1 + {M \over 2R}
 \right)^3 \phi^{(2)}_2(R) {R^2 \over r^2} \nonumber \\
{}^{(2)}K & = & {}^{(2)}H_2, \end{aligned}$$ and all first order time derivatives vanish.

The Zerilli function in an arbitrary gauge is given by [@moncrief],

$$\begin{aligned}
  \psi(t,r) & = & \frac{r(r-2M)}{3(2 r +3M)}\left[
{\ }^{(2)}{H}_2
-r\frac{\partial {\ }^{(2)}{K}}{\partial r}
-\frac{r-3M}{r-2M}{\ }^{(2)}{K}
\right] \nonumber \\
& & +{r^2 \over 2 r +3M} \left[{\ }^{(2)}{K}
+(r-2M)
\left(
\frac{\partial {\ }^{(2)}{G}}{\partial r}
-{2 \over r^2}{\ }^{(2)}{h}_1
\right) \right] .\end{aligned}$$

Substituting the above components yields the following initial data for the Zerilli function, $$\psi(t=0,r)  =  
P^2\left[ {(3M+8R)R^2 \over 6M^2r(2r+3M)} + {4 r (2M+3R)
\over 3 (2r+3M)} \phi^{(2)}_2(R)- {(4r-8M) R^{1/2} r^{3/2} \over 3
(2r+3M)} \left({\partial \phi^{(2)}_2(R) \over \partial r}\right)
\right]$$

$$\left.{\partial \psi(t,r) \over \partial t}\right|_{t=0} =0.$$

For the conformal factor we computed in (\[misnerbound\]), this results in, $$\begin{aligned}
\psi(t=0,r)  &=& 
{P^2 R^6 \over M (M+2R)^4 (M^2+10MR+4R^2)} \left(
{128 \over 3} +56 {M \over R} +{6113 \over 25} {M^2 \over R^2} + 
{32607 \over 100} {M^3 \over R^3} \right. \nonumber \\
& & \left. +{3882 \over 25} {M^4 \over R^4} -{235 \over 16} {M^5 \over R^5} 
- {691 \over 16} {M^6 \over R^6} - {26679 \over 1600} {M^7 \over R^7} -
{2047 \over 800} {M^8 