Question on the proof of Hawking radiation as done in "Quantum fields in curved space - Birrell,Davies" In their proof of Hawking radiation, Birrell and Davies consider two dimensional model of a collapsing star with two set of coordinates, one outside the star:
$$ds^2=C(r)du \space dv \tag{8.8}$$
$$u=t-r^*+R_0^* \tag{8.9}$$
$$v=t+r^*-R_0^*$$
$$r^*=\int C^{-1}dr \tag{8.10}$$
and one set inside the star:
$$ds^2=A(U,V)dU \space dV \tag{8.11}$$
$$U=\tau-r+R_0 \tag{8.12}$$
$$V=\tau+r-R_0$$
After assuming the star surface collapse along the world line $r=R(\tau)$ then they define two transformation between the coordinates:
$$ U = \alpha(u) \tag{8.13}$$
$$ v = \beta(V) \tag{8.14}$$
After expanding a Klein Gordon massless field in modes and getting a redshift dependent on $\alpha(u)$ and $\beta(v)$ they find an explicit expression of these function and I didn't understand how they did it: they write: "To determine the form of this redshift factor we have to match the interior and exterior metrics across the collapsing surface $r=R(\tau)$. This yields
$$\alpha^{\prime}(u)=\frac{\mathrm{d} U}{\mathrm{~d} u}=(1-\dot{R}) C\left\{\left[A C\left(1-\dot{R}^{2}\right)+\dot{R}^{2}\right]^{\frac{1}{2}}-\dot{R}^{\prime}\right\}^{-1} \\
\beta^{\prime}(V)=\frac{\mathrm{d} v}{\mathrm{~d} V}=C^{-1}(1+\dot{R})^{-1}\left\{\left[A C\left(1-\dot{R}^{2}\right)+\dot{R}^{2}\right]^{\frac{1}{2}}+\dot{R}\right\}$$
where $\dot{R}$ denotes $dR/d\tau$ and $U$,$V$ and $C$ are here evaluated at $r=R(\tau)$. Note $\dot{R}<0$ for a collapsing surface so $(\dot{R}^2)^{\frac{1}{2}} = -\dot{R}$". I don't understand how they computed $\alpha^{\prime}(u)$ and $\beta^{\prime}(V)$; I tried any combination of the coordinates I could think of but I didn't succeed in finding the expression they give (I found many other, all different though!) and all help in solving this conundrum is deeply appreciated.
Edit:
I can of course write
$$\frac{dU}{du}=\frac{dU}{d \tau} \frac{d \tau}{du}=(1-\dot{R}) \frac{d \tau}{du}$$
and
$$\frac{dv}{dV}=\frac{1}{1+\dot{R}}\frac{dv}{d\tau}$$
But i fail to understand how to compute $\frac{dv}{d\tau}$ and $\frac{d \tau}{du}$
 A: I'll derive the first relation for $\alpha'(u)$; $\beta'(V)$ is same except for some factors and sign difference. You have shown $\alpha'(u)=(1-\dot{R})\frac{d\tau}{du}$. Since the hypersurface where these two metric coincides is a $1$-d surface; visualize the hypersurface as $S^1$ in $R^2$.
Metric outside the hypersurface is given by $$ds^2=C(dt^2-\frac{dr^2}{C^2})$$
I have substituted $u,v$ as a function of $t,r$. At the hypersurface $r=R(\tau)$ and let $\color{red}{t=F(\tau)}$ which gives $$ds^2=C(\dot{F}^2d\tau^2-\frac{\dot{R}^2}{C^2}d\tau^2)$$
$$ds^2=\frac{1}{C}(\dot{F}^2C^2-\dot{R}^2)d\tau^2$$
as claimed the line element of hypersurface is one dimensional.
For the inside metric one gets $$ds^2=A(1-\dot{R}^2)d\tau^2$$ As mentioned these two metric should match as they represent the same surface. Solving for $F$ we get
$$\dot{F}=\pm\frac{1}{C}\sqrt{AC(1-\dot{R}^2)+\dot{R}^2}$$
Now $$\frac{du}{d\tau}=\frac{dt}{d\tau}-\frac{dr*}{d\tau}$$
$$\frac{du}{d\tau}=\dot{F}-\frac{\dot{R}}{C}$$
$$\implies\alpha'(u)=\pm C(1-\dot{R})\Bigg(\sqrt{AC(1-\dot{R}^2)+\dot{R}^2}-\dot{R}\Bigg)^{-1}$$
Plus sign is taken to match the flow of time direction between the interior and exterior.
