$\displaystyle\frac{1}{2\left(1-k^{2}\right)}\ln\left(\frac{\left(k^{2}-1\right)^{1-k}}{\left(k-1\right)^{2}}\frac{x^{1-k}}{y^{1+k}}\right)$
(9.92) $\displaystyle r$ $\displaystyle=$
$\displaystyle\sqrt{\left(k^{2}-1\right)xy}\left(\frac{\left(k^{2}-1\right)^{1-k}}{\left(k-1\right)^{2}}\frac{x^{1-k}}{y^{1+k}}\right)^{\frac{k}{2\left(k^{2}-1\right)}}$
(9.93)

and

for $k^{2}<1$

$\displaystyle\theta$ $\displaystyle=$
$\displaystyle\frac{1}{2\left(1-k^{2}\right)}\ln\left(\frac{\left(1-k^{2}\right)^{1-k}}{\left(1-k\right)^{2}}\frac{\bar{x}^{1-k}}{\bar{y}^{1+k}}\right)$
(9.94) $\displaystyle r$ $\displaystyle=$
$\displaystyle\sqrt{\left(1-k^{2}\right)\bar{x}\bar{y}}\left(\frac{\left(1-k^{2}\right)^{1-k}}{\left(1-k\right)^{2}}\frac{\bar{x}^{1-k}}{\bar{y}^{1+k}}\right)^{\frac{k}{2\left(k^{2}-1\right)}}$
(9.95)

Note that in the second case we have written $\bar{x},\bar{y}$ while we have
kept the $x,y$ notation for the case $k^{2}>1.$

###### The case $k^{2}>1$

Before we look for analytic solutions we transform the Lagrangian in the
canonical coordinates $x,y.$ Using the transformation relations (9.92), (9.93)
we find that in the coordinates $x,y$ the Lagrangian (9.43) takes the form

$L\left(x,y,\dot{x},\dot{y}\right)=\frac{N_{0}^{2}}{2}\dot{x}\dot{y}-U\left(x,y\right)$
(9.96)

where $U\left(x,y\right)=r^{2}V\left(\theta\right)$ where
$V\left(\theta\right)$ is one of the potentials computed above. In the
coordinates $x,y$ these potentials are as follows:

  * •

$U_{1}\left(x,y\right)=V_{0}r^{2}e^{2k\theta}=V_{0}\left(k^{2}-1\right)xy$
(9.97)

$U_{2}\left(x,y\right)=V_{0}r^{2}e^{+2\theta}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{}x^{\frac{{}^{2}}{1+k}}$
(9.98)
$U_{3}\left(x,y\right)=V_{0}r^{2}e^{-2\theta}=V_{0}\left(k-1\right)^{\frac{2}{1-k}}y^{\frac{2}{1-k}}$
(9.99)

  * •

$U_{4}\left(x,y\right)=V_{0}r^{2}e^{-2\frac{\left(k^{2}-2\right)}{k}\theta}=V_{0}\frac{\left(k^{2}-1\right)^{\frac{2}{k}-1}}{\left(k-1\right)^{\frac{4}{k}}}\frac{1}{y^{2}}\left(\frac{x}{y}\right)^{\frac{2}{k}-1}.$

To the potentials $U_{2},U_{3},$ $U_{4}$ we have to add three more which we
obtain by the addition of the potential of the harmonic oscillator. Therefore
finally we have 7 potentials. The extra potentials are

  * •

$\displaystyle U_{5}\left(x,y\right)$ $\displaystyle=$ $\displaystyle
V_{0}r^{2}e^{+2\theta}+mr^{2}e^{2k\theta}=\bar{V}_{0+}~{}x^{\frac{{}^{2}}{1+k}}+\bar{m}xy$
(9.100) $\displaystyle U_{6}\left(x,y\right)$ $\displaystyle=$ $\displaystyle
r^{2}e^{-2\theta}+mr^{2}e^{2k\theta}=\bar{V}_{0-}y^{\frac{2}{1-k}}+\bar{m}xy$
(9.101)

where
$\bar{V}_{0+}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{},~{}\bar{V}_{0-}=\left(k-1\right)^{\frac{2}{1-k}}~{},~{}\bar{m}=m\left(k^{2}-1\right)$

  * •

$U_{7}\left(x,y\right)=V_{0}r^{2}e^{-2\frac{\left(k^{2}-2\right)}{k}\theta}+mr^{2}e^{2k\theta}=\bar{V}_{0}\frac{1}{y^{2}}\left(\frac{x}{y}\right)^{\frac{2}{k}-1}+\bar{m}xy$
(9.102)

where
$\bar{V}_{0}=V_{0}\frac{\left(k^{2}-1\right)^{\frac{2}{k}-1}}{\left(k-1\right)^{\frac{4}{k}}}$.

  * •

And the free particle potential

$U_{8}\left(x,y\right)=0$ (9.103)

These expressions allow us to write for each potential the Lagrangian and the
Hamiltonian constraint in the coordinates $x,y$. This means that we obtain the
corresponding field equations in the coordinates $x,y$ to determine their
solution.

###### Analytic solutions

The solution of the field equations for each potential is a formal and lengthy
operation which adds nothing but unnecessary material to the matter. What is
interesting is of course the final answer for each case and this is what we
give below for each of the potentials above.

  * •

$U_{1}\left(x,y\right)=V_{0}\left(k^{2}-1\right)xy$

$\displaystyle x\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}\sin\left(\omega t\right)+x_{2}\cos\left(\omega t\right)$ (9.104)
$\displaystyle y\left(t\right)$ $\displaystyle=$ $\displaystyle
y_{1}\sin\left(\omega t\right)+y_{2}\cos\left(\omega t\right)$ (9.105)

where $\omega^{2}=\frac{2V_{0}\left(k^{2}-1\right)}{N_{0}^{2}}$ and the
Hamiltonian is

$E=V_{0}\left(k^{2}-1\right)\left(x_{1}y_{1}+x_{2}y_{2}\right)$

  * •

$U_{2}\left(x,y\right)=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{}x^{\frac{{}^{2}}{1+k}}$

When $k\neq-3$

$\displaystyle x\left(t\right)$ $\displaystyle=$ $\displaystyle x_{1}t+x_{2}$
(9.106) $\displaystyle y\left(t\right)$ $\displaystyle=$
$\displaystyle-\frac{2\bar{V}\left(k+1\right)\left(x_{1}t+x_{2}\right)^{\left(1+\frac{2}{1+k}\right)}}{x_{1}^{2}\left(3+k\right)N_{0}^{2}}+y_{1}t+y_{2}$
(9.107)

where $\bar{V}=V_{0}\left(k+1\right)^{\frac{2}{1+k}},$ and the Hamiltonian is

$E=\frac{y_{1}x_{1}N_{0}^{2}}{2}.$

When $k=-3$

$y\left(t\right)=-2\frac{\bar{V}}{N_{0}^{2}x_{1}^{2}}\ln\left(x_{1}t+x_{2}\right)+y_{1}t+y_{2}.$
(9.108)

$x(t),$ $E$ being the same.

  * •

$U_{3}\left(x,y\right)=V_{0}\left(k-1\right)^{\frac{2}{1-k}}y^{\frac{2}{1-k}}$

When $k\neq 3$

$\displaystyle x\left(t\right)$ $\displaystyle=$
$\displaystyle\frac{2\bar{V}\left(k-1\right)\left(y_{1}t+y_{2}\right)^{1+\frac{2}{k-1}}}{y_{1}^{2}\left(k-3\right)N_{0}^{2}}+x_{1}t+x_{2}$
(9.109) $\displaystyle y\left(t\right)$ $\displaystyle=$ $\displaystyle
y_{1}t+y_{2}$ (9.110)

where $\bar{V}=V_{0}\left(k-1\right)^{\frac{2}{1-k}}$ , $k\neq 3$and the
Hamiltonian is

$E=\frac{y_{1}x_{1}N_{0}^{2}}{2}.$

When $k=3~{};$

$x\left(t\right)=-\frac{2\bar{V}}{N_{0}^{2}y_{1}^{2}}\ln\left(y_{1}t+y_{2}\right)+x_{1}t+x_{2}.$
(9.111)

$y(t),E$ being the same

  * •

$U_{5}\left(x,y\right)=\bar{V}_{0+}~{}x^{\frac{{}^{2}}{1+k}}+\bar{m}xy$

$\displaystyle x\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}\sin\left(\omega t+\omega_{0}\right)$ (9.112) $\displaystyle
y\left(t\right)$ $\displaystyle=$ $\displaystyle\cos\left(\omega
t+\omega_{0}\right)\left(y_{1}+2\frac{\omega}{\bar{m}}\int\frac{y_{2}-x_{1}\bar{V}_{0+}\sin\left(\omega
t+\omega_{0}\right)^{\frac{2}{1+k}}}{x_{1}\left(\cos\left(\omega
t+\omega_{0}\right)+1\right)}dt\right)$ (9.113)

where $\omega^{2}=\frac{2\bar{m}}{N_{0}^{2}}$ and $E=y_{2}.$

  * •

$U_{6}\left(x,y\right)=r^{2}e^{-2\theta}+mr^{2}e^{2k\theta}=\bar{V}_{0-}y^{\frac{2}{1-k}}+\bar{m}xy$

$\displaystyle x\left(t\right)$ $\displaystyle=$
$\displaystyle\cos\left(\omega
t+\omega_{0}\right)\left(x_{1}+2\frac{\omega}{\bar{m}}\int\frac{x_{2}-y_{1}\bar{V}_{0-}\sin\left(\omega
t+\omega_{0}\right)^{\frac{2}{1-k}}}{y_{1}\left(\cos\left(\omega
t+\omega_{0}\right)+1\right)}dt\right)$ (9.114) $\displaystyle
y\left(t\right)$ $\displaystyle=$ $\displaystyle y_{1}\sin\left(\omega
t+\omega_{0}\right)$ (9.115)

where $\omega^{2}=\frac{2\bar{m}}{N_{0}^{2}}$and $E=x_{2}.$

  * •

$U_{4,7}\left(x,y\right)=\bar{V}_{0}\frac{1}{y^{2}}\left(\frac{x}{y}\right)^{\frac{2}{k}-1}+\bar{m}xy$.
$\left(U_{4}\text{ is for }\bar{m}=0\right)$

This is the Ermakov Pinney system. To solve it, it is convenient to go to
spherical coordinates. We consider the coordinate transformation

$x=ze^{w}~{},~{}y=ze^{-w}~{}$ (9.116)

and the Lagrangian takes the form

$L\left(z,w,\dot{z},\dot{w}\right)=\frac{N_{0}^{2}}{2}\left(\dot{z}^{2}-z^{2}\dot{w}^{2}\right)-\frac{\bar{V}_{0}}{z^{2}}e^{\frac{4}{k}w}-\bar{m}z^{2}$
(9.117)

whereas the Hamiltonian becomes

$E=\frac{N_{0}^{2}}{2}\left(\dot{z}^{2}-z^{2}\dot{w}^{2}\right)+\frac{\bar{V}_{0}}{z^{2}}e^{\frac{4}{k}w}+\bar{m}z^{2}.$
(9.118)

This system admits the Ermakov-Lewis invariant, which is

$J_{EL}=z^{4}\dot{w}^{2}-2\frac{\bar{V}_{0}}{N_{0}^{2}}e^{\frac{4}{k}w}.$
(9.119)

Using the Ermakov invariant the Hamiltonian becomes

$E=\frac{N_{0}^{2}}{2}\dot{z}^{2}-N_{0}^{2}\frac{J_{EL}}{2z^{2}}+\bar{m}z^{2}.$
(9.120)

This is the Hamiltonian of the Ermakov Pinney equation:

$\ddot{z}+2\bar{m}z+N_{0}^{2}\frac{J_{EL}}{z^{3}}=0$ (9.121)

whose solution is

$\displaystyle z\left(t\right)$ $\displaystyle=$
$\displaystyle\left(l_{0}z_{1}\left(t\right)+l_{1}z_{2}\left(t\right)+l_{3}\right)^{\frac{1}{2}}$
(9.122) $\displaystyle e^{\frac{4}{k}w\left(t\right)}$ $\displaystyle=$
$\displaystyle\frac{N_{0}^{2}J_{EL}}{2\bar{V}_{0}}\left[\tanh^{2}\left(\frac{2\sqrt{J_{EL}}}{k}\left(\int\frac{dt}{z^{2}\left(t\right)}+l_{4}\right)\right)-1\right]$
(9.123)

where $z_{1,2}\left(t\right)$ are solutions of the differential equation
$\ddot{z}+2\bar{m}z=0$ and $l_{0-4}$ are constants.

  * •

$U_{8}(x,y)=0$

This is the free particle whose solution is

$\displaystyle x\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}t+x_{2}~{},~{}y\left(t\right)=y_{1}t+y_{2}$ (9.124) $\displaystyle E$
$\displaystyle=$ $\displaystyle\frac{N_{0}^{2}}{2}x_{1}y_{1}$ (9.125)

###### The case $k^{2}<1$

In this case the canonical coordinates are the $\bar{x},\bar{y}.$ Using the
transformation equations (9.94), (9.95) we write the Lagrangian (9.43) as
follows:

$L\left(\bar{x},\bar{y},\overset{\cdot}{\bar{x},}\overset{\cdot}{\bar{y}}\right)=-\frac{N_{0}^{2}}{2}\overset{\cdot}{\bar{x}}\overset{\cdot}{\bar{y}}-\bar{U}\left(\bar{x},\bar{y}\right)$
(9.126)

where $U\left(\bar{x},\bar{y}\right)=r^{2}V\left(\theta\right)$ the potentials
$V\left(\theta\right)$ being as above. In the coordinates $\bar{x},\bar{y}$
the potentials $U\left(\bar{x},\bar{y}\right)$ are:

  * •

$\bar{U}_{1}\left(\bar{x},\bar{y}\right)=V_{0}r^{2}e^{2k\theta}=V_{0}\left(1-k^{2}\right)\bar{x}\bar{y}$
(9.127)

$\bar{U}_{2}\left(\bar{x},\bar{y}\right)=V_{0}r^{2}e^{+2\theta}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{}\bar{x}^{\frac{{}^{2}}{1+k}}$
(9.128)
$\bar{U}_{3}\left(\bar{x},\bar{y}\right)=V_{0}r^{2}e^{-2\theta}=V_{0}\left(1-k\right)^{\frac{2}{1-k}}\bar{y}^{\frac{2}{1-k}}$
(9.129)

  * •

$\bar{U}_{4}\left(\bar{x},\bar{y}\right)=V_{0}r^{2}e^{-2\frac{\left(k^{2}-2\right)}{k}\theta}=V_{0}\frac{\left(1-k^{2}\right)^{\frac{2}{k}-1}}{\left(1-k\right)^{\frac{4}{k}}}\frac{1}{\bar{y}^{2}}\left(\frac{\bar{x}}{\bar{y}}\right)^{\frac{2}{k}-1}$
(9.130)

As before we have three more potentials

  * •

$\displaystyle\bar{U}_{5}\left(\bar{x},\bar{y}\right)$ $\displaystyle=$
$\displaystyle
V_{0}r^{2}e^{+2\theta}=\bar{V}_{0+}~{}\bar{x}^{\frac{{}^{2}}{1+k}}+\bar{m}\bar{x}\bar{y}$
(9.131) $\displaystyle\bar{U}_{6}\left(\bar{x},\bar{y}\right)$
$\displaystyle=$ $\displaystyle
V_{0}r^{2}e^{-2\theta}=\bar{V}_{0-}\bar{y}^{\frac{2}{1-k}}+\bar{m}\bar{x}\bar{y}$
(9.132)

where
$\bar{V}_{0+}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{},~{}\bar{V}_{0-}=\left(1-k\right)^{\frac{2}{1-k}}~{},~{}\bar{m}=m\left(1-k^{2}\right).$

  * •

$\bar{U}_{7}\left(\bar{x},\bar{y}\right)=V_{0}r^{2}e^{-2\frac{\left(k^{2}-2\right)}{k}\theta}+mr^{2}e^{2k\theta}=\bar{V}_{0}\frac{1}{\bar{y}^{2}}\left(\frac{\bar{x}}{\bar{y}}\right)^{\frac{2}{k}-1}+\bar{m}\bar{x}\bar{y}$
(9.133)

where
$\bar{V}_{0}=V_{0}\frac{\left(1-k^{2}\right)^{\frac{2}{k}-1}}{\left(1-k\right)^{\frac{4}{k}}}$.

  * •

and the free particle potential.

$\bar{U}_{8}\left(\bar{x},\bar{y}\right)=0$ (9.134)

###### Analytic solutions

Working as in the case $k^{2}>1$ we find the following analytic solutions and
the associated Hamiltonian constraint for each of the potentials above.

  * •

$\bar{U}_{1}\left(\bar{x},\bar{y}\right)=V_{0}\left(1-k^{2}\right)\bar{x}\bar{y}.$

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}\cosh\left(\omega t\right)+x_{2}\sinh\left(\omega t\right)$ (9.135)
$\displaystyle\bar{y}\left(t\right)$ $\displaystyle=$ $\displaystyle
y_{1}\cosh\left(\omega t\right)+y_{2}\sinh\left(\omega t\right)$ (9.136)

and the Hamiltonian $E=V_{0}m\left(x_{1}y_{1}-x_{2}y_{2}\right)$ where
$\omega^{2}=\frac{2V_{0}\left(1-k^{2}\right)}{N_{0}^{2}}$.

  * •

$\bar{U}_{2}\left(\bar{x},\bar{y}\right)=V_{0}\left(k+1\right)^{\frac{2}{1+k}}~{}\bar{x}^{\frac{{}^{2}}{1+k}}$

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}t+x_{2}$ (9.137) $\displaystyle\bar{y}\left(t\right)$ $\displaystyle=$
$\displaystyle\frac{2\bar{V}\left(k+1\right)\left(x_{1}t+x_{2}\right)^{\left(1+\frac{2}{1+k}\right)}}{x_{1}^{2}\left(3+k\right)N_{0}^{2}}+y_{1}t+y_{2}$
(9.138)

where $\bar{V}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}$and the Hamiltonian
$E=-\frac{N_{0}^{2}}{2}x_{1}y_{1}$.

  * •

$\bar{U}_{3}\left(\bar{x},\bar{y}\right)=V_{0}\left(1-k\right)^{\frac{2}{1-k}}\bar{y}^{\frac{2}{1-k}}$

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$
$\displaystyle\frac{2\bar{V}\left(1-k\right)\left(y_{1}t+y_{2}\right)^{1+\frac{2}{k-1}}}{y_{1}^{2}\left(k-3\right)N_{0}^{2}}+x_{1}t+x_{2}$
(9.139) $\displaystyle\bar{y}\left(t\right)$ $\displaystyle=$ $\displaystyle
y_{1}t+y_{2}$ (9.140)

where $\bar{V}=V_{0}\left(k+1\right)^{\frac{2}{1+k}}$and the Hamiltonian
constrain $E=-\frac{N_{0}^{2}}{2}x_{1}y_{1}$.

  * •

$\bar{U}_{5}\left(\bar{x},\bar{y}\right)=\bar{V}_{0+}~{}\bar{x}^{\frac{{}^{2}}{1+k}}+\bar{m}\bar{x}\bar{y}$

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$
$\displaystyle\bar{x}_{1}\sinh\left(\omega t+\omega_{0}\right)$
$\displaystyle\bar{y}\left(t\right)$ $\displaystyle=$
$\displaystyle\cosh\left(\omega
t+\omega_{0}\right)\left(\bar{y}_{1}-\frac{2\omega}{\bar{m}}\int\frac{E-\bar{x}_{1}\bar{V}_{0+}\sinh\left(\omega
t+\omega_{0}\right)^{\frac{2}{1+k}}}{\bar{x}_{1}\left(\cosh\left(\omega
t+\omega_{0}\right)+1\right)}\right)$

where $\omega^{2}=\frac{2\bar{m}}{N_{0}^{2}}$

  * •

$\bar{U}_{6}\left(\bar{x},\bar{y}\right)=\bar{V}_{0-}\bar{y}^{\frac{2}{1-k}}+\bar{m}\bar{x}\bar{y}$

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$
$\displaystyle\cosh\left(\omega
t+\omega_{0}\right)\left(\bar{x}_{1}-\frac{2\omega}{\bar{m}}\int\frac{E-\bar{y}_{1}\bar{V}_{0-}\sinh\left(\omega
t+\omega_{0}\right)^{\frac{2}{1-k}}}{\bar{y}_{1}\left(\cosh\left(\omega
t+\omega_{0}\right)+1\right)}\right)$ $\displaystyle\bar{y}\left(t\right)$
$\displaystyle=$ $\displaystyle\bar{y}_{1}\sinh\left(\omega
t+\omega_{0}\right)$

where $\omega^{2}=\frac{2\bar{m}}{N_{0}^{2}}$

  * •

$\bar{U}_{4,7}\left(\bar{x},\bar{y}\right)=\bar{V}_{0}\frac{1}{\bar{y}^{2}}\left(\frac{\bar{x}}{\bar{y}}\right)^{\frac{2}{k}-1}+\bar{m}\bar{x}\bar{y}$.
$\left(\bar{U}_{4}\text{ is for }\bar{m}=0\right)$.

This is again the Ermakov Piney potential. To solve it we go to spherical
coordinates

$\bar{x}=ze^{w}~{}\ ,~{}\bar{y}=ze^{-w}~{}$ (9.141)

in which the Lagrangian takes the form

$L\left(z,w,\dot{z},\dot{w}\right)=\frac{N_{0}^{2}}{2}\left(-\dot{z}^{2}+z^{2}\dot{w}^{2}\right)-\frac{\bar{V}_{0}}{z^{2}}e^{\frac{4}{k}w}-\bar{m}z^{2}$
(9.142)

and the Hamiltonian

$E=\frac{N_{0}^{2}}{2}\left(-\dot{z}^{2}+z^{2}\dot{w}^{2}\right)+\frac{\bar{V}_{0}}{z^{2}}e^{\frac{4}{k}w}+\bar{m}z^{2}.$
(9.143)

The Ermakov-Lewis invariant is

$J_{EL}=z^{4}\dot{w}^{2}+2\frac{\bar{V}_{0}}{N_{0}^{2}}e^{\frac{4}{k}w}$
(9.144)

which when replaced in the Hamiltonian gives

$E=-\frac{N_{0}^{2}}{2}\dot{z}^{2}+N_{0}^{2}\frac{J_{EL}}{2z^{2}}+\bar{m}z^{2}.$
(9.145)

This is the Hamiltonian of the Ermakov Pinney equation:

$\ddot{z}-2\bar{m}z-N_{0}^{2}\frac{J_{EL}}{z^{3}}=0$ (9.146)

whose solution is

$\displaystyle z\left(t\right)$ $\displaystyle=$
$\displaystyle\left(l_{0}z_{1}\left(t\right)+l_{1}z_{2}\left(t\right)+l_{3}\right)^{\frac{1}{2}}$
(9.147) $\displaystyle e^{\frac{4}{k}w\left(t\right)}$ $\displaystyle=$
$\displaystyle\frac{N_{0}^{2}J_{EL}}{2\bar{V}_{0}}\left[1-\tanh^{2}\left(\frac{2\sqrt{J_{EL}}}{k}\left(\int\frac{dt}{z^{2}\left(t\right)}+l_{4}\right)\right)\right]$
(9.148)

where $z_{1,2}\left(t\right)$ are solutions of the ode $\ddot{z}-2\bar{m}z=0$
and $l_{0-4}$ are constants.

  * •

$\bar{U}_{8}\left(\bar{x},\bar{y}\right)=0$

  * •

This is the free particle whose solution is

$\displaystyle\bar{x}\left(t\right)$ $\displaystyle=$ $\displaystyle
x_{1}t+x_{2}~{},~{}\bar{y}\left(t\right)=y_{1}t+y_{2}$ (9.149) $\displaystyle
E$ $\displaystyle=$ $\displaystyle-\frac{N_{0}^{2}}{2}x_{1}y_{1}.$ (9.150)

#####  9.5.4 Case B: The 2d metric is conformally flat

In this case it is preferable to work with (9.40) which under the coordinate
transformation (9.41) becomes:

$L=N^{2}\left(\theta\right)\left[-\frac{1}{2}\dot{r}^{2}+\frac{1}{2}r^{2}\dot{\theta}^{2}\right]-r^{2}V\left(\theta\right).$
(9.151)

The kinetic metric in this case is not flat (i.e. $R_{\left(2\right)}\neq 0)$
but of course it is conformally flat being a two dimensional metric. Its
conformal algebra is infinity dimensional however it has a closed subalgebra
consisting of the following vectors (this is the special conformal algebra of
$M^{2}$):

$\displaystyle X^{1}$ $\displaystyle=$
$\displaystyle\cosh\theta\partial_{r}-\frac{1}{r}\sinh\theta\partial_{\theta}~{}~{},~{}X^{2}=\sinh\theta\partial_{r}-\frac{1}{r}\cosh\theta\partial_{\theta}$
$\displaystyle X^{3}$ $\displaystyle=$ $\displaystyle\partial_{\theta}~{}~{}\
,~{}X^{4}=r\partial_{r}~{}~{},~{}X^{5}=\frac{1}{2}r^{2}\cosh\theta\partial_{r}+\frac{1}{2}r\sinh\theta\partial_{\theta}$
$\displaystyle X^{6}$ $\displaystyle=$
$\displaystyle\frac{1}{2}r^{2}\sinh\theta\partial_{r}+\frac{1}{2}r\cosh\theta\partial_{\theta}$
(9.152)

Writing $L_{X^{I}}g_{ij}=2C_{I}\left(r,\theta\right)g_{ij}$ we find the
conformal factors of the CKVs $X^{I}$ $I=1,...6$ above in terms of the the
conformal function. The result is:

$\displaystyle C_{1}\left(r,\theta\right)$ $\displaystyle=$
$\displaystyle-\frac{1}{r}\sinh\left(\theta\right)\frac{N_{,\theta}}{N}~{},~{}C_{2}\left(r,\theta\right)=-\frac{1}{r}\cosh\left(\theta\right)\frac{N_{,\theta}}{N}$
$\displaystyle C_{3}\left(r,\theta\right)$ $\displaystyle=$
$\displaystyle\frac{N_{,\theta}}{N}~{}~{},~{}C_{4}\left(r,\theta\right)=1~{},~{}~{}C_{5}\left(r,\theta\right)=\frac{r}{2}\left(\frac{2N\cosh\theta+\sinh\theta
N_{\theta}}{N}\right)$ $\displaystyle C_{6}\left(r,\theta\right)$
$\displaystyle=$
$\displaystyle\frac{r}{2}\left(\frac{2N\sinh\theta+\cosh\theta
N_{\theta}}{N}\right)$ (9.153)

The $N\left(\theta\right)\neq e^{c\theta}$, otherwise the kinetic metric of
the Lagrangian (9.151) is flat (the Ricci scalar vanishes) and we return to
the previous case A. This means that the vectors $X^{I}$ $I=1,...6$ except the
$I=4$ are proper CKVs therefore they do not give (in general) a Noether
symmetry. The vector $X_{4}$ is a non-gradient HV which does not also produce
a Noether symmetry. Therefore according to theorem 4.3.2 only Killing vectors
are possible to serve as Noether symmetries. KVs do not exist but for special
forms of the conformal function $N(\theta)$. Each such form of $N(\theta)$
results in a potential $V(\theta)\ $hence to a scalar field potential which
admits Noether point symmetries. In the following we determine the possible
$N(\theta)$ which lead to a KV and give the corresponding Noether symmetry and
the corresponding Noether integral which will be used for the solution of the
field equations.

  1. 1.

If $N\left(\theta\right)=\frac{N_{0}}{\cosh\left(2\theta\right)-1}$ then
$X^{5}$ is a non gradient KV and a Noether symmetry for the Lagrangian (9.151)
for the potential

$V\left(\theta\right)=\frac{V_{0}}{\cosh\left(2\theta\right)-1}~{}\text{or
}V\left(\theta\right)=0$ (9.154)

The corresponding Noether integral is

$I_{X^{5}}=\frac{N_{0}^{2}r^{2}}{\left(\cosh\left(2\theta\right)-1\right)^{2}}\left(r\dot{\theta}\sinh\theta-\dot{r}\cosh\theta\right).$
(9.155)

  2. 2.

If $N\left(\theta\right)=\frac{N_{0}}{\cosh\left(2\theta\right)+1}$ then
$X^{6}$ is a non gradient KV, $X^{6}$ and a Noether symmetry for the
Lagrangian (9.151) if

$V\left(\theta\right)=\frac{V_{0}}{\cosh\left(2\theta\right)+1}~{}\text{or
}V\left(\theta\right)=0$ (9.156)

The corresponding Noether integral is

$I_{X^{6}}=\frac{N_{0}^{2}r^{2}}{\left(\cosh\left(2\theta\right)+1\right)^{2}}\left(r\dot{\theta}\cosh\theta-\dot{r}\sinh\theta\right)$
(9.157)

  3. 3.

If
$N\left(\theta\right)=\frac{N_{0}}{\cosh^{2}\left(\theta+\theta_{0}\right)}$
then the linear combination $X^{56}=c_{1}X^{5}+c_{2}X^{6}$ where
$c_{1}=\sinh\left(\theta_{0}\right)$ and $c_{2}=\cosh\left(\theta_{0}\right)$.
$X^{56}$ is a Noether symmetry for the Lagrangian (9.151) if

$V\left(\theta\right)=\frac{V_{0}}{\cosh^{2}\left(\theta+\theta_{0}\right)}~{}\text{or
}V\left(\theta\right)=0$ (9.158)

with Noether integral

$I_{X^{56}}=\frac{N_{0}^{2}r^{2}}{\cosh^{4}\left(\theta+\theta_{0}\right)}\left(r\dot{\theta}\cosh\left(\theta+\theta_{0}\right)-\dot{r}\sinh\left(\theta+\theta_{0}\right)\right)$

Obviously case 3 is the most general and contains cases 1 and 2 (and the
trivial case) as special cases. Therefore in the following we look for
analytic solutions for the vector $X^{56}$only.

We recall that
$\frac{1}{\sqrt{-2F\left(\theta\right)}}=N^{2}\left(\theta\right)$ from which
follows:

$~{}F\left(\theta\right)=-\frac{1}{2N_{0}^{4}}\cosh^{8}\left(\theta+\theta_{0}\right)~{},N_{0}\in\mathbb{R}.$
(9.159)

We may consider $\theta_{0}=0$ (e.g. by introducing the new variable
$\Theta=\theta+\theta_{0}).$

For the potential (9.158) the Lagrangian (9.151) becomes

$L=\frac{N_{0}^{2}}{\cosh^{4}\theta}\left[-\frac{1}{2}\dot{r}^{2}+\frac{1}{2}r^{2}\dot{\theta}^{2}\right]-r^{2}\frac{V_{0}}{\cosh^{2}\theta}$
(9.160)

and the Hamiltonian

$E=\frac{N_{0}^{2}}{\cosh^{4}\theta}\left[-\frac{1}{2}\dot{r}^{2}+\frac{1}{2}r^{2}\dot{\theta}^{2}\right]+r^{2}\frac{V_{0}}{\cosh^{2}\theta}.$
(9.161)

The equations of motion are:

$\displaystyle\ddot{r}+r\dot{\theta}^{2}-4\tanh\theta~{}\dot{r}\dot{\theta}-2\frac{V_{0}}{N_{0}^{2}}r\cosh^{2}\theta$
$\displaystyle=$ $\displaystyle 0$ (9.162)
$\displaystyle\ddot{\theta}-2\tanh\theta~{}\left(\frac{1}{r^{2}}\dot{r}^{2}+\dot{\theta}^{2}\right)+\frac{2}{r}\dot{r}\dot{\theta}-2\frac{V_{0}}{N_{0}^{2}}\cosh\theta\sinh\theta$
$\displaystyle=$ $\displaystyle 0$ (9.163)

and the Noether integral $I_{X^{56}}$ for $\theta_{0}=0$ becomes:

$I_{X^{56}}=\frac{N_{0}^{2}r^{2}}{\cosh^{4}\left(\theta+\theta_{0}\right)}\left(r\dot{\theta}\cosh\left(\theta\right)-\dot{r}\sinh\left(\theta\right)\right).$
(9.164)

In order to proceed with the solution of the system of equations (9.162),
(9.163) we change to the coordinates $x,y~{}$which we define by the relations

$r=\frac{x}{\sqrt{1-x^{2}y^{2}}}~{},~{}\theta=\arctan h\left(xy\right).$
(9.165)

In the coordinates $x,y$ the Lagrangian and the Hamiltonian become:

$L=\frac{N_{0}^{2}}{2}\left(-\dot{x}^{2}+x^{4}\dot{y}\right)-V_{0}x^{2}$
(9.166)
$E=\frac{N_{0}^{2}}{2}\left(-\dot{x}^{2}+x^{4}\dot{y}^{2}\right)+V_{0}x^{2}$
(9.167)

and the Noether integral (we write $I$ for $I_{X^{56}})$

$I=x^{4}\dot{y}.$ (9.168)

Let us assume that $I\neq 0.$In the new variables the Euler - Lagrange
equations read:

$\displaystyle\ddot{x}+2x^{3}\dot{y}^{2}-2V_{0}x$ $\displaystyle=$
$\displaystyle 0$ (9.169) $\displaystyle\ddot{y}+\frac{4}{x}\dot{x}\dot{y}$
$\displaystyle=$ $\displaystyle 0.$ (9.170)

From the Noether integral we have $\dot{y}=\frac{I}{x^{4}}~{}$which upon
substitution in the field equations gives the equation

$\displaystyle\ddot{x}-2V_{0}x+\frac{2I^{2}}{x^{5}}$ $\displaystyle=$
$\displaystyle 0$ (9.171)
$\displaystyle\frac{1}{2}\left(-\dot{x}^{2}+\frac{I^{2}}{x^{4}}\right)+V_{0}x^{2}$
$\displaystyle=$ $\displaystyle E.$ (9.172)

from which we compute

$\dot{x}=\sqrt{\frac{I^{2}}{x^{4}}+2V_{0}x^{2}-2E}$ (9.173)

and finally the analytic solution

$\int\frac{dx}{\sqrt{\frac{I^{2}}{x^{4}}+2V_{0}x^{2}-2E}}=t-t_{0}.$ (9.174)

From the Noether integral we find

$y\left(t\right)-y_{0}=\int\frac{I}{x^{4}}dt.$ (9.175)

If $I=0$ then the analytic solution is

$x=x_{0}\sinh\left(\sqrt{2V_{0}}t+x_{1}\right),~{}y=y_{0}$ (9.176)

with Hamiltonian constrain $E=-x_{0}^{2}V_{0}.$

If $V_{0}=0~{}$(i.e. free particle) and $I=0$ the analytic solution is

$x=x_{0}t+x_{1}~{},~{}y=y_{0}$ (9.177)

with Hamiltonian constrain $E=-\frac{1}{2}x_{0}^{2}.$

####  9.6 Noether point symmetries of a minimally coupled Scalar field.

In this section we study the Noether point symmetries of a minimally coupled
scalar field in a spatially flat FRW spacetime. The action of the field
equations is

$S_{M}=\int d\tau
dx^{3}\sqrt{-g}\left[R+\frac{1}{2}g_{ij}\phi^{;i}\phi^{;j}-V\left(\phi\right)\right]+\int
L_{m}d\tau dx^{3}.$ (9.178)

where $L_{m}$ is the Lagrangian of the dust matter fluid. For a spatially flat
FRW spacetime the Ricciscalar is

$R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right)$

hence, the Lagrangian of the field equation is (9.12) and the field equations
are

$E=-3a\dot{a}^{2}+\frac{\varepsilon}{2}a^{2}\dot{\phi}^{2}+a^{3}V\left(\phi\right)$
(9.179)
$\ddot{a}+\frac{1}{2a}\dot{a}^{2}+\frac{\varepsilon}{4}\dot{\phi}^{2}-\frac{1}{2}aV=0$
(9.180) $\ddot{\phi}+\frac{3}{a}\dot{a}\dot{\phi}+\varepsilon V_{,\phi}=0$
(9.181)

From the kinetic term of (9.12) we define the two dimensional metric

$ds^{2}=-6a~{}da^{2}+\varepsilon a^{3}d\phi^{2}.$ (9.182)

We find that the curvature of the $\\{a,\phi\\}$ space is $R=0$ implying
flatness (since all $2$ dimensional spaces are Einstein spaces hence
$\hat{R}=0$ implies that the space is flat). Also, the signature of the metric
eq.(9.182) is $0$, hence the space is the 2-d Minkowski space. In order to
simplify the field equations we apply the following coordinate transformation

$r=\sqrt{\frac{8}{3}}a^{3/2}\;\;\;\;\;\theta=\sqrt{\frac{3k\epsilon}{8}}\phi\;,$
(9.183)

in the new coordinates the two dimensional metric (9.182) is given by

$d{\hat{s}}^{2}=-dr^{2}+r^{2}d\theta^{2}$ (9.184)

that is, $(r,\theta)$ are hyperbolic spherical coordinates in the two
dimensional Minkowski space $\\{a,\phi\\}$. Next we introduce the new
coordinates $(x,y)$ with the transformation:

$\displaystyle x$ $\displaystyle=r\cosh\;(\theta)$ $\displaystyle y$
$\displaystyle=r\sinh\;(\theta)$ (9.185)

which implies that the metric (9.184) becomes $d\hat{s}^{2}=-dx^{2}+dy^{2}$.
We also point here that

$r^{2}=x^{2}-y^{2}\;\;\;\;\;\theta=\mathrm{arctanh}(y/x)\;.$ (9.186)

The scale factor ($a(t)>0$) is now given by:

$a=\left[\frac{3(x^{2}-y^{2})}{8}\right]^{1/3}$ (9.187)

which means that the new variables have to satisfy the following inequality:
$x\geq|y|$.

In the new coordinate system $(x,y)$ the Lagrangian (9.12) and the Hamiltonian
(9.179) are written:

$L=\frac{1}{2}\left(\dot{y}^{2}-\dot{x}^{2}\right)-V_{eff}(x,y)$ (9.188)
$E=\frac{1}{2}\left(\dot{y}^{2}-\dot{x}^{2}\right)+V_{eff}(x,y)$ (9.189)

where

$V_{eff}(x,y)=\left(x^{2}-y^{2}\right)\tilde{V}\left(\frac{y}{x}\right).$
(9.190)

Note that we have used

$\tilde{V}\left(\theta\right)=\frac{3k}{8}V(\theta)\;.$ (9.191)

We now proceed in an attempt to provide the Noether point symmetries of the
current dynamical problem using the results of chapter 4.

Since the Lagrangian (9.188) is autonomous admits the Noether point symmetry
$\partial_{t}$ with Noether integral the Hamiltonian (9.189). Lagrangian
(9.188) admits extra Noether point symmetries in the following cases.

#####  9.6.1 Hyperbolic - UDM Potential

Hyperbolic - UDM Potential: Generically, we use the following potential:

$\tilde{V}(\theta)=\frac{\omega_{1}\cosh^{2}\left(\theta\right)-\omega_{2}\sinh^{2}\left(\theta\right)}{2}$
(9.192)

or

$V_{eff}(x,y)=r^{2}\tilde{V}(\theta)=\frac{\omega_{1}x^{2}-\omega_{2}y^{2}}{2}$
(9.193)

The corresponding Noether symmetries, $X_{n}$, are known (see for example
[5]). These are:

$\displaystyle X_{n_{1}}$
$\displaystyle=\partial_{t}~{},~{}X_{n_{2}}=\sinh\left(\sqrt{\omega_{1}}t\right)\partial_{x}~{},~{}X_{n_{3}}=\cosh\left(\sqrt{\omega_{1}}t\right)\partial_{x}$
$\displaystyle X_{n_{4}}$
$\displaystyle=\sinh\left(\sqrt{\omega_{2}}t\right)\partial_{y}~{},~{}X_{n_{5}}=\cosh\left(\sqrt{\omega_{2}}t\right)\partial_{y}$

The Noether integrals are the Hamiltonian and the quantities:

$\displaystyle I_{n_{2}}$
$\displaystyle=\sinh\left(\sqrt{\omega_{1}}t\right)\dot{x}-\sqrt{\omega_{1}}\cosh\left(\sqrt{\omega_{1}}t\right)x$
$\displaystyle I_{n_{3}}$
$\displaystyle=\cosh\left(\sqrt{\omega_{1}}t\right)\dot{x}-\sqrt{\omega_{1}}\sinh\left(\sqrt{\omega_{1}}t\right)x$
$\displaystyle I_{n_{4}}$
$\displaystyle=\sinh\left(\sqrt{\omega_{2}}t\right)\dot{y}-\sqrt{\omega_{2}}\cosh\left(\sqrt{\omega_{2}}t\right)y$
$\displaystyle I_{n_{5}}$
$\displaystyle=\cosh\left(\sqrt{\omega_{2}}t\right)\dot{y}-\sqrt{\omega_{2}}\sinh\left(\sqrt{\omega_{2}}t\right)y$

Obviously the UDM potential is a particular case of the current general
hyperbolic potential. Indeed for $\omega_{1}=2\omega_{2}$ and with the aid of
eqs.(9.183), (9.191) we fully recover the UDM potential [153, 154, 155, 142]

$V(\phi)=V_{0}\left[1+\cosh^{2}\left(\frac{3k\epsilon}{8}\phi\right)\right]$
(9.194)

where $V_{0}=\frac{4\omega_{2}}{3k}$ modulus a constant.

#####  9.6.2 Exponential Potential

Exponential Potential: The exponential potential

$V_{eff}(r,\theta)=r^{2}\tilde{V}(\theta)=r^{2}e^{-d\theta}\;.$

admits the extra Noether symmetry

$X_{n}=2t\partial_{t}+\left(x+\frac{4}{d}y\right)\partial_{x}+\left(y+\frac{4}{d}x\right)\partial_{y}\;.$
(9.195)

In general the Noether integral for the vector
$X_{n}=2t\partial_{t}+\eta^{i}\partial_{i}$ is

$I=2tE+\left(x+\frac{4}{d}y\right)\dot{x}-\left(y+\frac{4}{d}x\right)\dot{y}$
(9.196)

where $E$ is the Hamiltonian. Using ${\tilde{V}}=e^{-d\theta}$ together with
eq.(9.183) and eq.(9.191) we write the potential to its nominal form [156]
which is

$V(\phi)=V_{0}\mathrm{exp}\left(-d\;\sqrt{\frac{3k\epsilon}{8}}\phi\right)$
(9.197)

where $V_{0}=\frac{8}{3k}$ modulus a constant333In the special case of $d=2$,
the system admits an additional Lie symmetry $\partial_{x}+\partial_{y}$, with
Noether integral $I=\dot{x}-\dot{y}.$.

We note that the analytic solutions of sections 9.5.3 and 9.5.4 are also
solutions for the minimally coupled Lagrangian (9.12) if and only if the space
does not admit dust.

####  9.7 The Lie and Noether symmetries of Bianchi class A homogeneous
cosmologies with a scalar field.

The class of Bianchi spatially homogeneous cosmologies contains many important
cosmological models, including the standard FRW model. In these models the
spacetime manifold is foliated along the time axis, with three dimensional
homogeneous hypersurfaces. Bianchi has classified all three dimensional real
Lie algebras and has shown that there are nine of them. This results in nine
types (two of them being families of spacetimes) of Bianchi spatially
homogeneous spacetimes. The principal advantage of Bianchi cosmological models
is that, in these models the physical variables depend on the time only,
reducing the Einstein and other governing equations to ordinary differential
equations.

The Bianchi models are studied in the well known ADM decomposition ([157,
158]) according to which the metric is written

$ds^{2}=-N^{2}(t)dt^{2}+g_{\mu\nu}\omega^{\mu}\otimes\omega^{\nu}$ (9.198)

where $N(t)$ is the lapse function and $\\{\omega^{a}\\}$ is the canonical
basis of 1-forms which satisfy the Lie algebra

$d\omega^{i}=C_{jk}^{i}\omega^{j}\wedge\omega^{k}.$ (9.199)

$C_{jk}^{i}$ are the structure constants of the algebra. The spatial metric
$g_{\mu\nu}$ splits so that

$g_{\mu\nu}=\exp(2\lambda)\exp(-2\beta)_{\mu\nu}$ (9.200)

where $\exp(2\lambda)$ is the scale factor of the universe and
$\beta_{\mu\nu}$ is a $3\times 3$ symmetric, traceless matrix, which can be
written in a diagonal form with two independent quantities, the so called
anisotropy parameters $\beta_{+},\beta_{-}$, as follows:

$\beta_{\mu\nu}=diag\left(\beta_{+},-\frac{1}{2}\beta_{+}+\frac{\sqrt{3}}{2}\beta_{-},-\frac{1}{2}\beta_{+}-\frac{\sqrt{3}}{2}\beta_{-}\right).$
(9.201)

The Bianchi models are grouped in classes A and B by means of a vector
$a^{\mu}$ and a symmetric 3$\times 3$ metric $n_{\mu\nu}$ which are
constrainted by the condition $n_{\mu\nu}a^{\nu}=0.$ Class A is defined by
$a^{\mu}=0$ and Class B by $a^{\mu}\neq 0.$ Each Class is divided into several
types according to the rank and (the modulus of the) signature of
$n^{\mu\nu}.$ Because of the difficulty in formulating the class B Bianchi
models in the ADM formalism, it is usually the case that one confines
attention to the class A models. Furthermore it is well known that for the
class A models there is a Lagrangian [159] whereas for the class B models, to
the author’s knowledge, no such Lagrangian seems to exist. Details on the
structure and the Physics of the Bianchi models can be found e.g. in [157,
159].

Research in Physics on inflationary models has shown the importance of scalar
fields in cosmology [160]. This has raised interest in the dynamics of Bianchi
spacetimes filled with a scalar field, with an arbitrary self interaction
potential, minimally coupled to the gravitational field [161]. The Lagrangian
leading to the full Bianchi scalar field dynamics is

$L=e^{3\lambda}\left[R^{\ast}+6\lambda-\frac{3}{2}(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2})-\dot{\phi}^{2}+V(\phi)\right]$
(9.202)

where $R^{\ast}$ is the Ricci scalar of the $3$ dimensional spatial
hypersurfaces given by the expression:

$\displaystyle R^{\ast}$
$\displaystyle=-\frac{1}{2}e^{-2\lambda}\left[N_{1}^{2}e^{4\beta_{1}}+e^{-2\beta_{1}}\left(N_{2}e^{\sqrt{3}\beta_{2}}-N_{3}e^{-\sqrt{3}\beta_{2}}\right)^{2}-2N_{1}e^{\beta_{1}}\left(N_{2}e^{\sqrt{3}\beta_{2}}-N_{3}e^{-\sqrt{3}\beta_{2}}\right)\right]$
$\displaystyle+\frac{1}{2}N_{1}N_{2}N_{3}(1+N_{1}N_{2}N_{3}).$

The constants $N_{1},N_{2},$ and $N_{3}$ are the components of the
classification vector $n^{\mu}$ and
$\beta_{1}=-\frac{1}{2}\beta_{+}+\frac{\sqrt{3}}{2}\beta_{-},$
$\beta_{2}=-\frac{1}{2}\beta_{+}-\frac{\sqrt{3}}{2}\beta_{-}$. It is important
to note that the curvature scalar $R^{\ast}$ does not depend on the
derivatives of the anisotropy parameters $\beta_{+},$ $\beta_{-}$ ,
equivalently on $\beta_{1},\beta_{2}.$

The Euler Lagrange equations due to the Lagrangian (9.202) are [86]:

$\displaystyle\ddot{\lambda}+\frac{3}{2}\dot{\lambda}^{2}+\frac{3}{8}(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2})+\frac{1}{4}\dot{\phi}^{2}-\frac{1}{12}e^{-3\lambda}\frac{\partial}{\partial\lambda}\left(e^{3\lambda}R^{\ast}\right)-\frac{1}{2}V(\phi)$
$\displaystyle=0$
$\displaystyle\ddot{\beta}_{1}+3\dot{\lambda}\dot{\beta}_{1}+\frac{1}{3}\frac{\partial
R^{\ast}}{\partial\beta_{1}}$ $\displaystyle=0$
$\displaystyle\ddot{\beta}_{2}+3\dot{\lambda}\dot{\beta}_{2}+\frac{1}{3}\frac{\partial
R^{\ast}}{\partial\beta_{2}}$ $\displaystyle=0$
$\displaystyle\ddot{\phi}+3\dot{\phi}\dot{\lambda}+\frac{\partial
V}{\partial\phi}$ $\displaystyle=0$

where a dot over a symbol indicates derivative with respect to $t.$

In the following we apply Theorem 4.2.2 and Theorem 4.3.2 and compute the Lie
and the Noether point symmetries of class A Bianchi models. The Lie and the
Noether point symmetries of Bianchi class A models with a scalar field have
also been computed in [86, 143, 139]. However, as it will be shown, these
studies are not complete, in the sense that they have not found all Noether
symmetries. Furthermore our approach is entirely different than the classical
Lie approach employed in these works. Finally it is general and can be applied
without difficulty to Class B spacetimes.

We consider the four dimensional Riemannian space with coordinates
$x^{i}=\left(\lambda,\beta_{1},\beta_{2},\phi\right)$ and metric

$ds^{2}=e^{3\lambda}\left(12d\lambda^{2}-3d\beta_{1}^{2}-3d\beta_{2}^{2}-2d\phi^{2}\right).$
(9.203)

The metric (9.203) is the conformally flat FRW spacetime whose special
projective algebra consists of the non gradient KVs [46, 45]

$\displaystyle Y^{1}$
$\displaystyle=\partial_{\beta_{1}},~{}Y^{2}=\partial_{\beta_{2}},~{}Y^{3}=\partial_{\phi},~{}Y^{4}=\beta_{2}\partial_{\beta_{1}}-\beta_{1}\partial_{\beta_{2}}$
$\displaystyle~{}Y^{5}$
$\displaystyle=\phi\partial_{\beta_{1}}-\frac{3}{2}\beta_{1}\partial_{\phi},~{}Y^{6}=\phi\partial_{\beta_{2}}-\frac{3}{2}\beta_{2}\partial_{\phi}$

and the gradient HV

$H^{i}=\frac{2}{3}\partial_{\lambda}~{},~{}\psi=1.$

The Lagrangian (9.202) is written:

$L=T-U(x^{i})$

where $T=\frac{1}{2}g_{ij}\dot{x}^{i}\dot{x}^{i}$ is the geodesic Lagrangian,
$U(x^{i})$ is the potential function

$U(x^{i})=-e^{3\lambda}\left(V\left(\phi\right)+R^{\ast}\right)$ (9.204)

and we have used the fact that the curvature scalar does not depend on the
derivatives of the coordinates $\beta_{1},\beta_{2}.$ Now we apply Theorem
4.2.2 and Theorem 4.3.2 to determine the Lie and the point Noether symmetries
of the dynamical system with Lagrangian (9.202).

In order to compute the potential $U(x^{i})$ we need the expression of
$R^{\ast}$ for each Bianchi type. We find for the Class A models

Bianchi I: $R^{\ast}=0$

Bianchi II: $R^{\ast}=-e^{\left(2\beta_{1}-\lambda\right)}$

Bianchi VI0: Class A. ,
$R^{\ast}=-\frac{1}{2}e^{-2\lambda}\left(e^{4\beta_{1}}+e^{-2\left(\beta_{1}-\sqrt{3}\beta_{2}\right)}+2e^{\beta_{1}+\sqrt{3}\beta_{2}}\right)$

Bianchi VII0: Class A.,
$R^{\ast}=-\frac{1}{2}e^{-2\lambda}\left(e^{4\beta_{1}}+e^{-2\left(\beta_{1}-\sqrt{3}\beta_{2}\right)}-2e^{\beta_{1}+\sqrt{3}\beta_{2}}\right)$

Bianchi VIII:
$R^{\ast}=-\frac{1}{2}e^{-2\lambda}\left(e^{4\beta_{1}}+e^{-2\beta_{1}}\left(e^{\sqrt{3}\beta_{2}}+e^{-\sqrt{3}\beta_{2}}\right)^{2}-2e^{\beta_{1}}\left(e^{\sqrt{3}\beta_{2}}-e^{-\sqrt{3}\beta_{2}}x\right)\right)$

Bianchi IX:
$R^{\ast}=-\frac{1}{2}e^{-2\lambda}\left(e^{4\beta_{1}}+e^{-2\beta_{1}}\left(e^{\sqrt{3}\beta_{2}}+e^{-\sqrt{3}\beta_{2}}\right)^{2}-2e^{\beta_{1}}\left(e^{\sqrt{3}\beta_{2}}-e^{-\sqrt{3}\beta_{2}}x\right)\right)+1$.

We determine the Lie and the Noether point symmetries in the following cases:

Case 1. Vacuum. In this case $\phi=$constant and the metric (9.203) reduces to
the three dimensional FRW metric.

Case 2. Zero potential $V\left(\phi\right)=0,$ $\dot{\phi}\neq 0$

Case 3. Constant Potential $V\left(\phi\right)=$constant, $\dot{\phi}\neq 0$

Case 4. Arbitrary Potential $V\left(\phi\right),~{}\dot{\phi}\neq 0.$

#####  9.7.1 Bianchi I

Case 1.

In this case $\dot{\phi}=0$, $V\left(\phi\right)=0$ and the Lagrangian becomes
$~{}L=e^{3\lambda}\left[6\dot{\lambda}^{2}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)\right]~{}$hence
the potential $U(x^{\mu})=0$ where $x^{\mu}=(\lambda,\beta_{1},\beta_{2}).$
The auxiliary metric is
$ds^{2}=e^{3\lambda}\left(12d\lambda^{2}-3d\beta_{1}^{2}-3d\beta_{2}^{2}\right).$
The special PCs of this metric are the non gradient KVs
$Y^{1},~{}Y^{2},~{}Y^{4}~{}$and the gradient HV $H^{i}$ .
From Theorem 4.2.2 we find that the Lie point symmetries are the vectors

$\partial_{t},~{}t\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4},H^{i},~{}t^{2}\partial_{t}+tH^{i}$

which coincide with those found in [86]. From Theorem 4.3.2 we find that the
Noether point symmetries are

$\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4},~{}2t\partial_{t}+H^{i},t^{2}\partial_{t}+tH^{i}$

i.e. we find two more Noether symmetries than [86].

Case 2. $V\left(\phi\right)=0\,,$ $\dot{\phi}\neq 0$

In this case the Lagrangian is
$~{}L=e^{3\lambda}\left[6\dot{\lambda}^{2}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)-\dot{\phi}^{2}\right]~{}$and
the potential function $U(x^{i})=0.$ The auxiliary metric is (9.203). From
Theorem 4.2.2 we find that the Lie point symmetries are

$\partial_{t},~{}~{}t\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6},~{}H^{i},~{}t^{2}\partial_{t}+tH^{i}~{}$

and coincide with those found in [86]. Application of Theorem 4.3.2 gives that
the Noether point symmetries are

$\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6},~{}2t\partial_{t}+H^{i},t^{2}\partial_{t}+tH^{i}~{}$

i.e. two more from the ones found in [86].

Case 3. $V\left(\phi\right)=C_{0},~{}\dot{\phi}\neq 0$

The Lagrangian
is$~{}L=e^{3\lambda}\left[6\dot{\lambda}^{2}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)-\dot{\phi}^{2}+C_{0}\right]~{}$hence
the potential $U(x^{i})=-C_{0}e^{3\lambda}.~{}$Using Theorem 4.2.2 we find
that the Lie point symmetries are

$\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6},~{}H^{i},~{}\frac{1}{C}e^{\pm
Ct}\partial_{t}\pm e^{\pm Ct}H^{i}$

where $C=\frac{\sqrt{6C_{0}}}{2}$ , and coincide with those found in [86].
Application of Theorem 4.3.2 gives the Noether point symmetries

$\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6},~{}\frac{1}{C}e^{\pm
Ct}\partial_{t}\pm e^{\pm Ct}H^{i}.~{}$

Again we find two more Noether symmetries than [86].

Case 4. $V\left(\phi\right)=$arbitrary $\dot{\phi}\neq 0$

In this case the Lagrangian is
$L=e^{3\lambda}\left[6\dot{\lambda}^{2}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)-\dot{\phi}^{2}+V\left(\phi\right)\right]~{}$and
the potential $U(x^{i})=-e^{3\lambda}V\left(\phi\right)$. Application of
Theorem 4.2.2 gives the Lie point symmetries
$~{}\partial_{t}~{},~{}Y^{1}~{},~{}Y^{2}~{},~{}Y^{4}~{}~{},~{}H^{i}~{}\ $and
application of Theorem 4.3.2 gives the Noether symmetries
$\partial_{t}~{},~{}Y^{1}~{},~{}Y^{2}~{},~{}Y^{4}.$

Working in a similar manner we compute the Lie and the Noether point
symmetries of all Bianchi class A homogenous spacetimes. The results of the
calculations are collected in the following Tables.

Table 9.1: Lie and Noether Symmetries of Bianchi I scalar field Bianchi I | Noether Symmetries | Lie Symmetries
---|---|---
Case 1 | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4}$ | $\partial_{t},~{}t\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4},~{}H^{i}$
| $2t\partial_{t}+H^{i},~{}t^{2}\partial_{t}+tH^{i}$ | $t^{2}\partial_{t}+tH^{i}$
Case 2 | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6}$ | $\partial_{t},~{}t\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},Y^{5},~{}Y^{6}~{}$
| $~{}2t\partial_{t}+H^{i},~{}t^{2}\partial_{t}+tH^{i}$ | $H^{i},~{}t^{2}\partial_{t}+tH^{i}$
Case 3 | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y~{},~{}Y^{6}$ | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{3},~{}Y^{4},~{}Y^{5},~{}Y^{6},~{}H^{i}$
| $\frac{1}{C}e^{\pm Ct}\partial_{t}\pm e^{\pm Ct}H^{i}$ | $\frac{1}{C}e^{\pm Ct}\partial_{t}\pm e^{\pm Ct}H^{i}$
Case 4 | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4}$ | $\partial_{t},~{}Y^{1},~{}Y^{2},~{}Y^{4}~{},~{}H^{i}$
Table 9.2: Lie and Noether Symmetries of Bianchi II scalar field Bianchi II | Noether Symmetries | Lie Symmetries
---|---|---
Case 1 | $\partial_{t},~{}Y^{2},~{}6t\partial_{t}+3H^{i}-5Y^{1}$ | $\partial_{t},~{}Y^{2},~{}\frac{1}{3}t\partial_{t}+H^{i},~{}t\partial_{t}-Y^{1}~{}$
Case 2 | $\partial_{t},~{}Y^{2},~{}Y^{3},~{}Y^{6},~{}6t\partial_{t}+3H^{i}-5Y^{1}$ | $\partial_{t},~{}Y^{2},~{}~{}Y^{3},~{}Y^{6},~{}\frac{1}{3}t\partial_{t}+H^{i},~{}t\partial_{t}-Y^{1}$
Case 3 | $\partial_{t},~{}Y^{2},~{}Y^{3},~{}Y^{6}$ | $\partial_{t},~{}Y^{2},~{}Y^{3},~{}Y^{6},~{}3H^{i}+Y^{1}$
Case 4 | $\partial_{t},~{}Y^{2}$ | $\partial_{t},~{}Y^{2},~{}3H^{i}+Y^{1}$
Table 9.3: Lie and Noether Symmetries of Bianchi VI/VII scalar field Bianchi VI${}_{0}~{}/$ VII0 | Noether Symmetries | Lie Symmetries
---|---|---
Case 1 | $\partial_{t},~{}~{}6t\partial_{t}+3H^{i}-2Y^{1}-2\sqrt{3}Y^{2}$ | $\partial_{t},~{}H^{i}+\frac{1}{3}Y^{1}+\frac{\sqrt{3}}{3}Y^{2},~{}2t\partial_{t}-Y^{1}-\sqrt{3}Y^{2}$
Case 2 | $\partial_{t},~{}Y^{3},~{}6t\partial_{t}+3H^{i}-2Y^{1}-2\sqrt{3}Y^{2}$ | $\partial_{t},~{}Y^{3},~{}H^{i}+\frac{1}{3}Y^{1}+\frac{\sqrt{3}}{3}Y^{2},~{}2t\partial_{t}-Y^{1}-\sqrt{3}Y^{2}$
Case 3 | $\partial_{t},~{}Y^{3}$ | $\partial_{t},~{}Y^{3},~{}H^{i}+\frac{1}{3}Y^{1}+\frac{\sqrt{3}}{3}Y^{2}$
Case 4 | $\partial_{t}~{}$ | $\partial_{t},~{}H^{i}+\frac{1}{3}Y^{1}+\frac{\sqrt{3}}{3}Y^{2}$
Table 9.4: Lie and Noether Symmetries of Bianchi VIII/IX scalar field Bianchi VIII | Noether Symmetries | Lie Symmetries
---|---|---
Case 1 | $\partial_{t}$ | $\partial_{t},~{}\frac{2}{3}t\partial_{t}~{}+H^{i}$
Case 2 | $\partial_{t},~{}Y^{3}$ | $\partial_{t},~{}Y^{3},~{}\frac{2}{3}t\partial_{t}~{}+H^{i}~{}$
Case 3 | $\partial_{t},~{}Y^{3}$ | $\partial_{t},~{}Y^{3}$
Case 4 | $\partial_{t}$ | $\partial_{t}$
|  |
Bianchi IX | Noether Symmetries | Lie Symmetries
Case 1 | $\partial_{t}$ | $\partial_{t}$
Case 2 | $\partial_{t},~{}Y^{3}$ | $\partial_{t},~{}Y^{3}$
Case 3 | $\partial_{t},~{}Y^{3}$ | $\partial_{t},~{}Y^{3}$
Case 4 | $\partial_{t}$ | $\partial_{t}$

From the above tables we infer that the Lie point symmetries we found coincide
with those of [86]. Some differences which appear are due to linear
combinations of symmetries from the other set. The same does not apply to the
Noether symmetries, for which we found a larger number than in [86].

We note that in Case 1 the Noether symmetry $2t+H$ is a combination of two Lie
point symmetries, which is peculiar since the Noether point symmetries are
considered to be a direct subset of Lie symmetries. This is explained as
follows. The addition of a Killing vector to a homothetic vector retains a
homothetic vector. Therefore a Lie symmetry due to a Killing vector and one
due to a homothetic vector is possible to give a Lie point symmetry due to a
homothetic vector. Concerning [143] from the examination of the Tables 9.1-9.4
and the results they present it can be seen that they loose the Noether
symmetries which have a component along $\partial_{t}~{}$direction.

###### The Bianchi I model with scalar field and exponential potential

We consider a scalar field described by an exponential potential
$V\left(\phi\right)=e^{-d\phi}$ in a Bianchi class A spacetime. For this
potential all the models admit the extra Lie symmetry
$t\partial_{t}+\frac{2}{d}Y^{3}.$ Concerning the Noether symmetries we have an
extra Noether symmetry only for the types I. II, VI${}_{0},$ VII 0 as follows:

Type I

$t\partial_{t}+\frac{1}{2}H^{i}+\frac{2}{d}Y^{3}$

Type II

$t\partial_{t}+\frac{1}{2}H^{i}-\frac{5}{6}Y^{1}+\frac{2}{d}Y^{3}$

Type VI${}_{0}~{}/$ VII0

$6t\partial_{t}+3H^{i}-2Y^{1}-2\sqrt{3}Y^{2}+\frac{6}{d}Y^{3}.$

In the following we concentrate on the Bianchi I model and make use of the
extra Noether integral to define a transformation [162] which allows the
determination of the analytic form of the metric.

The Lagrangian describing a scalar field with exponential potential in an
empty Bianchi I spacetime is

$L=e^{3\lambda}\left[6\dot{\lambda}^{2}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)-\dot{\phi}^{2}+V_{0}e^{-d\phi}\right]$
(9.205)

and the corresponding Hamiltonian vanishes. The metric defined by this
Lagrangian is (9.203).

Using the transformation $\lambda=\frac{1}{3}\ln\left(\frac{a^{3}}{2}\right)$
the Lagrangian becomes

$L=3a\dot{a}^{2}-\frac{1}{2}a^{3}\dot{\phi}-\frac{3}{4}a^{3}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)+\frac{1}{2}V_{0}a^{3}e^{-d\phi}.$

We change variables by means of the transformation

$u=\frac{\sqrt{6}}{4}\phi+\frac{1}{2}\ln\left(a^{3}\right)~{},~{}v=-\frac{\sqrt{6}}{4}\phi+\frac{1}{2}\ln\left(a^{3}\right)~{}$

and the Lagrangian takes the form

$L\left(u,v,\dot{u},\dot{v}\right)=e^{\left(u+v\right)}\left(\frac{8}{3}\dot{u}\dot{v}-\frac{3}{2}\left(\dot{\beta}_{1}^{2}+\dot{\beta}_{2}^{2}\right)+V_{0}e^{-2K\left(u-v\right)}\right).$

Next we change the time coordinate as follows

$\frac{d\tau}{dt}=\sqrt{\frac{3V_{0}}{8}}e^{-K\left(u-v\right)}.$

We make one more change
$\beta_{1}=\sqrt{\frac{9}{16}}B_{1}~{},~{}\beta_{2}=\sqrt{\frac{9}{16}}B_{2}~{}$and
in the coordinates $\tau,u,v,B_{1},B_{2}$ the Lagrangian is:

$L\left(\tau,u,v,B_{1},B_{2}\right)=e^{\left(u+v\right)}e^{-K\left(u-v\right)}\left(u^{\prime}v^{\prime}-\left(B_{1}^{\prime
2}+B_{2}^{\prime 2}\right)+1\right)$ (9.206)

where $u^{\prime}=\frac{du}{d\tau},$
$v^{\prime}=\frac{dv}{d\tau},~{}B_{1,2}^{\prime}=\frac{dB_{1,2}}{d\tau}.$ The
equations of motion are

$\displaystyle u^{\prime\prime}+\left(1-K\right)u^{\prime
2}+\left(1+K\right)B_{1}^{{}^{\prime}2}+\left(1+K\right)B_{2}^{{}^{\prime}2}-\left(1+K\right)$
$\displaystyle=0$ (9.207) $\displaystyle
v^{\prime\prime}+\left(1+K\right)v^{\prime
2}+\left(1-K\right)B_{1}^{{}^{\prime}2}+\left(1-K\right)B_{2}^{{}^{\prime}2}-\left(1-K\right)$
$\displaystyle=0$ (9.208) $\displaystyle
B_{1}^{{}^{\prime\prime}}+\left(1-K\right)B_{1}^{\prime}u^{\prime}+\left(1+K\right)B_{1}^{{}^{\prime}}v^{\prime}$
$\displaystyle=0$ (9.209) $\displaystyle
B_{2}^{{}^{\prime\prime}}+\left(1-K\right)B_{2}^{{}^{\prime}}u^{\prime}+\left(1+K\right)B_{2}^{\prime}v^{\prime}$
$\displaystyle=0$ (9.210)

with constrain (the zero Hamiltonian)
$u^{\prime}v^{\prime}-\left(B_{1}^{\prime 2}+B_{2}^{\prime 2}\right)-1=0.~{}$
These expressions are symmetric in $B_{1},B_{2}$ therefore we set
$B_{1}=B_{2.}=\frac{1}{2}B$ and the system of equations of motion becomes

$\displaystyle u^{\prime\prime}+\left(1-K\right)u^{\prime
2}+\left(1+K\right)B^{{}^{\prime}2}-\left(1+K\right)$ $\displaystyle=0$
(9.211) $\displaystyle v^{\prime\prime}+\left(1+K\right)v^{\prime
2}+\left(1-K\right)B^{{}^{\prime}2}-\left(1-K\right)$ $\displaystyle=0$
(9.212) $\displaystyle
B^{{}^{\prime\prime}}+\left(1-K\right)B^{\prime}u^{\prime}+\left(1+K\right)B^{{}^{\prime}}v^{\prime}$
$\displaystyle=0$ (9.213)

with constraint

$u^{\prime}v^{\prime}-B^{\prime 2}-1=0.$ (9.214)

We consider two cases $K=1$ and $K\neq 1$.

For $K=1$ the metric is $ds^{2}=2e^{2v}\left(dudv-dB^{2}\right).~{}$The
potential is the gradient KV $V\left(u,v\right)=-e^{2v}$ and the solution of
the system is

$\displaystyle u\left(\tau\right)$
$\displaystyle=\tau^{2}+2c_{1}^{2}\ln\left(\tau+c_{2}\right)+2tc_{2}$ (9.215)
$\displaystyle v\left(\tau\right)$
$\displaystyle=\frac{1}{2}\ln\left(2\tau+c_{2}\right)$ (9.216) $\displaystyle
B\left(\tau\right)$ $\displaystyle=c_{1}\ln\left(\tau+c_{2}\right).$ (9.217)

For $K\neq 1$ we make use of the extra Noether integral
$e^{\left(u+v\right)}e^{-K\left(u-v\right)}\dot{B}=C$ and solve the system. We
find that $K=\frac{1-C^{2}}{1+C^{2}}$ and that the system has two solutions.
The first is:

$\displaystyle u\left(\tau\right)$
$\displaystyle=\frac{1}{2\left(1-K\right)}\ln\left(\frac{K-1}{1+K}\sin\left(2\sqrt{K^{2}-1}\tau\right)\right)$
(9.218) $\displaystyle v\left(\tau\right)$
$\displaystyle=\frac{1}{2\left(1+K\right)}\ln\left(\sin\left(2\sqrt{K^{2}-1}\tau\right)\right)$
(9.219) $\displaystyle B\left(\tau\right)$
$\displaystyle=\frac{i}{2\sqrt{K^{2}-1}}\text{{\normalsize
arctanh}}\left(\cos\left(2\sqrt{K^{2}-1}\tau\right)\right)$ (9.220)

and the second:

$\displaystyle u\left(\tau\right)$
$\displaystyle=\frac{1}{2\left(1-K\right)}\ln\left(\frac{K-1}{1+K}\cos\left(2\sqrt{K^{2}-1}\tau\right)\right)$
(9.221) $\displaystyle v\left(\tau\right)$
$\displaystyle=\frac{1}{2\left(1+K\right)}\ln\left(\cos\left(2\sqrt{K^{2}-1}\tau\right)\right)$
(9.222) $\displaystyle B\left(\tau\right)$
$\displaystyle=\frac{i}{2\sqrt{K^{2}-1}}\text{{\normalsize
arctanh}}\left(\frac{1}{\sin\left(2\sqrt{K^{2}-1}\tau\right)}\right).$ (9.223)

These solutions complement the results of [163]

####  9.8 Conclusion

In this chapter we have studied conformally related metrics and Lagrangians in
the context of scalar–tensor cosmology. We have found that to every non-
minimally coupled scalar field we can associate a unique minimally coupled
scalar field in a conformally related space with an appropriate potential. The
existence of such a connection can be used in order to study the dynamical
properties of the various cosmological models, since the field equations of a
non-minimally coupled scalar field are the same, at the conformal level, of
the field equations of the minimally coupled scalar field. The above
propositions can be extended to general Riemannian spaces in n-dimensions.
Furthermore, we have identified the Noether point symmetries and the analytic
solutions of the equations of motion in the context of a minimally coupled and
a non minimally coupled scalar field in a FRW spacetime and we have classified
the Noether symmetries of the field equations in Bianchi class A models with a
minimally coupled scalar field. We found that there is a rather large class of
hyperbolic and exponential potentials which admit extra (beyond the
$\partial_{t}$) Noether pont symmetries which lead to integrals of motions.

In general, the Noether point symmetries play an important role in physics
because they can be used to simplify a given system of differential equations
as well as to determine the integrability of the system. The latter will
provide the necessary platform in order to solve the equations of motion
analytically and thus to obtain the evolution of the physical quantities. In
cosmology, such a method is extremely relevant in order to compare
cosmographic parameters, such as scale factor, Hubble expansion rate,
deceleration parameter, density parameters with observational constrains.

However, since the Noether point symmetries are generated from the kinetic
metric of the Lagrangian, they are not only a criterion for the integrability
of the system and a method to determine analytical solutions of the field
equations, but they are also a geometric criterion since by demanding the
existence of Noether symmetries we let the geometry to select the dynamics,
i.e. the dark energy model.

In the following chapters we study the Noether symmetries of the
$f\left(R\right)$ and the $f\left(T\right)$ theories of gravity.

###  Appendix 9.A Relating the ranges of the constants $F_{0}$ and
$\left|k\right|$

We consider the following ranges for the constants $F_{0}~{}$ and
$~{}\left|k\right|$

a. $F_{0}>0.~{}$In this case we have
$\left|k\right|=\frac{1}{3}\sqrt{\frac{F_{0}}{F_{0}+1}}$from which follows

$\left|k\right|^{2}<1~{},~{}F_{0}>0$

b. $-1<F_{0}<0.$ In this case we
have$\left|k\right|=\frac{1}{3}\sqrt{\frac{\left|F_{0}\right|}{F_{0}+1}}$from
which follows.

$\left|k\right|^{2}=1~{},~{}F_{0}=-\frac{9}{10}~{};~{}\left|k\right|^{2}<1,~{}F_{0}>-\frac{9}{10};~{}\left|k\right|^{2}>1~{},~{}F_{0}\in\left(-1,-\frac{9}{10}\right)$

c. $F_{0}<1$ . Then
$\left|k\right|=\frac{1}{3}\sqrt{\frac{\left|F_{0}\right|}{\left|F_{0}\right|-1}}$from
which follows

$\left|k\right|^{2}=1~{},~{}F_{0}=-\frac{9}{8}~{};~{}\left|k\right|^{2}<1~{},~{}F_{0}<-\frac{9}{8};\left|k\right|^{2}>1~{},~{}F_{0}\in\left(-\frac{9}{8},-1\right).$

The ranges of $\left|k\right|$ are needed because they select different groups
of Killing vectors of the metric (9.42).

###  Appendix 9.B Computation of the gradient functions
$S_{1}\left(r,\theta\right),$ $S_{2}\left(r,\theta\right)$

The functions $S_{1}\left(r,\theta\right),$ $S_{2}\left(r,\theta\right)$ are
the canonical coordinates $x,y$ for the KVs $K^{1},K^{2}.$ The canonical
coordinates are defined with the requirement $K^{1}=\frac{\partial}{\partial
x},K^{2}=\frac{\partial}{\partial y}$ and are computed as follows. We have the
system of differential equations:

$\displaystyle\frac{\partial}{\partial y}$ $\displaystyle=$
$\displaystyle\frac{e^{\left(1-k\right)\theta}r^{k}}{N_{0}^{2}}\left(-\partial_{r}+\frac{1}{r}\partial_{\theta}\right)$
$\displaystyle\frac{\partial}{\partial x}$ $\displaystyle=$
$\displaystyle\frac{e^{-\left(1+k\right)\theta}r^{-k}}{N_{0}^{2}}\left(\partial_{r}+\frac{1}{r}\partial_{\theta}\right).$

To solve it we consider the associated Lagrange system and write:

$\frac{dy}{1}=-N_{0}^{2}\frac{dr}{r^{k}e^{\left(1-k\right)\theta}}=N_{0}^{2}\frac{d\theta}{r^{k-1}e^{\left(1-k\right)\theta}}$

The first equation gives:

$y=-N_{0}^{2}e^{\left(-1+k\right)\theta}\int\frac{dr}{r^{k}}=-N_{0}^{2}e^{\left(-1+k\right)\theta}\frac{r^{1-k}}{1-k}+\Phi(\theta)$

The second equation gives:

$y=N_{0}^{2}r^{-k+1}\int\frac{d\theta}{e^{\left(1-k\right)\theta}}=N_{0}^{2}\frac{r^{1-k}}{-1+k}e^{\left(1-k\right)\theta}+\Phi_{1}(r)$

hence we have (this is the $S_{2}\left(r,\theta\right)$):

$y=N_{0}^{2}\frac{r^{1-k}}{k-1}e^{-\left(1-k\right)\theta}.$

For the other coordinate we have:

$\frac{dx}{1}=N_{0}^{2}\frac{dr}{r^{-k}e^{-\left(1+k\right)\theta}}=N_{0}^{2}\frac{d\theta}{r^{-k-1}e^{-\left(1+k\right)\theta}}$

The first equation gives:

$x=N_{0}^{2}e^{\left(1+k\right)\theta}\int\frac{dr}{r^{-k}}=N_{0}^{2}e^{\left(1+k\right)\theta}\frac{1}{1+k}r^{1+k}+\Phi(\theta)$

and the second equation gives:

$x=N_{0}^{2}r^{1+k}\int\frac{d\theta}{e^{-\left(1+k\right)\theta}}=\frac{1}{1+k}N_{0}^{2}r^{1+k}e^{\left(1+k\right)\theta}+\Phi_{1}(r)$

hence (this is the $S_{1}\left(r,\theta\right)$):

$x=\frac{1}{1+k}N_{0}^{2}r^{1+k}e^{\left(1+k\right)\theta}.$

Therefore, we have the canonical coordinates

$x=\frac{r^{1+k}e^{-\left(1+k\right)\theta}}{k+1}~{},~{}y=\frac{r^{1-k}e^{\left(-1+k\right)\theta}}{k-1}$

###  Chapter 10 Using Noether point symmetries to specify $f(R)$ gravity

####  10.1 Introduction

In chapter 9 we used the Noether point symmetries of the scalar tensor
theories in order to constrain the dark energy models. Except the sclarar
field cosmology there are other possibilities to explain the present
accelerating stage. For instance, one may consider that the dynamical effects
attributed to dark energy can be resembled by the effects of a nonstandard
gravity theory. In other words, the present accelerating stage of the universe
can be driven only by cold dark matter, under a modification of the nature of
gravity. Such a reduction of the so-called dark sector is naturally obtained
in the $f(R)$ gravity theories [87]. In the original nonstandard gravity
models, one modifies the Einstein-Hilbert action with a general function
$f(R)$ of the Ricci scalar $R$. The $f(R)$ approach is a relatively simple but
still a fundamental tool used to explain the accelerated expansion of the
universe. A pioneering fundamental approach was proposed long ago with
$f(R)=R+mR^{2}$ [164]. Later on, the $f(R)$ models were further explored from
different points of view in [89, 88, 165] and indeed a large number of
functional forms of $f(R)$ gravity is currently available in the literature
[140, 166, 167, 168, 169].

In the following, we will use the Lie and the Noether point symmetries in
order to specify the $f\left(R\right)$ gravity in a FRW spacetime and use the
first integrals of these models to determine analytic solutions of their field
equations.

The structure of this chapter is as follows. The basic theoretical elements of
the problem are presented in section 10.2, where we also introduce the basic
FRW cosmological equations in the framework of $f(R)$ models. The Noether
point symmetries and their relevance to the $f(R)$ models are discussed in
section 10.4. In section 10.5 we provide analytical solutions for those $f(R)$
models which are Liouville integrable via Noether point symmetries. In section
10.6 we study the Noether point symmetries in spatially non-flat $f(R)$
cosmological models. Finally, we draw our main conclusions in section 10.7.

####  10.2 Cosmology with a modified gravity

Consider the modified Einstein-Hilbert action:

$S=\int
d^{4}x\sqrt{-g}\left[\frac{1}{2k^{2}}f\left(R\right)+\mathcal{L}_{m}\right]$
(10.1)

where $\mathcal{L}_{m}$ is the Lagrangian of dust-like ($p_{m}=0$) matter and
$k^{2}=8\pi G$. Varying the action with respect to the metric111We use the
metric i.e. the Hilbert variational approach. we arrive at

$(1+f^{\prime})G_{\nu}^{\mu}\,-\,g^{\mu\alpha}f_{{}_{R},\,\alpha\,;\,\nu}+\left[\frac{2\Box
f^{\prime}-(f-Rf^{\prime})}{2}\right]\delta_{\;\nu}^{\mu}=k^{2}\,T_{\nu}^{\mu}$
(10.2)

where the prime denotes derivative with respect to $R$, $G_{\nu}^{\mu}$ is the
Einstein tensor and $T_{\nu}^{\mu}$ is the ordinary energy-momentum tensor of
matter. Based on the matter era we treat the expanding universe as a dust
fluid which includes only cold dark matter with comoving observers
$U^{\mu}=\delta_{0}^{\mu}$. Thus the energy momentum tensor becomes
$T_{\mu\nu}=\rho_{m}U_{\mu}U_{\nu}$, where $\rho_{m}$ is the energy density of
the cosmic fluid.

Now, in the context of a flat FRW model the metric is

$ds^{2}=-dt^{2}+a^{2}(t)(dx^{2}+dy^{2}+dz^{2}).$ (10.3)

The components of the Einstein tensor are computed to be:

$G_{0}^{0}=-3H^{2},\;G_{b}^{a}=-\delta_{b}^{a}\left(2\dot{H}+3H^{2}\right)\;.$
(10.4)

Inserting (10.4) into the modified Einstein’s field equations (10.2), for
comoving observers, we derive the modified Friedman’s equation

$3f^{{}^{\prime}}H^{2}=k^{2}\rho_{m}+\frac{f^{{}^{\prime}}R-f}{2}-3Hf^{{}^{\prime\prime}}\dot{R}$
(10.5)

$2f^{{}^{\prime}}\dot{H}+3f^{{}^{\prime}}H^{2}=-2Hf^{{}^{\prime\prime}}\dot{R}-\left(f^{{}^{\prime\prime\prime}}\dot{R}^{2}+f^{{}^{\prime\prime}}\ddot{R}\right)-\frac{f-Rf^{{}^{\prime}}}{2}.$
(10.6)

The contraction of the Ricci tensor provides the Ricci scalar

$R=g^{\mu\nu}R_{\mu\nu}=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right)=6(2H^{2}+\dot{H})\;.$
(10.7)

The Bianchi identity $\bigtriangledown^{\mu}\,{T}_{\mu\nu}=0$ leads to the
matter conservation law:

$\dot{\rho}_{m}+3H\rho_{m}=0\,$ (10.8)

whose solution is

$\rho_{m}=\rho_{m0}a^{-3}.$ (10.9)

Note that the over-dot denotes derivative with respect to the cosmic time $t$
and $H\equiv\dot{a}/a$ is the Hubble parameter.

If we consider $f(R)=R$ then the field equations (10.2) boil down to the
Einstein’s equations. On the other hand, the concordance $\Lambda$ cosmology
is fully recovered for $f(R)=R-2\Lambda$.

From the current analysis it becomes clear that unlike the standard Friedman
equations in Einstein’s GR, the modified equations of motion (10.5) and (10.6)
are complicated and thus it is difficult to solve them analytically.

We would like to stress here that within the context of the metric formalism
the above $f(R)$ cosmological models must obey simultaneously some strong
conditions [136]. These are: (i) $f^{{}^{\prime}}>0$ for $R\geq R_{0}>0$,
where $R_{0}$ is the Ricci scalar at the present time. If the final attractor
is a de Sitter point we need to have $f^{{}^{\prime}}>0$ for $R\geq R_{1}>0$,
where $R_{1}$ is the Ricci scalar at the de Sitter point, (ii)
$f^{{}^{\prime\prime}}>0$ for $R\geq R_{0}>0$, (iii) $f(R)\approx R-2\Lambda$
for $R\gg R_{0}$ and finally (iv)
$0<\frac{Rf^{{}^{\prime\prime}}}{f^{{}^{\prime}}}(r)<1$ at
$r=-\frac{Rf^{{}^{\prime}}}{f}=-2$

####  10.3 Modified gravity versus symmetries

In the last decade a large number of experiments have been proposed in order
to constrain dark energy and study its evolution. Naturally, in order to
establish the evolution of the dark energy (DE) (”geometrical” in the current
work) equation of state parameter a realistic form of $H(a)$ is required while
the included free parameters must be constrained through a combination of
independent DE probes (for example SNIa, BAOs, CMB etc). However, a weak point
here is the fact that the majority of the $f(R)$ models appeared in the
literature are plagued with no clear physical basis and/or many free
parameters. Due to the large number of free parameters many such models could
fit the data. The proposed additional criterion of Noether point symmetry
requirement is a physically meaning-full geometric ansatz.

According to the theory of general relativity, the space-time Killing and
homothetic symmetries via the Einstein’s field equations, are also symmetries
of the energy momentum tensor. Due to the fact that the $f(R)$ models provide
a natural generalization of GR one would expect that the theories of modified
gravity must inherit the symmetries of the space-time as the usual gravity
(GR) does.

Furthermore, besides the geometric symmetries we have to consider the
dynamical symmetries, which are the symmetries of the field equations (Lie
symmetries). If the field equations are derived from a Lagrangian then there
is the special class of Lie symmetries, the Noether symmetries, which lead to
conserved currents or, equivalently, to first integrals of the equations of
motion. The Noether integrals are used to reduce the order of the field
equations or even to solve them. Therefore a sound requirement, which is
possible to be made in Lagrangian theories, is that they admit extra Noether
symmetries. This assumption is model independent, because it is imposed after
the field equations have been derived, therefore it does not lead to conflict
with the geometric symmetries while, at the same time, serves the original
purpose of a selection rule. Of course, it is possible that a different method
could be assumed and select another subset of viable models. However, symmetry
has always played a dominant role in Physics and this gives an aesthetic and a
physical priority to our proposal.

In the Lagrangian context, the main field equations (10.5) and (10.6),
described in section 10.2, can be produced by the following Lagrangian:

$L\left(a,\dot{a},R,\dot{R}\right)=6af^{{}^{\prime}}~{}\dot{a}^{2}+6a^{2}f^{{}^{\prime\prime}}~{}\dot{a}\dot{R}+a^{3}\left(f^{{}^{\prime}}R-f\right)\qquad$
(10.10)

in the space of the variables $\\{a,R\\}$. Using eq.(10.10) we obtain the
Hamiltonian of the current dynamical system

$E=6af^{{}^{\prime}}~{}\dot{a}^{2}+6a^{2}f^{{}^{\prime\prime}}~{}\dot{a}\dot{R}-a^{3}\left(f^{{}^{\prime}}R-f\right)$
(10.11)

or

$E=6a^{3}\left[f^{{}^{\prime}}H^{2}-\frac{1}{6f^{{}^{\prime}}}\left(\left(f^{{}^{\prime}}R-f\right)-6\dot{R}Hf^{{}^{\prime\prime}}\right)\right]\;.$
(10.12)

Combining the first equation of motion (10.5) with eq.(10.12) we find

$\rho_{m}=\frac{E}{2k^{2}}\;a^{-3}\;.$ (10.13)

The latter equation together with $\rho_{m}=\rho_{m0}a^{-3}$ implies that

$\rho_{m0}=\frac{E}{2k^{2}}\Rightarrow\Omega_{m}\rho_{cr,0}=\frac{E}{2k^{2}}\Rightarrow
E=6\Omega_{m}H_{0}^{2}$ (10.14)

where $\Omega_{m}=\rho_{m0}/\rho_{cr,0}$, $\rho_{cr,0}=3H_{0}^{2}/k^{2}$ is
the critical density at the present time and $H_{0}$ is the Hubble constant.

We note that the current Lagrangian eq.(10.10) is time independent implying
that the dynamical system is autonomous hence the Hamiltonian $E$ is
conserved.

####  10.4 Noether point symmetries of $f\left(R\right)$ gravity

The Noether condition for the Lagrangian (10.10) is equivalent with the
following system of eight equations

$\displaystyle\xi_{,a}$ $\displaystyle=$ $\displaystyle 0~{}$ (10.15)
$\displaystyle~{}\xi_{,R}$ $\displaystyle=$ $\displaystyle 0$ (10.16)
$a^{2}f^{\prime\prime}\eta_{,R}^{\left(1\right)}=0$ (10.17)
$f^{\prime}\eta^{\left(1\right)}+af^{\prime\prime}\eta^{\left(2\right)}+2af^{\prime}\eta_{,a}^{\left(1\right)}+a^{2}f^{\prime\prime}\eta_{,a}^{\left(2\right)}-\frac{1}{2}af^{\prime}\xi_{,t}=0$
(10.18)
$2af^{\prime\prime}\eta^{\left(1\right)}+a^{2}f^{\prime\prime\prime}\eta^{\left(2\right)}+a^{2}f^{\prime\prime}\eta_{,a}^{\left(1\right)}+2af^{\prime}\eta_{,R}^{\left(1\right)}+a^{2}f^{\prime\prime}\eta_{,R}^{\left(2\right)}-\frac{1}{2}a^{2}f^{\prime\prime}\xi_{,t}=0$
(10.19)
$-3a^{2}Rf^{\prime}\eta^{\left(1\right)}+3a^{2}f\eta^{\left(1\right)}-a^{3}Rf^{\prime\prime}\eta^{\left(2\right)}+a^{3}\left(f-f^{{}^{\prime}}R\right)\xi_{,t}+g_{,t}=0$
(10.20)

$12af^{\prime}\eta_{,t}^{\left(1\right)}+6a^{2}f^{\prime\prime}\eta_{,t}^{\left(2\right)}+a^{3}\left(f^{{}^{\prime}}R-f\right)\xi_{,a}-g_{,a}=0$
(10.21)
$6a^{2}f^{\prime\prime}\eta_{,t}^{\left(1\right)}+a^{3}\left(f^{{}^{\prime}}R-f\right)\xi_{,R}-g_{,R}=0$
(10.22)

The solution of the system (10.15)-(10.22) determines the Noether symmetries.

Since the Lagrangian (10.10) is in the form
$L=T\left(a,\dot{a},R,\dot{R}\right)-V\left(a,R\right)$, the results of
chapter 4 can be used 222Where $T$ is the ”kinetic” term and $V$ is the
”potential. The kinematic term defines a two dimensional metric in the space
of $\\{a,R\\}$ with line element

$d\hat{s}^{2}=12af^{\prime}da^{2}+12a^{2}f^{\prime\prime}da~{}dR$ (10.23)

while the ”potential” is

$V(a,R)=-a^{3}(f^{{}^{\prime}}R-f)\;.$ (10.24)

The Ricci scalar of the two dimensional metric (10.23) is computed to be
$\hat{R}=0,$ therefore the space is a flat space333All two dimensional
Riemannian spaces are Einstein spaces implying that if $\hat{R}=const$ the
space is maximally symmetric [23] and if $\hat{R}=0,$ the space admit gradient
homothetic vector, i.e. is flat. with a maximum homothetic algebra. The
homothetic algebra of the metric (10.23) consists of the vectors

$\displaystyle\mathbf{K}^{1}$
$\displaystyle=a\partial_{a}-3\frac{f^{\prime}}{f^{\prime\prime}}\partial_{R}~{},~{}\mathbf{K}^{2}=\frac{1}{a}\partial_{a}-\frac{1}{a^{2}}\frac{f^{\prime}}{f^{\prime\prime}}\partial_{R}~{}$
$\displaystyle\mathbf{K}^{3}$
$\displaystyle=\frac{1}{a}\frac{1}{f^{\prime\prime}}\partial_{R}~{},~{}\mathbf{H}=\frac{a}{2}~{}\partial_{a}+\frac{1}{2}\frac{f^{\prime}}{f^{\prime\prime}}\partial_{R}$

where $\mathbf{K}$ are Killing vectors ($\mathbf{K}^{2,3}$ are gradient) and
$\mathbf{H}$ is a gradient Homothetic vector.

Therefore applying theorem 4.3.2 we have the following cases:

Case 1: If $f\left(R\right)$ is arbitrary the dynamical system admits as
Noether symmetry the $X^{1}=\partial_{t}~{}$with Noether integral the
Hamiltonian $E$.

Case 2: If $f\left(R\right)=R^{\frac{3}{2}}~{}$the dynamical system admits the
extra Noether symmetries

$X^{2}=\mathbf{K}^{2},~{}X^{3}=t\mathbf{K}^{2}$ (10.25)
$X^{4}=2t\partial_{t}+\mathbf{H+}\frac{5}{6}\mathbf{K}^{1}.$ (10.26)

with corresponding Noether Integrals

$I_{2}=\frac{d}{dt}\left(a\sqrt{R}\right)$ (10.27)
$I_{3}=t\frac{d}{dt}\left(a\sqrt{R}\right)-a\sqrt{R}$ (10.28)
$I_{4}=2tE-6a^{2}\dot{a}\sqrt{R}-6\frac{a^{3}}{\sqrt{R}}\dot{R}.$ (10.29)

the non vanishing commutators of the Noether algebra being

$\left[X^{1},X^{3}\right]=X^{2}~{}~{}~{}~{}\left[X^{1},X^{4}\right]=2X^{1}$
$\left[X^{2},X^{4}\right]=\frac{8}{3}X^{2}~{}~{}~{}~{}\left[X^{3},X^{4}\right]=\frac{2}{3}X^{3}$

Case 3: If $f\left(R\right)=R^{\frac{7}{8}}$ the dynamical system admits the
extra Noether symmetries

$X^{5}=2t\partial_{t}+\mathbf{H}~{},~{}X^{6}=t^{2}\partial_{t}+t\mathbf{H}$
(10.30)

with corresponding Noether Integrals

$I_{5}=2tE-\frac{21}{8}\frac{d}{dt}\left(a^{3}R^{-\frac{1}{8}}\right)$ (10.31)
$I_{6}=t^{2}E-\frac{21}{8}t\frac{d}{dt}\left(a^{3}R^{-\frac{1}{8}}\right)+\frac{21}{8}a^{3}R^{-\frac{1}{8}}.$
(10.32)

with non vanishing commutators

$\left[X^{1},X^{5}\right]=2X^{1}~{}~{}~{}~{}\left[X^{1},X^{6}\right]=X^{5}~{}~{}~{}~{}\left[X^{5},X^{6}\right]=2X^{6}$

From the time dependent integrals (10.31),(10.32) and the Hamiltonian we
construct the Ermakov-Lewis invariant (see chapter 5)

$\Sigma=4I_{6}E-I_{5}^{2}$ (10.33)

Case 4: If $f\left(R\right)=\left(R-2\Lambda\right)^{\frac{3}{2}}$ the
dynamical system admits the extra Noether symmetries

$\bar{X}^{2}=e^{\sqrt{m}t}\mathbf{K}^{2}~{},~{}\bar{X}^{3}=e^{-\sqrt{m}t}\mathbf{K}^{2}$
(10.34)

with corresponding Noether Integrals

$\bar{I}_{2}=e^{\sqrt{m}t}\left(\frac{d}{dt}\left(a\sqrt{R-2L}\right)-9\sqrt{m}a\sqrt{R-2\Lambda}\right)$
(10.35)
$\bar{I}_{3}=e^{-\sqrt{m}t}\left(\frac{d}{dt}\left(a\sqrt{R-2L}\right)+9\sqrt{m}a\sqrt{R-2\Lambda}\right)$
(10.36)

where $m=\frac{2}{3}\Lambda.$ The non vanishing commutators of the Noether
algebra are

$\left[X^{1},\bar{X}^{2}\right]=\sqrt{m}\bar{X}^{2}~{}~{}~{}~{}\left[\bar{X}^{3},X^{1}\right]=\sqrt{m}\bar{X}^{3}~{}$

From the time dependent integrals (10.35),(10.36) we construct the time
independent integral $\bar{I}_{23}=\bar{I}_{2}\bar{I}_{3}.$

Case 5: If $f\left(R\right)=\left(R-2\Lambda\right)^{\frac{7}{8}}$ the
dynamical system admits the extra Noether symmetries

$\displaystyle\bar{X}^{5}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{m}}e^{2\sqrt{m}t}\partial_{t}+e^{2\sqrt{m}t}~{}\mathbf{H}$
(10.37) $\displaystyle\bar{X}^{6}$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{m}}e^{-2\sqrt{m}t}\partial_{t}+e^{-2\sqrt{m}t}~{}\mathbf{H}$
(10.38)

with corresponding Noether Integrals

$\bar{I}_{5}=e^{2\sqrt{m}t}\left[\frac{1}{\sqrt{m}}E-\frac{21}{8}\frac{d}{dt}\left(a^{3}\left(R-2\Lambda\right)^{-\frac{1}{8}}\right)+\frac{21}{4}\sqrt{m}a^{3}\left(R-2\Lambda\right)^{-\frac{1}{8}}\right]$
(10.39)
$\bar{I}_{6}=e^{-2\sqrt{m}t}\left[\frac{1}{\sqrt{m}}E+\frac{21}{8}\frac{d}{dt}\left(a^{3}\left(R-2\Lambda\right)^{-\frac{1}{8}}\right)+\frac{21}{4}\sqrt{m}a^{3}\left(R-2\Lambda\right)^{-\frac{1}{8}}\right]$
(10.40)

and the non vanishing commutators of the Noether algebra are

$\left[X^{1},\bar{X}^{5}\right]=2\sqrt{m}\bar{X}^{5}~{}~{}~{}~{}\left[\bar{X}^{6},X^{1}\right]=2\sqrt{m}\bar{X}^{6}$
$\left[\bar{X}^{5},\bar{X}^{6}\right]=\frac{4}{\sqrt{m}}X^{1}$

From the time dependent integrals (10.31),(10.32) and the Hamiltonian we
construct the Ermakov-Lewis invariant.

$\phi=E^{2}-\bar{I}_{5}\bar{I}_{6}$ (10.41)

Case 6: If $f\left(R\right)=R^{n}$ (with $n\neq 0,1,\frac{3}{2},\frac{7}{8}$)
the dynamical system admits the extra Noether symmetry

$X^{7}=2t\partial_{t}+\mathbf{H+}\left(\frac{4n}{3}-\frac{7}{6}\right)\mathbf{K}^{1}$
(10.42)

with corresponding Noether Integral

$I_{7}=2tE-8na^{2}R^{n-1}\dot{a}\left(2-n\right)-4na^{3}R^{n-2}\dot{R}\left(2n-1\right)\left(n-1\right).$
(10.43)

and the commutator of the Noether algebra is
$\left[X^{1},X^{7}\right]=2X^{1}.$

We note that the Noether subalgebra of case 2,
$\left\\{X^{1},X^{2},X^{3}\right\\}$ and the algebra of case 4
$\left\\{X^{1},\bar{X}^{2},\bar{X}^{3}\right\\}$ is the same Lie algebra but
in different representation. The same observation applies to the subalgebra of
case 3 $\left\\{X^{1},X^{5},X^{6}\right\\}$ and the algebra of case 5
$\left\\{X^{1},\bar{X}^{5},\bar{X}^{6}\right\\}$. This connection between the
Lie groups is useful because it reveals common features in the dynamic
systems, as is the common transformation to the normal coordinates of the
systems.

For the cosmological viability of the models see [169, 136]

####  10.5 Analytic Solutions

Using the Noether symmetries and the associated Noether integrals we solve
analytically the differential eqs.(10.5), (10.6) and (10.7) for the cases
where the dynamical system is Liouville integrable, that is for cases 2-5.
Case 6 (i.e. $f\left(R\right)=R^{n}$) is not Liouville integrable via Noether
point symmetries, since the Noether integral (10.43) is time dependent444In
the appendix 10.A we present special solutions for the
$f\left(R\right)=R^{n}~{}$model, using the zero order invariants..

#####  10.5.1 Power law model $R^{\mu}$ with $\mu=\frac{3}{2}$

In this case the Lagrangian eq.(10.10) of the $f(R)=R^{\frac{3}{2}}$ model is
written as

$L=9a\sqrt{R}\dot{a}^{2}+\frac{9a^{2}}{2\sqrt{R}}\dot{a}\dot{R}+\frac{a^{3}}{2}R^{\frac{3}{2}}$
(10.44)

Changing the variables from $(a,R)$ to $(z,w)$ via the relations:

$a=\left(\frac{9}{2}\right)^{-\frac{1}{3}}\sqrt{z}\;\;\;\;R=\frac{w^{2}}{z}$
(10.45)

the Lagrangian (10.44) and the Hamiltonian (9.179) become

$L=\dot{z}\dot{w}+V_{0}w^{3}$ (10.46) $E=\dot{z}\dot{w}-V_{0}w^{3}$ (10.47)

where $V_{0}=\frac{1}{9}.~{}$The equations of motion in the new coordinate
system are

$\displaystyle\ddot{w}$ $\displaystyle=$ $\displaystyle 0$ (10.48)
$\displaystyle\ddot{z}-3V_{0}w^{2}$ $\displaystyle=$ $\displaystyle 0$ (10.49)

The Noether integrals (10.27),(10.28) in the coordinate system
$\left\\{z,y\right\\}$ are

$I_{1}^{{}^{\prime}}=\dot{w}~{}\ ,~{}~{}I_{2}^{{}^{\prime}}=t\dot{w}-w$
(10.50)

The general solution of the system is:

$y\left(t\right)=I_{1}^{\prime}t-I_{2}^{{}^{\prime}}$ (10.51)
$z\left(t\right)=\frac{1}{36\left(I_{1}^{{}^{\prime}}\right)^{2}}\left(I_{1}^{{}^{\prime}}t-I_{2}^{{}^{\prime}}\right)^{4}+z_{1}t+z_{0}$
(10.52)

The Hamiltonian constraint gives $E=z_{1}I_{1}^{{}^{\prime}}$ where
$z_{0,1}~{}$are constants and the singularity condition results in the
constraint

$\frac{1}{36\left(I_{1}^{{}^{\prime}}\right)^{2}}\left(I_{2}^{{}^{\prime}}\right)^{4}+z_{0}=0.$
(10.53)

#####  10.5.2 Power law model $R^{\mu}$ with $\mu=\frac{7}{8}$

In this case the Lagrangian eq.(10.10) is written as

$L=\frac{21a}{4R^{\frac{1}{8}}}\dot{a}^{2}-\frac{21}{16}\frac{a^{2}}{R^{\frac{9}{8}}}\dot{a}\dot{R}-\frac{1}{8}a^{3}R^{\frac{7}{8}}.$
(10.54)

Changing now the variables from $(a,R)$ to $(\rho,\sigma)$ via the relations:

$a=\left(\frac{21}{4}\right)^{-\frac{1}{3}}\sqrt{\rho
e^{\sigma}}\;\;\;\;R=\frac{e^{12\sigma}}{\rho^{4}}.$ (10.55)

The Lagrangian (10.82) and the Hamiltonian (9.179) become

$L=\frac{1}{2}\dot{\rho}^{2}-\frac{1}{2}\rho^{2}\dot{\sigma}^{2}+V_{0}\frac{e^{12\sigma}}{\rho^{2}}$
(10.56)
$E=\frac{1}{2}\dot{\rho}^{2}-\frac{1}{2}\rho^{2}\dot{\sigma}^{2}-V_{0}\frac{e^{12\sigma}}{\rho^{2}}.$
(10.57)

where$~{}V_{0}=-\frac{1}{42}.$ The Euler-Lagrange equations provide the
following equations of motion:

$\displaystyle\ddot{\rho}+\rho\dot{\sigma}^{2}+2V_{0}\frac{e^{12\sigma}}{\rho^{3}}$
$\displaystyle=0$ (10.58)
$\displaystyle\ddot{\sigma}+\frac{2}{\rho}\dot{\sigma}\dot{\rho}+12V_{0}\frac{e^{12\sigma}}{\rho^{2}}$
$\displaystyle=0.$ (10.59)

The Noether integrals (10.31), (10.32) and the Ermakov-Lewis invariant 10.33
in the coordinate system $\left\\{u,v\right\\}$ are

$\displaystyle I_{5}^{{}^{\prime}}$ $\displaystyle=2tE-\rho\dot{\rho}$ (10.60)
$\displaystyle I_{6}^{{}^{\prime}}$
$\displaystyle=t^{2}E-t\rho\dot{\rho}+\frac{1}{2}\rho^{2}.$ (10.61)
$\Sigma=\rho^{4}\dot{\sigma}^{2}+4V_{0}e^{12\sigma}.$ (10.62)

Using the Ermakov-Lewis Invariant, the Hamiltonian (10.56) and equation
(10.58) are written:

$\displaystyle\frac{1}{2}\dot{\rho}^{2}-\frac{1}{2}\frac{\Sigma}{\rho^{2}}$
$\displaystyle=E$ (10.63) $\displaystyle\ddot{\rho}+\frac{\Sigma}{\rho^{3}}$
$\displaystyle=0.$ (10.64)

And the analytical solution of the system is

$\rho\left(t\right)=\left(\rho_{2}t^{2}+\rho_{1}t+\frac{\left(\left(\rho_{1}\right)^{2}-4\Sigma\right)}{4\rho_{2}}\right)^{\frac{1}{2}}$
(10.65)
$\exp\left(\sigma\left(t\right)\right)=\left\\{\frac{21}{2}\Sigma\left[\left(\tanh\left[\sigma_{0}\rho_{2}\sqrt{\Sigma}-6\arctan
h\left(\frac{2\rho_{2}t+\rho_{1}}{2\sqrt{\Sigma}}\right)\right]\right)^{2}-1\right]\right\\}^{\frac{1}{12}}$
(10.66)

where
$B\left(t\right)=\left(\frac{1}{2}\frac{2\rho_{2}t+\rho_{1}}{\sqrt{\Sigma}}\right)$
and $\rho_{1,2}~{}$,$~{}\sigma_{0}$ are constants with Hamiltonian constrain
$E=\frac{1}{2}\rho_{2}~{}$. The singularity constraint gives
$\left(\rho_{1}\right)^{2}=4\Sigma$

In the case $\Sigma=0$ the analytical solution is

$\rho\left(t\right)=\left(\rho_{2}t^{2}+\rho_{1}t+\frac{1}{2}\frac{\left(\rho_{1}\right)^{2}}{\rho_{2}}\right)^{\frac{1}{2}}$
(10.67)
$\exp\sigma\left(t\right)=\left[\frac{1}{24\sqrt{V_{0}}}\frac{\left(2\rho_{2}t+\rho_{1}\right)}{\left(4\sigma_{0}\rho_{2}^{2}t+2\sigma_{0}\rho_{2}\rho_{1}-1\right)}\right]^{\frac{1}{6}}$
(10.68)

The singularity constraint gives $\rho_{1}=0$, then the solution is

$a\left(t\right)=\frac{a_{0}t^{\frac{7}{6}}}{\left(a_{2}t-1\right)^{\frac{1}{6}}}$
(10.69)

In contrast with the claim of [170] this model is analytically solvable and
there exists models which admit Noether integrals with time dependent gauge
functions.

#####  10.5.3 $\Lambda_{bc}$CDM model with $(b,c)=(1,\frac{3}{2})$

Inserting $f(R)=(R-2\Lambda)^{3/2}$ into eq.(10.10) we obtain

$L=9a\sqrt{R-2\Lambda}\dot{a}^{2}+\frac{9a^{2}}{2\sqrt{R-2\Lambda}}\dot{a}\dot{R}+\frac{a^{3}}{2}\left(R+4\Lambda\right)\sqrt{R-2\Lambda}$
(10.70)

Changing the variables from $(a,R)$ to $(x,y)$ via the relations:

$a=\left(\frac{9}{2}\right)^{-\frac{1}{3}}\sqrt{x}\;\;\;\;R=2\Lambda+\frac{y^{2}}{x}$
(10.71)

the Lagrangian (10.70) and the Hamiltonian (9.179) become

$L=\dot{x}\dot{y}+V_{0}\left(y^{3}+\bar{m}xy\right)$ (10.72)
$E=\dot{x}\dot{y}-V_{0}\left(y^{3}+{\bar{m}}xy\right)$ (10.73)

where $V_{0}=\frac{1}{9}$ and $\bar{m}=6\Lambda$.
The equations of motion, using the Euler-Lagrange equations, in the new
coordinate system are

$\ddot{x}-3V_{0}y^{2}-{\bar{m}V}_{0}x=0$ (10.74) $\ddot{y}-{\bar{m}}V_{0}y=0.$
(10.75)

The Noether integrals (10.35),(10.36) in the coordinate system
$\left\\{x,y\right\\}$ are

$\displaystyle\bar{I}_{1}^{\prime}$ $\displaystyle=e^{\omega t}\dot{y}-\omega
e^{\omega t}y$ (10.76) $\displaystyle\bar{I}_{2}^{\prime}$
$\displaystyle=e^{-\omega t}\dot{y}+\omega e^{-\omega t}y.$ (10.77)

where $\omega=\sqrt{2\Lambda/3}$. From these we construct the time independent
first integral

$\Phi=I_{1}I_{2}=\dot{y}^{2}-\omega^{2}y^{2}.$ (10.78)

The constants of integration are further constrained by the condition that at
the singularity ($t=0$), the scale factor has to be exactly zero, that is,
$x(0)=0$.
The general solution of the system (10.74)-(10.75) is:

$y\left(t\right)=\frac{I_{2}}{2\omega}e^{\omega
t}-\frac{I_{1}}{2\omega}e^{-\omega t}$ (10.79)
$x\left(t\right)=x_{1G}e^{\omega t}+x_{2G}e^{-\omega
t}+\frac{1}{4\bar{m}\omega^{2}}\left(I_{2}e^{\omega t}+I_{1}e^{-\omega
t}\right)^{2}+\frac{\Phi}{\bar{m}\omega^{2}}.$ (10.80)

The Hamiltonian constrain gives $E=\omega\left(x_{1G}I_{1}-x_{2G}I_{2}\right)$
where $x_{1G,2G}~{}$are constants and the singularity condition results in the
constrain

$x_{1G}+x_{2G}+\frac{1}{4\bar{m}\omega^{2}}\left(I_{1}+I_{2}\right)^{2}+\frac{\Phi}{\bar{m}\omega^{2}}=0.$
(10.81)

At late enough times the solution becomes $a^{2}(t)\propto e^{2\omega t}$

#####  10.5.4 $\Lambda_{bc}$CDM model with $(b,c)=(1,\frac{7}{8})$

In this case the Lagrangian eq.(10.10) of the $f(R)=(R-2\Lambda)^{7/8}$ model
is written as

$L=\frac{21a}{4\left(R-2\Lambda\right)^{\frac{1}{8}}}\dot{a}^{2}-\frac{21}{16}\frac{a^{2}}{\left(R-2\Lambda\right)^{\frac{9}{8}}}\dot{a}\dot{R}-\frac{1}{8}a^{3}\frac{\left(R-16\Lambda\right)}{\left(R-2\Lambda\right)^{\frac{1}{8}}}.$
(10.82)

Changing the variables from $(a,R)$ to $(u,v)$ via the relations:

$a=\left(\frac{21}{4}\right)^{-\frac{1}{3}}\sqrt{ue^{v}}\;\;\;\;R=2\Lambda+\frac{e^{12v}}{u^{4}}.$
(10.83)

the Lagrangian (10.82) and the Hamiltonian (9.179) become

$L=\frac{1}{2}\dot{u}^{2}-\frac{1}{2}u^{2}\dot{v}^{2}+V_{0}\frac{\bar{m}}{4}u^{2}+V_{0}\frac{e^{12v}}{u^{2}}$
(10.84)
$E=\frac{1}{2}\dot{u}^{2}-\frac{1}{2}u^{2}\dot{v}^{2}-V_{0}\frac{\bar{m}}{4}u^{2}-V_{0}\frac{e^{12v}}{u^{2}}.$
(10.85)

where $\bar{m}=-28\Lambda~{},~{}V_{0}=-\frac{1}{42}.$
The Euler-Lagrange equations provide the following equations of motion:

$\displaystyle\ddot{u}+u\dot{v}^{2}-\frac{V_{0}\bar{m}}{2}u+2V_{0}\frac{e^{12v}}{u^{3}}$
$\displaystyle=0$ (10.86)
$\displaystyle\ddot{v}+\frac{2}{u}\dot{u}\dot{v}+12V_{0}\frac{e^{12v}}{u^{4}}$
$\displaystyle=0.$ (10.87)

The Noether integrals (10.39),(10.40) and the Ermakov-Lewis invariant (10.41)
in the coordinate system $\left\\{u,v\right\\}$ are

$\displaystyle I_{+}$ $\displaystyle=\frac{1}{\lambda}e^{2\lambda
t}E-e^{2\lambda t}u\dot{u}+\lambda e^{2\lambda t}u^{2}$ (10.88) $\displaystyle
I_{-}$ $\displaystyle=\frac{1}{\lambda}e^{-2\lambda t}E-e^{-2\lambda
t}u\dot{u}+\lambda e^{-2\lambda t}u^{2}.$ (10.89)
$\phi=u^{4}\dot{v}^{2}+4V_{0}e^{12v}.$ (10.90)

where $\lambda=\frac{1}{2}\sqrt{\frac{2}{3}\Lambda}.$

Using the Ermakov-Lewis Invariant (10.90), the Hamiltonian (10.85) and
equation (10.86) are written:

$\displaystyle\frac{1}{2}\dot{u}^{2}-V_{0}\frac{m}{8}u^{2}-\frac{1}{2}\frac{\phi}{u^{2}}$
$\displaystyle=E$ (10.91)
$\displaystyle\ddot{u}-\frac{V_{0}m}{4}u+\frac{\phi}{u^{3}}$
$\displaystyle=0.$ (10.92)

The solution of (10.92) has been given by Pinney [79] and it is the following:

$u\left(t\right)=\left(u_{1}e^{2\lambda t}+u_{2}e^{-2\lambda
t}+2u_{3}\right)^{\frac{1}{2}}$ (10.93)

where $u_{1-3}$. From the Hamiltonian constraint (10.91) and the Noether
Integrals (10.88),(10.89) we find

$E=-2\lambda u_{3}~{},~{}I_{+}=2\lambda u_{2}~{},~{}I_{-}=2\lambda u_{1}.$

Replacing (10.93) in the Ermakov-Lewis Invariant (10.90) and assuming
$\phi\neq 0~{}$we find:

$\exp\left(v\left(t\right)\right)=2^{\frac{1}{6}}\phi^{\frac{1}{12}}e^{-A\left(t\right)}\left(4V_{0}+e^{-12A\left(t\right)}\right)^{-\frac{1}{6}}$
(10.94)

where

$A\left(t\right)=\arctan\left[\frac{2\lambda}{\sqrt{\phi}}\left(u_{1}e^{2\lambda
t}+u_{3}\right)\right]+4\lambda^{2}u_{1}\sqrt{\phi}.$ (10.95)

Then the solution is

$a^{2}\left(t\right)=2^{-\frac{1}{3}}\phi^{\frac{1}{12}}e^{-A\left(t\right)}\left(4V_{0}+e^{-12A\left(t\right)}\right)^{-\frac{1}{6}}\left(u_{1}e^{2\lambda
t}+u_{2}e^{-2\lambda t}+2u_{3}\right)^{\frac{1}{2}}$ (10.96)

where from the singularity condition $x\left(0\right)=0~{}$we have the
constrain $u_{1}+u_{2}+2u_{3}=0$ , or

$2E-\left(I_{+}+I_{-}\right)=0.$ (10.97)

At late enough time we find $A\left(t\right)\simeq A_{0}$, which implies
$a^{2}(t)\propto e^{\lambda t}.$

In the case where $\phi=0$ equations (10.91),(10.92) describe the hyperbolic
oscillator and the solution is

$u\left(t\right)=\sinh\lambda t~{},~{}2E=\lambda^{2}.$ (10.98)

From the Ermakov-Lewis Invariant we have

$\exp\left(v\left(t\right)\right)=\left(\frac{\lambda\sinh\lambda t}{\lambda
v_{1}\sinh\lambda t-12\sqrt{\left|V_{0}\right|}e^{-2\lambda
t}}\right)^{\frac{1}{6}}$ (10.99)

where $v_{1}$ is a constant. The analytic solution is

$a^{2}\left(t\right)=\left(\frac{\lambda\sinh^{7}\lambda t}{\lambda
v_{1}\sinh\lambda t-12\sqrt{\left|V_{0}\right|}e^{-2\lambda
t}}\right)^{\frac{1}{6}}$ (10.100)

####  10.6 Noether point symmetries in spatially non-flat $f(R)$ models

In this section we study further the Noether point symmetries in non flat
$f(R)$ cosmological models. In the context of a FRW spacetime the Lagrangian
of the overall dynamical problem and the Ricci scalar are

$L=6f^{\prime}a\dot{a}^{2}+6f^{\prime\prime}\dot{R}a^{2}\dot{a}+a^{3}\left(f^{\prime}R-f\right)-6Kaf^{\prime}$
(10.101)

$R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}+K}{a^{2}}\right)$ (10.102)

where $K$ is the spatial curvature. Note that the two dimensional metric is
given by eq.(10.23) while the ”potential” in the Lagrangian takes the form

$V_{K}(a,R)=-a^{3}(f^{\prime}R-f)+Kaf^{\prime}.$ (10.103)

Based on the above equations and using the theoretical formulation presented
in section 10.4, we find that the $f(R)$ models which admit non trivial
Noether symmetries are the $f(R)=(R-2\Lambda)^{3/2}$, $~{}f(R)=R^{3/2}$ and
$f(R)=R^{2}$. The Noether symmetries can be found in section 10.4.

In particular, inserting $f(R)=(R-2\Lambda)^{3/2}$ into the Lagrangian
(10.101) and changing the variables from $(a,R)$ to $(x,y)$ [see section
10.5.3] we find

$L=\dot{x}\dot{y}+V_{0}\left(y^{3}+\bar{m}xy\right)-\bar{K}y$ (10.104)
$E=\dot{x}\dot{y}-V_{0}\left(y^{3}+{\bar{m}}xy\right)+\bar{K}y$ (10.105)

where $\bar{K}=3(6^{1/3}K)$. Therefore, the equations of motion are

$\displaystyle\ddot{x}-3V_{0}y^{2}-\bar{m}V_{0}x+\bar{K}$ $\displaystyle=$
$\displaystyle 0$ $\displaystyle\ddot{y}-\bar{m}V_{0}y$ $\displaystyle=$
$\displaystyle 0\;.$

The constant term $\bar{K}$ appearing in the first equation of motion is not
expected to affect the Noether symmetries (or the integrals of motion). Indeed
we find that the corresponding Noether symmetries coincide with those of the
spatially flat $f(R)=(R-2\Lambda)^{3/2}$ model. However, in the case of $K\neq
0$ (or $\bar{K}\neq 0$) the analytic solution for the $x$-variable is written
as

$x_{K}(t)\equiv x(t)+\frac{{\bar{K}}}{\omega^{2}}$ (10.106)

where $x(t)$ is the solution of the flat model $K=0$ (see section 10.5.3).
Note that the solution of the $y$-variable remains unaltered.

Similarly, for the $f(R)=R^{3/2}~{}$model the analytic solution is

$z_{K}\left(t\right)=z\left(t\right)+\bar{K}$ (10.107)

where $z\left(t\right)$ is the solution of the spatially flat model (see
section 10.5.1).

####  10.7 Conclusion

In the literature the functional forms of $f(R)$ of the modified $f(R)$
gravity models are mainly defined on a phenomenological basis. In this article
we use the Noether symmetry approach to constrain these models with the aim to
utilize the existence of non-trivial Noether symmetries as a selection
criterion that can distinguish the $f(R)$ models on a more fundamental level.
Furthermore the resulting Noether integrals can be used to provide analytic
solutions.

In the context of $f(R)$ models, the system of the modified field equations is
equivalent to a two dimensional dynamical system moving in $M^{2}$ (mini
superspace) under the constraint $\bar{E}=$constant. Following the general
methodology of chapter 4, we require that the two dimensional system admits
extra Noether point symmetries. This requirement fixes the $f\left(R\right)$
function and the corresponding analytic solutions are computed. It is
interesting that two well known dynamical systems appear in cosmology: the
anharmonic oscillator and the Ermakov-Pinney system. We recall that the field
equations of the $\Lambda-$cosmology is equivalent with that of the hyperbolic
oscillator.

###  Appendix 10.A Special solutions for the power law model $R^{n}$

The case $f\left(R\right)=R^{n}$ is not Liouville integrable via Noether point
symmetries. However the zero order invariant will be used in order to find
special solutions. Inserting $f(R)=R^{n}~{}~{}\left(n\neq
0,1,\frac{3}{2},\frac{7}{8}\right)~{}$into eq.(10.10) we obtain

$L\left(a,\dot{a},R,\dot{R}\right)=6naR^{n-1}\dot{a}^{2}+6n\left(n-1\right)a^{2}R^{n-2}\dot{a}\dot{R}+\left(n-1\right)a^{3}R^{n}$
(10.108)

and this leads to the modified field equations

$\ddot{a}+\frac{1}{a}\dot{a}^{2}-\frac{1}{6}aR=0$ (10.109)
$\ddot{R}+\frac{n-2}{R}\dot{R}^{2}-\frac{1}{n-1}\frac{R}{a^{2}}\dot{a}^{2}+\frac{2}{a}\dot{a}\dot{R}-\frac{\left(n-3\right)}{6n\left(n-1\right)}R^{2}=0$
(10.110)
$E=6naR^{n-1}\dot{a}^{2}+6n\left(n-1\right)a^{2}R^{n-2}\dot{a}\dot{R}-\left(n-1\right)a^{3}R^{n}.$
(10.111)

The Noether point symmetry (10.42) is also and a Lie symmetry, hence we have
the zero order invariants

$a_{0}=at^{-N}~{},~{}R_{0}=Rt^{-2}.$ (10.112)

Applying the zero order invariants in the field equations (10.109)-(10.111)
and in the Noether integral (10.43) we have the following results.

The dynamical system admits a special solution of the form

$a=a_{0}t^{N}~{},~{}R=6N\left(2N-1\right)t^{-2}$ (10.113)

where the constants $N,$ $E$ and $I_{7}$ are

$N=\frac{1}{2}~{},~{}E=0~{},~{}I_{7}=0$

or

$N=-\frac{\left(2n-1\right)\left(n-1\right)}{n-2}~{},~{}E=0~{},~{}I_{7}=0$

or

$N=\frac{2}{3}n~{},~{}E=\left(\frac{12n}{9}\right)^{n}\left(4n-3\right)^{n-1}\left(13n-8n^{2}-3\right)a_{0}^{3}~{},~{}I_{7}=0.$

Another special solution is the deSitter solution for $n=2$

$a=a_{0}e^{H_{0}t}~{},~{}R=12H_{0}^{2}$ (10.114)

where $I_{7}=0~{}$ and the spacetime is empty i.e. $E=0$.

###  Chapter 11 Noether symmetries in $f(T)$ gravity

####  11.1 Introduction

In this chapter we continue our analysis on the application of Noether point
symmetries in alternative theories of gravity and specifically the
$f\left(T\right)$ modified theory of gravity. $f\left(T\right)$ gravity it is
based on the old formulation of Teleparallel Equivalent of General Relativity
(TEGR) [171, 172, 173, 174] which instead of the torsion-less Levi-Civita
connection uses the curvatureless Weitzenbock connection [175] in which the
corresponding dynamical fields are the four linearly independent vierbeins.
Therefore, all the information concerning the gravitational field is included
in the Weitzenbock tensor. Within this framework, considering invariance under
general coordinate transformations, global Lorentz-parity transformations, and
requiring up to second order terms of the torsion tensor, one can write down
the corresponding Lagrangian density $T$ by using some suitable contractions.

Furthermore $f\left(T\right)$ gravity which is based on the fact that we allow
the Lagrangian to be a function of $T$ [176, 177, 178], inspired by the
extension of $f\left(R\right)~{}$Einstein-Hilbert action. However,
$f\left(T\right)$ gravity does not coincide with $f\left(R\right)$ extension,
but it rather consists of a different class of modified gravity models. It is
interesting to mention that the torsion tensor includes only products of first
derivatives of the vierbeins, giving rise to second-order field differential
equations in contrast to the $f\left(R\right)$ gravity that provides fourth-
order equations, which potentially may lead to some problems, for example in
the well position and well formulation of the Cauchy problem [179]. Moreover,
as we showed in chapter 10 the Lagrangian of the field equations in
$f\left(R\right)$ gravity described a regular dynamical system; however, in
$f\left(T\right)$ the dynamical system is singular and the $T$ variable can be
seen as a Lagrange multiplier.

In section 11.2 we discuss the role of unholonomic frames in the context of
teleparallel gravity and its straightforward extension. In section 11.3, we
briefly present $f(T)$ gravity, while in section 11.4 we construct the
corresponding generalized Lagrangian formulation. In section 11.5, we analyze
the main properties of the Noether Symmetry Approach for $f(T)$ gravity. Then,
in section 11.6 and 11.7, we apply these results in the FRW and the static
spherically symmetric spaces.

####  11.2 Unholonomic frames and Connection Coefficient

In an $n-$dimensional manifold $M$ consider a coordinate neighborhood $U$ with
a coordinate system $\\{x^{\mu}\\}.$ At each point $P\epsilon U$ we have the
resulting holonomic frame $\\{\partial_{\mu}\\}.$ We define in $U\ \ $a new
frame $\\{e_{a}(x^{\mu})\\}$ which is related to the holonomic frame
$\\{\partial_{a}\\}$ as follows:

$e_{a}(x^{\mu})=h_{a}^{\mu}\partial_{\mu}~{}\ \ a,\mu=1,2,...,n$ (11.1)

where the quantities $h_{a}^{\mu}(x)$ are in general functions of the
coordinates (i.e. depend on the point $P$). Notice that Latin indexes count
vectors, while Greek indexes are tensor indices. We assume that $\det
h_{a}^{\mu}\neq 0$ which guaranties that the vectors $\\{e_{a}(x^{\mu})\\}$
form a set of linearly independent vectors. We define the ”inverse” quantities
$h_{a}^{\mu}$ by means of the following ”orthogonality” relations:

$h_{a}^{\mu}h_{\nu}^{a}=\delta_{\nu}^{\mu},h_{b}^{\mu}h_{\mu}^{c}=\delta_{b}^{c}.$
(11.2)

The commutators of the vectors $\\{e_{a}\\}$ are not in general all zero. If
they are, then there exists a new coordinate system in $U$, the $\\{y^{b}\\}$
say so that $e_{b}=\frac{\partial}{\partial y^{b}}$ i.e. the new frame is
holonomic. If there are commutators $[e_{a},e_{b}]\neq 0$ then the new frame
$\\{e_{a}\\}$ is called unholonomic and the vectors $e_{a}$ cannot be written
in the form $e_{b}=\partial_{b}.$ The quantities which characterize an
unholonomic frame are the objects of unholonomicity or Ricci rotation
coefficients $\Omega_{\text{ }bc}^{a}$ defined by the relation

$[e_{a},e_{b}]=\Omega_{\text{ }ab}^{c}e_{c}.$ (11.3)

We compute:

$[e_{a},e_{b}]=[h_{a}^{\mu}\partial_{\mu},h_{b}^{\nu}\partial_{\nu}]=\left[h_{a}^{\mu}h_{b,\mu}^{\nu}h_{\nu}^{c}-h_{b}^{\nu}h_{a,\nu}^{\mu}h_{\mu}^{c}\right]e_{c}$

from which follows that the Ricci rotation coefficients of the frame
$\\{e_{a}\\}$ are:

$\Omega_{\text{ }bc}^{a}=2h_{[b}^{\mu}h_{c],\mu}^{\nu}h_{\nu}^{a}.$ (11.4)

The condition for $\\{e_{a}\\}$ to be a holonomic basis is $\Omega_{\text{
}bc}^{a}=0$ at all points $P\epsilon U.$ This is a set of linear partial
differential equations whose solution defines all holonomic frames and all
coordinate systems in $U.$ One obvious solution is $h_{b}^{c}=\delta_{b}^{c}$.
The set of all coordinate systems in $U\ $equipped with the operation of
composition of transformations has the structure of an infinite dimensional
Lie group which is called the Manifold Mapping Group.

We consider now the special unholonomic frames which satisfy Jacobi’s
identity:

$[[e_{a},e_{b}],e_{c}]+[[e_{b},e_{c}],e_{a}]+[[e_{c},e_{a}],e_{b}]=0.$ (11.5)

These frames are the generators of a Lie Algebra, therefore they have an extra
role to play. Replacing the commutator in terms of the unholonomicity objects
we find the following identity:

$\Omega_{\text{ }ab,c}^{d}+\Omega_{\text{ }ba,a}^{d}+\Omega_{\text{
}ca,b}^{d}-\Omega_{\text{ }ab}^{l}\Omega_{cl}^{d}-\Omega_{\text{
}bc}^{l}\Omega_{al}^{d}-\Omega_{\text{ }ca}^{l}\Omega_{bl}^{d}=0.$ (11.6)

Using the definition of the covariant derivative we write:

$\nabla_{e_{i}}e_{j}=\Gamma_{ij}^{k}e_{k}$ (11.7)

where $\Gamma_{ij}^{k}$ are the connection coefficients in the frame
$\\{e_{i}\\}.$ Let us compute these $\Gamma_{ij}^{k}$ assuming that

$[e_{i},e_{j}]=C_{.ij}^{k}e_{k}$

from which follows

$C_{.ij}^{k}=\Omega_{.jk}^{k}.$

Consider three vector fields $X,Y,Z$ and the covariant derivative of the
metric wrt $X.$ Then we have:

$\nabla_{X}g(Y,Z)=X(g(Y,Z))-g(\nabla_{X}Y,Z)-g(Y,\nabla_{Y}Z)$ (11.8)

and by interchanging the role of $X,Y,Z:$

$\nabla_{Y}g(Z,X)=Y(g(Z,X))-g(\nabla_{Y}Z,X)-g(Z,\nabla_{Z}X)$ (11.9)

$\nabla_{Z}g(X,Y)=Z(g(X,Y))-g(\nabla_{Z}X,Y)-g(X,\nabla_{X}Y).$ (11.10)

Adding (11.8), (11.9) and subtracting (11.10) we obtain:

$\displaystyle\nabla_{X}g(Y,Z)+\nabla_{Y}g(Z,X)-\nabla_{Z}g(X,Y)$
$\displaystyle=$ $\displaystyle X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))+$
$\displaystyle-\left[g(\nabla_{X}Y,Z)+g(\nabla_{Y}Z,X)-g(\nabla_{Z}X,Y)\right]+$
$\displaystyle-\left[g(Y,\nabla_{X}Z)+g(Z,\nabla_{Y}X)-g(X,\nabla_{Z}Y)\right]$

then

$\displaystyle\nabla_{X}g(Y,Z)+\nabla_{Y}g(Z,X)-\nabla_{Z}g(X,Y)$
$\displaystyle=$ $\displaystyle X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))+$
$\displaystyle-\left[g(\nabla_{X}Y,Z)+g(Z,\nabla_{Y}X)\right]+$
$\displaystyle-\left[g(\nabla_{Y}Z,X)-g(X,\nabla_{Z}Y)\right]+$
$\displaystyle-\left[g(Y,\nabla_{X}Z)-g(\nabla_{Z}X,Y)\right]/$

that is

$\displaystyle\nabla_{X}g(Y,Z)+\nabla_{Y}g(Z,X)-\nabla_{Z}g(X,Y)$
$\displaystyle=$ $\displaystyle X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))+$
$\displaystyle-\left[g(Z,\nabla_{X}Y+\nabla_{Y}X)+g(X,\nabla_{Y}Z-\nabla_{Z}Y)+g(Y,\nabla_{X}Z-\nabla_{Z}X)\right].$

The term

$g(Z,\nabla_{X}Y+\nabla_{Y}X)=2g\left(Z,\nabla_{X}Y\right)+g\left(Z,\nabla_{Y}X-\nabla_{X}Y\right).$

Replacing in the last relation and solving for $2g\left(Z,\nabla_{X}Y\right)$
we find

$\displaystyle 2g\left(Z,\nabla_{X}Y\right)$ $\displaystyle=$
$\displaystyle\left[X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))\right]+$
$\displaystyle-\left[\nabla_{X}g(Y,Z)+\nabla_{Y}g(Z,X)-\nabla_{Z}g(X,Y)\right]+$
$\displaystyle-\left[g\left(Z,\nabla_{Y}X-\nabla_{X}Y\right)+g(X,\nabla_{Y}Z-\nabla_{Z}Y)+g(Y,\nabla_{X}Z-\nabla_{Z}X)\right]$

or

$\displaystyle 2g\left(Z,\nabla_{X}Y\right)$ $\displaystyle=$
$\displaystyle\left[X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))\right]+$
$\displaystyle-\left[\nabla_{X}g(Y,Z)+\nabla_{Y}g(Z,X)-\nabla_{Z}g(X,Y)\right]+$
$\displaystyle-\left[g\left(Z,\nabla_{Y}X-\nabla_{X}Y-\left[Y,X\right]\right)+g(X,\nabla_{Y}Z-\nabla_{Z}Y-\left[Y,Z\right])+g(Y,\nabla_{X}Z-\nabla_{Z}X-\left[X,Z\right])\right]+$
$\displaystyle-\left[g\left(Z,\left[Y,X\right]\right)+g\left(X,\left[Y,Z\right]\right)+g\left(Y,\left[X,Z\right]\right)\right].$

Define the quantities

$\displaystyle T_{\nabla}(X,Y)$ $\displaystyle=$
$\displaystyle\nabla_{X}Y-\nabla_{Y}X-[X,Y]$ $\displaystyle A_{\nabla}(X,Y,Z)$
$\displaystyle=$ $\displaystyle\nabla_{X}g(Y,Z)$

The tensors $T_{\nabla}$ and $A_{\nabla}$ are called the torsion and the
metricity of the connection $\nabla$ respectively. Last relation in terms of
the fields $T_{\nabla}$ and $A_{\nabla}$ is written as follows$:$

$\displaystyle 2g\left(Z,\nabla_{X}Y\right)$ $\displaystyle=$
$\displaystyle\left[X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))\right]+$ (11.11)
$\displaystyle-\left[A_{\nabla}\left(X,Y,Z\right)+A_{\nabla}\left(Y,Z,X\right)-A_{\nabla}\left(Z,X,Y\right)\right]+$
$\displaystyle-\left[g\left(Z,T_{\nabla}\left(Y,X\right)\right)+g\left(X,T_{\nabla}\left(Y,Z\right)\right)+g\left(Y,T_{\nabla}\left(X,Z\right)\right)\right]$
$\displaystyle-\left[g\left(Z,\left[Y,X\right]\right)+g\left(X,\left[Y,Z\right]\right)+g\left(Y,\left[X,Z\right]\right)\right].$

Let $X=e_{l}~{},~{}Y=e_{j}$ and $Z=e_{k}.$ Contracting with
$\frac{1}{2}g^{il}$ we have

$2g\left(Z,\nabla_{X}Y\right)\rightarrow\Gamma_{jk}^{i}$
$\left[X(g(Y,Z))+Y(g(Z,X))-Z(g(X,Y))\right]\rightarrow\left\\{{}_{jk}^{i}\right\\}$
$g\left(X,T_{\nabla}\left(Y,Z\right)\right)\rightarrow Q_{.kj}^{i}$
$g\left(Z,T_{\nabla}\left(Y,X\right)\right)+g\left(Y,T_{\nabla}\left(X,Z\right)\right)\rightarrow
g^{il}(g_{tj}Q_{kl}^{t}+g_{tk}Q_{jl}^{t})=-\bar{S}_{.kj}^{i}$
$g\left(X,\left[Y,Z\right]\right)\rightarrow\frac{1}{2}C_{.jk}^{i}$
$g\left(Z,\left[Y,X\right]\right)+g\left(Y,\left[X,Z\right]\right)=\frac{1}{2}g^{il}(g_{tj}C_{lk}^{t}+g_{tk}C_{jl}^{t})=-S_{.kj}^{i}$

and

$A_{\nabla}\left(X,Y,Z\right)+A_{\nabla}\left(Y,Z,X\right)-A_{\nabla}\left(Z,X,Y\right)\rightarrow\frac{1}{2}g^{il}\Delta_{jkl}$

Replacing in (11.11) we find the connection coefficients in the frame
$\\{e_{i}\\}$

$\Gamma_{jk}^{i}=\left\\{{}_{jk}^{i}\right\\}+\bar{S}_{.kj}^{i}+S_{.kj}^{i}-\frac{1}{2}g^{il}\Delta_{jkl}+Q_{jk}^{i}-\frac{1}{2}C_{.jk}^{i}$
(11.12)

where $\left\\{{}_{jk}^{i}\right\\}$ is the standard Levi Civita connection
coefficients (Christofell symbols). This is the most general expression for
the connection coefficients in terms of the fields
$\left\\{{}_{jk}^{i}\right\\},$ $T_{\nabla},$ $A_{\nabla}$ and $C_{jk}^{i}$.

Concerning the symmetric and antisymmetric part we have:

$\displaystyle\Gamma_{.(jk)}^{i}$ $\displaystyle=$
$\displaystyle\left\\{{}_{jk}^{i}\right\\}+\bar{S}_{.jk}^{i}+S_{.jk}^{i}-\frac{1}{2}g^{il}\Delta_{jkl}$
(11.13) $\displaystyle\Gamma_{.[jk]}^{i}$ $\displaystyle=$ $\displaystyle
Q_{.jk}^{i}-\frac{1}{2}C_{.jk}^{i}.$ (11.14)

From the above we draw the following conclusions:

1\. The connection coefficients in a frame $\\{e_{i}\\}$ are determined from
the metric, the torsion, the metricity and the unholonomicity objects
(equivalently the commutators) of the frame vectors

2\. The symmetric part $\Gamma_{.(jk)}^{i}$ of $\Gamma_{jk}^{i}$ depends on
all fields. This means that the geodesics and the autoparallels in a given
frame depend on the geometric properties of the space (fields $g_{ij},$
$Q_{.kj}^{i},g_{ij|k})$ and the unholonomicity of the frame (field
$C_{.jk}^{i})$

3\. The antisymmetric part $\Gamma_{.[jk]}^{i}$ of $\Gamma_{jk}^{i}$ depends
only on the fields $Q_{.kj}^{i}$ and $C_{.jk}^{i}.$

4\. The objects of unholonomicity $C_{.jk}^{i}$ behave in the same way as the
components of the torsion. This means that even in a Riemannian space where
$Q_{.kj}^{i}=0,g_{ij|k}=0$ in an unholonomic basis the antisymmetric part
$\Gamma_{.[jk]}^{i}=-\frac{1}{2}C_{.jk}^{i}\neq 0.$ This result has lead to
the misunderstanding that when one works in an unholonomic frame then one has
introduced torsion, which is not correct! This is the case with the TEGR. This
misunderstanding has important consequences because the effects one will
observe in an unholonomic frame will be frame dependent effects and not
covariant effects. Therefore all conclusions made in a specific unholonomic
frame are restricted to that frame only.

####  11.3 $f(T)$-gravity

Teleparallelism uses as dynamical objects special unholonomic frames in
spacetime, called vierbeins, which are defined by the requirement

$g(e_{i},e_{j})=e_{i}.e_{j}=\eta_{ij}$ (11.15)

where $\eta_{ij}=\mathrm{diag}(1,-1,-1,-1)$ is the Lorentz metric in canonical
form. Obviously $g_{\mu\nu}(x)=\eta_{ij}h_{\mu}^{i}(x)h_{\nu}^{j}(x)$ where
$e^{i}(x)=h_{\mu}^{i}(x)dx^{i}$ is the dual basis. Differing from GR, which
uses the torsionless Levi-Civita connection, Teleparallelism utilizes the
curvatureless Weitzenböck connection [175], where Weitzenböck non-null torsion
is

$T_{\mu\nu}^{\beta}=\hat{\Gamma}_{\nu\mu}^{\beta}-\hat{\Gamma}_{\mu\nu}^{\beta}=h_{i}^{\beta}(\partial_{\mu}h_{\nu}^{i}-\partial_{\nu}h_{\mu}^{i})\;.$
(11.16)

Notice that we assume that the Ricci rotation coefficients obey
$\Omega_{jk}^{i}=T_{jk}^{i}$ and encompass all the information concerning the
gravitational field. The TEGR Lagrangian for the gravitational field equations
(Einstein equations) is taken to be:

$T=S_{\beta}^{\mu\nu}T_{\mu\nu}^{\beta}$ (11.17)

where

$S_{\beta}^{\mu\nu}=\frac{1}{2}(K_{\beta}^{\mu\nu}+\delta_{\beta}^{\mu}T_{\theta}^{\theta\nu}-\delta_{\beta}^{\nu}T_{\theta}^{\theta\mu})$
(11.18)

and $K_{\beta}^{\mu\nu}$ is the tensor

$K_{\beta}^{\mu\nu}=-\frac{1}{2}(T_{\beta}^{\mu\nu}-T_{\beta}^{\nu\mu}-T_{\beta}^{\mu\nu}),$
(11.19)

which equals the difference of the Levi Civita connection in the holonomic and
the unholonomic frame.

In this work the gravitational field will be driven by a Lagrangian density
which is a function of $T$. Therefore, the corresponding action of $f(T)$
gravity reads as

$\mathcal{A}=\frac{1}{16\pi G}\int{d^{4}xef(T)}$ (11.20)

where $e=det(e_{\mu}^{i}\cdot e_{\nu}^{i})=\sqrt{-g}$ and $G$ is Newton’s
constant. TEGR and thus General Relativity is restored when $f(T)=T$. First of
all, in order to construct a realistic cosmology we have to incorporate the
matter and radiation sectors too. Therefore, the total action is written as

$S_{\mathrm{tot}}=\mathcal{A}+\frac{1}{16\pi G}\int
d^{4}xe\left(L_{m}+L_{r}\right),$ (11.21)

If matter couples to the metric in the standard form then the variation of the
action (11.21) with respect to the vierbein leads to the equations [180]

$\displaystyle
e^{-1}\partial_{\mu}(e{S}_{i}^{\mu\nu})f^{\prime}(T)-h_{i}^{\lambda}T_{\mu\lambda}^{\beta}S_{\beta}^{\nu\mu}f^{\prime}(T)$
$\displaystyle+S_{i}^{\mu\nu}\partial_{\mu}(T)f^{\prime\prime}(T)+\frac{1}{4}h_{i}^{\nu}f(T)=4\pi
Gh_{i}^{\beta}T_{\beta}^{\nu}$ (11.22)

where a prime denotes differentiation with respect to $T$,
${S_{i}}^{\mu\nu}={h_{i}}^{\beta}S_{\beta}^{\mu\nu}$ and $T_{\mu\nu}$ is the
matter energy-momentum tensor.

####  11.4 Generalized Lagrangian formulation of $f(T)$ gravity

In this section, we provide a generalized Lagrangian formulation in order to
construct a theory of $f(T)$ gravity. Specifically, the gravitational field is
driven by the Lagrangian density $f(T)$ in (11.20), which can be generalized
through the use of a Lagrange multiplier. In particular, we can write it as

$L\left(x^{k},x^{\prime
k},T\right)=2f_{T}\bar{\gamma}_{ij}\left(x^{k}\right)x^{\prime i}x^{\prime
j}+M\left(x^{k}\right)\left(f-Tf_{T}\right)$ (11.23)

where $x^{\prime}=\frac{dx}{d\tau}$, $M(x^{k})$ is the Lagrange multiplier and
$\bar{\gamma}_{ij}$ is a second rank tensor which is related to the frame [one
can use $eT(x^{k},x^{\prime k})$] of the background spacetime. In the same
lines, the Hamiltonian of the system is written as

$H\left(x^{k},x^{\prime
k},T\right)=2f_{T}\bar{\gamma}_{ij}\left(x^{k}\right)x^{\prime i}x^{\prime
j}-M\left(x^{k}\right)\left(f-Tf_{T}\right)=0\;.$ (11.24)

In this case, the system is autonomous hence $\partial_{\tau}$ is a Noether
symmetry with corresponding Noether integral the Hamiltonian $H$.

In this framework, considering $\\{x^{k},T\\}$ as the canonical variables of
the configuration space, we can derive, after some algebra, the general field
equations of $f(T)$ gravity. Indeed, starting from the Lagrangian (11.23), the
Euler-Lagrange equations

$\frac{\partial L}{\partial T}=0,\;\;\;\;\frac{d}{d\tau}\left(\frac{\partial
L}{\partial x^{\prime k}}\right)-\frac{\partial L}{\partial x^{k}}=0\,$
(11.25)

give rise to

$f_{TT}\left(2\bar{\gamma}_{ij}x^{\prime i}x^{\prime j}-MT\right)=0,$ (11.26)
$x^{i\prime\prime}+{\bar{\Gamma}}_{jk}^{i}x^{j\prime}x^{k\prime}+\frac{f_{TT}}{f_{T}}x^{i\prime}T^{\prime}-M^{,i}\frac{\left(f-Tf_{T}\right)}{4f_{T}}=B_{m}^{i}\;.$
(11.27)

The functions ${\bar{\Gamma}}_{jk}^{i}$ are considered to be the Christoffel
symbols for the metric $\bar{\gamma}_{ij}$. Therefore, the system is
determined by the two independent differential equations (11.26),(11.27) and
the Hamiltonian constraint $H=C_{m}$ where $H$ is given by equation (11.24)
and $C_{m},$ $B_{m}^{i}$ are the components of the energy momentum tensor
$T_{\mu\nu.}$.

The point-like Lagragian (11.23) determines completely the related dynamical
system in the minisuperspace $\\{x^{k},T\\}$, implying that one can easily
recover some well known cases of cosmological interest. In brief, these are:

  * •

The static spherically symmetric spacetime:

$ds^{2}=-a^{2}\left(\tau\right)dt^{2}+\frac{1}{N^{2}\left(a\left(\tau\right),b\left(\tau\right)\right)}d\tau^{2}\,+b^{2}\left(\tau\right)\left(d\theta^{2}+\sin^{2}\theta
d\phi^{2}\right)$ (11.28)

arising from the diagonal vierbein 111Note that, in general, one can choose a
non-diagonal vierbein, giving rise to the same metric through (11.15).

$e_{i}^{A}=\left(a\left(\tau\right),\frac{1}{N\left(a\left(\tau\right),b\left(\tau\right)\right)},b\left(\tau\right),b\left(\tau\right)\sin\theta\right)\;$
(11.29)

where $a(\tau)$ and $b(\tau)$ are functions which need to be determined.
Therefore, the line element of $\bar{\gamma}_{ij}$ and $M\left(x^{k}\right)$
are given by

$ds_{\bar{\gamma}}^{2}=N\left(2b~{}da~{}db+a~{}db^{2}\right)~{},~{}~{}M(a,b)=\frac{ab^{2}}{N}.$
(11.30)

  * •

The flat FRW spacetime with Cartesian coordinates:

$ds^{2}=-dt^{2}+a^{2}\left(t\right)\left(dx^{2}+dy^{2}+dz^{2}\right)$ (11.31)

arising from the vierbein

$e_{i}^{A}=\left(1,a\left(t\right),a\left(t\right),a\left(t\right)\right)\;$
(11.32)

where $t$ is the cosmic time and $a(t)$ is the scale factor of the universe.
In this case we have

$ds_{\bar{\gamma}}^{2}=3a~{}da^{2}~{},~{}M(a)=a^{3}(t).$ (11.33)

  * •

The Bianchi type I spacetime:

$\\!ds^{2}=-\frac{1}{N^{2}\left(a\left(t\right),\beta\left(t\right)\right)}dt^{2}+a^{2}\left(t\right)\left[e^{-2\beta\left(t\right)}dx^{2}+e^{\beta\left(t\right)}\left(dy^{2}+dz^{2}\right)\right]\
$ (11.34)

arising from the vierbein

$e_{i}^{A}=\left(\frac{1}{N\left(a\left(t\right),\beta\left(t\right)\right)},a(t)e^{-\beta(t)},a(t)^{\frac{\beta(t)}{2}},a(t)^{\frac{\beta(t)}{2}}\right)\;.$
(11.35)

In this case, we obtain

$ds_{\bar{\gamma}}^{2}=N\left(-4ada^{2}+a^{3}d\beta^{2}\right)~{}~{},~{}~{}M(a,\beta)=\frac{a^{3}(t)}{N}.$
(11.36)

In the present work we will focus on the static spherically-symmetric metric
deriving new spherically symmetric solutions for $f(T)$ gravity. In
particular, we look for Noether symmetries in order to reveal the existence of
analytical solutions.

####  11.5 The Noether Symmetry Approach for $f(T)$ gravity

The aim is now to apply the Noether Symmetry Approach to a general class of
$f(T)$ gravity models where the corresponding Lagrangian of the field
equations is given by equation (11.23). First of all, we perform the analysis
for arbitrary spacetimes, and then we focus on the spatially flat FRW
spacetime and on the static spherically-symmetric spacetime.

#####  11.5.1 Searching for Noether point symmetries in general spacetimes

The Noether symmetry condition for Lagrangian (11.23) is given by

$X^{\left[1\right]}L+L\xi^{\prime}=g^{\prime}~{},~{}g=g\left(\tau,x^{i}\right).$
(11.37)

Notice that the Lagrangian (11.51) is a singular Lagrangian (the Hessian
vanishes), hence the jet space is
$\bar{B}_{M}=\left\\{\tau,x^{i},T,\dot{x}^{i}\right\\}$ and thus the first
prolongation of $X$ in the jet space $\bar{B}_{M}~{}$is [103, 181, 182]

$\displaystyle
X^{\left[1\right]}=\xi\left(\tau,x^{k},T\right)\partial_{\tau}+\eta^{k}\left(\tau,x^{k},T\right)\partial_{i}$
$\displaystyle\ \ \ \ \ \ \ \ \
+\mu\left(\tau,x^{k},T\right)\partial_{T}+\left(\eta^{\prime
i}-\xi^{\prime}x^{\prime i}\right)\partial_{x^{\prime i}}.$ (11.38)

For each term of the Noether condition (11.37) for the Lagrangian (11.23) we
obtain

$\displaystyle X^{\left[1\right]}L$ $\displaystyle=$ $\displaystyle
2f_{T}\bar{g}_{ij,k}\eta^{k}x^{\prime i}x^{\prime
j}+M_{,k}\eta^{k}\left(f-Tf_{T}\right)$
$\displaystyle+2f_{TT}\mu\bar{g}_{ij}x^{\prime i}x^{\prime j}-Mf_{TT}\mu$
$\displaystyle+4f_{T}\bar{g}_{ij}x^{\prime
i}\left(\eta_{,\tau}^{j}+\eta_{,k}^{j}x^{\prime
k}+\eta_{,T}^{j}T^{\prime}\right.$ $\displaystyle\left.-\xi_{,\tau}x^{\prime
j}-\xi_{,k}x^{\prime j}x^{\prime k}-\xi_{,T}x^{\prime j}T^{\prime}\right),$
$L\xi^{\prime}=\left[2f_{T}\bar{g}_{ij}x^{\prime i}x^{\prime
j}+M\left(x^{i}\right)\left(f-Tf_{T}\right)\right]\left(\xi_{,\tau}+\xi_{,k}x^{\prime
k}+\xi_{,T}T^{\prime}\right),$ $g^{\prime}=g_{,\tau}+g_{,k}x^{\prime
k}+g_{,T}T^{\prime}\;.$

Inserting these expressions into (11.37) we find the Noether symmetry
conditions

$\xi_{,k}=0~{},~{}\xi_{,T}=0~{},~{}g_{,T}=0~{},~{}\eta_{,T}=0,$ (11.39)
$4f_{T}\bar{\gamma}_{ij}\eta_{,\tau}^{k}=g_{,k},$ (11.40)
$M_{,k}\eta^{k}\left(f-Tf_{T}\right)-MTf_{TT}\mu+\xi_{,\tau}M\left(f-Tf_{T}\right)-g_{,\tau}=0,$
(11.41)
$2f_{T}\bar{\gamma}_{ij,k}\eta^{k}+2f_{TT}\mu\bar{\gamma}_{ij}+4f_{T}\bar{\gamma}_{ij}\eta_{,k}^{j}-2f_{T}\bar{\gamma}_{ij}\xi_{,\tau}=0\;.$
(11.42)

Conditions $\eta_{,T}=g_{,T}=0$ imply, through equation (11.40), that
$\eta_{,\tau}^{k}=g_{,k}=0$. Also, equation (11.42) takes the form

$L_{\eta}\bar{\gamma}_{ij}=\left(\xi_{,\tau}-\frac{f_{TT}}{f_{T}}\mu\right)\bar{\gamma}_{ij},$
(11.43)

where $L_{\eta}\bar{\gamma}_{ij}$ is the Lie derivative with respect to the
vector field $\eta^{i}(x^{k})$. Furthermore, from (11.43) we deduce that
$\eta^{i}$ is a CKV of the metric $\bar{\gamma}_{ij}$, with conformal factor

$2\bar{\psi}\left(x^{k}\right)=\xi_{,\tau}-\frac{f_{TT}}{f_{T}}\mu=\xi_{,\tau}-S(\tau,x^{k})\;.$
(11.44)

Finally, utilizing simultaneously equations (11.41), (11.43), (11.44) and the
condition $g_{,\tau}=0$, we rewrite (11.41) as

$M_{,k}\eta^{k}+\left[2\bar{\psi}+\left(1-\frac{Tf_{T}}{f-Tf_{T}}\right)S\right]M=0\;.$
(11.45)

Considering that $S=S(x^{k})$ and using the condition $g_{,\tau}=0$, we
acquire $\xi_{,\tau}=2\bar{\psi}_{0},\bar{\psi}_{0}\in\mathbb{R}$ with
$S=2(\bar{\psi}_{0}-\bar{\psi})$. At this point, we have to deal with the
following two situations:

Case 1. In the case of $S=0$, the symmetry conditions are

$\displaystyle L_{\eta}\bar{\gamma}_{ij}=2\bar{\psi}_{0}\bar{\gamma}_{ij}$
$\displaystyle M_{,k}\eta^{k}+2\bar{\psi}_{0}M=0$ (11.46)

implying that the vector $\eta^{i}(x^{k})$ is a homothetic Vector of the
metric $~{}\bar{\gamma}_{ij}$. The latter means that for arbitrary
$f\left(T\right)\neq T^{n}$ functional forms, the dynamical system could
possibly admit extra (time independent) Noether symmetries.

Case 2. If $S\neq 0$ then equation (11.45) leads to the differential equation

$\frac{Tf_{T}}{f-Tf_{T}}=C$ (11.47)

which has the solution

$f(T)=T^{n},\;\;\;\;C\equiv\frac{n}{1-n}\;.$ (11.48)

In this context, $\eta^{i}(x^{k})$ is a CKV of $\bar{\gamma}_{ij}$, and the
symmetry conditions become

$\displaystyle L_{\eta}\bar{\gamma}_{ij}=2\bar{\psi}\bar{\gamma}_{ij},$
$\displaystyle M_{,k}\eta^{k}+\left[2\bar{\psi}+\left(1-C\right)S\right]=0,$
(11.49)

with $S=2(\bar{\psi}_{0}-\bar{\psi})$.

Collecting the above results we have the following result

###### Lemma 11.5.1

The general autonomous Lagrangian

$L\left(x^{k},x^{\prime
k},T\right)=2f_{T}\bar{\gamma}_{ij}\left(x^{k}\right)x^{\prime i}x^{\prime
j}+M\left(x^{k}\right)\left(f-Tf_{T}\right)$

admits extra Noether point symmetries as follows

a) If $f\left(T\right)$ is an arbitrary function of $T$, then the symmetry
vector is written as

$X=\left(2\psi_{0}\tau+c_{1}\right)\partial_{\tau}+\eta^{i}\left(x^{k}\right)\partial_{i}$

where $\eta^{i}\left(x^{k}\right)$ is a HV/KV of the metric
$\bar{\gamma}_{ij}$ and the following condition holds

$M_{,k}\eta^{k}+2\bar{\psi}_{0}M=0.$

b) If $f\left(T\right)$ is a power law, i.e. $f\left(T\right)=T^{n}$, then we
have the extra symmetry vector

$X=\left(2\bar{\psi}_{0}\tau\right)\partial_{\tau}+\eta^{i}\left(x^{k}\right)\partial_{i}+\frac{2\bar{\psi}_{0}-2\bar{\psi}\left(x^{k}\right)}{C}T\partial_{T}$

where $C=\frac{n}{1-n}$, $\eta^{i}\left(x^{k}\right)$ is a CKV of the metric
$\bar{\gamma}_{ij}$ with conformal factor $\bar{\psi}\left(x^{k}\right)$ and
the following condition holds

$M_{,k}\eta^{k}+\left[2\bar{\psi}+\left(1-C\right)S\right]M=0$

where $S=2\left(\bar{\psi}_{0}-\bar{\psi}\right)$.

In both cases the corresponding gauge function is a constant.

In the following we apply the above Lemma in the case of FRW cosmology and
static spherical symmetric spacetimes.

####  11.6 Spatially flat FRW

The FRW in the holonomic (commoving) frame $\\{\partial t,\partial x,\partial
y,\partial z\\}$ has the form

$ds^{2}=-dt^{2}+a^{2}(t)(dx^{2}+dy^{2}+dz^{2})$

where $a(t)$ is the cosmological scale factor. In this spacetime we define the
vierbein (unholonomic frame) $\\{e_{i}\\}$ with the requirement:

$h_{\mu}^{i}(t)=\mathrm{diag}(-1,a(t),a(t),a(t)).$ (11.50)

In order to derive the cosmological equations in a FRW metric, we need to
deduce a point-like Lagrangian from the action (11.20). As a consequence, the
infinite number of degrees of freedom of the original field theory will be
reduced to a finite number. In this framework considering $(a,T)$ as canonical
variables the corresponding $f(T)$ Lagrangian becomes:

$\mathcal{L}=a^{3}\left[f(T)-Tf^{\prime}(T)\right]-6\dot{a}^{2}af^{\prime}(T)\,.$
(11.51)

Therefore the field equations are

$T=-6\left(\frac{\dot{a}^{2}}{a^{2}}\right)=-6H^{2}$ (11.52)
$12H^{2}f^{\prime}(T)+f(T)=16\pi G\rho\;$ (11.53)
$48H^{2}f^{\prime\prime}(T)\dot{H}-f^{\prime 2}+4\dot{H}-f(T)=16\pi Gp$
(11.54)

where $H$ is the Hubble parameter, $\rho=\rho_{m}+\rho_{r}$ and
$p=p_{m}+p_{r}$ are the total energy density and (isotropic) pressure
respectively, which are measured in the unholonomic frame. It is interesting
to mention that using the conservation equation $\dot{\rho}+3H(\rho+p)=0$ one
can rewrite equations (11.53) and (11.54) in the usual form

$H^{2}=\frac{8\pi G}{3}(\rho+\rho_{T})$ (11.55) $2\dot{H}+3H^{2}=-\frac{8\pi
G}{3}(p+p_{T})$ (11.56)

where

$\rho_{T}=\frac{1}{16\pi G}[2Tf^{\prime}(T)-f(T)-T]$ (11.57)
$p_{T}=\frac{1}{16\pi
G}[2\dot{H}(4Tf^{\prime\prime}(T)+2f^{\prime}(T)-1)]-\rho_{T}.$ (11.58)

are the unholonomicity contributions to the energy density and pressure.
Finally, a basic question here is the following: under which circumstances
$f(T)$ gravity can resemble that of the scalar field dark energy? In order to
address this crucial question we need to calculate the effective equation-of-
state parameter $w(a)$ for the $f(T)$ cosmology. Indeed, utilizing equations
(11.57) and (11.58), we can easily obtain the effective unholonomicity
equation of state as

$\omega_{T}\equiv\frac{p_{T}}{\rho_{T}}=-1+\frac{4\dot{H}(4Tf^{\prime\prime}(T)+2f^{\prime}(T)-1)}{4Tf^{\prime}(T)-2f(T)-T}\;.$
(11.59)

#####  11.6.1 Noether symmetries

From Lemma 11.5.1 for the Lagrangian (11.51) we find that for

  * •

For arbitrary $f\left(T\right)$ the Lagrangian (11.51) admits only the Noether
symmetry $\partial_{t}$

  * •

For $f\left(T\right)=f_{0}T^{n}~{}$where $f_{0}$ is the integration constant
we have the following extra Noether symmetries:
\- For $n\neq\frac{1}{2},\frac{3}{2}$ the Noether point symmetry vector is

$\displaystyle X_{1}$ $\displaystyle=$
$\displaystyle\left(\frac{3C}{2n-1}t\right)\partial_{t}+\left(Ca+c_{3}a^{1-\frac{3}{2n}}\right)\partial_{a}+$
$\displaystyle+\left[\frac{1}{n}\left(\left(c-m\right)n+3c_{3}a^{-\frac{3}{2n}}\right)+\frac{3C}{2n-1}+c\right]T\partial_{T}$

with corresponding Noether integral

$I_{1}=\left(\frac{3C}{2n-1}t\right)\mathcal{H}-12f_{0}n\left(Ca^{2}+c_{3}a^{2-\frac{3}{2n}}\right)T^{n-1}\dot{a}$

where $C=\frac{m\left(1-n\right)+nc}{3}$.
\- For $n=\frac{3}{2},$ the Noether point symmetry is

$\displaystyle X_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{5}\left(3c-2m\right)t\partial_{t}+\left[\left(\frac{c}{2}-\frac{m}{6}\right)a+c_{4}\right]\partial_{a}+$
(11.60)
$\displaystyle+\left[\left(m+11c\right)-\frac{c_{4}}{a}+\frac{2}{5}\left(8c-2m\right)\right]T\partial_{T}$

with corresponding Noether integral

$I_{2}=\frac{1}{5}\left(3c-2m\right)t\mathcal{H}-18f_{0}\left[\left(\frac{c}{2}-\frac{m}{6}\right)a^{2}+c_{4}a\right]T^{\frac{1}{2}}\dot{a}\;.$

\- For $n=\frac{1}{2}$ the Noether point symmetry becomes

$X_{3}=c_{1}t\partial_{t}+\left(-2c_{1}+c_{3}a^{\frac{1}{4}}\right)\partial_{a}+\left(4c_{1}+c_{2}+\frac{3c_{3}}{2}a^{-\frac{3}{4}}\right)T\partial_{T}$

with Noether integral

$I_{3}=c_{1}t\mathcal{H}-6f_{0}\left(-2c_{1}a+c_{3}a^{\frac{3}{4}}\right)T^{-\frac{1}{2}}\dot{a}.$

We would like to stress that our results are in agreement with those of [147]
but they are richer because we have considered the term $\xi\partial_{t}$ in
the generator which is not done in [147]. To this end it becomes evident that
$f\left(T\right)=f_{0}T^{n}$ is the only form that admits extra Noether point
symmetries implying the existence of exact analytical solutions.

#####  11.6.2 Exact cosmological solutions

In this section we proceed in an attempt to analytically solve the basic
cosmological equations of the $f\left(T\right)=f_{0}T^{n}$ gravity model. In
particular from the Lagrangian (11.51) we obtain the main field equation

$\ddot{a}+\frac{1}{2a}\dot{a}^{2}+\frac{f^{\prime\prime}}{f^{\prime}}\dot{a}\dot{T}-\frac{1}{4}a\frac{f^{\prime}T-f}{f^{\prime}}=0\;.$
(11.61)

Also differentiating equation (11.52) we find

$\dot{T}=12\left[\left(\frac{\dot{a}}{a}\right)^{3}-\frac{\dot{a}\ddot{a}}{a^{2}}\right]\;.$
(11.62)

Finally, inserting $f\left(T\right)=f_{0}T^{n}$, $H=\dot{a}/a$, equation
(11.52) and equation(11.62) into equation(11.61) we derive after some algebra
that

$\left(2n-1\right)\left[\ddot{a}-\frac{\dot{a}^{2}}{2a}\frac{\left(2n-3\right)}{n}\right]=0$
(11.63)

a solution of which is

$a(t)=a_{0}t^{2n/3}\;\;\;\;H(t)=\frac{2n}{3t}$ (11.64)

or

$H=H_{0}a^{-3/2n}=H_{0}(1+z)^{3/2n}$ (11.65)

where $n\in\mathcal{R}_{+}^{\star}-\\{\frac{1}{2}\\}$, $a(z)=(1+z)^{-1}$ and
$H_{0}$ is the Hubble parameter. We note that the above analytic solution
confirms that of [147].

From equation(11.64) it is evident that this cosmological models have no
inflection point (that is the deceleration parameter does not change sign).
Therefore, the main drawback of the $f(T)=f_{0}T^{n}$ gravity model is that
the deceleration parameter preserves sign, and therefore the universe always
accelerates or always decelerates depending on the value of $n$. Indeed, if we
consider $n=1$ (TEGR) then the above solution boils down to the Einstein de
Sitter model as it should. On the other hand, the accelerated expansion of the
universe ($q<0$) is recovered for $n>\frac{3}{2}$. The latter means that even
if we admit $n>\frac{3}{2}$ as a mere phenomenological possibility, we would
be also admitting that the universe has been accelerating forever, which is of
course difficult to accept.

#####  11.6.3 Cosmological analogue to other models

In this section (assuming flatness) we present the cosmological equivalence at
the background level between the current $f(T)$ gravity with $f(R)$ modified
gravity and dark energy, through a specific reconstruction of the $f(R)$ and
vacuum energy density namely, $f(R)=R^{n}$ and $\Lambda(H)=3\gamma H^{2}$. In
particular, in the case of $f(R)=R^{n}$ it has been found (see chapter 10)
that the corresponding scale factor obeys equation (11.64), where
$n\in\mathcal{R}_{+}^{\star}-\\{2,\frac{3}{2},\frac{7}{8}\\}$.222The
Lagrangian here is
$\mathcal{L}_{R}=6naR^{n-1}\dot{a}^{2}+6n(n-1)a^{2}R^{n-2}\dot{a}\dot{R}+(n-1)a^{3}R^{n}$,
where $R$ is the Ricci scalar. For $n=1$ the solution of the Euler-Lagrange
equations is the Einstein de-Sitter model [$a(t)\propto t^{2/3}$] as it
should. Note, that for $n=2$ one can find a de-Sitter solution [$a(t)\propto
e^{H_{0}t}$, see chapter 10].

On the other hand, considering a spatially flat FRW metric and in the context
of GR the combination of the Friedmann equations with the total
(matter+vacuum) energy conservation in the matter dominated era, provides (for
more details see [183, 184])

${\dot{H}}+\frac{3}{2}H^{2}=\frac{\Lambda}{2}\;.$ (11.66)

Solving equation (11.66) for $\Lambda(H)=3\gamma H^{2}$ (see [185, 186, 187])
we end up with

$H=H_{0}a^{-3(1-\gamma)/2}=H_{0}(1+z)^{3(1-\gamma)/2}\;.$ (11.67)

Now, comparing equations (11.65), (11.67) and connecting the above
coefficients as $n^{-1}=1-\gamma$, we find that the $f(T)=f_{0}T^{n}$ and the
flat $\Lambda(H)=3\gamma H^{2}$ models can be viewed as equivalent cosmologies
as far as the Hubble expansion is concerned, despite the fact that the time
varying vacuum model is inside GR. However, when the $\Lambda(H)=3\gamma
H^{2}$ cosmological model is confronted with the current observations it
provides a poor fit [183, 184]. Because the current time varying vacuum model
shares exactly the same Hubble parameter with the $f(T)=f_{0}T^{n}$ gravity
model, it follows that the latter is also under observational pressure when
compared against the background cosmological data. The same observational
situation holds also for the $f(R)=R^{n}$ modified gravity.

####  11.7 Static spherically symmetric spacetimes

We apply now the results of the general Noether analysis of the previous
subsection, to the specific case of static spherically-symmetric geometry
given by the metric (11.28), that is the vierbein (11.29). Armed with the
general expressions provided above, we can deduce the Noether algebra of the
metric (11.28).

In this metric the Lagrangian (11.23) and the Hamiltonian (11.24) become

$L=2f_{T}N\left(2ba^{\prime}b^{\prime}+ab^{\prime
2}\right)+M(a,b)\left(f-f_{T}T\right)$ (11.68)
$H=2f_{T}N\left(2ba^{\prime}b^{\prime}+ab^{\prime
2}\right)-M(a,b)\left(f-f_{T}T\right)\equiv 0\;$ (11.69)

where $M(a,b)$ is given by (11.30). As one can immediately deduce, TEGR and
thus General Relativity is restored when $f(T)=T$, while if $N=1$, $\tau=r$
and $ab=1$ we fully recover the standard Schwarzschild solution.

Applying Lemma 11.5.1 in the case of static spherically-symmetric geometry, we
determine all the functional forms of $f(T)$ for which the above dynamical
system admits Noether point symmetries beyond the trivial one
$\partial_{\tau}.$ We summarize the results in Tables 11.1, 11.2 and 11.3.
Furthermore, we can use the obtained Noether integrals in order to classify
the analytic solutions for each case.

Table 11.1: Noether Symmetries and Noether Integrals for arbitary f(T) $\mathbf{N}\left(a,b\right)$ | Noether Symmetry | Noether Integral
---|---|---
$\frac{1}{a^{3}}N_{1}\left(a^{2}b\right)$ | $X_{1}=-\frac{a}{2b^{3}}\partial_{a}+\frac{1}{b^{2}}\partial_{b}$ | $I_{1}=\frac{N_{1}\left(a^{2}b\right)}{2a^{3}b^{2}}\left(2ba^{\prime}+ab^{\prime}\right)f_{T}$
$N_{2}\left(b\sqrt{a}\right)$ | $X_{2}=-2a\partial_{a}+b\partial_{b}$ | $I_{2}=N_{2}\left(b\sqrt{a}\right)\left(b^{2}a^{\prime}-abb^{\prime}\right)f_{T}$
$aN_{3}\left(b\right)$ | $X_{3}=\frac{1}{ab}\partial_{a}~{}$ | $I_{3}=N_{3}\left(b\right)b^{\prime}f_{T}$

In the case of $f\left(T\right)=T^{n}$ we have the additional extra Noether
symmetries of Table 11.1

Table 11.2: Extra Noether Symmetries and Noether Integrals for $f(T)=T^{n}$ $\mathbf{N}\left(a,b\right)$ | Noether Symmetry333Where $\bar{\psi}_{5-7}$ are the conformal factors of the generators of the symmetry vectors $\eta^{i}$, i.e. $\psi=\frac{1}{\dim\gamma_{ij}}\eta_{;i}^{i}$. | Noether Integral
---|---|---
arbitrary | $X_{4}=2\bar{\psi}_{0}\tau+\frac{2\bar{\psi}_{0}\left(C-1\right)}{2C+1}a\partial_{a}+\frac{2\bar{\psi}_{0}-2\bar{\psi}_{4}}{C}T\partial_{T}$ | $I_{4}=2\psi_{0}n\frac{C-1}{1+2C}abN\left(a,b\right)T^{n-1}b^{\prime}~{}$
arbitrary | $X_{5}=-2a\partial_{a}+b\partial_{b}-\frac{2\bar{\psi}_{5}}{C}T\partial_{T}$ | $I_{5}=nN\left(a,b\right)T^{n-1}\left(b^{2}a^{\prime}-abb^{\prime}\right)$
arbitrary | $X_{6}=-\frac{a}{2}b^{-\frac{3\left(1+2C\right)}{4C}}\partial_{a}+b^{-\frac{3+2C}{4C}}\partial_{b}-\frac{2\bar{\psi}_{6}}{C}T\partial_{T}$ | $I_{6}=\frac{n}{2}N\left(a,b\right)T^{n-1}\left(2b^{\frac{2C-3}{4C}}a^{\prime}+ab^{-\frac{3+2C}{4C}}b^{\prime}\right)$
arbitrary | $X_{7}=a^{-\frac{1}{2C}}b^{-\frac{1+2C}{4C}}\partial_{a}-\frac{2\bar{\psi}_{7}}{C}T\partial_{T}~{}$ | $I_{7}=N\left(a,b\right)na^{-\frac{1}{2C}}b^{-\frac{1+2C}{4C}}T^{n-1}b^{\prime}$
Table 11.3: Extra Noether Symmetries and Noether Integrals for $f(T)=T^{\frac{1}{2}}$ $\mathbf{N}\left(a,b\right)$ | Noether Symmetry | Noether Integral
---|---|---
arbitrary | $\bar{X}_{4}=2\bar{\psi}_{0}\tau+\frac{3\bar{\psi}_{0}}{2}a\ln\left(a^{2}b\right)\partial_{a}+\frac{2\bar{\psi}_{0}-2\bar{\psi}_{4}^{\prime}}{C}T\partial_{T}$ | $\bar{I}_{4}=\frac{3}{2}\psi_{0}N\left(a,B\right)T^{-\frac{1}{2}}ab\ln\left(a^{2}b\right)b^{\prime}$
arbitrary | $\bar{X}_{5}=b\partial_{b}-\frac{2\bar{\psi}_{5}}{C}T\partial_{T}$ | $\bar{I}_{5}=\frac{1}{2}N\left(a,b\right)T^{-\frac{1}{2}}\left(b^{2}a^{\prime}+abb^{\prime}\right)$
arbitrary | $\bar{X}_{6}=-a\ln\left(ab\right)\partial_{a}+b\ln b\partial_{b}-\frac{2\bar{\psi}_{6}}{C}T\partial_{T}$ | $\bar{I}_{2}=\frac{1}{2}N\left(a,B\right)T^{-\frac{1}{2}}b\left(b\ln b~{}a^{\prime}-a\ln a~{}b^{\prime}\right)$
arbitrary | $\bar{X}_{7}=a\partial_{a}-\frac{2\bar{\psi}_{7}}{C}T\partial_{T}~{}$ | $\bar{I}_{3}=\frac{1}{2}N\left(a,b\right)T^{-\frac{1}{2}}ab~{}b^{\prime}$

#####  11.7.1 Exact Solutions

Using the Noether symmetries and the corresponding integral of motions
obtained in the previous section, we can extract all the static spherically-
symmetric solutions of $f(T)$ gravity. Without loss of generality, we choose
the conformal factor $N(a,b)$ as $N(a,b)=ab^{2}$ [or equivalently444Since the
space is empty, the field equations are conformally invariant, therefore the
resutls are similar for an arbitrary function $N(a,b)~{}$(see chapter 9)
$M(a,b)=1$]. In order to simplify the current dynamical problem, we consider
the coordinate transformation

$b=\left(3y\right)^{\frac{1}{3}}~{}~{}\;\;a=\sqrt{\frac{2x}{\left(3y\right)^{\frac{1}{3}}}}\;.$
(11.70)