Extrinsic Curvature Formulas in General Relativity: Are They Equivalent?

[craigthone]

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 $Δ\phi$ 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!

 

[Charles Link]

A google of the subject of curvature in differential geometry gave some rather complex results, but one of the more fundamental topics that came up was Gaussian curvature and principal curvatures. Maybe there is a simple solution to this one, but the google did not seem to yield any simple explanations.

 

[craigthone]

The denominators are $(x'^2+y'^2)^{3/2}$ ratherthan $\sqrt{(x'^2+y'^2)}$

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 $Δ\phi$ 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
$$K=\frac{x'y''-x''y'}{(x'^2+y'^2)^{3/2}}$$
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
$$K=\partial_x n^x+\partial_y n^y=\frac{1}{x'} \partial_t n^x+\frac{1}{y'} \partial_t n^y=2\frac{x'y''-x''y'}{(x'^2+y'^2)^{3/2}}$$

Is there anything wrong here? Thanks in advance!

 

[Charles Link]

For motion in a plane, $|\frac{d \hat{N}}{ds}|$ could also be used to define the curvature, since $\hat{N}$ is always perpendicular to $\hat{T}$. I was looking for a similar concept for a surface covering 3 dimensions, and I think it should be of the form $\vec{N}_{ij} =\frac{\partial{N_i}}{\partial{x_j}}$, but I couldn't find it in a google search. This would make $d \hat{N}=\vec{N}_{ij} \cdot d \vec{s}$, but I couldn't find it in this form. $\\$ Meanwhile, the derivatives in the second case are quite lengthy. If I get some free time, I may try to compute them and see if they match your first expression.

 

[Charles Link]

I worked the through the calculation in the second case, and got exactly what you did with the extra factor of 2. The next step is to see if $K=d \phi/ds$ in the first expression is computed correctly...And a google of it shows the formula is correct for $d \phi/ds$ without the 2. $\\$ Additional comment: It was my first instincts that this definition $K= \nabla_{\alpha} n^{\alpha}$ is not precisely $|\frac{d \hat{N}}{ds}|$. It's good to see that there is a simple factor of 2 that connects them for the two dimensional case.