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?


1 Answer 1


The FLRW spacetime has a singularity at $t=0$. That means that $t=0$ must actually be removed from the manifold. So there is no object that emitted a photon at $t=0$ because there is no $t=0$ at all.

I know that this seems like avoiding the question, and it is in a way. But there is no mathematically consistent way to do it otherwise with our current physics theories. However, our current physics theories are expected to break down even before going all the way back to $t=0$.

  • $\begingroup$ So does this mean the whole calculation of "horizon" distance is not valid either then, since it essentially integrates from t=0 to t=today (or equivalently setting z= $\infty$ in the result if expressed in redshift rather than scale factor and time)? $\endgroup$
    – ABC
    Oct 13, 2022 at 22:38
  • $\begingroup$ @ABC even though you cannot say anything about what happens actually at t=0, there are many integrals that converge in the limit $\lim_{\epsilon \to 0} \int_\epsilon^{t'} f(t) dt$. Those are often sloppily written as integrals from t=0. $\endgroup$
    – Dale
    Oct 14, 2022 at 13:59

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