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Achieving normal derivative in spheroidal coordinate

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi PhysicsForums,
    I am calculating something related to the spheroidal membrane and want to ask you a question.

    I consider a oblate spheroid (Oblate spheroidal coordinates can also be considered as a limiting case of ellipsoidal coordinates in which the two largest semi-axes are equal in length.)

    In spheroidal coordinate, the relationship to Cartesian coordinates is
    [tex]x=a\sqrt((1+u^2) (1-v^2))\cos(\phi)[/tex]
    [tex]y=a\sqrt((1+u^2) (1-v^2))\sin(\phi)[/tex]
    [tex]z=a u v[/tex]

    Now, I want to know how to achieve the normal derivative to the surface of a spheroid (u = const), in terms of the derivatives of u, v and [tex]\phi[/tex].

    2. Relevant equations



    3. The attempt at a solution
    I firstly think that the normal derivative in this case is the partial derivative to u. Because in limit cases (v =1,-1) this lead to the derivative to z.

    Thank you very much.
     
  2. jcsd
  3. Jan 14, 2009 #2
    writing in Cartesian coordinate, the normal derivative is just
    [tex]\hat{n}\cdot \nabla[/tex]


    However, I'm pretty sure you want to work in Oblate spheroidal coordinate, just apply the chain rule and you should be fine.


    More intuitively, you could consider the normal derivative as a unit tangent vector (in terms of manifold) acted on a function.
    Simply write down the tangent vector and then normalize it using the metric.
     
  4. Jan 14, 2009 #3
    Thank you tim_lou,
    In fact, I am totally an amateur in this type of calculation.

    So, can you please recommend me any basic text books or simple introductories.

    For example, if I have the equation of a membrane in xyz coordinate, what can I do to get the normal vector [tex]\vec(n)[/tex].
    Moreover, the chain rule seems to be rather stenuous.
    Is there any different ways. I need some types in practical ways.
    Is there any function in Mathematica to do that?

    Thank you again.
     
    Last edited: Jan 14, 2009
  5. Jan 15, 2009 #4
    to find the normal vector in xyz coordinates, take the gradient...

    to be more precise, suppose you want to find the normal vector to a surface described by the set:
    A={(x,y,z)| f(x,y,z)=c}

    then the normal is in the direction of n=∇f(x,y,z)
    to see why that is, take a curve, (x(t), y(t), z(t)) such that the curve lies in A.
    Hence, f(x(t), y(t), z(t))=c and take derivative with respect to t,
    ∇f(x,y,z)· (x'(t), y'(t), z'(t))=0. We see that n must be always orthogonal to the derivative or any curve lying in A so we conclude n=∇f(x,y,z)

    Any tangent vector (b,c,d) attached to a point in R3 can be considered as a tangent vector on the manifold, so,
    (b,c,d)=b∂_x + c∂_y + d∂_z
    so when you find n·∇=b∂_x + c∂_y + d∂_z, you want to express it in the basis
    {∂_a, ∂_u, ∂_v}. Changing basis is a linear algebra problem and those two basis are related via Jacobian.

    things simplify when the {∂_a, ∂_u, ∂_v} are orthogonal, in which case, the gradient becomes
    f ∂_a + g ∂_u + h ∂_v where f,g,h are functions.
     
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