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- Homework Statement:
- The solution to the KG equation is assumed to take the form$$\Phi = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{1}{r} \phi_{lm}(t,r) Y_{lm}(\theta, \phi)$$

- Relevant Equations:
- N/A

It's required to show first that $$\left[ \frac{\partial^2}{\partial t^2} - \frac{d^2}{dr_*^2} + V_l(r_*)\right] \phi_{lm} =0$$ where ##V_l(r_*) = \left( 1- \frac{2M}{r}\right)\left( \frac{l(l+1)}{r^2} + \frac{2M}{r^3} + \mu^2 \right)## and ##r_*## is the tortoise coordinate. I have worked out that\begin{align*}

\frac{d^2 \phi_{lm}}{dr_*^2} = \frac{d}{dr_*} \left( \frac{dr}{dr_*} \frac{d\phi_{lm}}{dr} \right) = \frac{d^2 r}{dr_*^2} \frac{d\phi_{lm}}{dr} + \left( \frac{dr}{dr_*} \right)^2 \frac{d^2 \phi_{lm}}{dr^2}

\end{align*}and since ##r_* := r + 2M \ln{ \left| \frac{r}{2M} - 1 \right|}## we have\begin{align*}

\frac{dr_*}{dr} = \frac{1}{1- \frac{2M}{r}} \implies \frac{dr}{dr_*} = 1- \frac{2M}{r}

\end{align*}and consequently \begin{align*}

\frac{d^2 r}{dr_*^2} = \frac{dr}{dr_*} \frac{d}{dr} \left( \frac{dr}{dr_*} \right) = \frac{2M}{r^2} \left( 1- \frac{2M}{r} \right)

\end{align*}Then ##\frac{d^2 \phi_{lm}}{dr_*^2}## can be re-written as\begin{align*}

\frac{d^2 \phi_{lm}}{dr_*^2} = \left( 1- \frac{2M}{r} \right) \left[ \frac{2M}{r^2} \frac{d\phi_{lm}}{dr} + \left(1 - \frac{2M}{r} \right) \frac{d^2 \phi_{lm}}{dr^2} \right]

\end{align*}But I'm stuck on the rest; I don't know how to actually evaluate those derivatives of ##\phi_{lm}## with respect to ##r##, or what to write for ##\partial^2 \phi_{lm} / \partial t^2##. What do I need to do next? Thanks!\begin{align*}

\nabla_{\mu} \nabla^{\mu} \Phi - \mu^2 \Phi = 0 \\

\frac{1}{r} \phi_{lm}(t,r) \nabla_{\mu} \nabla^{\mu} Y_{lm} (\theta, \phi) + Y_{lm} (\theta, \phi) \nabla_{\mu} \nabla^{\mu} \left( \frac{1}{r} \phi_{lm}(t,r) \right) - \mu^2 \Phi = 0

\end{align*}

\frac{d^2 \phi_{lm}}{dr_*^2} = \frac{d}{dr_*} \left( \frac{dr}{dr_*} \frac{d\phi_{lm}}{dr} \right) = \frac{d^2 r}{dr_*^2} \frac{d\phi_{lm}}{dr} + \left( \frac{dr}{dr_*} \right)^2 \frac{d^2 \phi_{lm}}{dr^2}

\end{align*}and since ##r_* := r + 2M \ln{ \left| \frac{r}{2M} - 1 \right|}## we have\begin{align*}

\frac{dr_*}{dr} = \frac{1}{1- \frac{2M}{r}} \implies \frac{dr}{dr_*} = 1- \frac{2M}{r}

\end{align*}and consequently \begin{align*}

\frac{d^2 r}{dr_*^2} = \frac{dr}{dr_*} \frac{d}{dr} \left( \frac{dr}{dr_*} \right) = \frac{2M}{r^2} \left( 1- \frac{2M}{r} \right)

\end{align*}Then ##\frac{d^2 \phi_{lm}}{dr_*^2}## can be re-written as\begin{align*}

\frac{d^2 \phi_{lm}}{dr_*^2} = \left( 1- \frac{2M}{r} \right) \left[ \frac{2M}{r^2} \frac{d\phi_{lm}}{dr} + \left(1 - \frac{2M}{r} \right) \frac{d^2 \phi_{lm}}{dr^2} \right]

\end{align*}But I'm stuck on the rest; I don't know how to actually evaluate those derivatives of ##\phi_{lm}## with respect to ##r##, or what to write for ##\partial^2 \phi_{lm} / \partial t^2##. What do I need to do next? Thanks!\begin{align*}

\nabla_{\mu} \nabla^{\mu} \Phi - \mu^2 \Phi = 0 \\

\frac{1}{r} \phi_{lm}(t,r) \nabla_{\mu} \nabla^{\mu} Y_{lm} (\theta, \phi) + Y_{lm} (\theta, \phi) \nabla_{\mu} \nabla^{\mu} \left( \frac{1}{r} \phi_{lm}(t,r) \right) - \mu^2 \Phi = 0

\end{align*}

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