1
$\begingroup$

From what I understand, people have measured the rate of expansion of the universe by observing the red shift of galaxies moving away. Then, they took the size of the observable universe and used the measured expansion rate to extrapolate backwards to when the size was very tiny. They found that it took 13.7 billion years to get to the current size. (Please correct me if I am wrong about any part of this)

Why is this process not flawed due to the possibility that the entire universe may be larger than the observable universe? So if in a long time, light from a further point than we can see now arrives at Earth, this will mean that we can observe more of the universe and we will extrapolate a greater age.

$\endgroup$
2
  • $\begingroup$ (1) I've edited your question to make sense; (2) The modern thinking is that the universe was never tiny, but always infinite (whether you believe it or not is a separate question); (3) Consider a galaxy is at the distance $d$ from us now flying away with the velocity $v$. If you assume that in the past this galaxy was very close, then it was back at the time $t=\dfrac{d}{v}$ . Do you need the entire universe for this (perhaps oversimplified) calculation? $\endgroup$
    – safesphere
    Commented Nov 3, 2017 at 3:54
  • $\begingroup$ Yes but at that point in time the universe will be indeed older than now $\endgroup$
    – Alchimista
    Commented Nov 4, 2017 at 10:37

1 Answer 1

3
$\begingroup$

Hubble's law says that something that is twice as far away is receding twice as fast. Therefore it does not matter what volume you consider, if you run the clock backwards, it all converges to a point at the same time.

A galaxy at a distance $d$, apparently moving away from us$^{1}$ at velocity $v$ would have been on top of us at a time $t = d/v$ ago.

But Hubble's law says that $v= H_0d$ in general, where $H_0$ is the Hubble parameter. Thus a galaxy $2d$ away, recedes at $2v$, and the time when it is on top of us (and the first galaxy) is $t = 2d/2v = d/v = H_0^{-1}$. The same applies to every galaxy, so every galaxy was at the same point a time $H_0^{-1}$ ago.

NB: The age of the universe is only near to $H_0^{-1}$ since this simple calculation ignores deceleration/acceleration. Also note that the calculation does not need to know the size of the observable universe, nor is it relevant since we can't measure the redshift of something outside the observable universe...

1: This is not really the correct way to think about redshift, but it will do here.

$\endgroup$

Your Answer

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

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