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Math Help - Gaussian Curvature of a Sphere

  1. #1
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    Gaussian Curvature of a Sphere

    Hi all,

    I am trying to find the Gaussian Curvature of the sphere with equation:

    X(theta, phi) = (Rcos(theta)sin(phi) , Rsin(theta)sin(phi) , Rcos(theta)

    I have found the coefficients of the first fundamental form:

    E = R^2*sin(phi)^2

    F = 0

    G = R^2

    But where should I go from here?

    My (very unhelpful!) notes just jump to k = 1/R^2

    Thanks for the help!
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  2. #2
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    Re: Gaussian Curvature of a Sphere

    What is the definition of "Gaussian Curvature"?
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  3. #3
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    Re: Gaussian Curvature of a Sphere

    k = eg - f^2 / (EG - F^2) ?
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  4. #4
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    Re: Gaussian Curvature of a Sphere

    the easiest way to get from where you are to where you want to be is to compute the coefficients of the second fundamental form (see here: Second fundamental form - Wikipedia, the free encyclopedia)

    to do this, you are first going to have to compute the (unit) normal vector field, given by:

    \mathbf{n} = \frac{\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}}{\left|\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right|}

    and then compute the coefficients of the second fundamental form as follows:

    L = \mathbf{n} \cdot \frac{\partial^2X}{\partial \theta^2}

    M = \mathbf{n} \cdot \frac{\partial^2X}{\partial \theta \partial \phi}

    N = \mathbf{n} \cdot \frac{\partial^2X}{\partial \phi^2}

    you can then calculate the Gaussian curvature as:

    K = \frac{LN - M^2}{EG - F^2}.
    Last edited by Deveno; December 27th 2012 at 06:50 AM.
    Thanks from xxp9 and PStudent175
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  5. #5
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    Re: Gaussian Curvature of a Sphere

    Thanks for the help so far Deveno.

    I am trying to calculate the unit normal and am unsure where to go from here:

    dX/dtheta x dX/dphi = i j k
    -Rsin(theta)sin(phi) Rcos(theta)sin(phi) 0
    Rcos(theta)cos(phi) Rsin(theta)cos(phi) -Rsin(phi)

    Of which the determinant is:

    =i(-R^2cos(theta)sin(phi)^2) - j(R^2sin(theta)sin(phi)^2) + k(-R^2sin(theta)^2sin(phi)cos(phi) - R^2cos(theta)^2sin(phi)cos(phi))

    I am not sure how to simplify this to get a final answer for N.

    Thanks again for your help.
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  6. #6
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    Re: Gaussian Curvature of a Sphere

    for the cross-product i always use the formula:

    (u_1,u_2,u_3) \times (v_1,v_2,v_3) = (u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1).

    the way i remember this is:

    1st coordinate, leave out "1" subscripts, 2-->3 is positive (the positive cycle goes like this: 1-->2-->3-->1-->2-->3 etc.) (so 3-->2 is negative) so the u2v3 term is positive.

    2nd coordinate, leave out "2", the positive term is u3v1 (see above).

    3rd coordinate, well, there's only one possible way to "balance it out" now.

    i find this easier to remember than the "determinant formula":

    \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1  &v_2&v_3 \end{vmatrix}.

    (by the way, you wrote down the wrong "third coordinate" in your parametrization, it should be: R\cos(\phi) not theta. you did get the correct E,F and G, though)

    a lengthy calculation shows that:

    \frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi} = \begin{bmatrix}(R\cos\theta\sin\phi)(-R\sin\phi) - (0)(R\sin\theta\cos\phi)\\(0)(R\cos\theta\cos\phi) - (-R\sin\theta\sin\phi)(-R\sin\phi)\\ (-R\sin\theta\sin\phi)(R\sin\theta\cos\phi) - (R\cos\theta\sin\phi)(R\cos\theta\cos\phi)) \end{bmatrix}

    = \begin{bmatrix}-R^2\cos\theta\sin^2\phi\\ -R^2\sin\theta\sin^2\phi\\ -R^2\sin\phi\cos\phi \end{bmatrix}

    which is what you got (you can simplfy your "k" coordinate by taking out the common factor of -R^2\sin\phi\cos\phi).

    this is a normal vector, but it's not a UNIT vector, so we need to "normalize" it (how's that for an odd pun?) by dividing by its norm, which is:

    \left\| \frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi} \right\| = \sqrt{\left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right) \cdot \left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right)}

    this is almost as much work as finding the cross-product, first we find that:

    \left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}\right) \cdot \left(\frac{\partial X}{\partial \theta} \times \frac{\partial X}{\partial \phi}}\right)

     = R^4\cos^2\theta\sin^4\phi + R^4\sin^2\theta\sin^4\phi + R^4\sin^2\phi\cos^2\phi

     = R^4(\sin^4\phi + \sin^2\cos^2\phi) = R^4\sin^2\phi

    taking the square root, we get that the magnitude of our normal vector is R^2\sin\phi so that:

    \mathbf{n} = \begin{bmatrix}-\cos\theta\sin\phi \\ -\sin\theta\sin\phi \\ -\cos\phi \end{bmatrix}

    (a word about the signs, here: our parameterization takes horizontal rays vectors going left-to-right to "lattitude" arcs going counter-clockwise (west-to-east), and vertical rays going down-to-up to "longitude" arcs going from the north pole to the south pole, so the "outward" normal (using a right-hand-rule for the cross-product) in this case points "towards" the center of the sphere (inwards). we could have chosen a different parameterization to make the normal vector n simply be:

    \frac{1}{R}X(\theta,\phi)

    which would have been more intuitively obvious, geometrically (using -θ instead of θ, for example)).

    i leave it to you to compute the second derivatives, and thus L,M and N.
    Last edited by Deveno; December 28th 2012 at 10:08 AM.
    Thanks from xxp9 and PStudent175
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