Vector Calculus: Divergence of Dyadic AB

[Mr. Cosmos]

So I have a quick question that will hopefully yield some clarification. So the divergence of a dyadic $\bf{AB}$ can be written as,
$$\nabla \cdot (\textbf{AB}) = (\nabla \cdot \textbf{A}) \textbf{B} + \textbf{A} \cdot (\nabla \textbf{B})$$
where $\textbf{A} = [a_1, a_2, a_3]$ and $\textbf{B} = [b_1, b_2, b_3]$. However, it seems to me that the first term of the identity yields a vector (the vector $\bf{B}$ scaled by the divergence of $\bf{A}$), while the second term yields a scalar. Since one cannot simply add a scalar quantity to a vector quantity, it would appear the identity is false, or perhaps I am missing something. I know the dyadic of $\bf{AB}$ yields a tensor, and the divergence of a tensor is a vector, but it doesn't appear that the second term on the right hand side of the identity is a vector. Any help would be greatly appreciated.

 

[Lucas SV]

The second term does not mean a scalar. You are taking the laplacian of each component of $\mathbf{B}$. The $j$ component of the last term is $A^{i}\partial_i B^{j}$ (using Einstein summation convention). The easiest way to see the identity is to write the equation in components, and it will all make sense - it is just the Leibnitz rule.

Think of $A^i\partial_i=\mathbf{A}\cdot\mathbf{\nabla}$ as a differential operator acting on $\mathbf{B}$ componentwise.

 

[Mr. Cosmos]

The second term does not mean a scalar. You are taking the laplacian of each component of $\mathbf{B}$. The $j$ component of the last term is $A^{i}\partial_i B^{j}$ (using Einstein summation convention). The easiest way to see the identity is to write the equation in components, and it will all make sense.

Thanks for the quick response. I now see my mistake. Thanks!