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In any introductory book to GR where the Kerr metric is reviewed, they all apply transformations of the form:

$$cu' = cu + ia\cos\theta;\text{ } r' = r + ia\cos\theta$$

In the end, the result describes the gravitational field of an axisymmetrically rotating mass.

But why? I guess it has something to do with $z=r\cos\theta$, but I haven’t found any explanation on the Internet. I have read in the original paper from Boyer and Lindquist that you can presume $a$ to be the angular momentum per unit mass comparing the resulting equations to the Lense-Thirring effect ones, but I can’t see the connection between it and the transformation with $\cos\theta$.

Furthermore, I am curious to know what the transformations for $cdu$ and $r$ would be if the axis of rotation was not $z$ but rather $x$ or $y$; or even how a rotation along two axis would look like in terms of a Kerr transformation (for example, to describe a mass rotating like a pulsar, that could be described as rotating along the $z$ and $y$ axis at different speeds).

Antoniou
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1 Answers1

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The $~du\,d\phi~$ term in the Kerr metric is crucial because it signifies the coupling between time $~u~$ and the azimuthal angle $~\phi~$. This coupling is a direct consequence of the rotating nature of the Kerr black hole, and it manifests the phenomenon known as frame dragging or the Lense-Thirring effect.

to obtain the $~du\,d\phi~$ term in the Kerr metric you use this "complex transformation" .

\begin{align*} \begin{bmatrix} r' \\ u' \\ \end{bmatrix}= \begin{bmatrix} r \\ u \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 0 & i\,\cos(\theta) \\ i\,\cos(\theta) & 0 \\ \end{bmatrix}}_{\rm Transformation}\, \begin{bmatrix} a \\ a \\ \end{bmatrix} \end{align*}

where the parameter a is the angular momentum per mass $~a= \frac JM~$

Eli
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