[Differential Geometry] Angle between Curves that meet on the Surface of a Torus

How can we find the angle at which two parametrised differentiable curves α:IS\alpha: I \rightarrow S and β:IS\beta : I \rightarrow S make on the surface SS of a torus? We can take a parametrisation of a torus, ϕ(u,v)=(sinucosv,sinusinv,cosu) \phi(u,v) = ( \sin u \cos v, \sin u \sin v, \cos u) .

and let cosθ=<ϕu,ϕv>ϕuϕv=FEG=cos \theta = \frac{\left< \phi_u, \phi_v\right>}{\lvert \phi_u \rvert \lvert \phi_v \rvert} = \frac{F}{\sqrt{EG}} = . The coefficients of the first fundamental form can be calculated as follows: The differential of the map ϕ\phi with respect to uu is ϕu \frac{\partial \phi}{ \partial u} , written as ϕu \phi_u for convenience, ϕu=(cosucosv,cosusinv,sinu). \phi_u = ( \cos u \cos v, \cos u \sin v, -\sin u). Similarly, ϕv=(sinusinv,sinucosv,0). \phi_v = ( -\sin u \sin v, \sin u \cos v , 0). Hence E=<ϕu,ϕu>=cos2(u)+sin2(u)=1E= \left< \phi_u, \phi_u \right> = \cos^2(u) + \sin^2(u) = 1, F=<ϕu,ϕv>=0F= \left< \phi_u, \phi_v \right> = 0, and G=<ϕv,ϕv>=sin2(u).G= \left< \phi_v, \phi_v \right> = \sin^2(u). Hence the angle at which the two coordinate curves of the torus meet are: FEG=0. \frac{F}{\sqrt{EG} = 0.} Now the coordinate curves of a parametrisation are orthogonal if and only if F(u,v)=0F(u,v) = 0 for all (u,v) (u,v). And such a parametrisation is called an orthogonal projection. Therefore, we see that the above parametrisation ϕ\phi is an orthogonal projection.

The angle at which the two curves meet at t=t0t= t_0 can be expressed as cosθ1=<α(t0),β(t0)>α(t0)α(t0) cos \theta_1 = \frac{ \left< \alpha'(t_0), \beta' (t_0) \right>}{\lvert \alpha'(t_0) \rvert \lvert \alpha'(t_0) \rvert} where α(t)=ϕua+ϕvb \alpha ' (t) = \phi_u a + \phi_v b and β(t)=ϕuc+ϕvd\beta '(t) = \phi_u c + \phi_v d . Hence, we have

E(ac)+F(bc)+F(ad)+G(bd)a2+b2sin2(u)c2+d2sin2(u)=ac+sin2(u)(bd)(a2+b2sin2(u))(c2+d2sin2(u)). \frac {E (ac) + F (bc) + F (ad) + G ( bd)}{\sqrt{a^2 + b^2\sin^2(u)} \sqrt{c^2 + d^2\sin^2(u) }} =\frac { ac+ \sin^2(u)( bd)}{\sqrt{(a^2 + b^2\sin^2(u))(c^2 + d^2\sin^2(u)) }} .

Note by Tasha Kim
1 year, 4 months ago

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