3.5: Orientation of Surfaces

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One thing that made the flux integrals of the last section possible is that we could choose sensible unit normal vectors $\hat{\textbf{n}}\text{.}$ In this section, we explain this more carefully.

Consider the sphere $x^2+y^2+z^2=1\text{.}$ We can think of this surface as having two sides — an inside (the side you see when you are living inside the sphere) and an outside (the side you see when you are living outside the sphere). Concentrate on one point $(x_0,y_0,z_0)$ on the sphere. The surface $x^2+y^2+z^2=1$ has precisely two unit normal vectors at $(x_0,y_0,z_0)\text{,}$ namely

\[ \hat{\textbf{n}}_+ = +(x_0,y_0,z_0)\qquad\text{and}\qquad \hat{\textbf{n}}_- = -(x_0,y_0,z_0) \nonumber \]

We can view $\hat{\textbf{n}}_+$ as being associated to (or attached to) the outside of the sphere and $\hat{\textbf{n}}_-$ as being associated to (or attached to) the inside of the sphere. Note that, as we move over the sphere, both $\hat{\textbf{n}}_+$ and $\hat{\textbf{n}}_-$ change continuously.

Definition 3.5.1

An oriented surface is a surface together with a continuous function

\[ \hat{\textbf{N}}: S\rightarrow\mathbb{R} ^3 \nonumber \]

such that, for each point $p$ of $S\text{,}$ $\hat{\textbf{N}}(p)$ is a unit normal to $S$ at $p\text{.}$

Example 3.5.2. Sphere

One orientation of the sphere $S=\left \{(x,y,z)|x^2+y^2+z^2=1 \right \}$ is

\[ \hat{\textbf{N}}(x,y,z) = (x,y,z) \nonumber \]

It associates to each point $p$ of $S$ the outward pointing unit normal to $S$ at $p\text{.}$ We can think of $S$ with this orientation as being the outer side of $S\text{.}$

The other orientation of the sphere $S=\left \{(x,y,z)|x^2+y^2+z^2=1 \right \}$ is

\[ \hat{\textbf{N}}(x,y,z) = -(x,y,z) \nonumber \]

It associates to each point $p$ of $S$ the inward pointing unit normal to $S$ at $p\text{.}$ We can think of $S$ with this orientation as being the inner side of $S\text{.}$

While this discussion might seem inordinately picky, it turns out that not all surfaces can be oriented. Our next example exhibits one.

Example 3.5.3. Optional — The Möbius Strip

There are some surfaces $S$ for which it is not possible to choose a continuous orientation map $\hat{\textbf{N}}: S\rightarrow\mathbb{R} ^3\text{.}$ Such surfaces are said to be non-orientable. The most famous non-orientable surface is the Möbius¹ strip², which you can construct as follows. Take a rectangular strip of paper.

$mobiusC.svg$

Lay it flat and then introduce a half twist so that the arrow on the right hand end points upwards, rather than downwards. Then glue the two ends of the strip together, with the two arrows coinciding. That's the Möbius strip.

$mobiusB.svg$

Let's parametrize it. Think of the strip of paper that we used to construct it as consisting of a backbone (the horizontal black line in the figure below) with a bunch of ribs (like the thick blue line in the figure) emanating from it.

$mobiusD.svg$

When we glue the two ends of the strip together, the black line forms a circle. If the strip has length $\ell\text{,}$ the circle will have circumference $\ell$ and hence radius $\frac{\ell}{2\pi}\text{.}$ We'll parametrize it as the circle

\[\begin{gather*} \frac{\ell}{2\pi}\hat{\textbf{r}}(\theta) \qquad\text{where } \hat{\textbf{r}}(\theta) = \cos(\theta)\,\hat{\pmb{\imath}} + \sin(\theta)\,\hat{\pmb{\jmath}} \end{gather*}\]

This circle is in the $xy$-plane. It is the black circle in the figure below. (The figure only shows the part of the circle in the first octant, i.e. with $x,y,z\ge 0\text{.}$)

$mobiusE.svg$

Now we'll add in the blue ribs. We'll put the blue rib, that is attached to the backbone at $\frac{\ell}{2\pi}\hat{\textbf{r}}(\theta)\text{,}$ in the plane that contains the vectors $\hat{\textbf{r}}(\theta)$ and $\hat{\mathbf{k}}\text{.}$

A side view of the plane that contains the vectors $\hat{\textbf{r}}(\theta)$ and $\hat{\mathbf{k}}$ is sketched in the figure below.

$mobiusF.svg$

To put the half twist into the strip of paper, we want the blue rib to rotate about the backbone by $180^\circ\text{,}$ i.e. $\pi$ radians, as $\theta$ runs from $0$ to $2\pi\text{.}$ That will be the case if we pick the angle $\varphi$ in the figure to be $\frac{\theta}{2}\text{.}$ The vector that is running along the blue rib in the figure is

\[ \textbf{u}(v,\theta,\varphi)=v\cos(\varphi)\,\hat{\textbf{r}}(\theta) + v\sin(\varphi)\,\hat{\mathbf{k}} \nonumber \]

where the length, $v\text{,}$ of the vector is a parameter. If the width of our original strip of paper is $w\text{,}$ then as the parameter $v$ runs from $-\frac{w}{2}$ to $+\frac{w}{2}\text{,}$ the tip of the vector $\textbf{u}(v,\theta,\varphi)$ runs over the entire blue rib. So, choosing $\varphi=\frac{\theta}{2}\text{,}$ our parametrization of the Möbius strip is

\[\begin{align*} \vecs{r} (\theta,v) &= \frac{\ell}{2\pi}\hat{\textbf{r}}(\theta) + \textbf{u}\left(v,\theta,\frac{\theta}{2}\right)\\ &= \frac{\ell}{2\pi}\hat{\textbf{r}}(\theta) + v\cos\left(\frac{\theta}{2}\right) \hat{\textbf{r}}(\theta) + v\sin\left(\frac{\theta}{2}\right) \hat{\mathbf{k}}\\ & 0\le\theta \lt 2\pi,\ -\frac{w}{2}\le v\le \frac{w}{2} \end{align*}\]

where $\hat{\textbf{r}}(\theta) = \cos(\theta)\,\hat{\pmb{\imath}} + \sin(\theta)\,\hat{\pmb{\jmath}}\text{.}$

Now that we have parametrized the Möbius strip, let's return to the question of orientability. Recall, from Definition 3.5.1, that, if the Möbius strip were orientable, there would exist a continuous function $\hat{\textbf{N}}$ which assigns to each point $\vecs{r} $ of the strip a unit normal vector $\hat{\textbf{N}}(\vecs{r} )$ at $\vecs{r} \text{.}$ First, we'll find the normal vectors to the surface using 3.3.1. The partial derivatives

\[\begin{align*} \frac{\partial\vecs{r} }{\partial\theta}(\theta,v) &=\frac{\ell}{2\pi}\hat{\textbf{r}}'(\theta) + v\cos\left(\frac{\theta}{2}\right) \hat{\textbf{r}}'(\theta) - \frac{v}{2}\sin\left(\frac{\theta}{2}\right) \hat{\textbf{r}}(\theta) + \frac{v}{2}\cos\left(\frac{\theta}{2}\right) \hat{\mathbf{k}}\\ \frac{\partial\vecs{r} }{\partial v}(\theta,v) &=\cos\left(\frac{\theta}{2}\right) \hat{\textbf{r}}(\theta) + \sin\left(\frac{\theta}{2}\right) \hat{\mathbf{k}} \end{align*}\]

are relatively messy, so let's just consider the case $v=0$ (i.e. find the normal vectors on the backbone). Then

\[\begin{align*} \frac{\partial\vecs{r} }{\partial\theta}(\theta,0) &=\frac{\ell}{2\pi}\hat{\textbf{r}}'(\theta)\\ \frac{\partial\vecs{r} }{\partial v}(\theta,0) &=\cos\left(\frac{\theta}{2}\right) \hat{\textbf{r}}(\theta) + \sin\left(\frac{\theta}{2}\right) \hat{\mathbf{k}} \end{align*}\]

Since

\[\begin{align*} \hat{\textbf{r}}'(\theta)\times \hat{\textbf{r}}(\theta) &=\big(-\sin(\theta)\,\hat{\pmb{\imath}} + \cos(\theta)\,\hat{\pmb{\jmath}}\big)\times \big(\cos(\theta)\,\hat{\pmb{\imath}} + \sin(\theta)\,\hat{\pmb{\jmath}}\big) =-\hat{\mathbf{k}}\\ \hat{\textbf{r}}'(\theta)\times \hat{\mathbf{k}} &=\big(-\sin(\theta)\,\hat{\pmb{\imath}} + \cos(\theta)\,\hat{\pmb{\jmath}}\big)\times \hat{\mathbf{k}} =\hat{\textbf{r}}(\theta) \end{align*}\]

we have

\[\begin{gather*} \frac{\partial\vecs{r} }{\partial\theta}(\theta,0) \times \frac{\partial\vecs{r} }{\partial v}(\theta,0) =-\frac{\ell}{2\pi}\Big[\cos\left(\frac{\theta}{2}\right)\,\hat{\mathbf{k}} -\sin\left(\frac{\theta}{2}\right)\,\hat{\textbf{r}}(\theta) \Big] \end{gather*}\]

As $\hat{\mathbf{k}}$ and $\hat{\textbf{r}}(\theta)$ are mutually perpendicular unit vectors, $\cos\big(\frac{\theta}{2}\big)\,\hat{\mathbf{k}} -\sin\big(\frac{\theta}{2}\big)\,\hat{\textbf{r}}(\theta)$ has length one, and the two unit normal vectors to the Möbius strip at $\vecs{r} (\theta,0)$ are

\[ \pm \Big[\cos\left(\frac{\theta}{2}\right)\,\hat{\mathbf{k}} -\sin\left(\frac{\theta}{2}\right)\,\hat{\textbf{r}}(\theta) \Big] \nonumber \]

So, for each $\theta\text{,}$ $\hat{\textbf{N}}\big(\vecs{r} (\theta,0)\big)$ must be either

\[ \cos\left(\frac{\theta}{2}\right)\,\hat{\mathbf{k}} -\sin\big(\frac{\theta}{2}\big)\,\hat{\textbf{r}}(\theta) \qquad\text{or}\qquad -\Big[\cos\left(\frac{\theta}{2}\right)\,\hat{\mathbf{k}} -\sin\left(\frac{\theta}{2}\right)\,\hat{\textbf{r}}(\theta) \Big] \nonumber \]

Imagine walking along the Möbius strip. The normal vector $\hat{\textbf{N}}\big(\vecs{r} (\theta,v)\big)$ is our body when we are at $\vecs{r} (\theta,v)$ — our feet are at the tail of the vector $\hat{\textbf{N}}\big(\vecs{r} (\theta,v)\big)$ and our head is at the arrow of $\hat{\textbf{N}}\big(\vecs{r} (\theta,v)\big)\text{.}$ We start walking at $\vecs{r} (0,0)=\frac{\ell}{2\pi}\hat{\pmb{\imath}}\text{.}$ Our body, $\hat{\textbf{N}}\big(\frac{\ell}{2\pi}\hat{\pmb{\imath}}\big)=\hat{\textbf{N}}\big(\vecs{r} (0,0)\big)$ has to be one of $\pm \big(\cos(0)\,\hat{\mathbf{k}}-\sin(0)\, \hat{\textbf{r}} (0) \big)=\pm\hat{\mathbf{k}}\text{.}$ Let's suppose that $\hat{\textbf{N}}\big(\vecs{r} (0,0)\big)=+\hat{\mathbf{k}}\text{.}$ (We start upright.) Now we start walking along the backbone of the Möbius strip, increasing $\theta\text{.}$ Because $\hat{\textbf{N}}\big(\vecs{r} (\theta,0)\big)$ has to be continuous, $\hat{\textbf{N}}\big(\vecs{r} (\theta,0)\big)$ has to be $+\big(\cos\big(\frac{\theta}{2}\big)\,\hat{\mathbf{k}} -\sin\big(\frac{\theta}{2}\big)\, \hat{\textbf{r}} (\theta) \big)\text{.}$ We keep increasing $\theta\text{.}$ By continuity, $\hat{\textbf{N}}\big(\vecs{r} (\theta,0)\big)$ has to be $+\big(\cos\big(\frac{\theta}{2}\big)\,\hat{\mathbf{k}} -\sin\big(\frac{\theta}{2}\big)\, \hat{\textbf{r}} (\theta) \big)$ for bigger and bigger $\theta\text{.}$ Eventually we get to $\theta=2\pi\text{,}$ i.e. to

\[ \vecs{r} (2\pi,0)= \frac{\ell}{2\pi} \hat{\textbf{r}} (2\pi) = \frac{\ell}{2\pi}\hat{\pmb{\imath}} =\frac{\ell}{2\pi}\hat{\textbf{r}}(0)=\vecs{r} (0,0) \nonumber \]

We are back to our starting point. Continuity has forced

\[ \hat{\textbf{N}}\big(\vecs{r} (2\pi,0)\big) =\hat{\textbf{N}}\big(\vecs{r} (\theta,0)\big)\Big|_{\theta=2\pi} =+\Big[\cos\left(\frac{\theta}{2}\right)\,\hat{\mathbf{k}} -\sin\left(\frac{\theta}{2}\right)\,\hat{\textbf{r}}(\theta) \Big] \Big|_{\theta=2\pi} =-\hat{\mathbf{k}} \nonumber \]

So we have arrived back upside down. That's a problem — $\hat{\textbf{N}}\big(\vecs{r} (2\pi,0)\big) =\hat{\textbf{N}}\big(\frac{\ell}{2\pi}\hat{\pmb{\imath}}\big)$ and we have already defined $\hat{\textbf{N}}\big(\frac{\ell}{2\pi}\hat{\pmb{\imath}}\big)=+\hat{\mathbf{k}}\text{,}$ not $-\hat{\mathbf{k}}\text{.}$ So the Möbius strip is not orientable. The interested reader should look up M. C. Escher's Möbius Strip II (Red Ants).

August Ferdinand Möbius (1790--1868) was a German mathematician and astronomer. He was a descendant of Martin Luther and a student of Gauss.
Another famous non-orientable surface is the Klein bottle. You can easily find discussions of it using your favourite search engine.