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I am working through a fluid dynamics paper and came across this equation: $$ \frac{\partial \vec{v}}{\partial t} + \vec{v}\cdot\nabla\vec{v}=\nabla\cdot T - \frac{1}{\rho}\nabla \phi\tag1$$ where T is a tensor defined by $$ T_{ij} = -\frac{p}{\rho}\delta_{ij} + \nu\left(\frac{\partial v_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i}\right) - \beta\gamma\frac{\partial c}{\partial x_i}\frac{\partial c}{\partial x_j} $$ The paper is concerned with 2D so $\vec{x}=(x,z)$ and $\vec{v}=(u,w)$. I started by writing out the individual components of the tensor $T$ and could pretty easily see that it is symmetric (not sure if this matters). I wanted to then write out the component-wise equations of $(1)$ but to do that I needed to expand $\nabla\cdot T$ but I honestly have no idea how to do that. A quick google search says that it should be: $$ \nabla \cdot T = T_{ik,i}\cdot e_k $$ This doesn't really seem to make any sense to me though. For instance, in my case (for 2D) what are the values of $i$ and $k$ that I should be ranging over? Also what is the meaning of that comma in the index for the tensor - it is not anywhere included in the definition of $T$. An explanation of how to generally find the divergence of a tensor would be much appreciated. Also perhaps a simpler example worked out.

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    $\begingroup$ The comma means partial derivative $\endgroup$
    – John
    Commented Sep 7, 2021 at 21:34
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    $\begingroup$ So $T_{ik,i}$ is $\frac{\partial T_{ik}}{\partial x_i}$? $\endgroup$
    – Mjoseph
    Commented Sep 7, 2021 at 22:07
  • $\begingroup$ Yes, that is correct $\endgroup$
    – John
    Commented Sep 7, 2021 at 23:14

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The notation you found through google does not really make sense with the dot, but is ok without the dot. It means (in dimension $3$) $$ \nabla\cdot T = T_{ik,i}\,e_k = \sum_{i=1}^3 \sum_{k=1}^3 \partial_{x_i}(T_{ik})\, e_k, $$ by the convention of summation over repeated indices. Hence, it is the vector with $k$-th coordinate $$ (\nabla\cdot T)_k = \sum_{i=1}^3 \partial_{x_i}(T_{ik}). $$

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  • $\begingroup$ Small misprint, the summation in the last step is over i, not k. $\endgroup$ Commented Sep 8, 2021 at 0:52

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