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I have the following operator, where $\rho$ is a scalar and $u$ is a vector: $$ \nabla (\rho u) - (\nabla \rho)u - u(\nabla \rho) $$ My book writes this in index notation as $$ \partial_\alpha(\rho u_\beta) - u_\beta\partial_\alpha\rho-u_\alpha\partial_\beta\rho $$ My question is, why does the book use the indices $\alpha$ and $\beta$ again for the second and third term? Shouldn't they have their own set of indices?

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  • $\begingroup$ Is the nabla a gradient or a divergence? Notation seems to indicate gradient, so you are computing the Jacobi matrix of $ρu$. If so, then the original difference is a difference of three matrices that you now compute elementwise, you are taking the same elements with the same indices from all three matrices. $\endgroup$ – Lutz Lehmann Jan 2 '14 at 11:50
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    $\begingroup$ If what you say is true, then you would've got 6th order tensor. $\endgroup$ – Kaster Jan 2 '14 at 11:55
  • $\begingroup$ @Lutzl It is the gradient. My question is much more simple than that, why not use three different pairs of indices, one for each term? $\endgroup$ – BillyJean Jan 2 '14 at 11:55
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    $\begingroup$ Because you subtract matrices by subtracting terms at the same position. The formula is for column $α$ and row $β$, located as in the ordinary Jacobian $$\begin{bmatrix}\partial_1f_1&\dots&\partial_nf_1\\\vdots&&\vdots\\\partial_1f_m&\dots&\partial_nf_m\end{bmatrix}.$$ $\endgroup$ – Lutz Lehmann Jan 2 '14 at 12:00
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What you are really doing when introducing indices is replacing $\nabla$ by $e_α\partial_α$ (using Einstein summation) and $u$ by $u_βe_β$, $e_α$ being the canonical basis vectors. Then

$$∇(ρu)=∂_α(ρu_β)\cdot e_α\otimes e_β$$

and in the full expression you combine the coefficients of the same basis vector $e_α⊗e_β$ of the tensor product in all three terms.

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