Coordinate transformation to find the Minkowski metric

Question

I have the following cosmology exercise:

i) calculate the evolution scale factor for an open empty universe and write down its spacetime metric in terms of coordinate $\chi$ \begin{equation} d\chi = \frac{dr}{\sqrt{1-Kr^2}}. \end{equation}

ii) show that this is equivalent to the Minkowski metric $$ds^2 = d\tau^2 -dr^2 -r^2d\Omega^2 $$ using the following transformation: $$\tau=t \text{cosh}(|k|^{1/2}\chi) \quad \text{and} \quad r = t \text{sinh}(|k|^{1/2}\chi).$$

For i) I computed the scale factor from the I Friedman equation with $\Lambda = \rho = 0 $ for a empty universe:

\begin{equation} a(t)=\int \dot{a}(t) dt= \int_0^t c dt = [ct]^t_0=ct_o \end{equation}

where the metric is a Roberston-Walker metric:

\begin{eqnarray} ds^2&=&(cdt)^2 -a^2(t)\bigg (\frac{dr^2}{1-Kr^2} + r^2d\Omega^2 \bigg) \\ &=&(cdt)^2 -a^2(t)\bigg (d\chi^2 + S_k(r)^2d\Omega^2 \bigg) \end{eqnarray}

is that correct?

For the ii) However I have no ideas.. I noticed that the Minkowsi metric is given in spherical coordinate but I don't understand how to do the transform. Could anyone help me?

I just tried to derive the given coordinate and replace the result in Minkowski metric formula in order to obtain the RW metric but its not working...

\begin{equation} dr=dt \text{sinh}(\sqrt{|k|}\chi) + t \text{cosh}(\sqrt{|k|}\chi) \sqrt{|k|} d\chi \end{equation}

\begin{equation} d\tau=dt \text{cosh}(\sqrt{|k|}\chi) + t \text{sinh}(\sqrt{|k|}\chi) \sqrt{|k|} d\chi \end{equation}

Answer 1

So I think that my problem was that I forgot to replace $r=t\text{sinh}(\sqrt{|k|}\chi)$ in my Minkowski metric ! But after replacing it, and replacing the derivative I kind of get the Robertson-Walker metric again, with $a(t)=t$ and with $S_k(r)=R\text{sinh}(\frac{\chi}{R})$ for an open universe. I don't know if this can help anyone but I just let the post here in case :)