

We consider the existence of a Noether symmetry in the scalar-tensor theory of gravity in flat Friedman Robertson Walker (FRW) cosmology. The forms of coupling function $\omega(\phi)$ and generic potential $V(\phi)$ are obtained by requiring the existence of a Noether symmetry for such theory. We derive exact cosmological solutions of the field equations from a point-like Lagrangian.

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Introduction
============

Scalar-tensor theories of gravity have been widely used in recent years. In principle, they are related to many fundamental theories as super-string theories, grand unified theories and quantum gravity.${}^{1,2}$ These theories allow the gravitational coupling to vary and act as a dynamical field.${}^{3,4}$ In the Kaluza-Klein models this arises from a variation of the size of the internal dimensions. In a cosmological context such theories allow to look for dynamical answers for equations.

The general form of the extended gravitational actions in scalar-tensor theories can be written in terms of the Brans-Dicke field $\phi$ and the strength of the coupling between the scalar field and gravity which is represented by the dynamical coupling function $\omega(\phi)$. Furthermore, a nontrivial potential $V(\phi)$ may be introduced in terms of a scalar field which clearly affect the dynamics. Higher-order gravitational theories are equivalent to a scalar-tensor theory when $V(\phi)$ is non-zero but $\omega(\phi)=0$, which introduces Yukawa-type corrections to the Newtonian potential.${}^{5}$ The case of $V(\phi)=0$ ensures a strictly Newtonian weak field limit to lowest order. When the coupling functional $\omega(\phi)$ is considered to be a constant parameter, the scalar-tensor theory reduces to the Brans-Dicke theory.

In general, there is no unique method to determine the functional forms of $\omega(\phi)$ and $V(\phi)$. In this paper, we first prove the existence of the Noether symmetry and then we use this symmetry to find these functionals. The resulting functional forms for $\omega(\phi)$ and $V(\phi)$ are not independent of each other. One may note that the Lagrangian in the action becomes point-like if we impose a flat Friedman Robertson Walker (FRW) metric.

The paper is organized as follows: the existence of the Noether symmetry is proved in section 2. In section 3 the exact solutions of the field equations are derived from a point-like Lagrangian for a quartic potential. The concluding remarks appear in the section 4.

Noether Symmetry 
=================

We consider the action of scalar-tensor theory in the form $$\begin{aligned}
S[\Phi]={ \int d^4x {\sqrt {-g}}  \left \{ \Phi R+ \frac{
\omega(\Phi)}{\Phi} g^{\mu\nu} {\nabla}_\mu \Phi {\nabla}_\nu\Phi
-V(\Phi) \right \} }\end{aligned}$$ where $ R$ is the scalar curvature, $\Phi$ denotes a real scalar field, non-minimally coupled to gravity, and ${\omega(\Phi)}$ and ${V(\Phi)}$ represent the coupling function and generic potential, respectively. The action (1) can be rewritten, with redefinition $\Phi=\phi^2$, as $$\begin{aligned}
S[\phi]={ \int  d^4x  \sqrt{-g} \left \{\phi^2 R+4{ \bf \omega (\phi)} 
g^{\mu \nu} {\nabla}_\mu \phi{\nabla}_\nu\phi-V(\phi)\right\} }. \end{aligned}$$ In the cosmological case, when the space-time manifold is described by a flat FRW metric, the scalar curvature has the expression $R=-6({\ddot{a}}/{a}+{\dot{a}^2}/{a^2})$, in which $a$ is the scale factor of the universe and the dot denotes the derivative with respect to time. One can show simply that in this space-time manifold the Lagrangian density related to the action (2) takes the point-like form${}^{6,7}$ $$\begin{aligned}
{\cal{L}}= 12 a^2\dot{a} \phi \dot{\phi}+ 6a \dot{a}^2  \phi^2+a^3
(4\omega(\phi) \dot{\phi}^2-V(\phi)). \end{aligned}$$ The corresponding Euler-Lagrange equations are given by $$\begin{aligned}
[\frac{\ddot{a}}{a}+\frac{1}{2}(\frac{\dot{a}}{a})^2]\phi^2+
[\ddot{\phi}+2(\frac{\dot{a}}{a})\dot{\phi}]\phi+\dot{\phi}^2-
(\omega(\phi) \dot{\phi}^2-\frac{1}{4}V(\phi))=0,\end{aligned}$$ and $$\begin{aligned}
\dot{\phi} \omega(\phi)+3\dot{\phi}
(\frac{\dot{a}}{a})\omega(\phi)+\frac{3}{2}[\frac{\ddot{a}}{a}+
(\frac{\dot{a}}{a})^2]\phi +\frac{1}{2}
(\omega^{'}(\phi) {\dot{\phi}}^2+\frac{1}{4}V^{'}(\phi))=0.\end{aligned}$$ These equations are equivalent to the second order Einstein equation and the Klein-Gordon equation, in the flat FRW space-time, respectively.

We examine Noether symmetry of the Lagrangian (3) in order to determine the two unknown functions $\omega(\phi)$ and $V(\phi)$, as well as solving its dynamical field equations.\
The infinitesimal generator of the Noether symmetry, namely the lift vector field $X$, is $$\begin{aligned}
X=\alpha (\phi, a)\frac{\partial}{\partial a}+\beta
(\phi, a) \frac{\partial}
{\partial \phi}+\frac{d \alpha}{dt}\frac{\partial}{\partial \dot{a}} 
+\frac{d\beta}{dt}\frac{\partial}{\partial \dot{\phi}}.\nonumber\end{aligned}$$ We try to determine $\alpha (\phi, a)$ and $\beta (\phi, a)$ by imposing the following condition${}^8$ $$\begin{aligned}
L_{X} {\cal L}=0,\end{aligned}$$ where $L_{X}$ is the Lie derivative with respect to the vector field $X$. Condition (6) leads to the following differential equations $$\begin{aligned}
&{}&{3\alpha V(\phi)+a\beta V^{'}(\phi)=0}\\
&{}&{3\alpha\omega(\phi)+\beta
a\omega^{'}(\phi)+3\phi\frac{\partial\alpha}{\partial\phi}+
2a\omega(\phi)\frac{\partial \beta}{\partial\phi}=0}\\
&{}&{\alpha\phi^2+2\beta a \phi +2a\phi^2\frac{\partial \alpha}
{\partial a}+2a^2\phi \frac{\partial \beta}{\partial a}=0}\\
&{}&{2\phi\alpha+a\beta+a\phi\frac{\partial \alpha}{\partial a} 
+\phi^2\frac{\partial \alpha}{\partial\phi}+
a\phi\frac{\partial \beta}{\partial \phi}+\frac{2}{3}
a^2\omega(\phi)\frac{\partial \beta}{\partial a}=}0. \end{aligned}$$ From Eq. (7) we have $$\begin{aligned}
\alpha=-a\beta U(\phi),\end{aligned}$$ where we have used the definition $$\begin{aligned}
U(\phi)=\frac{1}{3}\frac{V^{'}(\phi)}{V(\phi)}.\end{aligned}$$ Substituting (11) into (8) and (9), we find that the variables $\phi$ and $a$ in $\beta(\phi,a)$ must separate as $$\begin{aligned}
\beta(\phi,a)=f(\phi)a^n,\end{aligned}$$ where the separation constant $n$ is given by $$\begin{aligned}
n=\frac{\frac{3}{2}\phi U(\phi)-1}{1-\phi U(\phi)},\end{aligned}$$ and functional $f(\phi)$ has the following form $$\begin{aligned}
f(\phi)=exp \left\{ \int { \frac{\omega(\phi) U(\phi)+\phi U^{'}(\phi)
-\frac{1}{3}\omega{'}(\phi) }{\frac{2}{3}\omega(\phi)-\phi U(\phi)} }d\phi
\right \}.\end{aligned}$$ From the definition (12) and using (14) we obtain $$\begin{aligned}
V(\phi)=\lambda \phi^m,~~~~~~~~~~~m=\frac{3(n+1)}{(n+3/2)}\end{aligned}$$ and $$\begin{aligned}
U(\phi)=\frac{1}{3}m\phi^{-1}\end{aligned}$$ where $\lambda$ is an arbitrary integration constant. Putting these results into Eq. (10) one has $$\begin{aligned}
(m-3)\phi
\omega^{'}(\phi)+n\omega^2(\phi)-[m^2+m(1+4n)-6]\omega(\phi)+4(nm+3m-6)m=0.\end{aligned}$$ This is a first-order differential equation for the dynamical coupling function $\omega(\phi)$ with the solution $$\begin{aligned}
\omega (\phi)=k_1 [\frac{(\phi/\phi_0)^{k_2} -1}{(\phi/\phi_0)^{k_2} +1}]
+\omega_0,\end{aligned}$$ where $\phi_0$ is an integration constant and $k_1$, $k_2$ and $\omega_0$ are defined as $$\begin{aligned}
k_1=\frac{3}{(2n+3)^{2}},~~~~~k_2=\frac{-2n}{2n+3},~~~~~ 
\omega_0=\frac{3(8n^2+24n+17)}{(2n+3)^2}.\nonumber\end{aligned}$$ Substituting (17) and (19) into (15) and using (13) and (11), one gets $$\begin{aligned}
\alpha (\phi,a) =-\frac{1}{3}{\beta_0} m \phi_0^{-1} a^n
(\phi/\phi_0)^{{l_1}-1}[(\phi/\phi_0)^{k_2}-\xi]^{l_2}[(\phi/\phi_0)^{k_2}+1]^{l_3}\end{aligned}$$ and $$\begin{aligned}
\beta (\phi,a) =\beta_0
a^n{(\phi/\phi_0)^{l_1}}[(\phi/\phi_0)^{k_2}-\xi]^{l_2}[(\phi/\phi_0)^{k_2}+1]^{l_3},\end{aligned}$$ where $\beta_0$ is an integration constant, and we have used the definitions $$\begin{aligned}
&{}&{l_1=\frac{m(\omega_0-k_1-1)}{2\omega_0-2k_1-m}} \nonumber\\
&{}&{l_2=\frac{2{k_1}^2k_2-k_2(m-2\omega_0)^2/2-2m(m-2)}{4{k_1}^2-(m-2\omega_0)^2}}\nonumber\\
&{}&{l_3=-\frac{2{k_1}^2k_2-k_2(m-2\omega_0)^2/2
-2m(m-2)^2(k_1+1) }{4{k_1}^2-(m-2\omega_0)^2}}\nonumber\\
&{}&{\xi=\frac{2k_1-(m-2\omega_0)}{2k_1+(m-2\omega_0)}}.\nonumber\end{aligned}$$ Therefore, the vector field $X$ exists, and the existence of the Noether symmetry leads to the determination of the class of potentials (16) and the dynamical coupling function (19), for arbitrary values of $n$, except $n=-3/2$. In principle, the existence of the Noether symmetry means that there must exist a constant of motion. In this case one may compute it using the Cartan one-form associated with the Lagrangian (3) $$\begin{aligned}
\theta_{\cal{L}} \equiv \frac{ \partial {\cal{L}}}{\partial \dot{a}}da+
\frac{ \partial {\cal{L}}}{\partial \dot{\phi}}d\phi.\nonumber\end{aligned}$$ Contracting $\theta_{\cal{L}}$ with $X$ gives the following required constant of motion $$\begin{aligned}
i_{X} \theta_{\cal{L}}=12a \phi\alpha(\phi,a) \{a\dot{\phi}+\phi\dot{a}\}
+4a^2 \beta(\phi,a) \{3\dot{a}\phi+2a \dot{\phi}\omega(\phi)\} \nonumber\end{aligned}$$ where $\omega(\phi)$, $\alpha(\phi,a)$ and $\beta(\phi,a)$ are given by (19), (20) and (21), respectively.

Dynamical Field Equations and Solutions
=======================================

Scalar-tensor theory reduces to Brans-Dicke theory when the coupling function $\omega(\phi)$ is taken to be a constant. For mathematical simplicity, we analyze the solutions of the field equations (4) and (5) for $n=-3$ and in the case that $\omega(\phi)$ is a constant parameter such as $\omega$. In this case the Lagrangian (3) takes the form $$\begin{aligned}
{\cal{L}}= 12\dot{a}a^2 \phi \dot{\phi}+ 6\dot{a}^2 a \phi^2+a^3
(4\omega\dot{\phi}^2-\lambda \phi^4).\end{aligned}$$ The corresponding field equations are given by $$\begin{aligned}
[\frac{\ddot{a}}{a}+\frac{1}{2}
(\frac{\dot{a}}{a})^2]\phi^2+[\ddot{\phi}+2(\frac{\dot{a}}{a})\dot{\phi}]\phi+{\dot{\phi}}^2
+\frac{1}{4}\lambda \phi^4=0\end{aligned}$$ and $$\begin{aligned}
\omega\ddot{\phi}+3\omega\dot{\phi}(\frac{\dot{a}}{a})+\frac{3}{2}
[\frac{\ddot{a}}{a}+(\frac{\dot{a}}{a})^2]\phi +\frac{1}{2}\lambda \phi^3=0.\end{aligned}$$ One may select the initial conditions of the field equations (23) and (24) such that the energy function associated with the Lagrangian (22) vanishes $$\begin{aligned}
E_{\cal{L}}&=&{\frac{\partial {\cal{L}}}{\partial \dot{a}}\dot{a}+
\frac{\partial {\cal{L}}}{\partial\dot{\phi}}\dot{\phi}-{\cal{L}}}\nonumber\\ 
&=&{ 12\dot{a}a^2 \phi \dot{\phi}+ 6\dot{a}^2 a \phi^2+a^3 (4\omega
{ \dot{\phi}}^2+\lambda \phi^4)=0}.\end{aligned}$$ In fact, by choosing $E_{\cal{L}}=0$, one imposes a constraint on the integration constants. From (20) and (21) we have $$\begin{aligned}
\alpha=-\frac{4}{3} \frac{\beta_0}{a^2\phi^2}, 
~~~~~~\beta=\frac{\beta_0}{a^3 \phi}\end{aligned}$$ for $n=-3$. Now, the existence of the vector field $X$ can be used to find a non-degenerate point transformation $$\begin{aligned}
 (a, \phi) \longrightarrow (p, q),\nonumber\end{aligned}$$ such that the transformed Lagrangian is cyclic in one of the new variables. A possible way is to compute the Cartan one-form associated with Lagrangian $\cal{L}$, namely $$\begin{aligned}
\left\{ \matrix{ i_X(dp)=1 \nonumber \\i_X(dq)=0,}\right. \nonumber\end{aligned}$$ or explicitly $$\begin{aligned}
\left\{ \matrix{\alpha \frac{\partial p}{\partial a}+\beta
\frac{\partial p}{\partial \phi}=1 \nonumber \\ \alpha 
\frac{\partial q}{\partial a}+\beta \frac{\partial q}{\partial \phi}=0.}\right.\end{aligned}$$ Substituting (26) into (27), yields the solutions $$\begin{aligned}
\left\{\matrix {p=-\frac{1}{2\beta_0}a^3\phi^2 \nonumber \\
q=\gamma_0 a^{3/\gamma} \phi^{4/\gamma},}\right.\end{aligned}$$ where $\gamma_0$ and $\gamma$ are constant. Under these transformations, the Lagrangian (22) takes the form $$\begin{aligned}
{\cal{L}}=2\beta_0 \left \{2\gamma(\omega-5/3)
\frac{\dot{p}\dot{q}}{q}+(\omega-4/3)
(\frac{{\dot{p}}^2}{p}+\gamma^2 \frac{p{\dot{q}}^2}{q^2})+
\frac{\lambda}{2\beta_0 \gamma_0}q^{\gamma}\right \}.\end{aligned}$$ One can easily check that Eq. (18) has two solutions: $\omega=3/2$ and $\omega=4/3$ for $n=-3$. Now, if one computes the Hessian determinant ${\cal{H}} 
=|{\partial^2 {\cal{L}}}/{\partial \dot{\phi}}{\partial \dot{a}}|$, one finds that $\omega=3/2$ leads to the degeneration of $\cal{L}$. In other words, this value of $\omega$ does not allow a Hamiltonian formulation for the theory and leads to pathological dynamics. Thus, one can disregard $\omega =3/2$. For $\omega=4/3$ the second term of (29) disappears and $\cal{L}$ takes the form $$\begin{aligned}
{\cal{L}}=-\frac{4}{3} \beta_0 \gamma\frac{\dot{p}\dot{q}}{q}+
\frac{\lambda}{\gamma_0}q^{\gamma}.\end{aligned}$$ This Lagrangian clearly does not depend on $p$. Therefore, in new configuration space $(p,q)$, the variable $p$ is cyclic. This implies the existence of the Noether symmetry.\
The corresponding field equations for the last Lagrangian are given by $$\begin{aligned}
\ddot{p}+\frac{3\lambda}{4\beta_0 \gamma_0}q^{\gamma}=0\nonumber\end{aligned}$$ and $$\begin{aligned}
\dot{q}-r_0 q=0\nonumber\end{aligned}$$ where $r_0$ is constant of motion. The solutions of these equations are $$\begin{aligned}
p=s_0(e^{r_0\gamma t}+p_0)+\dot{{p_0}}t\end{aligned}$$ and $$\begin{aligned}
q=q_0 e^{r_0 t}, \end{aligned}$$ where $p_0, \dot{{p_0}}$ and $q_0$ are integration constants and $s_0$ is given by $$\begin{aligned}
s_0=-\frac{3\lambda {q_0}^{\gamma}}{4\beta_0 \gamma_0 {\gamma}^2
{r_0}^2}.\nonumber\end{aligned}$$ In new configuration space $(p, q)$, the condition (25) can be rewritten as $$\begin{aligned}
E_{\cal{L}}=-\frac{4}{3} \beta_0 \gamma\frac{\dot{p}\dot{q}}{q}-
\frac{\lambda}{\gamma_0}q^{\gamma}=0.\end{aligned}$$ Substituting (31) and (32) into (33), leads to $\dot{{p_0}}=0$, and so the solution (31) reduces to $$\begin{aligned}
p=s_0(e^{r_0\gamma t}+p_0).\end{aligned}$$ Putting (32) and (34) into (28), one gets the following solutions $$\begin{aligned}
\phi=\psi_0e^{r_0\gamma t/2}(e^{r_0\gamma t}+p_0)^{-1/2},~~~~\psi_0 
=(-\frac{{q_0}^\gamma}{2s_0\beta_0 \gamma_0^\gamma})^{1/2}\nonumber\end{aligned}$$ and $$\begin{aligned}
a=a_0e^{-r_0\gamma t}(e^{r_0\gamma t}+p_0)^{2/3},~~~~a_0
=(\frac{4s_0^2 \beta_0^2 \gamma_0^\gamma}{q_0^\gamma})^{1/3}. \nonumber\end{aligned}$$ In the limit $t \rightarrow \infty$ the scalar field $\phi$ approaches a constant value $\psi_0$, and then the Einstein gravity is recovered and the Newtonian gravitational constant is identified as $G_N=\frac{1}{\psi_0}^2$.

Concluding Remarks
==================

We have studied scalar-tensor theories of gravity in which the coupling function $\omega(\phi)$ and the generic potential $V(\phi)$ are unknown. The form of the coupling function and the potential are determined using Noether symmetry in a flat FRW background. In the special case $n=-3$, we showed that Noether symmetry exists for $\omega=3/4$, and we derived the corresponding constant of motion, $r_0$. Furthermore, the exact solutions of the field equations were derived from a point-like Lagrangian for a quartic potential. It is also shown that in the case of $\omega=3/2$, there is no Noether symmetry for the Lagrangian. However, it is interesting to note that the theory is exactly conformal invariant.${}^{9}$ Further attempts seems to be necessary to understand this relationship more deeply.\

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{'source': 'arXiv_out/hep-th0012119'}