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Consider a flat universe, here, proper distance can be given by R-W Metric:

$$d_p (t_0) = c\int_{t_e}^{t_0}\frac{dt}{a(t)},$$ $t_e$ is the time when a photon is emitted from a distant galaxy, $t_0$ is the time right now.

In a flat universe, scale factor and time have a relation of:

$a(t) = (\frac{t}{t_e})^\frac{2}{3+3w}$, where $w$ is the dimensionless quantity in the Equation of State $P = w\varepsilon$, $P$ is pressure and $\varepsilon$ is energy density.

Also note that

$\frac{1}{1+z} = a(t) = (\frac{t_e}{t_0})^\frac{2}{3+3w}$ , where $z$ is redshift, and

$t_0 = \frac{2}{3+3w} H_0^{-1}$ (given by integrating Friedmann Equation), where $H_0$ is Hubble constant.

Substitute the above equations into the integral and solve for it:

$d_p(t_o) = \frac{c}{H_0} \frac{2}{1+3w} [ 1-(1+z)^{-\frac{1+3w}{2}}]$

Now we define horizon distance, or the farthest distance one can see in a flat universe, by definition $t_e = 0$ or $z = \infty$, substitute that into the proper distance equation:

$d_{hor} (t_0) = \frac{c}{H_0} \frac{2}{1+3w}$

In my textbook, it discuss the possibility of $w > -{\frac{1}{3}}$ and $w\le -{\frac{1}{3}}$. In the second possibility of $w\le -{\frac{1}{3}}$, it says that that brings a infinite horizon distance. Here is the problem:

I understand if $w = -{\frac{1}{3}}$, the resulting horizon distance will be infinity, but I don't get the situation when $w < -{\frac{1}{3}}$, this will bring a negative value to the horizon distance, how can a distance be negative? What does this negative distance mean?

(sorry if this is too long)

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The integral $$\int_0^t\frac{\mathrm{d}t^\prime}{a(t^\prime)}\propto \int_0^t\frac{\mathrm{d}t^\prime}{t^{\prime \frac{2}{3+3w}}}$$ diverges if $w\leq-1/3$. It doesn't give a negative result (nor could it -- the integrand is everywhere positive).

With regard to your expression

$d_p(t_o) = \frac{c}{H_0} \frac{2}{1+3w} [ 1-(1+z)^{-\frac{1+3w}{2}}]$

note that $$\lim_{z\to\infty}(1+z)^{-\frac{1+3w}{2}}=\begin{cases}0, \ \ \text{if} \ w>-1/3\\\infty, \ \ \text{if}\ w<-1/3\end{cases}.$$ So this expression implies an infinite horizon distance if $w<-1/3$.

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  • $\begingroup$ my math is a bit rusty here, could you explain why the integrand is positive everywhere? $\endgroup$ Commented Jun 11 at 5:45
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    $\begingroup$ The integrand is $1/a$, which is always positive since $a$ is always positive (or you could note that $1/t^{2/(3+3w)}$ is positive for all $t>0$). The integral is essentially adding up the values of $1/a$ (times a time interval) at a bunch of different times, so it's adding a bunch of positive numbers, and the result has to be positive. $\endgroup$
    – Sten
    Commented Jun 11 at 5:47
  • $\begingroup$ so are there restrictions on values of $w$, because from the integral's results, I got a function that is related to $w$ $\endgroup$ Commented Jun 11 at 5:49
  • $\begingroup$ I haven't worked through exactly where this is, but I think that at some step in deriving the expression, you assumed $w>-1/3$. $\endgroup$
    – Sten
    Commented Jun 11 at 5:53
  • $\begingroup$ I don't see any assumptions in the book that says $w>−\frac{1}{3}$ , also in my book it says for $w>−\frac{1}{3}$ , there is a infinite horizon distance, but according to the divergence of the function, there will be a finite value for this distance $\endgroup$ Commented Jun 11 at 6:29

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