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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 '14 at 18:36
@twistor59: Would you Explain the relationship between extrinsic curvature defined here and the mean curvature? –  Yuan Sun Dec 5 '14 at 10:08
I'm agree with @twistor59. At the horizon the normal vector and Killing vector coincide, then the Lie derivative of normal vector which by definition is called extrinsic curvature, will vanish. By the way I think it is not necessary to the horizon be two dimensional. –  MEDVIS Dec 5 '14 at 11:05

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