# Non-minimally coupled scalar field and scaling symmetry in a cosmological
background

Malik<EMAIL_ADDRESS>
(Department of Theoretical Physics, Faculty of Science, University of
Mazandaran, 47416-95447, Babolsar, Iran)

###### Contents

  1. 1 Abstract
  2. 2 Introducing a Scaling Symmetric Lagrangian in Non-Minimally Coupled Scalar Field in FRW spacetime
  3. 3 Critical points and scaling symmetry breaking
  4. 4 Slow Rolling Solutions
  5. 5 Stability of ground state value of scalar field and energy
  6. 6 Summary and Conclusion

##  1 Abstract

We study scaling symmetry in a class of non-minimally coupled scalar field in
a background of Friedmann-Robertson-Walker (FRW) spacetime. We use a non-
minimally coupling $RL^{(\varphi)}$. We find the corresponding conserved
charge of that symmetry and see its role in cosmology, and search for its
possible breaking down and its outcomes. A suitable potential
$V(\varphi)=\varphi^{2}/2$ of scalar field is adopted which is necessary to
get a scaling symmetric Lagrangian of the system including scalar field, non-
minimally coupling to Ricci scalar $RL^{(\varphi)}$ and dark matter dust. We
study evolution of the scalar field in the phase space of the model and
explore the stability of the obtained critical point. In this manner we derive
a relation that relates the cosmological constant and gravitational constant
via a unique identity which reflects the scaling symmetry breaking in the
space $(a,\varphi)$. And relate the cosmological constant to the vacuum
expectation value of the potential energy of $\varphi$. Finally we study the
stability of that vacuum expectation value.

Keywords: Scalar Field Cosmology, Non-minimal Gravitational Coupling; Noether
Symmetry, Cosmic Speed Up.

##  2 Introducing a Scaling Symmetric Lagrangian in Non-Minimally Coupled
Scalar Field in FRW spacetime

A scaling symmetry in the space $(a,\varphi)$ is a kind of unification of real
scalar field $\varphi(t)$(represents dark energy) with the universal scale
factor $a(t)$ (represents spatial homogenous FRW metric), that is, existence
one of them implies existence the other, by that the energy and geometry can
be accompanied in one identity, i.e, we can put them in one field such as
$\Phi=(a,\varphi)$ on a manifold without needing introducing any geometry and
then we can establish a Lagrangian in terms of $\Phi$ that respect the
corresponding symmetry(but in this paper we do not do that). By that we can
explain the relation between energy and geometry in more general concept( i.e,
symmetry concept), by which the gravity is explained by concept of scaling
symmetry group. Note that there is dissimilarity with symmetry of general
relativity in which the transformations do not change $\varphi$ regarding it
as a scalar field, unlike the scaling symmetry, in which the scalar field
changes with changing the metric.

In this paper, we use a non-minimally coupled scalar field to gravity by the
term $RL^{(\varphi)}$, we use FRW metric and find a global scaling symmetry
Lagrangian in the space $(a,\varphi)$, whose symmetry breaking yields the
usual Lagrangian of the non-minimally coupled scalar field. The scaling
symmetry Lagrangian in the space $(a,\varphi)$ implies existence a globally
conserved quantity (charge) which can be used for global classification of the
cosmological solutions, i.e, two solutions with unequal charges can not be
related to each other by any coordinates transformation. We treat the role of
the charge in the solutions of $\varphi$ and we show that by the universal
positively accelerated expansion (increasing the scale factor $a$
exponentially) the field $\varphi$ is always exponentially decreasing until
reaching a critical point in $\dot{\varphi}=0$ with $\varphi=\varphi_{0}\neq
0$, in which the global scaling symmetry breaks and the universal expansion is
approximately in a constant rate $H=H_{0}$.

The evidence of existence of symmetry breaking is seen by violating
conservation of the corresponding charge; $dQ/dt\neq 0$. We will find that
symmetry breaking occurs in the critical point $\dot{\varphi}=0$,
$\varphi=\varphi_{0}\neq 0$. The existence of a non-vanishing constant value
$\varphi_{0}$ at that critical point $\dot{\varphi}=0$ is needed for
satisfying the constraint equation $\delta S/\delta N=0$. We find that the
critical point $\dot{\varphi}=0$, $\varphi=\varphi_{0}\neq 0$ is unique and
stable (there are no other critical points). As a result, we can relate the
cosmological constant and gravitational constant to a same identity, which is
scaling symmetry breaking in the space $(a,\varphi)$. And relate the
cosmological constant to the vacuum expectation value of the potential energy
of $\varphi$. If we thank that the vacuum expectation value of $\varphi$ does
not depend on any metric and it is a quantum phenomena, we obtain universal
constant vacuum energy(cosmological constant). By that we relate the
cosmological constant to a quantum phenomena, but here the field is specified
by a scaling symmetry Lagrangian (5).

The Lagrangian of the gravity plus the scalar field can be written in the
background of the spatially flat FRW metric
$ds^{2}=-N(t)dt^{2}+a^{2}(t)(dx^{2}+dy^{2}+dz^{2})$ as a one-point Lagrangian
up to boundary terms as [1]

$\begin{split}\sqrt{g}L(N,a,\dot{a},\varphi,\dot{\varphi})&=\frac{1}{{16\pi
G}}\sqrt{-g}R+\sqrt{-g}L^{(\varphi)}\\\
&=-3m_{pl}^{2}Na^{3}\left(\frac{{\dot{a}^{2}}}{N^{2}a^{2}}\right)+Na^{3}\left({\frac{1}{2}\frac{{\dot{\varphi}^{2}}}{N^{2}}-V\left(\varphi\right)}\right)+{\text{boundary
terms}}\,.\end{split}$ (1)

Where $N(t)$ is lapse function and $a(t)$ is cosmic scale factor. We note that
$a$ and $\varphi$ are dynamical variables, while $N(t)$ is a non-dynamical
variable, it just represents the symmetry in direction of time, therefore we
choose $N(t)=1$ after deriving the equations of motion.

In this paper we study the scalar fields coupled non-minimally to gravity as
$RL^{(\varphi)}$, where $R$ is Ricci scalar and $L^{(\varphi)}$ is Lagrangian
of scalar field $\varphi$. Let us introduce a non-minimally coupled scalar
field to gravity by

$\begin{split}\sqrt{-g}L(N,a,\dot{a},\varphi,\dot{\varphi})=-3m_{pl}^{2}a\frac{{\dot{a}^{2}}}{N}&+a^{3}\left({\frac{1}{2}\frac{{\dot{\varphi}^{2}}}{N}-NV\left(\varphi\right)}\right)\\\
&+3kNa\frac{{\dot{a}^{2}}}{{N^{2}}}\left({\frac{1}{2}\frac{{\dot{\varphi}^{2}}}{N^{2}}-V\left(\varphi\right)}\right)\,,\end{split}$
(2)

in which we have used some constant $k>0$ for satisfying the units. Here the
non-minimal interaction(third term) of the scalar field with gravity is
represented by product of the scalar $Na\left({\dot{a}/N}\right)^{2}$ with the
Lagrangian $\dot{\varphi}^{2}/2N^{2}-V\left(\varphi\right)$ of scalar field.
We note that there is no problem with that coupling since both
$Na\left({\dot{a}}/N\right)^{2}$ and $L^{(\varphi)}$ are scalars, therefore
their product is also scalar and preserves all of their symmetries.

For more general case, we add dust matter (visible and dark) density
$\rho_{m}\left(a\right)=\rho_{0}^{(m)}/a^{3}$, where $\rho_{0}^{(m)}$ is a
constant as matter density at some scale factor $a_{0}=1$, and a cosmological
constant $\Lambda$ which will just become as a result of global scaling
symmetry breaking of the Lagrangian (5), while the dust matter does not effect
on the results of that global symmetry and its breaking, and setting
$\rho_{0}^{(m)}=0$ is possible, but we add it just to notify that dark matter
dust can exist in the phase of global scaling symmetry. By adopting the scalar
field potential of the form $V\left(\varphi\right)=\varphi^{2}/2$ which is
needed to obtain a scaling symmetric Lagrangian (5), we get

$\begin{split}\sqrt{-g}L(N,a,\dot{a},\varphi,\dot{\varphi},\rho_{m})=&-3m_{pl}^{2}a\frac{{\dot{a}^{2}}}{N}+a^{3}\left({\frac{1}{2}\frac{{\dot{\varphi}^{2}}}{N}-\frac{1}{2}N\varphi^{2}}\right)\\\
&+3ka\frac{{\dot{a}^{2}}}{{N^{2}}}\left({\frac{1}{2}\frac{{\dot{\varphi}^{2}}}{N}-\frac{1}{2}N\varphi^{2}}\right)-Na^{3}\rho_{m}\left(a\right)-Na^{3}\Lambda\,.\end{split}$
(3)

Thus we obtain a one point Lagrangian of gravity + NMC term + scalar field +
matter density in the minimum super-space $(a,\varphi)$ of the model. If we
let the variables to be measured in units of Planck mass, we set $m_{pl}=1$ to
get

$\begin{split}\sqrt{-g}L(N,a,\dot{a},\varphi,\dot{\varphi},\rho_{m})=&-3a\frac{{\dot{a}^{2}}}{N}+\frac{{a^{3}}}{2}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)\\\
&+\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{{N^{2}}}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)-N\rho_{0}^{(m)}-Na^{3}\Lambda\,.\end{split}$
(4)

We can let this Lagrangian be produced from another Lagrangian which has a
global scaling symmetry in the space $(a,\varphi)$, like the Lagrangian

$\sqrt{-g}L\left({N,a,\dot{a},\varphi,\dot{\varphi}}\right)=\frac{{a^{3}}}{2}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)+\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{{N^{2}}}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)-N\rho_{0}^{(m)}\,.$
(5)

This Lagrangian includes a scalar field Lagrangian with interaction with
gravity in addition to a dark matter density term
$\rho_{m}(a)=\rho_{0}^{(m)}/a^{3}$. Thus we have a global scaling symmetry in
the space of dynamical variables $a$ and $\varphi$ represented by the
transformations

$a\to e^{2\alpha}a,\quad\textrm{and}\quad\varphi\to e^{-3\alpha}\varphi\,,$
(6)

for an arbitrary real constant parameter $\alpha$ which can be either positive
or negative. Thus the Lagrangian (5) has global scaling symmetry
$L\left({e^{2\alpha}a,e^{-3\alpha}\varphi}\right)=L\left({a,\varphi}\right)$
in the space $(a,\varphi)$, but this symmetry is broken when there is a non-
vanishing ground state value of $\varphi^{2}$, like
$\left\langle\Omega\right|\varphi^{2}\left|\Omega\right\rangle=\varphi_{0}^{2}\neq
0$, for a ground state wave function $\left|\Omega\right\rangle$ of the
Lagrangian (5). This implies a replacing $\varphi^{2}$ with
$\varphi^{2}+\varphi^{2}_{0}$ nearby the minimum energy state
$\left|\Omega\right\rangle$ in the Lagrangian (5) to get

$\begin{split}L&\left({N,a,\dot{a},\varphi,\dot{\varphi}}\right)\\\
&=\frac{{a^{3}}}{2}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}-N\varphi_{0}^{2}}\right)+\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{{N^{2}}}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}-N\varphi_{0}^{2}}\right)-N\rho_{0}^{(m)}\\\
&=-\varphi_{0}^{2}\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{N}+\frac{{a^{3}}}{2}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)+\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{{N^{2}}}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)\\\
&-N\rho_{0}^{(m)}-\frac{{\varphi_{0}^{2}}}{2}Na^{3}\,.\end{split}$ (7)

If we choose $k$ and $\varphi_{0}$ to get

$k\varphi_{0}^{2}/2=1,\quad and\quad\varphi_{0}^{2}/2=\Lambda,\quad for\quad
N(t)=1\,,$ (8)

thus we get the Lagrangian (4) with scaling symmetry breaking. In terms of
Planck mass, we get $k\varphi_{0}^{2}/2=m^{2}_{pl}$ so $k\Lambda=m^{2}_{pl}$
which unifies the gravitational constant with the cosmological constant via
the scaling symmetry breaking. While $\varphi_{0}^{2}/2=\Lambda$ relates
cosmological constant $\Lambda$ to vacuum energy $\varphi_{0}^{2}/2$ of scalar
field. Actually the equation $k\varphi_{0}^{2}/2=1$ ensures that $k>0$,
otherwise we will not get the usual general relativity of FRW metic as a
result of scaling symmetry breaking in the space $\left(a,\varphi\right)$. As
we will see that symmetry breaking occurs in the critical point
$\dot{\varphi}=0$, $\varphi=\varphi_{0}\neq 0$ and this critical point is
stable and unique.

Now we derive the conserved charge and the equations of motions of the
Lagrangian (5). Since
$L\left({e^{2\alpha}a,e^{-3\alpha}\varphi}\right)=L\left({a,\varphi}\right)$,
the action $S=\int Ldt$ is also invariant. Therefore,

$\delta_{\alpha}S=\int{dt\delta_{\alpha}L}=\int{dt\left({L\left({e^{2\alpha}a,e^{-3\alpha}\varphi}\right)-L\left({a,\varphi}\right)}\right)}=0\,.$

If we use an infinitesimal transformation $\alpha\ll 1$, we get

$\delta_{\alpha}a=e^{2\alpha}a-a\approx\left({1+2\alpha}\right)a-a=2\alpha
a\,,$

and

$\delta_{\alpha}\varphi=e^{-3\alpha}\varphi-\varphi\approx\left({1-3\alpha}\right)\varphi-\varphi=-3\alpha\varphi\,.$

Using these results in the following relation

$\begin{split}\delta_{\alpha}S&=\int{dt\delta_{\alpha}L}\\\
&=-\int{dt\left(\frac{d}{{dt}}\frac{{\partial
L}}{{\partial\dot{a}}}-\frac{{\partial L}}{{\partial
a}}\right)}-\int{dt\left(\frac{d}{{dt}}\frac{{\partial
L}}{{\partial\dot{\varphi}}}-\frac{{\partial L}}{{\partial\varphi}}\right)}\\\
&+\int{dt\frac{d}{{dt}}\left({\frac{{\partial
L}}{{\partial\dot{a}}}\delta_{\alpha}a+\frac{{\partial
L}}{{\partial\dot{\varphi}}}\delta_{\alpha}\varphi}\right)}=0\,,\end{split}$
(9)

with regarding the equations of motions, we obtain a conserved charge as

$Q=\frac{{\partial L}}{{\partial\dot{a}}}\left({2a}\right)+\frac{{\partial
L}}{{\partial\dot{\varphi}}}\left({-3\varphi}\right),\quad\frac{dQ}{{dt}}=0\,.$

Therefore we get(With using the gauge $N(t)=1$)

$\begin{split}Q&=3ka\dot{a}\left({\dot{\varphi}^{2}-\varphi^{2}}\right)\left({2a}\right)+\left({a^{3}+3ka\dot{a}^{2}}\right)\dot{\varphi}\left({-3\varphi}\right)\\\
&=6ka^{3}H\left({\dot{\varphi}^{2}-\varphi^{2}}\right)-3a^{3}\left({1+3kH^{2}}\right)\frac{d}{{dt}}\left({\frac{{\varphi^{2}}}{2}}\right)\\\
&=12ka^{3}Hp-\frac{3}{2}a^{3}\left({1+3kH^{2}}\right)\left({\dot{\rho}-\dot{p}}\right)=constant\,.\end{split}$
(10)

In which we have used $H\equiv\dot{a}/a$, the energy density
$\rho\equiv\dot{\varphi}^{2}/2+\varphi^{2}/2$ and the momentum density
(pressure density) $p\equiv\dot{\varphi}^{2}/2-\varphi^{2}/2$ of $\varphi$.

We note that for a solution like $H=H_{0}$, $\dot{\rho}=\dot{p}=0$ and
$\varphi=\varphi_{0}=constant\neq 0$ (that is, $\dot{\varphi}=0$), we have

$\left.{\frac{{dQ}}{{dt}}}\right|_{c}=-12ka^{3}\left({3H_{0}^{2}+\dot{H}_{0}}\right)\rho_{0}\neq
0\,,$ (11)

where the non-vanishing value in the right side comes from slow-rolling
condition $\dot{H}_{0}\approx 0$. Thus in this case, the scaling symmetry of
the Lagrangian (5) breaks and by that we get the Lagrangian (4). Actually we
will find that the point $H=H_{0}=constant$, $\varphi=\varphi_{0}=constant\neq
0$ is a stable critical point for the dynamical system of the Lagrangian (5)
and it is a unique critical point.

The equation of motions of $a$ from the Lagrangian (5), $\delta S/\delta
a=0$(With using the gauge $N(t)=1$), is

$\frac{d}{{dt}}\left({\frac{{\partial
L}}{{\partial\dot{a}}}}\right)-\frac{{\partial L}}{{\partial a}}=0\,,$

which yields

$\begin{split}&3k\dot{a}\dot{a}\left({\dot{\varphi}^{2}-\varphi^{2}}\right)+3ka\ddot{a}\left({\dot{\varphi}^{2}-\varphi^{2}}\right)+6ka\dot{a}\frac{d}{{dt}}\left({\frac{{\dot{\varphi}^{2}}}{2}-\frac{{\varphi^{2}}}{2}}\right)\\\
&-\frac{{3a^{2}}}{2}\left({\dot{\varphi}^{2}-\varphi^{2}}\right)-\frac{{3k}}{2}\dot{a}^{2}\left({\dot{\varphi}^{2}-\varphi^{2}}\right)=0\,,\end{split}$
(12)

and by using $H=\dot{a}/a$, $\ddot{a}/a=\dot{H}+H^{2}$ and the momentum
density $p=\dot{\varphi}^{2}/2-\varphi^{2}/2$, the last equation becomes

$\left({6k\dot{H}+9kH^{2}-3}\right)p+6kH\frac{{dp}}{{dt}}=0\,.$

Using a dimensionless time parameter defined as
$\eta=\ln\left({a/a_{0}}\right)$ which regards the scale factor $a$ as a
cosmological time, we have $d/dt=Hd/d\eta$. The last equation becomes

$\left({6kHH^{\prime}+9kH^{2}-3}\right)p+6kH^{2}p^{\prime}=0\,,$

or

$\left({h^{\prime}+3h-3}\right)p+2hp^{\prime}=0\,,$ (13)

where the prime indicates the derivative with respect to the dimensionless
time $\eta$, and we used $h=3kH^{2}$ as a dimensionless function.

The equation of motion of $\varphi$ from the Lagrangian (5), $\delta
S/\delta\varphi=0$(With using the gauge $N(t)=1$), is

$\left({3a^{2}\dot{a}+3k\dot{a}\dot{a}^{2}+6ka\dot{a}\ddot{a}}\right)\dot{\varphi}+\left({a^{3}+3ka\dot{a}^{2}}\right)\ddot{\varphi}+a^{3}\varphi+3ka\dot{a}^{2}\varphi=0\,.$

Following the same steps as for $a$, we obtain

$\left({h^{\prime}+3h+3}\right)\left(\rho+p\right)+\left({1+h}\right)\rho^{\prime}=0\,,$
(14)

where we used the energy density $\rho=\dot{\varphi}^{2}/2+\varphi^{2}/2$ of
$\varphi$.

We note that the previous equations of $\rho$ and $p$ include $h^{\prime}$, so
we need to omit it, by that the two equations (13) and (14) give one equation.

The constraint equation of the Lagrangian (5), $\delta S/\delta N=0$, implies

$\frac{{a^{3}}}{2}\left({-\frac{{\dot{\varphi}^{2}}}{{N^{2}}}-\varphi^{2}}\right)-3k\frac{{a\dot{a}^{2}}}{{N^{3}}}\left({\frac{{\dot{\varphi}^{2}}}{N}-N\varphi^{2}}\right)+\frac{{3k}}{2}\frac{{a\dot{a}^{2}}}{{N^{2}}}\left({-\frac{{\dot{\varphi}^{2}}}{{N^{2}}}-\varphi^{2}}\right)-\rho_{0}^{(m)}=0\,.$

Using the gauge $N(t)=1$, we obtain

$\frac{{a^{3}}}{2}\left({\dot{\varphi}^{2}+\varphi^{2}}\right)+\frac{{9k}}{2}a\dot{a}^{2}\dot{\varphi}^{2}-\frac{{3k}}{2}a\dot{a}^{2}\varphi^{2}+\rho_{0}^{(m)}=0\,,$

or

$2a^{3}\rho+3a^{3}h\left({\rho+p}\right)-a^{3}h\left({\rho-p}\right)+2\rho_{0}^{(m)}=0\,.$

Therefore we obtain the energy constraint equation

$\rho+h\left({\rho+2p}\right)+\frac{{\rho_{0}^{(m)}}}{{a^{3}}}=0\,.$ (15)

We note that the energy constraint (15) does not include any critical
energy(such as $3H^{2}$). But it imposes some conditions, since
$\rho=\dot{\varphi}^{2}/2+\varphi^{2}/2$, $h=3kH^{2}$ and $\rho_{0}^{(m)}\neq
0$ are always positive, we have $\rho+2p<0$ and it must be always satisfied.
Therefore the pressure $p=\dot{\varphi}^{2}/2-\varphi^{2}/2$ must be always
negative, $p<0$, and does not vanish. However negative pressure is needed for
getting universal expansion.

This means that the potential energy $\varphi^{2}/2$ is always larger than the
kinetic energy $\dot{\varphi}^{2}/2$, therefore there is no possibility to
increase the kinetic energy and vanishing the potential energy, while the
opposite is possible, that is increasing in potential energy while decreasing
in kinetic energy until it vanishes. Thus the solution $\dot{\varphi}=0$,
$\varphi=\varphi_{0}=constant\neq 0$ is possible.

Since $\rho+2p<0$ and $p<0$, we obtain $\rho+3p<0$ which according to
Friedmann equations implies an universal accelerated expansion.

We also note that the case $\rho=0$ ($\varphi=0$) does not exist since it
implies $p=0$. So, we have $0+{{\rho_{0}^{(m)}}}/{{a^{3}}}=0$ which is not
satisfied unless $\rho_{0}^{(m)}=0$. Therefore the acceptable minimum energy
is $\rho_{0}\neq 0$ (for $\dot{\varphi}=0$) and this value corresponds to the
vacuum expectation value of $\varphi^{2}$, as discussed just after equation
(6).

The energy constraint equation (15) does not give $h$ as a function only of
$\rho$ and $p$, in addition it includes $a(t)$. Therefore we need to omit
$h^{\prime}$ from the two equations (13) and (14) to get

$\left({1+h}\right)p\rho^{\prime}-2h\left({\rho+p}\right)p^{\prime}+6p\left({\rho+p}\right)=0\,.$
(16)

The same equation we will obtain if we get $h^{\prime}$ from the constraint
equation (15) and use it in the equations (13) and (14).

In order to get another equation for $\rho^{\prime}$ and $p^{\prime}$, we omit
$1/a^{3}$ from the charge equation (10) and constraint equation (15). We
obtain

$12kHp-\frac{3}{2}\left({1+h}\right)H\left({\rho^{\prime}-p^{\prime}}\right)+c\rho+ch\left({\rho+2p}\right)=0,\quad\textrm{for}\quad
c=\frac{Q}{{\rho_{0}^{(m)}}}\,,$

which gives

$\rho^{\prime}-p^{\prime}=\frac{{8kp}}{{\left({1+h}\right)}}+\frac{{2c\rho}}{{3H\left({1+h}\right)}}+\frac{{2ch}}{{3H\left({1+h}\right)}}\left({\rho+2p}\right)\,.$
(17)

Since both $H$ and $h>0$ can not vanish for any solution, there is no problem
with $H\left({1+h}\right)$ in the denominator of the last equation.

By that we have two equations, (16) and (17), that include $\rho^{\prime}$,
$p^{\prime}$, $\rho$, $p$ and $h=3kH^{2}$. From these equations, we obtain

$\begin{split}&\left[2h\left({\rho+p}\right)-\left({1+h}\right)p\right]\rho^{\prime}\\\
&=\left({\rho+p}\right)\left[{\frac{{16khp}}{{\left({1+h}\right)}}+\frac{{4ch\rho}}{{3H\left({1+h}\right)}}+\frac{{4ch^{2}}}{{3H\left({1+h}\right)}}\left({\rho+2p}\right)+6p}\right]\\\
&=\left({\rho+p}\right)\left[{{4ckH}\rho+\frac{{16khp}}{{\left({1+h}\right)}}+\frac{{8ch^{2}}}{{3H\left({1+h}\right)}}p+6p}\right]\,,\end{split}$
(18)

and

$\left[2h\left({\rho+p}\right)-\left({1+h}\right)p\right]p^{\prime}=8kp^{2}+\frac{{2c}}{{3H}}p\rho+2ckHp\left({\rho+2p}\right)+6p\left({\rho+p}\right)\,.$
(19)

We have $p<0$, $h>0$ and $\rho+p=\dot{\varphi}^{2}\geq 0$, therefore it is
always $[2h\left({\rho+p}\right)-\left({1+h}\right)p]>0$ and does not vanish.
Thus there is no problem with multiplying $\rho^{\prime}$ and $p^{\prime}$ by
$[2h\left({\rho+p}\right)-\left({1+h}\right)p]$.

##  3 Critical points and scaling symmetry breaking

We note that the constraint equation (15) does not imply any critical energy
(such as $3H^{2}$), so we do not need to divide $\rho$ and $p$ by any energy
and since we set $m_{pl}=1$, the variables $\rho$, $p$, $H$, $a$ and
$\eta=\ln(a)$ are dimensionless. Thus the critical points of the equations
(18) and (19) can be obtained by finding the points of
$\rho^{\prime}=p^{\prime}=0$, at a time $\eta_{0}=\ln(a_{0})$, in the space
$(\rho,p)$, where $H$ can be written in terms of these quantities. We note
that the time $\eta_{0}=\ln(a_{0})$ does not mean to stop universal expansion,
but it is just point in the space $(\rho,p)$, and nearby that point the
velocity $(\rho^{\prime}(\eta),p^{\prime}(\eta))$ decreases till finish at the
point $\eta_{0}=\ln(a_{0})$. So this does not mean stop universal expansion,
but it is just a moment of it(at $a_{0}=a(t_{0})$). And since velocity
$(\rho^{\prime},p^{\prime})$ is infinitesimal in vicinity of the point
$\rho^{\prime}=p^{\prime}=0$, thus the evolution of the system nearby that
point needs largest times, so the time is most spent in vicinity of critical
points $\rho^{\prime}=p^{\prime}=0$. Therefore the solutions near by the
critical points characterizes the solutions of the system in good accepted
approximation, i.e, solutions in $t=\pm\infty$ or at $t=t_{0}$.

We note that since the scale factor $a(t)$ is assumed always in increasing, so
indeed the energy density $\rho$ of the scalar field is in decreasing till
reaching a smallest possible value at
$\rho^{\prime}=p^{\prime}=0$($\eta_{0}=\ln(a_{0})$). We denote
$(\rho_{0},p_{0})$ as a critical point ($\rho^{\prime}=p^{\prime}=0$) and this
critical point belongs to a trajectory in the space $(\rho,p)$ where this
trajectory is parameterized by the time parameter $\eta=\ln(a)$. Therefore,
the critical point $(\rho_{0},p_{0})$ is determined by the time
$\eta_{0}=\ln(a_{0})$ on that trajectories. Thus, for each critical point
($\rho^{\prime}=p^{\prime}=0$), we have the quantities of $\rho_{0}$, $p_{0}$,
$H_{0}$ and $\eta_{0}=\ln(a_{0})$. As we will show there is only one critical
point associated with the scaling symmetry breaking of the Lagrangian (5).

The condition $\rho^{\prime}=0$(equation (18)) gives the following two
equations,

$4ckH\rho+\frac{{16khp}}{{\left({1+h}\right)}}+\frac{{8ch^{2}}}{{3H\left({1+h}\right)}}p+6p=0\,,$
(20)

and

$\rho+p=0\,.$ (21)

While the condition $p^{\prime}=0$(equation (19)) gives only one equation
(with $p\neq 0$) as

$8kp^{2}+\frac{{2c}}{{3H}}p\rho+2ckHp\left({\rho+2p}\right)+6p\left({\rho+p}\right)=0\,.$
(22)

While the energy constraint (15) implies(at $\rho^{\prime}=p^{\prime}=0$)

$\left.{h^{\prime}}\right|_{c}\left({\rho_{0}+2p_{0}}\right)-\frac{{3\rho_{0}^{(m)}}}{{a_{0}^{4}}}\left.{a^{\prime}}\right|_{c}=0\,,$

and by using

$a^{\prime}=\frac{{\partial a}}{{\partial\eta}}=\frac{{\partial
a}}{{\partial\ln\left(a\right)}}=a\frac{{\partial a}}{{\partial a}}=a\,,$

we get the equation

$\left.{h^{\prime}}\right|_{c}\left({\rho_{0}+2p_{0}}\right)-\frac{{3\rho_{0}^{(m)}}}{{a_{0}^{3}}}=0\,,$
(23)

which determines $h^{\prime}$ at the critical point
$\rho^{\prime}=p^{\prime}=0$. Note that $\left.{h^{\prime}}\right|_{c}=0$ is
satisfied only when $\rho_{0}^{(m)}=0$(so getting de Sitter solution). However
if we assume that ${\rho_{0}^{(m)}}/{a_{0}^{3}}$ is small enough, which
implies $\left.{h^{\prime}}\right|_{c}\approx 0$ (so $\dot{H}\approx 0$), we
obtain solutions close to de Sitter solution(we will find that in slow-rolling
condition).

In fact, the two equations (20) and (22) disagree, therefore the critical
points are given only by the two equations (21) and (22). We can see this
disagreement if we multiply the equation (20) by $3H\left({1+h}\right)/2\neq
0$, to get

$2c\rho h+2ch^{2}\left({\rho+2p}\right)+24khpH+9pH\left({1+h}\right)=0\,.$
(24)

While, multiplying equation (22) by $3Hh\neq 0$ and dividing it by $p\neq 0$
with using $h=3kH^{2}$, we find

$24kHhp+2c\rho h+2ch^{2}\left({\rho+2p}\right)+18Hh\left({\rho+p}\right)=0\,.$
(25)

Now subtracting equation (24) from equation (25), we obtain

$-9pH\left({1+h}\right)+18Hh\left({\rho+p}\right)=0\,.$ (26)

But as we saw, the pressure $p$ in this setup is always negative and non-
vanishing, $p<0$ (which comes from the conditions $\rho\neq 0$ and
$(\rho+2p)<0$), and also $H>0$ does not vanish, while $(\rho+p)\geq 0$, thus
the last equation is sum of two positive terms and one of them does not
vanish, so their sum also does not vanish. Therefore the last equation can not
be satisfied as required. So, the two equations (20) and (22) disagree and the
critical points $\rho^{\prime}=p^{\prime}=0$ are described only by two
equations (21) and (22).

From the equation (21), we get $p_{0}=-\rho_{0}<0$, using it in equation (22),
we get

$3ckH^{2}_{0}+12kH_{0}-c=0\,.$

Its positive solution is

$H_{0}=\frac{{-2}}{c}+\sqrt{\frac{4}{{c^{2}}}+\frac{1}{{3k}}}=\frac{{-2\rho_{0}^{(m)}}}{Q}+\sqrt{\left({\frac{{2\rho_{0}^{(m)}}}{Q}}\right)^{2}+\frac{1}{{3k}}}\,.$

From the equation of the charge (10), we get

$Q=12ka^{3}Hp-\frac{3}{2}a^{3}\left({1+3kH^{2}}\right)\left({\dot{\rho}-\dot{p}}\right)=-12ka_{0}^{3}H_{0}\rho_{0}\,.$

But the quantities $a_{0}$, $H_{0}$, and $\rho_{0}$ are all positive,
therefore $Q$ is negative. Thus we replace $Q\to-Q$ to get a positive quantity
for our forthcoming purpose. In this manner we obtain the expansion rate at
the critical point as follows

$H_{0}=\frac{{2\rho_{0}^{(m)}}}{Q}+\sqrt{\left({\frac{{2\rho_{0}^{(m)}}}{Q}}\right)^{2}+\frac{1}{{3k}}}>0\,.$

We note that for $\rho_{0}^{(m)}\ll\rho_{0}$, this expansion rate approximates
to $H_{0}=1/\sqrt{3k}$ which agrees with slow rolling solution.

Now we show that the conservation of the charge (10) is broken at this
critical point. We have

$\begin{split}\left.{Q^{\prime}}\right|_{c}&=\left.{\frac{{dQ}}{{d\eta}}}\right|_{c}=-12ka^{3}\left.{\left({3H+H^{\prime}}\right)}\right|_{c}\rho_{0}=-\frac{{12ka^{3}}}{{3kH_{0}}}\left.{\left({9kH^{2}+3kHH^{\prime}}\right)}\right|_{c}\rho_{0}\\\
&=-\frac{{12ka^{3}}}{{3kH_{0}}}\left.{\left({3h+\frac{1}{2}h^{\prime}}\right)}\right|_{c}\rho_{0}\,,\end{split}$
(27)

where we have used $\rho^{\prime\prime}=p^{\prime\prime}=0$ because
$\rho^{\prime}\sim(\rho-\rho_{0})$ and $p^{\prime}\sim(p-p_{0})$, so
$\rho^{\prime\prime}\sim(\rho^{\prime}-\rho_{0})$ and
$p^{\prime\prime}\sim(p^{\prime}-p_{0})$, therefore
$\rho^{\prime\prime}=p^{\prime\prime}=0$ at the critical point
$(\rho_{0},p_{0})$.

From the equations (15) and (23), we obtain

$\left.h\right|_{c}=1+\frac{{\rho_{0}^{(m)}}}{{\rho_{0}a^{3}_{0}}},\quad\textrm{and}\quad\left.{h^{\prime}}\right|_{c}=-\frac{{3\rho_{0}^{(m)}}}{{\rho_{0}a^{3}_{0}}}\,.$
(28)

Using these relations in $Q^{\prime}$, we get

$\left.{Q^{\prime}}\right|_{c}=-\frac{{12ka_{0}^{3}}}{{3kH_{0}}}\left({3+\frac{{3\rho_{0}^{(m)}}}{{2\rho_{0}a_{0}^{3}}}}\right)\rho_{0}\neq
0\,.$

In this situation, the scaling symmetry of the Lagrangian (5) is broken at the
critical point $\dot{\varphi}=0$, $\varphi(a_{0})=\varphi_{0}\neq 0$, at time
$\eta_{0}=\ln(a_{0})$, thus we get the Lagrangian (4). We note that for
$\rho_{0}^{(m)}\ll\rho_{0}$, we have $\left.{h^{\prime}}\right|_{c}\approx 0$
implying $H=H_{0}=constant$, which agrees with the slow rolling solution and
indicates that nearby the critical point $\dot{\varphi}\approx 0$,
$\varphi_{0}\neq 0$, the universal expansion rate becomes constant and we
obtain a de Sitter solution.

We note that the quantities $\varphi_{0}$ and $H_{0}$ do not need to depend on
$\eta_{0}=\ln(a_{0})$, so only indeed the matter dust $\rho^{(m)}\sim 1/a^{3}$
will depend on $a_{0}$, thus we are free in choosing $a_{0}$ to get a suitable
$\rho_{0}^{(m)}$ at point of scaling symmetry breaking.

Now we show that the critical point $\dot{\varphi}=0$, $\varphi=\varphi_{0}>0$
is stable. We find first order approximation of $\rho^{\prime}$ and
$p^{\prime}$ nearby the critical point $(\rho_{0},p_{0})$; $\rho_{0}+p_{0}=0$.
Actually according to the equations (28), and with
$\rho_{0}^{(m)}\ll\rho_{0}$, we can neglect perturbations on $h$ and so on
$H$; $\delta H\sim 1/a_{0}^{3}<<1$.

We have, the equations (18) and (19),

$\begin{split}&\left[2h\left({\rho+p}\right)-\left({1+h}\right)p\right]\rho^{\prime}\\\
&=\left({\rho+p}\right)\left[{{4ckH}\rho+\frac{{16khp}}{{\left({1+h}\right)}}+\frac{{8ch^{2}}}{{3H\left({1+h}\right)}}p+6p}\right]\,,\end{split}$
(29)

and

$\left[2h\left({\rho+p}\right)-\left({1+h}\right)p\right]p^{\prime}=8kp^{2}+\frac{{2c}}{{3H}}p\rho+2ckHp\left({\rho+2p}\right)+6p\left({\rho+p}\right)\,.$
(30)

Multiplying first equation by ${3H\left({1+h}\right)/2}$ and using
$h=3kH^{2}$, we obtain

$\begin{split}&\frac{{3H\left({1+h}\right)}}{2}\left[{2h\left({\rho+p}\right)-\left({1+h}\right)p}\right]\rho^{\prime}\\\
&=\left({\rho+p}\right)\left[{2ch\left({1+h}\right)\rho+24kHhp+4ch^{2}p+9H\left({1+h}\right)p}\right]\\\
&=\left({\rho+p}\right)\left[{2ch\rho+2ch^{2}\rho+24kHhp+4ch^{2}p+9H\left({1+h}\right)p}\right]\\\
&=\left({\rho+p}\right)\left[{24kHhp+2ch\rho+2ch^{2}\left({\rho+2p}\right)+9H\left({1+h}\right)p}\right]\,.\end{split}$
(31)

Thus, nearby $\rho_{0}+p_{0}=0$ and by using the equation (25) (equation of
$p^{\prime}=0$), we get first order approximation

$\begin{split}\left({1+h_{0}}\right)^{2}\rho_{0}\rho^{\prime}&=\left({\Delta\rho+\Delta
p}\right)\left[{-12h\left({\rho_{0}+p_{0}}\right)-6\left({1+h_{0}}\right)\rho_{0}}\right]\\\
&\\\ &\to\left({\Delta\rho+\Delta
p}\right)\left[{-6\left({1+h_{0}}\right)\rho_{0}}\right]\,,\end{split}$ (32)

so

$\left({1+h_{0}}\right)\rho_{0}\rho^{\prime}=\left({\Delta\rho+\Delta
p}\right)\left({-6\rho_{0}}\right)\,\Rightarrow\rho^{\prime}=\frac{{-6}}{{1+h_{0}}}\left({\Delta\rho+\Delta
p}\right)\,,$

for $\Delta\rho=\rho-\rho_{0}\ll 1$ and $\Delta p=p-p_{0}\ll 1$. Using this
equation in the first order perturbation of equation (17), we get

$p^{\prime}=\frac{{-2}}{{1+h_{0}}}\left[{3\Delta\rho+\left({3+4k}\right)\Delta
p}\right]\,.$

From last two equations, we obtain $(\lambda_{1},\lambda_{2})$ the eigenvalues
of the velocities $(\rho^{\prime},p^{\prime})$ nearby $(\rho_{0},p_{0})$, we
get

$\lambda_{1}=-\frac{1}{{1+h_{0}}}\left({6+4k-2\sqrt{4k^{2}+9}}\right)\approx-\left({3+2k-\sqrt{4k^{2}+9}}\right)\,,$

and

$\lambda_{2}=-\frac{1}{{1+h_{0}}}\left({6+4k+2\sqrt{4k^{2}+9}}\right)\approx-\left({3+2k+\sqrt{4k^{2}+9}}\right)\,.$

Since $k>0$(regarding equation (8)), it is always $(6+4k-2\sqrt{4k^{2}+9})>0$,
therefore both $\lambda_{1}$ and $\lambda_{2}$ are negative, thus the critical
point $(\rho_{0},p_{0})$; $\rho_{0}+p_{0}=0$, $\rho_{0}>0$ is stable.
Therefore the global scaling symmetry breaking is inevitable matter, and it is
global critical point since it depends on vacuum energy of the scalar field
$\varphi(t)$, which can be related to quantum phenomena(i.e, quantization,
bosonic fields,…).

##  4 Slow Rolling Solutions

According to the equation (23), in all critical points, we have
$\left.{h^{\prime}}\right|_{c}\approx 0$ when ${\rho_{0}^{(m)}}/{a_{0}^{3}}$
is small enough(such ${\rho_{0}^{(m)}}/{a_{0}^{3}}<<1$). This condition yields
to the slow-rolling conditions
$\left|{\ddot{\varphi}}\right|\ll\left|{\varphi}\right|$ and
$\left|{\dot{\varphi}}\right|\ll\left|{\varphi}\right|$ which take place
nearby the critical point $\dot{\varphi}=0$, $\varphi(a_{0})=\varphi_{0}\neq
0$ of the scaling symmetry Lagrangian (equation (5)), that yields to solutions
close to de Sitter solution(universal expansion with constant rate
$H=constant$). The necessity of slow-rolling solutions is in their obtaining
the behaviour of all variables nearby the critical point $\dot{\varphi}=0$,
$\varphi(a_{0})=\varphi_{0}\neq 0$ and before the scaling symmetry breaking.
As usual, we get the equation of expansion rate $H$ from the energy constraint
equation (equation (15)). We obtain

$\begin{split}h=3kH^{2}&=\frac{{-\rho-\frac{{\rho_{0}^{(m)}}}{{a^{3}}}}}{{\rho+2p}}=\frac{{-\dot{\varphi}^{2}-\varphi^{2}-\frac{{\rho_{0}^{(m)}}}{{a^{3}}}}}{{\dot{\varphi}^{2}+\varphi^{2}+2\dot{\varphi}^{2}-2\varphi^{2}}}\\\
&=\frac{{-\dot{\varphi}^{2}-\varphi^{2}-\frac{{\rho_{0}^{(m)}}}{{a^{3}}}}}{{3\dot{\varphi}^{2}-\varphi^{2}}}\Rightarrow\frac{{-\varphi^{2}-\frac{{\rho_{0}^{(m)}}}{{a^{3}}}}}{{-\varphi^{2}}}=1+\frac{{\rho_{0}^{(m)}}}{{\varphi^{2}a^{3}}}\approx
1+\frac{{\rho_{0}^{(m)}}}{{\varphi_{0}^{2}a^{3}}}\,,\end{split}$ (33)

where we have used the slow rolling condition
$\left|{\dot{\varphi}}\right|\ll\left|{\varphi}\right|$. If we impose a
condition as ${{\rho_{0}^{(m)}}}/{{\varphi_{0}^{2}a^{3}}}\ll 1$ which takes
place at large scale factor values $a\gg 1$ and with
$\rho_{0}^{(m)}\ll\varphi_{0}^{2}/2$ for which the universe is dominated by
the ground state energy of $\varphi$(vacuum energy) which has the role of
cosmological constant, however, we identified the energy $\varphi_{0}^{2}/2$
with cosmological constant, formulas (8). We obtain $3kH^{2}\approx 1$,
therefore we get approximately constant expansion rate $H_{0}=1/\sqrt{3k}$.
Note that phase occurs at late times $a\gg 1$ of universal expansion. Using
$h^{\prime}=0$, $h=1$ in the equation (14), we obtain

$6\dot{\varphi}^{2}+2\left({\dot{\varphi}\ddot{\varphi}+\varphi\dot{\varphi}}\right)=0\,\,\,\Rightarrow\,\,\,6\dot{\varphi}+2\left({\ddot{\varphi}+\varphi}\right)=0\,,$

which has the solution

$\varphi\left(t\right)=Ae^{-0.4t}+Be^{-2.6t}\,.$

For some real constants $A$, $B$. It is clear that in this approximation, the
field $\varphi$ decreases in time until vanishes.

However, $t$ is measured in unit of Plank mass, so $t=1$ is the time of value
$m_{pl}^{-1}$ which is a large value, thus the slow rolling period is long,
but if a vacuum expectation value $\left\langle
0\right|\varphi^{2}\left|0\right\rangle=\varphi_{0}^{2}\neq 0$ appears, the
scaling symmetry breaks and a new Lagrangian (equation (4)) takes place
instead.

##  5 Stability of ground state value of scalar field and energy

222This section is not included in the published edition.

We have seen that there is a non-zero positive value of the energy density of
the scalar field $\varphi$, this value $\rho_{0}>0$ is given in the critical
point $\dot{\varphi}=0$. But in order to relate $\varphi_{0}\neq 0$ to quantum
phenomena(i.e, vacuum expectation value), we need $\rho_{0}$ be stable and do
not depend on time $\eta=\ln\left(a(t)\right)$. So we can regard
$\varphi_{0}\neq 0$ as a global constant value that can be given by
$\varphi_{0}^{2}=\left\langle\Omega\right|\hat{\varphi}^{2}\left|\Omega\right\rangle>0$,
for a ground state function $\left|\Omega\right\rangle$. But we need to relate
$\hat{\varphi}$ and $\left|\Omega\right\rangle$ to a quantum phenomena which
is global and does not depend on any geometry.

From the charge equation (10) and constraint equation (15), we obtain at the
critical point $\dot{\varphi}=0$, $\varphi=\varphi_{0}\neq 0$ the relations

$\left.Q\right|_{c}=Q=-12ka_{0}^{3}H_{0}\rho_{0}\Rightarrow\rho_{0}=-\frac{Q}{{12ka_{0}^{3}H_{0}}}\,;\quad
Q<0\,,$

and

$\rho_{0}-h_{0}\rho_{0}+\frac{{\rho_{0}^{(m)}}}{{a^{3}_{0}}}=0\Rightarrow\rho_{0}=\frac{{\rho_{0}^{(m)}}}{{\left({h_{0}-1}\right)a_{0}^{3}}}\,.$
(34)

These two equations imply

$-\frac{Q}{{12kH_{0}}}=\frac{{\rho_{0}^{(m)}}}{{\left({h_{0}-1}\right)}}\,,$

and by using $h=3kH^{2}$, we obtain

$H_{0}=-\frac{{2\rho_{0}^{(m)}}}{Q}+\sqrt{\left({\frac{{2\rho_{0}^{(m)}}}{Q}}\right)^{2}+\frac{1}{{3k}}}>0\,;\quad-Q>0\,.$
(35)

It is clear that $H_{0}$ does not depend on the scale factor $a_{0}$, also it
is global by its dependence only on the constants $k$, $Q$ and
$\rho_{0}^{(m)}$ which are global by the meaning that they classify the
solutions(do not depend on time).

Therefore $H_{0}$ is global constant value. But in other side, we have

$a\left(t\right)=a\left(0\right)e^{\int{H\left(t\right)dt}}\,.$

Regarding the scaling symmetry, transformations (6), and before reaching the
critical point $\dot{\varphi}=0$, $\varphi_{0}=\varphi(a_{0})\neq 0$(in
vicinity of it), we have the more general solution

$a\left(t\right)=a\left(0\right)e^{2\alpha+\int{H\left(t\right)dt}}\,,$

for any real arbitrary constant $\alpha$. And according to equation (28),
$h^{\prime}\approx 0$ and so $H^{\prime}\approx 0$ when
$\rho^{(m)}_{0}/a^{3}_{0}\rho_{0}<<1$. Thus in vicinity of the critical point
$\dot{\varphi}=0$, $\varphi_{0}=\varphi(a_{0})\neq 0$, we use the value (35)
of $H_{0}$ to approximate $a\left(t\right)$ to

$a\left(t\right)=Ae^{2\alpha+H_{0}t}\,,$

for some constant $A>0$. If we let the critical point $\dot{\varphi}=0$,
$\varphi(a_{0})=\varphi_{0}\neq 0$ be reached in time $t=t_{0}$, we obtain

$a_{0}=a\left({t_{0}}\right)=Ae^{2\alpha+H_{0}t_{0}}\,.$

Now we can write

$2\alpha+H_{0}t_{0}=H_{0}T_{0}$

and choose $\alpha$ such that $T_{0}=1$, by that we obtain

$a_{0}=Ae^{H_{0}}\,,$

in the critical point $\dot{\varphi}=0$, $\varphi(a_{0})=\varphi_{0}\neq 0$.
But according to the equation (35), $H_{0}=H(k,Q,\rho_{0}^{(m)})$ which
implies that $a_{0}$ depends only on the globally constants $k$, $Q$ and
$\rho_{0}^{(m)}$. Thus $a_{0}(k,Q,\rho_{0}^{(m)})$ is also globally constant
value and it also classifies the solutions, by that the energy density
$\rho_{0}$, equation (34), depends only on the globally constants $k$, $Q$ and
$\rho_{0}^{(m)}$, so it is also a globally constant value, not geometrical,
thus it does not change under the universal expansion after passing the
critical point $a=a_{0}$($\dot{\varphi}=0$). Therefore
$\varphi_{0}^{2}=\left\langle\Omega\right|\hat{\varphi}^{2}\left|\Omega\right\rangle$
and $\left|\Omega\right\rangle$ are global structures, where
$\rho_{0}=\varphi_{0}^{2}/2$.

According to this discussion, we can think that $\rho_{0}$ is the vacuum
expectation value of $\hat{\varphi}^{2}$, where $\hat{\varphi}$ is quantum
field that does not depend on any geometry, as well as the quantum ground
state $\left|\Omega\right\rangle$. By that the equality
$\varphi_{0}^{2}/2=\Lambda$(equations (8)) is well defined and the
cosmological constant $\Lambda$ in this view is global stable value, that it
does not relate with the universal expansion, i.e, does not change under the
universal expansion after passing the critical point
$a=a_{0}$($\dot{\varphi}=0$).

##  6 Summary and Conclusion

In this paper, we have studied some novel aspects of cosmological dynamics of
a quintessence scalar field non-minimally coupled to gravity in a spatially
flat FRW background via the Noether Symmetry approach. We considered the non-
minimal coupling between the scalar field and gravitational sector as
$RL^{(\varphi)}$, that is essentially a subclass of the general Horndeski
gravity and reduces to non-minimal derivative coupling in the case of kinetic
dominance of the scalar field. We applied the Noether symmetry approach to the
Lagrangian of the model and derived the corresponding Noether charge by
exploring the status of the scaling symmetry in this framework. We adopted a
suitable potential of the scalar field $\varphi$ and estimated the behaviour
of the scale factor via scaling symmetry breaking in this setup. We treated
the role of the Noether charge in the solutions of the scalar field and we
have shown that by the universal positively accelerated expansion (especially
an exponential expansion), the field $\varphi$ is always exponentially
decreasing until reaching a critical point at $\dot{\varphi}=0$, that is, when
$\varphi=\varphi_{0}\neq 0$, in which the global scaling symmetry breaks and
the universal expansion is approximately in a constant rate $H=H_{0}$.
Existence of scaling symmetry breaking violates the conservation of the
corresponding charge, that is, $dQ/dt\neq 0$ in the critical point
$\dot{\varphi}=0$, $\varphi=\varphi_{0}\neq 0$. The existence of a non-
vanishing constant positive value $\varphi_{0}$ at the critical point
$\dot{\varphi}=0$ is necessary for fulfilling the constraint equation $\delta
S/\delta N=0$. We have demonstrated that the critical point $\dot{\varphi}=0$,
$\varphi=\varphi_{0}\neq 0$ is unique and stable in this setup and as an
important result, we were able to relate the cosmological constant and
gravitational constant via an identity, which is scaling symmetry breaking in
the space $(a,\varphi)$. Finally we tried to show that the ground state energy
density $\rho_{0}$ relates to quantum phenomena and globally stable.

Funding and/or Conflicts of interests/Competing interests:

There is no funding and/or conflicts of interests/ompeting interests regarding
this manuscript.

Data Availability Statement:

No Data associated in the manuscript.

## References

  * [1] G. N. Remmen, S. M. Carroll, _Attractor Solutions in Scalar-Field Cosmology_ , Phys. Rev. D 88 (2013) 083518.