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

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Xepto
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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
 
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bapowell said:
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?
 
If the universe grows by [itex]N[/itex] e-folds of expansion during inflation, then [itex]a_{end} = e^N a_i[/itex]. 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, [itex]N[/itex] is defined as the number of e-folds before the end of inflation. This means that [itex]dN = -Hdt[/itex] -- the number [itex]N[/itex] gets smaller as inflation progresses, and becomes [itex]N=0[/itex] at the end.