1
$\begingroup$

I'm just having a bit of difficulty understanding the last couple of steps made in the paper Horowitz & Hubeny - Quasinormal Modes of AdS Black Holes and the Approach to Thermal Equilibrium (p.8) which can be found at this link where the following is stated

$$\int_{r_+}^{\infty}dr[f|\psi'|^2+2i\omega\bar{\psi}\psi'+V|\psi|^2]=0$$

and taking the imaginary part gives

$$\int_{r_+}^{\infty}dr[\omega \bar{\psi}\psi'+\bar{\omega}\psi\bar{\psi'}]=0.$$

Integration by parts of the second term yields

$$(\omega - \bar{\omega})\int_{r_+}^{\infty}dr \bar{\psi} \psi'$$

$$=\bar{\omega}|\psi(r_+)|^2$$

given that $\psi(\infty)=0$ and $\psi'$ denotes differentiation w.r.t $r$. Substituting this final result back into the first equation we obtain

$$\int_{r_+}^{\infty}dr[f|\psi'|^2+V|\psi|^2]=\frac{|\omega^2|\psi(r_+)|^2}{Im\omega}.$$

My problem lies in showing these last two results; namely finding $\bar{\omega}|\psi(r_+)|^2$ from the previous equation and then showing the final substitution.

I've been trying for a very long time with integration by parts, using some complex identities involving the conjugate etc., but I can't arrive at the final result.

This is simply a case of me trying to fully understand a paper I'm interested in. Any help would be greatly appreciated.

$\endgroup$
3
$\begingroup$

Adding more intermediate steps: \begin{align*} \int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]&= \int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]=\\ &=\int_{r_+}^{\infty}dr[\color{red}{\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'}+\color{blue}{\bar{\omega}(\bar{\psi}\psi)'}]=\\ &=\color{red}{(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'}+ \color{blue}{\bar{\omega}|\psi|^2\biggl|_{r_+}^{\infty}}=\\ &=(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'-\bar{\omega}|\psi(r_+)|^2. \end{align*} Since the initial expression was previosly shown to be zero, we obtain the first result.

To show the second result, we substitute $\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'$ by $\displaystyle \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}$ in the 2nd term of your first equation: $$2i\omega \int_{r_+}^{\infty}dr\,\bar{\psi}\psi'=2i\omega\cdot \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}=\frac{2i}{\omega-\bar{\omega}}|\omega|^2|\psi(r_+)|^2 =\frac{1}{\Im\omega}|\omega|^2|\psi(r_+)|^2=\frac{|\omega^2||\psi(r_+)|^2}{\Im\omega}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.