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= NEW Problems in Dark Energy Category =
 
= NEW Problems in Dark Energy Category =
  
==Single Scalar Cosmology==
 
[[Single Scalar Cosmology|'''Single Scalar Cosmology''']]
 
 
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 $\varphi$.
 
 
----
 
 
New subsection in this section [[Single_Scalar_Cosmology#Exact_Solutions_for_the_Single_Scalar_Cosmology| '''Exact Solutions for the Single Scalar Cosmology''']]
 
 
----
 
<div id="SSC_0"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 1'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_0</p>
 
Show that the Hubble parameter cannot increase with time in the single scalar cosmology.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_00"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 2'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_00</p>
 
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.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">\[\dot H=-\frac12(\rho+p).\]
 
If $\rho+p\ge0$ (null energy condition), $H$ is a semi-monotonically decreasing function of time.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_0_1"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 3'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_0_1</p>
 
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.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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 \ref{SSC_0}) 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 $H[\varphi(t)]$.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_1"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
 
'''Problem 4'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_1</p>
 
Obtain first-order differential equation for the Hubble parameter $H$ as function of $\varphi$ and find its stationary points.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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: $H'' \propto 1/H'$.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_2"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 5'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_2</p>
 
Consider eternally oscillating scalar field of the form $\varphi(t) = \varphi_0 \cos \omega t$ and analyze stationary points in such a model.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_3"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 6'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_3</p>
 
Obtain explicit solution for the Hubble parameter in the model considered in the previous problem.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">\begin{align}
 
H & =  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} \\
 
& = H_0 - \frac{1}{4} \omega^2 \varphi_0^2 t + \frac{1}{8} \omega \varphi_0^2 \sin 2 \omega t.
 
\end{align}</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_4"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 7'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_4</p>
 
Obtain explicit time dependence for the scale factor in the model of problem [[#SSC_2]].
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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).
 
[[File:Osc phi.png|center|thumb|400px|Scalefactor $a(t)$ for an eternally oscillating scalar field.]]</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_5"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 8'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_5</p>
 
Reconstruct the scalar field potential $V(\varphi)$ needed to generate the model of problem [[#SSC_2]].
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_6_00"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 9'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_6_00</p>
 
Describe possible final states for the Universe governed by
 
a single scalar field at large times.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_6_0"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 10'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_6_0</p>
 
Formulate conditions for existence of end points of evolution in terms of the potential $V(\varphi)$.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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 \ref{SSC_1} 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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_6_1"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 11'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_6_1</p>
 
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.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.
 
 
<gallery widths=400px heights=300px perrow=2 caption="Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (left), $v_0 = 0$ (middle) and $v_0 < 0$ (right).">
 
File:Eps gt 0v2.png|a)
 
File:Eps eq 0v2.png|b)
 
File:Eps lt 0v2.png|c)
 
</gallery>
 
'''Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (a)), $v_0 = 0$ (b)) and $v_0 < 0$ (c)).
 
'''
 
<br/>
 
 
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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_7"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
 
'''Problem 12'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_7</p>
 
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 $v_0 < 0$.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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).
 
 
[[File:Epsgt0v2.png|center|thumb|400px|Solutions $H[\varphi]$ for quadratic potentials (problem [[#SSC_6_1]]) with $v_0 > 0$.]]
 
[[File:Epslt0v2.png|center|thumb|400px|Solutions $H[\varphi]$ for quadratic potentials (problem [[#SSC_6_1]]) with $v_0 < 0$.]]
 
</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_8"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
 
'''Problem 13'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_8</p>
 
Estimate main contribution to total expansion factor of the Universe.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_9_0"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 14'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_9_0</p>
 
Explain difference between end points and turning points of the scalar field evolution.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_9"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 15'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_9</p>
 
Show that the exponentially decaying scalar field
 
\[
 
\varphi(t) = \varphi_0 e^{-\omega t}
 
\]
 
can give rise to unstable end points of the evolution.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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).
 
 
[[File:Quarticv2.png|center|thumb|400px|Critical curves of stationary points and solutions $H[\varphi]$ for a quartic potential with spontaneous symmetry breaking.]]
 
</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_10"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
 
'''Problem 16'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_10</p>
 
Analyze all possible final states in the model of previous problem.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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 ).
 
\]</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_11"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 17'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_11</p>
 
Express initial energy density of the model of problem [[#SSC_9]] in terms of the $e$-folding number $N$.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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 $N$.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_12"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 18'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_12</p>
 
Estimate mass of the particles corresponding to the exponential scalar field considered in problem [[#SSC_9]].
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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.</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_13"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 19'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_13</p>
 
Calculate the deceleration parameter for flat Universe filled with the scalar field in form of quintessence.
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">There are several ways to obtain the result:
 
<br/>
 
'''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)''' \[q=\frac{d}{dt}\frac 1 H-1=-\frac{\dot H}{H^2}-1=\frac{3H^2-V}{H^2}-1=\frac{2H^2-V}{H^2}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V}\]
 
</p>
 
  </div>
 
</div></div>
 
 
 
<div id="SSC_14_"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
'''Problem 20'''
 
<p style= "color: #999;font-size: 11px">problem id: SSC_14_</p>
 
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)
 
<div class="NavFrame collapsed">
 
  <div class="NavHead">solution</div>
 
  <div style="width:100%;" class="NavContent">
 
    <p style="text-align: left;">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
 
\[\frac12\frac{df(\varphi)}{d\varphi}+\sqrt3\sqrt{\frac12f(\varphi+V(\varphi)}\sqrt{f(\varphi)}+\frac{dV}{d\varphi}=0.\]</p>
 
  </div>
 
</div></div>
 
  
 
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Revision as of 21:57, 18 June 2015




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Exactly Integrable n-dimensional Universes

Exactly Integrable n-dimensional Universes


UNSORTED NEW Problems

The history of what happens in any chosen sample region is the same as the history of what happens everywhere. Therefore it seems very tempting to limit ourselves with the formulation of Cosmology for the single sample region. But any region is influenced by other regions near and far. If we are to pay undivided attention to a single region, ignoring all other regions, we must in some way allow for their influence. E. Harrison in his book Cosmology, Cambridge University Press, 1981 suggests a simple model to realize this idea. The model has acquired the name of "Cosmic box" and it consists in the following.

Imaginary partitions, comoving and perfectly reflecting, are used to divide the Universe into numerous separate cells. Each cell encloses a representative sample and is sufficiently large to contain galaxies and clusters of galaxies. Each cell is larger than the largest scale of irregularity in the Universe, and the contents of all cells are in identical states. A partitioned Universe behaves exactly as a Universe without partitions. We assume that the partitions have no mass and hence their insertion cannot alter the dynamical behavior of the Universe. The contents of all cells are in similar states, and in the same state as when there were no partitions. Light rays that normally come from very distant galaxies come instead from local galaxies of long ago and travel similar distances by multiple re?ections. What normally passes out of a region is reflected back and copies what normally enters a region.

Let us assume further that the comoving walls of the cosmic box move apart at a velocity given by the Hubble law. If the box is a cube with sides of length $L$, then opposite walls move apart at relative velocity $HL$.Let us assume that the size of the box $L$ is small compared to the Hubble radius $L_{H} $ , the walls have a recession velocity that is small compared to the velocity of light. Inside a relatively small cosmic box we use ordinary everyday physics and are thus able to determine easily the consequences of expansion. We can even use Newtonian mechanics to determine the expansion if we embed a spherical cosmic box in Euclidean space.


Problem 1

problem id:

As we have shown before (see Chapter 3): $p(t)\propto a(t)^{-1} $ , so all freely moving particles, including galaxies (when not bound in clusters), slowly lose their peculiar motion and ultimately become stationary in expanding space. Try to understand what happens by considering a moving particle inside an expanding cosmic box


Problem 2

problem id:

Show that at redshift $z=1$ , when the Universe is half its present size, the kinetic energy of a freely moving nonrelativistic particle is four times its present value, and the energy of a relativistic particle is twice its present value.


Problem 3

problem id:

Let the cosmic box is filled with non-relativistic gas. Find out how the gas temperature varies in the expanding cosmic box.


Problem 4

problem id:

Show that entropy of the cosmic box is conserved during its expansion.


Problem 5

problem id:

Consider a (cosmic) box of volume V, having perfectly reflecting walls and containing radiation of mass density $\rho $. The mass of the radiation in the box is $M=\rho V$ . We now weigh the box and find that its mass, because of the enclosed radiation, has increased not by M but by an amount 2M. Why?


Problem 6

problem id:

Show that the jerk parameter is \[j(t)=q+2q^{2} -\frac{\dot{q}}{H} \]


Problem 7

problem id:

We consider FLRW spatially flat Universe with the general Friedmann equations \[\begin{array}{l} {H^{2} =\frac{1}{3} \rho +f(t),} \\ {\frac{\ddot{a}}{a} =-\frac{1}{6} \left(\rho +3p\right)+g(t)} \end{array}\] Obtain the general conservation equation.


Problem 8

problem id:

Show that for extra driving terms in the form of the cosmological constant the general conservation equation (see previous problem) transforms in the standard conservation equation.


Problem 9

problem id:

Show that case $f(t)=g(t)=\Lambda /3$ corresponds to $\Lambda (t)CDM$ model.