[Differential Geometry] Computing the Gaussian curvature of an Orthogonal Parametrisation

Let's derive a formula for the Gaussian curvature of an orthogonal parametrisation X(u,v)X(u,v) of a coordinate neighbourhood at point pp on a smooth, orientable surface, we have that <Xu,Xv>p=0 \left<X_u, X_v\right>_p = 0, hence F=0F=0 of the first fundamental form at pp.

First, we consider the Gauss formula expressed in Christoffel symbols: EK=(Γ122)u(Γ112)v+Γ121Γ112+Γ122Γ122Γ112Γ222Γ111Γ122 -EK = (\Gamma_{12}^2)_u - (\Gamma_{11}^2)_v + \Gamma_{12}^1\Gamma_{11}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2

How do we compute the Christoffel symbols?

If we assign each point of X(U)X(U) a natural trihedron given by the vectors Xu X_u , XvX_v , NN, and express the derivatives of the vectors XuX_u, XVX_V and NN in the basis {Xu,Xv,N}\{ X_u, X_v, N\},

Xuu=Γ111Xu+Γ112Xv+L1NX_{uu} = \Gamma^1_{11}X_u + \Gamma^2_{11}X_v+L_1N Xuv=Γ121Xu+Γ122Xv+L2NX_{uv} = \Gamma^1_{12}X_u + \Gamma^2_{12}X_v+L_2N Xvu=Γ211Xu+Γ212Xv+L2NX_{vu} = \Gamma^1_{21}X_u + \Gamma^2_{21}X_v + \overline{L_2}N Xvv=Γ221Xu+Γ222Xv+L3NX_{vv} = \Gamma^1_{22}X_u + \Gamma^2_{22}X_v + L_3N

And by taking the inner product of the first four relations below with XuX_u and XvX_v we get the following:

Xu,Xuu=Γ111Xu,Xu+Γ112Xv,Xu+0=Γ111E+Γ112F=12Eu, \langle X_u, X_{uu} \rangle = \Gamma^1_{11}\cdot \langle X_u, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_u \rangle + 0 = \Gamma^1_{11}E + \Gamma^2_{11}F = \frac{1}{2} E_u , Xv,Xuu=Γ111Xv,Xu+Γ112Xv,Xv+0=Γ111F+Γ112G=Fu12Ev.\langle X_v, X_{uu}\rangle = \Gamma^1_{11}\cdot \langle X_v, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_v \rangle + 0 = \Gamma^1_{11}F + \Gamma^2_{11}G = F_u-\frac{1}{2}E_v. Xu,Xuv=Γ121Xu,Xu+Γ122Xu,Xv+0=Γ121E+Γ122F=12Ev, \langle X_u, X_{uv} \rangle = \Gamma^1_{12}\cdot\langle X_u, X_u \rangle + \Gamma^2_{12}\cdot \langle X_u, X_v\rangle + 0 = \Gamma^1_{12}E + \Gamma^2_{12}F = \frac{1}{2}E_v, Xv,Xuv=Γ121Xv,Xu+Γ122Xv,Xv+0=Γ121F+Γ122G=12Gu.\langle X_v, X_{uv}\rangle = \Gamma^1_{12}\cdot \langle X_v, X_u \rangle + \Gamma^2_{12}\cdot \langle X_v, X_v\rangle +0= \Gamma^1_{12}F+ \Gamma^2_{12}G = \frac{1}{2} G_u. Xu,Xvv=Γ221Xu,Xu+Γ222Xu,Xv+0=Γ221E+Γ222F=Fv12Gu, \langle X_u, X_{vv}\rangle = \Gamma^1_{22}\cdot \langle X_u, X_u\rangle + \Gamma^2_{22}\cdot \langle X_u, X_v \rangle + 0 = \Gamma^1_{22}E + \Gamma^2_{22}F = F_v - \frac{1}{2}G_u, Xv,Xvv=Γ221Xv,Xu+Γ222Xv,Xv+0=Γ221F+Γ222G=12Gv.\langle X_v, X_{vv} \rangle = \Gamma^1_{22}\cdot \langle X_v, X_u \rangle + \Gamma^2_{22}\cdot \langle X_v, X_v\rangle + 0 = \Gamma^1_{22}F + \Gamma^2_{22}G = \frac{1}{2}G_v.

Using the condition that xx is an orthogonal parametrisation, that is, F=0F=0, then the above reduces to Γ111=12EuEΓ112=12EvG,Γ121=12EvE,Γ122=12GuG,Γ222=12GuE,Γ122=12GuG \Gamma_{11}^1 = -\frac{1}{2}\frac{E_u}{E}\Gamma_{11}^2 = -\frac{1}{2}\frac{E_v}{G}, \Gamma_{12}^1 = \frac{1}{2}\frac{E_v}{E}, \Gamma_{12}^2 = -\frac{1}{2}\frac{G_u}{G}, \Gamma_{22}^2 = -\frac{1}{2}\frac{G_u}{E}, \Gamma_{12}^2 = \frac{1}{2}\frac{G_u}{G}

We substitute these values into the Gauss formula: EK=(12GuG)u(12EvG)v14EuEGuG14EvEEvG+14EvGGvG+14GuGGuG-EK = \left(\frac{1}{2}\frac{G_u}{G}\right)_u - \left(-\frac{1}{2}\frac{E_v}{G}\right)_v - \frac{1}{4}\frac{E_u}{E}\frac{G_u}{G} - \frac{1}{4}\frac{E_v}{E}\frac{E_v}{G} +\frac{1}{4}\frac{E_v}{G}\frac{G_v}{G} +\frac{1}{4}\frac{G_u}{G}\frac{G_u}{G} =(12GuG)u(12EvG)v14EuGuEG14(Ev)2EG+14EvGvG2+14(Gu)2G2 = \left(\frac{1}{2}\frac{G_u}{G}\right)_u - \left(-\frac{1}{2}\frac{E_v}{G}\right)_v - \frac{1}{4}\frac{E_u G_u}{EG} - \frac{1}{4}\frac{(E_v)^2}{EG} + \frac{1}{4}\frac{E_v G_v}{G^2} +\frac{1}{4}\frac{(G_u)^2}{G^2}     K=12EG[EvvEGEv(EvG+EG2(EG)32+GuuEGGu(EuG+EGu2(EG)32]\implies K = -\frac{1}{2\sqrt{EG}}\left[\frac{E_{vv}}{\sqrt{EG}}- \frac{E_v(E_vG +EG}{2(EG)^{\frac{3}{2}}}+ \frac{G_{uu}}{\sqrt{EG}} - \frac{G_u(E_uG+EG_u}{2(EG)^{\frac{3}{2}}}\right] Therefore, simplifying gives the Gaussian curvature K=12EG[(EvEG)v+(GuEG)u].K = -\frac{1}{2\sqrt{EG}}\left[ \left(\frac{E_v}{\sqrt{EG}}\right)_v + \left(\frac{G_u}{\sqrt{EG}}\right)_u \right].

Note by Bright Glow
3 years, 2 months ago

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