f)}{df}\equiv U^{\prime}. \label{con2} \ee This
relation could have been obtained by requiring that eq.
(\ref{eom1}) be satisfied for constant fields. The existence of
such a solution for eq. (\ref{eom3}) leads to
\be
e f^2 g = - e' N \frac{k_F^3}{3 \pi^2}. \label{con3} \ee This
implies
 \be
e'Q_\psi+eQ_\phi=0 \label{neutrality} \ee and guarantees the
electric neutrality of the interior of the Q-ball.

We can also obtain the equilibrium volume of our large fermion
Q-ball. It is given by \be V = \frac{Q_\phi}{ \sqrt{f^3
U^{\prime}}}\label{vol}\ee Minimization of eq. (\ref{enthin}) with
respect to $V$ results in the relation
\be
U(f)=\frac{1}{2}f^2 g^2 + \frac{1}{3} N \frac{k_F^4}{4 \pi^2}=
\frac{1}{2}f^2 g^2 + \frac{1}{3} \epsilon_\psi. \label{con1} \ee
It can be put in a more convenient equivalent form, for which the
scaling between $Q_\phi$ and $V$ is explicit. Expressed solely in
terms of $f$ it takes the form
\be
2U = f U^{\prime} + \frac{\left(3\pi^2 \right)^{4/3}}{6
\pi^2}\left| \frac{e}{e'}\right|^{4/3} \frac{1}{N^{1/3}}f^2 \left[
U^{\prime} \right]^{2/3}. \label{final}
 \ee

 Equations (\ref{con2}),(\ref{con3}) and (\ref{final})  uniquely determine the values
of $f$, $g$, $k_F$ in the interior of a large Q-ball. As a
consequence, $\mut$ can be also specified  through eq.
(\ref{mutpsi}). It is now obvious that the total energy scales
linearly with $Q_\phi$ for a given set of values for $f$ and $g$
\be
\frac{E}{Q_\phi} = \left[  \frac{1}{2}\sqrt{\frac{U^\prime}{f}} +
\frac{U}{\sqrt{f^3 U^\prime}} + \frac{\left(3\pi^2
\right)^{4/3}}{4\pi^2}\left| \frac{e}{e'}\right|^{4/3}
\frac{1}{N^{1/3}} \left(f^3 U^\prime\right)^{1/6}\right].
\label{Energy}\ee

For massless fermions, the stability condition $\min (E/Q_\phi) <
\sqrt{U''(0)}$ guarantees that a large gauged Q-ball cannot
disintegrate into a collection of free particles. One could also
consider the possibility that a $\phi$ particle is surrounded by a
``cloud'' of fermions (or the other way around), so that the
resulting ``atom'' is approximately neutral. A collection of such
states would probably be energetically favourable to a collection
of free particles, due to the electrostatic attraction. However,
for couplings $e,|e'| \lta 1$, the electrostatic binding energy is
expected to be much smaller than the mass of the free scalars,
similarly to the situation in normal atoms. For this reason, the
above relation gives a sufficiently accurate criterion for the
classical stability of Q-balls.



 In the limit $e \to 0$, eqs. (\ref{con2})--(\ref{con1}) give
$k_F=0$ and the well-known conditions for the existence of global
Q-balls are reproduced : $\omega^2= E^2/Q^2= 2U(f)/f^2=
U^{\prime}/f$ \cite{coleman}.



\begin{figure}[t]

%\vspace{1.cm}
\centerline{\psfig{figure=pplot02.eps,width=12cm}}
\caption{The electric
field $E$ as a function of the radial distance $r$ for
Q-balls of increasing size.
}
\label{fig2}
\end{figure}

\section{Numerical solutions}

In this section we present numerical solutions of the
equations of motion (\ref{eom3}), (\ref{eom1}) and
(\ref{mupsi}). The two differential equations require
four boundary conditions. We impose
$f'(r=0)=g'(r=0)=0$, so that there are no singularities at
the center of the Q-balls. We also impose
$f(r=\infty)=0$ and $g'(r=\infty)=0$, so that the solutions
outside the Q-balls
correspond to the normal vacuum.
We use a potential of the form
$U(f)=f^2/2-f^4/4+\lx^2 f^6/6$, in order to make comparisons with
the results of ref. \cite{local}. For the same reason
we choose $\lx^2=0.2$ and $e=0.1$. We assume that there are $N=10$
fermionic species of unit charge $e'=-0.2$. A large number of
species results in a small fermionic kinetic
energy that helps to keep the Q-balls classically stable.
Moreover, values $N={\cal O}(10)$ are typical of realistic
theories, such as the MSSM.
All dimensionful quantities are
considered to be renormalized with respect to the mass term
in the potential (set equal to 1).

Q-ball solutions of various sizes are obtained by
fixing the value of $\mu$ and varying $\omega$.
The chemical potential $\mu$ is assumed to be negative. The reason
is apparent through eq. (\ref{mupsi}). If we would like to interpret
$A_0$ as the electrostatic potential, we must choose a gauge
such that $A_0(r) \propto r^{-1}$ for large $r$.
By taking $\mu$ negative, we expect that $k_F$ will become 0 at a finite
radial distance. We assume that there are no fermions at larger distances, so
that eq. (\ref{mupsi}) is inapplicable.
Instead we impose $k_F=0$ in eq. (\ref{eom3}), which results in the expected
behaviour for $A_0(r)$. Positive values of $\mu$ would result in a non-zero
fermionic density at arbitrary distances from the center of the Q-ball.


In figs. 1--3 we present a series of Q-ball solutions of increasing
size. In fig. 1 we plot the scalar field $f$ as a function of the
radial distance from the center of the Q-ball. We observe that $f(r)$
behaves as a step function to a very good approximation,
even for fairly small Q-balls. In fig. 2 we depict the magnitude of
the electric field for the same solutions. The smallest Q-ball has
the strongest electric field. The field vanishes at the center for
symmetry reasons, but quickly grows with $r$. For large enough $r$ it
falls $\propto r^{-2}$. For larger Q-balls the electric field is zero in
the interior, because of the cancellation of the charge of the
scalar field by that of the fermionic gas. The electric field is non-zero
near the surface, while it falls again $\propto r^{-2}$ for large
$r$.




\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=pplot03.eps,width=12cm}}
\caption{The charge densities of
the scalar condensate and the fermionic gas
as a function of the radial distance $r$ for
Q-balls of increasing size.
}
\label{fig3}
\end{figure}

In fig. 3 we plot the scalar and fermionic charge densities as a function
of the radial distance. For the smallest Q-ball the fermions
are not capable to neutralize the interior. There is a mismatch between the
scalar and fermionic densities. Moreover, there is a large concentration
of scalar charge near the surface. This is a result of the electrostatic
repulsion that forces the positive unit charges to maximize the distance
between themselves. As the size of the Q-ball increases, the charge densities
in the interior become opposite to each other. For large Q-balls
their magnitude is independent of the radius.
This is the ``thin-wall'' limit we discussed in the previous section.
The values of $f$, $g$ and $k_F$ in the interior (and, therefore,
the charge densities) should be uniquely determined
by eqs. (\ref{con2})--(\ref{con1}).
We have checked that $f(r=0)$, $g(r=0)$ and $k_F(r=0)$ for the large
Q-ball solutions
depicted in figs. 1--3 satisfy eqs. (\ref{con2})--(\ref{con1}) with an
accuracy better than 1\%.

A particular question merits some discussion at this point. From figs. 1--3
one could infer naively
that the profile of the surface is constant for large Q-balls
and merely displaced at different radii $R$. This would mean that the electric
field is the same near the surface for all large Q-balls. Such a field can
only be produced if the net surface charge density is constant and the
net surface charge scales $Q_s \propto R^2$. One implication
would be that the electrostatic contribution to the total energy of the
system
$\sim Q_s^2/R \propto R^3$ would scale proportionally to the
volume. As a result, our assumption
that the surface effects are negligible in the ``thin-wall'' limit would
be invalid.
However, the numerical solutions do not confirm this picture.
The shape of the numerical solution varies slightly at the surface
even for large Q-balls. The electric field at the surface
becomes smaller for increasing radius, while the fermionic density is modified
appropriately. Numerically
we have not identified any residual surface effect. Moreover,
we expect that a more rigorous treatment of the fermionic cloud that surrounds
the Q-ball would support this conclusion. Our simple approximation of the
fermions as a non-interacting gas is adequate for the interior but very crude
near the surface. We expect
that, in a more careful treatment of the surface, a surrounding
fermionic cloud will neutralize completely the Q-ball
(similarly to the neutralization of atoms).
In this picture, the surface effects would be even less important in the
``thin-wall'' limit.



\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=pplot05.eps,width=12cm}}
\caption{The energy to charge ratio $E/Q_\phi$ as a function of the
scalar charge $Q_\phi$ of the Q-ball.
}
\label{fig4}
\end{figure}



In fig. 4 we plot the energy to charge ratio as a function of
the charge $Q_\phi$ of the scalar condensate.
The fermionic charge $Q_\psi$ may differ substantially from $Q_\phi$ for
small Q-balls. As we have assumed that the fermions are massless and
normalized everything with respect to the scalar mass term,
the classical stability requirement is
$E/Q_\phi < 1$. In fig. 4 we observe a series of curves that correspond to
different (negative) values of $\mu$. The fermionic content of a
small Q-ball is controlled through $\mu$ and, for
the same $Q_\phi$, the various curves have
different ratios $Q_\psi/Q_\phi$.
In fig. 5 we plot $Q_\psi/Q_\phi$ as a function of $Q_\phi$
for the same range values of $\mu$ as in fig. 4.
We observe that $Q_\psi/Q_\phi$
tends to increase with decreasing $|\mu|$.
As the ratio $E/Q_\phi$ decreases for decreasing $|\mu|$,
we conclude that, for fixed $Q_\phi$,
the Q-balls become more stable by absorbing fermions
and increasing their fermionic content.
A limit to the process of fermion accretion
is set by the requirement of a positive chemical
potential, so that the fermions are bound to the Q-ball.


\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=pplot07.eps,width=12cm}}
\caption{The ratio of scalar to fermionic charge
$Q_\psi/Q_\phi$ as a function of the
scalar charge $Q_\phi$ of the Q-ball.
}
\label{fig5}
\end{figure}



For $Q_\phi\lta 10^4$, the fermionic content becomes negligible
and we obtain the gauged Q-balls of ref. \cite{local}. For
$Q_\phi\lta 10^3$ the ratio $E/Q_\phi$ increases because of the
contribution of the derivative terms to the energy \cite{local}.
For $Q_\phi\gta 10^7$, $Q_\psi/Q_\phi=1$ and the energy to charge
ratio has a very weak dependence on $\mu$ and $Q_\phi$. In this
region the "thin-wall" approximation is valid. The gauge invariant
quantity $\mut$ of eq. (\ref{mutpsi}) is almost constant.
Asymptotically for $Q_\phi \to \infty$, the properties of the
Q-balls are completely determined by the values of $f$, $g$ and
$k_F$ in the interior as given by the solution of eqs.
(\ref{con2})--(\ref{con1}).

\begin{table}[b]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
& $N=1$ & $N=2$  & $N=4$  & $N=6$   & $N=10$ & $N=15$ & $N=20$ \\ \hline
$e'=-0.1$ & 1.954 & 1.654 & 1.403 & 1.277 & 1.135 & 1.036 & 0.972 \\ \hline
$e'=-0.2$ & 1.021 & 0.878 & 0.759 & 0.700 & 0.635 & 0.589 & 0.560 \\ \hline
$e'=-0.3$ & 0.724 & 0.633 & 0.559 & 0.522 & 0.481 & 0.453 & 0.435 \\ \hline
\end{tabular}
\caption{
The energy to charge ratio $E/Q_\phi$ in the ``thin-wall'' limit, for
various values of the parameters $N$ and $e'$.
}
\label{param}
\end{center}
\end{table}


For our choice of parameters, the biggest Q-balls are the most
stable. Moreover, as we discussed above, the stability is enhanced
by the absorption of fermions. These results suggest an efficient
accretion mechanism for large Q-balls with important astrophysical
implications \cite{starwreck}. For different parameters it is
possible that small Q-balls with $Q_\phi \sim 10^3$ and without
fermions become the most stable states. However, it is apparent
from fig. 4 that a barrier would still separate the large from the
small Q-balls. The decay of large Q-balls into smaller fragments
and free fermions would involve tunnelling and probably would
proceed at a very slow rate.


The dependence of the ratio $E/Q_\phi$ on $N$ and $e'$ in the ``thin-wall''
limit is given in table 1. The various values have been obtained
through the numerical solution of the algebraic system of equations
(\ref{con2})--(\ref{con1}) for $e=0.1$.
We observe that a small number $N$ of fermionic species with $|e'|=e$
results in a high energy to charge ratio and, therefore, unstable
Q-balls. This behaviour is caused by the big contribution from
the fermionic kinetic energy. Large values of $N$ permit the
distibution of the compensating charge among various
species, thus reducing the fermionic energy $\propto N^{-1/3}$.



\section{Large Gauged Q-Balls with two Scalar Condensates}

Another possibility is that two scalar condensates with opposite charges
form in the interior of a gauged Q-ball.
An appropriate Lagrangian density is
\be
{\cal L} = \frac{1}{2} \partial_\mu f\partial^\mu f + \frac{1}{2}
f^2 \left( \partial_\mu\theta_1-e A_\mu \right)^2 +\frac{1}{2}
\partial_\mu \chi \partial^\mu \chi + \frac{1}{2} \chi^2 \left(
\partial_\mu\theta_2-e' A_\mu \right)^2 - U(f,\chi) -\frac{1}{4}
F_{\mu\nu} F^{\mu\nu} \label{toyfxlag} \ee
The two scalar fields
carry independent conserved $U(1)$ charges, a linear combination
of which is gauged. We assume a time dependence for the two
condensates of the form: $\theta_1=\omega_1 t$, $\theta_2=\omega_2
t$. The resulting equations of motion are \beq
f''+\frac{2}{r}f'+f(\omega_1-e A_0)^2-\frac{\partial
U(f,\chi)}{\partial f}&=&0 \label{eomn1} \\
\chi''+\frac{2}{r}\chi'+\chi (\omega_2-e' A_0)^2 -\frac{\partial
U(f,\chi)}{\partial \chi}&=&0 \label{eomn2} \\
A_0''+\frac{2}{r}A_0'+ef^2 (\omega_1-e A_0)+e'\chi^2 (\omega_2-e'
A_0)&=&0. \label{eomn3} \eeq The energy of the system is given by
the expression \beq E= 4 \pi \int r^2\, dr &\Biggl\{& \frac{1}{2}
A_0'^2 +\frac{1}{2} f'^2 +\frac{1}{2} f^2 (\omega_1-e A_0)^2
+\frac{1}{2} \chi'^2 +\frac{1}{2} \chi^2 (\omega_2-e' A_0)^2
+U(f,\chi) \Biggr. \nonumber \\ &+& \Biggl. \left( \vec{E} \cdot
\vec{\nabla} A_0 + e f^2 (\omega_1-e A_0) A_0 +e' \chi^2
(\omega_2-e' A_0) A_0 \right) \Biggr\}. \label{energyfx} \eeq
Similarly to the discussion in section 3, the equations of motion
(\ref{eomn1})--(\ref{eomn3}) can be obtained by minimizing the
total energy under constant total charges of the two scalar
condensates. The expression in the second line of eq.
(\ref{energyfx}) vanishes through application of Gauss' law, eq.
(\ref{eomn3}).

The numerical solution of the three second-order differential
equations (\ref{eomn1})--(\ref{eomn3})
is more difficult than in the case of a scalar condensate with
compensating fermions. In that case we had to integrate two
second-order differential equations and an algebraic one.
Moreover, we expect a qualitative behaviour very similar to
the one studied in sections 3 and 4. For a large Q-ball to remain
classically stable, the net charge in its interior must be zero.
A possible mis-match at the surface could result in non-zero
electrostatic energy. However, in the ``thin-wall'' limit
this contribution is expected to become negligible.
For this reason, we limit our discussion to the analytical treatment
of the ``thin-wall'' limit, which is more useful for practical applications.

In this limit, the total energy is given by
\be
E = \left( \frac{1}{2}f^2 g_1^2 +\frac{1}{2}\chi^2 g_2^2 + U(f,\chi)
\right) V,
\label{enthinx} \ee
with $V$ the volume of the Q-ball and $g_1=\omega_1-e A_0$,
$g_2=\omega_2-e' A_0$.
The charges of the two scalar condensates are
\be
Q_\phi = f^2 g_1 V, ~~~~~~~~~~~~~~~~~~~ Q_\chi = \chi^2 g_2 V
\label{chargesx} \ee

They are taken to satisfy an electric charge neutrality condition
\be  e Q_{\phi} + e' Q_{\chi}=0 \label{neutral}\ee Keeping each of
them fixed means that we must minimize the quantity
\be
E= \frac{1}{2} \frac{Q_\phi^2}{f^2 V} +\frac{1}{2}
\frac{Q_\chi^2}{\chi^2 V} + U(f,\chi) V
 \label{enthin2x} \ee

Minimization with respect to $V$ results in the relation
\be
U(f,\chi)=\frac{1}{2}f^2 g_1^2+ \frac{1}{2}\chi^2 g_2^2.
\label{con1x} \ee Three more constraints can be obtained by
requiring that eqs. (\ref{eomn1})--(\ref{eomn3}) be satisfied for
constant fields. They are
 \beq fg_1^2 &=& \frac{\partial
U(f,\chi)}{\partial f}, \label{con2x} \\ \chi g_2^2 &=&
\frac{\partial U(f,\chi)}{\partial \chi}, \label{con3x} \\ e f^2
g_1 &+& e'\chi^2 g_2=0 \label{con4x} \eeq The first two could have
been obtained through the minimization of the total energy of eq.
(\ref{enthin2x}) with respect to $f$ and $\chi$. The last equation
guarantees the electric neutrality of the interior of the Q-ball.
The four equations (\ref{con1x})--(\ref{con4x}) uniquely determine
the gauge-invariant quantities $f$, $\chi$, $g_1$, $g_2$ in the
interior of a large Q-ball. As a consequence, $E$, $Q_\phi$,
$Q_\chi$ are also specified through eqs. (\ref{enthinx}),
(\ref{chargesx}). Similarly to the fermionic case, the fixed
charges $ Q_\phi, Q_\chi $ scale linearly with the volume for
fixed values of their interior field variables
$f,\chi,g_1,g_2$.One reason for this is the charge neutrality
condition in eq.(\ref{neutral}) which they satisfy. Hence the
total energy of the double condensate configuration scales
linearly with respect to the scalar charge $Q_{\phi} = |e'/e|
Q_{\chi}$. The classical stability condition becomes $\min(E) <
m_\phi Q_\phi + m_\chi Q_\chi$, with $m_\phi^2=\partial^2
U(0,0)/\partial \phi^2$, $m_\chi^2=\partial^2 U(0,0)/\partial
\chi^2$ the masses of the two scalars at the vacuum at
$\phi=\chi=0$.



\section{Conclusions}

The main emphasis in the studies of Q-balls has been on
theories with global $U(1)$ symmetries. Theories with local $U(1)$
symmetries can support Q-balls as well. However, in the absence of
a neutralizing mechanism, the electrostatic repulsion destabilizes the
Q-balls with significant charge.
In the main part of this paper
we pointed out that gauged Q-balls can be stabilized through the
neutralization of the scalar condensate by fermions of
opposite charge. The total energy is increased because of the kinetic energy
of the fermions. However, the resulting configuration can be
stable even for arbitrarily large charge of the scalar condensate.

>From a cosmological perspective, the neutralization of gauged
Q-balls is expected.
For example, one could envision the existence of electric Q-balls, that could
be produced during phase
transitions \cite{formation} when the Universe passes through
an electric-charge breaking vacuum \cite{zarikas}.
It seems likely that several fermionic species could be trapped within
the Q-ball during its formation. The ones with charge of similar sign to
the scalar condensate will be expelled, so that the resulting object
will remain approximately neutral.

The existence of large electric fields can lead to spontaneous
pair creation. The presence of a
strong electrostatic field at the surface of the Q-ball can
separate a virtual fermion-antifermion pair
and bring the particles on mass shell \cite{morris}.
The fermion will be attracted towards
the surface, while the antifermion will be expelled.
In the vacuum, the critical field strength is $E_{crit}=m_\psi^2/|e'|$.
Our assumption that the fermion mass is much smaller than the typical
scale of the potential of the scalar field implies that this mechanism
is very efficient. In the interior of
a Q-ball, the pair creation stops only when the fermionic energy levels
are populated
up to a Fermi momentum comparable to the scale of the scalar field potential.
It seems, therefore, likely that
large gauged Q-balls can be neutralized through this mechanism, instead of
disintegrating.

We mention that evaporation from the surface is possible if the
scalar field has decay channels into light species \cite{evaporate}.
In this case, simultaneous evaporation of the decay products and fermions
maintains the approximate neutrality of the Q-ball.

The tendency of gauged Q-balls to trap fermions in their interior could have
interesting experimental consequences.
Even though we concentrated on masslees fermions, heavy exotic species may
have found their way to the interior of gauged Q-balls. Thus the
discovery of a Q-ball may lead to the additional discovery
of the exotic species trapped in its interior. The fact that
the energy per charge of a Q-ball is reduced when its fermionic
content is increased
(up to neutralization) indicates an efficient accretion mechanism with
important astrophysical implications \cite{starwreck}.

%{\bf Discuss}

We also discussed the possibility of neutralization of a gauged
Q-ball through the presence of two scalar condensates of opposite
charge in its interior. In this case the formation of Q-balls seems less
likely than in the case of one scalar condensate with compensating
fermions.
For neutral Q-balls to be produced, two
condensates with the appropriate properties (values of $f$, $\chi$,
$\omega_1$, $\omega_2$) must be assumed to be generated dynamically after a
phase transition \cite{formation}. This should be more difficult than
the trapping of fermions from the thermal bath in the region with a
non-zero charged condensate.

Finally, we point out that the neutralization mechanism is expected to work
for general potentials of the scalar field. In particular, we expect it
to be applicable to the case of potentials with flat directions, such
as the ones appearing in supersymmetric extensions of the Standard Model.
In this case
the global Q-balls do not approach the ``thin-wall'' limit, even though
they can become very big, with energy that scales $E \propto Q^{3/4}$
\cite{dvali}. The gauged Q-balls with similar potentials cannot reach
large sizes, unless a neutralization mechanism (through trapping of
fermions for example) eliminates the electrostatic repulsion.







\paragraph{Acknowledgements:} We  would like to thank
L. Perivolaropoulos for helpful discussions.  The work of K.N.A. and
N.T. was supported by the European Commission under the RTN programs
HPRN--CT--2000--00122, HPRN--CT--2000--00131 and
HPRN--CT--2000--00148. The work of K.N.A. was also supported by the RTN
program HPRN--CT--1999-00161, a National Fellowship Foundation of
Greece (IKY) and the INTAS contract N 99 0590.




%\newpage

%\small


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\begin{table}[t]
\begin{center}
\begin{tabular}{||l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l||}
\hline\hline
$\frac{\lambda _{1}}{g^{2}}$ & \multicolumn{1}{||l|}{$\frac{\lambda _{2}}{%
g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{3}}{g^{2}}$} &
\multicolumn{1}{||l|}{$\frac{\lambda _{4}}{g^{2}}$} & \multicolumn{1}{||l|}{$%
\frac{\lambda _{5}}{g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{6}}{%
g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{7}}{g^{2}}$} &
\multicolumn{1}{||l|}{$\beta $} & \multicolumn{1}{||l|}{$\omega _{0}$} &
\multicolumn{1}{||l|}{$F_{a0}$} & \multicolumn{1}{||l|}{$F_{b0}$} &
\multicolumn{1}{||l|}{$F_{q0}$}& \multicolumn{1}{||l|}{$\left( \frac{E}{Q}%
\right) _{g}$} & \multicolumn{1}{||l|}{$\left( \frac{E}{Q}\right) _{\psi }$}
& \multicolumn{1}{||l|}{$M^l_{H}$} & \multicolumn{1}{||l||}{$M^c_{H}$}
\\ \hline\hline
4.5 & 4.5 & \phantom{-}4.5 & -4.5 & -4.5 & -0.45 & 0.45 & 100 & 1.82 &
\phantom{-}0.99995 &
\phantom{-}0.01 & \phantom{-}0.0001 & 1.82 & 1.82 & 1.87 & 1.82 \\ \hline
4.5 & 4.5 & -0.45 & -4.5 & -0.45 & -0.45 & 0.45 & 100 & 1.70 & -0.99995 &
-0.01 & -0.000094 & 1.70 & 1.70 & 1.87 & 1.70\\ \hline
4.5 & 4.5 & \phantom{-}0.45 & -4.5 & -0.45 & -0.45 &
0.45 & 100 & 1.70 & -0.99995 &
-0.01 & \phantom{-}0.000098 & 1.70 & 1.70 & 1.87 & 1.70\\ \hline
4.5 & 4.5 & \phantom{-}4.5 & -4.5 & -0.45 & -0.45 & 0.45 &
\phantom{1}60 & 1.39 & \phantom{-}0.99986 &
\phantom{-}0.0167 & -0.000135 & 1.39 & 1.40 & 1.41 & 1.39\\ \hline
4.5 & 4.5 & -0.45 & \phantom{-}0.45 & -4.5 & -0.45 & 0.45 &
\phantom{1}80 & 1.52 & \phantom{-}0.99992 &
\phantom{-}0.0125 & -0.0001 & 1.52 & 1.53 & 1.80 & \\ \hline
4.5 & 4.5 & \phantom{-}0.45 & \phantom{-}0.45 & -4.5 & -0.45 & 0.45 & 150 &
1.99 & \phantom{-}0.99997 &
\phantom{-}0.0066 & \phantom{-}0.00013 & 1.99 & 2.00 & 1.88 & \\ \hline
4.5 & 4.5 & \phantom{-}4.5 & \phantom{-}0.45 &
-4.5 & -0.45 & 0.45 & 120 & 1.81 & \phantom{-}0.99996 &
\phantom{-}0.0083 & -0.00013 & 1.81 & 1.81 & 1.88 & \\ \hline\hline
\end{tabular}
\caption{
Another class of Q-balls with and without quarks.
}
\label{param2}
\end{center}
\end{table}








\section{Electric Q-balls}
We look for Q-balls by means of the ansatz
\begin{equation}
\Phi _{1}=\frac{1}{\sqrt{2}}
\left( \begin{array}{c}
0 \\ f_a \end{array}
\right),~~~~~
\Phi _{2}=\frac{1}{\sqrt{2}}
\left( \begin{array}{c}
f_q \exp(i\omega t) \\ f_b \end{array}
\right).
\label{ansatz} \end{equation}
We work in the ``thin-wall'' limit of large Q-balls.
The fields $f_a$, $f_b$, $f_q$ take constant values $F_a$, $F_b$, $F_q$
in the interior of
the Q-ball, while at the surface they switch quickly to
the values $f_a=u$, $f_b=v$, $f_q=0$.

According to our discussion in sections 3 and 4,
we need the values $F_{a0}$, $F_{b0}$, $F_{q0}$ and $\omega_0$
in the ``global thin-wall limit'', when the electric charge $e$ is set to
zero. For their determination it is convenient to define the
``effective'' potential
\be
V_{eff}(f_a,f_b,f_q) =
V(f_a,f_b,f_q) - \frac{1}{2}\omega^2 f_q^2.
\label{veff} \ee
Then, $F_{a0}$, $F_{b0}$, $F_{q0}$ and $\omega_0$
can be obtained from the algebraic relations
\be
\frac{\partial V_{eff}}{\partial f_a}=0,~~~~~~
\frac{\partial V_{eff}}{\partial f_b}=0,~~~~~~
\frac{\partial V_{eff}}{\partial f_q}=0,~~~~~~
V_{eff}=0.
\label{solution} \ee

The ratio $(E/Q)_g$ in the absence of fermions in the interior of the
Q-ball is given by eqs. (\ref{check1}),
(\ref{check2}) with $F_0=F_{q0}$ and $e=0.3$.
In the presence of fermions the relevant ratio $(E/Q)_\psi$
is given by eq. (\ref{eqf}) with $F_0=F_{q0}$. The parameters $N$ and $e'$
appearing in eq. (\ref{eqf}) are determined by the nature of the fermions
that can exist in the interior of the solutions. These must
carry baryonic charge in order to prevent their annihilation
against the charged Higgs background.
The typical energy scale of the electric Q-balls is set by the electroweak
scale. The resulting charge and energy densities exceed typical
nuclear densities by several orders of magnitude. Therefore, we expect
baryonic charge to be carried by almost freely propagating quarks.
A possible scenario involves a Higgs condensate with positive
electric charge ($\omega_0>0$) and $d$, $s$, $b$
%$u$, $c$, $t$
quarks filling up
separate Fermi seas. This scenario implies $e'=-1/3\,e$ and
$N=(3 {\rm ~flavours})\times (3{\rm ~colours})=9$.
Another scenario could involve a negatively charged condensate ($\omega_0<0$)
and $u$, $c$, $t$ quarks with $e'=2/3\,e$.
(One may also consider gases of anti-quarks in the interior of the Q-balls.)
We also point out that we neglect the Yukawa couplings and assume that
the (anti-)quarks are massless. Significant corrections are expected only for
the contributions from the $t$ (anti-)quark.



\begin{table}[t]
\begin{center}
\begin{tabular}{||l|l|l|l|l|l|l|l|l|l|c|l|l||}
\hline\hline
$\frac{\lambda _{1}}{g^{2}}$ & \multicolumn{1}{||l|}{$\frac{\lambda _{2}}{%
g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{3}}{g^{2}}$} &
\multicolumn{1}{||l|}{$\frac{\lambda _{4}}{g^{2}}$} & \multicolumn{1}{||l|}{$%
\frac{\lambda _{5}}{g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{6}}{%
g^{2}}$} & \multicolumn{1}{||l|}{$\frac{\lambda _{7}}{g^{2}}$} &
\multicolumn{1}{||l|}{$\beta $} & \multicolumn{1}{||l|}{$\omega _{0}$} &
\multicolumn{1}{||l|}{$F_{q0}$} & \multicolumn{1}{||c|}{$\left( \frac{E}{|Q|}%
\right) _{\psi }$} & \multicolumn{1}{||l|}{$M^l_{H}$}
& \multicolumn{1}{||l||}{$M^c_{H}$} \\ \hline\hline
4.5 & 4.5 & \phantom{-}4.5 & -4.5 & -4.5 & 0.45 & -0.45 & 0.2 & 0.071 & 0.966 &
0.84 & 0.40 & 1.03\\ \hline
4.5 & 4.5 & \phantom{-}4.5 & -4.5 & -4.5 & 0.225 & -0.225 & 0.05 & 0.076 & 0.990
& 0.88 & 1.28 & 1.82\\ \hline
4.5 & 4.5 & -0.45 & -4.5 & -0.45 & 0.45 & -0.45 & 0.01 & 0.015 &
0.999 & 0.49 & 1.37 & 1.70\\ \hline
4.5 & 4.5 & \phantom{-}0.45 & -4.5 & -0.45 & 0.45 & -0.45 & 0.01 & 0.015 & 0.999
& 0.49 & 1.37 & 1.70\\ \hline
4.5 & 4.5 & \phantom{-}4.5 & -4.5 & -0.45 & 0.45 & -0.45 & 0.01 & 0.015 & 0.999
& 0.49 & 1.37 & 1.70 \\ \hline
4.5 & 4.5 & -0.45 & \phantom{-}0.45 & -4.5 & 0.45 & -0.45 & 0.01 & 0.015 & 0.999
& 0.49 & 1.37 & 1.67\\ \hline
4.5 & 4.5 & \phantom{-}0.45 & \phantom{-}0.45 & -4.5 & 0.45 & -0.45 & 0.05 & 0.034 & 0.997
& 0.66 & 0.69 & 0.95\\ \hline
4.5 & 4.5 & \phantom{-}0.45 & \phantom{-}0.45 & -4.5 & 0.45 & -0.45 & 0.01 & 0.015 & 0.999
& 0.49 & 1.37 & 1.67 \\ \hline
4.5 & 4.5 & \phantom{-}4.5 & \phantom{-}0.45 & -4.5 & 0.45 & -0.45 & 0.07 & 0.040 & 0.996
& 0.70 & 0.59 & 0.87 \\ \hline
4.5 & 4.5 & \phantom{-}4.5 & \phantom{-}0.45 & -4.5 & 0.45 & -0.45 & 0.01 & 0.015 & 0.999
& 0.49 & 1.37 & 1.67 \\ \hline\hline
\end{tabular}
\caption{
The parameters of a class of electric Q-balls with quarks. $F_{a0}=F_{b0}=0$.
Dimensionful
parameters are normalized with respect to 246 GeV.
}
\label{param}
\end{center}
\end{table}


In table \ref{param} we present the properties of
a class of electric Q-balls with quarks. We assume that, in the interior
of the Q-ball, a negatively charged
Higgs condensate is neutralized by quarks with $e'= 2/3 \, e.$
All dimensionful quantities
are given in units of 246 GeV. In the first eight columns
we give the parameters of the two-Higgs potential. All the solutions
are characterized by small values of the parameter $\beta$.
The next columns give $\omega_0$ and the charged Higgs field
expectation value in the ``global thin-wall limit''. The neutral Higgs fields
have zero expectation values ($F_{a0}=F_{b0}=0$).
The energy per unit charge $\left({E}/{|Q|}\right) _{\psi }$ for
large electric Q-balls with quarks
is obtained from eq. (\ref{eqf}) with $F_0=F_{q0}$.
Large Q-balls without quarks have
a large energy per unit charge and are unstable. For this reason we
have not included them in the table.
The last two columns give the masses of
the neutral and charged Higgs particles
in the standard vacuum. As $\left({E}/{|Q|}\right) _{\psi } < M^c_H$
the Q-balls cannot disintegrate into charged Higgs particles.

These solutions do not require fine tuning. One can
always produce stable electric Q-balls with quarks for moderately small
values of $\beta$. In general, if $\beta \lta 0.2$ the Q-balls possess energy
per charge smaller than the mass of the charged Higgs particle.
If $\beta \lta 0.05$
the energy per charge is smaller than the mass of the lightest
Higgs particle. We point out that
the couplings $\lambda _{6}$ and $\lambda _{7}$ must have
opposite sign. Their non-zero values
also ensure that, for small values of $\beta$,
the neutral Higgs particle does not have a mass smaller than the
experimential lower bound.


We have also searched unsuccessfully for parameter regions with stable Q-balls
without quarks. The electrostatic repulsion is very strong and destabilizes
all Q-balls with $Q| \gta 1000$, for which the approximate analytical
treatment of section 2 is valid. We have only found
a region with large $\beta$, $\lx_6$ and $\lx_7$ of opposite sign, and
very small $F_{q0}$, for which the upper bound of eqs. (\ref{check1}),
(\ref{check2}) gives a value slightly larger than the
mass of the charged Higgs particle.
A typical example has
$\lx_1/g^2=\lx_2/g^2=\lx_3/g^2=4.5$,
$\lx_4/g^2=-4.5$,
$\lx_5/g^2=\lx_6/g^2=-\lx_7/g^2=-0.45$,
$\beta=60$, $\omega_0=1.3915$,$F_{a0}\simeq 1$,
$F_{b0}= 0.0167$,
$F_{q0}= -0.000135$,
$(E/|Q|)_\psi=1.3972$,
$M^c_H=1.3917$.
It is possible that the actual ratio is sufficiently smaller than the
upper bound $(E/|Q|)_\psi$ so that the Q-balls are stable. However, such small
Q-balls evaporate very fast and probably do not have any significant
cosmological or experimental consequences.



\section{The two-Higgs potential}
Extensions of the standard model that involve more than one Higgs doublets
provide additional sources of $CP$ violation \cite{cp} that can be important
for the scenaria of electroweak baryogenesis \cite{baryog}. The presence
of a second Higgs doublet is automatic in supersymmetric extensions of the
Standard Model. However, the existence of electric Q-balls does not require
supersymmetry. The two-Higgs potential has been analysed
extensively in the literature. (For a review see ref. \cite{sher}.) Its phase
structure has also been investigated at high temperature, and the
possibility of electric-charge violating phases has been demonstrated
\cite{zarikas}.

The potential can be written using weak isospin doublets
both of weak hypercharge\footnote{%
We follow here the notation of Ref.~\cite{sher} in which both Higgs doublet
fields have the same hypercharge.} $Y_{{\rm weak}}=+1$ as follows \cite{sher}
\begin{eqnarray}
V &=&\mu _{1}^{2}\Phi _{1}^{\dagger }\Phi _{1}+\mu _{2}^{2}\Phi
_{2}^{\dagger }\Phi _{2}+\lambda _{1}(\Phi _{1}^{\dagger }\Phi
_{1})^{2}+\lambda _{2}(\Phi _{2}^{\dagger }\Phi _{2})^{2}+\ \lambda
_{3}(\Phi _{1}^{\dagger }\Phi _{1})(\Phi _{2}^{\dagger }\Phi _{2})  \nonumber
\\
&&+\lambda _{4}(\Phi _{1}^{\dagger }\Phi _{2})(\Phi _{2}^{\dagger }\Phi
_{1})+{\frac{1}{2}}\lambda _{5}[(\Phi _{1}^{\dagger }\Phi _{2})^{2}+(\Phi
_{2}^{\dagger }\Phi _{1})^{2}]+V_{{\rm D}}\,,
\label{pot1}
\end{eqnarray}
where $\lambda _{i}$ are real numbers and
\begin{equation}
\Phi _{1}={\frac{1}{\sqrt{2}}}\left(
\begin{array}{c}
\phi _{1}+i\phi _{2}
\\
\phi_{3}+i\phi _{4}
\end{array}
\right), ~~~~~~~~~~~~~
\Phi _{2}={\frac{1}{\sqrt{2}}}\left(
\begin{array}{c}
\phi _{5}+i\phi _{6}
\\
\phi_{7}+i\phi _{8}
\end{array}
\right).
\label{fields} \end{equation}
The above potential is the most general one satisfying the following
discrete symmetries:
\begin{equation}
\Phi _{2}\rightarrow -\Phi _{2},\;\;\Phi _{1}\rightarrow \Phi
_{1},\;\;d_{R}^{i}\rightarrow -d_{R}^{i},\;\;u_{R}^{i}\rightarrow
u_{R}^{i}\,,  \label{disc}
\end{equation}
where $u_{R}^{i}$ and $d_{R}^{i}$ represent the right-handed weak
eigenstates with charges ${\frac{2}{3}}$ and $-{\frac{1}{3}}$ respectively.
This symmetry forces all the quarks of a given charge to interact with only
one doublet, and thus Higgs-mediated flavour changing neutral currents are
absent.

When the discrete symmetry is broken during a cosmological phase
transition, it produces stable domain walls via the Kibble mechanism \cite
{kibble}. One can overcome this problem by adding terms that break the
symmetry, providing at the same time the required explicit $CP$ violation
for baryogenesis. The most general form of the part of the potential that
breaks the discrete symmetry is
\begin{equation}
V_{{\rm D}}=-\mu _{3}^{2}\Phi _{1}^{\dagger }\Phi _{2}+\lambda _{6}(\Phi
_{1}^{\dagger }\Phi _{1})(\Phi _{1}^{\dagger }\Phi _{2})+\lambda _{7}(\Phi
_{2}^{\dagger }\Phi _{2})(\Phi _{1}^{\dagger }\Phi _{2})\,+\,h.c.
\label{soft}
\end{equation}
The parameters
$\mu_{3} $, $\lambda _{6}$ and $\lambda _{7}$ are in general complex numbers
providing explicit {\em CP} violation at the tree level. However, the phases
do not play a significant role in our considerations and we assume here that
they are zero.

In order to study the structure of the vacua we can perform an $SU(2)$
rotation that sets $\phi _{1,2,4}=0$.
The potential must have a
minimum
that respects the $U(1)$ of electromagnetism. We neglect possible
phases of the fields and parametrize this minimum as
\begin{equation}
\Phi _{1s}=\frac{1}{\sqrt{2}}
\left( \begin{array}{c}
0 \\ u \end{array}
\right),~~~~~
\Phi _{2s}=\frac{1}{\sqrt{2}}
\left( \begin{array}{c}
0 \\ v \end{array}
\right),
\label{min1} \end{equation}
with $u^2+v^2 = (246{\,\rm GeV})^2$.
The acceptable parameter values of the model are those ensuring that the above
stationary point becomes the absolute minimum at zero temperature.

The free parameters of the model can be taken to be the quartic couplings $%
\lambda _{i}$, the ratio $\beta =u/v$ and the mass parameter $\mu _{3}$,
because the following conditions hold at the stationary points
\begin{eqnarray}
\mu _{1}^{2} &=&-\lambda _{1}\beta ^{2}v^{2}-\frac{1}{2}\left( \lambda
_{3}+\lambda _{4}+\lambda _{5}\right) v^{2}+m_{3}^{2}\ \beta^{-1}
-\frac{1}{4}l_{7}\ v^{2}\beta^{-1}
-\frac{3}{4}l_{6}\ v^{2}\beta,
\label{ex1} \\
\mu _{2}^{2} &=&-\lambda _{2}v^{2}-\frac{1}{2}\left( \lambda _{3}+\lambda
_{4}+\lambda _{5}\right) \beta ^{2}v^{2}+m_{3}^{2} \ \beta
-\frac{3}{4}l_{7}\ v^{2}\beta
-\frac{1}{4}l_{6}\ v^{2}\beta^{3}.
\label{ex2}
\end{eqnarray}

Scanning the whole parameter space would be too time consuming. Therefore in
our analysis we consider values of the quartic couplings such that
the ratios $\left| \lambda _{i}\right|/g^{2}$, with $i=1,2,...,5$,
vary in the range [0.1,\,4.5], where $g^2=0.21$ is the $SU(2)$ coupling
constant. The ratios $\lambda _{6,7}/g^2$
vary in the range
[0.1,\,0.5].
%, while we concentrate on the
%regions $\beta \ll 1$ and $\beta \gg 1$.
We always make sure that the couplings
respect the experimental limits on the Higgs-boson
masses, the condition that the potential is bounded from below, and the
requirement that
the vacuum of eq. (\ref{min1}) is the absolute minimum of the potential.




We show that,
in extensions of the Standard Model with two Higgs doublets,
Q-balls can exist
with a non-zero expectation value for the charged Higgs field in their
interior. Large Q-balls are destabilized by the electrostatic
repulsion, unless they are neutralized by quarks of opposite
charge trapped inside them.
We give examples of such Q-balls and verify their
stability. We briefly discuss the evaporation of
quarks or Higgs particles from their surface and
comment on their cosmological role.

Most of the above ingredients exist within the Standard Model. There is
a local $U(1)$ symmetry that mediates the electromagnetic force.
There are also charged particles, quarks or their bound states (nucleons or
even nuclei), that carry an additional conserved quantum number,
baryonic charge. One only needs a charged scalar field in order to
complete the list of necessary ingredients for a gauged Q-ball.
In order to find such a scalar one must consider extensions of the
Standard Model with two doublets in the Higgs sector. Supersymmetry is
not required, even though the presence of a second Higgs doublet
is automatic in supersymmetric models, such as the MSSM.



Minimization with respect to $R$ then gives
\be
R=\left(2 \bar{U} F^2_0 \right)^{-1/6}
\left( \frac{eQ_s}{8 \pi} \right)^{2/3}
\label{rr} \ee
and
\be
E_s= \frac{3}{4 \pi^{1/3}} \left(2 \bar{U} F^2_0 \right)^{1/6}
\left( e Q_s\right)^{4/3}
=12 \pi \left(2 \bar{U} F^2_0 \right)^{1/2} R^2.
\label{surfee} \ee



For very large Q-balls gravitational effects become important. The
resulting astrophysical objects are characterized
as Q-stars \cite{gravferm}.
The gravitational attraction leads to an increase in the pressure in the
interior. The Thomas-Fermi approximation can still be applied, but
the Fermi momentum obeys (in the absence of an electrostatic field)
\be
\left( k_F^2 + m_\psi^2 \right)^{1/2} g_{00}^{1/2} = \mu_\psi.
\label{} \ee

The purpose of this letter was to indicate the possible presence of
electric Q-balls in extensions of the Standard Model with a second
Higgs doublet. The setting is very simple and can be incorporated easily in
supersymmetric extensions of the Standard Model.
We showed that electric Q-balls can exist, especially if
their charge is neutralized by
quarks that are trapped in their interior.

We verified the stability of these Q-balls with respect
to disintegration into charged Higgs particles.
We also pointed out that they tend to lose the trapped quarks
through evaporation from the surface. Preservation of
the neutrality of the Q-ball requires the simultaneous evaporation of
charged Higgs particles. In our discussion we neglected the
effect of the Yukawa couplings.
When they are switched on,
it becomes possible for the Q-balls to decay into
light particles of the Standard Model.
This decay process takes place only at the surface of
the Q-ball, because the decay products are fermions \cite{evaporate},
and can be characterized
as evaporation as well.
The total evaporation rate depends on the Q-ball size and requires
a detailed numerical calculation \cite{evaporate,iiro}
that is beyond the scope of this
letter. It is obvious, however, that
the lifetime of a Q-ball is strongly dependent on its size.


and discard the
contribution $\sim f'^2$ to the total energy.
The solution for $g$ reads \cite{local}
\beq
g(r)&=&\left(\omega-\frac{e^2 Q}{4\pi R} \right)
\frac{R \sinh(eFr)}{r \sinh(eFR)}~~~~~~~{\rm for}~~r\leq R
\label{gr1} \\
g(r)&=&\omega-\frac{e^2 Q}{4\pi r}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\rm for}~~r > R,
\label{gr2} \eeq
where $Q$ is the total charge.
Consistency with eq. (\ref{charge}) implies that
\be
\omega = \frac{e^2 Q}{4 \pi R}
\left[1-\frac{\tanh x}{x} \right]^{-1},
\label{omega} \ee
Minimization of the energy with respect to $R$ for fixed $Q$
and $F$ gives
\be
x \left[\frac{x}{\tanh x} -1\right] =
\frac{e^3 Q F^2}{4 \pi \sqrt{2 \, U(F)}}.
\label{minimum} \ee

The limit $e\to 0$ reproduces the well-known properties of
large global Q-balls.
Eqs. (\ref{gr1}), (\ref{gr2}) give $g=\omega_0$ and eq. (\ref{eom1})
is satisfied for
$\omega_0^2 F_0= [{dU}/{df}](F_0).$
Eqs. (\ref{omega}), (\ref{minimum}) give
$\omega_0^2 F_0^2= 2\,U(F_0),$ while
eqs. (\ref{charge}), (\ref{energy}) lead to
${E_0}/{Q_0}=\omega_0.$

We are interested in the case of small $x$.
Through an expansion of the lhs of eq. (\ref{minimum})
up to terms $\sim x^2$, one finds \cite{local}


For $e=0$ the energy for fixed $Q$ is minimized for
$\sqrt{2U(F_0)/F^2_0}=\omega_0$.
We are interested in the range
$C\lta 1$. In this range and for $R$ given by eq. (\ref{approx1a}))
we have $x\simeq C^{1/3} \lta 1$, consistently with our earlier assumption.
We can
obtain an upper bound for the energy through
\beq
\frac{E}{Q} &\leq&\left( \frac{E}{Q} \right)_g =
\omega_0\left(1+\frac{C_0^{2/3}}{5} \right)
\label{check1} \\
C_0 &=&\frac{3e^3Q}{4\pi} \frac{F_0}{\omega_0}.
\label{check2} \eeq
The minimization of eq. (\ref{approx2}) with respect to $F$ can provide
a more accurate determination of $E/Q$. However, $(E/Q)_g$ provides
an upper bound that is easier to use when looking for Q-balls.

The total energy must now be minimized for conserved $\rho_\psi$.

This can be implemented through a Lagrange multiplier $\mu_\psi$.
Minimization with respect to $k_F$ gives














Similarly to the previous section, we look for large Q-balls in
the ``thin-wall'' limit. First we neglect the effect of
the surface and examine if infinitely large Q-balls can exist.
Their interior can be characterized as Q-matter.
Our crucial assumption is
that the most favourable state involves a zero electrostatic field $A_0=0$.
Otherwise, the repulsion would increase the total energy and render
Q-matter unstable. Eq. (\ref{mupsi}) gives $\mu_\psi =k_F$, while
eq. (\ref{eom3}) requires the cancellation of the opposite charge
densities $e\rho$=$eF^2\omega=-Ne'\rho_\psi$.
Eq. (\ref{eom1}) implies that $\omega^2 F= [{dU}/{df}](F).$
For fixed charge, the energy is minimized with respect to the radius for
$\omega^2 F^2= 2U(F)$. We see that $\omega$ and $F$ are given by the
same values $\omega_0$, $F_0$ as in the case of global Q-balls.
Finally, the ratio $E/Q$ becomes
\be
\left( \frac{E}{Q} \right)_\psi = \omega_0 + N \frac{k_F^4}{4 \pi^2}
\frac{1}{F_0^2\omega_0} =
\omega_0 \left[1+
\frac{(3\pi^2)^{4/3}}{4\pi^2}
\frac{1}{N^{1/3}}
\left(-\frac{e\omega_0}{e'}\right)^{4/3}
\frac{F_0^{2/3}}{\omega_0^2}
\right].
\label{eqf} \ee
The interpretation is clear: If the configuration does not involve
an electrostatic field, the energy cost compared to the global Q-ball case
is increased only by the kinetic energy of the fermions.

We have assumed that the fermions are massless. Thus
the requirement for the stability of the Q-matter is
$(E/Q)_\psi<[d^2U/df^2](f=0)$.
This equation can be compared to eq. (\ref{approx2}),
in which the energy increase relative to the global case is proportional to
a certain power of the total charge $Q$. As a result large Q-balls become
unstable without fermions, and Q-matter does not exist.
The structure of the Q-ball surface is important for processes such as
charge evaporation.
In the previous section we concluded
that the chemical potential associated with the global
charge carried by the fermions is positive. (It is equal to $k_F$.)
This means that, despite being stable against disintegration into
free scalars and fermions, the Q-balls can lose fermions through evaporation
from the surface. As a result, a net electric charge appears at the surface.
When the electrostatic field reaches a certain critical value,
further evaporation is blocked by the electrostatic attraction of the fermions
by the Q-ball.

Let us assume that there are no fermions within a
surface layer of thickness $\dr$ and total charge $Q_s$.
The total surface energy can be approximated by \cite{local}
\be
E_s=2 \pi R^2 \frac{F_0^2}{\dr} + 4 \pi R^2 \bar{U} \dr +
\frac{e^2 Q_s^2}{8 \pi R}.
\label{surfe} \ee
The first term is the gradient energy of the scalar field, the second
the average potential energy and the third originates in the electrostatic
repulsion.
Minimizing with respect to $\dr$ we find $\dr=\sqrt{F^2_0/2 \bar{U}}$.
The magnitude of the surface charge is determined by the requirement that
the fermions in the interior cannot evaporate because of the
electrostatic field.
For this we need $|e'|A_0(r=R)=|e'|eQ_s/4\pi R=k_F$.

It is clear from the above that the energy associated with the scalar
condensate scales $\sim R^2$, while the energy of the required electrostatic
field is $\sim R$.
As the energy from the interior scales $\sim Q \sim R^3$,
the contribution from the surface to the total energy
of the Q-ball always becomes negligible for sufficiently large $R$.
We conclude that large Q-balls are stable and do not evaporate fermions
if Q-matter is stable.

It must be pointed out, however, that the state of minimal energy is likely
to be different than the one we have just described.

\documentstyle[11pt,equation,
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\noindent
%July 1999
\begin{flushright}
%SNS--PH/1999--11
%\\
%hep--ph/9908209
\end{flushright}
\vspace{3cm}
\begin{center}
{ \Large \bf
On Brane Stabilization and the Cosmological Constant
}
\\ \vspace{1cm}
{\large
N. Tetradis
}
\\
\vspace{1cm}
{\it
Department of Physics, University of Crete,
710 03 Heraklion, Crete, Greece
}
\\
%\vspace{0.3cm}
and
\\
%\vspace{0.3cm}
{\it
Department of Physics, University of Athens,
157 71 Athens, Greece
}
\\
\vspace{2cm}
\abstract{
Abstract
\\
\vspace{1cm}
%PACS number: 98.80.Cq
}
\end{center}
\vspace{4cm}
\noindent
%CERN--TH/97--XXX \\
%July 1999


\newpage

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\paragraph{Introduction:}
The smallness of the cosmological constant has defied explanation
despite a long history of attempts \cite{weinberg}. The problem is
further complicated by the recent observational evidence that the present
cosmological constant may
be of the order of the critical density of the Universe \cite{obsc}.
An interesting proposal




\paragraph{The fine-tuning:}
We consider a system of two branes in the background of
a bulk scalar field $\phi$. The action is given by
\be
S=\int d^4x\,dy \, \sqrt{\left| \det g_{\mu\nu}\right|} \,
\left[ - 2 M^3 R + \frac{1}{2} \left( \partial \phi \right)^2
- V(\phi) \right] -\sum_{\alpha=1,2}
\int d^4x \, \sqrt{\left| \det g_{ij}\right|}\, \lx_\alpha(\phi).
\label{action0}
\ee
By rescaling all dimensionful
quantities by $2M$ (which is of the order of the fundamental Planck's constant)
we obtain
\be
S=\int d^4x\,dy \, \sqrt{\left| \det g_{\mu\nu}\right|} \,
\left[ -\frac{1}{4} R + \frac{1}{2} \left( \partial \phi \right)^2
- V(\phi) \right] -\sum_{\alpha=1,2}
\int d^4x \, \sqrt{\left| \det g_{ij}\right|}\, \lx_\alpha(\phi),
\label{action}
\ee
consistently wity the notation of ref. \cite{dewolfe}.
We emphasize that all quantities in eq. (\ref{action}) are dimensionless,
even though they are denoted by the same symbols as in eq. (\ref{action0}).

For the metric we assume the ansatz
\be
ds^2 = e^{2 A(y)}\, \eta_{ij} \,dx^i dx^j - dy^2
\label{metr1} \ee
with space-time topology $R^{3,1}\times S^1/Z_2$ \cite{rs1}.
The two branes are located at the boundaries of the fifth dimension.
Einstein's equations and the equation of motion of the field are
\cite{dewolfe}
\beq
\phi''+4 A' \phi'= &\frac{\partial V(\phi)}{\partial \phi} +
\sum_{\alpha=1,2}
\frac{\partial \lx_\alpha (\phi)}{\partial \phi}\, \delta (y-y_\alpha)
\label{eoma1} \\
A'' = &-\frac{2}{3}\phi'^2-\frac{2}{3}
\sum_{\alpha=1,2}
\lx_\alpha (\phi) \,\delta (y-y_\alpha)
\label{eoma2} \\
A'^2 = &-\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
\label{eoma3}
\eeq
We set $y_1=0$ and $y_2=R$.

The solutions of the above equations for general potentials $V(\phi)$
predict a fixed distance $R$ between the two branes. They generalize
the stabilization mechanism of ref. \cite{goldwise} by taking into account
the backreaction of the scalar field on the gravitational background.
The functions $\lx_\alpha (\phi)$ are characterized as the brane tensions.
Their form is determined by the vacuum energy of the fields that live
on the brane. We have also allowed for
an interaction of these fields with the bulk field, so
that the tensions depend on $\phi$.
The presence of the branes
imposes boundary conditions for $A'(y)$ and $\phi(y)$ at $y=0,R$.
The integration of eq. (\ref{eoma1}), (\ref{eoma2})
around the $\delta$-functions and use of the $S^1/Z_2$ symmetry
leads to
\beq
y=0
~~~~~~~~~~~~~~~~
\phi'=&~\frac{1}{2}\frac{\partial \lx_1(\phi)}{ \partial \phi}
~~~~~~~~~~~~~~~~
A'=-\frac{1}{3} \lx_1(\phi)
\label{bound1} \\
y=R
~~~~~~~~~~~~~~~~
\phi'=&~-\frac{1}{2}\frac{\partial \lx_2(\phi) }{\partial \phi}
~~~~~~~~~~~~
A'=\frac{1}{3} \lx_2(\phi).
\label{bound2} \eeq
By imposing these conditions, we have only to solve
eqs. (\ref{eoma1})--(\ref{eoma3}), neglecting the
$\delta$-function contributions.

The three equations (\ref{eoma1})--(\ref{eoma3}) are not independent, as
they are related through the Bianchi identities. We look for a solution of
eqs. (\ref{eoma1}), (\ref{eoma3}), which automatically satisfies
eq. (\ref{eoma2}). Our ansatz for the metric, eq. (\ref{metr1}), indicates
that we should expect a fine-tuning for the existence of
a static solution \cite{dewolfe}. The reason is that our choice of
four-dimensional Minkowski metric $\eta_{ij}$ requires the vanishing of
the effective cosmological constant on the branes.

A simple way to understand the fine-tuning is the following:
We can substitute $A'$ as given by eq. (\ref{eoma3}) into
eq. (\ref{eoma1}).
Without loss of generality we choose the negative root for $A'$. We
assume that our universe corresponds to the positive-tension brane located
at $y=0$ (with $\lx_1(\phi)>0$.). This means that $A<0$ in the bulk.
Eq. (\ref{eoma1}) now becomes a
non-linear second-order differential equation for $\phi(y)$, whose
solution requires two boundary conditions. These are obtained
by substituting eqs. (\ref{bound1}) into eq. (\ref{eoma3}).
The resulting algebraic equation in general has a discrete number
of solutions that give the allowed values
of $\phi$ at the location of the first brane.
For each of them the corresponding
value of $\phi'$ is given by the first of eqs. (\ref{bound1}).
Let us denote generically these solutions by $(\phi_1$, $\phi_1')$.
Now we can integrate eq. (\ref{eoma1}), with $A'$ expressed
in terms of eq. (\ref{eoma3}) and the initial conditions
$\phi(0)=\phi_1$, $\phi'(0)=\phi_1'$. The resulting trajectory
$(\phi(y),\phi'(y))$ determines the form of the field and the
metric in the bulk.

In analogy with above, the subsitution of the conditions
(\ref{bound2}) into eq. (\ref{eoma3}) leads to a discrete
number of possible values of $\phi$ and $\phi'$ at the location
of the second brane. Let us denote them generically by $(\phi_2,\phi'_2)$.
The fine-tuning is now apparent: The trajectory
$(\phi(y),\phi'(y))$ must pass through $(\phi_2,\phi'_2)$.
This can be achieved only through a careful choice of
$\lx_2(\phi)$ (assuming that $\lx_1(\phi)$, $V(\phi)$) are chosen
arbitrarily). However, a possible change of $\lx_1(\phi)$, through
a phase transition on the first brane for example, destabilizes the
solution. The trajectory corresponding to the new initial condition
$(\tilde{\phi}_1,\tilde{\phi}'_1$) does not pass through
$(\phi_2,\phi'_2)$. Some unknown mechanism must
modify the tension $\lx_2(\phi)$ of the second brane
for a new static solution to exist.

However, there is another possibility:
A new static configuration can still
exist without modification of $\lx_2(\phi)$
if the
new boundary conditions $(\tilde{\phi}_1,\tilde{\phi}'_1$) lie on the
initial trajectory $(\phi(y),\phi'(y))$. Then the new solution of eqs.
(\ref{eoma1}), (\ref{eoma3}) is the part of the original one between
$(\tilde{\phi}_1,\tilde{\phi}'_1$) and $(\phi_2,\phi'_2)$.
Physically it corresponds to a small displacement of the positive-tension
brane in a way that the second brane remains unaffected.
Since the values of $(\tilde{\phi}_1,\tilde{\phi}'_1$) are determined
by the initial function $\lx_1(\phi)$ and its change through a phase
transition, it seems that this scenario is not possible in general.
We discuss in the following to what extent it may work.

\paragraph{A first attempt:}
We start by assuming an initial configuration with a metric given by
eq. (\ref{metr1}). As we explained above this requires an initial fine-tuning
of the brane tensions $\lx_1(\phi)$, $\lx_2(\phi)$. We do not address this
issue in this work,
even though we comment on its possible resolution later on.
We are concerned with the requirement of a new fine-tuning every time
$\lx_1(\phi)$ changes. We consider a small change $\lx_1(\phi) \to
\lx_1(\phi) + c(\phi)$, with $|c(\phi)| \ll 1$. There are no constraints
on the form of the function $c(\phi)$, apart from the assumption that it
is small. If we restore the dimensions of the brane tension
(see eq. (\ref{action0})), our assumption is that $|c(\phi)| \ll (2 M)^4$
for the relavant values of $\phi$.

For the new brane tension,
the boundary conditions (\ref{bound1})
when substituted into eq. (\ref{eoma3})
lead to a new value for the bulk field, which we denote by
$\phit_1=\phi_1+\done$. Assuming $\done = {\cal O} (c(\phi_1))$ and
keeping terms up to order $c(\phi_1)$, we find
\be
\frac{\partial}{\partial \phi}
\left[ V -\frac{1}{8} \left( \frac{\partial \lx_1}{\partial \phi}\right)^2
+ \frac{1}{3} \lx_1^2 \right](\phi_1)\,\, \done=
-\frac{2}{3} \lx_1(\phi_1)c(\phi_1)
+\frac{1}{4} \frac{\partial \lx_1}{\partial \phi}(\phi_1)
\frac{\partial c}{\partial \phi}(\phi_1)+ {\cal O}(c^2(\phi_1)).
\label{dphi1} \ee
For the field derivative we find from the first of eqs. (\ref{bound1})
\be
\done'= \frac{1}{2} \frac{\partial^2 \lx_1}{\partial \phi^2}(\phi_1)\,\, \done
+ \frac{1}{2} \frac{\partial c}{\partial \phi} (\phi_1)
+ {\cal O}(c^2(\phi_1)).
\label{dphi1p} \ee
As we discussed above, we are looking for modifications of the brane tension
that lead to a new solution lying on the initial trajectory from
$(\phit_1,\phit'_1)$ to $(\phit_2,\phit'_2)$.
At a small distance $y$ from the initial location of the positive-tension
brane, assuming $y= {\cal O} (c(\phi_1))$, we have up to order $y$
\beq
\delta \phi'(y)=\phi''(0) \dy + {\cal O}(y^2)
=& \left[ -4 A'(\phi_1) \phi'_1 +
\frac{\partial V}{\partial \phi}(\phi_1) \right] \dy + {\cal O}(c^2(\phi_1))
\nonumber \\
=&
\left[ \frac{1}{3} \frac{\partial\lx_1^2}{\partial \phi}(\phi_1)+
\frac{\partial V}{\partial \phi}(\phi_1) \right] \dy
+ {\cal O}(c^2(\phi_1))
\label{dphi2} \\
\delta \phi (y)=\phi'(0) \dy + {\cal O}(y^2)=& \frac{1}{2}
\frac{\partial \lx_1}{\partial \phi}(\phi_1) \dy+ {\cal O}(c^2(\phi_1)).
\label{dphi2p}
\eeq

We identify $\delta \phi(y)$ and $\delta \phi'(y)$
with $\done$ and $\done'$ respectively.
Solving eq. (\ref{dphi2p}) for $\dy$, substituting into eq. (\ref{dphi2})
and employing eq. (\ref{dphi1p}) we find
\be
\frac{\partial}{\partial \phi}
\left[ V -\frac{1}{8} \left( \frac{\partial \lx_1}{\partial \phi}\right)^2
+ \frac{1}{3} \lx_1^2 \right](\phi_1)\,\, \done =
\frac{1}{4} \frac{\partial \lx_1}{\partial \phi}(\phi_1)
\frac{\partial c}{\partial \phi}(\phi_1)+ {\cal O}(c^2(\phi_1)).
\label{sol1} \ee
Comparison with eq. (\ref{dphi1}) leads to the requirement
$\lx_1(\phi_1)\, c(\phi_1)=0$ up to order $c(\phi_1)$.
As we would like to keep the form of $c(\phi)$ arbitrary, we are led to
the constraint $\lx_1(\phi_1)=0$. The interpretation is the following:
If the brane is initially located at a position where the value of the
bulk field is such that the brane tension vanishes, subsequent small
modifications of the brane tension can be absorbed in small displacements of
the brane with the metric retaining its four-dimensional Minkowski form.
We emphasize that it is not necessary for $\lx_1(\phi)$ to vanish for
any $\phi$. On the contrary, it may be of order 1 in general. Only the
presence of a zero is necessary. This is possible because the constraint
we derived is independent of the derivatives of $\lx(\phi)$, contrary to
the naive expectation.

There are two unsatisfactory elements in our solution: Firstly,
it seems inconsistent to introduce the brane tension as a
$\delta$-function source in
Einstein's equations and then require it to vanish. Notice, however, that
the derivatives of $\lx_1$ at $\phi_1$ are not constrained to be zero. Instead
they may be of order 1.
Secondly, it is questionable if a static solution can be obtained when
we take into account corrections of order $c^2(\phi_1)$. In order to
resolve these issues we need to expand the framework of our discussion
to include the possibility of more general metrics.
More specifically, we consider
the possibility that the four-dimensional ($y=$const.) part of the metric
may have (anti-)deSitter form.

\paragraph{Non-zero cosmological constant:}
We consider the ansatz \cite{dewolfe}
\be
ds^2 = e^{2 A(y)}\, g_{ij} \,dx^i dx^j - dy^2,
\label{metr2} \ee
with
\beq
dS_4:~~~~
g_{ij}dx^i dx^j=&dt^2-e^{2\sqrt{\Lx}t}\left( dx_1^2+dx_2^2+dx_3^2 \right)
\label{ds4} \\
AdS_4:~~~~
g_{ij}dx^i dx^j=&e^{-2\sqrt{-\Lx}t}\left( dt^2-dx_1^2-dx_2^2 \right)-dx_3^2.
\label{ads4} \eeq
Positive (negative) values of $\Lx$ correspond to deSitter (anti-deSitter)
four-dimensional metrics.
There is an arbitrariness in the choice of $\Lx$ and $A$, which we
remove by setting $A(y=0)=0$.
The presence of an non-zero effective cosmological constant $\Lx$
leads to the replacement of eq. (\ref{eoma3}) by \cite{dewolfe}
\be
A'^2 -\Lx e^{-2A}= -\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
\label{eomb3}
\ee
The boundary conditions (\ref{bound1}), (\ref{bound2}) remain unaffected.
Eqs. (\ref{bound1}), (\ref{eomb3}) then demonstrate
that the value of $\Lx$ is
determined by the mis-match at the location of the first brane
between the brane tension and the
potential of the bulk field that plays the role of the bulk cosmological
constant. The limit $\Lx \to 0$ reproduces the Minkowski metric we considered
earlier.

Within this framework, no fine-tuning is required. General choices of
$\lx_1(\phi)$, $\lx_2(\phi)$ are expected to lead to a solution with some value
of $\Lx$ \footnote
{It is still possible that no solution exists
for certain choices of $\lx_1(\phi)$, $\lx_2(\phi)$. Our intention is
to show that there are large continuously connected
families of $\lx_1(\phi)$, $\lx_2(\phi)$ for which solutions can be found.
}.
There is a graphic way to see this: Substitution of
eqs. (\ref{bound1}) into eq. (\ref{eomb3}) with $\Lx=\Lx_1$ and
$A_1=0$ gives the value
$\phi_1$ at the location of the positive-tension brane as a function of $\Lx$.
The first of eqs. (\ref{bound1}) gives $\phi'_1$. The set of initial
conditions forms a curve $C_1$ on the $A=0$ plane, parametrized by $\Lx_1$.
We assume
that $\Lx_1$ grows in the direction of the arrow in fig. 1.
For given $\Lx_1$,
the initial conditions $(\phi_1,\phi_1',0)$ result in a
unique solution for eqs. (\ref{eoma1}), (\ref{eomb3}). This corresponds to
a trajectory in $(\phi,\phi',A)$ space. Varying $\Lx_1$ results in a surface
formed by the various trajectories.

At the location of the second brane, eq. (\ref{eomb3}) must be satisfied
after substitution of eqs. (\ref{bound2}). For given $\Lx=\Lx_2$, the
possible values $(\phi_2,\phi'_2,A_2)$ form a curve in $(\phi,\phi',A)$ space.
In general, this curve meets the surface of trajectories at some point A. The
trajectory going through A corresponds to a value $\Lx_1$ that is
not necessarily equal to $\Lx_2$. However, identification of
$\Lx_1$ and $\Lx_2$ is expected
to be possible in general through variation of $\Lx_2$. The
point A moves with changing $\Lx_2$, in a way that it creates a curve $C_2$
on
the trajectory surface. On $C_2$ trajectories characterized by $\Lx_1$
meet boundary conditions characterized by $\Lx_2$.
For large families of functions $\lx_2(\phi)$,
the identification of $\Lx_1$ and $\Lx_2$ should be possible at
some point on $C_2$  without the necessity of fine-tuning.
For example, one may consider some function $\lx_2(\phi)$ for which
$\Lx_2$ increases on $C_2$ in the direction indicated by the arrow
in fig. 1, opposite to the direction of increase of $\Lx_1$.
The location of A with $\Lx_1=\Lx_2=\Lx$ determines the trajectory
that corresponds to a static solution of eqs. (\ref{eoma1}), (\ref{eomb3})
under the boundary conditions (\ref{bound1}), (\ref{bound2}).

The presence of the term proportional to $\Lx$ in eq. (\ref{eomb3})
modifies the nature of the solutions relative to the case with
four-dimensional Minkowski metric.
The nature of the change
can be seen by solving eq. (\ref{eomb3}) for
a flat potential
$V(\phi)=-\bar{\Lx}$=const. and $\phi$=const.
For $0< |\Lx| \ll 1$ one finds
that $A'$ remains constant (as in the case with $\Lx=0$),
but then quickly diverges or goes to zero
for $\Lx >0$ or $\Lx <0$ respectively, at
a distance
\be
R_1 \simeq -\frac{\sqrt{3}}{2\sqrt{\bar{\Lx}}} \ln |\Lx|
= {\cal O} \left(\left| \ln  |\Lx|  \right| \right)
\label{div} \ee
from the positive-tension brane \cite{dewolfe}.
The negative-tension brane must exist at $y=R < R_1$ in both cases.
For a $y$-dependent $\phi$, one expects that
the solutions with $\Lx=0$  will not be modified significantly
if $|A|$
is sufficiently small for the second term to be negligible relative
to the first one in the lhs of eq. (\ref{eomb3}). This implies that
the second brane must be located at a distance
$y= R < R_1 \sim {\cal O} \left(\left| \ln  |\Lx|  \right| \right)$.
This is confirmed by numerical studies.
If the negative-tension
brane in the solution with $\Lx=0$ was
located at $R > R_1 $, the solution for $\Lx \not= 0$ must
change drastically in order to accomodate the second brane
much closer to the first one than before \cite{far}.
This observation may provide a link between the cosmological constant
and the brane location. Configurations with branes far apart seem to be
possible only
if the effective cosmological constant is exponentially small.

\paragraph{Small cosmological constant:}
We now return to the problem of finding a static configuration after
the tension of the first brane has changed from $\lx_1(\phi)$ to
$\lx_1(\phi)+c(\phi)$.
With the new brane tension the boundary conditions
at $y=0$ are given by a new curve $\Ct_1$ on the $A=0$ plane.
Let us consider the points $(\phit_1(\Lt_1),\phit'_1(\Lt_1))$
on $\Ct_1$  that are close to
the point $(\phi_1,\phi'_1)$ on $C_1$ at $\Lx_1=0$.
Denoting $\phit_1=\phi_1+\done$ and repeating the calculation that led
to eqs. (\ref{dphi1}), (\ref{dphi1p}) we find
\be
\frac{\partial}{\partial \phi}
\left[ V -\frac{1}{8} \left( \frac{\partial \lx_1}{\partial \phi}\right)^2
+ \frac{1}{3} \lx_1^2 \right](\phi_1)\,\, \done + 3 \Lt_1=
-\frac{2}{3} \lx_1(\phi_1)c(\phi_1)
+\frac{1}{4} \frac{\partial \lx_1}{\partial \phi}(\phi_1)
\frac{\partial c}{\partial \phi}(\phi_1)+ {\cal O}(c^2(\phi_1)).
\label{dphi1t} \ee
and
\be
\done'=~ \frac{1}{2} \frac{\partial^2 \lx_1}{\partial \phi^2}(\phi_1)\,\, \done
+ \frac{1}{2} \frac{\partial c}{\partial \phi} (\phi_1)
+ {\cal O}(c^2(\phi_1)).
\label{dphi1pt} \ee

Let us consider the variations of $(\phi,\phi')$ on the
trajectory that starts at
the point $\Lx_1=0$ on $C_1$.
They are given by eqs. (\ref{dphi2}), (\ref{dphi2p}).
It is possible now to identify them with $(\done,\done')$ of eqs.
(\ref{dphi1t}), (\ref{dphi1pt}) if
\be
\Lt_1=-\frac{2}{9} \lx_1(\phi_1)c(\phi_1) + {\cal O}(c^2(\phi_1)).
\label{cosm} \ee
In the limit $\lx_1(\phi_1)\to 0$, we obtain
$\Lt_1={\cal O}(c^2(\phi_1))$. Moreover, as $A'(y=0)=-\lx_1(\phi_1)/3$, we
conclude that $A(y=\dy) \to 0$. This means that, for
$\lx_1(\phi_1)\to 0$, the trajectory that starts at
the point $\Lx_1=0$ on $C_1$ passes through the
curve $\Ct_1$ at some point with $\Lt_1={\cal O}(c^2(\phi_1))$.
We also know where this trajectory ends:
on the curve $C_2$ at $\Lx_2=0$. (This is the initial fine-tuning
that we assumed to have been achieved.)

If we concentrate only on the part of the trajectory from
$\Ct_1$ to $C_2$, we have the situation we analyzed earlier. A
slight mis-match between $\Lt_1$ and $\Lx_2$. But, as we argued before,
there should be a nearby trajectory for which $\Lt_1$ and $\Lx_2$ can be
matched, especially if $\Lt_1$ and $\Lx_2$ grow in opposite directions
on $C_2$.
For the case of interest $\Lt_1={\cal O}(c^2(\phi_1))$
and $\Lx_2=0$, this trajectory is expected to have
$\Lx_f={\cal O}(c^2(\phi_1))$.

The upshot of this complicated reasoning is that we identified
a static solution of Einstein's equations with tensions
$\lx_1(\phi)+c(\phi)$ for the first brane and
$\lx_2(\phi)$ for the second. It is a solution with an
effective four-dimensional constant $\Lx$.
Contrary to expectations, $\Lx$ is not of order $c(\phi_1)$, but of
order $c^2(\phi_1)$. The reason is that the location of the first
brane has been shifted by an amount $dy={\cal O}(c(\phi_1))$, so
as to absorb the leading effect from the change of the brane tension.
The necessary condition is $\lx_1(\phi_1) \approx 0$. However, it
is apparent from eq. (\ref{cosm}) that even for
$\lx_1(\phi_1) = {\cal O} (c(\phi_1))$ our conclusions remain valid.
In order not to change the positivity of the tension for any sign of
$c(\phi_1)$,  it is
preferable to take
$\lx_1(\phi_1)$ somewhat larger than $|c(\phi_1)|$.


\paragraph{Discussion:}

The basic objective of our approach was to compensate a possible
change in a brane tension with a slight modification of the
two-brane configuration in a way that keeps the effective cosmological
constant $\Lx$ small. We considered
variations of the positive tension
$\lx_1(\phi)$ of the first
brane. The reason is that we identify our Universe with the positive-tension
brane, while we view the negative-tension one as a regulator that cuts
off possible singularities in Einstein's equations.  We allowed arbitrary
small changes $c(\phi)$ of $\lx_1(\phi)$, while we kept the form of the
negative tension $\lx_2(\phi)$ and the potential $V(\phi)$ of the bulk field
fixed. We view $\lx_1(\phi)$, $\lx_2(\phi)$ and $V(\phi)$ as effective
low-energy quantities. In particular, $\lx_1(\phi)$ represents the vacuum
energy on the first brane, induced by the fields localized on it (such as the
Standard Model fields) and their interactions with
the bulk field. The changes $c(\phi)$ originate in
variations of the characteristic
energy scale on the first brane, or possible phase transitions
on the brane. We assumed that, apart from isolated points where it
may approach zero,
$\lx_1(\phi)$ is of order
1 in units of the fundamental Planck's constant, while $|c(\phi)| \ll 1$.

We started from an initial configuration with zero effective cosmological
constant. This requires an initial fine-tuning of
$\lx_1(\phi)$ and  $\lx_2(\phi)$ for which we do not have a convincing
explanation. We can only speculate that, since configurations with
the two branes far apart can exist only if $\Lx$ is exponentially
small, the fine-tuning
may be a consequence of the initial location of the branes.
Our concern was the destabilization of this configuration every time
$\lx_1(\phi)$ changes. We looked for possible new static configurations,
similar to the initial one. We did not address the question of
the evolution of the system from one configuration to the other.
This requires a time-dependent solution of Einstein's equations, with
ansatzes for the metric much more general than the ones we employed.
The technical difficulties involved in obtaining such solutions
are a significant obstacle in this direction.

We demonstrated in a graphic way the known fact that, for
general $\lx_1(\phi)$, $\lx_2(\phi)$, a static configuration exists
with some value of the cosmological constant. Consequently, every
change $c(\phi)$ of $\lx_1(\phi)$ from its initial fine-tuned form
results in a new static configuration with non-zero $\Lx$.
In general, one expects $\Lx = {\cal O}(c(\phi_1))$, where $\phi_1$ is
the value of the bulk field at the initial location of the first brane.
We showed that this is not always the case.
Our main result is that one can have $\Lx = {\cal O}(c^2(\phi_1))$.
The only requirement is that $\phi_1$ is near a zero of
$\lx_1(\phi)$, so that $\lx_1(\phi_1) = {\cal O}(c(\phi_1))$.
We point out that the derivatives of $\lx_1(\phi)$ at $\phi_1$ are
in general of order 1. If $\lx_1(\phi)$ can become negative we assume
that $\lx_1(\phi_1)$ is somewhat larger than $|c(\phi_1)|$,
so that the positivity of the tension is maintained for any sign of
$c(\phi_1)$. A promising possibility is that $\lx_1(\phi)$
has minimum near $\phi_1$, where $\lx_1(\phi)$ and $c(\phi)$ are
comparable.


The implications of our result for a possible
realistic scenario are interesting. In order to be
more specific, we return to dimensful quantities
We take the fundamental constant $M$ defined in
the beginning to be $M={\cal O} (10^{19}$ GeV). For the four-dimensional
Planck's constant we expect \cite{dewolfe}
\be \frac{M^2_4}{M^2} \approx \int_0^{R} dr\, e^{2A(r)} = {\cal O} (1).
\label{fourd} \ee
Recent observations are consistent with a cosmological constant of the
order of the critical density of the Universe \cite{obsc}
\be
\frac{\Lx}{M^4} = {\cal O} \left(10^{-120} \right).
\label{obscosm} \ee
If we assume an initial two-brane configuration with zero effective
cosmological constant, subsequent changes in the tension of the
first brane are consistent with the above constraints if
\be
c(\phi_1)={\cal O} \left( 10^{-60} M^4 \right)
={\cal O} \left( \left( 10{\rm~ TeV}\right)^4 \right).
\label{cbound} \ee

Let us assume that, for a
certain value $\phi_1$ of the bulk field,
there is an ultraviolet cutoff of order 10 TeV
for the vacuum energy associated with the fields of the first brane.
Then $\lx(\phi_1)$ and $c(\phi_1)$ are comparable and of order
$\left( 10{\rm~ TeV}\right)^4$.
Our result implies that
the resulting
cosmological constant can be consistent with observational data.
Moreover, cosmological phase transitions at scales below 10 TeV do not
modify substantially this conclusion. All phase transitions predicted by
known physics (such as the electroweak or the QCD phase transitions)
fall in this category. The nature of the cutoff cannot be specified by
our considerations. However, the possibility that supersymmetry provides
the necessary mechanism for the cancellation of quantum contributions
to the energy density at energy scales above 10 TeV is intriguing.

Before concluding, we would like to emphasize that the mechanism
we presented should be viewed only as a simple example of how
a non-trivial topology can ameliorate the cosmological constant problem.
We find that its most important merit relative to alternative proposals
within the same framework \cite{adks} is the absence of singularities and
strong assumptions about the form of the interactions of the
brane fields with the bulk field. Our initial ansatz for the effective action
of the system is general and the only assumption about the changes
of the brane tension is that they are small in units of Planck's constant.
Moreover, our scenario allows for a non-zero cosmological constant of
the right order of magnitude \cite{obsc}.



\vspace{0.5cm}
\noindent
{\bf Acknowledgements}:
I would like to thank E. Kiritsis, A. Krause, R. Rattazzi and A. Strumia
for useful discussions.
This research was supported by the E.C.
under TMR contract No. ERBFMRX--xxx.


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L. Randall and R. Sundrum,
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\bibitem{adks}
N. Arkani-Hamed, S. Dimopoulos, N. Kaloper and R. Sundrum,
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\end{thebibliography}



%\newpage






\end{document}


\bibitem{dewolfe2}
D.Z. Freedman, S.S. Gubser K. Pilch and N.P. Warner,
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U. Ellwanger, hep-th/0001126.

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A. Lukas, B.A. Ovrut, K. Stelle and D. Waldram,
Phys. Rev. D {\bf 59}, 086001 (1999), hep-th/9803235;
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Nucl. Phys. B {\bf 5559}, 71 (1999), hep-ph/ 9811133.


 \begin{figure}[t]
 %\vspace{1.cm}
 \centerline{\psfig{figure=fig2.eps,width=12cm}}
 \caption{
 The y-dependence of $\phi$, $\phi'$ and $A'$ near the
 negative-tension brane.}
 \label{fig2}
 \end{figure}


 Follow ref. \cite{dewolfe} and consider potentials
 Eqs. (\ref{eoma1})--(\ref{eoma3}) are solved by
 if
 \be
 W(\phi(y_\alpha))=\lx_{\alpha}(\phi(y_\alpha))
 ~~~~~~~~~~~~~~~~~~~~~
 \frac{\partial W}{\partial \phi}(\phi(y_\alpha))=
 \frac{\partial \lx_{\alpha}}{\partial \phi}(\phi(y_\alpha)).
 \label{bc1} \ee
 Now consider the ansatz \cite{dewolfe}
 \beq
 ds^2 = \; &e^{2 A(y)}\, \bar{g}_{ij} \,dx^i dx^j - dy^2
 \label{metr2a} \\
 dS_4(\Lx > 0): ~~~~~\bar{g}_{ij} dx^idx^j
 = \; &dt^2 -e^{2\sqrt{\Lx}t} \left(dx_1^2+dx_2^2+dx_3^2 \right)
 \label{metr2b} \\
 AdS_4(\Lx < 0): ~~~~~\bar{g}_{ij} dx^idx^j
 = \; &e^{-2\sqrt{-\Lx}t} \left(dt^2-dx_1^2-dx_2^2 \right) -dx^2_3
 \label{metr2c} \eeq
 For the potential of eq.~(\ref{ex2}) with $b=0$
 the solution of eq.~(\ref{eomb3}) is
 \beq
 \Lx > 0:
 ~~~~~~
 e^A = &\sqrt{\Lx}L \sinh \frac{y_1-y}{L}
 ~~~~~~
 \lx_1=\frac{3}{L} \coth\frac{y_1}{L}
 ~~~~~~
 \lx_2=-\frac{3}{L} \coth\frac{y_1-y_0}{L}
 \label{sing1} \\
 \Lx < 0:
 ~~~~~~
 e^A = &\sqrt{-\Lx}L \cosh \frac{y_1-y}{L}
 ~~~~~~
 \lx_1=\frac{3}{L} \tanh\frac{y_1}{L}
 ~~~~~~
 \lx_2=-\frac{3}{L} \tanh\frac{y_1-y_0}{L}.
 \label{sing2} \eeq



 \beq
 \phi''+4 A' \phi'= &\frac{\partial V(\phi)}{\partial \phi} +
 \sum_\alpha
 \frac{\partial \lx_\alpha (\phi)}{\partial \phi}\, \delta (y-y_\alpha)
 \label{eomb1} \\
 A'' +\Lx e^{-2A}= &-\frac{2}{3}\phi'^2-\frac{2}{3}
 \sum_\alpha
 \lx_\alpha (\phi) \,\delta (y-y_\alpha)
 \label{eomb2} \\
 A'^2 -\Lx e^{-2A}= &-\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
 \label{eomb3}
 \eeq


 \begin{figure}[t]
 %\vspace{1.cm}
 \centerline{\psfig{figure=fig2.eps,width=12cm}}
 \caption{
 The y-dependence of $\phi$ and $A'$ between the two branes.}
 \label{fig2}
 \end{figure}

 In fig. \ref{fig2} we




 It is usually assumed that the form of $\lx_\alpha (\phi)$
 constraints $\phi$ to have specific values (for example
 near deep minima of $\lx_\alpha (\phi)$ \cite{goldwise}) at the
 location of the branes, and this stabilizes the system.
 However, this requires significant interactions of the bulk
 field with the observable brane fields.
 We are interested in the opposite possibility, i.e.
 that the bulk field
 interacts only weakly with the positive-tension brane
 (where we assume that our universe is located).
 We consider $\lx_1(\phi)$ as a slowly varying function of
 $\phi$ that describes the brane vacuum energy due to the
 self-interactions of the brane fields and their weak
 coupling to the bulk field.
 First we show that stabilized solutions can be obtained within
 this approach.

 We consider the extreme case of a brane tension $\lx_1$ that
 is independent of $\phi$.
 Eq. (\ref{eoma1}) is the equation of motion of a ``particle''
 rolling down a potential $-V(\phi)$ with a ``friction'' term
 determined through eq. (\ref{eoma3}).
 We consider a brane with positive tension $\lx_1 >0$ located
 at $y=0$. This implies that we must choose the negative root of
 eq. (\ref{eoma3}) for $A'$. As a result the ``friction'' term
 tends to accelerate the ``particle'' instead of slowing it down.
 Because we have assumed that $\lx_1$ is $\phi$-independent, the
 ``particle'' starts rolling with zero initial ``velocity'' $\phi'(0)$.
 The second brane can be located at a point $y=R$ where $A'$, $\phi'$
 take values corresponding to
 the negative tension of the second brane $\lx_2(\phi)$.
 Thus, the distance $R$
 between the branes is uniquely determined.
 An appropriate form of $\lx_2(\phi)$
 must be employed for this to happen, as eq. (\ref{eoma3}) must
 be satisfied at $y=R$. This is the expected fine-tuning associated
 with the vanishing of the effective four-dimensional cosmological constant.

 Similarly to the situation near $y=0$, it is interesting to look for
 solutions with $\phi'(R)=0$. This
 can be achieved only if
 the ``particle'' rolls within a convex region of $-V(\phi)$. Moreover, in order
 to compensate the accelerating effect of the ``friction'' term near the
 second brane, the potential must become very steep for
 the relevant values of $\phi$. In this way, ``solitonic'' configurations can
 be obtained with a definite distance $R$ between the branes. However,
 the part for the potential $V(\phi)$ that is relevant for these solutions
 has negative curvature. Even though the complete stability analysis
 is very difficult to perform,
 it is probable that they are unstable. For this reason we concentrate
 on solutions with positively curved potentials.

 \begin{figure}[t]
 %\vspace{1.cm}
 \centerline{\psfig{figure=fig1.eps,width=12cm}}
 \caption{
 The y-dependence of $\phi$, $\phi'$ and $A'$ near the
 negative-tension brane.}
 \label{fig1}
 \end{figure}

 We consider
 a toy model with a potential
 \be
 V(\phi) = -\Lx + c \phi^n.
 \label{pot} \ee
 In fig. \ref{fig1} we plot the solution of eqs. (\ref{eoma1})--(\ref{eoma3})
 for the choice $\Lx=1$, $c=1$, $n=6$.
 In practice a numerical solution can be obtained as follows:
 We start by choosing values of $\lx_1$ and $\phi(0)$ that satisfy
 eq. (\ref{eoma3}) with $\phi'(0)=0$.
 The numerical integration of eq. (\ref{eoma1}) is then straightforward.
 In fig. \ref{fig1} we depict a solution with $\phi(0) = 10^{-4}$.
 The brane tension $\lx_1$ has a value very close to $\sqrt{3}$.
 As the initial value of the field corresponds to the very flat part
 of the potential, the evolution is very slow: $\phi$ and $A'$ remain
 constant for a long distance from the brane. Slowly $\phi$
 grows to values near 1.
 Eventually the ``particle'' rolls down the steep part of the
 potential $-V(\phi)$. This leads to a singularity with
 diverging $\phi$ and $A'$. The negative-tension brane can be placed
 at any point $y=R$ on the left of the singularity.
 Its tension must be chosen as
 \be
 \lx_2(\phi)= 3A'(R)-\phi'(R)\left[\phi-\phi(R) \right] +
 \gamma_2 \left[\phi-\phi(R) \right]^2 +~...
 \label{endpoint} \ee
 The parameter $\gamma_2$ is not determined by our considerations, but
 we can assume that it is small.
 This behaviour is generic for all the stabilizing potentials $V(\phi)$.
 At least one of the branes must be located in the flat part
 of $V(\phi)$, so that a large distance can separate them.

 In fig. \ref{fig1} we display two possibilities:
 The brane (a) has
 $R=18$,
 $\phi(R)=1.28 \times 10^{-2}$,
 $\phi'(R)=2.94 \times 10^{-2}$,
 $A'(R)=-0.577$. If we assume that $\gamma_2$ is small as well, the
 brane is weakly coupled to the bulk field.
 The brane (b) has
 $R=19.8$,
 $\phi(R)=1.02$,
 $\phi'(R)=4.03$,
 $A'(R)=-1.63$.
 It is coupled much more strongly to $\phi$.
 It is located close to the singularity and one can speculate
 that its presence is connected to the existence of a strong gravitational
 background. In both the above scenarios the positive-tension brane
 has no direct
 interactions with $\phi$ and its location is determined only through
 gravitational interactions. The total configuration is expected to
 be stable as the potential has positive curvature. We also point out that
 a (probably unstable) configuration with $\lx_2$ independent of $\phi$ can
 be obtained for potential given by eq. (\ref{pot}) with
 $\Lx=1$, $c=-1$, $n=6$, $\lx_1=0.577$.

 If the branes are weakly coupled to the bulk field, it is probable that
 their location can be shifted in order to accomodate changes in
 their tension. These can be caused by
 possible phase transitions on the brane that change the value of the
 brane cosmological constant. We consider a simplified version of this
 scenario, in which we omit the matter on the brane (and its effects, such as
 standard cosmological expansion). Instead, we concentrate
 on the possibility that the four-dimensional ($y=$const.) part of the metric
 has (anti-)deSitter form.

 The presence of an non-zero effective cosmological constant leads to
 the replacement of eq. (\ref{eoma3}) by \cite{dewolfe}
 \be
 A'^2 -\Lx e^{-2A}= -\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
 \label{eomb3}
 \ee
 Positive (negative) values of $\Lx$ correspond to deSitter (anti-deSitter)
 four-dimensional metric.
 There is an arbitrariness in the choice of $\Lx$ and $A$, which we
 remove by setting $A(y=0)=0$ or $A(y=R)=0$, depending on which
 brane we are considering.
 Eq. (\ref{eomb3}) then demonstrates that the value of $\Lx$ is
 determined by the mis-match between the brane tension and the
 bulk cosmological constant.

 The presence of the new terms, proportional to $\Lx$,
 modifies the solutions we discussed earlier.
 The nature of the change can be seen by assuming that $\phi$=const. and
 solving eq. (\ref{eomb3}).
 For $0< \Lx \ll 1$, one finds
 that $A'$ remains constant, but then quickly diverges or goes to zero,
 for $\Lx >0$ or $\Lx <0$ respectively, at
 a distance
 \be
 R_1 = -\frac{1}{4} \ln \left( |\Lx| \right)
 \label{div} \ee
 from the brane.
 The negative-tension brane must exist at $y=R < R_1$ in both cases.
 If one allows for a $y$-dependent $\phi$,
 the solutions we discussed earlier are not modified significantly
 for $|\Lx|$ sufficiently small. However, if the brane was initially
 located at $R \gta -\ln \left( |\Lx| \right)/4$, the solution must
 change drastically in order to accomodate the second brane
 much closer to the first one than before.

 One can imagine the following situation: A system of two branes
 is stabilized initially at a large distance $R = {\cal{O}}(100)$ with
 zero effective cosmological constant $\Lx$.
 Subsequently, the positive tension of the first brane is modified through, for
 example, a phase transition on the brane. We assume that a new stable
 configuration is approached. There are two possibilites: Either the
 effective cosmological constant on the brane becomes non-zero, or
 its previous zero value is maintained through an appropriate shift of
 the brane in the fifth dimension.
 We would like to demonstrate that the second scenario
 leads to very small modifications of the initial configuration. Thus,
 it seems more likely.

 In order not to rely exclusively on numerical solutions, we
 work within the class of models discussed in ref. \cite{dewolfe}, for
 which exact analytical solutions exist.
 They involve potentials of the special form
 \be
 V(\phi) = \frac{1}{8} \left[ \frac{\partial W(\phi)}{\partial \phi}\right]^2
 -\frac{1}{3} \left[ W(\phi) \right]^2,
 \label{ww} \ee
 with $W(\phi)$ a general function of $\phi$.
 The solution of eqs. (\ref{eoma1})--(\ref{eoma3}) is \cite{dewolfe}
 \beq
 \phi'=&~\frac{1}{2} \frac{\partial W(\phi)}{\partial \phi}
 \label{sol1a} \\
 A'=&-\frac{1}{3} W(\phi),
 \label{sol1b} \eeq
 with
 \beq
 \lx_1(\phi)= &~W(\phi(0))+\frac{ \partial W(\phi(0))}{\partial \phi}
 \left[
 \phi-\phi(0)\right] +\gamma_1 \left[\phi-\phi(0)\right]^2 +~...
 \label{sol1c} \\
 \lx_2(\phi)= &-W(\phi(R))-\frac{ \partial W(\phi(R))}{\partial \phi}
 \left[
 \phi-\phi(R)\right] +\gamma_2\left[ \phi-\phi(R) \right]^2 +~...
 \label{sol1d} \eeq
 We point out that this class of models is too restrictive
 to contain the possibility $\lx_1 =$ const. that we discussed earlier.
 However, it can accomodate branes whose tension depends only weakly on
 the bulk field $\phi$.

 For the choice
 \beq
 W(\phi)= &3-b\phi^2
 \label{ex1} \\
 V(\phi)= &-3 + \left( \frac{b^2}{2}
 +2b \right) \phi^2 -\frac{b^2}{3} \phi^4,
 \label{ex4} \eeq
 the solution is \cite{dewolfe}
 \beq
 \phi(y)= &\phi_1\; e^{-by}
 \label{solex1} \\
 A(y)=&-y + \frac{1}{6}\phi_1^2 \left(1- e^{-2b y}\right).
 \label{solex2} \eeq
 The distance between the branes is
 \be
  R=\frac{1}{b}\ln \left( \frac{\phi_1}{\phi_2} \right),
 \label{dist} \ee
 with $\phi_2=\phi(R)$.
 For values
 $b= {\cal O} \left( 10^{-2} \right)$,
 $\phi_1 /\phi_2= {\cal O} \left(1 \right)$, such as the ones
 considered in ref. \cite{dewolfe},
 one finds $R={\cal O} \left( 10^{2} \right)$.

 For the choice $\gamma_1=-b$ the tension of the first
 brane is
 \be
 \lx_1(\phi)=3-b\phi^2= W(\phi)
 \label{choice} \ee
 for any value of $\phi_1$. The fact that $\lx_1(\phi)$ has
 negative curvature does not indicate an instability.
 This function appears in the equation of motion of $\phi$, or
 its perturbations around a given background, always multiplied
 by  a $\delta$-function. As a result, it only determines the cusp in the
 first derivative of the bulk field near the brane.
 The potential $V(\phi)$ has positive curvature for the values of
 $\phi$ that we are considering.

 Let us now consider the possibility that a phase
 transition modifies the brane vacuum energy. This can be modelled
 by adding a constant $C$ to the tension $\lx_1(\phi)$. We
 would like to find a static solution that can accomodate this change.
 One of the possibilities that we mentioned earlier is that
 the bulk field retains its value at the location of the
 brane. This implies that $A'$, given by the second of eqs. (\ref{bound1}),
 is shifted by $-C/3$. A non-zero effective cosmological constant
 $\Lx$ appears, determined through eq. (\ref{eomb3}) with $A=0$.
 However, a static solution cannot be obtained unless
 the second brane is displaced at a distance $R \lta R_1$, with
 $R_1$ given by eq. (\ref{div}).

 The alternative possibility is that the value of the bulk field
 $\phi$ changes so that the cosmological constant remains zero.
 The new value is given by







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\noindent
%July 1999
\begin{flushright}
%SNS--PH/1999--11
%\\
%hep--ph/9908209
\end{flushright}
\vspace{3cm}
\begin{center}
{ \Large \bf
On Brane Stabilization and the Cosmological Constant
}
\\ \vspace{1cm}
{\large
N. Tetradis
}
\\
\vspace{1cm}
{\it
Department of Physics, University of Athens,
157 71 Athens, Greece
}
\\
\vspace{2cm}
\abstract{
Abstract
\\
\vspace{1cm}
%PACS number: 98.80.Cq
}
\end{center}
\vspace{4cm}
\noindent
%CERN--TH/97--XXX \\
%July 1999


\newpage

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We consider a system of two branes in the background of
a bulk scalar field $\phi$. The action is given by
(we follow the notation of ref. \cite{dewolfe} and rescale all dimensionful
quantities by Planck's constant)
\be
S=\int d^4x\,dy \, \sqrt{\left| \det g_{\mu\nu}\right|} \,
\left[ -\frac{1}{4} R + \frac{1}{2} \left( \partial \phi \right)^2
- V(\phi) \right] -\sum_{\alpha=1,2}
\int d^4x \, \sqrt{\left| \det g_{ij}\right|}\, \lx_\alpha(\phi).
\label{action}
\ee
For the metric we assume the ansatz
\be
ds^2 = e^{2 A(y)}\, \eta_{ij} \,dx^i dx^j - dy^2
\label{metr1} \ee
with space-time topology $R^{3,1}\times S^1/Z_2$ \cite{rs1}.
The two branes are located at the boundaries of the fifth dimension.
Einstein's equations and the equation of motion of the field are
\cite{dewolfe}
\beq
\phi''+4 A' \phi'= &\frac{\partial V(\phi)}{\partial \phi} +
\sum_\alpha
\frac{\partial \lx_\alpha (\phi)}{\partial \phi}\, \delta (y-y_\alpha)
\label{eoma1} \\
A'' = &-\frac{2}{3}\phi'^2-\frac{2}{3}
\sum_\alpha
\lx_\alpha (\phi) \,\delta (y-y_\alpha)
\label{eoma2} \\
A'^2 = &-\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
\label{eoma3}
\eeq
We set $y_1=0$ and $y_2=R$.

The solutions of the above equations for general potentials $V(\phi)$
predict a fixed distance $R$ between the two branes. They generalize
the stabilization mechanism of ref. \cite{goldwise} by taking into account
the backreaction of the scalar field on the gravitational background.
The functions $\lx_\alpha (\phi)$ determine the brane tensions. Their
presence
imposes boundary conditions for $A'(y)$ and $\phi(y)$ at $y=0,R$.
The integration of eq. (\ref{eoma1}), (\ref{eoma2})
around the $\delta$-functions and use of the $S^1/Z_2$ symmetry
leads to
\beq
y=0
~~~~~~~~~~~~~~~~
\phi'=&~\frac{1}{2}\frac{\partial \lx_1(\phi)}{ \partial \phi}
~~~~~~~~~~~~~~~~
A'=-\frac{1}{3} \lx_1(\phi)
\label{bound1} \\
y=R
~~~~~~~~~~~~~~~~
\phi'=&~-\frac{1}{2}\frac{\partial \lx_2(\phi) }{\partial \phi}
~~~~~~~~~~~~
A'=\frac{1}{3} \lx_2(\phi).
\label{bound2} \eeq
By imposing these conditions, one has only to solve
eqs. (\ref{eoma1}) and (\ref{eoma3}), neglecting the
$\delta$-function contributions.

It is usually assumed that the form of $\lx_\alpha (\phi)$
constraints $\phi$ to have specific values (for example
near deep minima of $\lx_\alpha (\phi)$ \cite{goldwise}) at the
location of the branes, and this stabilizes the system.
However, this requires significant interactions of the bulk
field with the observable brane fields.
We are interested in the opposite possibility, i.e.
that the bulk field
interacts only weakly with the positive-tension brane
(where we assume that our universe is located).
We consider $\lx_1(\phi)$ as a slowly varying function of
$\phi$ that describes the brane vacuum energy due to the
self-interactions of the brane fields and their weak
coupling to the bulk field.
First we show that stabilized solutions can be obtained within
this approach.

We consider the extreme case of a brane tension $\lx_1$ that
is independent of $\phi$.
Eq. (\ref{eoma1}) is the equation of motion of a ``particle''
rolling down a potential $-V(\phi)$ with a ``friction'' term
determined through eq. (\ref{eoma3}).
We consider a brane with positive tension $\lx_1 >0$ located
at $y=0$. This implies that we must choose the negative root of
eq. (\ref{eoma3}) for $A'$. As a result the ``friction'' term
tends to accelerate the ``particle'' instead of slowing it down.
Because we have assumed that $\lx_1$ is $\phi$-independent, the
``particle'' starts rolling with zero initial ``velocity'' $\phi'(0)$.
The second brane can be located at a point $y=R$ where $A'$, $\phi'$
take values corresponding to
the negative tension of the second brane $\lx_2(\phi)$.
Thus, the distance $R$
between the branes is uniquely determined.
An appropriate form of $\lx_2(\phi)$
must be employed for this to happen, as eq. (\ref{eoma3}) must
be satisfied at $y=R$. This is the expected fine-tuning associated
with the vanishing of the effective four-dimensional cosmological constant.

Similarly to the situation near $y=0$, it is interesting to look for
solutions with $\phi'(R)=0$. This
can be achieved only if
the ``particle'' rolls within a convex region of $-V(\phi)$. Moreover, in order
to compensate the accelerating effect of the ``friction'' term near the
second brane, the potential must become very steep for
the relevant values of $\phi$. In this way, ``solitonic'' configurations can
be obtained with a definite distance $R$ between the branes. However,
the part for the potential $V(\phi)$ that is relevant for these solutions
has negative curvature. Even though the complete stability analysis
is very difficult to perform,
it is probable that they are unstable. For this reason we concentrate
on solutions with positively curved potentials.

\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=fig1.eps,width=12cm}}
\caption{
The y-dependence of $\phi$, $\phi'$ and $A'$ near the
negative-tension brane.}
\label{fig1}
\end{figure}

We consider
a toy model with a potential
\be
V(\phi) = -\Lx + c \phi^n.
\label{pot} \ee
In fig. \ref{fig1} we plot the solution of eqs. (\ref{eoma1})--(\ref{eoma3})
for the choice $\Lx=1$, $c=1$, $n=6$.
In practice a numerical solution can be obtained as follows:
We start by choosing values of $\lx_1$ and $\phi(0)$ that satisfy
eq. (\ref{eoma3}) with $\phi'(0)=0$.
The numerical integration of eq. (\ref{eoma1}) is then straightforward.
In fig. \ref{fig1} we depict a solution with $\phi(0) = 10^{-4}$.
The brane tension $\lx_1$ has a value very close to $\sqrt{3}$.
As the initial value of the field corresponds to the very flat part
of the potential, the evolution is very slow: $\phi$ and $A'$ remain
constant for a long distance from the brane. Slowly $\phi$
grows to values near 1.
Eventually the ``particle'' rolls down the steep part of the
potential $-V(\phi)$. This leads to a singularity with
diverging $\phi$ and $A'$. The negative-tension brane can be placed
at any point $y=R$ on the left of the singularity.
Its tension must be chosen as
\be
\lx_2(\phi)= 3A'(R)-\phi'(R)\left[\phi-\phi(R) \right] +
\gamma_2 \left[\phi-\phi(R) \right]^2 +~...
\label{endpoint} \ee
The parameter $\gamma_2$ is not determined by our considerations, but
we can assume that it is small.
This behaviour is generic for all the stabilizing potentials $V(\phi)$.
At least one of the branes must be located in the flat part
of $V(\phi)$, so that a large distance can separate them.

In fig. \ref{fig1} we display two possibilities:
The brane (a) has
$R=18$,
$\phi(R)=1.28 \times 10^{-2}$,
$\phi'(R)=2.94 \times 10^{-2}$,
$A'(R)=-0.577$. If we assume that $\gamma_2$ is small as well, the
brane is weakly coupled to the bulk field.
The brane (b) has
$R=19.8$,
$\phi(R)=1.02$,
$\phi'(R)=4.03$,
$A'(R)=-1.63$.
It is coupled much more strongly to $\phi$.
It is located close to the singularity and one can speculate
that its presence is connected to the existence of a strong gravitational
background. In both the above scenarios the positive-tension brane
has no direct
interactions with $\phi$ and its location is determined only through
gravitational interactions. The total configuration is expected to
be stable as the potential has positive curvature. We also point out that
a (probably unstable) configuration with $\lx_2$ independent of $\phi$ can
be obtained for potential given by eq. (\ref{pot}) with
$\Lx=1$, $c=-1$, $n=6$, $\lx_1=0.577$.


In order not to rely exclusively on numerical solutions, we
work within the class of models discussed in ref. \cite{dewolfe}, for
which exact analytical solutions exist.
They involve potentials of the special form
\be
V(\phi) = \frac{1}{8} \left[ \frac{\partial W(\phi)}{\partial \phi}\right]^2
-\frac{1}{3} \left[ W(\phi) \right]^2,
\label{ww} \ee
with $W(\phi)$ a general function of $\phi$.
The solution of eqs. (\ref{eoma1})--(\ref{eoma3}) is \cite{dewolfe}
\beq
\phi'=&~\frac{1}{2} \frac{\partial W(\phi)}{\partial \phi}
\label{sol1a} \\
A'=&-\frac{1}{3} W(\phi),
\label{sol1b} \eeq
with
\beq
\lx_1(\phi)= &~W(\phi(0))+\frac{ \partial W(\phi(0))}{\partial \phi}
\left[
\phi-\phi(0)\right] +\gamma_1 \left[\phi-\phi(0)\right]^2 +~...
\label{sol1c} \\
\lx_2(\phi)= &-W(\phi(R))-\frac{ \partial W(\phi(R))}{\partial \phi}
\left[
\phi-\phi(R)\right] +\gamma_2\left[ \phi-\phi(R) \right]^2 +~...
\label{sol1d} \eeq
We point out that this class of models is too restrictive
to contain the possibility $\lx_1 =$ const. that we discussed earlier.
However, it can accomodate branes whose tension depends only weakly on
the bulk field $\phi$.

For the choice
\beq
W(\phi)= &3-b\phi^2
\label{ex1} \\
V(\phi)= &-3 + \left( \frac{b^2}{2}
+2b \right) \phi^2 -\frac{b^2}{3} \phi^4,
\label{ex4} \eeq
the solution is \cite{dewolfe}
\beq
\phi(y)= &\phi_1\; e^{-by}
\label{solex1} \\
A(y)=&-y + \frac{1}{6}\phi_1^2 \left(1- e^{-2b y}\right).
\label{solex2} \eeq
The distance between the branes is
\be
 R=\frac{1}{b}\ln \left( \frac{\phi_1}{\phi_2} \right),
\label{dist} \ee
with $\phi_2=\phi(R)$.
For values
$b= {\cal O} \left( 10^{-2} \right)$,
$\phi_1 /\phi_2= {\cal O} \left(1 \right)$, such as the ones
considered in ref. \cite{dewolfe},
one finds $R={\cal O} \left( 10^{2} \right)$.

For the choice $\gamma_1=-b$ the tension of the first
brane is
\be
\lx_1(\phi)=3-b\phi^2= W(\phi)
\label{choice} \ee
for any value of $\phi_1$. The fact that $\lx_1(\phi)$ has
negative curvature does not indicate an instability.
This function appears in the equation of motion of $\phi$, or
its perturbations around a given background, always multiplied
by  a $\delta$-function. As a result, it only determines the cusp in the
first derivative of the bulk field near the brane.
The potential $V(\phi)$ has positive curvature for the values of
$\phi$ that we are considering.

Let us now consider the possibility that a phase
transition modifies the brane vacuum energy. This can be modelled
by adding a constant $C$ to the tension $\lx_1(\phi)$. We
would like to find a static solution that can accomodate this change.
One of the possibilities that we mentioned earlier is that
the bulk field retains its value at the location of the
brane. This implies that $A'$, given by the second of eqs. (\ref{bound1}),
is shifted by $-C/3$. A non-zero effective cosmological constant
$\Lx$ appears, determined through eq. (\ref{eomb3}) with $A=0$.
However, a static solution cannot be obtained unless
the second brane is displaced at a distance $R \lta R_1$, with
$R_1$ given by eq. (\ref{div}).

The alternative possibility is that the value of the bulk field
$\phi$ changes so that the cosmological constant remains zero.
The new value is given by



\begin{thebibliography}{99}


\bibitem{rs1}
L. Randall and R. Sundrum,
Phys. Rev. Lett. {\bf 83}, 3370 (1999), hep-ph/9905221;
Phys. Rev. Lett. {\bf 83}, 4690 (1999), hep-ph/9906064.


\bibitem{dewolfe}
O. DeWolfe, D.Z. Freedman, S.S. Gubser and A. Karch,
hep-th/9909134.

\bibitem{goldwise}
W.D. Goldberger and M.B. Wise,
Phys. Rev. Lett. {\bf 83}, 4922 (1999), hep-ph/9907447.



\end{thebibliography}



%\newpage






\end{document}





\vspace{0.5cm}
\noindent
{\bf Acknowledgements}:
I would like to thank E. Kiritsis and R. Rattazzi
for useful discussions.
This research was supported by the E.C.
under TMR contract No. ERBFMRX--xxx.
\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=fig2.eps,width=12cm}}
\caption{
The y-dependence of $\phi$, $\phi'$ and $A'$ near the
negative-tension brane.}
\label{fig2}
\end{figure}


Follow ref. \cite{dewolfe} and consider potentials
Eqs. (\ref{eoma1})--(\ref{eoma3}) are solved by
if
\be
W(\phi(y_\alpha))=\lx_{\alpha}(\phi(y_\alpha))
~~~~~~~~~~~~~~~~~~~~~
\frac{\partial W}{\partial \phi}(\phi(y_\alpha))=
\frac{\partial \lx_{\alpha}}{\partial \phi}(\phi(y_\alpha)).
\label{bc1} \ee
Now consider the ansatz \cite{dewolfe}
\beq
ds^2 = \; &e^{2 A(y)}\, \bar{g}_{ij} \,dx^i dx^j - dy^2
\label{metr2a} \\
dS_4(\Lx > 0): ~~~~~\bar{g}_{ij} dx^idx^j
= \; &dt^2 -e^{2\sqrt{\Lx}t} \left(dx_1^2+dx_2^2+dx_3^2 \right)
\label{metr2b} \\
AdS_4(\Lx < 0): ~~~~~\bar{g}_{ij} dx^idx^j
= \; &e^{-2\sqrt{-\Lx}t} \left(dt^2-dx_1^2-dx_2^2 \right) -dx^2_3
\label{metr2c} \eeq
For the potential of eq.~(\ref{ex2}) with $b=0$
the solution of eq.~(\ref{eomb3}) is
\beq
\Lx > 0:
~~~~~~
e^A = &\sqrt{\Lx}L \sinh \frac{y_1-y}{L}
~~~~~~
\lx_1=\frac{3}{L} \coth\frac{y_1}{L}
~~~~~~
\lx_2=-\frac{3}{L} \coth\frac{y_1-y_0}{L}
\label{sing1} \\
\Lx < 0:
~~~~~~
e^A = &\sqrt{-\Lx}L \cosh \frac{y_1-y}{L}
~~~~~~
\lx_1=\frac{3}{L} \tanh\frac{y_1}{L}
~~~~~~
\lx_2=-\frac{3}{L} \tanh\frac{y_1-y_0}{L}.
\label{sing2} \eeq



\beq
\phi''+4 A' \phi'= &\frac{\partial V(\phi)}{\partial \phi} +
\sum_\alpha
\frac{\partial \lx_\alpha (\phi)}{\partial \phi}\, \delta (y-y_\alpha)
\label{eomb1} \\
A'' +\Lx e^{-2A}= &-\frac{2}{3}\phi'^2-\frac{2}{3}
\sum_\alpha
\lx_\alpha (\phi) \,\delta (y-y_\alpha)
\label{eomb2} \\
A'^2 -\Lx e^{-2A}= &-\frac{1}{3} V(\phi) + \frac{1}{6} \phi'^2.
\label{eomb3}
\eeq


\begin{figure}[t]
%\vspace{1.cm}
\centerline{\psfig{figure=fig2.eps,width=12cm}}
\caption{
The y-dependence of $\phi$ and $A'$ between the two branes.}
\label{fig2}
\end{figure}

In fig. \ref{fig2} we






\newpage



The necessary fine-tuning of $\lx_1(\phi)$, $\lx_2(\phi)$ for
this to occur may be provided by the dynamical mechanism that
determines the location of the branes.

Our first observation is that the situation can be
ameliorated if the potential $V(\phi)$ and the brane tensions
are related in a certain manner. More specifically, we can demand that
eq. (\ref{eoma3}), after the substitution of eqs. (\ref{bound1}),
be satisfied for all $\phi$. This means that the potential
must be given by
\be
V(\phi) = \frac{1}{8} \left[
\frac{\partial\lx_1(\phi)}{\partial \phi} \right]^2
- \frac{1}{3} \left[ \lx_1(\phi) \right]^2.
\label{v1} \ee
As a result, the field $\phi_1$ at the location of the
first brane can have a continuous range of values,
instead of a discrete spectrum.
The solution of eqs. (\ref{eoma1}), (\ref{eoma3})
for the potential (\ref{v1}) is given by \cite{dewolfe}
\be
\phi'(y)=\frac{1}{2}\frac{\partial \lx_1(\phi(y))}{ \partial \phi}
~~~~~~~~~~~~~~~~
A'(y)=-\frac{1}{3} \lx_1(\phi(y)),
\label{sol11} \ee
and the initial conditions (\ref{bound1}) are trivially satisfied for
any $\phi(0)=\phi_1$.
We may demand that the same apply to the second brane as well.
This would require $\lx_2(\phi)=-\lx_1(\phi)$.
However, we may impose the weaker constraint
that $\lx_2(\phi_2)=-\lx_1(\phi_2)$
only for a finite number of field values $\phi_2$.

For the potential of eq. (\ref{v1}) the value $\phi_1$ of $\phi$ at the
location of the positive-tension brane is arbitrary.
Each choice corresponds
to a solution with different tension $\lx_1(\phi_1)$. Also the distance
between the two branes is different.





1) Gauged supergravity

3) Effective potentials



2) Initial fine-tuning because the branes are far apart

4) Time-dependent scenario not easy to discuss.
   Coupling of brane fields to $\phi$ not elaborated

5) Changes of Newton's constant.
   Everything as in the Goldberger-Wise scenario

6) Matter on the branes


In the following we
assume the latter constraint
and keep the form of $\lx_2(\phi)$ fixed. Thus, we implicitly assume that
some mechanism has enforced the initial fine-tuning of the configuration.
(We comment below on a possible scenario.) We are concerned with the
destabilization of the configuration through subsequent changes of the
tension of the first brane (within a cosmological scenario, for example).


It seems reasonable, therefore, to speculate
that an arbitrary
change in the vacuum energy of the brane (its tension) does not
destabilize the system. Instead, it leads to a new configuration
with the first brane shifted from its previous position and
a different value of $\phi_1$. It is important that this can be realized for
arbitrary tension changes only
if the potential has the form of eq. (\ref{v1}).

The modification of the initial configuration takes place only in the
vicinity of the location of the positive-tension brane.
Let us assume that
the change in the tension requires the positive-tension brane to
move from the point ($\phi_1,\phi_1'$) to
$(\tilde{\phi}_1,\tilde{\phi}'_1$) on the trajectory of eqs. (\ref{sol1}).
The part of the trajectory from $(\tilde{\phi}_1,\tilde{\phi}'_1$) to
$(\phi_2,\phi'_2)$ remains unaffected. Thus the negative-tension brane
does not play a significant role in our considerations. It can be
viewed as a regulator brane that cuts off possible naked singularities
that appear in the solutions of eqs. (\ref{eoma1})--(\ref{eoma3}) \cite{cosm}.


We are not able to provide
a dynamical mechanism that determines the location of the branes,
as this requires the solution of Einstein's equations for more
general ansatzes with full time dependence.
Instead we are concentrating here on possible static solutions.



One can imagine the following situation: A system of two branes
is stabilized initially at a large distance $R = {\cal{O}}(100)$ with
zero effective cosmological constant $\Lx$.
Subsequently, the positive tension of the first brane is modified through, for
example, a phase transition on the brane. We assume that a new stable
configuration is approached. There are two possibilites: Either the
effective cosmological constant on the brane becomes non-zero, or
its previous zero value is maintained through an appropriate shift of
the brane in the fifth dimension.
We would like to demonstrate that the second scenario
leads to very small modifications of the initial configuration. Thus,
it seems more likely.


\newpage

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