In order to get dark energy to dominate, wouldn't you first need another form of energy to push the expansion until dark energy could dominate? Otherwise I don't understand how the universe could shift from having a decelerating expansion to an accelerating expansion. Is there any analogy that could help understand this?


4 Answers 4


The title and the text actually ask two different questions. While Kyle Oman and Thriveth answer the title excellently, I'll address the question in the text which asks "Why did the Universe expand in the first place, before dark energy (DE) started to dominate".

The answer to this is inflation (we think). The first fraction of a second after the creation of space, it was dominated by "something" that mimicked the effect of DE, causing space to expand by a factor of $\sim e^{60}$. The epoch of inflation lasted until the Universe was some $10^{-32}\,\mathrm{s}$ old.

The expansion continued, but were slowed down by the mutual attraction of radiation, and later matter. If the ratio of DE-to-matter had been smaller, this attraction might have slowed it down sufficiently to halt the expansion before DE started dominating, but that was just not the case in our Universe.

Now what caused the inflation is another question, which someone else than me is better at answering. But I think the most accepted theory, or rather hypothesis, is some scalar field consisting of inflatons.


You request an analogy. I can give you the following:

Throw a rock into the air. Your push is inflation. The distance from Earth to the rock is the size of the Universe. The gravitational force between Earth and the rock is the mutual attraction between various forms of energy in the Universe. The speed of the rock is the expansion rate of the Universe. Now if your pitch was too weak, the rock will eventually fall back (Big Crunch), while if you throw hard enough (11 km/s), the rock will escape Earth's pull (Big Freeze). But even if the initial speed was less than 11 km/s, if the rock comes sufficiently close to the Moon (dark energy), it will start picking up speed and eventually escape.

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    $\begingroup$ Good call covering the other part of the question. And that's a decent analogy :) +1 $\endgroup$
    – Kyle Oman
    May 24, 2016 at 6:44
  • $\begingroup$ @KyleOman: I didn't realize until you said "the other part of the question", that the title asks a different question than the text, and that you and Thriveth actually answer the title perfectly well. I'll edit a bit… $\endgroup$
    – pela
    May 24, 2016 at 7:03
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    $\begingroup$ And I, on the other hand, was too tired to see that there was actually more to the question than the title. Oops :-) $\endgroup$
    – Thriveth
    May 24, 2016 at 8:34

Let's start partway through the expansion of the Universe in the matter dominated epoch. At this time the energy density is dominated by matter, but the dark energy and radiation components are still present, just relatively small. The Universe is expanding, but the expansion is gradually slowing down.

As the Universe expands, the density of matter scales as:

$$\rho_m \propto a^{-3}$$

This is intuitive: space is expanding, which goes as the cube of the scale, but the amount of matter remains constant, so the density drops.

Provided we're talking about the usual cosmological constant dark energy model, the energy density for that component scales as:

$$\rho_\Lambda \propto {\rm constant}$$

I.e. the energy density doesn't change as the Universe expands. Roughly speaking, if you think of the DE as "vacuum energy", then since there is more and more "vacuum" as the Universe expands, there is proportionally more and more DE. In absolute terms, then, the energy in DE increases with expansion. This is what allows the relative contribution from $\Lambda$ to eventually dominate over the matter density. As this happens, the dynamics gradually change to an accelerating expansion. They key is that the Universe gets "big enough" for there to be "enough vacuum" for $\Omega_\Lambda$ to overtake $\Omega_m$. The fact that the Universe is decelerating during matter domination doesn't matter (as long as it doesn't stop and start to recollapse too early); the Universe is still getting bigger, just at a slower rate.

While I'm at it, I may as well mention that the radiation energy density scales as:

$$\rho_r \propto a^{-4}$$

This is intuitive; there's a drop of $a^{-3}$ because of the increasing volume, and a further drop of $a^{-1}$ because of the redshift - the photon energy is $E=h\nu$, and $\nu\propto a^{-1}$.


I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $\mathrm dr/\mathrm dt = x ~\mathrm da/\mathrm dt$, and the second derivative $\mathrm d^2/\mathrm dt^2 = x ~\mathrm d^2a/\mathrm dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that depends on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \begin{align}\dot p = -\frac{\partial {\cal H}}{\partial a} &= \frac{16\pi G\rho a}{3} \\ \dot a &= \frac{\partial {\cal H}}{\partial p} \end{align}$$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0\exp\left[t \sqrt{8\pi G\rho/3}\right] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe. enter image description here

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    $\begingroup$ This chart would benefit from some units listed for the axes. $\endgroup$
    – Beska
    May 24, 2016 at 12:59

The short version: The amount of matter in the Universe is fixed, so as the Universe expands, matter density will drop because the same amount of matter will be spread out on more space.

Dark Energy, on the other hand, is (by definition) constant or almost constant in density. This means that no matter how dilute the Dark Energy is, if it waits long enough, the Matter density will drop below it and never get back up above it again. From this point, the Universe will be Dark Energy dominated, rather than matter dominated.

The energy to get to this points comes from the leftover momentum of the Epoch of Inflation. The Universe did indeed decelerate in the beginning, but not enough to prevent getting into the Dark Energy dominated epoch.


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