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2. Imagine a type of phantom energy with w<-1. (a) Write down the Friedmann equation for a flat universe containing matter
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(a) Consider the generalised time-like Friedman equation in this form:

Ωρα + (1 – Ωfotal)α 2 Π=-

where H_0 is the Hubble parameter, and \Omega is the ratio between density and critical density. Now, consider an universe with only matter and phantom energy. The Friedman equation then reduces to:

= Ωma + (1 – Ωδα 2

where the subscript denotes that this parameter corresponds to ordinary matter (the ordinary Friedman equation pops out if we neglect the second term), and since the total energy comes from matter and phantom energy, the contribution of the latter is just one minus the matter contribution.

Now, one of the space-like Friedman equation says:

à = Ha-(3w+1)/2

Integrating this and remembering that for phantom energy w<-1, we get:

towe {m) +1}}=>

Putting this in the time-like Friedman equation () and using that HP = (/a) , we have:

H2 = H (m/d3 + (1 - 2m)a-3(w+1)

(b) For this, we just equalize the parameters corresponding to matter and phantom energy respectively and get:

acy (- 1) = = (1+m}/(3 – 1) = /ury

(c) Now consider a phantom energy dominated universe, and hence we can drop the first term in the Friedman equation, and then we integrate (a/a)2 = HŽ {(1 -m)a=3(w+1) from t_0 to some time t_{rip} and get for a → 00:

a = HV1 -2 -3-1)/2

\Rightarrow \frac{1}{H_0\sqrt{1-\Omega_m}}\int_{a_0}^{\epsilon\to\infty}a^{(3w+1)/2}=\int_{t_0}^{t_{rip}}dt

\Rightarrow t_{rip}-t_0=\frac{2}{3H_0|1+w|\sqrt{1-\Omega_m}}

(d) Now, consider the ripping time. Putting in values of \Omega_m=0.04 and w = -1.1, it is easily seen that:

\Rightarrow t_{rip}-t_0=6.8H_0^{-1}

\Rightarrow t_{rip}=7.8H_0^{-1}\approx 8H_0^{-1}\approx 111\ Gyr

(e) It's easy to see that we cannot see them after the specified time, because the curvature of the universe is definitely not positive (else note that big rip is not possible), and hence there will always be some distant galaxies which will escape from our visible universe (horizon), just because of the curvature of the universe cause by the phantom energy. And also, after ripping time, nothing will be gravitationally bound and one can extrapolate that the very structure of space-time falls.

(f) To answer this, note that the source for the gravitational potential is the volume integral of p+ 3p and hence a planet in an orbit of radius R around a star of mass M will become unbound roughly when -(4\pi/3)(\rho+3p)R^3\approx M . With w<-1. such a system would gravitationally decouple after time t\approx P\sqrt{2|1+3w|}/(6\pi|1+w|) where P is the period of revolution and in most practical cases t\leq t_{rip}, and hence the solar system would no longer be gravitationally bound at t_{rip}.

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