On pages 19-20 of Kerr's conference about how he discovered his metric, he basically performs several coordinate transformations until reaching:
$$ds^2 = dx^2 + dy^2 + dz^2 - dt^2 + \dfrac{2mr^3}{r^4 + a^2z^2}[dt + \dfrac{z}{r}dz \\ + \dfrac{r}{r^2 + a^2}(xdx + ydy) - \dfrac{a}{r^2 + a^2}(xdy -ydx)]^2$$
Where the surfaces of constant $r$ are confocal ellipsoids of revolution about the $z$-axis:
$$\dfrac{x^2 + y^2}{r^2 + a^2} + \dfrac{z^2}{r^2}= 1$$
Now, immediately after this he excitedly got together with Schild to perform some calculations (Schild was mostly an expectator) and found:
$$ds^2 = dx^2 + dy^2 + dz^2 - dt^2 + \dfrac{2m}{R}(dt + dR)^2 \\ - \dfrac{4ma}{R^3}(xdy - ydx)(dt - dR) + O(R^{-3})$$
Where $R = x^2 + y^2 + z^2$.
Furthermore, he continues with yet another coordinate transformation and says that, if $x^\mu \rightarrow x^\mu + a^\mu$ is an inifinitesimal transformation, then $ds^2 \rightarrow ds^2 + 2da_\mu dx^\mu$, and he chooses:
$$a_\mu dx^\mu = -\dfrac{am}{R^2}(xdy - ydx) \implies \\ 2da_\mu dx^\mu = -4m\dfrac{4am}{R^3}(xdy - ydx)dR$$
Thus simplifying it to:
$$ds^2 = dx^2 + dy^2 + dz^2 - dt^2 + \dfrac{2m}{R}(dt + dR)^2 \\ - \dfrac{4ma}{R^3}(xdy - ydx)dt + O(R^{-3})$$
And then he comes up with the final formula:
$$4R^{-3}\epsilon_{ijk}J^ix^jdx^kdt$$
Which describes the contribution from the angular momentum $\textbf{J}$. Finally, he finds $\textbf{J}=(0, 0, ma)$, proving the black hole rotates around the $z$ axis.
Here are my doubts:
- How does he go from the first $ds^2$ of this post to the next $ds^2$ (involving $dR$)?
- Are the transformations with $a^\mu$ really necessary to find the contribution of the angular momentum?
- Where does the formula for $J^i$ come from? Is it a projection of the momentum to the $z$-axis or how exactly does he know that $xdy - ydx$ (and $dt$) is related to $\mathbf{J}$?
