# Limits FRW universe , rate of expansion, k=-1,0.

1. Dec 28, 2014

### binbagsss

Hi,

I'm looking at deriving the limits of $\dot{a}$ as $a-> \infty$ , using the Friedmann equation and conservtion of the $_{00}$ component of the energy momentum tensor for a perfect fluid. Both of these equations respectively are:

$\dot{a^{2}}=\frac{8\pi G}{3} \rho a^{2} + | k |$ (from Friedmann and using $k \leq 0$ [1]

$0= -\partial_{0} \rho - 3\frac{\dot{a}}{a}(\rho + p)$ [2]

$a$ is the scale factor, $\rho$ is the density

I've just seen a proof when modelling this perfect fluid as dust, where dust obeys $\rho a^{3}$ is a constant. I'm looking at how you generalise this proof to, radiation say - which obeys $\rho a^{4}$ is constant. I will now state the solution proof for dust, and my attempt for the radiation proof. My question is whether my radiation proof is valid - I doubt it because the last step looks a bit dodgy - can you simply divided by $a$ and have the limits still hold - that is $\dot{a^{2}} -> | k |$ ...

Dust Proof:

$\partial_{0} \rho a^{3} = a^{3} (\dot{\rho} + 3\frac{\rho\dot{a}}{a})= -3pa^{2}\dot{a}$.
RHS $< \leq 0$ => $\rho a^{2} -> 0$ as $a -> \infty$
=> $\dot{a^{2}} -> | k |$ from eq [1].

My Radiation Proof:
- Simply multiply eq [1] by a, eq[2] by $a^{4}$ as a pose to $a^{3}$ as done in the worked dust proof. I then have similar conclusions , so $\rho a^{3} -> 0$ as $a -> \infty$ and so by eq[1] multiplied by a I conclude that $a\dot{a^{2}} -> a | k |$ .

This is the line I'm concerned with, dividing by $a$ am I then ok to claim that $\dot{a^{2}} -> | k |$

Thanks in advance.

Last edited: Dec 28, 2014
2. Dec 28, 2014

### Staff: Mentor

$\rho a^4 = c$ implies $\dot {a^2} = \frac{c'}{a^2} + |k|$ with constants c and c'. As $a \to \infty$, $\dot {a^2} \to |k|$?

Pressure will be different for dust and photons, that could be relevant in the derivation of those equations.

3. Dec 28, 2014

### binbagsss

Thanks. That looks like a better way to do it. Looking att the full derivation, apologie I don't have time to include it all here, pressure varying will not affect the derivation. So I conlude that $\dot {a^2} -> 1$ for an open universe with k=-1 and $\dot {a^2} -> 0$ when k=0. Are these the known limiting expansion values for the dust and radiation case? That they have the same limit, does this sound correct?

4. Dec 28, 2014

### Staff: Mentor

You are considering the evolution of a universe with negligible mass and radiation and without cosmological constant. There is nothing that would accelerate or slow expansion, the solution looks natural.

5. Dec 30, 2014

### binbagsss

Sorry to re-bump. So post 2 proves the limits without needing the energy-momentum tensor or the conservation of energy? I'm just wondering why the source I used , used them if it can be shown without.

Is what I did generally mathematically incorrect?

6. Dec 30, 2014

### Staff: Mentor

In an empty universe ($\rho \approx 0$) without cosmological constant, the constant $\dot a$ is a direct consequence of the first Friedmann equation.
The evolution until the universe is nearly empty is different and can need a more detailed analysis. If the initial matter density is too high, you never reach that point.

7. Dec 31, 2014

### DEvens

Not really relevant to the discussion above but...

FRW cosmology is "introductory physics homework?" Wow. I wish I had gone to a school where that was the case.:)

8. Dec 31, 2014

### binbagsss

ha. it says undergrad in the description. I'm a final year :)

9. Dec 31, 2014

### binbagsss

Is the last line of what I did mathematially incorrect?

10. Dec 31, 2014

### Staff: Mentor

If you know $a$ has a lower bound (at least after some time T), it works.

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