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Derivative of a tensor

  1. Jun 1, 2005 #1
    the derivative of a tensor

    [tex]
    a_{ij}x^ix^j
    [/tex]

    with respect to [tex]x^k[/tex], k=2 and i,j = 1,2,3.

    solution:

    [tex]
    \frac {d} {dx^k}a_{ij}x^ix^j =
    a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} =
    a_{2j}x^j + a_{i2}x^i =
    a_{21}x^1 + a_{22}x^2 + a_{23}x^3 + a_{12}x^1 + a_{22}x^2 + a_{32}x^3

    [/tex]

    is that correct?
     
  2. jcsd
  3. Jun 1, 2005 #2

    dextercioby

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    That's not a tensor,but a scalar.

    Are u familiar with the delta-Kronecker invariant tensor ??

    Daniel.
     
  4. Jun 1, 2005 #3
    My misstake...the problem was to calculate the derivative of that _expression_...not tensor.
    Did i do it correct?
    yes...i am familiar with the delta-Kronecker invariant tensor.
     
  5. Jun 1, 2005 #4

    dextercioby

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    [tex] \frac{\partial x^{i}}{\partial x^{k}} =...? [/tex]

    If you know that,u could just apply the Leibniz rule and then the summation convention.

    Daniel.
     
  6. Jun 1, 2005 #5

    HallsofIvy

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    Are you given that the components of the tensor aij are constant? You seem to be assuming that.
     
  7. Jun 2, 2005 #6
    yes the components of [tex]a_{ij}[/tex] is constant.

    i used that

    [tex] \frac{\partial x^{i}}{\partial x^{k}} = \delta^i{}_{k}[/tex]

    and Leibnitz rule and the summation convention.

    i thought i had the right answer?
     
  8. Jun 2, 2005 #7

    robphy

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    Note that
    [tex]\frac {d} {dx^k}a_{ij}x^ix^j [/tex] has two "dummy-indices [involved in summation]", namely, i and j, and one "free-index [not involved in summation]", k. So, the right-hand side must also have the free-index k.
     
  9. Jun 2, 2005 #8

    dextercioby

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    That's actually [itex] \delta_{i}^{k} [/itex],because it is a symmetric invariant tensor.

    Daniel.
     
  10. Jun 2, 2005 #9

    robphy

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    I think the indices are flipped. They should be
    [tex]\displaystyle\frac{\partial x^{i}}{\partial x^{k}} = \delta_{k}^{i} [/tex]
    ... but it's a good idea to preserve the "slots" and write [itex] \delta_{k}{}^{i} [/itex] or [itex] \delta^{i}{}_{k} [/itex].
     
  11. Jun 2, 2005 #10
    k=2...but maybe that doesnt matter.
     
  12. Jun 2, 2005 #11

    robphy

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    Ok, I see now that it's a issue of clarity of presentation.
    You are correct, but I would have written your initial post as

    [tex]
    \begin{align*}
    \frac {d} {dx^k}a_{ij}x^ix^j &=
    a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} \\
    &=a_{ij} \delta^i{}_k x^j + a_{ij}x^i \delta^j{}_k\\
    &=a_{kj} x^j + a_{ik}x^i \mbox{ which is a co-vector.}
    \\
    \intertext{Now, with k=2,}
    \frac {d} {dx^2}a_{ij}x^ix^j
    &=
    a_{2j}x^j + a_{i2}x^i
    \\&=
    a_{21}x^1 + a_{22}x^2 + a_{23}x^3 + a_{12}x^1 + a_{22}x^2 + a_{32}x^3 \mbox{ which is the 2-component of that co-vector}
    \end{align*}
    [/tex]
     
  13. Jun 2, 2005 #12

    dextercioby

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    Incidentally that matrix "a" is symmetrical (or if it isn't,the contracted tensor product selects the symmetrical part),so in the final result there should be a 2...

    Daniel.
     
  14. Jun 2, 2005 #13

    robphy

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    I think what you mean is this:

    [tex]\begin{align*}\frac {d} {dx^k}a_{ij}x^ix^j &=a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} \\
    &=a_{ij} \delta^i{}_k x^j + a_{ij}x^i \delta^j{}_k\\
    &=a_{kj} x^j + a_{ik}x^i\\
    &=a_{k\color{red}{i}} x^{\color{red}{i}} + a_{ik}x^i\\
    &=\left( a_{ki} + a_{ik} \right) x^i\\
    &=\left( 2 a_{ik}^S \right) x^i = 2 a_{ik}^Sx^i \\
    \end{align*}[/tex]
    So, with k=2,
    [tex]\begin{align*}
    \frac {d} {dx^2}a_{ij}x^ix^j
    &= 2 a_{i2}^Sx^i = 2\left( a_{12}^Sx^1 + a_{22}^Sx^2 + a_{32}^Sx^3\right)
    \end{align*}[/tex]
    which is equal to what is in the original post.
     
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