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Is there a good interpretation of what the normal vector (and its magnitude) $$\mathbf{N}=\frac{\partial \mathbf{X}}{\partial s}\times\frac{\partial\mathbf{X}}{\partial t}$$ to the parametric surface $\mathbf{X}(s,t)$ represents?

In line integrals the quantity $||\mathbf{x}'(t)||$ is the speed of the curve, and the way to "normalize" things is to use the arc-length parameterization.

So does there always exist (for the usual "nice" surfaces) a similar parameterization where $||\mathbf{N}||=1$? Perhaps more stringently, does there always exist a parameterization where $$\left|\left|\frac{\partial \mathbf{X}}{\partial s}\right|\right|=\left|\left|\frac{\partial \mathbf{X}}{\partial t}\right|\right|=1\;\;\text{and}\;\;\frac{\partial \mathbf{X}}{\partial s}\cdot \frac{\partial \mathbf{X}}{\partial t}=0$$ and if so, what would this mean? Is this some kind of "orthogonal unit speed" parameterization or something? Cheers!

  • You can always get $| \mathbf{N} | = 1$ through $$| \mathbf{N} | = \frac{\partial_t \mathbf{X} \times \partial_t \mathbf{X}}{| \partial_s \mathbf{X} \times \partial_t \mathbf{X} | }.$$ – IAmNoOne Apr 25 '14 at 03:10
  • In line integrals the quantity ||x′(t)|| is the speed of the curve, and the way to "normalize" things is to use the arc-length parameterization. Can you elaborate on "things"? – IAmNoOne Apr 25 '14 at 03:11
  • @Nameless For example, scalar line integrals become just $\int_C{f(\mathbf{x}(t));\mathrm{d}t}$ without the "adjustment factor" $||f(\mathbf{x})||$. In general the derivative is the unit tangent vector, which is the "normalized" velocity. –  Apr 25 '14 at 03:14
  • Is there a good interpretation of what the normal vector (and its magnitude) to the parametric surface? We can identify whether the surface is orientable through its normal vector; and as far as I know, we are more concerned with finding the unit normal than just the regular normal. – IAmNoOne Apr 25 '14 at 03:14
  • @OP (your name keeps disappearing) sorry I am still not understanding you. You still haven't told me what the "thing" is. – IAmNoOne Apr 25 '14 at 03:16
  • @Nameless The 'thing' that is normalized is the velocity vector/derivative $x'$. I am wondering if one can parameterize surfaces so that their tangents are of unit length and perpendicular to one another. –  Apr 25 '14 at 03:20
  • You can always choose "orthogonal coordinates" (or isothermal coordinates) to parametrize your surface, but there is no canonical choice for parametrizing a surface such as the case for curves. In fact, the lack of a canonical parametrization serves to make the study of surfaces considerably more difficult than the study of curves. – THW Apr 25 '14 at 05:17
  • @THW Could you elaborate please (perhaps as answer)? –  Apr 25 '14 at 05:18

2 Answers2

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The direction is obvious ($\bf N$ is orthogonal to the tangent plane), but the magnitude is definitely nontrivial: is the local factor of measure change of the parametrization. Is like $\|x'(t)\|$ in curves and abs(jacobian) in the change of variables formula.

EDIT:

The easier case in the change of variables formula is: $A\subset\Bbb R^n$ open, $T:\Bbb R^n\longrightarrow\Bbb R^n$ linear, $m=$Lebesgue measure. $$|\det T|\,m(A)=\int_A|\det T|=\int_{TA}1=m(TA),$$ i.e., $$|\det T|={m(TA)\over m(A)}.$$

Martín-Blas Pérez Pinilla
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  • Could you elaborate on what that means? I'm not sure what "measure change of the parameterization" is, could you perhaps be more detailed please? –  Apr 27 '14 at 03:56
  • Start thinking what says the change of variables formula when the integrand is 1. – Martín-Blas Pérez Pinilla Apr 27 '14 at 08:20
  • I am perfectly familiar with the Jacobian, but I am sorry I really don't see what you're getting at. Could you please be more explicit, maybe add an edit to your answer? –  Apr 27 '14 at 17:46
  • See the edit and my answer in http://math.stackexchange.com/questions/627926/understanding-integration-techniques. – Martín-Blas Pérez Pinilla Apr 28 '14 at 06:19
  • I am really sorry for all these questions - you are taking $T:\mathbb{R}^n\to\mathbb{R}^n$ but I am interested in $T:\mathbb{R}^2\to\mathbb{R}^3$ so I don't think taking the determinant works.... out of curiosity, would the formula $\sqrt{\det (A^T A)}$ work in general as the "Jacobian" for integrals of functions from $\mathbb{R}^n$ to $\mathbb{R}^m$?? (Where $A$ would be the derivative matrix of the parameterization of the space over which integration was happening) - and I haven't learned what Lebesgue measures are yet. –  Apr 28 '14 at 15:46
  • Lebesgue measure is simply length/area/volume... technical details are unimportant. About the dimensions: a parametrized curve is a function $c:[a,b]\longrightarrow\Bbb R^3$, but the image set (excluding the usual suspects) is much smaller: a 1D subset of $\Bbb R^3$. The length is $\int_a^b|c'(t)|$dt; think what says this formula in an small interval $[t_0,t_0+\epsilon]$. I will think in your generalized "jacobian". – Martín-Blas Pérez Pinilla Apr 29 '14 at 06:25
  • Ok, so direction is normal to surface, and magnitude is the amount of "area distortion" that a small rectangle $[s_0,s_0+\epsilon]\times [t_0,t_0+\epsilon]$ receives under the transformation $\mathbf{X}(s,t)$. Does that look right? And thank you very much for your help, cheers! –  Apr 29 '14 at 17:04
  • You sound doubtful :) Why the .......? Is that wrong? –  Apr 29 '14 at 20:14
  • Isn't doubt. A comment requires 15 characters of more. – Martín-Blas Pérez Pinilla Apr 29 '14 at 20:15
  • I see........ :) –  Apr 29 '14 at 20:15
  • Oh and I forgot to ask, can one always "normalize" by re-parameterizing the surface? What is the answer to the second part of my question? –  Apr 29 '14 at 20:17
  • Yes, is possible re-parametrize with the condition $|{\bf N}|=1$, but I haven't find a reference. About the othogonatily condition I do not know if it will be possible. – Martín-Blas Pérez Pinilla Apr 30 '14 at 11:58
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Given a parametric representation $(s,t)\mapsto{\bf x}(s,t)$ of a surface $S\subset{\mathbb R}^3$ one usually reserves the letter ${\bf n}$ for the normalized surface normal vector: $${\bf n}(s,t):={{\bf x}_s\times{\bf x}_t\over|{\bf x}_s\times{\bf x}_s|}\ .$$ The vector product ${\bf x}_s\times{\bf x}_t$ itself, resp., the $2$-form $$d\omega\!\!\!\!\!\omega:={\bf x}_s\times{\bf x}_t\ {\rm d}(s,t)\ ,$$ plays a rôle in the computation of flow integrals. When the surface $S$ is embedded in a flow field ${\bf v}$ defined in ${\mathbb R}^3$ then $${\bf v}\cdot d\omega\!\!\!\!\!\omega={\bf v}\cdot ({\bf x}_s\times{\bf x}_t)\ {\rm d}(s,t)=\bigl[{\bf v},{\bf x}_s,{\bf x}_t\bigr]\ {\rm d}(s,t)$$ can be interpreted as the amount of fluid traversing the "surface element" $\>dS\>$ spanned by ${\bf x}_s\>ds$ and ${\bf x}_t\>dt$ (an infinitesimal parallelogram) per second, and the flow integral $$\int_S{\bf v}\cdot d\omega\!\!\!\!\!\omega=\int_S{\bf v}\cdot ({\bf x}_s\times{\bf x}_t)\ {\rm d}(s,t)$$ is the total amount of fluid crossing $S$ per second. Note that this integral is free of any square roots.

A parametrization $(s,t)\mapsto{\bf x}(s,t)$ with $|{\bf x}_s\times{\bf x}_s|\equiv1$ is area preserving, i.e., pieces $B$ of the $(s,t)$-parameter domain are mapped to pieces of $S$ having the same area as $B$. When $S$ is a rotational surface this can be realized in a rotationally symmetric way; whereas for an "arbitrary" $S$ nothing nice comes out of such an endeavor.

Christian Blatter
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  • So can a surface always be re-parameterized as described in the post? See the second part of my question. –  Apr 29 '14 at 20:20