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. $\endgroup$ – Rob Jeffries Sep 18 at 18:41
  • $\begingroup$ In the image it is seen clearly that the universe is decelerating. But in reality it is accelerating @ Rob Jeffries $\endgroup$ – Soumanti Chakraborty Sep 18 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. $\endgroup$ – Rob Jeffries Sep 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. $\endgroup$ – Soumanti Chakraborty Sep 19 at 15:23
  • $\begingroup$ Are you still interested for an answer ? $\endgroup$ – Reign 3 hours ago

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