Signed geodesic curvature

In a local $x$ chart write $x(t)=x\circ \gamma (t)$ and $T={\dot{x}}^j({\mathrm{\partial }}_j\circ \gamma )$. By the pullback rule [F8], the affine-connection rules [F7], and the coordinate coefficients [F3], for a smooth curve $D_{t}T={\ddot{x}}^k{\mathrm{\partial }}_k+{\dot{x}}^i{\dot{x}}^j{\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_j=({\ddot{x}}^k+{\mathrm{\Gamma }}^k{}_{ij}{\dot{x}}^i{\dot{x}}^j){\mathrm{\partial }}_k\mathrm{.}$ The same coordinate expression is defined for a $C^2$ curve. In a second chart $y=y(x)$, the chain rule gives ${\ddot{y}}^a=\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^k}{\ddot{x}}^k+\frac{{\mathrm{\partial }}^2{y}^a}{\mathrm{\partial }{x}^i\mathrm{\partial }{x}^j}{\dot{x}}^i{\dot{x}}^j\mathrm{.}$ Substitute ${\dot{y}}^b=(\mathrm{\partial }{y}^b\mathrm{/}\mathrm{\partial }{x}^i){\dot{x}}^i$ and [F4] into ${\ddot{y}}^a+{\tilde{\mathrm{\Gamma }}}^a{}_{bc}{\dot{y}}^b{\dot{y}}^c$. The second-derivative terms cancel by differentiating $y^a(x(y))=y^a$ twice, leaving ${\ddot{y}}^a+{\tilde{\mathrm{\Gamma }}}^a{}_{bc}{\dot{y}}^b{\dot{y}}^c=\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^k}({\ddot{x}}^k+{\mathrm{\Gamma }}^k{}_{ij}{\dot{x}}^i{\dot{x}}^j)\mathrm{.}$ This is the tangent-vector coordinate transformation rule. Hence the formula defines a chart-independent covariant acceleration for $C^2$ curves; for smooth curves it is the usual $D_{t}T$, and [F5] supplies the regularity of the boundary arcs. At included endpoints all derivatives are one-sided.

Along a positively oriented boundary arc with regular-speed tangent $T$, let ${\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be the outward unit conormal. It is perpendicular to $T$ and has unit length, so it equals either $JT\mathrm{/}\mathrm{\mid }T\mathrm{\mid }$ or $-JT\mathrm{/}\mathrm{\mid }T\mathrm{\mid }$. In the positive basis $(T,JT)$, the ordered pair $(-JT,T)$ has positive determinant. Thus the outward-normal-first rule [F6] selects ${\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}=-JT\mathrm{/}\mathrm{\mid }T\mathrm{\mid }$; consequently $JT\mathrm{/}\mathrm{\mid }T\mathrm{\mid }$ is the inward unit conormal. For unit speed this reduces to $JT$.

In coordinates, metric compatibility [F2] says ${\mathrm{\partial }}_k{g}_{ij}={\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ki}{g}_{\mathrm{\ell }j}+{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{kj}{g}_{i\mathrm{\ell }}\mathrm{.}$ Differentiate the unit-speed identity $g_{ij}(x(t)){\dot{x}}^i{\dot{x}}^j=1$. Substituting this expression for ${\mathrm{\partial }}_k{g}_{ij}$ and using $g_{ij}=g_{ji}$ gives $0=2g_{\mathrm{\ell }j}({\ddot{x}}^{\mathrm{\ell }}+{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ki}{\dot{x}}^k{\dot{x}}^i){\dot{x}}^j=2g(A_{\gamma },T)\mathrm{.}$ So $A_{\gamma }$ is perpendicular to $T$, including when $A_{\gamma }=0$.

By [F1], $(T,JT)$ is a positive orthonormal basis. Since $g(A_{\gamma },T)=0$, write $A_{\gamma }=bJT$; taking the inner product with $JT$ forces the unique coefficient $b=g(A_{\gamma },JT)$. This defines the signed scalar $k_g$ and proves $A_{\gamma }=k_{g}JT$. If the curve is geodesic at a point, then $A_{\gamma }=0$ there and $k_g=0$.

In standard Cartesian coordinates on the Euclidean plane, $g_{ij}={\delta }_{ij}$. Lower the first index of ${\mathrm{\Gamma }}^k{}_{ij}$ using $\delta$. Torsion freeness in [F2] makes the resulting ${\mathrm{\Gamma }}_{aij}$ symmetric in its last two indices, while metric compatibility with constant $g_{ij}$ makes it skew in its first two. Hence ${\mathrm{\Gamma }}_{aij}={\mathrm{\Gamma }}_{aji}=-{\mathrm{\Gamma }}_{jai}=-{\mathrm{\Gamma }}_{jia}={\mathrm{\Gamma }}_{ija}={\mathrm{\Gamma }}_{iaj}=-{\mathrm{\Gamma }}_{aij}$, so all these coefficients vanish. Now take $\gamma (s)=(R\cos (s\mathrm{/}R),R\sin (s\mathrm{/}R))$ with $R>0$. Its unit tangent is $T=(-\sin (s\mathrm{/}R),\cos (s\mathrm{/}R))$, the positive quarter-turn is $JT=(-\cos (s\mathrm{/}R),-\sin (s\mathrm{/}R))$, and its ordinary acceleration is $T^{\mathrm{\prime }}=(-\cos (s\mathrm{/}R),-\sin (s\mathrm{/}R))\mathrm{/}R=(1\mathrm{/}R)JT$. Hence $k_g=1\mathrm{/}R>0$, consistent with the inward normal of the counterclockwise circle.

Reversing the curve parameter preserves $A_{\gamma }$ in local coordinates, because the second derivative and the product of the two first derivatives are unchanged, but replaces $T$ by $-T$ and $JT$ by $-JT$; hence it replaces $k_g$ by $-k_g$. Reversing the surface orientation also changes $J$ to $-J$ and so changes the sign of $k_g$ for the same curve. These checks include $k_g=0$.