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A Tensor Calculus and Divergence

  1. May 25, 2016 #1

    joshmccraney

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    Hi PF!

    I have a question on the dyadic product and the divergence of a tensor. I've never formally leaned this, although I'm sure it's published somewhere, but this is how I understand the operators. Can someone tell me if this is right or wrong? Lets say I have some vector ##\vec{V} = v_x i + v_y j##. Then ##\vec{V} \otimes \vec{V} = (v_x i + v_y j)^2 = v_x v_x ii + v_x v_y ij+v_yv_xji+v_yv_yjj## (which is a 2 by 2 matrix). Now if I take $$\nabla \cdot (\vec{V} \otimes \vec{V}) = \partial_x (v_x v_x) (i\cdot i)i +\partial_x( v_x v_y)(i\cdot i)j+\partial_x (v_yv_x)(i\cdot j)i+\partial_x(v_yv_y)(i\cdot j)j+\\ \partial_y (v_x v_x) (j\cdot i)i +\partial_y( v_x v_y)(j\cdot i)j+\partial_y (v_yv_x)(j\cdot j)i+\partial_y(v_yv_y)(j\cdot j)j=\\
    \partial_x (v_x v_x)i +\partial_x( v_x v_y)j+\partial_y (v_yv_x)i+\partial_y(v_yv_y)j$$ where I could then use the product rule, simplify, factor out ##\nabla \cdot \vec{V}## and set equal to zero by continuity (if ##\vec{V}## was velocity of an incompressible fluid). Do these operations look correct?

    Again, I've never been showed this but it looks intuitive and feels correct.
     
    Last edited: May 25, 2016
  2. jcsd
  3. May 25, 2016 #2
    This all looks correct up to here.
    This part, I'm not so sure about. Why don't you expand it out using the product rule and see what you get?
     
  4. May 25, 2016 #3

    joshmccraney

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    I did it on paper and it works! The result is the convective term for the Navier-Stokes Eq! Did you want me to post it?

    Since you're here on this thread, I do have a question on Navier-Stokes. If I need to post a different thread I'm happy to, but my question is, why use the dyadic product when considering the momentum flux? Why ##\vec{V}\otimes \vec{V}## rather than, say, the inner/outer product ##\vec{V}:\vec{V}## or some other product that gives a second rank tensor from two vectors?
     
  5. May 26, 2016 #4

    haushofer

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    Just my 2 cents:

    If you take the inner product $V\cdot V$, you lose information about the direction of the momentum flux. If you want to construct a 2-tensor, there is not much other choice then the direct product, since the direct product of a vector with itself is symmetric; the only other choice I can think of is the traceless part of it,

    [tex]
    V_i V_j - \frac{1}{2} (V \cdot V) \delta_{ij} \,.
    [/tex]

    Note that this transforms irreducibly under the group of spatial rotations SO(2). Btw, your first expression is correct, because in components it reads (using your incompressibility)

    [tex]
    \partial_i (V^i V^j) = (\partial_i V^i) V^j + V^i \partial_i V^j = V^i \partial_i V^j
    [/tex]
     
  6. May 26, 2016 #5
    There is only one way of doing it that gives the momentum fluxes correctly. ##\vec{V}:\vec{V}## is not a 2nd order tensor. In fact, it is not anything.

    In elementary treatments, the dyadic form of the momentum flux is typically derived by using a control volume, and evaluating the momentum flux terms in the differential force balance first, and then recognizing that it can also be expressed in terms of the divergence of the dyadic product. Alternately, in relativistic developments, it is obtained by taking the divergence of the stress-energy tensor.
     
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