Question

2. Imagine a type of "phantom energy" with w<-1. (a) Write down the Friedmann equation for a flat universe containing matter and phantom energy 1/(3 ) (b) Show that the matter-phantom equality scale factor is amp = (1- 0) (c) In the phantom energy dominated epoch show that a → at a finite time trip-to = 200-311+wl This is called a "big rip". Hint: integrate over the time range from to to trip. (d) Current data is consistent with w = -1.1, use benchmark values for H, and mo to show that we might have 8 ' time left. Fortunately you can show this is 111 Gyr. (e) What happens to distant galaxies during the big rip? Can we still see them after trip? Determine this by comparing their distance from us to the horizon size after trip (f) What would happen to the Milky Way during a big rip? (Consider the scale on which Friedmann equations apply).

Answer 1

(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 , we get:

w<-1

towe {m) +1}}=>

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

HP = (/a)

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 , it is easily seen that:

w = -1.1

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

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

(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 around a star of mass 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 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 $t_{rip}$.