# Eikonal equation and double null coordinates

+ 1 like - 0 dislike
476 views

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.

This post imported from StackExchange MathOverflow at 2015-12-17 17:23 (UTC), posted by SE-user GregVoit
retagged Dec 17, 2015
You mean, besides the fact that the coordinate functions $u$ and $v$ by definition solves the eikonal equation?

This post imported from StackExchange MathOverflow at 2015-12-17 17:23 (UTC), posted by SE-user Willie Wong
@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

This post imported from StackExchange MathOverflow at 2015-12-17 17:23 (UTC), posted by SE-user GregVoit

+ 2 like - 0 dislike

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.

This post imported from StackExchange MathOverflow at 2015-12-17 17:23 (UTC), posted by SE-user Ben Whale
answered Nov 30, 2015 by (20 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.