I came to know that dark energy is constant.As mentioned in the Friedmann equations

How can a constant energy cause an accelerating rate of expansion in the universe?

  • $\begingroup$ There's a constant mass directly beneath you, yet if you jump in the air, it will produce an acceleration. I would not recommend taking this analogy any further, but do keep in mind that the thing causing an acceleration doesn't have to change in order for the acceleration to exist. $\endgroup$
    – HDE 226868
    Dec 3, 2016 at 14:48
  • 3
    $\begingroup$ Dark energy density is constant. More space = more total dark energy = accelerated expansion. $\endgroup$
    – zephyr
    Dec 3, 2016 at 15:22
  • $\begingroup$ @zephyr: You should make that an answer. $\endgroup$
    – pela
    Dec 3, 2016 at 18:22
  • $\begingroup$ Is this a homework question? $\endgroup$ Dec 3, 2016 at 18:39
  • $\begingroup$ @zephyr, you need to be careful about what you mean by "more space". Volume does not feature in the Friedmann equations and indeed many cosmological models are infinite so the total volume is always infinite. There is a sense that the increase of the proper volume of a sphere of constant comoving volume contribute to the increase of $\dot{a}$, which is why $\dot{a}$ can be increasing even when $H$ is decreasing. $\endgroup$
    – John Davis
    Dec 3, 2016 at 19:02

1 Answer 1


This is a very good question and it comes down to what is meant by "accelerating expansion".

In cosmology we define something called the scale factor $a$. By definition at the present time $a=1$, billions of years ago when linear distances on a cosmic scale (e.g. the distance to a faraway galaxy) were half what they are now $a$ was equal to $0.5$ and many years in the future when cosmic linear distances have increased to twice their current size the $a$ will equal $2$.

$a$ is a function of cosmological time and we also define its 1st derivative and 2nd derivatives, with respect to cosmological time, $\dot{a}$ and $\ddot{a}$.

$\dot{a}$ is a measure of the Universe's rate of expansion and so when $\dot{a}$ is increasing with time we say that the expansion of the Universe is accelerating. $\ddot{a}$ is a measure of the change of the rate of the Universe's expansion so equivalently we say if $\ddot{a}$ is positive then the expansion of the Universe is accelerating.

However note we have defined $a$ at the present time to be $1$, but the present time is entirely arbitrary. This means that $a$, $\dot{a}$ and $\ddot{a}$ depend on when we define $a$ to be equal to $1$. To get round this can instead think of the Hubble Parameter $H = \frac{\dot{a}}{a}$ as a better measure of the rate of expansion as it does not depend on when $a=1$.

In a de Sitter Universe (spatially flat, empty, positive cosmological constant) the Friedmann equations reduce to:

$$ \frac{\ddot{a}}{a} = H^2 = \frac{\Lambda c^2}{3}$$

From this we can see that $H$ and $\ddot{a}$ are both positive constants. So in one sense (the sense of $H$) the rate of expansion of the Universe is constant, but in another sense (in the sense of $\dot{a}$) the rate of expansion is increasing. So we can see a link between constant energy desnity (the energy density of the cosmological constant is constant) and a constant rate of expansion as long as we define the rate of expansion in a certain way.

In the currently-favoured cosmological model dark energy is a cosmological constant and we are currently in an era when dark energy dominates the dynamics of the Universe. (Looking at the full Friedmann equations) the contribution of matter reduces $\ddot{a}$, but as the cosmological constant dominates it is still positive and we say that the Universe's acceleration is expanding. Looking at $H$ however we say that, because the energy density of matter decreases with time asymptotically to zero, $H$ is decreasing asymptotically to the value given above for the de Sitter Universe.

  • $\begingroup$ I'm not entirely sure this really answers the question. You defined $a$, $\dot{a}$, and $H$ and how those relate to expansion, but don't really talk about dark energy except for your last sentence and it's not clear to me how that sentence answers the question. $\endgroup$
    – zephyr
    Dec 3, 2016 at 21:41
  • $\begingroup$ @zephyr My point is the important thing is how you define expansion, so that is why on concentrated more on the definition, but I've added more detail to how dark energy fits in. $\endgroup$
    – John Davis
    Dec 3, 2016 at 22:43

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