Why do parametrizations to the normal of a sphere sometimes fail?

Question

If I take the upper hemisphere of a sphere, $x^2 + y^2 + z^2 = 1$, to be $\sqrt{1 - x^2 - y^2}$, then the normal is given by $\langle -f_x, - f_y, 1\rangle$ at any point.

This leads to an odd result: on the plane $z = 0$, while one might expect all normals of the sphere to not have any $z$ component (i.e, to only point radially outwards), the $\vec{k}$ component is still 1.

A similar parametrization in cylindrical coordinates is:

$\langle \sin(\phi)^2 \cos(\theta), \sin(\phi)^2 \sin(\theta), \sin(\phi) \cos(\phi)\rangle$.

At $\phi = 0$ and $\theta = 0$, which corresponds to the point $(0,0,1)$ in cartesian coordinates, the normal is $\langle 0,0,0 \rangle$ while one would expect $(0,0,1)$.

Answer 1

I think a large part of the difficulty here is the variation in the magnitudes of your "normals" in general.

When you give a "normal" in the form $\langle -f_x, - f_y, 1\rangle$, basically what you have is a radial projection of the hemisphere parameterized by $\langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi\rangle$ (for $0 \leq \phi \leq \frac\pi2$) onto the plane $z = 1$. That is, you have a projection that takes a point $P$ on the surface of the hemisphere to a projected point $P'$ on the plane such that $P$ and $P'$ are collinear with the center of the sphere. (This is also called a gnomonic projection of the sphere.) The coordinates of your "normals" are the coordinates of the projected points.

As long as $0 \leq \phi < \frac\pi2$, each point of your sphere does in fact project onto that plane, although as $\phi$ approaches $\frac\pi2$ the magnitudes of your "normals" grow without bound. And obviously the boundary of your hemisphere, where $\phi = \frac\pi2$, does not project on to the plane $z = 1$ at all.

If you change your choice of $f$ as recommended in the answer by H.R., this problem goes away. All your normals then will have the same magnitude, and they will be defined at all points of the hemisphere.

When you give your "normals" in the form $\langle\sin^2\phi\cos\theta, \sin^2\phi\sin\theta, \sin\phi\cos\phi\rangle$, you again have non-uniform magnitudes, but this time the magnitudes go to zero as $\phi$ approaches zero. In effect, you are radially projecting the hemisphere onto a kind of degenerate torus given by $r = \sin\phi$.

But notice that the three components of your "normals" all have the common factor $\sin\phi$. You can normalize the magnitudes of all your "normals" (except for the case $\phi = 0$) by multiplying by the scalar $1/\sin\phi$. If you do this, you get the vectors $\langle\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi\rangle$, that is, the coordinates of each vector (for $0 < \phi \leq \frac\pi2$) are simply the coordinates of the point on the sphere. If you define these vectors as the normals for all points such that $0 < \phi \leq \frac\pi2$, you may see that you can use the same formula to define the normal for $\phi = 0$ as well, and it works very nicely.

The formula ${\bf r}_\phi \times {\bf r}_\theta$ works nicely for the measurement of area (which is what it is used for in the example you quoted) precisely because its magnitude does go to zero as $\phi$ goes to zero and does so in the same way that the area element $r \,d\phi\,d\theta$ goes to zero. But as you observed, this property is not so desirable when you are trying to construct a set of normal vectors rather than trying to integrate some scalar function over area.

Answer 2

Cartesian Coordinates

I think you should look closer to understand why this happens. If we write the implicit equation of the upper semi-sphere as

$$f(x,y,z)=z-\sqrt{1-x^2-y^2}$$

and then compute the $\nabla f$ to get

$$\nabla f=\frac{x}{\sqrt{1-x^2-y^2}}{\bf{i}}+\frac{y}{\sqrt{1-x^2-y^2}}{\bf{i}}+{\bf{k}}$$

then it is evident that this is not defined on the circle $x^2+y^2=1$ which lie on the plane $z=0$. However, there is a better way to define $f(x,y,z)$

$$f(x,y,z)=x^2+y^2+z^2-1$$

and hence

$$\nabla f=2x{\bf{i}}+2y{\bf{i}}+2z{\bf{k}}=2{\bf{x}}$$

Spherical Coordinates

The radial unit vector in spherical coordinate is normal to the surface of unit sphere and it is given by

$${\bf{e}}_r= \sin\phi \cos\theta {\bf{i}}+ \sin\phi \sin\theta {\bf{j}}+ \cos\phi {\bf{k}}$$

and when $\theta=0$ and $\phi=0$ we get

$${\bf{e}}_r= {\bf{k}}$$

so in this case you we have no problems and everything is well-defined.

About the Example in the Book

You should note that

$${\bf{r}}_{\phi} \times {\bf{r}}_{\theta}=a^2 \sin\phi {\bf{e}}_r $$

and when just $\phi=0$ you will get ${\bf{r}}_{\phi} \times {\bf{r}}_{\theta}={\bf{0}}$. This is because the intersection of the $\phi=0$ (the positive part of $z$ axis) with the sphere is just a point and evaluating tangent for a point is meaning-less. You see that ${\bf{r}}_{\theta}$ which is meant to be the tangent to the curves of $\theta=\text{Const}$ at $\phi=0$ is $\bf{0}$ because there is no curves there actually! You can observe the same thing for $\phi=\pi$!

Answer 3

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$Your intuition is probably based on unit normal vectors. The (outward) unit normal vector $\Vec{n}$ to the unit sphere at the point $(x, y, z)$ is $x\Vec{i} + y\Vec{j} + z\Vec{k}$, and indeed, the unit normals along the equator $\{z = 0\}$ have vanishing $z$ component while the unit normal at the north pole $(0, 0, 1)$ is $\Vec{k}$.

What happens in your examples is this: Let $\Vec{X}$ be a smooth parametrization of a surface $S$ in $\Reals^{3}$, and let $\Vec{X}_{u}$ and $\Vec{X}_{v}$ denote the partial derivatives of $\Vec{X}$. If the partials are linearly independent for parameter values $(u, v)$, their cross product $$ \Vec{X}_{u} \times \Vec{X}_{v} = \|\Vec{X}_{u} \times \Vec{X}_{v}\|\, \underbrace{\frac{\Vec{X}_{u} \times \Vec{X}_{v}}{\|\Vec{X}_{u} \times \Vec{X}_{v}\|}}_{\Vec{n}} $$ is normal to $S$ at $\Vec{X}(u, v)$, but has length $\|\Vec{X}_{u} \times \Vec{X}_{v}\|$.

Write $f(u, v) = \sqrt{1 - u^{2} - v^{2}}$. For the graph parametrization $\Vec{X}(u, v) = \bigl(u, v, f(u, v)\bigr)$, you have $$ \Vec{X}_{u} \times \Vec{X}_{v} = -f_{u}\Vec{i} - f_{v} \Vec{j} + \Vec{k} = \frac{1}{f(u, v)} \underbrace{\bigl(-u \Vec{i} -v \Vec{j} + f(u, v)\Vec{k}\bigr)}_{\Vec{n}}; $$ the $z$ component of the unit normal approaches $0$ along the equator, as expected, while the magnitude of the cross product itself grows without bound in such a way that the $z$ component is identically equal to $1$.

For the spherical coordinates parametrization $$ \Vec{X}(\theta, \phi) = a\cos\theta \sin\phi \Vec{i} + a\sin\theta \sin\phi \Vec{j} + a\cos\phi \Vec{k}, $$ the partial $\Vec{X}_{\theta} = a\sin\phi(-\sin\theta \Vec{i} + \cos\theta \Vec{j})$ vanishes at the poles $\phi = 0$ and $\phi = \pi$ (geometrically, longitudes all meet at the poles, and so latitude circles shrink to points), so the cross product vanishes at the poles, too: $$ \Vec{X}_{\phi} \times \Vec{X}_{\theta} = a^{2} \sin\phi \underbrace{(\cos\theta \sin\phi \Vec{i} + \sin\theta \sin\phi \Vec{j} + \cos\phi \Vec{k})}_{\Vec{n}}. $$ The unit normal, however, has the "expected" behavior even at the poles.