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Divergence of product of tensor and vector

  1. Mar 12, 2013 #1
    I am new to tensor algebra. I have an expression involving a 2nd rank tensor (actually a dyadic) and a vector. I want to take divergence of the product

    i.e. ∇. (T.V)

    However, I am not sure if the simple product rule would work here. If I use that
    ∇. (T.V)= (∇.T).V + T. (∇.V)

    However, the first term in the rhs is a vector but the 2nd term is a tensor. So, it seems this product rule is not applicable here. Can someone tell me what is the correct way?

    thanks,

    PB
     
  2. jcsd
  3. Mar 13, 2013 #2
    Sorry, there was a mistake in my post. The 1st term in the rhs is a scalar (because div T is a vector and vector. vector =scalar).
     
  4. Mar 13, 2013 #3

    Fredrik

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    I don't completely understand what you want to do. Is the first period a dot product? Then what is the second period? If Tv is a vector, then how about this?
    $$\nabla\cdot(Tv)=\partial_i(Tv)_i =\partial_i (T_{ij}v_j) =(\partial _i T_{ij})v_j+T_{ij}\partial_i v_j.$$
     
  5. Mar 13, 2013 #4
    Thanks, Fredrik. The first dot i.e. the one after ∇ is for the divergence and the second dot is dot product. Is your formula still valid?

    Praban
     
  6. Mar 13, 2013 #5

    Fredrik

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    How do you define the dot product of a tensor and a vector?
     
  7. Mar 13, 2013 #6

    dextercioby

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    It's not the dot product, it's the contracted product of a dyad and a vector

    [tex] \mbox{contraction of} \left(\bar{\bar{T}}\otimes\vec{v}\right) = \left(\sum_{j}T_{ij} v_{j}\right) \vec{e}_{i} [/tex]
     
  8. Mar 13, 2013 #7
    Yes, Dextercioby - it is the contraction.

    Is the formula given by Fredrik valid in that case?

    thanks
     
  9. Mar 13, 2013 #8

    WannabeNewton

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    If [itex]\triangledown _{a}[/itex] is the derivative operator on your smooth manifold then by definition of the derivative operator, the Leibniz rule is valid: [itex]\triangledown _{c }(T_{ab}v^{b}) = T_{ab}\triangledown _{c}v^{b} + v^{b}\triangledown _{c}(T_{ab})[/itex]. Apply it however you like.
     
  10. Mar 14, 2013 #9

    Fredrik

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    I don't even recall how "divergence" is defined in this context. Praban, you may want to clarify what exactly the expression you posted means. But almost certainly, the answer to your question is going to be yes, you can use the product rule. But the details of the proof is going to depend on what exactly you meant.
     
  11. Mar 15, 2013 #10

    WannabeNewton

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    I think he may be talking about something like [itex]\triangledown ^{a}(T_{ab}v^{b}) = v^{b}\triangledown ^{a}T_{ab} + T_{ab}\triangledown ^{a}v^{b}[/itex] but without further clarification from the OP one can only make assumptions :p. In physics books anyways the term "divergence free" is used for higher order tensors than just vectors to characterize things like the fact that [itex]\triangledown ^{a}T_{ab} = 0[/itex] for the energy momentum tensor and that [itex]\triangledown ^{a}T_{abcd} = 0[/itex] for the Bel - Robinson tensor etc.
     
  12. Mar 17, 2013 #11
    Sorry for the late reply. Let me be more specific

    I have ∇.v (1) i.e. divergence of a vector v.

    then v is expressed as v=T.v0 (T is a tensor and v0 is another vector. The book I am using
    - Happel and Brenner - Hydrodynamics does say that the T and v0 can have dot product and end result is a vector).

    so, eq. (1) becomes ∇.(T.v0) (2)

    My problem is with eq. (2). simple product rule gives
    ∇.(T.v0) = (∇.T).(v0) + T.(∇.v0) (3) (if it is correct)
    however, LHS is a scalar (since divergence of a vector is a scalar)

    1st term of RHS is a scalar (divergence of T is a vector and its dot product with another vector is a scalar). 2nd term of RHS - there is a problem it is going to be a tensor, which is not correct.

    Obviously there is something wrong with my understanding. Can someone help?

    p.s. here tensor is actually a dyadic
     
  13. Mar 17, 2013 #12

    Fredrik

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    It's still not clear to me how this author defines "dot product" and "divergence". Can you post the book's definitions?
     
  14. Mar 17, 2013 #13

    dextercioby

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    No, no, the 2nd term in the RHS is also a scalar. It's a double contracted tensor product between T and Grad v_0, see the post 10 by WBN using the fancy notation of (riemannian) differential geometry.
     
  15. Mar 18, 2013 #14
    Dextercioby,

    Thanks. Now I think I have understood. So, in my notation it would be
    .
    ∇.(T.v0) = (∇.T).(v0) + T.(∇v0)

    where dot in the 2nd term in the rhs is double contraction of tensors and ∇v0 is the gradient of the vector v0 (which is a tensor).


    Fredrik,

    the dot product here is same as contraction as written by Dextercioby in post 6.
    The book I mentioned uses the standard definition of divergence of a dyadic.
     
  16. Mar 18, 2013 #15

    Fredrik

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    Does that mean that the ∇ is the covariant derivative, or is it a partial derivative operator defined using a coordinate system?
     
  17. Mar 18, 2013 #16

    dextercioby

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    Usually hydrodynamics is done in a Galilean geometry, so that the x's have one type of index, either upper, or lower, the nabla is defined as [itex]\vec{e}_{i}\partial_{i} [/itex] and with it the divergence and gradients of tensors of arbitrary degree.
     
  18. Mar 20, 2013 #17
    Yes, Fredrik, Dextercioby is right.

    Praban
     
  19. Mar 20, 2013 #18

    Fredrik

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    Still not sure I get it. Is this the correct interpretation of the notation?
    $$\nabla\cdot(T\cdot v) =\nabla\cdot\big((T^i{}_j e_i\otimes e^j)\cdot(v^k e_k)\big) =\nabla\cdot(T^i{}_j v^j e_i) =\partial_i (T^i{}_j v^j).$$ In that case, the next step is a rather trivial application of the product rule and the linearity of ##\partial_i##.
    $$\partial_i (T^i{}_j v^j) =(\partial_i T^i{}_j)v^j+T^i{}_j \partial_i v^j.$$ We also have
    $$(\partial_i T^i{}_j)v^j =(\partial_i T^i{}_j e^j)\cdot(v^k e_k).$$ The second factor is ##v##, and the first one is what I would guess that ##\nabla\cdot T## means.

    Not sure what to make of the term ##T^i{}_j \partial_i v^j##. I suppose that we could generalize the dot product notation so that this is equal to the dot product of T and something. That something would be ##\partial_i v^j e^i\otimes e_j##. I suppose it makes sense to denote that by ##\nabla v##. With these notations we have
    $$\nabla\cdot(T\cdot v)=(\nabla\cdot T)\cdot v+T\cdot\nabla v.$$
     
  20. Mar 23, 2013 #19
    The first term is correct, and is a scalar. The second terms is not correct. It too, of course, is supposed to be a scalar, and is given by the double dot product: [tex]\vec T:(\nabla \vec V)[/tex]

    See Bird, Stewart, and Lightfoot, "Transport Phenomena"
     
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