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

Question

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)?

Answer 1

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:

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

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.

Answer 2

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:

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

Answer 3

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):

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

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 $z=1.474 \approx 1.5$

Best regards.