# Shape of the observable universe or Cosmic Horizon

My question is about the shape of the observable universe or Cosmic Horizon.

In literature it is described as having a radius, constant.

But in an accelerating expanding universe this seems impossible to me. I will explain my doubt.

The universe started at a certain location with the Big Bang, lets call that location X. A galaxy at the same radial distance from that point as our, but at another angle (i'm imagining some kind of spherical coordinate system here) should move away from us much more slowly than instead a galaxy on a grater or lower radial distance from the point X but with the same angle as our. This since our velocity in the first case will be similar to the other galaxy which is getting away from us only due to expansion of the universe itself (or increasing angle) while in the second case the velocity of the other galaxy will be much greater or slower.

That said the distance for which the recession speed is higher than light speed and therefore we cannot see more far than this should be different depending on direction and will not be a fixed radius.

Maybe it is fault of my total misunderstanding, but if so, let me know, it will be appreciated.

• The universe did not start at a certain location in space, it started everywhere. There is no center of the universe. – Kosm Aug 9 '17 at 15:53

Your intuition is incrorrect. Suposse you have a point in space from which expansion looks radial, so $r(t)=\alpha(t) r_0$, where $r$ is the distance of some object with initial distance $r_0$, and $\alpha(t)$ is the expansion rate, which might depend on $t$ but it is uniform across space (for instance, $\alpha(t)=at$, with $a$ a constant). Notice that the change in distance, or the apparent velocity, will increase with the radius.
Now you can ask what another observer will see, let us say, somebody located at some position $\vec{r_c}$. This observer will see that the coordinates of the objects are $\vec{r'}=\vec{r}-\vec{r_c}$. It is trivial to show that $r'=\alpha(t)r'_0$, thus any other observer will also see a radial expansion independent of angle. I can post the demonstration if you can not figure it out by yourself.