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

[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.

 

[mfb]

##\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.

 

[binbagsss]

##\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]

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.

 

[binbagsss]

##\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]

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]

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.:)

 

[binbagsss]

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 :)

 

[binbagsss]

Is the last line of what I did mathematially incorrect?

 

[mfb]

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