# Big Bang singularity

Does a FRW universe with the equation of state:

$$p = -\frac{\rho c^2}{3}$$

have a singularity at the Big Bang?

I was looking at:

http://en.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

and trying to decide if such a Universe obeys the "dominant energy condition" for the Penrose-Hawking theorem (energy greater than pressure) or not.

Last edited:

Simon Bridge
Homework Helper
What would it mean to have "a singularity at the big bang"?
Crunch the numbers and see.

Chalnoth
Does a FRW universe with the equation of state:

$$p = -\frac{\rho c^2}{3}$$

have a singularity at the Big Bang?

I was looking at:

http://en.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

and trying to decide if such a Universe obeys the "dominant energy condition" for the Penrose-Hawking theorem (energy greater than pressure) or not.
The dominant energy condition basically states that mass-energy can never be observed to be flowing faster than light:
http://en.wikipedia.org/wiki/Energy_condition#Dominant_energy_condition

So yes, it holds necessarily.

But it's also pretty easy to show it using the Friedmann equations.

First, we can get a(t) using the second derivative equation with $\Lambda=0$:

$${\ddot{a} \over a} = -{4 \pi G \over 3} \left( \rho + {3p \over c^2}\right) = 0$$

Thus, we can write:
$$a(t) = a(t=0) + H_0 t$$

This is important because it shows that $a = 0$ at some finite time.

Second, energy conservation shows that the energy density of this type of matter scales as $\rho \propto 1/a^2$. Thus, when $a=0$, the energy density is infinte, so there's your singularity.

The only universe in which there doesn't appear, at first glance, to be a singularity is the de Sitter universe, where $\rho$ is a constant. But this doesn't quite work out because even a single photon causes a singularity to happen somewhere in the finite past, and a de Sitter universe produces Hawking radiation at its horizon.