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.

   <p style="text-align: left;"></p>