Difference between revisions of "New problems"

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.

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: $H'' \propto 1/H'$.

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.

\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}

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.

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.

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.

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.

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.

• Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0
• 
• 
• 

Critical curves $H'[\varphi] = 0$ for quadratic potentials with $v_0 > 0$ (left), $v_0 = 0$ (middle) and $v_0 < 0$ (right).

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 $H < 0$.

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 $v_0 > 0$ (left), and $v_0 < 0$ (right).

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.

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.

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 \ref{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.

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 \[ \langle V \rangle = v_0 - \frac{\mu^4}{4\lambda} = \frac{\omega}{3} ( 6h - \omega ). \]

The energy density for the solution of problem \ref{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 $N$.

Taking the final energy density $\langle V \rangle$ (see problem \ref{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.

There are several ways to obtain the result: \begin{description}

• • \[q=-\frac{\ddot a }{aH^2};\quad \frac{\ddot a}{a}=-\frac16(\rho+3p);\quad H^2=\frac13\rho;\]

• • \[q=\frac12\Omega_{tot}+\frac32\sum\limits_iw_i\Omega_i.\] For a flat single-component Universe one obtains

• • \[q=\frac{d}{dt}\frac 1 H-1=-\frac{\dot H}{H^2}-1=\frac{3H^2-V}{H^2}-1=\frac{2H^2-V}{H^2}=\frac{\dot\varphi^2-V}{\frac{\dot\varphi^2}2+V}\]