<p>You answered your question actually. You have said</p>
<blockquote>
<p>Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1</p>
</blockquote>
<p>then you are saying</p>
<blockquote>
<p>But if 'w' is negative-one then 'a' is proportional to tt</p>
</blockquote>
<p>The <span class="math-container">$w=-1$</span> does not apply to <span class="math-container">$a(t) \propto t^{2/3(1+w)}$</span></p>

CC BY-SA 4.0
2021-06-16T07:27:16.833Z
Soo, what solution applies then to the case of a universe dominated by the cosmological constant (what is the expression for $a(t)$)? :)
NeStack
CC BY-SA 4.0
2020-10-27T08:58:12.427Z
2020-10-27T08:58:12.427Z
seVenVo1d
<p>The Friedmann equations can be solved exactly in presence of a perfect fluid with equation of state</p>
<p><span class="math-container">$${\displaystyle p=w\rho c^{2}} \qquad p=w\rho c^2$$</span></p>
<p>where <span class="math-container">${\displaystyle p}$</span> is the pressure, <span class="math-container">${\displaystyle \rho }$</span>  is the mass density of the fluid in the co-moving frame and <span class="math-container">$w$</span> is some constant.</p>
<p>In spatially flat case (<span class="math-container">$k = 0$</span>), the solution for the scale factor is</p>
<p><span class="math-container">$${\displaystyle a(t)=a_{0}\,t^{\frac {2}{3(w+1)}}}$$</span>
where <span class="math-container">${\displaystyle a_{0}}$</span> is some integration constant to be fixed by the choice of initial conditions. This family of solutions labelled by <span class="math-container">${\displaystyle w}$</span> is extremely important for cosmology. E.g. <span class="math-container">${\displaystyle w=0}$</span> describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as</p>
<p><span class="math-container">$${\displaystyle a(t)\propto t^{2/3}}$$</span> matter-dominated
Another important example is the case of a radiation-dominated universe, i.e., when <span class="math-container">${\displaystyle w=1/3}$</span>. This leads to</p>
<p><span class="math-container">$${\displaystyle a(t)\propto t^{1/2}}$$</span> radiation dominated
Note that this solution is not valid for domination of the cosmological constant, which corresponds to an <span class="math-container">${\displaystyle w=-1}$</span>. In this case the energy density is constant and the scale factor grows exponentially.</p>
<p>So, '<span class="math-container">$a$</span>' is proportional to <span class="math-container">$t^{2/3}$</span> or <span class="math-container">$t^{1/2}$</span> for matter- or radiation-dominated universes, respectively... But if '<span class="math-container">$w$</span>' is negative-one then '<span class="math-container">$a$</span>' is proportional to <span class="math-container">$t^t$</span>?
I mean, what is the exponent in this 'exponential growth' phase where the '<span class="math-container">$w$</span>' 'constant' is <span class="math-container">$-1$</span>?</p>

CC BY-SA 4.0
2020-10-16T07:48:48.293Z
Exponential means $a \propto \exp(t/t_0)$. Your question title mentions dark matter, but the body is asking about dark energy? They aren't the same thing.
ProfRob
<astrophysics><cosmology><big-bang-theory><theories>
What is the 'scale factor' equation for a dark-matter dominated universe?
Kurt Hikes
# What is the 'scale factor' equation for a dark-matter dominated universe?

(Asked by: username_0 on 2020-10-15)

The Friedmann equations can be solved exactly in presence of a perfect fluid with equation of state

$${\displaystyle p=w\rho c^{2}} \qquad p=w\rho c^2$$

where ${\displaystyle p}$ is the pressure, ${\displaystyle \rho }$ is the mass density of the fluid in the co-moving frame and $w$ is some constant.

In spatially flat case ($k = 0$), the solution for the scale factor is

$${\displaystyle a(t)=a\_{0}\,t^{\frac {2}{3(w+1)}}}$$
where ${\displaystyle a\_{0}}$ is some integration constant to be fixed by the choice of initial conditions. This family of solutions labelled by ${\displaystyle w}$ is extremely important for cosmology. E.g. ${\displaystyle w=0}$ describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as

$${\displaystyle a(t)\propto t^{2/3}}$$ matter-dominated
Another important example is the case of a radiation-dominated universe, i.e., when ${\displaystyle w=1/3}$. This leads to

$${\displaystyle a(t)\propto t^{1/2}}$$ radiation dominated
Note that this solution is not valid for domination of the cosmological constant, which corresponds to an ${\displaystyle w=-1}$. In this case the energy density is constant and the scale factor grows exponentially.

So, '$a$' is proportional to $t^{2/3}$ or $t^{1/2}$ for matter- or radiation-dominated universes, respectively... But if '$w$' is negative-one then '$a$' is proportional to $t^t$?
I mean, what is the exponent in this 'exponential growth' phase where the '$w$' 'constant' is $-1$?

**Tags:** astrophysics, cosmology, big-bang-theory, theories
> **Comments:**
> **username_1**: Exponential means $a \propto \exp(t/t_0)$. Your question title mentions dark matter, but the body is asking about dark energy? They aren't the same thing.  

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## 1 answers

### Answer by username_2 on 2020-10-27. Score: 1

You answered your question actually. You have said

> Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1

then you are saying

> But if 'w' is negative-one then 'a' is proportional to tt

The $w=-1$ does not apply to $a(t) \propto t^{2/3(1+w)}$

> **Comments:**
> **username_3**: Soo, what solution applies then to the case of a universe dominated by the cosmological constant (what is the expression for $a(t)$)? :)  

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