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.

  • 1
    $\begingroup$ You mean, besides the fact that the coordinate functions $u$ and $v$ by definition solves the eikonal equation? $\endgroup$ Commented Oct 20, 2015 at 12:52
  • $\begingroup$ @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 $\endgroup$
    – GregVoit
    Commented Oct 21, 2015 at 4:51

1 Answer 1


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).

Long story short: global solutions to the eikonal inequality will only exist if the Lorentzian distance is finite. This imposes conditions on the causal structure as well as the conformal class of the metric. My guess, and this is really only a guess, is that global solutions of the eikonal equation will require finiteness of the Lorentzian distance plus some extra condition on the causal structure.

If you happen to be working with globally hyperbolic manifolds then the global existence is trivial.


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