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Can the scale factor in a Robertson-Walker metric with $k=0$ (flat universe, with only matter, no cosmological constant or quintessence) depend on a polynomial function?

For example:

$$a(t)=-t^3+4t^2-t+3.$$

Because the strange thing is that this type of scale factor become negative on the infinite. So here we have that $a(t)$ become 0 at max 3 times.

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  • $\begingroup$ When $k=0$, the solution is $$a(t)=a_0t^{\frac{2}{3(w+1)}}$$, which is certainly not a polynomial like the one you describe. $\endgroup$
    – HDE 226868
    Oct 4, 2015 at 19:02
  • $\begingroup$ Yes, but it depends on the behaviour of the matter. For now it doesn't count much the real solution, but the meaning of a (possible)solution as the one I wrote. $\endgroup$
    – Saladinos
    Oct 4, 2015 at 19:09
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    $\begingroup$ I'm not sure I follow. That's the solution for $k=0$. In a homogeneous FLRW universe with $k=0$, that's the solution that exists. $\endgroup$
    – HDE 226868
    Oct 4, 2015 at 19:13
  • $\begingroup$ Ok, if I ask you can, in general, the scale factor be negative or be as the polynomial that I wrote? $\endgroup$
    – Saladinos
    Oct 4, 2015 at 19:23
  • $\begingroup$ It can be negative, but I don't know if a polynomial form like the one you wrote can arise. $\endgroup$
    – HDE 226868
    Oct 4, 2015 at 19:28

1 Answer 1

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The only approach I know relating $H(t)$ and $a(t)$ is from

https://en.wikipedia.org/wiki/Friedmann_equations#Detailed_derivation .

Your assumptions leave only one non-zero omega density term under the square root, the one for matter. Also, the sum of the radiation, matter, and dark energy omegas must equal unity. Solving the differential equation yields:

$a(t) = A t^{2/3}$.

where A is a constant. I have no idea where a polynomial solution can come from. Also, a scale factor cannot be negative.

The usual convention is that $a = 1$ for the corresponding value of t being the current age of the universe, say $t=t_{now}$. Then

$A = t_{now}^{-2/3}$.

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