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I have read that the extrinsic curvature at the horizon of a euclidean black hole is zero? Does anybody know how this can be shown?

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Not sure how to answer your question fully, but since the horizon is two dimensional, there are two normals with respect to which extrinsic curvatures can be defined. If $n$ is one of the normals, the ext. curvature for that direction is $\mathcal{L}_ng_{\mu\nu}$ and so vanishes if $n$ is a Killing vector for your metric. In your Euclidean case this Killing property is what you'd need to show presumably. –  twistor59 Nov 1 '13 at 8:38
    
@twistor59: I'd probably define the horizon to be the "stack" of two dimensional horizons along the null tangent, which would be a 3-d null surface in the static case. –  Jerry Schirmer Jun 2 at 18:36

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