So solving for the scale factor from the Friedmann equation we can then use it to calculate proper distance via $$d_p(t_o) = c \int_{t_e}^{t_o} \frac{dt}{a(t)}$$

For a particular universe a(t) is proportional to t, and therefore at the end we have

$$d_p(t_o) = \frac{c}{H_0}\ln\Big(\frac{t_o}{t_e}\Big)$$

Now, if we want to know the today's horizon distance, we'd set time of emission to zero. But that means the denominator goes to zero and the whole expression blows up to infinity. In such a universe is it meaningful to say that the object that emitted a photon at $t_e =0 $ now has a proper distance of infinity at the current time?

So solving for the scale factor from the Friedmann equation we can then use it to calculate proper distance via $$d_p(t_o) = c \int_{t_e}^{t_o} \frac{dt}{a(t)}$$

For a particular universe $a(t)$ is proportional to $t$, and therefore at the end we have

$$d_p(t_o) = \frac{c}{H_0}\ln\Big(\frac{t_o}{t_e}\Big)$$

Now, if we want to know the today's horizon distance, we'd set time of emission to zero. But that means the denominator goes to zero and the whole expression blows up to infinity. In such a universe is it meaningful to say that the object that emitted a photon at $t_e =0 $ now has a proper distance of infinity at the current time?

So solving for the scale factor from the Friedmann equation we can then use it to calculate proper distance via $d_p(t_o) = c \int_{t_e}^{t_o} \frac{dt}{a(t)}$.

For a particular universe a(t) is proportional to t, and therefore at the end we have

$d_p(t_o) = \frac{c}{H_o}ln(\frac{t_o}{t_e})$.

Now, if we want to know the today's horizon distance, we'd set time of emission to zero. But that means the denominator goes to zero and the whole expression blows up to infinity. In such a universe is it meaningful to say that the object that emitted a photon at $t_e =0 $ now has a proper distance of infinity at the current time?

So solving for the scale factor from the Friedmann equation we can then use it to calculate proper distance via $$d_p(t_o) = c \int_{t_e}^{t_o} \frac{dt}{a(t)}$$

For a particular universe a(t) is proportional to t, and therefore at the end we have

$$d_p(t_o) = \frac{c}{H_0}\ln\Big(\frac{t_o}{t_e}\Big)$$

Now, if we want to know the today's horizon distance, we'd set time of emission to zero. But that means the denominator goes to zero and the whole expression blows up to infinity. In such a universe is it meaningful to say that the object that emitted a photon at $t_e =0 $ now has a proper distance of infinity at the current time?