Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gradients of Vectors and Dyadic Products

  1. Oct 18, 2006 #1
    I'm encountering the gradient of a vector field in a problem at the moment. Not the divergence, specifically the vector.

    My problem at the moment is the represenation of this using the "nabla" notation. Some authors seem to be defining this as [tex]\nabla \otimes \vec{v}[/tex], the tensor or dyadic product. But this doesn't seem to give the correct answer.

    Could someone please confirm for me that the dyadic product [tex]\vec{a} \otimes \vec{b} = \vec{a} \vec{b}^T[/tex] if a and b are column vectors? What way is the gradient of a vector normally represented?
     
  2. jcsd
  3. Oct 20, 2006 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The gradient is an operator which, when applied on a tensor field of rank "n" (let's say it's covariant), increases the rank by one unit. That mean that, when acting on a scalar, it produces a vector field. And when acting on a vector field it produces a 2-nd rank tensor with mixed components. The "dyadic" product is an ancient name for the tensor product of 2 vectors (vector fields).

    Daniel.
     
  4. Oct 20, 2006 #3
    So just be be specific, what would the gradient in 3d of a vector [tex]\vec{v}=(a,b,c)[/tex] be?

    Is it
    [tex]\left( \begin{array}{ccc}
    \partial_x a & \partial_y a & \partial_z a \\
    \partial_x b & \partial_y b & \partial_z b \\
    \partial_x c & \partial_y c & \partial_z c
    \end{array} \right) [/tex]

    Or is it instead
    [tex]\left( \begin{array}{ccc}
    \partial_x a & \partial_x b & \partial_x c \\
    \partial_y a & \partial_y b & \partial_y c \\
    \partial_z a & \partial_z b & \partial_z c
    \end{array} \right) [/tex]

    The first gives the right answer when right multiplied by a column vector and the second when left multiplied by a row vector. Personally, I favour the first representation, but the standard tensor product gives [tex]\nabla \otimes \vec{v}[/tex] as the second of these two matrices. And that gives the wrong answer when right multiplied by a column vector.

    Is there anything canonical about the final result of a tensor product. I mean lets say you had.
    [tex]
    \vec{u}=(u_1,u_2,u_3)
    \vec{v}=(v_1,v_2,v_3)
    [/tex]
    The standard tensor product seems to give:
    [tex]\vec{u} \otimes \vec{v} =
    \left( \begin{array}{ccc}
    u_1 v_1 & u_1 v_2 & u_1 v_3 \\
    u_2 v_1 & u_2 v_2 & u_2 v_3 \\
    u_3 v_1 & u_3 v_2 & u_3 v_3
    \end{array} \right)
    [/tex]

    But is there anything fundamentally wrong with defining the product to instead be:
    [tex]\vec{u} \otimes \vec{v} =
    \left( \begin{array}{ccc}
    v_1 u_1 & v_1 u_2 & v_1 u_3 \\
    v_2 u_1 & v_2 u_2 & v_2 u_3 \\
    v_3 u_1 & v_3 u_2 & v_3 u_3
    \end{array} \right)
    [/tex]?
     
    Last edited: Oct 20, 2006
  5. Oct 23, 2006 #4
    Bumping to request that this be moved to the tensor analysis subforum. Thanks in advance.
     
  6. Nov 1, 2006 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The 3*3 matrix

    [tex] (\nabla\otimes \vec{v})_{ij} =\partial_{i}v_{j} [/tex]

    is the matrix of the gradient of the vector field [itex] \hat{v} [/itex] in the tensor space basis.

    Daniel.
     
  7. Nov 11, 2006 #6
    The gradient of a vector is not the same that the dyadic product between the "nabla vector" and the vector itself.
    In fact, the gradient of a vector is the dyadic product transpossed !
     
  8. Nov 11, 2006 #7

    gvk

    User Avatar

    Gradient of a vector component as discribed here does not create a tensor. Strictly speaking gradient is only applied to the scalar. If you applied it to whole vector, you have to differentiate the components of the vector and the unitary vectors too. This gives the Christoffel's symbols. Historicaly, this problem of covariant derivative was a starting point of tensor analysis.
     
  9. Nov 13, 2006 #8
    .... Go on.
     
  10. Nov 14, 2006 #9

    gvk

    User Avatar

    Well, the finding the differential of the vector is a way of finding connection between the vector in one point of manifold and the vector in infinitesimally close another point of manifold. This difference comprises of two parts: one is the change the component of vector itself, and another is the change of reference coordinate system.
    What was amazing that the first part is antisymmetric and second part is symmetric in regard to the path from one point to the another. This means that if you reach the neighbor points of manifold by two different paths and substract from first differential the second, you receive the "pure" change of the component of vector. If you add two different paths, you receive change related to curvature of coordinate system, which characterize manifold itself. This is the basical idea. The math you can find in many diff. geometry books. I would recomend the old, but deep and profound source: Levi-Civita Absolute Differential Calculus 1927, Chapter IV.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gradients of Vectors and Dyadic Products
Loading...