<blockquote>
  <p>AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe.</p>
</blockquote>

<p>If some articles on the web suggest that, they are mistaken. Dark energy affects the acceleration of the cosmic expansion, but it is not necessary for the universe to be expanding, so it cannot be "the reason behind the expansion".</p>

<p>How the dark energy density changes as the universe expands depends on its current energy density and its equation of state--the relationship between its energy density and pressure. For cosmological constant provides the simplest dark energy model in cosmological models--it's the $\Lambda$ in the <a href="http://en.wikipedia.org/wiki/Lambda-CDM_model" rel="nofollow noreferrer">ΛDCM model</a>--for which the pressure is exactly the negative of its energy density: $p = -\rho$. In general relativity, both energy density and pressure gravitate, and in three spatial dimensions expansion the acceleration is proportional to $\rho + 3p$ (cf. also <a href="http://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow noreferrer">Friedmann equations</a>). As a result, a cosmological constant with <em>positive</em> energy density $\rho&gt;0$ leads to <em>accelerated expansion</em>, because the contribution of its negative pressure means the overall effect is repulsive.</p>

<blockquote>
  <p>If so, why isn't it getting used up in doing so.</p>
</blockquote>

<p>Somewhat ironically the answer to your question for cosmological constant can be interpreted like so: it doesn't "get used up" <em>because</em> of a kind of energy conservation.</p>

<p>As a given small volume expands, it keeps a constant energy density $\rho$ and a constant pressure $p$. Is this situation sensible? Suppose its volume increased by $\Delta V$, so its energy content increased by $\rho\Delta V$. If the pressure is constant, this is some kind of <a href="http://en.wikipedia.org/wiki/Isobaric_process" rel="nofollow noreferrer">isobaric process</a>, so the work done is $W = p\Delta V$. But if $p = -\rho$, then everything is consistent because the dark energy does negative work as it expands to exactly balance the increase in its energy.</p>

<p>In reality, you shouldn't take the above reasoning too literally. There are many issues in interpreting anything like "how much energy" there is in general relativity and whether or not it's conserved, some of which I've covered in <a href="https://astronomy.stackexchange.com/a/1500/661">this</a> answer, though some of which can be alleviated by the stressing of the "<em>small</em> volume" qualifier. Still, "it doesn't get used up because the equations of the theory say it shouldn't be getting used up", while perfectly true, would not be a very interesting answer.</p>

<p>Additionally, it should be clear from the above that the cosmological constant case of $p = -\rho$ is very itself special. The reason for its simplicity is that it's completely Lorentz-invariant: in different local inertial frames, the energy density and pressure of a perfect fluid get intermixed and transformed into different values... except in the case where the pressure is exactly the negative of energy density. Every observer would agree about the cosmological constant, so one can interpret it as an intrinsic energy density of the vacuum. In fact that's how it was originally introduced; other forms of dark energy are a generalization of $\Lambda$.</p>

<blockquote>
  <p>And consequently, the rate of expansion should reduce! </p>
</blockquote>

<p>With a slightly more general equation of state $p = w\rho$, the cosmological constant corresponds to $w = -1$. Other cases give a perfect fluid, which might indeed "get used up" in the sense of getting more dilute, and for positive energy density, $w&lt;-1/3$ is necessary to outwardly accelerate cosmic expansion. But something curious happens if $w&lt;-1$: we should expect from the work done argument for the energy density to <em>accumulate</em> instead of remaining constant. This kind of "phantom energy" would, if positive, produce a <a href="http://en.wikipedia.org/wiki/Big_Rip" rel="nofollow noreferrer">Big Rip</a> scenario, in which the cosmic expansion accelerates to an infinite value in finite cosmological time, ending the universe by tearing everything apart.</p>

CC BY-SA 3.0
2014-03-24T10:39:50.123Z
2014-03-24T10:39:50.123Z
2017-04-13T12:59:07.017Z
Stan Liou
<p>AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe. If so, why isn't it getting used up in doing so. And consequently, the rate of expansion should reduce! </p>

CC BY-SA 3.0
2014-03-24T09:17:43.373Z
First of all, what article?
Py-ser
CC BY-SA 3.0
2014-03-24T15:06:43.487Z
No - Dark energy is the cause of the acceleration of the universe. There are other combinations of $\Omega_{i}$ which cause the universe to expand and contract which do not contain dark energy.
astromax
<dark-energy>
Why isn't the dark energy getting decreased?
Spacetrekker
# Why isn't the dark energy getting decreased?

(Asked by: username_0 on 2014-03-24)

AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe. If so, why isn't it getting used up in doing so. And consequently, the rate of expansion should reduce!

**Tags:** dark-energy

> **Comments:**
> **username_1**: First of all, what article?  
> **username_2**: No - Dark energy is the cause of the acceleration of the universe. There are other combinations of $\Omega_{i}$ which cause the universe to expand and contract which do not contain dark energy.  

---

## 1 answers

### Answer by username_3 on 2014-03-24. Score: 6

> AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe.

If some articles on the web suggest that, they are mistaken. Dark energy affects the acceleration of the cosmic expansion, but it is not necessary for the universe to be expanding, so it cannot be "the reason behind the expansion".

How the dark energy density changes as the universe expands depends on its current energy density and its equation of state--the relationship between its energy density and pressure. For cosmological constant provides the simplest dark energy model in cosmological models--it's the $\Lambda$ in the [ΛDCM model](http://en.wikipedia.org/wiki/Lambda-CDM_model)--for which the pressure is exactly the negative of its energy density: $p = -\rho$. In general relativity, both energy density and pressure gravitate, and in three spatial dimensions expansion the acceleration is proportional to $\rho + 3p$ (cf. also [Friedmann equations](http://en.wikipedia.org/wiki/Friedmann_equations)). As a result, a cosmological constant with *positive* energy density $\rho>0$ leads to *accelerated expansion*, because the contribution of its negative pressure means the overall effect is repulsive.

> If so, why isn't it getting used up in doing so.

Somewhat ironically the answer to your question for cosmological constant can be interpreted like so: it doesn't "get used up" *because* of a kind of energy conservation.

As a given small volume expands, it keeps a constant energy density $\rho$ and a constant pressure $p$. Is this situation sensible? Suppose its volume increased by $\Delta V$, so its energy content increased by $\rho\Delta V$. If the pressure is constant, this is some kind of [isobaric process](http://en.wikipedia.org/wiki/Isobaric_process), so the work done is $W = p\Delta V$. But if $p = -\rho$, then everything is consistent because the dark energy does negative work as it expands to exactly balance the increase in its energy.

In reality, you shouldn't take the above reasoning too literally. There are many issues in interpreting anything like "how much energy" there is in general relativity and whether or not it's conserved, some of which I've covered in [this](https://astronomy.stackexchange.com/a/1500/661) answer, though some of which can be alleviated by the stressing of the "*small* volume" qualifier. Still, "it doesn't get used up because the equations of the theory say it shouldn't be getting used up", while perfectly true, would not be a very interesting answer.

Additionally, it should be clear from the above that the cosmological constant case of $p = -\rho$ is very itself special. The reason for its simplicity is that it's completely Lorentz-invariant: in different local inertial frames, the energy density and pressure of a perfect fluid get intermixed and transformed into different values... except in the case where the pressure is exactly the negative of energy density. Every observer would agree about the cosmological constant, so one can interpret it as an intrinsic energy density of the vacuum. In fact that's how it was originally introduced; other forms of dark energy are a generalization of $\Lambda$.

> And consequently, the rate of expansion should reduce!

With a slightly more general equation of state $p = w\rho$, the cosmological constant corresponds to $w = -1$. Other cases give a perfect fluid, which might indeed "get used up" in the sense of getting more dilute, and for positive energy density, $w<-1/3$ is necessary to outwardly accelerate cosmic expansion. But something curious happens if $w<-1$: we should expect from the work done argument for the energy density to *accumulate* instead of remaining constant. This kind of "phantom energy" would, if positive, produce a [Big Rip](http://en.wikipedia.org/wiki/Big_Rip) scenario, in which the cosmic expansion accelerates to an infinite value in finite cosmological time, ending the universe by tearing everything apart.


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