2
$\begingroup$

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.]

$\endgroup$
  • 1
    $\begingroup$ @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. $\endgroup$ – Ritesh Singh Dec 7 '19 at 23:41
  • 1
    $\begingroup$ Oops, I missed that, thanks! $\endgroup$ – uhoh Dec 7 '19 at 23:45
  • 1
    $\begingroup$ Duplicate at Physics SE and it has an answer there physics.stackexchange.com/questions/518543/… $\endgroup$ – Alchimista Dec 9 '19 at 9:30
  • 1
    $\begingroup$ It is in there. You can't further simplify things. You have equation and various plots depending on parameters. $\endgroup$ – Alchimista Dec 9 '19 at 9:37
  • 4
    $\begingroup$ 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. $\endgroup$ – Chappo Hasn't Forgotten Monica Dec 9 '19 at 23:36
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! I am sure no rules are broken by trying to help the community :) $\endgroup$ – Ritesh Singh Dec 10 '19 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.