# A Help with covariant differentiation

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1. Jun 26, 2016

### redtree

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
2. Jun 26, 2016

### Orodruin

Staff Emeritus
Getting the first step correct. Your expression even has different free indices on both sides.

3. Jun 26, 2016

### wrobel

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 $$\nabla_ke_i=\Gamma_{ik}^se_s,\quad \nabla_ke^i=-\Gamma_{sk}^ie^s$$

4. Jun 26, 2016

### redtree

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
5. Jun 26, 2016

### redtree

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. Jun 26, 2016

### Orodruin

Staff Emeritus
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. Jun 26, 2016

### redtree

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. Jun 26, 2016

### Orodruin

Staff Emeritus
No, I mean a general type (1,1) tensor $T_i^j$.

9. Jun 26, 2016

### redtree

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

The tensor $T^{i}_{j}$ can be contracted $T^{i}_{j} z^{j} = T^{i}$?

10. Jun 26, 2016

### redtree

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

$\frac{\delta T }{\delta z^{j}} = \frac{\delta (T^{k} z_{k})}{\delta z^{j}}$ and then solve for $\frac{\delta T^{k}}{\delta z^{j}}$?

11. Jun 26, 2016

### redtree

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. Jun 26, 2016

### Orodruin

Staff Emeritus
No, put $T^i_j = \nabla_j T^i$.

13. Jun 26, 2016

### redtree

Ahhh. Thanks!!