<p>When including a cosmological constant, the acceleration equation reads
$$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\epsilon+3P)+\frac{\Lambda}{3}$$
In a pressureless universe, $P=0$, and in a static universe $\frac{\ddot{a}}{a}=0$. The equation becomes:
$$\frac{4\pi G}{3c^2}\epsilon=\frac{\Lambda}{3}$$Energy density is always positive and non-zero, so $\Lambda&gt;0$.</p>

<p>The solution is unstable since after a slight perturbation, if $a$ grows, $\epsilon$ will go down ($\epsilon_m\propto a^{-3}$ for matter for example), which will result in an increase of the acceleration, making $a$ go up even more.</p>

CC BY-SA 4.0
2018-06-07T13:35:17.357Z
2018-06-07T13:42:39.643Z
2018-06-07T13:42:39.643Z
user22391
<p>This is the question:</p>

<p><em>By considering both the Friedmann and acceleration equations, and assuming a pressureless
Universe, demonstrate that in order to have a static Universe we must have
a closed Universe with a positive vacuum energy. Using either physical arguments
or mathematics, demonstrate that this solution must be unstable.</em></p>

<p>I started off with the friedmann equation and equated H^2=0 and K>0. How to proceed further and prove that lamda is positive?</p>

<universe><astrophysics><cosmology>
Question on Friedman equation involving cosmological constant
user193469
# Question on Friedman equation involving cosmological constant

(Asked by: username_0 on 2018-06-05)

This is the question:

*By considering both the Friedmann and acceleration equations, and assuming a pressureless
Universe, demonstrate that in order to have a static Universe we must have
a closed Universe with a positive vacuum energy. Using either physical arguments
or mathematics, demonstrate that this solution must be unstable.*

I started off with the friedmann equation and equated H^2=0 and K>0. How to proceed further and prove that lamda is positive?

**Tags:** universe, astrophysics, cosmology


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## 1 answers

### Answer by username_1 on 2018-06-07. Score: 1

When including a cosmological constant, the acceleration equation reads
$$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\epsilon+3P)+\frac{\Lambda}{3}$$
In a pressureless universe, $P=0$, and in a static universe $\frac{\ddot{a}}{a}=0$. The equation becomes:
$$\frac{4\pi G}{3c^2}\epsilon=\frac{\Lambda}{3}$$Energy density is always positive and non-zero, so $\Lambda>0$.

The solution is unstable since after a slight perturbation, if $a$ grows, $\epsilon$ will go down ($\epsilon\_m\propto a^{-3}$ for matter for example), which will result in an increase of the acceleration, making $a$ go up even more.


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