At what cosmological redshift $z$, does the recession speed equal the speed of light?

What equations are used to calculate this number (since at large redshifts, $z=v/c$ won't apply)?

[The interested reader may refer to answers on Physics SE.]

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    $\begingroup$ @uhoh Thanks for the link. However, the equations for relativistic redshift do not apply to cosmological redshift, where $v \ge c$ is quite well known and commonly observed. $\endgroup$ Dec 7, 2019 at 23:41
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    $\begingroup$ Oops, I missed that, thanks! $\endgroup$
    – uhoh
    Dec 7, 2019 at 23:45
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    $\begingroup$ Duplicate at Physics SE and it has an answer there physics.stackexchange.com/questions/518543/… $\endgroup$
    – Alchimista
    Dec 9, 2019 at 9:30
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    $\begingroup$ It is in there. You can't further simplify things. You have equation and various plots depending on parameters. $\endgroup$
    – Alchimista
    Dec 9, 2019 at 9:37
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    $\begingroup$ I'm voting to close this question as off-topic because the OP asked the same question on Physics.SE where it has been answered. Cross-posting the same question on different SE sites is discouraged. $\endgroup$ Dec 9, 2019 at 23:36

1 Answer 1


From Friedmann Equation, distance as a function of redshift is:

$$d(z)=\frac{c}{H_0}\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}$$

The Hubble-Lemaître Law:

$$v=H_0 \cdot d$$

We want $\boxed{v=c}$ The distance that fulfils this condition is known as Hubble Distance, (or Hubble Radius, or Hubble Length):


Combining both, we obtain the condition:

$$\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}=1$$

For $\Omega_{R_0}\approx 0 \quad \Omega_{K_0}\approx 0 \quad \Omega_{M_0}\approx 0.31 \quad \Omega_{\Lambda_0}\approx 0.69$

The condition is:

$$\int_0^z \frac{dx}{\sqrt{0.31(1+x)^3+0.69}}=1$$

Searching by trial and error, we find that the value of redshift that fulfils the condition is: $$z=1.474 \approx 1.5$$

I hope I am not breaking any rules by repeating here the solution I wrote on Physics StackExchange I did it because the creator of the thread asked me to do it, there.

Best regards.

  • $\begingroup$ Thank you! I am sure no rules are broken by trying to help the community :) $\endgroup$ Dec 10, 2019 at 10:19

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