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I am trying to prove following vector identity using index notation. $$ \nabla \cdot ( A \otimes B ) = ( \nabla \cdot A ) B + A \cdot \nabla B $$

Where A and B are ordinary vectors (rank-1 tensors).

I found the identity here. https://en.wikipedia.org/wiki/Derivation_of_the_Navier%E2%80%93Stokes_equations#Continuity_equations

Unfortunately I got stuck at the starting point itself. I think $ A \otimes B $ will be represented as $ A^i B^j $, but I do not know how to take its inner product with $ \nabla $. I mean, I am at loss to figure out how to contract both superscripts i and j with $ \nabla $. Help will be greatly appreciated.

dionysus
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  • Just use the "usual" product rule for the covariant derivative. $(A^jB^k){|j}=A^{j}{|j}B^k+A^jB^k_{|j}$ – ContraKinta Feb 07 '22 at 20:17
  • @ContraKinta, You have written the expression as $ (A^jB^k)_{|j} $. This is precisely my problem. Why j? Why should the contraction be over the subscript j and not k? – dionysus Feb 07 '22 at 21:41
  • As we discussed in another thread, most "dot" products of higher order are not well defined. By convention we contract the first upper index. This convention is used in some literature on continuum mechanics and electromagnetic field theory. Note that this means $\nabla \cdot T\neq \nabla \cdot T^T$ unless $T=T^T$. Most of the time those guys use an Euclidean flat space with affine tensors and only lower indices, then it is $(\nabla \cdot\mathbf{S})k=S{ik,i}$ (with $\Gamma\equiv0$). And if $S$ is symmetric $S_{ki,i}$ – ContraKinta Feb 07 '22 at 23:03
  • I found a link for you! https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics) – ContraKinta Feb 07 '22 at 23:39
  • @mathcounterexamples.net, yes, the link you posted does answer the question, but it is more cumbersome than necessary. This identity can be proved in a more natural way by following the dot product rules for higher order tensors. – dionysus Feb 09 '22 at 09:14
  • @ContraKinta, it is clear now. Actually, the dot product of higher order tensors is quite well defined. I found multiple sources where they all agree that it is the contraction of last index of the left side tensor with the first index of the right side tensor, with the added condition that these indices must be up/down or down/up. Using this convention the identity can be easily proved. – dionysus Feb 09 '22 at 09:21
  • You do understand $\nabla$ is not a tensor, right? And no, its not well defined. To cite one source "The difference stems from whether the differentiation is performed with respect to the rows or columns and is conventional." – ContraKinta Feb 09 '22 at 10:48
  • Also, read this https://mathoverflow.net/questions/230988/divergence-of-a-second-order-tensor – ContraKinta Feb 09 '22 at 11:00
  • And, some sources even use a transpose of higher order tensors in their definitions. So you should be aware that even the transpose is a matter of convention. Like this: https://sbrisard.github.io/posts/20140219-on_the_double_dot_product.html – ContraKinta Feb 09 '22 at 11:09
  • You do understand ∇ is not a tensor, right?. Well, the del operator ∇ translates to covariant derivative $ \nabla_i $, which IS a tensor. So there shouldn't be any issues in applying the tensor rules to $ \nabla_i $, isn't it? – dionysus Feb 09 '22 at 18:04
  • No, for several reasons. Violation of the multiplication rule (does not commute) $\nabla_i\nabla_j\neq\nabla_j\nabla_i$. Violation of symmetric transformation rules, the list is long. You need to specify what it acts on and keep the order intact and so on. – ContraKinta Feb 09 '22 at 19:12
  • @ContraKinta, $ \nabla_i $ IS commutative under multiplication. I can cite multiple sources on this. I have to state this, lest someone gets wrong ideas reading this exchange. – dionysus Feb 09 '22 at 20:34
  • $(\nabla_k\nabla_h-\nabla_h\nabla_k)X^j=K_{l,,,,hk}^{,,j}X^l-S_{h,,,k}^{,,l}X^j_{|l}$, where $K$ is the curvature tensor and $S$ is the torsion tensor. – ContraKinta Feb 10 '22 at 03:28
  • I suggest you actually try to study the symmetry properties- or rather the lack thereof - of the tensor $X^j_{|h|k}$ in the indices $h$ and $k$. – ContraKinta Feb 10 '22 at 03:36

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