# Event Horizon in a closed, matter (dust) dominated universe

## Main Question or Discussion Point

Hi!

It is stated in V. Mukhanov's book "Physical foundations of Cosmology" the following (page 44, after equation 2.25): "In contrast, for the dust dominated universe, where ηmax=2π, the event horizon exists only during the contraction phase when η>π." could someone please explain why is this true? the equation for the event horizon in a closed, dust dominated universe is (again from Mukhanov): de(t)=am(1-cosη)(ηmax-η),(am is a constant) which clearly has non-vanishing values for η≤π

Thanks!

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George Jones
Staff Emeritus
For a closed, dust dominated universe, $0 < \eta < 2\pi$. During the expansion phase, $0 < \eta < \pi$. Combining this with
$$\chi_e \left( \eta \right) = \eta_\mathrm{max} - \eta = 2\pi - \eta$$
\begin{align} 0 > -\eta > -\pi \\ 2\pi> 2\pi - \eta > \pi \\ 2\pi> \chi_e \left( \eta \right) > \pi \end{align}
But this "solution" is unphysical, because, in a closed universe, $\chi$ is never larger that $\pi$.