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Principle curvature: min and max curvature are always perpendicular?

  1. Apr 27, 2013 #1
    I am trying to understand Gaussian curvature. This led me in to looking at principle curvature. Now If one takes a look at the picture of the "Saddle Surface" on Wikipedia here: http://en.wikipedia.org/wiki/Principal_curvature

    I see that at the point p on the saddle where curvature goes both positive and negative in different directions that there is a normal vector at point p. There are two planes of "principle curvature" which intersect at the normal vector at p. But furthermore that each normal plane (containing the normal p) intersects the surface and that at the point p a tangent plane is made which intersects any particular normal plane through p. That also for each normal plane going through p, the intersection of the normal plane with the surface gives rise to a curve on the normal plane - each curve different depending on the normal plane chosen. Now According to Wikipedia:

    "The directions of the normal plane where the curvature takes its maximum and minimum values are always perpendicular, if k1 does not equal k2"

    This part I don't understand.. I see that this result is due to Euler, but don't see why (or where to find an explanation on as to why this is true.

    Any thoughts?

    Thanks,

    Brian
     
  2. jcsd
  3. Apr 27, 2013 #2

    quasar987

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    Hi,

    It is easier to see what's going on if you define the principal curvatures as the eigenvalues of the second fundamental form II. Since II is symmetric, the spectral theorem guarantees that the eigenvalues k1,k2 of II are real and that there exists an orthonormal basis t1, t2 with respect to which II is diagonal (with diagonal entries its 2 eigenvalues). Define the principal curvatures as k1, k2.

    To see now that they correspond to the values that minimize/maximize the normal curvature, just pick a curve c(t) (unit speed), and write c' in terms of t1,t2 and the angle µ between t1 and c': c'=cos(µ)t1+sin(µ)t2. Then use that the normal curvature of c is (c')^T II c' (by definition of II). Write this matrix equation wrt to the basis t1,t2. It yields the following expression for the normal curvature of c: k_n(c)=k1cos²(µ)+k2sin²(µ). Use calculus to minimize/maximize this function. You will find that the min/max values of k_n(c) are when µ=0 and pi/2, which correspond to k_n=k1 and k2.
     
  4. Apr 28, 2013 #3
    Thanks Quasar! I am a little confused and am hoping you might clarify

    I am guessing you mean maybe c and c' as t1 and t2 are orthonormal? Also that the normal curvature of c is (c')^T II c'. I don't see what that could be.

    Perhaps you might be able to clarify?

    Thanks much though for your insight!

    Brian
     
  5. Apr 29, 2013 #4

    quasar987

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    No, I do mean c' and t1, why? What seems weird about that to you? (Note by the way that c and c' does not make sense as c is a curve, not a vector, so you can't really compare the two).

    All I'm saying is that since t1,t2 is a basis of the tangent plane of the surface (vector space), you can write c' as a linear combination of them. Moreover, you can use plane trigonometry to express the coefficients in terms of the angle µ between t1 and c'.

    This is a matrix equation, with (c')^T being the transpose of c'. This is a compact form of the expression

    [itex]L\dot{u}^2+2M\dot{u}\dot{v}+N\dot{v}^2[/itex]

    with which you are perhaps more familiar? In this expression, it is implicit that we have chosen local coordinates (u,v) for the surface that that wrt those coordinates, the curve c(t) is given by (u(t),v(t)). Dots mean derivatives wrt t. That the second fundamental form gives the normal curvature via this expression is practically by definition of the second fundamental form.
     
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