My Question: In his GR text, Robert Wald claims that solutions, $\phi$, to the Klein-Gordon equation, $\nabla_a\nabla^a \phi = m^2 \phi$, in Schwarzschild spacetime can be expanded in modes of the form $\phi(r,t,\theta,\varphi) = r^{-1}f(r,t)Y_{lm}(\theta,\varphi)$ to obtain

\begin{equation} \frac{\partial^2 f}{\partial t^2}-\frac{\partial^2f}{\partial r_*}+\left(1-\frac{2M}{r}\right)\left[\frac{l(l+1)}{r^2}+\frac{2M}{r^3}+m^2 \right]f=0, \label{eq 1} \tag{Wald 14.3.1} \end{equation}

where we have the Regge-Wheeler tortoise coordinate $r_{*}\equiv r+2M\ln(r/2M-1)$, with $r$ the usual Schwarzschild "radial" coordinate, and $Y_{lm}(\theta,\varphi)$ the spherical harmonics.

I simply want to verify \eqref{eq 1}, using the definition of the tortoise coordinate and spherical harmonics.

My Attempt: I first note that the Schwarzschild metric in its usual "radial" coordinate, $r$, is

\begin{equation} ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2\left(d\theta^2+\sin^2\theta d\varphi^2\right). \end{equation}

Now, the definiton of the tortoise coordinate $r_*$ gives

\begin{equation} \frac{dr_*}{dr}=\left(1-\frac{2M}{r}\right)^{-1}, \tag{*} \label{dr} \end{equation} so one can express the Schwarzschild metric instead as

\begin{equation} ds^2=\left(1-\frac{2M}{r}\right)\left(-dt^2+dr_*^2\right)+r^2d\Omega^2. \end{equation}

Therefore, the operator $\nabla_a\nabla^a$ in Schwarzschild is given by

\begin{align} g^{ab}\nabla_a\nabla_b &= g^{tt} \partial_t^2 + g^{r_*r_*} \partial^2_{r_*} + g^{\theta \theta} \partial^2_{\theta} +g^{\varphi \varphi} \partial^2_{\varphi} \\ &= \left(1-\frac{2M}{r}\right)^{-1}\left(-\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial r_* ^2}\right)+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\\ \end{align}

Hence, plugging this back into the K-G equation and multiplying through by $-(1-2M/r)$ yields an equation which seems on the right track toward \eqref{eq 1}:

\begin{equation} \frac{\partial^2\phi}{\partial t^2}-\frac{\partial^2\phi}{\partial r_*^2}+\left(1-\frac{2M}{r}\right)\left[-\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}-\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}+m^2\right]\phi^2=0.\tag{**}\label{promising} \end{equation}

However, two things aren't working out quite right for me: 1) When I plug in the mode expansion $\phi=r^{-1}f(r,t)Y_{lm}(\theta,\varphi)$, I would like to invoke the relation \begin{equation} \Delta Y_{lm}(\theta,\varphi)=-\frac{l(l+1)}{r^2}Y_{lm}(\theta,\varphi), \end{equation} where $\Delta$ is the Laplacian on the 2-sphere, \begin{equation}\Delta\equiv \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}. \end{equation} BUT, this looks to differ from the angular part of \eqref{promising} by a term proportional to $\partial/\partial\theta$.

And 2) My second issue is in the derivative with respect to the tortoise coordinate. In particular, I find (using the inverse of \eqref{dr} when necessary)

\begin{align} \frac{\partial^2}{\partial r_*^2}\left[\frac{f(r,t)}{r}\right] &=\frac{\partial}{\partial r_*}\left[-\frac{f}{r^2}\frac{\partial r}{\partial r_*}+\frac{1}{r}\frac{\partial f}{\partial r_*}\right]\\ &=\left[2\left(1-\frac{2M}{r}\right)\left(\frac{1}{r^3}-\frac{3M}{r^4}\right)-\frac{2}{r^2}\left(1-\frac{2M}{r}\right)\frac{\partial }{\partial r_*}+\frac{1}{r}\frac{\partial^2}{\partial r_*^2}\right]f,\\ \end{align} which differs from terms in \eqref{eq 1} that I needed to come from the tortoise derivatives by terms proportional to both $f$ and $\partial f/\partial r_*$ (and even overall factors on terms with correct derivative number and $r$ order).

  • 2
    $\begingroup$ Your expression for $\nabla^2$ looks funny. It should in general be $(1/\sqrt{g}) \partial_a \sqrt{g} g^{ab} \partial_b$ where $\sqrt{g}$ is short hand for the square root of the determinant of the metric. $\endgroup$ Feb 20, 2017 at 3:28
  • $\begingroup$ @user2309840 If you are referring to $\nabla_a \nabla^a$, then $\nabla_a$ is merely the covariant derivative compatible with the metric---i.e., $\nabla_c g_{ab} =0 $. If you are instead protesting about the form of the "K-G equation" in curved spacetime, I am using Wald's definition $\nabla_a\nabla^a-m^2\phi=0$, equation 14.2.4 in his text. Call it whatever you wish; that's the equation of motion I'm considering here. $\endgroup$
    – user143410
    Feb 20, 2017 at 4:34
  • $\begingroup$ I would suggest trying instead with my definition and seeing if the situation improves. $\endgroup$ Feb 20, 2017 at 13:15
  • $\begingroup$ @user2309840 No, that is not a legal move. If I were to change the derivative operator/the equations of motion, I would be considering a different physical theory. Besides, the claim is not unique to Wald's textbook: Wald equation 14.3.1 can also be found in J. Dimock's '85 paper on clasical scattering, for instance. $\endgroup$
    – user143410
    Feb 20, 2017 at 14:37
  • 2
    $\begingroup$ I'm using Wald (3.4.10). $\endgroup$ Feb 20, 2017 at 16:49

1 Answer 1


The resolution to both my issues is that I was not correctly applying the covariant derivative in $\nabla_a \nabla^a$ (as was mentioned in a comment by user2309840). More specifically, my expression should have been

\begin{equation} \nabla_a\nabla^a\phi= \partial_a\nabla^a\phi+\Gamma^a_{\;ab}\nabla^b\phi \tag{*}\label{1} \end{equation}

Now, the contracted Levi-Civita can be expressed succinctly (Wald 3.4.9) in terms of the metric determinant $g\equiv \det{g_{\mu\nu}}$ as

\begin{equation} \Gamma^a_{\;ab}=\frac{\partial}{\partial x^\mu}\ln{\sqrt{|g|}}. \end{equation}

Thus, \eqref{1} becomes

\begin{align} \nabla_a\nabla^a&=\sum_{\mu,\nu}\frac{1}{\sqrt{|g|}}\partial_\mu\left[\sqrt{|g|}g^{\mu\nu}\partial_\nu \phi\right] \\ &= g^{tt}\partial^2_t\phi+\frac{1}{\sqrt{|g|}}\partial_{r_*}\left[\sqrt{|g|}g^{r_*r_*}\partial_{r_*} \phi\right]+\frac{g^{\theta\theta}}{\sqrt{|g|}}\partial_\theta\left[\sqrt{|g|}\partial_\theta\phi\right]+g^{\varphi\varphi}\partial^2_\varphi\phi, \label{2}\tag{**}\\ \end{align}

where the square root of the metric determinant here is

\begin{equation} \sqrt{|g(r,\theta)|}=\sqrt{-g_{tt}g_{r_*r_*}g_{\theta\theta}g_{\varphi\varphi}}=\left(1-\frac{2M}{r}\right)r^2\sin{\theta} \end{equation}

Therefore, the angular part of \eqref{2} is now precisely the Laplacian on the 2-sphere as I sought to show

\begin{equation} \Delta\equiv\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left[\sin\theta\frac{\partial}{\partial\theta}\right]+\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}. \end{equation}

And we note that

\begin{align} \partial_{r_*}\left[\sqrt{|g|}g^{r_*r_*}\partial_{r_*}\phi\right]=\frac{\partial}{\partial r_*}\left[r^2\frac{\partial\phi}{\partial r_*}\right]&=2r\frac{\partial r}{\partial r_*}\frac{\partial\phi}{\partial r_*}+r^2\frac{\partial^2\phi}{\partial r^2_*}\\ &=rY_{lm}\frac{\partial^2 f}{\partial r_*^2}-\frac{2M}{r^2}\left(1-\frac{2M}{r}\right)Y_{lm}f, \label{3}\tag{***} \end{align} where in the second line I have taken $\phi(r,t,\theta,\varphi)=r^{-1}f(r,t)Y_{lm}(\theta,\varphi)$ and applied $\partial r/\partial r_*=(1-2M/r)$ .

Wald 14.3.1 now follows directly from \eqref{2} and \eqref{3} (plus the definition of the spherical harmonics, $r^2\Delta Y=l(l+1)Y$).


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