How do I convert the number of efolds to conformal time during inflation?

[Xepto]

Hi everyone,
I need to convert the number of efolds to confromal time during inflation in order to do numerical integrations. Suppose expansion is De Sitter and you have to calculate the integral $$\int^0_{-\infty} d\tau F(N)$$ where F is a function of the efolds number N (from the beginning of inflation). All issues arise at the integration limits. I know that $$a = - \frac{1}{H \tau} $$ but I cannot obtain the correct relationship between N and τ.
(My guess is $$\tau = - \frac{1}{H e^N} $$ but this doesn't seem to be correct. In fact for $N\rightarrow \infty$ we get $\tau \rightarrow 0$, but for $N\rightarrow 0$ we get $\tau\rightarrow - \frac{1}{H}$)
Can anyone help me?
Thanks

 

[bapowell]

What is the relationship between a and N? Hint: N is not defined with respect to the start of inflation, but the end.

 

[Xepto]

What is the relationship between a and N? Hint: N is not defined with respect to the start of inflation, but the end.

From $dN = H dt$, we get $$N = \text{ln} \frac{a(t_{end})}{a(t)}$$. Tthen $a(t) = a(t_{end}) e^{-N}$, that can be rewritten in terms of the number of efolds as $a(N(t)) = a(N_{end}) e^{-N}.$ If expansion is DeSitter, $H$ is a constant and $$ \tau = -\frac{1}{a(\tau) H} = - \frac{1}{a(N_{end}) e^{-N} H},$$ that can be inverted $$ N=\text{ln} \Biggl( - \frac{1}{\tau H a(N=0) } \Biggl). $$ If this is correct, how can I calculate the scale factor $a$ at the end of inflation?

 

[bapowell]

If the universe grows by N e-folds of expansion during inflation, then a_{end} = e^N a_i. What's important is not the value of the scale factor at a particular time, because it can always be renormalized (e.g. the scale factor is often defined to be equal to 1 today). What's generally important in cosmology is the ratio of scale factors at two different times because this gives the amount of expansion.

Also, N is defined as the number of e-folds before the end of inflation. This means that dN = -Hdt -- the number N gets smaller as inflation progresses, and becomes N=0 at the end.