I wonder if there a difference between horizon and comoving distances in cosmology ?

I mean both correspond to a null geodesic $(ds^2=0)$ right ?

$$d_{horizon} (t) =a(t) \int\limits_{0}^{r} \frac{{\rm d}r'}{\sqrt{1-kr'^2}}= a(t) \int\limits_{0}^{t}\frac{ c \, {\rm d}t'}{a(t')}$$

with $a(t)$ the scale factor.

Is the difference just that for a comoving distance the integration bounderies changes ? like $r_e$, $r_0$ and $t_0$, $t_e$

Thanks in advance for help


1 Answer 1


The comoving distance is a distance. You can calculate the comoving distance from your head to your toes (though it's more useful in cosmological contexts). It is defined such that today it coincides with the actual, physical distance you would measure if you froze the Universe and laid out meter sticks. In 11 billion years from now, when the Universe has doubled in size, physical (cosmological) distances have increased by a factor of two, but the comoving distances are, by definition, the same.

A "horizon" is a term used for a (non-physical) boundary, e.g. the "particle horizon", which is the boundary between the observable and the un-observable Universe. Or the "(cosmological) event horizon", which marks the boundary between regions from which we may at some point receive a signal emitted today, and regions from which we cannot.

Just like on Earth it makes sense to talk about "the distance to the horizon", you can say, e.g., "The comoving distance to the particle horizon is 46 billion lightyears", or "The comoving distance to the event horizon is 17 billion lightyears".

As you say, to calculate the comoving distance you use the Robertson-Walker metric (assuming here a flat Universe for simplicity) $$ ds^2 = -c^2 dt^2 + a(t)^2 \big[ dr^2 + r^2 d\Omega^2 \big], $$ and set $ds=d\Omega=0$ to measure along a radial null geodesic. Then $dr = c\,dt/a(t)$, or $$ r = c \int_{t_\mathrm{em}}^{t_\mathrm{obs}} \frac{dt}{a(t)}, $$ where $t_\mathrm{em}$ and $t_\mathrm{obs}$ is the time when the light is emitted and observed, respectively.

To calculate the comoving distance to the particle horizon, you set $t_\mathrm{em} = 0$ and $t_\mathrm{obs} =$ today, because the particle horizon corresponds to light emitted at Big Bang and observed today.

To calculate the comoving distance to the event horizon, you set $t_\mathrm{em} =$ today and $t_\mathrm{obs} = \infty$, because the event horizon corresponds to light emitted today and observed in the (almost) infinitely distant future.

  • $\begingroup$ Thanks for your answer ! So to get comoving distance we take $ds^2=0$ right ? And when we talk of "paticle horizon" is it normal to do it in comoving distance ? Why not physical, or luminosity distance ? $\endgroup$ May 16, 2018 at 9:01
  • 1
    $\begingroup$ @AlbertBranson Yes to the "$ds^2=0$", see update. The distance to the particle horizon is used both in physical and comoving distance, depending on the context. But luminosity distance has to do with the observed luminosity of objects, which is rarely if ever of interest when talking about the horizon, I'd say. $\endgroup$
    – pela
    May 16, 2018 at 21:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .