Re: Re: Urgent Help needed

*To*: mathgroup at smc.vnet.net*Subject*: [mg20578] Re: [mg20523] Re: [mg20390] Urgent Help needed*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Sat, 30 Oct 1999 14:54:53 -0400*Sender*: owner-wri-mathgroup at wolfram.com

One can certainly do this in some cases: for example for surfaces of the form a*x^k+b*y^k+c*z^k==1 (a,b,c constant) this is worked out in detail on page 411 in Alfred Grey's book "Modern Differential Geometry of Curves and Surfaces with Mathematica" using exactly the method described below. However, I do not think you can write a "general formula" for an arbitrary surface like the one that Daniel Lichtblau gave for a curve. Of course the meaning of "formula" is vague so one may dispute the point. Perhaps the question could be reformulated as follows: can one give an algorithm that can be turned into a Mathematica program which, given an equation in x,y, and z and a point in R^3, computes the Gaussian curvature of th corresponding surface at this point. I do not have O'Neill's book and this but I would be surprised if a practical algorithm of this kind could be given. My reasons are only intuitive: Gaussian curvature is an intrinsic local property of a surface, while the equation depends on the exact position of the curve in R^3. This is of course not a proof, but it makes me feel unlikely that an algorithm of the above kind could exist. Still, I may well be wrong: I am a topologist, not a geometer. -- Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp > From: "Richard I. Pelletier" <bitbucket at home.com> > Organization: @Home Network > Date: Sat, 30 Oct 1999 00:13:46 -0400 > To: mathgroup at smc.vnet.net > Subject: [mg20578] [mg20523] Re: [mg20390] Urgent Help needed > >> Vladimir Tsyrlin wrote: >>> >>> Given the implcit form of a curve, i.e. F(x,y,z) = 0, do you know how to >>> find the curvature of F at a point in 3D space? All the references I have >>> assume F is in parametric form and take the standard differential geometry >>> approach. >>> >>> -- >>> ************************************************** >>> *************Vladimir Tsyrlin ******************* >>> vtsyrlin at cs.rmit.edu.au vtsyrlin at ozemail.com.au >>> ****************************************************** > > I have seen one post suggesting that you _meant_ the curvature of a > curve in 3-space, and therefore, need 2 equations. Let me assume you > _meant_ one equation, and therefore, the curvatures of a surface in > 3-space. > > And rather than try to show you the gory details, let me give you the > name of the answer, and one reference. > > What you want is called _the shape operator_. It is the covariant > derivative of a unit normal along a tangent to the surface. The nice > thing about the equation F(x,y,z)=0 is that the gradient of F _is_ > normal to the surface, so you just need to compute a covariant > derivative. > > Once you have the shape operator, in this case a 2x2 matrix, the > principal curvatures are its eigenvalues, so the mean and Gaussian > curvatures are their average and product resp. > > Chapter V of Barrett O'Neill Elementary Differential Geometry (1966 > ed.) is entitled _Shape Operators_, and pp. 216-219 work out the > details for precisely this problem. > > It does seem very common to focus on the case where you have a > parametric representation, but you don't need one. > > Vale, > Rip > -- > Multiplication is not commutative before breakfast. > > Richard I. Pelletier > NB eddress: r i p 1 [at] h o m e [dot] c o m >

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