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 $\ddot{a}<0$.

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 $\ddot{a}<0$.

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 $$\ddot{a}=-\frac{4 \pi G}{3} \left(\rho + \frac{p}{c^2} \right)=\frac{4 \pi G}{3} \times 5\rho = \frac{4 \pi G}{3c^2} 5\varepsilon$$ Since $\varepsilon <0$, this implies $\ddot{a}<0$.

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 $\ddot{a}<0$.