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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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