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So I'm having problems finding the second order covariant derivitive in index notation. My teacher said to just find the covariant derivative of a covariant derivative, so I first started with finding just for the lower indexes. $$ v_{i,j}=\frac{{\partial {v_i}}}{{\partial {x^j}}} - {v_k}{\Gamma }_{ij}^k $$ \begin{align} v_{i,jk}&=\frac{{\partial {v_{ij}}}}{{\partial {x^k}}} - {v_{im}}{\Gamma }_{jk}^m - {v_{mj}}{\Gamma }_{ik}^m\\ &=\frac{\partial }{{\partial {x^k}}}\left[ {\frac{{\partial {v_i}}}{{\partial {x^j}}} - {v_l}{\Gamma }_{ij}^l} \right] - {v_{i,m}}{\Gamma }_{jk}^m - {v_{m,j}}{\Gamma }_{ik}^m\\ & = \frac{\partial }{{\partial {x^k}}}\left[ {\frac{{\partial {v_i}}}{{\partial {x^j}}} - {v_l}{\Gamma }_{ij}^l} \right] - {v_{i,m}}{\Gamma }_{jk}^m - {v_{m,j}}{\Gamma }_{ik}^m\\ & = \frac{{{\partial ^2}{v_i}}}{{\partial {x^k}\partial {x^j}}} - \frac{\partial }{{\partial {x^k}}}\left[ {{v_l}{\Gamma }_{ij}^l} \right] - \left[ {\frac{{\partial {v_i}}}{{\partial {x^m}}} - {v_l}{\Gamma }_{im}^l} \right]{\Gamma }_{jk}^m - \left[ {\frac{{\partial {v_m}}}{{\partial {x^j}}} - {v_l}{\Gamma }_{mj}^l} \right]{\Gamma }_{ik}^m\\ & = \frac{{{\partial ^2}{v_i}}}{{\partial {x^k}\partial {x^j}}} - \frac{{\partial {v_l}}}{{\partial {x^k}}}{\Gamma }_{ij}^l - {v_l}\frac{\partial }{{\partial {x^k}}}\left[ {{\Gamma }_{ij}^l} \right] - \frac{{\partial {v_i}}}{{\partial {x^m}}}{\Gamma }_{jk}^m + {v_l}{\Gamma }_{im}^l{\Gamma }_{jk}^m - \frac{{\partial {v_m}}}{{\partial {x^j}}}{\Gamma }_{ik}^m - {v_l}{\Gamma }_{mj}^l{\Gamma }_{ik}^m\\ & = \frac{{{\partial ^2}{v_i}}}{{\partial {x^k}\partial {x^j}}} - \frac{{\partial {v_l}}}{{\partial {x^k}}}{\Gamma }_{ij}^l - \frac{{\partial {v_i}}}{{\partial {x^m}}}{\Gamma }_{jk}^m - \frac{{\partial {v_m}}}{{\partial {x^j}}}{\Gamma }_{ik}^m - {v_l}\left[ {\frac{\partial }{{\partial {x^k}}}\left[ {{\Gamma }_{ij}^l} \right] - {\Gamma }_{mj}^l{\Gamma }_{ik}^m - {\Gamma }_{im}^l{\Gamma }_{jk}^m} \right] \end{align}

So I can use the same logic to find for the mixed indexes which is what I need. This is what I got

\begin{align} {v^i}_{,j}&= \frac{{\partial {v^i}}}{{\partial {x^j}}} + {v^k}{\Gamma }_{kj}^i\\ {v^i}_{j,k} &= \frac{{\partial {v^i}_j}}{{\partial {x^k}}} + {v^m}_j{\Gamma }_{mk}^i - {v^i}_m{\Gamma }_{jk}^m\\ {v^i}_{jk} &= \frac{\partial }{{\partial {x^k}}}\left[ {\frac{{\partial {v^i}}}{{\partial {x^j}}}\,\, + {v^l}{\Gamma }_{lj}^i} \right] + {v^m}_{,j}{\Gamma }_{mk}^i - {v^i}_{,m}{\Gamma }_{jk}^m \\ {v^i}_{,jk} &= \frac{{{\partial ^2}{v^i}}}{{\partial {x^k}\partial {x^j}}} + \frac{\partial }{{\partial {x^k}}}\left[ {{v^l}{\Gamma }_{lj}^i} \right] + \left[ {\frac{{\partial {v^m}}}{{\partial {x^j}}} + {v^l}{\Gamma }_{lj}^m} \right]{\Gamma }_{mk}^i - \left[ {\frac{{\partial {v^i}}}{{\partial {x^m}}}\,\, + {v^l}{\Gamma }_{lm}^i} \right]{\Gamma }_{jk}^m \\ &= \frac{{{\partial ^2}{v^i}}}{{\partial {x^k}\partial {x^j}}} + \frac{{\partial {v^l}}}{{\partial {x^k}}}{\Gamma }_{lj}^i + \frac{\partial }{{\partial {x^k}}}\left[ {{\Gamma }_{lj}^i} \right]{v^l} + \frac{{\partial {v^m}}}{{\partial {x^j}}}{\Gamma }_{mk}^i + {v^l}{\Gamma }_{lj}^m{\Gamma }_{mk}^i - \frac{{\partial {v^i}}}{{\partial {x^m}}}{\Gamma }_{jk}^m - {v^l}{\Gamma }_{lm}^i{\Gamma }_{jk}^m \\ &= \frac{{{\partial ^2}{v^i}}}{{\partial {x^k}\partial {x^j}}} + \frac{{\partial {v^l}}}{{\partial {x^k}}}{\Gamma }_{lj}^i + \frac{{\partial {v^m}}}{{\partial {x^j}}}{\Gamma }_{mk}^i - \frac{{\partial {v^i}}}{{\partial {x^m}}}{\Gamma }_{jk}^m + {v^l}\left[ {\frac{\partial }{{\partial {x^k}}}\left[ {{\Gamma }_{lj}^i} \right] + {\Gamma }_{lj}^m{\Gamma }_{mk}^i - {\Gamma }_{lm}^i{\Gamma }_{jk}^m} \right] \end{align}

I need to find this to use in this formula $$ \Delta {v^j} = v_{\,,11}^j{g^{11}} + v_{\,,22}^j{g^{22}} + v_{\,,33}^j{g^{33}} $$

in my Navier-Stokes equation problem

$$ {\rho _0}\left( {\frac{{\partial {v^j}}}{{\partial t}} + {v^j}_{,k}{v^k}} \right) = \, - {p_{,i}}{g^{ij}} + \mu \Delta {v^j} + {\rho _0}{f^j} $$

Am I doing this right? Also is there an easier way to do this?

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closed as off-topic by Kyle Kanos, Jon Custer, sammy gerbil, JMac, Mitchell Dec 30 '17 at 18:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Please note that array environments are for matrices, not aligned equations. You also seem to over-use the braces (.e.g, \frac{{\partial y}}{{\partial x} can be done as \frac{\partial y}{\partial x}); while rendered the same, it can save you some characters to type. $\endgroup$ – Kyle Kanos Dec 23 '17 at 13:44
  • $\begingroup$ i typed this in MathType so i can use it in word,and it has an option to convert it for use on this website,so i just copy pasted the code from the MatyhType editor. $\endgroup$ – Andrej Licanin Dec 23 '17 at 13:52
  • $\begingroup$ Okay, then I would recommend not using MathType because its LaTeX/MathJax conversion is entirely inappropriate. $\endgroup$ – Kyle Kanos Dec 23 '17 at 13:53
  • $\begingroup$ Of course, that assumes you are doing the MathType correctly & not using the aforementioned wrong environments $\endgroup$ – Kyle Kanos Dec 23 '17 at 13:55
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    $\begingroup$ Well, for what it's worth, one usually denotes the covariant differentiation by a semicolon sign, not by a comma. This is a notation I am not a fan of, but it helped PAM Dirac to write a splendid brochure on field-theoretical GR. $\endgroup$ – DanielC Dec 23 '17 at 18:34