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Suppose that we could measure the current acceleration with respect to us of a distant galaxy chosen at random in our observable universe (Ignore the peculiar motion of the galaxy). Is the acceleration constant, is it increasing or is it decreasing?

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I think the correct answer is that nobody knows - it all depends on the behaviour, or equation of state, of the dark energy.

If the dark energy takes the form of a cosmological constant, then the energy density due to dark energy will end up completely dominant as the universe expands.

The acceleration of the scale factor $a$ is given by $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3P}{c^2}\right) $$ where $\rho$ is the energy density, and $P$ is the pressure. A cosmological constant has $P = -\rho c^2$, which leads to $$ \ddot{a} = \frac{8\pi G \rho a}{3}$$

So, if dark energy dominates, then $\rho a$ increases with time, so the acceleration is positive and increasing with $a$ and hence time.

If matter dominates, which it did in the past, then $P \simeq 0$ and $\rho a \propto a^{-2}$, so the acceleration is negative, with a magnitude that becomes smaller.

The transition from deceleration to acceleration happens when $\rho_{\Lambda} + \rho_m + 3P/c^2= 0$. If we assume a cosmological constant with $P = -\rho_{\Lambda} c^2$, where $\rho_{\Lambda}$ is the dark energy density (which is constant), and $\rho_m$ is the matter density, where $\rho_m = \rho_0 a_0^{3}/a(t)^3$ and $\rho_0$ and $a_0$ are the present-day matter density and scale factor respectively. Thus $$ \rho_{\Lambda} + \rho_m -3\rho_{\Lambda} = 0$$ and the transition point occurs when $\rho_m = 2 \rho_{\Lambda}$. But because $\rho_{\Lambda}/\rho_0 = 0.7/0.3$, then $$ \rho_{\Lambda} = \frac{7\rho_0}{3} = \frac{7}{3}\left(\frac{a}{a_0}\right)^3 \rho_m$$ and so the scale factor at the time of transition from decleration to acceleration is found from $$ \rho_m = \frac{14}{3}\left(\frac{a}{a_0}\right)^3 \rho_m,$$ $$ a = \left(\frac{3}{14}\right)^{1/3} a_0$$ i.e. when the scale factor was 60% of the present day value. According to this cosmology calculator, this happened roughly 6 billion years ago.

Thus the universe was decelerating then accelerated. So clearly the acceleration now is increasing with time.

On the other hand, if for whatever reason, the dark energy started to "decay" then it might be that the acceleration would decrease.

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  • $\begingroup$ Hubble's law itself (v = HoD) implies some slight acceleration, as distant galaxies become more distant with each passing second, though I suspect that's not what's being asked about here: en.wikipedia.org/wiki/Hubble%27s_law $\endgroup$ – Wayfaring Stranger Aug 18 '15 at 12:29
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    $\begingroup$ @Rob shouldn't your first equation contain the scale factor a in the right side? I don't understand why it then appears in your second equation. $\endgroup$ – set5 Aug 19 '15 at 19:58
  • $\begingroup$ @Rob To introduce your first equation, you talk about the acceleration of the scale factor. Which you simbolize as the second derivative of a with respect to time. I would be interested to know the form of the function a, if it is known, I mean, is it a polynomial? is it a exponential function? $\endgroup$ – set5 Aug 19 '15 at 20:26
  • $\begingroup$ @mick It's messy and I think has to be numerically solved. But you are correct about the missing a. $\endgroup$ – ProfRob Aug 19 '15 at 20:38
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    $\begingroup$ @mick See eagle.phys.utk.edu/guidry/astro421/lectures/lecture490_ch19.pdf for a discussion of solutions for $a(t)$ in various approximations. A numerical solution is discussed at the end. $\endgroup$ – ProfRob Aug 19 '15 at 20:53

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