Difficulties with recollapsing models in closed isotropic loop quantum cosmology

The use of techniques from loop quantum gravity for cosmological models may solve some difficult problems in quantum cosmology. The solutions under a number of circumstances have been well studied. We will analyze the behavior of solutions in the closed model, focusing on the behavior of a universe containing a massless scalar field. The asymptotic behavior of the solutions is examined, and is used to determine requirements of the initial conditions.

Figure 1

This is the exact solution to the closed $\Gamma =\frac{1}{2}$ with $\omega =55$ and $j=20$. The initial value was set at $\mu =4{\mu }_o$ with ${\mu }_o=\sqrt{3}/4$. The state clearly diverges as we go to large scales. Since large $\mu$ is classically disallowed, we would like not to admit these states as physical possibilities. The factor $\exp (i\mu \Gamma /2)$ has been left out of the solution.

Figure 2

This is the exact solution to the closed model $\Gamma =1/2$ with $\omega =55$ and $j=20$. The initial value was set at $\mu =3{\mu }_o$ with ${\mu }_o=1$. The initial conditions have chosen to eliminate the growing solution in the $-\mu$ domain. However, it is clear we cannot eliminate the growing solutions in both positive and negative domains for the asymmetric model (i.e., the model where we do not hit zero volume). The factor $\exp (i\mu \Gamma /2)$ has been left out of the solution.

Figure 3

This is the exact solution to the closed model $\Gamma =1/2$ with $\omega =55$, $j=20$ and $\gamma =\frac{\log 2}{\sqrt{3}\pi }$. The crosses represent the solution for $l=.45$ and solid line is for $l=.4475$. The initial value is set at $\mu =4{\mu }_o$ and is equal in both cases. The change in the sign of the growing exponential part of the solution suggests that we can fix $l$ such that only the decaying solution remains.

Figure 4

This is the exact solution to the closed model $\Gamma =1/2$ with $\omega =50$, $j=20$ and $\gamma =\frac{\log 2}{\sqrt{3}\pi }$. The crosses represent the solution for $l=.45$ and solid line is for $l=.4475$. The initial value is set at $\mu =4{\mu }_o$ and is equal in both cases. These solutions are nearly identical for all $\mu$. This suggests that there is no single value of $l$ that will produce bounded solutions for all choices of $\omega$.

Figure 5

This is a plot of the log of modulus of $s_{\mu }$ for the closed model $\Gamma =1/2$ with $\omega =55$, $j=20$ and $\gamma =\frac{\log 2}{\sqrt{3}\pi }$. The crosses represent the critical solution for $l=.447....$ in which the solution is bounded at large scales. The state is normalized such that the amplitude of the oscillation is of the same order as the other solutions. Thus, the value at $\mu =4{\mu }_o$ is nearly ${10}^{60}$ compared to ${10}^{-50}$ in the other solutions shown. Clearly, the behavior near $\mu =0$ dominates the solution.