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

<p>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.</p>

<p>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
$$ \ddot{a} = \frac{8\pi G \rho a}{3}$$</p>

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

<p>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.</p>

<p>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 <a href="http://home.fnal.gov/~gnedin/cc/" rel="nofollow">cosmology calculator</a>, this happened roughly 6 billion years ago.</p>

<p>Thus the universe was decelerating then accelerated. So clearly the acceleration now is <em>increasing</em> with time.</p>

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

CC BY-SA 3.0
2015-08-18T12:29:01.760Z
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: https://en.wikipedia.org/wiki/Hubble%27s_law
Wayfaring Stranger
CC BY-SA 3.0
2015-08-19T19:58:42.853Z
@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.
set5
CC BY-SA 3.0
2015-08-19T20:26:55.247Z
@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?
set5
CC BY-SA 3.0
2015-08-19T20:38:11.823Z
@mick It's messy and I think has to be numerically solved. But you are correct about the missing a.
ProfRob
CC BY-SA 3.0
2015-08-19T20:53:20.360Z
@mick See http://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.
ProfRob
CC BY-SA 3.0
2015-08-22T22:54:52.010Z
@Rob that's interesting. So it would be right to say that the expansion of the Universe accelerated during inflation, then decelerated until 6 billion  years ago, then accelerated again?
set5
CC BY-SA 3.0
2015-08-22T23:31:58.550Z
@mick For the usual inflation model and a cosmological constant, yes.
ProfRob
CC BY-SA 3.0
2015-08-17T22:41:03.697Z
2015-08-19T20:39:01.457Z
2015-08-19T20:39:01.457Z
ProfRob
<p>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?</p>

<cosmology><expansion>
Are galaxies moving away with constant acceleration?
set5
# Are galaxies moving away with constant acceleration?

(Asked by: username_0 on 2015-08-17)

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?

**Tags:** cosmology, expansion

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## 1 answers

### Answer by username_1 on 2015-08-17. Score: 3 | Accepted answer

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](http://home.fnal.gov/~gnedin/cc/), 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.

> **Comments:**
> **username_2**: 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: https://en.wikipedia.org/wiki/Hubble%27s_law  
> **username_0**: @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.  
> **username_0**: @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?  
> **username_1**: @mick It's messy and I think has to be numerically solved. But you are correct about the missing a.  
> **username_1**: @mick See http://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.  
> **username_0**: @Rob that's interesting. So it would be right to say that the expansion of the Universe accelerated during inflation, then decelerated until 6 billion  years ago, then accelerated again?  
> **username_1**: @mick For the usual inflation model and a cosmological constant, yes.  

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