# Higher order terms in Big Bang derivation

You can easilty proof that an SEC fluid gives a big bang by looking at the second Friedmann equation: $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3P) \le 0$$ This implies that $$\ddot{a} \le 0$$ and thus $$a$$ continues to get smaller and smaller for smaller t, so at some point a(t)=0. Now we'll get the time relation by looking at Hubble's law: $$a(t) = a(t_0)[1+(t-t_0)H_0 + ...]\\ 0 = 1 + (t-t_0)H_0 + ...\\ (t-t_0)H_0 < -1\\ (t_0-t)H_0 < 1\\ (t_0 - t)< H_0^{-1}$$ Why, however, can I drop the higher order terms?

As put by this course, when using the linear approximation to estimate the age of the universe:

"This result of 14 billion years is surprisingly close to the currently accepted value of around 13.8 billion years. However, there is a large dose of luck in this agreement, since the linear approximation is not very good when extrapolated over the full age of the universe."

So you can drop the higher order terms when considering $$t$$ near the present day; the Taylor expansion is about the present day. However, this expansion is not meant to be used out to $$a(t)=0$$.

It is only luck that gives a reasonable age for the universe.

• although this seems like a good answer my professor seemed to imply that there is a way to deal with the higher order terms
– nemo
Jun 16, 2022 at 16:21
• Got it. Check out the claim at the bottom of page 41 (of the pdf) in the notes I linked. What they show is the fact that a double dot over a is negative implies the bound you want. It's an argument of concavity. Jun 16, 2022 at 16:32
• that does seem to be the answer, thanks :)
– nemo
Jun 16, 2022 at 19:27