I am trying to follow the calculation of the paper Local Approach to Hawking Radiation.
Given the Klein Gordon equation
$$g^{\mu\nu} D_\mu D_\nu \phi=0$$
where $D_\mu$ is the covariant derivative and considering the metric $$ds^2=-f(r)dt^2+f(r)dr_*^2+r^2(d\theta^2+\sin^2\theta d\Phi^2)$$
in tortoise coordinates, $f(r)$ a generic function, I want to find equations $(7)-(8)$ of the paper.
In particular, they use the approximation $\phi=\phi(t,r)$ and set $\phi=1/rR(t,r)$.
In my calculation I get something like this:
$$\frac{1}{r}\left[-\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial r_*^2}\right]R(t,r)+\frac{\partial^2}{\partial r_*^2}\left(\frac{1}{r}\right)R(t,r)+\frac{1}{r^2}\frac{\partial}{\partial r_*}R(t,r)=0$$
How can I get eqs. $(7)-(8)$ from this?
Answer
The original metric in the paper is given by,
$$ds^2 = -f(r)dt^2 + \frac{dr^2}{f(r)} +r^2 d \theta^2 + r^2 \sin^2 \theta \, d\phi^2.$$
Since the metric is diagonal, $g_{ab}$ and $g^{ab}$ follow straightforwardly. In curved space, the wave operator is given by the Laplace-Beltrami operator, $\nabla_a \nabla^a u$ which for a scalar can be simplified to,
$$\nabla^2 u = \frac{1}{\sqrt{|g|}}\partial_a (\sqrt{|g|}g^{ab}\partial_b u).$$
In our case, $\sqrt{|g|} = r^2 \sin \theta$. We can expand the operator as,
$$\nabla^2 u = \frac{1}{r^2 \sin \theta} \left[ \partial_t \left( -\frac{r^2\sin\theta}{f(r)} \partial_t u\right) + \partial_r\left( r^2 f(r) \sin\theta\, \partial_r u\right)+\partial_\theta \left( \sin\theta \, \partial_\theta u\right) + \partial_\phi \left(\csc \theta \, \partial_\phi u \right)\right]$$ $$=-\frac{1}{f(r)}\partial^2_tu + \frac{1}{r} \left\{ \left( 2f(r) + r f'(r) \right)\partial_r u + r f(r)\partial^2_r u\right\} + \frac{1}{r^2} \left( \cot \theta \, \partial_\theta u + \partial^2_\theta u\right) + \frac{1}{r^2}\csc^2\theta \, \partial^2_\phi u.$$
Thus, we obtain an explicit differential equation for $u(t,r,\theta,\phi)$. If we assume, $u = u(t,r)$ only, then the equation greatly simplifies to,
$$\nabla^2 u(t,r) = -\frac{1}{f(r)}\frac{\partial^2 u}{\partial t^2} + \left( \frac{2f(r)}{r} + f'(r)\right)\frac{\partial u}{\partial r} + f(r)\frac{\partial^2 u}{\partial r^2}.$$
It now suffices to plug in, $u(t,r) = \frac{1}{r}R(t,r)$ and then make the change of coordinates to $r_*$ defined by the relation,
$$\frac{dr_*}{dr}=\frac{1}{f(r)}.$$
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