I am trying to calculate the Hubble Radius, or the distance from which an object will recede from an observer at the speed of light. I initially tried using $v=H_0d$ which gives $d=1.424\times10^{26}$ for $H_0 = 65km/s/Mpc$ however I have been told that a different equation should be used at relativistic speeds.

After some research I have found the following formula:

$z = \sqrt{\dfrac{c + v}{c - v}} - 1 \implies v = c \cdot \dfrac{(z+1)^2-1}{(z+1)^2+1}$

Therefore: $d = \dfrac{c}{H_0} \cdot \dfrac{(z+1)^2-1}{(z+1)^2+1}$

However, I have no idea how to get a decent answer out of this equation. Obviously $z \rightarrow \infty$ as $v \rightarrow c$ and so $d \rightarrow \infty$ also. Is this the expected result? How does one interpret this?


1 Answer 1


You might find the following paper useful: http://arxiv.org/abs/astro-ph/0310808v2 In the figure 2 on page 7 you see the plots of different $v(z)$ relations, among them the classical and the special relativistic ones that you have used in your calculations. You can see that the classical $v(z)$ relation intersects the $v=1c$ line for some redshift $z$, this corresponds to the $d$ value which you got in your first calculation. You can also see that the special relativistic $v(z)$ relation asymptotically approaches the $v=1c$ line which corresponds to the fact, that you found in your second calculation, that v reaches the speed of light only in the limit of $z\rightarrow \infty$. This is quite natural since in special relativity $v$ cannot become larger than $c$.

If you want to describe the space-time properly, you need to use the general relativistic $v(z)$ relation. It depends crucially on the matter content of the universe described by the relative densities $\Omega_M$ and $\Omega_\Lambda$ of matter and dark energy (if one neglects the contribution of radiation, of course). You can see some possible plots in the same figure. For obtaining the redshift value of the point where $v=c$ you would need to solve the general relativistic $v(z)$ relation for $z$, which is unfortunately only possible numerically for the general case.

But I guess, what you are interested in is the distance of the point from which on signals cannot reach Earth. This is not the same point as the one given by $v(z)=c$! It is rather the point given by the so called particle horizon. You can see a nice plot in figure 1 on page 3 of the above paper. If you are interested in how to calculate this, just drop a comment.

For further reading I recommend the first chapters of the excellent book "Physical Foundations of Cosmology" by V. Mukhanov.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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