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Curvature for surfaces?

  1. May 24, 2007 #1
    How do you define curvature for curves on three-dimensional surfaces when the surface is given in the form z=f(x,y)?

    The resulting formula should be a lot simpler than the one for parametric curves of the form r(t)=(x(t),y(t),z(t)), like it becomes for two-dimensional curves given by y=f(x).

    I cannot figure it out! Please help.
     
  2. jcsd
  3. May 24, 2007 #2

    quasar987

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    Well it seems to me that x and y serve as the parameters (i.e. set x=s, y=t). Then, Q(s,t)=(s,t,f(s,t)) is the parametrization of the surface, where s and t take all the values in the domain of f.
     
  4. May 24, 2007 #3
    Yes, so true. But what I am looking for is the equivalent to curvature as defined in vector calculus. What I have found are results where you only have one free parameter, usually termed "t". I do not know how to generalize to two or more free variables, while still being mathematically rigid, :(
     
  5. May 24, 2007 #4

    quasar987

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    oh! Well have you tried wikipedia? http://en.wikipedia.org/wiki/Curvature#Curvature_of_2-dimensional_surfaces

    There are many types of curvatures one can be interested in: gaussian curvature, mean curvature, normal curvature, principal curvature.

    For instance, the gaussian curvature is interesting because it characterizes surfaces up to isometry (see gauss' theorema egregium). Meaning in layman terms that if a surface is deformed without stretching and ripping (ex: turning a sheet of paper into a cylinder), then the resulting surface will have the same gaussian curvature at each point.

    The mean curvature is more like the equivalent of a curve's curvature applied to surfaces in that it will say that the curvature of a cylinder is not the same as that of a plane sheet, as intuition would suggest.
     
    Last edited: May 24, 2007
  6. May 24, 2007 #5

    Chris Hillman

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    Such a surface is often called a "Monge patch" in the theory of surfaces. So are you are seeking an expression for path curvature [itex]\kappa[/itex] and (torsion [itex]\tau[/itex]) for a space curve which belongs to a Monge patch?

    Moderators: I suggest moving this thread to the Differential Geometry forum, where it obviously belongs.
     
  7. May 25, 2007 #6

    quasar987

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    (Chris, if you wanna talk to the mods about moving a thread, I suggest pressing the little "REPORT" button uner the OP's name)
     
  8. May 25, 2007 #7
    THANK YOU GUYS!!!!

    For days I have been working hard to prove that a sphere has the smallest surface area, volume fixed, of all solids. I ended up with something that looked like a curvature in 3 dimensions, and that something had to be constant! I wasn't able to show that it represented some kind of curvature though.

    Due to the Monge Patch hint from Chris Hillman I immediately found out that what I had was exactly the mean curvature for Monge Patches, and therefore the solid must be part of a sphere...

    Thanks! I am sooooo happy right now, :)

    Now to the next problem: Can I use my proof for curves given by z=f(x,y) and extend the result to more general parametric surfaces? I think one has to do that in order to get a proof for full spheres and not only parts of them, or is there some simple trick?

    Man, am I happy or what, :)
     
  9. May 25, 2007 #8
    I have some more;

    Suppose that you showed what the optimum solution must look like, if it exists, but cannot fulfill the boundary conditions. Does it mean that there does not exist any optimal solution then? How to resolve it?

    One example: Suppose that you have showed that part of a sphere must be an optimal solution, but you have a circular boundary for the domain on the x-y plane where z(x,y) varies sinusoidally along the boundary. I suppose no sphere could fix that. What does that mean?
     
  10. May 31, 2007 #9
    Since this forum is labeled TensorAnalysis & Differential Geometry where do you suggest moving it to since this is the exact forum where one asks questions such as the one asked by the OP??


    Pete
     
  11. Jun 1, 2007 #10
    It might have been moved from somewhere else since Chris posted that sometime ago.
     
  12. Jun 1, 2007 #11
    I see. Thanks. Makes perfect sense. In fact I woke up with that thought this morning and came here to check on that possibility.

    Pete
     
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