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I'm having trouble getting the scale factor for conformal time. In de Sitter we've $$a(t)=\sinh(H_It), -\infty <t <\infty$$

and the relation: $dt=ad\eta.$

I don't see how this relation implies: $a(\eta)=-\frac{1}{\sinh(H_I\eta)}, -\infty < \eta < \infty$.

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    $\begingroup$ could you provide a source of the formulas above? $\endgroup$
    – Mateo
    Commented Apr 2, 2020 at 16:54

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So the ideas is writing $t$ in terms of $\eta$ and put it into the $sinh(Ht)$.

By using $dta=d\eta$ we can write

$d\eta = \frac{dt}{da}$ hence $$\eta = \int \frac{dt}{sinh(Ht)} = \frac{1}{H}ln(tanh(\frac{Ht}{2}))$$

Thus

$$\eta = \frac{ln(tanh(Ht/2)}{H}$$

if you take the the inverse of it you will see that it's

$$t = \frac{2arctanh(e^{H\eta})}{H}$$

Then I inserted this into the $sinh(Ht)$ to get a function in terms of $\eta$.

So $$a(\eta) = sinh(2arctanh(e^{H\eta}))$$

Now at this point I kinda got stuck so I ask for a help at math stack exchange and here is the link

https://math.stackexchange.com/questions/3607046/simplifying-sinh-left2-operatornamearctanh-leftehx-right-right

And they proved that

$$a(\eta) = sinh(2arctanh(e^{H\eta})) = -csch(H\eta)$$

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