The discovery of the Higgs particle has confirmed that scalar fields play a fundamental
role in subatomic physics. Therefore they must also have been present in the early Universe and played a part
in its development. About scalar fields on present cosmological scales nothing is known, but in view of the observational evidence for accelerated expansion it is quite well possible that they take part
in shaping our Universe now and in the future. In this section we consider the evolution of a flat, isotropic and homogeneous Universe in the presence of a single cosmic
scalar field. Neglecting ordinary matter and radiation, the evolution of such a Universe is described by two
degrees of freedom, the homogeneous scalar field $\varphi(t)$ and the scale factor of the Universe $a(t)$. The
relevant evolution equations are the Friedmann and Klein-Gordon equations,
reading (in the units in which $c = \hbar = 8 \pi G = 1$)
\[
\frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2, \quad \ddot{\varphi} + 3 H \dot{\varphi} + V' = 0,
\]
where $V[\varphi]$ is the potential of the scalar fields, and $H = \dot{a}/a$ is the Hubble parameter.
Furthermore, an overdot denotes a derivative w.r.t.\ time, whilst a prime denotes a derivative w.r.t.\ the
scalar field $\varphi$.
Show that the Hubble parameter cannot increase with time in the single scalar cosmology.
solution
Let the scalar field $\varphi(t)$ is a single-valued function of time, then
it is possible to reparametrize the Hubble parameter in terms of $\varphi$:
\[
H(t) = H[\varphi(t)].
\]
Taking time derivatives in the Friedman equation
\[
\frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2,
\]
one arrives at the results:
\[
\dot{\varphi} ( \ddot{\varphi} + V' ) = 6 H \dot{H},\quad \dot{H} \equiv H' \dot{\varphi}.
\]
Taking into account the Klein Gordon equation
\[
\ddot{\varphi} + 3 H \dot{\varphi} + V' = 0,
\]
it follows, that for $\dot{\varphi} \neq 0$ and $H \neq 0$ one gets
\[
\dot{\varphi} = - 2 H', \quad \dot{H} = - \frac{1}{2}\, \dot{\varphi}^2 \leq 0.
\]
Thus the Hubble parameter is a semi-monotonically decreasing function of time.
Show that if the Universe is filled by a substance which satisfies the null energy condition then the Hubble parameter is a semi-monotonically decreasing function of time.
solution
\[\dot H=-\frac12(\rho+p).\]
If $\rho+p\ge0$ (null energy condition), $H$ is a semi-monotonically decreasing function of time.
For single-field scalar models express the scalar field potential in terms of the Hubble parameter and its derivative with respect to the scalar field.
solution
For single-field models in which the scalar field is a single-valued function of time in some interval, it is possible to reparametrize the Hubble parameter in terms of $\varphi$:
\[H(t)=H[\varphi(t)]\]
Replacing the time derivatives $\dot{\varphi} = - 2 H'$ (see the problem 1) in the Friedmann equation we find
\[V=3H^2-2H'^2.\]
The latter expression can be used to reconstruct the potential if the evolution history $H[\varphi(t)]$ is known, or for given $V(\varphi)$ this is a first-order differential equation for $H[\varphi(t)]$.
Obtain first-order differential equation for the Hubble parameter $H$ as function of $\varphi$ and find its stationary points.
solution
Replacing the time derivatives in the Friedmann equation using the results of the previous problem, one finds
\[
2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0.
\]
There are two kinds of stationary points; a point where $\dot{\varphi} = H' = 0$ is an end point of the evolution if
\[
\ddot{\varphi} = 4 H' H'' = 0,
\]
which happens if $H''$ is finite. In contrast, if
\[
\ddot{\varphi} = 4 H' H'' \neq 0,
\]
$H''$ necessarily diverges in such a way as to make $\ddot{\varphi}$ finite: $H'' \propto 1/H'$.
Consider eternally oscillating scalar field of the form $\varphi(t) = \varphi_0 \cos \omega t$ and analyze stationary points in such a model.
solution
For such a scalar field to exist it is required that
\[
H' = - \frac{1}{2}\, \dot{\varphi} = \frac{\omega \varphi_0}{2} \sin \omega t = \frac{\omega}{2} \sqrt{\varphi_0^2 - \varphi^2}.
\]
There are infinitely many stationary points
\[
\omega t_n = n \pi, \quad \varphi(t_n) = (-1)^n \varphi_0,
\]
where $H' = 0$. Now
\[
H'' = - \frac{1}{2} \frac{\omega \varphi}{\sqrt{\varphi_0^2 - \varphi^2}},
\]
and therefore $H''$ diverges at all stationary points $t_n$, but in such a way that
\[
4 H' H'' = - \omega^2 \varphi = \ddot{\varphi}.
\]
Then all stationary points in the considered model are turning points.
Obtain explicit time dependence for the scale factor in the model of problem #SSC_2.
solution
The corresponding solution for the scale factor is
\[
a(t) = a(0) \exp\left\{H_0 t - \frac{1}{8} \omega^2 \varphi_0^2 t^2 + \frac{1}{16} \left( 1 - \cos 2 \omega t \right)\right\}.
\]
which is a gaussian, slightly modulated by an oscillating function of time (see figure).
Scalefactor $a(t)$ for an eternally oscillating scalar field.
Reconstruct the scalar field potential $V(\varphi)$ needed to generate the model of problem #SSC_2.
solution
The potential giving rise to this behavior reads
\begin{align}
V & = 3 H^2 - 2 H^{\prime\, 2} \\
& = 3 \left( H_0 - \frac{1}{4}\, \omega \varphi_0^2 \arccos \left( \frac{\varphi}{\varphi_0} \right) + \frac{1}{4} \omega \varphi
\sqrt{ \varphi_0^2 - \varphi^2} \right)^2 - \frac{\omega^2}{2} \left( \varphi_0^2 - \varphi^2 \right).
\end{align}
Observe, that this potential keeps track of the number of oscillations the scalar field has performed through
the arccos-function, so ultimately $V$ increases indefinitely as a function of time, whilst the volume of a
representative domain of space decreases rapidly.
Describe possible final states for the Universe governed by
a single scalar field at large times.
solution
If $H$ becomes negative then the Universe inevitably collapses.If $H$ never becomes negative, it must tend to a vanishing or positive final minimum, which can be reached either in finite or infinite time. The universe then ends up in a Minkowski or in a de Sitter state. These conclusions are a consequence of the non-positivity of $\dot{H}$ (see problem #SSC_0, which implies that a negative $H$ can never return to larger values at later times.
Formulate conditions for existence of end points of evolution in terms of the potential $V(\varphi)$.
solution
In order to establish the existence of end points or asymptotic end points of evolution at non-negative
values of $H$, we first consider the locus of all stationary points, defined by
\[
\dot{\varphi} = - 2 H' = 0 \quad \Rightarrow \quad V = 3 H^2 \geq 0 .
\]
It follows that stationary points can occur only in the region of positive or vanishing potential.
In particular this holds for end points, which therefore do not occur in a region of negative potential.
Moreover, it is clear that a Minkowski final state occurs only at a stationary point where $V = 0$, whereas
all stationary points with $V > 0$ correspond to de Sitter states. To correspond to an end point of the
evolution, $H''$ must be finite at these stationary points to guarantee that $\ddot{\varphi} = 0$ as well.
From the results of the problem 4 it follows that
\[
V' = 2 H' (3 H - 2 H''),
\]
and therefore $V' = 0$ if $H' = 0$ and $H$ and $H''$ are finite. As a result end points of
the evolution necessarily occur at an extremum of $V$, but only if $V \geq 0$ there.
Consider a single scalar cosmology described by the quadratic potential
\[
V = v_0 + \frac{m^2}{2}\, \varphi^2.
\] Describe all possible stationary points and final states of the Universe in this model.
solution
We distinguish the cases $v_0 > 0$, $v_0 = 0$ and $v_0 < 0$. The stationary points
are represented graphically by the curves in the $\varphi$-$H$-plane in figure.
a)
b)
c)
Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (a)), $v_0 = 0$ (b)) and $v_0 < 0$ (c)).
For $v_0 > 0$ there exists a stationary point for any value of $\varphi$, but the potential has a unique
minimum at $\varphi = 0$, which is the only stationary point where $V' = 0$, and therefore the only end point.
Indeed, once this point is reached $H$ can not decrease anymore and we have final state of de
Sitter type.
For $v_0 = 0$ the critical curves become straight lines, crossing at the origin where $H = 0$ at
$V = 0$. This is still a stationary point with $\ddot{\varphi} = 0$ representing a Minkoswki state, but as $V'$ is not defined there it is really to be interpreted as a limit of the previous case. There are no evolution curves flowing from the domain $H > 0$ to the domain $H < 0$.
For $v_0 < 0$ there are no stationary points in the region $\varphi^2 < 2 |v_0| /m^2$, and the solutions can cross into the domain of negative $H$ there.
Obtain actual solutions for the model of previous problem using the power series expansion
\[
H[\varphi] = h_0 + h_1 \varphi + h_2 \varphi^2 + h_3 \varphi^3 + ...
\] Consider the cases of $v_0 > 0$ and $v_0 < 0$.
solution
Substitution into equation $2 H^{\prime\, 2} - 3 H^2 + V(\varphi) = 0
$ then leads to the equalities
\[
3 h_0^2 - 2 h_1^2 = v_0, \quad h_1(3h_0 - 4 h_2) = 0, \quad 4h_1(h_1 - 4 h_3)
+ \frac{8}{3}\, h_2 ( 3 h_0 - 4 h_2 ) = m^2, \quad ...,
\]
from which the solutions $H[\varphi]$ can be reconstructed (see figure below).
Solutions $H[\varphi]$ for quadratic potentials (problem #SSC_6_1) with $v_0 > 0$.Solutions $H[\varphi]$ for quadratic potentials (problem #SSC_6_1) with $v_0 < 0$.
Estimate main contribution to total expansion factor of the Universe.
solution
Using the results of previous problem, one can define the number of $e$-folds in some interval of time:
\[
N = \int_{t_1}^{t_2} dt H = - \int_{\varphi_1}^{\varphi_2} d\varphi\, \frac{H}{2H'} = - \frac{1}{2}\, \int_{\varphi_1}^{\varphi_2} d\varphi\,
\frac{h_0 + h_1 \varphi + h_2 \varphi^2 + ...}{h_1 + 2 h_2 \varphi + 3 h_3 \varphi^2 + ...}.
\]
This number can get sizeable contributions only in regions where the slow-roll condition is satisfied:
\[
\varepsilon = - \frac{\dot{H}}{H^2} = \frac{2H^{\prime\, 2}}{H^2} < 1 \quad \Rightarrow \quad
3H^2 - V < H^2.
\]
Thus we simultaneously have
\[
V < 3 H^2 \quad \mbox{and} \quad V > 2H^2 \quad \Leftrightarrow \quad
0 \leq \frac{V}{3} < H^2 < \frac{V}{2}.
\]
In most cases this holds only for a relatively narrow range of field values.
Explain difference between end points and turning points of the scalar field evolution.
solution
In both cases $\dot{\varphi} = 0$, but at end points in addition $\ddot{\varphi} = 0$, which can happen only at extrema of the
potential $V[\varphi]$. However, if the end point occurs at a relative maximum or saddle point of the potential, this end
point will be classically unstable. Indeed, the field can remain there for an indefinite period of time, but any slight
change in the initial conditions will cause it to move on to lower values of the Hubble parameter. Nevertheless, such
a period of temporary slow roll of the field creates the right conditions for a period of inflation.
Show that the exponentially decaying scalar field
\[
\varphi(t) = \varphi_0 e^{-\omega t}
\]
can give rise to unstable end points of the evolution.
solution
The Hubble parameter and potential giving rise to this solution can be constructed following the same procedure as for the eternally oscillating field (see problem #SSC_2), with the result
\[
H = h + \frac{1}{4}\, \omega \varphi^2, \quad V[\varphi] = v_0 - \frac{\mu^2}{2}\, \varphi^2 + \frac{\lambda}{4}\, \varphi^4,
\]
where
\[
v_0 = 3 h^2, \quad
\mu^2 = \omega^2 - 3 \omega h, \quad \lambda = \frac{3 \omega^2}{4}.
\]
Thus we obtain a quartic potential; for $\mu^2 > 0$ it has minima in which reflection symmetry is spontaneously
broken. The exponential solution ends asymptotically at the unstable maximum of the potential where
$\dot{\varphi} = \ddot{\varphi} = 0$. As such it represents an end point of the evolution, but a minimal change in the
initial conditions for the scalar field will turn the end point into a reflection point (if it starts at lower $H$), or
it will overshoot the maximum (if it starts at higher $H$). Thus the end point is unstable, but the exponential
decay will still provide a good approximation to first part of the evolution of the universe for
solutions $H[\varphi]$ coming close to the maximum of the potential (see figure below).
Critical curves of stationary points and solutions $H[\varphi]$ for a quartic potential with spontaneous symmetry breaking.
Analyze all possible final states in the model of previous problem.
solution
The exponential scalar field leads to a behavior of the scale factor given by
\[
a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}.
\]
Thus for $h > 0$ this epoch in the evolution of the Universe ends in an asymptotic de Sitter state with
Hubble constant $h$. Afterwards, the scalar field will roll further down the potential; provided $3h \leq \omega \leq 6h$
it will oscillate around the minimum until it comes to rest in another de Sitter or a Minkowski state, again depending
on the value of $h$. In particular, for $\omega \geq 3h$ the model has a final de Sitter or Minkowski state in which
$\dot{\varphi} = 0$ and
\[
\langle \varphi^2 \rangle = \frac{\mu^2}{\lambda} = \frac{4}{3} \left(1 - \frac{3h}{\omega} \right).
\]
In this final state the energy density is
\[
\langle V \rangle = v_0 - \frac{\mu^4}{4\lambda} = \frac{\omega}{3} ( 6h - \omega ).
\]
Express initial energy density of the model of problem #SSC_9 in terms of the $e$-folding number $N$.
solution
The energy density for the solution of problem #SSC_9 is
\[
\rho_s(t) = \frac{1}{2}\, \dot{\varphi}^2 + V = 3 H^2 = 3 \left( h + \frac{1}{4}\, \omega\, \varphi_0^2\, e^{-2 \omega t} \right)^2.
\]
Now the solution for the scale factor
\[
a(t) = a_0\, e^{ht + \frac{1}{8} \varphi^2_0\, (1 - e^{-2\omega t})}.
\]
shows, that before reaching the first turning point
at $\varphi = 0$ the scale factor increases by an additional number of $e$-folds given by
\[
N = \frac{1}{8}\, \varphi_0^2.
\]
Therefore the initial energy density at $t = 0$ can be written as
\[
\rho_s(0) = 3 ( h + 2N \omega )^2.
\]
If we take this initial energy density to equal the Planck density: $\rho_s(0) = 1$, this establishes a
simple relation between $h$, $\omega$ and $N$.
Estimate mass of the particles corresponding to the exponential scalar field considered in problem #SSC_9.
solution
Taking the final energy density $\langle V \rangle$ (see problem #SSC_10) equal to the observed energy density of
the Universe today:
\[
\langle V \rangle = 3 H_0^2 = 1.04 \times 10^{-120}
\]
in Planck units, being so close to zero, one can set to an extremely good approximation $\omega = 6h$,
and
\[
3 h^2 ( 1 + 12 N )^2 = 1, \quad \mu^2 = 18 h^2, \quad \lambda = 27 h^2.
\]
The lower limit on $N$ for inflation as derived from the CMB observations is $N \geq 60$, which requires
\begin{equation}
h \leq 0.8 \times 10^{-3}.\label{h_estimate}
\end{equation}
Now expanding $\varphi$ around its vacuum expectation value
\[
\varphi = \frac{\mu}{\sqrt{\lambda}} + \chi,
\]
the potential becomes
\[
V = \frac{1}{2}\, m_{\chi}^2 \chi^2 + \frac{\alpha}{3}\, m_{\chi} \chi^3 + \frac{\lambda}{4}\, \chi^4,
\]
where
\[
m_{\chi} = 6h, \quad
\alpha = 9h , \quad \lambda = 27 h^2.
\]
According to the estimate (\ref{h_estimate}) the upper limits on these parameters are
\[
m_{\chi} = 0.48 \times 10^{-2}, \quad \alpha = 0.72 \times 10^{-2}\approx1/137, \quad \lambda = 0.17 \times 10^{-4}.
\]
Converting to particle physics units, the upper limit on the mass is $m_{\chi} \leq 1.2 \times 10^{-16}$ GeV.
This suggests that the inflaton could be associated with a GUT scalar of Brout-Englert-Higgs type.
Calculate the deceleration parameter for flat Universe filled with the scalar field in form of quintessence.
solution
There are several ways to obtain the result:
1) \[q=-\frac{\ddot a }{aH^2};\quad \frac{\ddot a}{a}=-\frac16(\rho+3p);\quad H^2=\frac13\rho;\]
\[q=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\]
2) \[q=\frac12\Omega_{tot}+\frac32\sum\limits_iw_i\Omega_i.\] For a flat single-component Universe one obtains
\[q=\frac12+\frac32w=\frac12+\frac32\frac{\dot\varphi^2-2V}{\dot\varphi^2+2V}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V};\]
3) \[q=\frac{d}{dt}\frac 1 H-1=-\frac{\dot H}{H^2}-1=\frac{3H^2-V}{H^2}-1=\frac{2H^2-V}{H^2}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V}\]
When considering dynamics of scalar field $\varphi$ in flat Universe, let us define a function $f(\varphi)$ so that $\dot\varphi=\sqrt{f(\varphi)}$. Obtain the equation describing evolution of the function $f(\varphi)$. (T. Harko, F. Lobo and M. K. Mak, Arbitrary scalar field and quintessence cosmological models, arXiv: 1310.7167)
solution
By substituting the Hubble function
\[H^2=\frac13\left(\frac12\dot\varphi^2+V(\varphi)\right)\]
into the Klein-Gordon equation we obtain
\[\ddot\varphi+\sqrt3\sqrt{\frac12\dot\varphi^2+V(\varphi)}\dot\varphi + \frac{dV}{d\varphi}=0\]
Introducing $\dot\varphi=\sqrt{f(\varphi)}$ and changing the independent variable from $t$ to $\varphi$, transform last equation into
\[\frac12\frac{df(\varphi)}{d\varphi}+\sqrt3\sqrt{\frac12f(\varphi+V(\varphi)}\sqrt{f(\varphi)}+\frac{dV}{d\varphi}=0.\]
Obtain solution of the equation (\ref{fin}) with
\begin{equation}
V=V_{0}\exp \left( \sqrt{6}\alpha _{0}\phi \right). \label{pp}
\end{equation}
in the case $\alpha _0\neq \pm 1$.
Depicted is the time variation of the scale factor (in arbitrary units), in the first plot, and the time variation of the cosmological scalar field, in the second plot, with an exponential potential for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.
Plots of the time variation of the exponential scalar field potential, depicted in the fist figure, and the time variation of the deceleration parameter of the Universe filled with an exponential potential scalar field, depicted in the second figure, for different values of $\alpha _0$: $\alpha _0=1.5 $ (solid curve), $\alpha _0=2.5$ (dotted curve), $\alpha _0=3.5$ (short dashed curve), $\alpha _0=4.5$ (dashed curve), and $\alpha _0=5.5$ (long dashed curve), respectively. The arbitrary integration constants $\phi _0$ and $V_0$ have been normalized so that $\exp \left(-\sqrt{3/2}\alpha _0\phi _0\right)=\sqrt{3V_0}$.
Obtain solution of the equation (\ref{fin}) with
\begin{equation}
\frac{1}{2V}\frac{dV}{d\phi }=\sqrt{\frac{3}{2}}\;\alpha _{1}\,\tanh G,
\end{equation}
where $\alpha _{1}$ is an arbitrary constant.
solution
With this choice, the
evolution equation (\ref{fin}) takes the simple form
\begin{equation}
\frac{dG}{d\phi }=\sqrt{\frac{3}{2}}\left( 1+\alpha _{1}\right) ,
\end{equation}
with the general solution given by
\begin{equation}
G\left( \phi \right) =\sqrt{\frac{3}{2}}\left( 1+\alpha _{1}\right) \left(
\phi -\phi _{0}\right) ,
\end{equation}
where $\phi _{0}$ is an arbitrary constant of integration.
\begin{equation}
V\left( \phi \right) =V_{0}\cosh ^{\frac{2\alpha _{1}}{1+\alpha _{1}}}\left[
\sqrt{\frac{3}{2}}\left( 1+\alpha _{1}\right) \left( \phi -\phi _{0}\right)
\right] . \label{kkk}
\end{equation}
\begin{equation}
t-t_{0}=\frac{1}{\sqrt{2V_{0}}} \int \frac{d\phi }{\cosh ^{\frac{\alpha _1}{
1+\alpha _1}}\left[ \sqrt{\frac{3}{2}}\left( 1+\alpha _1\right) \left( \phi
-\phi _{0}\right) \right] \sinh \left[ \sqrt{\frac{3}{2}}\left( 1+\alpha _1
\right) \left( \phi -\phi _{0}\right) \right] }.
\end{equation}
\begin{equation}
a=a_{0}\sinh ^{\frac{1}{3\left( 1+\alpha _1\right) }}\left[ \sqrt{\frac{3}{2}
}\left( 1+\alpha _1\right) \left( \phi -\phi _{0}\right) \right] ,
\end{equation}
\begin{equation}
q=3\tanh ^{2}\left[ \sqrt{\frac{3}{2}}\left( 1+\alpha _1\right) \left( \phi
-\phi _{0}\right) \right] -1.
\end{equation}
Depicted is the variation of the generalized hyperbolic cosine scalar field potential as a function of $\phi $, in the first plot, and as the variation in time, in the second plot, for different values of $\alpha _1$: $\alpha _1=0.1 $ (solid curve), $\alpha _1=0.15$ (dotted curve), $\alpha _1=0.20$ (short dashed curve), $\alpha _1=0.25$ (dashed curve), and $\alpha _1=0.30$ (long dashed curve), respectively.
Depicted is the time variation of the scale factor, in the first plot, and of the deceleration parameter, in the second plot, of the Universe filled with a scalar field with a generalized hyperbolic cosine self-interaction potential for different values of $\alpha _1$: $\alpha _1=0.1 $ (solid curve), $\alpha _1=0.15$ (dotted curve), $\alpha _1=0.20$ (short dashed curve), $\alpha _1=0.25$ (dashed curve), and $\alpha _1=0.30$ (long dashed curve), respectively.
Obtain solution of the equation (\ref{fin}) for the case
\begin{equation}
G=\mathrm{arccoth}\left( \sqrt{\frac{3}{2}}\frac{\phi }{\alpha _{2}}\right)
,\qquad \alpha _{2}=\mathrm{constant}.
\end{equation}
solution
\begin{equation}
V\left( \phi \right) =V_{0}\left( \frac{\phi }{\alpha _{2}}\right)
^{-2\left( \alpha _{2}+1\right) }\left[ \frac{3}{2}\left( \frac{\phi }{
\alpha _{2}}\right) ^{2}-1\right] , \label{mmm}
\end{equation}
where $V_{0}$ is an arbitrary constant of integration.
\begin{equation}
\frac{\phi (t)}{\alpha _{2}}=\left[ \frac{\sqrt{2V_{0}}\left( \alpha
_{2}+2\right) }{\alpha _{2}}\right] ^{\frac{1}{\alpha _{2}+2}}\left(
t-t_{0}\right) ^{\frac{1}{\alpha _{2}+2}}.
\end{equation}
\begin{equation}
a=a_{0}\exp \left( \frac{\phi ^{2}}{4\alpha _{2}}\right) =
a_0\exp \left\{
\frac{1}{4\alpha _{2}}\left[ \frac{\left( \alpha _{2}+2\right) \sqrt{2V_{0}}
}{\alpha _{2}}\right] ^{\frac{2}{\alpha _{2}+2}}\left( t-t_{0}\right) ^{
\frac{2}{\alpha _{2}+2}}\right\} ,
\end{equation}
with $a_{0}$ an arbitrary constant of integration. The deceleration
parameter is given by
\begin{equation}
q=2\left( \frac{\phi }{\alpha _{2}}\right) ^{-2}-1=
2\left[ \frac{\sqrt{2V_{0}
}\left( \alpha _{2}+2\right) }{\alpha _{2}}\right] ^{-\frac{2}{\alpha _{2}+2}
}\left( t-t_{0}\right) ^{-\frac{2}{\alpha _{2}+2}}-1.
\end{equation}
Obtain a consistency integral relation between the
separation function $M(\phi )$ and the self-interaction potential $V(\phi )$, corresponding to the equations for the variable $w$, obtained in the previous problem.
solution
Equation~(\ref{n2}) can be integrated to provide
\begin{equation}
w=\sqrt{V}e^{-\alpha _{3}\phi }\left[ C_{0}+\int \frac{M\left( \phi \right)
e^{\alpha _{3}\phi }}{\sqrt{V}}d\phi \right] , \label{n4}
\end{equation}
where $C_{0}$ is an arbitrary constant of integration. Equation~(\ref{n3})
can be integrated to give
\begin{equation}
w=\frac{e^{-\alpha _{3}\phi }}{\sqrt{V}}\left[ C_{1}+3\int M\left( \phi
\right) V^{3/2}e^{3\alpha _{3}\phi }d\phi \right] ^{1/3}, \label{n5}
\end{equation}
where $C_{1}$ is an arbitrary constant of integration.
Then the consistency relation reads
\begin{equation}
C_{1}+3\int M\left( \phi \right) V^{3/2}e^{3\alpha _{3}\phi }d\phi =V^{3}
\left[ C_{0}+\int \frac{M\left( \phi \right) e^{\alpha _{3}\phi }}{\sqrt{V}}
d\phi \right] ^{3}. \label{n7}
\end{equation}
Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \phi \right) =\sqrt{V}$.
solution
The consistency relation (\ref{n7}) now reads
\begin{equation}
C_{1}+3\int V^{2}e^{3\alpha _{3}\phi }d\phi =V^{3}A\left( \phi \right) ,
\label{n8}
\end{equation}
where we have denoted $A(\phi )=\left( C_{0}+e^{\alpha _{3}\phi }/\alpha
_{3}\right) ^{3}$.
In order to solve the integral Eq.~(\ref{n8}), we rewrite
it as a linear first order differential equation for $V\left( \phi \right) $
\begin{equation}
\frac{dV}{d\phi }+\left[ \frac{d}{d\phi }\left( \ln A^{1/3}\right) \right] V=
\frac{e^{3\alpha _{3}\phi }}{A\left( \phi \right) }, \label{n9}
\end{equation}
with the general solution given by
\begin{equation}
V(\phi )=A^{-1/3}\left( \phi \right) \left[ C_{2}+\int e^{3\alpha _{3}\phi
}A^{-2/3}\left( \phi \right) d\phi \right] , \label{nn}
\end{equation}
where $C_{2}$ is an arbitrary constant of integration. Now by inserting $
A\left( \phi \right) $ into Eq.~(\ref{nn}) yields the expression of the
scalar field potential as
\begin{equation}
V(\phi )=\frac{\alpha _{3}^{2}\left\{ e^{3\alpha _{3}\phi }+2C_{0}\alpha
_{3}e^{2\alpha _{3}\phi }-C_{0}^{3}\alpha _{3}^{3}+\left( C_{0}\alpha
+e^{\alpha _{3}\phi }\right) ^{2}\left[ \frac{C_{2}}{\alpha _{3}}
-2C_{0}\alpha _{3}\ln \left| C_{0}\alpha _{3}+e^{\alpha _{3}\phi }\right|
\right] \right\} }{\left( C_{0}\alpha _{3}+e^{\alpha _{3}\phi }\right) ^{3}}.
\label{pot1}
\end{equation}
Therefore the general solution of Eq.~(\ref{fin}) is given
by
\begin{equation}
G=\mathrm{arccoth}\left( \frac{ 1+w^{2}} { 1-w^{2}}
\right) =\ln \left| \frac{1}{w}\right| ,
\end{equation}
where
\begin{equation}
w(\phi )=\sqrt{V(\phi )}
e^{-\alpha _{3}\phi }\left( C_{0}+\frac{e^{\alpha _{3}\phi }}{\alpha _{3}}\right) .
\end{equation}
Obtain exact solution of the equation (\ref{fin}) in the case $M\left( \phi \right) =V^{-3/2}$.
solution
The consistency relation (\ref{n7}) now reads
\begin{equation}
C_{0}+\int \frac{e^{\alpha _{3}\phi }}{V^{2}}d\phi =\frac{\left( C_{1}+\frac{
1}{\alpha _{3}}e^{3\alpha _{3}\phi }\right) ^{1/3}}{V}. \label{b1}
\end{equation}
In order to solve Eq.~(\ref{b1}), we rewrite it as a linear first order
differential equation for $V\left( \phi \right) $
\begin{equation}
\frac{dV}{d\phi }+\left[ \frac{d}{d\phi }\ln \left| \frac{1}{\left( C_{1}+
\frac{1}{\alpha _{3}}e^{3\alpha _{3}\phi }\right) ^{1/3}}\right| \right] V =-\frac{e^{\alpha _{3}\phi }}{\left( C_{1}+\frac{1}{\alpha _{3}}e^{3\alpha
_{3}\phi }\right) ^{1/3}}. \label{b2}
\end{equation}
Equation~(\ref{b2}) can be easily integrated, and yields the following
solution
\begin{equation}
V\left( \phi \right) =\left( C_{1}+\frac{1}{\alpha _{3}}e^{3\alpha _{3}\phi
}\right) ^{1/3}
\times \left[ C_{3}-\int \frac{e^{\alpha _{3}\phi }}{\left( C_{1}+
\frac{1}{\alpha _{3}}e^{3\alpha _{3}\phi }\right) ^{2/3}}d\phi \right] .
\label{pot2}
\end{equation}
where $C_{3}$ is an arbitrary constant of integration.
Therefore the general solution of Eq.~(\ref{fin}) is given
by
\begin{equation}
G=\mathrm{arccoth}\left( \frac{ 1+w^{2}}{1-w^{2}}
\right) =\ln \left| \frac{1}{w}\right| ,
\end{equation}
where
\begin{equation}
w(\phi )=\left[ \frac{e^{-\alpha
_{3}\phi }}{\sqrt{V(\phi )}}\right] \left( C_{1}+\frac{e^{3\alpha _{3}\phi }}{\alpha
_{3}}\right) ^{1/3}.
\end{equation}
