1

Given S -- a smooth surface in 3 dimensional space, p in S.

Needham doesn't really define things like curves and surfaces. He takes the Euclidean approach that you know what they are. So smooth will have to do as a description of the surface.

In Tristan Needham's book Visual Differential Geometry and Forms p. 109 Euler's Curvature formula is derived. It all depends on a plane containing the normal vector (called n_p) to a point p on S. No method is given as to how such a vector is found.

I am well aware that if we have a parameterized surface, that there is a formula for the normal vector involving the cross product.

Is there a geometric way to find n_p, or did Euler just assume a normal exists, or did Euler know about parameterized surfaces and use the cross product formula

lewis robinson
  • 29
  • For a parametric surface $\vec{S}(u, v)$, you can take $\frac{∂ \vec{S}(u,v)}{∂u} × \frac{∂\vec{S}(u,v)}{∂v}$ to get a normal vector — e.g., see this article. – Jollywatt Mar 08 '22 at 23:19
  • @Jollywatt I think the OP already knows this: "I am well aware that if we have a parameterized surface, that there is a formula for the normal vector involving the cross product." – Arthur Mar 08 '22 at 23:19
  • There is no such thing as an non-parameterized smooth surface. The very definition of a smooth surface in $\mathbb{R}^n$ is the existence of a smooth parameterization. – junjios Mar 08 '22 at 23:22
  • @Arthur Oh, whoops. I misread. OP, what do you mean by a “geometric way” to find a normal vector? Geometrically, you could just… choose a vector which is perpendicular. – Jollywatt Mar 08 '22 at 23:22
  • 1
    Here's how Needham (obliquely) tells you that his curves in 3 dimensional space are differentiable. On p. 106 he says that “each infinitesimal segment (of a curve) nevertheless lies in a plane.” This tells you that the curve has a tangent, and a normal to it at 90 degrees (but not necessarily in the plane). So it must be differentiable (oblique no?). On p. 107 he differentiates the tangent in the infinitesimal plane to get the principal normal (which DOES lie in the plane). – lewis robinson Mar 08 '22 at 23:36

1 Answers1

1

I don't know the history well enough to make written claims, but here are some observations from a modern viewpoint.

We speak of curves and surfaces as sets, but our analytic tools in differential geometry usually rely on mappings. We're then obliged to show our definitions do not depend on the mapping. In order to calculate anything, we're generally obliged to pick a mapping of some sort. Even to describe many geometric objects, we're obliged to refer to mappings.

To use surfaces as an example, if $U$ is a non-empty open set in the Euclidean plane and $x$ is a smooth, injective mapping to Euclidean three-space whose differential $Dx$ has rank two at each point of $U$, and if the inverse $x^{-1}:x(U) \to U$ is continuous, then $x$ is a parametrization and the set $x(U)$ is a smooth surface patch.

We can now define a subset $S$ of Euclidean three-space to be a smooth surface if for each point $p$ of $S$, there exists an open set $V$ in Euclidean three-space and a surface patch whose image is $x(U) = S \cap V$. It's a theorem that if $S$ is a smooth surface and $p$ is a point of $S$, there exist precisely two unit vectors at $p$ normal to $S$.

We have other ways to identify subsets as smooth surfaces. Particularly, the implicit function theorem guarantees that if $F$ is a smooth, real-valued function on some non-empty open subset of Euclidean three-space, if $S = F^{*}(\{c\})$ is a level set of $F$, and if $DF(p)$ is not zero at each point $p$ of $S$, then $S$ is a smooth surface. The normal vectors at $p$ turn out to be the two unit vectors proportional to the gradient of $F$ at $p$.

From either perspective, incidentally, it's straightforward to check that the graph of a smooth, real-valued function in $U$ is a surface, and to calculate its normal vector at an arbitrary point. But note carefully that derivatives can be visually-surprising. Suppose $S$ looks like the $(x, y)$-plane, and suppose $P$ is an arbitrary plane through the origin. Mathematically, $S$ might have a microscopic bump making $P$ tangent to $S$! On closer inspection (as it were), we can't generally define, much less find, tangent planes of smooth surfaces geometrically, which means we can't find normal vectors geometrically either. (By contrast, we can define tangent planes geometrically for geometrically-defined surfaces such as planes, cylinders, and spheres.)

This is a lot of technical fine print and conceptual baggage to chew when one first encounters surfaces! Small wonder that instructors and authors often wave our hands and refer to soap films or membranes or similar physical metaphors, and assume students and readers have a naive geometric idea what is a normal vector.

Andrew D. Hwang
  • 78,195
  • To be fair, we don't really define number until we get to university in general, and even then, it's glossed over, yet we're happy to expect calculations and even field extensions for a long time thence. Using some naïve intuition is almost requisite - you have to draw the line of defining things somewhere, otherwise it must be circular reasoning. – Nij Mar 09 '22 at 00:54
  • @Nij You're absolutely right about both logic and pedagogy. The point of this answer is not to express disapproval with Needham's treatment, but to note that it is qualitative and conceptual, not strictly computational. – Andrew D. Hwang Mar 09 '22 at 11:54