0
$\begingroup$

The age of the Universe can be estimated from taking the inverse of the Hubble constant: $t_\text{universe} = 1/H_0 =d/v.$

enter image description here

It seems to me this method assumes that any given galaxy has been receding at a constant apparent velocity for the lifetime of the Universe. However, by including data from larger & larger distances, it can be seen that Hubble's law doesn't hold. The data suggest that the rate of expansion was actually slower in the past.

enter image description here

Why is it appropriate to use only nearby galaxies to estimate the age of the Universe using the Hubble time when the rate of expansion isn't constant?

$\endgroup$
  • 1
    $\begingroup$ It isn't, really. As you say, it's just an estimation, an order-of-magnitude age. To get the correct age, you have to integrate the Friedmann equation. $\endgroup$ – pela Jul 26 '16 at 20:51
2
$\begingroup$

Who told you that $H_0^{-1}$ is the age of the universe? It isn't, and you have correctly identified the reasons why not.

A correct age estimation relies on knowing $H_0$ and the densities of matter and dark energy, so that the past expansion history of the universe can be correctly modelled. Even this relies on an assumption about how dark energy behaves.

$\endgroup$
  • 2
    $\begingroup$ I think you're taking what BMS has said a bit too far. He specifically mentioned that $H_o^{-1}$ is an estimate, which it is. $H_o^{-1}=14\:Gyr$ which is a pretty good estimate of the age of the universe, though of course not exact. I think his question can be better interpreted to be, why does using $H_o^{-1}$ produce such a good estimate, relatively speaking. $\endgroup$ – zephyr Jul 27 '16 at 14:05
  • $\begingroup$ @zephyr Then that is a different question and it is because we live in a universe that is flat and where the dynamics are not presently dominated by either matter or dark energy. $\endgroup$ – Rob Jeffries Jul 27 '16 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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