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, $z=\dfrac{v}{c}$ won't apply)?

  • 1
    $\begingroup$ It doesn't make sense in general to talk about the velocity of one object relative to another when the distance between them is on the cosmological scale. See physics.stackexchange.com/q/400457 $\endgroup$
    – user4552
    Dec 10, 2019 at 3:34

3 Answers 3


The exact number depends on the cosmological model and its parameters. In special relativistic models (e.g. the Milne model), the redshift at the speed of light is of course infinite. However, in all viable cosmological models, recession velocities exceed the speed of light for objects with redshifts greater than $z\sim 1.5$.

The general relativistic relation between recession velocity and cosmological redshift is:

$$ v_{rec}(t,z)=c\dfrac{\dot{R(t)}}{R(0)}\int_0^z{\dfrac{dz^{'}}{H(z^{'})}} $$

enter image description here

The solid dark lines and gray shading in the graph show a range of FLRW models.

For more details, please see: Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the universe.

  • $\begingroup$ Thank you for your informative answer. I had read about this $z \sim 1.5$ factor. But wanted to know how is it arrived at. $\endgroup$ Dec 9, 2019 at 7:04

If you have a redshift associated with a comoving object you get two answers; one for the recessional velocity it had when it emmited its light, and one for the recessional velocity it has now when its light reaches you.

One is calculated by the present distance times the Hubble constant, and the other by multiplying the former distance with the Hubble parameter at that time.

The higher the redshift, the larger the difference (for example, the last scattering surface with z=1089 had a recessional velocity of 63c when it emmited its light, and now has around 3c, since the Hubble parameter was higher in the past).

On this plot the red curve is the recessional velocity when the light was emmited, and the brown curve when the light reaches the observer (as you can see at z=10 there is already a difference of a factor ≈2, and like the previous speakers already mentioned c is at z≈1.5

At z≈1.9 the curves cross and it was the same recessional velocity then as it is now again, so objects with z<1.9 are faster now than they were then, and objects with z>1.9 are slower now then they were at the time they emmited their light:

enter image description here

x-axis: redshift, y-axis: recessional velocity, parameters: Planck 2013


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 $\boxed{v=c}$ now. The distance that fulfils this condition is known as current 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 $z=1.474 \approx 1.5$

Best regards.

  • $\begingroup$ Thank you! While the other answers are also highly informative and very valuable for the community, this is the most direct illustration. May I request you to also post this on Astronomy SE for the benefit of the community there as well? (Link: astronomy.stackexchange.com/questions/34218/…) $\endgroup$ Dec 10, 2019 at 3:37
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    $\begingroup$ Done. I hope I am not breaking any rules by repeating there the solution $\endgroup$
    – Albert
    Dec 10, 2019 at 10:10
  • $\begingroup$ Thank you! I am sure no rules are broken by trying to help the community :) $\endgroup$ Dec 10, 2019 at 10:18
  • $\begingroup$ @eshaya please review my post again, everything in it is correct within the framework of the concordant ΛCDM cosmological model. Regards. $\endgroup$
    – Albert
    Feb 24, 2023 at 8:00
  • $\begingroup$ This is the redshift of a galaxy that today has recessional velocity of c. The galaxy had a lower recessional velocity when the light we see now was emitted, as Yukterez shows above. Usually, astronomers are interested in the properties of galaxies as they were at the time and age that they are viewed $\endgroup$
    – eshaya
    Feb 25, 2023 at 6:52

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