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Assume the Robertson-Walker metric: $$g = -d\tau^2 + a^2(\tau)\gamma$$ where $\gamma$ is the flat, spherical or hyperbolic spatial metric and $a$ is the scale factor. Wald seems to calculate the age of the universe strictly from the relationship $$\frac{dR}{d\tau} = \frac{R}{a}\frac{da}{d\tau} = HR$$ where $R$ is the spatial distance measured between two isotropic observers at time $\tau$ and $H$ is Hubble's constant. Wald says that

"If the universe had always expanded at its present rate, then at the time $T = \frac{a}{\dot a} = H^{-1}$ ago, we would have had $a = 0$".

For some reason I am having trouble seeing that $a = 0$ when $T = \frac{a}{\dot a}$

I would appreciate some help.

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I wonder of you are overthinking this. Wald says:

If the universe had always expanded at its present rate

that is, $\dot{a}$ is a constant and independent of time. In that case the value of $a$ at time $t$ after the Big Bang is simply:

$$ a = \dot{a} t $$

So if you define $T$ by $T = a/\dot{a}$ then $T$ is necessarily the age of the universe.

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  • $\begingroup$ Well, this is embarrassing... $\endgroup$
    – Jonathan Gafar
    Jun 21, 2016 at 4:43
  • $\begingroup$ Quick question - $\dot a$ being constant implies that $a = \dot a t + c$ for some constant $c$, in which case $t = 0$ implies $a = c$ where $c$ may be non-zero. Are there physical reasons which imply that $c$ must be $0$? $\endgroup$
    – Jonathan Gafar
    Jun 21, 2016 at 15:09
  • $\begingroup$ @JonathanGafar: we normally take $t$ (i.e. comoving time) to be zero at the Big Bang so that for every comoving observer $t$ is simply the time since the Big Bang. And of course at the Big Bang $a$ is zero, and that means your constant $C$ must be zero. There's nothing to stop you moving the time origin, i.e. have $t \ne 0$ at the Big Bang, but it's not obvious this is a useful thing to do. $\endgroup$
    – John Rennie
    Jun 21, 2016 at 15:15
  • $\begingroup$ I guess I am looking for proof (or evidence) that $a = 0$ at the Big Bang. I was looking at Wald's statements as, given the Robertson-Walker model, we can mathematically show that $a = 0$ corresponds to $t = 0$ (which is the Big Bang). But it seems like we can show only that $ a = c$ at the Big Bang, but there may be other physical evidence for why we must have $c = 0$. Is this correct? For example, going back in time, how do we know that spacetime doesn't contract to a smaller size where $a \neq 0$? $\endgroup$
    – Jonathan Gafar
    Jun 21, 2016 at 15:19
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    $\begingroup$ @JonathanGafar The behaviour of $a(t)$ depends on how the energy density behaves. For photons and relativistic matter $\rho\propto a^{-4}$ and for non-relativistic matter $\rho\propto a^{-3}$. In both cases solving the Friedmann equation shows that as we go back in time $a$ reaches zero in a finite time. However if there is only dark energy/a cosmological constant present then $\rho$ is constant, and in this case $a$ does indeed approach zero asymptotically. This is the de Sitter geometry, and a de Sitter universe has no Big Bang. This would be best explored in the chat room or a new question. $\endgroup$
    – John Rennie
    Jun 21, 2016 at 15:26

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