📂Geometry

Formulas for Curves on a Sphere

Formula ¹

Let’s say a unit speed curve $\alpha : I \to \mathbb{R}^{3}$ is placed on a sphere with center $m$ and radius $r$. That is, $$\alpha (I) \subset S_{r,m} = \left\{ x \in \mathbb{R}^{3} : \left< x - m , x - m \right> = r^{2} \right\}$$ then it follows that $\kappa \ne 0$. If $\tau \ne 0$, then with respect to $\rho = 1/\kappa$ and $\sigma = 1 / \tau$, $$\alpha - m = - \rho N - \rho^{\prime} \sigma B$$ and, arranging for the radius, $$r^{2} = \rho^{2} + \left( \rho^{\prime} \sigma \right)^{2}$$

Derivation

Lemma: In an $n$-dimensional inner product space $V$, if $E = \left\{ e_{1} , \cdots , e_{n} \right\}$ is an orthogonal set, then $E$ form a basis of $V$, and for all $v \in V$, $$v = \sum_{k=1}^{n} \left< v , e_{k} \right> e_{k}$$

Differentiation of Inner Products: $$\left< f, g \right>^{\prime} = \left< f^{\prime}, g \right> + \left< f, g^{\prime} \right>$$

Frenet-Serret Formulas: If $\alpha$ is a unit speed curve with $\kappa (s) \ne 0$, $$\begin{align*} T^{\prime}(s) =& \kappa (s) N(s) \\ N^{\prime}(s) =& - \kappa (s) T(s) + \tau (s) B(s) \\ B^{\prime}(s) =& - \tau (s) N(s) \end{align*}$$

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$$\left< \alpha (s) - m , \alpha (s) - m \right> = r^{2}$$ Differentiating, if $r^{2}$ is constant and because of $T = \alpha^{\prime}$, $$\begin{equation} 0 = 2 \left< T , \alpha (s) - m \right> \label{1} \end{equation}$$ After dividing both sides by $2$ and differentiating once more, since $\alpha$ is a unit speed curve, according to the Frenet-Serret formula $T^{\prime} = \kappa N$, $$\begin{align*} 0 =& \left< T , \alpha (s) - m \right>^{\prime} \\ =& \left< T^{\prime} , \alpha (s) - m \right> + \left< T ,T \right> \\ =& \left< \kappa N , \alpha (s) - m \right> + 1 \end{align*}$$ Arranging, $$\kappa \left< N, \alpha (s) - m \right> = -1$$ Expressing in terms of $\rho = 1 / \kappa$, $$\begin{equation} \left< N, \alpha (s) - m \right> = - {{ 1 } \over { \kappa }} = - \rho \label{2} \end{equation}$$ According to the lemma, $$\begin{align*} & \alpha (s) - m \\ =& \left< \alpha (s) - m , T \right> T + \left< \alpha (s) - m , N \right> N + \left< \alpha (s) - m , B \right> B & \\ =& 0 - \rho N + \left< \alpha (s) - m , B \right> B & \because (1), (2) \end{align*}$$ Now, we only need to find $\left< \alpha (s) - m , B \right>$. Differentiating both sides of $\eqref{2}$ gives, $$\begin{align*} - \rho^{\prime} =& \left< N , \alpha (s) - m \right>^{\prime} \\ =& \left< N^{\prime} , \alpha (s) - m \right> + \left< N, T \right> \\ =& \left< -\kappa T + \tau B , \alpha (s) - m \right> + 0 \\ =& \tau \left< B , \alpha (s) - m \right> \end{align*}$$ Expressing in terms of $\sigma = 1/\tau$, $$\left< \alpha (s) - m , B \right> = - {{ \rho^{\prime} } \over { \tau }} = - \sigma \rho^{\prime}$$ Lastly, we obtain the following: $$\alpha (s) - m = - \rho N - \sigma \rho^{\prime} B$$

The formula for $r^{2}$ is derived as follows: $$\begin{align*} r^{2} =& \left< \alpha (s) - m , \alpha (s) - m \right> \\ =& \left< -\rho N - \rho^{\prime} \sigma B , -\rho N - \rho^{\prime} \sigma B \right> \\ =& \rho^{2} \left< N, N \right>^{2} + 2 \rho \rho^{\prime} \sigma \left< N, B \right> + \left( \rho^{\prime} \sigma \right)^{2} \left< B, B \right>^{2} \\ =& \rho^{2} \cdot 1 + 2 \rho \rho^{\prime} \sigma \cdot 0 + \left( \rho^{\prime} \sigma \right)^{2} \\ =& \rho^{2} + \left( \rho^{\prime} \sigma \right)^{2} \end{align*}$$

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