the origin of the null coordinates by the convention that the incoming ray $v=V=0$ goes through the point $(\tau = 0, R_0)$ and that $U=u=0$ denotes the very same ray on its way out (see Fig. \[scene\]). Thus $v_0 = V_0 = 0$ and at $r=0$ we have $U + U_0 = V + 2R_0$. With our choice for $U$ we finally have $U_0 = 2R_0$.

We define the expansion velocity of the shell by $$\begin{aligned}
\label{nu}
\nu := \frac{\mathrm{d}R}{\mathrm{d}\tau}\; ,\end{aligned}$$ which is actually the coordinate velocity as seen from the inside of the shell. In principle $\nu$ is a function of time, but we will see below that it is a good approximation to work with a constant expansion velocity. In that case the world line of a point on the shell is given by $$\begin{aligned}
 \label{RvonV}
R(\tau)=\left\{\begin{array}{lcr}
R_0 + \frac{\nu}{1+\nu}V =  R_0 + \frac{\nu}{1-\nu}(U+U_0)
&\mbox{when}& \tau >0\\
R_0 & \mbox{when} & \tau <0
\end{array}\right. \; .\end{aligned}$$ Our result will finally depend on three parameters: $R_0$, $M$ and $\nu$.

From the coordinate transformations (\[trafouv\]) and (\[trafoUV\]) and from the definition of the expansion velocity (\[nu\]) we find $$\begin{aligned}
\label{ugenU}
\left.\frac{\mathrm{d}u}{\mathrm{d}U}\right|_{r = R} &=&
\left\{\begin{array}{ccr}
       \left(\frac{\mathrm{d}t}{\mathrm{d}\tau} - \frac{\nu}{\gamma(R)}\right)
       \biggr/(1-\nu) 
         &\mbox{when}& 