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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]

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.

## \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].

- 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 | ## .

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.**
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