I asked this question first on physics forum, but there is no response. I post it here and hope someone can help me out.
I know two kinds formulas to calculate extrinsic curvature. But I found they do not match.
One is from "Calculus: An Intuitive and Physical Approach"$K=\frac{d\phi}{ds}$ where $\Delta \phi$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 $$ K=\frac{x'y''−x''y'}{(x'^2+y'^2)^{3/2}} $$ 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 (my interpretation of Possion's formula) $$K=\frac{\partial}{\partial x}n^x+\frac{\partial}{\partial y}n^y=\frac{1}{x'}\frac{\partial}{\partial t}n^x+\frac{1}{y'}\frac{\partial}{\partial t}n^y=2\frac{x'y''−x''y'}{(x'^2+y'^2)^{3/2}}$$
Is there anything wrong here? Thanks in advance!