# Eikonal equation and double null coordinates

I"m trying to understand the exact/technical link between the Eikonal equation and a double-null form of the metric (if such a direct link even exists). R. Wald, in his "General Relativity", doesn't say anything about it.

A function $f$ satisfies (per definition) the Eikonal equation if

$g(\nabla f,\nabla f)=0$

i.e. the gradient field $\nabla f$ is a null vector field.

On the other hand, a metric $g$ has double-null coordinates $(u,v)$ if $g=h+F\dot dudv$ (i.e. no $du^{2}$ and $dv^{2}$ terms appear).

I was wondering if there is a direct link between this function $f$ and the metric double null form?

Thank you for any hints.

• You mean, besides the fact that the coordinate functions $u$ and $v$ by definition solves the eikonal equation? Commented Oct 20, 2015 at 12:52
• @WillieWong yes, that's what I meant. At the first glance, they look like they might be more like the same thing written in a slightly different form, but I was wondering if there is more to it Commented Oct 21, 2015 at 4:51

Per Willie's answer, locally this is the same thing. The global situation is, predictably, very different. I don't know anything specifically about the eikonal equation, but I do know about global solutions to the eikonal inequality $g(\nabla f, \nabla f)\leq 0$ (self plug: http://arxiv.org/abs/1412.5652).