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

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## 

## 0= -\partial_{0} \rho - 3\frac{\dot{a}}{a}(\rho + p)## 

##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 .

- Simply multiply eq  by a, eq 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 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 | ##

Last edited:

mfb
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.

##\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.

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?

mfb
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.

##\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.

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?

mfb
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.

DEvens
Gold Member
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.:)

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.:)

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

Is the last line of what I did mathematially incorrect?

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