Dark energy with equation of state parameter $\omega<-1$ is called phantom dark energy. Its energy density increases with time.

My question is whether this model violates the concept of accelerating universe, because the energy density turns out to be negative. If we use these values in the acceleration equation, we see the acceleration of the universe (second derivative of the scale factor) is negative.

My derivation is as follows. Let's set $\omega=-2$, so the equation of state is $p=-2\rho c^2$. From the fluid equation, $$\dot{\rho}+\frac{3\dot{a}}{a} \left(\rho + \frac{p}{c^2} \right)=0\Rightarrow\rho\propto a^3$$ From the Friedman equations, $a=-k_1 t^{-2/3}$, so the energy density is $\varepsilon=\rho c^2=-kt^{-2}<0$. Hence the acceleration is $$\ddot{a}=-\frac{4 \pi G}{3} \left(\rho + \frac{3p}{c^2} \right)=\frac{4 \pi G}{3} \times 5\rho = \frac{4 \pi G}{3c^2} 5\varepsilon$$ Since $\varepsilon <0$, this implies $\ddot{a}<0$.

  • 2
    $\begingroup$ This question needs more work to be understood. $\endgroup$
    – ProfRob
    Sep 18, 2019 at 18:41
  • $\begingroup$ In the image it is seen clearly that the universe is decelerating. But in reality it is accelerating @ Rob Jeffries $\endgroup$
    – user28809
    Sep 18, 2019 at 19:13
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    $\begingroup$ I don't follow where your negative solution to the Friedman equation comes from (and it obvously doesn't make sense in an expanding universe) or why the energy density is negative, or where the $5\rho$ comes from in the last line. $\endgroup$
    – ProfRob
    Sep 19, 2019 at 10:09
  • 1
    $\begingroup$ @SoumantiChakraborty Why would phantom energy density be negative ? $\endgroup$
    – seVenVo1d
    Nov 20, 2019 at 20:43
  • 2
    $\begingroup$ Why is DM (dark matter) relevant to this question? $\endgroup$
    – PM 2Ring
    Apr 19, 2020 at 9:55

2 Answers 2


For phantom energy model we have $$H^2 = H_0^2\Omega_pa^{-3-3w_p}$$ such that $w_p < -1$

$$\dot{a}^2 = H_0^2\Omega_pa^{-1-3w_p}$$

$$\dot{a} = H_0 \sqrt{\Omega_p} a^{-1/2(1+3w_p)}$$

$$\int a^{1/2(1+3w_p)}da = \int H_0 \sqrt{\Omega_p}dt$$

Now at this part we should define the integrals upper and lower bounds. When we set time from $t → t_0$

$$\int_a^{1} a^{1/2(1+3w_p)}da = \int_{t}^{t_0} H_0 \sqrt{\Omega_p}dt$$

We get something like,

$$\frac{2} {3(1+w_p)}[ 1 - a^{3/2(1+w_p)}] = H_0 \sqrt{\Omega_p}(t_0 - t)$$

$$a^{3/2(1+w_p)} =1 - \frac{3(1+w_p)}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$

Let us take $w_p = -2$

$$a^{-3/2} =1 + \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$


$$a =(1 + \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t))^{-2/3} $$

enter image description here

here $y$ axis is the scale factor $(a)$ and $x$ axis is the time $(t)$

For $C = H_0 \sqrt{\Omega_p}$ and $a=t_0$

As you can see scale factor goes to infinity

For the acceleration equation

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2}\epsilon_p(1+3w_p)$$

For $w_p=-2$

$$\frac{\ddot{a}}{a} = \frac{4\pi G}{3c^2}5\epsilon_p$$

From the graph and also from here its clear that $\ddot{a} > 0$.


To explain more plainly what is going on here, the (now deleted) asker noticed that solving the first Friedmann equation for a phantom dark energy model leads to a universe whose size scales as a negative power of the time. For the equation of state parameter $w=2$, the dark energy density grows as $\rho\propto a^3$, where $a$ is the scale factor. The Hubble rate then grows as $$H\equiv\frac{\mathrm{d}a/\mathrm{d}t}{a}\propto a^{3/2}$$ or $$\frac{\mathrm{d}a}{a^{5/2}}\propto \mathrm{d}t.$$ Integrating this without paying attention to integration constants leads to $t\propto a^{-3/2}$ and hence $a\propto t^{-2/3}$. That is odd -- apparently the universe must shrink? Alternatively, if the proportionality constant is negative, the universe must have negative size and energy density?

The resolution is to pay attention to the integration constant. Integrating the equation above really leads to $a\propto (\tilde t-t)^{-2/3}$, where $\tilde t$ is the integration constant. This universe grows and accelerates for $t<\tilde t$, with $a\to\infty$ as $t\to\tilde t$. $\tilde t$ is the time of the Big Rip!


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