# [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 $$\alpha: I \rightarrow S$$ and $$\beta : I \rightarrow S$$ make on the surface $$S$$ of a torus? We can take a parametrisation of a torus, $\phi(u,v) = ( \sin u \cos v, \sin u \sin v, \cos u)$.

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

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

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

Note by Tasha Kim
10 months, 2 weeks ago

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