# 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

## 2 Answers

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.

• Well, you are assuming that you are at a point where expansion looks radial, if you postulate that everywhere an observer looks the universe expands with "r(t)=α(t)r0" of course you can show it is radial! – J.Gili Aug 9 '17 at 15:38
• My question is (very much summarized) how can that be if we are not at the centre, which we are not. – J.Gili Aug 9 '17 at 15:39
• No, I postulated that there was radial for one observer (what you call the center). Then is easy to show that every observer will see that too. In the real case there is no center though, but I did not want to comment on that issue, only on your wrong intuition. – Wolphram jonny Aug 9 '17 at 16:11

If our universe is a 3-sphere expanding at the speed of light it answers a range of questions, including why there is no centre. A 3-sphere has a centre but the centre is not in the 3-sphere just as the surface of the earth has a centre which is not on the surface of the earth. See modeloftheuniverse@wordpress.com for more detail.(you may need to paste the address into the address bar)

• Link doesn't exist. – AccidentalFourierTransform Jun 1 '18 at 0:54
• That link appears to be an email address? – Nat Jun 1 '18 at 1:31