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The conservation form of the Cauchy momentum equationis$$\frac{\partial}{\partial t}(\rho\vec{u})+\nabla\cdot(\rho\vec{u}\vec{u}^T)=-\nabla p+\nabla\cdot\vec{\tau}+\rho\vec{g}$$The second term on the LHS is the dot product of $\nabla$ with the $3\times3$ matrix. But since $\nabla$ is a column vector and $\mathbf{a}\cdot{\mathbf{b}}$ is defined as $\mathbf{a}^T\mathbf{b}$ then the result of the second term is a $1\times3$ 'row' instead of a vector, even though it has the correct terms. Does this mean we should take the transpose of the second term in order to technically have the correct equation?

J.G.
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jg mr chapb
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Since we're taking a dot product with a matrix, the result is a vector viz.$$(A\cdot B)_j=A_iB_{ij}.$$This is a matrix multiplication with $A$ a row matrix; I understand your point there. If we want a transpose anywhere, it'll have to be on the $\nabla$ alone, not the whole term.

Of course, that's only if we think $\nabla$ is meant to be a row vector in the first place. A dot product of vectors implies an invisible transpose anyway, as with vectors in $\vec{V}\cdot\vec{W}=V^TW$ (except not really).

If we replace $\nabla$ with an "ordinary" vector, entries of $A$ commute with those of $B$, so the above dot product is $(B^TA)_j=((A^TB)^T)_{ij}$ if we take $A$ to denote a row vector, or $(B^TA^T)_j=((AB)^T)_{ij}$ if we take $A$ to denote a column vector. But since $\nabla$ can't be commuted in that way, it has to stay on the left.

Personally, I'd write it all with explicit indices viz.$$\partial_t(\rho u_j)+\partial_i(\rho u_iu_j)=-\partial_jp+\partial_i\tau_{ij}+\rho g_j.$$

J.G.
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  • I was actually thinking of all vectors as column vectors, including nabla and defining the dot product as stated in the original question. Using that, the dot product should be a row vector times a matrix, which is then another row vector. Not a column vector as intended? – jg mr chapb Jun 19 '20 at 15:02