Why doesn't the Kerr-Newman metric reduce to the Schwarzschild metric?

Question

I hope this is not just an algebra mistake (and I'm very sorry if it is). Also, note I use the convention that the time argument is negative.

Setting $G=M=c=1$ for the Schwarzschild metric, it becomes $$g_{\mu \nu} = \left(\begin{matrix}-\left(1-2/r \right)&0&0&0\\0&\left(1-2/r \right)^{-1}&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2\theta \end{matrix}\right)$$

But, when I set $G=M=c=\epsilon_0=1$ and $Q=L=0$ in the Kerr-Newman metric (from Wikipedia), I get $$g_{\mu \nu} = \left(\begin{matrix}-\left(1-2/r \right)&0&0&0\\0&\left(r-2 \right)^{-2}&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2\theta \end{matrix}\right)$$

It seems to me that this metric should reduce to the Schwarzschild metric when there is no angular momentum or net charge. Can someone tell me why this isn't the case?

Answer 1

Setting $J = 0$, $Q = 0$ in the metric of Kerr-Newman spacetime expressed in Boyer-Lindquist coordinates gives you Schwarschild metric in Schwarzschild coordinates. I don't see how you arrived at the metric you propose.