B How can we see 45B ly away if the universe is only 13B years

1. Feb 13, 2017

unwillingly ignorant

I'm sure this is probably a stupid question with an obvious answer.. But i can't come up with it..

So, the observable universe is ment to be 90 billion lightyears across, right?
And it's 13.8 billion years old, right?
And a lightyear is the distance it takes for light and fields and stuff to reach us in a year, right?
So the light and stuff from the things of 13.8 billion lightyears away should be 13.8 billion lightyears old, and as it was at the bang in all directions around us...right..?
How could we see further than that?

2. Feb 13, 2017

Staff: Mentor

It is because the universe is expanding. We can only see what used to be some 13 billion ly away from us, but that thing we are seeing is now much farther away from us because of the expansion.

3. Feb 13, 2017

unwillingly ignorant

that was my first thought but it doesn't seem to make sence in my head as the light would still have to be over 13B years old even if that stuff it's coming from used to be near us.. Am I just being dense?

4. Feb 13, 2017

Bandersnatch

Imagine an ant walking on some surface. Say, it walks at $V=1 cm/s$. It starts at point A and wants to go to point B, which are 100 cm apart. The distance it covers is $D=V*t$. After $t=100 s$, it will have arrived at B. This is analogous to light travelling through non-expanding space.

Now, imagine the surface the ant is walking on is a rubber band that is slowly being stretched as the ant walks. Two things will be different than the previous case:
1) the and will not have reached point B in 100 s, as during its walk point B was continuously receding. It needs more than 100 s to get there - how long depends on the rate of expansion of the rubber band
2) by the time the ant reaches the target, the distance between A and B will be larger than both the original distance (when the ant started walking) and than the distance the ant 'covered' - i.e. the distance you'd get from calculating $D=V*t$

This means, that in the expanding universe, of which the second case described above was an analogy, there are three distances to the source of observed light that one needs to consider:
- the 'light travel distance' (I've put it in scare quotes, because it's not really a useful measure of anything physical, it's just what you'd get from $D=V*t$)
- the distance at the time of emission
- the distance at the time of observation

In the particular case of our universe and the light from the farthest observable objects (CMBR) took ~14 billion years to get here, so its light travel distance is 14 billion light-years.
The distance at the time of emission was approx. 40 million light-years (note, million, not billion).
The distance at the time of reception is ~45 billion light-years.

Ah, but we can't (yet). Of all the light currently observed, the farthest distance at emission was approx. 5 Gly, if I'm eyeballing the following graph correctly:

5. Feb 13, 2017

unwillingly ignorant

Yup, i was being pretty dense.. But i totally get it now! Thank you!

6. Feb 13, 2017

Staff: Mentor

I wrote "some 13 GLy" to give me some wiggle room, but I was far off the mark. Thanks for the clarification!

7. Feb 14, 2017

Jorrie

Correct. Here is a better resolution with the latest data. The past light-cone (Dthen) max distance is about 5.8 Gly, at time 4.2 Gy, which one can read off most accurately form the tabular output of the calculator.