# At what cosmological redshift $z$, does the recession speed equal the speed of light? How is it calculated?

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)?

• @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. – Ritesh Singh Dec 7 '19 at 23:41
• Oops, I missed that, thanks! – uhoh Dec 7 '19 at 23:45
• Duplicate at Physics SE and it has an answer there physics.stackexchange.com/questions/518543/… – Alchimista Dec 9 '19 at 9:30
• It is in there. You can't further simplify things. You have equation and various plots depending on parameters. – Alchimista Dec 9 '19 at 9:37
• 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. – Chappo Says SE Dudded Monica Dec 9 '19 at 23:36

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):

$$d_H=\frac{c}{H_0}$$

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.

• Thank you! I am sure no rules are broken by trying to help the community :) – Ritesh Singh Dec 10 '19 at 10:19