I have the following metric

$$(ds)^2 = A(r) dt^2 + 2B(r) drdt - C(r)dr^2 - r^2d\Omega^2,$$

where $d\Omega^2 = d\theta^2 + \sin^2 \theta \;d\phi^2$.

Is it possible to write this metric in isotropic form without performing a coordinate transformation in the time variable to remove the non-orthogonal components of the metric tensor?

I have seen this answer. However, by following the method I get stuck with the $drdt$ term. I notice in Cheng, Relativity, Gravitation and Cosmology they remove the off diagonal term also.

Any suggestions?


You need to get rid of the $dr dt$ term first before you get a chance to make the coordinates isotropic. The most general method is just the reduction of a quadratic form, here $2\times 2$. Le't forget about the angular part:

$$(ds)^2 = \begin{pmatrix}dt & dr\end{pmatrix} \begin{pmatrix}A(r) & B(r) \\ B(r) & -C(r)\end{pmatrix}\begin{pmatrix}dt\\dr\end{pmatrix}.$$

You just need to diagonalise that matrix. That will give you a linear transformation $(dt,dr)\to(dt',dr')$ such that

$$ds^2 = \mathcal{A}dt'^2 - \mathcal{B}dr'^2$$

Then you can further try to change variables to get an isotropic form.


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