Weyl–Lewis–Papapetrou coordinates

Remarks

In these coordinates the Gram matrix of the Killing fields has determinant $$g_{tt} g_{\phi\phi} - g_{t\phi}^2 = (-f)\left(-f\omega^2 + \frac{\rho^2}{f}\right) - f^2\omega^2 = -\rho^2.$$ So $\rho$ is fixed by the geometry: $\rho^2 = -\left(g(\xi,\xi) g(\eta,\eta) - g(\xi,\eta)^2\right)$.

Existence: let $(M, g)$ be a vacuum solution, and assume that $\eta$ vanishes somewhere, on a rotation axis. Then the planes orthogonal to $\xi$ and $\eta$ are integrable, and the function $\rho$ just defined is harmonic on the two-dimensional orbit space. Where $d\rho \ne 0$ one can take $\rho$ and a harmonic conjugate $z$ as coordinates. This gives the form above.

Example: Minkowski space in cylindrical coordinates, $g = -dt^2 + d\rho^2 + dz^2 + \rho^2 d\phi^2$, has this form with $f = 1$, $\omega = 0$, $\gamma = 0$.

Non-example: the usual coordinates $(t, r, \theta, \phi)$ of the Schwarzschild metric are not of this form with $\rho = r\sin\theta$. Indeed $-g_{tt} g_{\phi\phi} = (1 - 2m/r) r^2\sin^2\theta$, so the Weyl coordinate is $\rho = \sqrt{r^2 - 2mr}\,\sin\theta$, with $z = (r - m)\cos\theta$.

If $\omega = 0$ the spacetime is static and these are Weyl coordinates. For vacuum with $\omega = 0$, write $f = e^{2U}$. Then the Einstein equations say that $U$ is an axisymmetric harmonic function on flat $\R^3$ with cylindrical coordinates $(\rho, z, \phi)$. The function $\gamma$ is then found by a line integral.