# [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 Bright Glow
2 years, 6 months ago

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