Prove that the Kruskal solution is stable to scalar field pertubations

[etotheipi]

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

To first show that $$\left[ \frac{\partial^2}{\partial t^2} - \frac{d^2}{dr_*^2} + V_l(r_*)\right] \phi_{lm} =0$$ where $V_l(r_*) = \left( 1- \dfrac{2M}{r}\right)\left( \dfrac{l(l+1)}{r^2} + \dfrac{2M}{r^3} + \mu^2 \right)$ and $r_*$ is the tortoise coordinate. I have worked out that $\dfrac{d^2 \phi_{lm}}{dr_*^2} = \dfrac{d}{dr_*} \left( \dfrac{dr}{dr_*} \dfrac{d\phi_{lm}}{dr} \right) = \dfrac{d^2 r}{dr_*^2} \dfrac{d\phi_{lm}}{dr} + \left( \dfrac{dr}{dr_*} \right)^2 \dfrac{d^2 \phi_{lm}}{dr^2}$ and since $r_* := r + 2M \ln{ \left| \frac{r}{2M} - 1 \right|}$ then write the following:\begin{align*}

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

\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) \\ \\

\implies \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*}What do I need to do next? \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*}What is the functional form of $\phi_{lm}(t,r)$? Should I try to look for a solution $\phi_{lm}(t,r) = T(t)R(r)$?

 

[Ambitwistor]

Thank you for your question. It seems like you have already made some progress in solving the problem. Here are a few suggestions to help you continue:

1. You have correctly used the chain rule to find the expression for $\frac{d^2 \phi_{lm}}{dr_*^2}$. However, in the next step, you have incorrectly distributed the terms. The correct expression is:
$$\frac{d^2 \phi_{lm}}{dr_*^2} = \frac{2M}{r^2} \frac{d\phi_{lm}}{dr} + \left( 1- \frac{2M}{r} \right) \frac{d^2 \phi_{lm}}{dr^2}$$
You can verify this by using the product rule and the expression for $\frac{dr}{dr_*}$ that you have already found.

2. Next, you need to substitute this expression for $\frac{d^2 \phi_{lm}}{dr_*^2}$ into the given equation and simplify the resulting equation. This will give you an equation in terms of $\phi_{lm}$ and its derivatives with respect to $r$.

3. Once you have this equation, you can use the fact that $\phi_{lm}$ is a solution to the KG equation and simplify the equation further. This will eventually lead you to the desired expression for $\phi_{lm}$ in terms of $r$ and $t$.

4. As for the functional form of $\phi_{lm}$, you are on the right track by looking for a solution of the form $\phi_{lm}(t,r) = T(t) R(r)$. You can substitute this into the simplified equation that you have found in step 3 and use the properties of $Y_{lm}$ to further simplify the equation.

I hope this helps. Keep up the good work and don't hesitate to ask for further clarification if needed.