I know two kinds formulas to calculate extrinsic curvature. But I found they do not match.(adsbygoogle = window.adsbygoogle || []).push({});

One is from "Calculus: An Intuitive and Physical Approach"##K=\frac{d\phi}{ds}## where ##Δ\phi## is the change in direction and ##Δs## is the change in length. For parametric form curve ##(x(t),y(t))## the extrinsic curvature is given by

$$K=\frac{x'y''-x''y'}{\sqrt{x'^2+y'^2}}$$

where ##'≡\frac{d}{dt}##

The extrinsic curvature formula in general relativity from "Eric Possion, A Relativist's Toolkit" is given by ##K=∇_\alpha n^\alpha##. For a plane curve ##(x(t),y(t))## in flat space, the outgoing unit normal vector is ##(n^x,n^y)=(\frac{y'}{\sqrt{x'^2+y'^2}},\frac{-x'}{\sqrt{x'^2+y'^2}})##, and the extrinsic curvature is

$$K=\partial_x n^x+\partial_y n^y=\frac{1}{x'} \partial_t n^x+\frac{1}{y'} \partial_t n^y=2\frac{x'y''-x''y'}{\sqrt{x'^2+y'^2}}$$

Is there anything wrong here? Thanks in advance!

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# I About extrinsic curvature

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