# Vector Calculus?

• I
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
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!

Lucas SV