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Setting constant curvature $k = 0$, the Friedmann equation takes the simple form: $\dot{a}^{2} = \dfrac{8 \pi G \epsilon_o}{3 c^2}a^{-(1+3\omega)}$ If We suppose: $a \propto t^q$. We have $a(t) = (\dfrac{t}{t_o})^{2/(3+3\omega)}$. Then the horizon distance take the form: $d_{hor}(t_o) = c \int_{0}^{t_o} \dfrac{dt}{a(t)} \ = \ \dfrac{c}{H_o} \dfrac{2}{1+ 3\omega}$.

If we consider a universe with only non-relativistic matter, then $ \omega = 0$ and $ d_{hor}(t_o) = 3ct_o$.

MY doubt is: What is the physical meaning of the horizontal distance in the universe with only matter being greater than the distance that light travels?

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The horizon distance, and any other distance defined in this way, is the distance from the horizon (or something else) right now. When the light you observe was emitted, the object was closer to you, but in the meantime space has expanded.

Hence, whereas in a static universe the horizon distance would be $ct_0$, in an expanding universe it will be larger.

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