In this book the kerr metric was given by

enter image description here

I am confused because of the appearance of the $2d\nu dr$ term because in the standard Kerr metric I know this term doesn't appear. Does anyone know what coordinates is this metric given by and the relationship between these coordinates and the standard Kerr coordinates?



Sorry but have you read that book/section you referenced? The line element you gave is given in eq. (19.45) and in the text directly below it says

"The coordinates $(\nu,r,\theta,\bar \phi)$ are the Kerr coordinates."

The section 19.4.1 in your reference discusses in detail how to get from Boyer-Lindquist coordinates to this so called Kerr coordinates. The term you are confused about come from the $-dt^2$ and the $dr^2$ terms in Boyer-Lindquist coordinates, using the coordinate change $d\nu=dt +\frac{r^2+a^2}{\Delta}dr$, which is described in eq. (19.42).

  • $\begingroup$ I did read it and that is why I am confused because I thought that the metric in Wikipedia is given by the Kerr coordinates. So which of the metrics is given by Boyer-Lindquist coordinates and which is given by Kerr coordinates? $\endgroup$ – gbd Sep 14 '16 at 22:56
  • $\begingroup$ Do you mean that if I change coordinate by $d\nu=dt +\frac{r^2+a^2}{\Delta}dr$ then the metric I gave changes to the metric in Wikipedia? $\endgroup$ – gbd Sep 14 '16 at 22:58
  • $\begingroup$ The wikipedia article does not mention "Kerr coordinates". Boyer-Lindquist are coordinates are described in the wikipeda article and in the eq. (19.1) of the book. $\endgroup$ – M. J. Steil Sep 14 '16 at 22:59
  • $\begingroup$ If they both use the same coordinates then why does the $2d\nu dr$ term only appears in one of them? $\endgroup$ – gbd Sep 14 '16 at 23:02
  • $\begingroup$ You can write the line element $ds^2$ in Boyer-Lindquist coordinates $(t,r,\theta,\phi)$ or in Kerr coordinates $(\nu,r,\theta,\bar\phi)$. The line element in those two coordinates looks differently and you can get from one to another using the coordinate changes. $\endgroup$ – M. J. Steil Sep 14 '16 at 23:04

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