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I'm currently working through a chapter on tensor analysis. One of the end-of-chapter questions asks me to show that, given a vector field $\mathbf{v}$ and a second order tensor field $\mathbf{S}$, $$ \int_{\partial B} (\mathbf{S}\mathbf{n})\otimes\mathbf{v}\ dA = \int_{B}(\nabla \cdot\mathbf{S})\otimes\mathbf{v} + \mathbf{S}\nabla\mathbf{v}^{T} dV $$ Here, $B$ is the volume over which the integral is taken, with $\partial B$ representing its boundary. The vector $\mathbf{n}$ represents the outward normal field on $\partial B$. The question directs me to use the divergence theorem, which for second order tensors is $$ \int_{\partial B} \mathbf{S}\mathbf{n}\ dA = \int_{B}(\nabla \cdot\mathbf{S})\ dV $$


This leads me to believe that, in some way, I have to rewrite $(\mathbf{Sn})\otimes\mathbf{v}$ in a way that $\mathbf{v}$ is included in the bracket, with $\mathbf{n}$ outside of it.

Is there any standard identity for this kind of representation? How does the transpose of the vector gradient of $\textbf{v}$ come about?

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In index notation, the integrand is a matrix but it can also be written as the normal vector times a third-order tensor, i.e. $$\eqalign{ M &= (S\cdot n)\otimes v \quad&\implies\quad M_{ij} = S_{ik} n_k v_j \\ &= n\cdot T \quad&\implies\quad T_{kij} = S_{ik}v_j \\ }$$ where $T$ is a third-order tensor which can be used in the divergence theorem $$\eqalign{ Q &= \int_{\partial B} n\cdot T\;\,dA = \int_B \left(\nabla\cdot T\right)\,dV \\ Q_{ij} &= \int_B \big(\partial_k T_{kij}\big)\,dV \\ &= \int_B \big(\partial_k S_{ik}\big)v_j + S_{ik}\big(\partial_k v_j\big)\;dV \\ Q &= \int_B \left(\nabla\cdot S^T\right)\otimes v + S\cdot\left(\nabla\otimes v\right)\;\;dV \\ }$$

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  • $\begingroup$ Wonderful! Now that I see it, it seems so straightforward... thank you very much! :) $\endgroup$
    – user685194
    Commented Jul 28, 2020 at 6:53

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