We know that dark energy with value of state parameter $\omega<-1$ is called phantom dark energy model. In this model if we assume dark energy to be a fluid with value of state parameter $\omega=-2$, we can say that the energy density of such dark matter increases with time.

My question is whether the above statement violates the concept of accelerating universe, because both the energy density and matter density turn 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. We have $\omega=-2$, 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 $\ddot{a}<0$.

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    $\begingroup$ This question needs more work to be understood. $\endgroup$ – ProfRob Sep 18 '19 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 '19 at 19:13
  • $\begingroup$ 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 '19 at 10:09
  • $\begingroup$ It's very easy. Just integrating the Friedmann equation you can derive the relation that the scale factor is proportional to t^-2/3 as it is shown above. Now If you simply put the value of p in the acceleration equation you will get (5×rho) in the last line. $\endgroup$ – user28809 Sep 19 '19 at 15:23
  • $\begingroup$ @SoumantiChakraborty Why would phantom energy density be negative ? $\endgroup$ – Layla Nov 20 '19 at 20:43

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 $\ddot{a} > 0$.


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