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I'm having kind of a problem on calculating the normal vector to a sphere using a parameterization. Consider a unit-radius sphere centered at the origin.

One can parameterize it using the following: $$P(\phi, \theta)=(\sin(\phi)\cos(\theta),\,\sin(\phi)\sin(\theta),\,\cos(\phi)) $$

My Vector Calculus book says that the Vector Product between the two partial derivates of the parameterized surface gives a Normal Vector to the surface.

I found that $$\frac {\partial P(\phi, \theta)}{\partial \phi} \times \frac {\partial P(\phi, \theta)}{\partial \theta} = (\sin^2(\phi)\cos(\theta))\hat i + (\sin^2(\phi)\sin(\theta))\hat j + (\sin(\phi)\cos(\phi))\hat k$$

Which, as one can easily verify, yields $(0, 0,0)$ for $(\phi,\theta) = (0,0)$ which means that the sphere is not regular at the point $(0,0,1)$. That is obviously wrong, so I would like to know where is my mistake.

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    $\begingroup$ As a level surface $x^2+y^2+z^2=R^2$ is a sphere. The normal vector is given by $<2x,2y,2z>$ which is clearly non-trivial provided $R \neq 0$. As mentioned, the trouble you find is due to the coordinates chosen, it's not a genuine defect of the space. $\endgroup$ Jun 18 '13 at 2:46
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For a sphere, the surface normal is exactly your $P(\phi,\theta)$, since the normal is just the vector from the origin. Your vector $\vec N$ is indeed normal to the surface, but it's not normalized: $$|\vec N| = |\sin \phi \cdot \vec P| \neq1$$ The correct definition of the surface normal , coupled with the limit $\theta \to 0$ should give you the correct result: $$\vec N=\frac{\vec a \times \vec b}{\sqrt{|\vec a|^2|\vec b|^2-|\vec a\cdot\vec b|^2}}$$ Where: $$\vec a = \vec P_\phi, \ \ \vec b =\vec P_\theta$$

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  • $\begingroup$ But why does the vector product give the wrong result? Shouldn't it work for every surface? $\endgroup$
    – Thiago
    Jun 18 '13 at 2:32
  • $\begingroup$ @Thiago - fixed (I hope). $\endgroup$
    – nbubis
    Jun 18 '13 at 2:35
  • $\begingroup$ So I can only talk about surface regularity when evaluating the normalized normal vector? $\endgroup$
    – Thiago
    Jun 18 '13 at 2:39
  • $\begingroup$ No, it's just that the Jacobian of the parameterization you chose is singular at $\theta=n\pi$, so that your formula doesn't work there. The surface Normal obviously exists as you pointed out yourself. $\endgroup$
    – nbubis
    Jun 18 '13 at 2:42
  • $\begingroup$ Now I understand it. Thank you. $\endgroup$
    – Thiago
    Jun 18 '13 at 2:44
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This is a typical example of singular point(s) arising from the parametrization, they are called the artificial singularities.

For a parametrized surface: $S: \mathbf{r} = \mathbf{r}(u,v)$, like you said: $$ \mathrm{If } \;\; \mathbf{r}_u \times \mathbf{r}_v \neq 0, $$ then it indeed is a normal vector (un-normalized) to the surface $S$. Now: $$ \frac {\partial P(\phi, \theta)}{\partial \phi} \times \frac {\partial P(\phi, \theta)}{\partial \theta} = \begin{vmatrix} \mathbf{i}& \mathbf{j} &\mathbf{k} \\ \cos\phi \cos\theta & \cos\phi \sin\theta & -\sin\phi \\ -\sin\phi \sin\theta & \sin\phi \cos\theta & 0 \end{vmatrix}, $$ notice the cross product is 0 for $\phi = 0,\pi$, this happens due to the choice of parametrization. For example, if you choose the angle between $y$-axis as the polar angle, then this singular point will be gone (other arises though), and this gives you a heuristics of one of the reasons why we wanna cut a manifold into pieces and establish a local coordinate system...

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