Ernst equation

Remarks

The twist potential exists locally when $\partial_\rho(\rho^{-1} f^2 \partial_\rho\omega) + \partial_z(\rho^{-1} f^2 \partial_z\omega) = 0$, which is one of the vacuum equations; sign conventions for $\psi$ vary between texts. The metric $g$ is a vacuum solution if and only if $\mathcal{E}$ solves the Ernst equation and $\gamma$ solves a pair of first-order equations. These can be integrated once $\mathcal{E}$ is known.

Example: in the static case $\omega = 0$ one has $\psi = 0$ and $\mathcal{E} = f = e^{2U}$. Then $\nabla f = 2f\nabla U$ and $\nabla^2 f = 2f\nabla^2 U + 4f\abs{\nabla U}^2$, so the Ernst equation becomes $2f^2\nabla^2 U = 0$, that is, Laplace's equation $\nabla^2 U = 0$. For instance $U = -M / \sqrt{\rho^2 + z^2}$ with $M > 0$ is harmonic away from the origin, so $\mathcal{E} = e^{2U}$ solves the Ernst equation (the Curzon solution).

Non-example: $\mathcal{E} = z^2$. Here $\operatorname{Re}\mathcal{E}\, \nabla^2\mathcal{E} = 2z^2$ while $\nabla\mathcal{E}\cdot\nabla\mathcal{E} = 4z^2$, so the equation fails wherever $z \ne 0$.

An equivalent form uses $\xi$ with $\mathcal{E} = \frac{\xi - 1}{\xi + 1}$, which gives $\left(\xi\bar\xi - 1\right)\nabla^2\xi = 2\bar\xi\, \nabla\xi\cdot\nabla\xi$.