Help with covariant differentiation

  • #1
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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)
 
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  • #2
Orodruin
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What are the next steps to complete the covariant differentiation?
Getting the first step correct. Your expression even has different free indices on both sides.
 
  • #3
wrobel
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If you will
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)

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 $$\nabla_ke_i=\Gamma_{ik}^se_s,\quad \nabla_ke^i=-\Gamma_{sk}^ie^s$$
 
  • #4
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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}##
 
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  • #5
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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}##.
 
  • #6
Orodruin
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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}##.

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##?
 
  • #7
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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##?

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}##?
 
  • #8
Orodruin
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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}##?
No, I mean a general type (1,1) tensor ##T_i^j##.
 
  • #9
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##\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} ##?
 
  • #10
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##\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}}##?
 
  • #11
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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}##
 
  • #12
Orodruin
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##\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} ##?
No, put ##T^i_j = \nabla_j T^i##.
 
  • #13
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Ahhh. Thanks!!
 

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