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 fig.\ref{ssc_f2}. [[File:Eps gt 0v2.png.png File:Eps eq 0v2.png File:Eps lt 0v2.png |center|thumb|400px|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 $H < 0$. 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.