# Hubble time and its derivation? [duplicate]

I know the derivation of Hubble time goes something like this (I am an a-level student so this may not be the actual derivation): Two galaxy that is moving away from each other at speed v are now D distance apart assuming the time when they where together is t=0 i.e. at the Big Bang the time since the Big Bang is given by $$T= \frac{D}{V}$$ $$V=\frac{D}{T}$$ Hubble's law is given by $$V=HD$$ therefore subbing the first expression into the latter gives $$\frac{1}{T}=H$$ and therefore $$T=\frac{1}{H}$$ The thing I do not understand is if we are assuming the rate of expansion is constant (as is required for this) why does that mean we can use $$t=\frac{d}{v}$$ Because as the galaxies get further apart together the space between them gets more and more hence they will move away from each other faster the further the are apart. This equation needs v to be constant which is not the case. Please can someone explain?

In the linked qestion they use the formula t=d/v just saying it is a linear extrapolation. This does not help me with my qestion as i can still not understand why this can be used as it assumes v is constant (i think) which it would not be. I am looking for this to be explained in more detail then the linked question.

• possible duplicate of Universe Expansion as an absolute time reference – John Rennie May 29 '14 at 12:03
• The link I've suggested may not be an obvious duplicate, but the answers to it explain how $H$ is calculated and explain why it isn't constant. – John Rennie May 29 '14 at 12:04
• I have just looked at this question and it does not explain my main question of why v can be considered constant, I am aware that H is not constant as time progresses. The answer in the question also links to t=d/v but does not expalin it in detail (i.e. why we can use it) – user43487 May 29 '14 at 12:28
• Who says that $V$ is a constant? – Kyle Kanos May 29 '14 at 12:34
• It has to be to use $$t=d/v$$ does it not? d is there distance apart if you just use v is it is now the whole derivation breaks down does it not? – user43487 May 29 '14 at 13:01

## 1 Answer

In Hubble's mind, the universe was a static frame of reference. You can define an absolute rest frame, no need for all this badabada relativity stuff. In this frame, galaxies are moving in straight lines at constant speeds.

Now, what I think is confusing you is the value of $H$. As you have correctly derived, in this simplified model of the universe, $H=\frac{1}{t}$: the Hubble parameter is changing over time! We talk of the Hubble constant because Hubble's original data reached only 2 Mpc away, or 6.5 millions light years, that is a tiny fraction of the total age. So, even for the furthest galaxy he could observe, that was 6.5 million years old, the universe was pretty much as it is here today.

You can see in this other question more details about the evolution of this parameters over time.

• Are you saying taking the universe as a refrence frame a galaxy moving away from us today at v m/s was moving away from us at v m/s e.g. 5000 million years ago? Is it just our refreance frame that more distant once seem to be moving away faster?? – user43487 May 29 '14 at 13:50