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A Help with covariant differentiation

  1. Jun 26, 2016 #1
    I'm having trouble evaluating the following expression (LATEX):

    ##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{i m} \frac{\delta T^{m}}{\delta z^{i}} + \Gamma^{k}_{i m} \Gamma^{m}_{i l} T^{l}##

    What are the next steps to complete the covariant differentiation? (This is not a homework assignment)
     
    Last edited by a moderator: Jun 26, 2016
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  3. Jun 26, 2016 #2

    Orodruin

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    Getting the first step correct. Your expression even has different free indices on both sides.
     
  4. Jun 26, 2016 #3
    If you will
    Just use basis and these formulas will be easy to write. For example let ##A=a_i^je^i\otimes e_j## then $$(\nabla_k A)e^k=(\frac{\partial a_i^j}{\partial x^k}e^i\otimes e_j+a_i^j(\nabla_ke^i)\otimes e_j+a_i^je^i\otimes(\nabla_k e_j))\otimes e^k$$ and remember that $$\nabla_ke_i=\Gamma_{ik}^se_s,\quad \nabla_ke^i=-\Gamma_{sk}^ie^s$$
     
  5. Jun 26, 2016 #4
    Sorry. Better?

    ##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##
     
    Last edited: Jun 26, 2016
  6. Jun 26, 2016 #5
    The first covariant derivative I get. I'm having trouble with the second covariant differentiation ##\nabla_{i}## of ##\nabla_{j} T^{k}##, where ##\nabla_{j} T^{k} = \frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}##.
     
  7. Jun 26, 2016 #6

    Orodruin

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    How does the covariant derivative act on a general type (1,1) tensor? How can you apply that to the type (1,1) tensor ##\nabla_j T^k##?
     
  8. Jun 26, 2016 #7
    You mean the invariant ##T##? Where ##T=T^{k} z_{k}##?


    ##\nabla_{j} T = \nabla_{j} T^{k} z_{k} = (\frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}) z_{k}##?
     
  9. Jun 26, 2016 #8

    Orodruin

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    No, I mean a general type (1,1) tensor ##T_i^j##.
     
  10. Jun 26, 2016 #9
    ##\nabla_{k} T^{i}_{j} = \frac{ \delta T^{i}_{j}}{\delta z^{k}} + \Gamma^{i}_{k m} T^{m}_{j} - \Gamma^{m}_{j k} T^{i}_{m} ##

    The tensor ##T^{i}_{j}## can be contracted ## T^{i}_{j} z^{j} = T^{i} ##?
     
  11. Jun 26, 2016 #10
    ##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##

    Is this your suggestion?

    ##\frac{\delta T }{\delta z^{j}} = \frac{\delta (T^{k} z_{k})}{\delta z^{j}} ## and then solve for ##\frac{\delta T^{k}}{\delta z^{j}}##?
     
  12. Jun 26, 2016 #11
    One thought I had was to consider ##T^{k}## a function of ##z_{k}##, i.e., ##T^{k} = T^{k}(z_{k})##, such that ##\frac{\delta T^{k}}{\delta z^{j}} = \frac{\delta T^{k}}{\delta z^{k}} \frac{\delta z_{k}}{\delta z^{j}}## and ##\frac{\delta z_{k}}{\delta z^{j}} = \Gamma^{m}_{k j} z_{m}##
     
  13. Jun 26, 2016 #12

    Orodruin

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    No, put ##T^i_j = \nabla_j T^i##.
     
  14. Jun 26, 2016 #13
    Ahhh. Thanks!!
     
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