# Wave operator in Kerr spacetime: change of coordinates

The wave equation for a scalar field, in Kerr geometry and in Boyer-Lindquist coordinates, reads:

$$-\left[\frac{(r^2 + a^2)^2 }{\Delta} - a^2 \sin^2\theta \right] \partial^2_t \Phi - \frac{4Mar}{\Delta}\partial_t\partial_{\phi}\Phi + \left[\frac{1}{\sin^2 \theta} - \frac{a^2}{\Delta} \right]\partial^2_\phi \Phi + \partial_r(\Delta\partial_r \Phi) + \frac{1}{\sin\theta} \partial_\theta (\sin \theta \partial_\theta \Phi) = 0$$

where $$\Delta = r^2 - 2Mr - a^2 .$$

In many papers dealing with the time evolution of the scalar field, besides the tortoise coordinates $$\frac{dr_*}{dr} = \frac{r^2 + a^2}{\Delta}$$, a specific change of variables is introduced to cure unphysical pathologies near the horizon, which is $$d\phi_* = d\phi + \frac{a}{\Delta} dr.$$ Assuming the ansatz: $$\Phi = \Psi(t,r,\theta)e^{im\phi_*}$$ the wave equation in $$(t,r_*,\theta,\phi_*)$$ coordinates is: $$-\partial^2_t \Psi - \frac{(r^2 + a^2)^2}{\sigma}\partial^2_{r*}\Psi + \frac{4imarM}{\sigma}\partial_t \Psi - \frac{2\left[r\Delta + iam(r^2 + a^2)\right]}{\sigma}\partial_{r*}\Psi - \frac{\Delta}{\sigma}\left[\partial^2_\theta \Psi - \cot \theta \partial_{\theta} \Psi + \frac{m^2}{\sin^2\theta}\Psi\right] = 0.$$ where $$\sigma = -(a^2 + r^2)^2 + a^2\Delta \sin^2\theta$$.

I tried to re-do the change of variables by myself but I got some "spurious" terms in the wave equation like for example a cross term $$\partial_t\partial_{r*}$$. Can someone explicitly show how to get the last wave equation through the changes of variables shown above?

The solution to this problem comes back to expressing the partial derivatives in the old coordinates $$(t, r, \theta, \phi)$$ in partial derivatives of the new coordinates $$(t, r_*, \theta, \phi_*)$$.
Let's consider the differential of some function $$f$$ (which could be your scalar field $$\Phi$$) in the old coordinate basis $$df = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial \theta} d\theta + \frac{\partial f}{\partial \phi} d\phi \qquad (1)$$ and in the new coordinate basis $$df = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial r_*} dr_* + \frac{\partial f}{\partial \theta} d\theta + \frac{\partial f}{\partial \phi_*} d\phi_*.$$ With the two coordinate transformations you mentioned, i.e. $$dr_* = \frac{r^2 + a^2}{\Delta} dr\quad\text{and}\quad d\phi_* = d\phi + \frac{a}{\Delta} dr,$$ the latter becomes \begin{align*} & = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial r_*} \frac{r^2+a^2}{\Delta} dr + \frac{\partial f}{\partial \theta} d\theta + \frac{\partial f}{\partial \phi_*} d\phi + \frac{\partial f}{\partial \phi_*} \frac{a}{\Delta} dr \\ & = \frac{\partial f}{\partial t} dt + \left[\frac{\partial f}{\partial r_*} \frac{r^2+a^2}{\Delta} + \frac{\partial f}{\partial \phi_*} \frac{a}{\Delta} \right]dr + \frac{\partial f}{\partial \theta} d\theta + \frac{\partial f}{\partial \phi_*} d\phi. \qquad (2) \end{align*}
As the 1-forms $$dt, dr, d\theta, d\phi$$ form a basis, their coefficients in both (1) and (2) must be identical; hence \begin{align*} \frac{\partial f}{\partial r} & = \frac{r^2+a^2}{\Delta} \frac{\partial f}{\partial r_*} + \frac{a}{\Delta} \frac{\partial f}{\partial \phi_*}, \\ \frac{\partial f}{\partial \phi} & = \frac{\partial f}{\partial \phi_*}. \end{align*}