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Gradient of a tensor? Del operator on tensor?

  1. Apr 26, 2009 #1
    hi all,
    do you know what is the gradient of a tensor looks like?
    I mean the del operator on a second order tensor, not the divergence of the tensor.
    And actually I need them in polar coordinates..
    I have been searching so hard in web, but I cant find anything useful.
    Please help.
     
  2. jcsd
  3. Apr 27, 2009 #2
    I think what you're looking for is the covariant derivative, which is often written using the del notation. The covariant derivative is an operator acting on tensors in the tangent spaces of a differentiable manifold, defined by means of a construction called a connection. Given an arbitrary connection, there's no relation between the covariant derivative and the familiar del operator on Euclidean space. However, if the connection is the Levi-Civita connection (see <http://en.wikipedia.org/wiki/Levi-Civita_connection>), [Broken] then the covariant derivative can be viewed as a generalization of the flat-space del operator.

    If you're working in flat space, finding a usable expression for the covariant derivative of a (0,2) tensor is simple, even in spherical coordinates. If Tab is a (0,2) tensor, then D(T), its covariant derivative, has three indices, and we denote D(T) by Tab;c. Regular partial differentiation is denoted by Tab,c (i.e., the partial derivative of the ab-component of T wrt the variable numbered c). Then
    Tab;c = Tab,c - CkacTkb - CkbcTak ,
    where Cabc is a Christoffel symbol and the Einstein summation convention is being used. (This looks much more complicated than it really is.) In spherical coordinates (I'm using p = phi, t = theta, and r = r), the nonvanishing Christoffel symbols are
    Crpp = -r
    Crtt = -r sin2(t)
    Cptt = -sin(p) cos(p)
    Ctrt = 1/r
    Ctpt = cot(p)
    If T happens to be totally antisymmetric (i.e., a differential form), then there are other nice notions of "differentiation" for T. The most important of these is the exterior derivative, denoted dT. (A nice formula for the exterior derivative is dT = Alt(D(T)), where D(T) is the covariant derivative.) The exterior derivative can be combined with the Hodge dual operator to yield more del-like operators; in particular, the natural generalization of the Laplacian on flat space is the Laplace-Beltrami operator is given by (bd + db), where d is the exterior derivative and b is the codifferential, defined by b = *d* (* is the Hodge operator). The Laplace-Beltrami operator occurs in the curved-space version of the wave equation for electromagnetic fields. You can learn more about these operators here: <http://en.wikipedia.org/wiki/Codifferential#The_codifferential> (or in any good book on general relativity or differential geometry).
     
    Last edited by a moderator: May 4, 2017
  4. May 4, 2009 #3
    Hello, can I just use the normal product rules to differentiate the tensor? I mean treat the tensor as a "dyandic product" of the two basis vectors and differentiate those vectors?

    And I am working in a 2D polar coordinates...
    I am wondering if you have the formula?

    I have derived it but I am not sure if it is correct...
     
    Last edited by a moderator: May 4, 2017
  5. May 11, 2009 #4
    [itex]
    \nabla \equiv \mathbf{e^j} \frac{\partial }{\partial u^j}[/itex]
    Hence

    [itex]
    \nabla \mathbf{T} = \mathbf{e^j} \frac{\partial \mathbf{T}}{\partial u^j}[/itex]

    If T is a second order tensor, then

    [itex]\nabla \mathbf{T} = \mathbf{e^j} \frac{\partial \mathbf{T}}{\partial u^j}=\mathbf{e^j} \frac{\partial ( T^m ,_n \mathbf{e_me^n} ) }{\partial u^j}[/itex]

    In Polar coordinates, being an orthogonal reference system, the covariant and contravariant bases coincide, and you don't have to worry about sub- or superindices, so you can just write:

    [itex]\nabla \mathbf{T} = \mathbf{e_j} \frac{\partial \mathbf{T}}{\partial u_j}=\mathbf{e_j} \frac{\partial ( T_m _n \mathbf{e_me_n} ) }{\partial u_j}[/itex]

    By all means use the product rule to differentiate the term in brackets, just remember that for Polar coordinates, the basis vectors are themselves functions of the coordinates and so must be differentiated.

    Also note that the result is a third order tensor.

    Lol can you tell me what this is for, I'm curious . . . I've only ever had to consider gradients of scalars or vectors before:P

    Hope this helps.

    O: I don't even know what a manifold is :S I wish I could follow VKint's post, who seems to know so much more . . .
     
    Last edited: May 11, 2009
  6. May 11, 2009 #5
    NB I forgot to say, some authors define

    [itex]
    \nabla
    [/itex]

    as
    [itex]

    \nabla \equiv \frac{\partial ( . ) }{\partial u^j}\mathbf{e^j}
    [/itex]

    in which case you'd have

    [itex]
    \nabla \mathbf{T} = \frac{\partial \mathbf{T}}{\partial u^j} \mathbf{e^j}
    [/itex]

    which is not the same unless the tensor is symmetric.

    Note that

    [itex]
    \nabla
    [/itex]

    is a symbolic notation, as is vector notation in general (vector product, etc), a sort of shorthand notation. Writing things out in indicial notation leaves no room for ambiguity :)
     
  7. Sep 22, 2009 #6
    Knowing the gradient of a tensor in cylindrical or spherical coordinates is extremely useful. Are you still in need of the answer? I have had to work it in both systems, both using physical components and using covariant/contravariant index notation. I can search my notes to find it if anybody wants it.
     
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