Parametric representation of the motion of a particle in context of kruskal coordinates In "N. Straumann - General Relativity" talking about the Kruskal continuation of the Schwarschild solution it is considered for radial timelike geodesic the following:
$$d\tau=(\frac{2m}{r}-\frac{2m}{R})^{\frac{-1}{2}}dr$$
with the following parametric solution:
$$r=\frac{R}{2}(1+\cos{\eta})$$
$$\tau=(\frac{R^3}{8m})^{\frac{1}{2}}(\eta+\sin{\eta})$$
I would like to verify these parametric representation of the motion. So I have computed
$$dr=-\frac{R}{2}\sin{\eta}d\eta$$
$$d\tau=(\frac{R^3}{8m})^{\frac{1}{2}}(1+\cos{\eta})d\eta$$
And so I should have:
$$(\frac{R^3}{8m})^{\frac{1}{2}}(1+\cos{\eta})d\eta=-\Big(\frac{2m}{\frac{R}{2}(1+\cos{\eta})}-\frac{2m}{R}\Big)^{\frac{-1}{2}}\frac{R}{2}\sin{\eta}d\eta$$
So from this seems that the identity does not hold...thus I am doing some mistakes. Can you help me?
 A: Correct me if I am wrong, but I think it should be
$$\text{d}\tau = -\left(\frac{r_s}{r}-\frac{r_s}{R}\right)^{-\frac 12}\text{d}r$$
with $r_s = 2M$.
This would be consistent for an infalling motion with the changes in the proper time and radius when compared to the changes in the parameter $\eta\,$ ($\text{d}r <0$ while we want $\text{d}\tau > 0$).
With this, you can show that the parametrization is indeed equal.
One has
\begin{align*}\left(\frac{r_s}{r}-\frac{r_s}{R}\right)^{-\frac 12} &= \frac{1}{\sqrt{r_s}}\left(\frac{1-\frac 12 (1+\cos\eta)}{\frac R2 (1+\cos\eta)}\right)^{-\frac 12} \\ 
&= \frac{1}{\sqrt{r_s}} \left(\frac{\frac 12 - \frac 12 \cos\eta}{\frac R2 (1+\cos\eta)}\right)^{-\frac 12} \\
&= \sqrt{\frac{R}{r_s}} \sqrt{\frac{1+\cos\eta}{1-\cos\eta}} \\
&= \sqrt{\frac{R}{r_s}} \frac{1}{\tan \frac \eta 2}
\end{align*}
Plugging this in leads to
\begin{align*}
\text{d}\tau &= -\left(\frac{r_s}{r}-\frac{r_s}{R}\right)^{-\frac 12}\text{d}r \\
&= \sqrt{\frac{R}{r_s}} \frac{1}{\tan \frac \eta 2} \frac{R}{2}\sin\eta \text{d}\tau \\
&= \sqrt{\frac{R^3}{4r_s}} \frac{\sin\eta}{\tan \frac \eta 2}\text{d}\eta
\end{align*}
Notice that
\begin{align*}
\frac{\sin\eta}{\tan \frac \eta 2} &= \sin\eta \frac{\cos \frac \eta 2}{\sin \frac \eta 2} \\
&= 2\sin\frac \eta 2\,\cos\frac \eta 2\, \frac{\cos \frac \eta 2}{\sin \frac \eta 2} \\
&= 2\,\cos^2\frac \eta 2 \\
&= 2 \left(\frac{1+\cos\eta}{2}\right) \\
&= 1+\cos\eta
\end{align*}
Hence
\begin{equation*}
\text{d}\tau = \sqrt{\frac{R^3}{4r_s}} (1+\cos\eta)\text{d}\eta = \left(\frac{R^3}{8m}\right)^{\frac 12} (1+\cos\eta) \text{d}\eta
\end{equation*}
