2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well, this is embarrassing... $\endgroup$ – Jonathan Gafar Jun 21 '16 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 '16 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 '16 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 '16 at 15:19
  • 1
    $\begingroup$ @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 '16 at 15:26

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.