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I am tasked to prove that $\nabla (\phi {\bf a}) = \phi \nabla {\bf a} + \nabla \phi \cdot {\bf a}$, where $\phi$ and ${\bf a}$ are scalar and vector fields respectively. I have a few questions regarding this proof. The first is that on inspection, there seems to be some sort of dimensional inconsistency. On the LHS, we are given the gradient of a vector (which I am unsure of how to interpret, but I’ll leave that for another time), while the last term of the RHS is supposedly a scalar – am I interpreting the equation correctly?

I also tried to prove this using index notation, which gives me

$$ \partial_i (\phi a_j) = \phi \partial_i a_j + a_j \partial_i \phi$$

My logic for using different indices for the partial derivative and the vector field was because we were taking the gradient. Is this correct? Furthermore, how do i interpret $\partial_i a_j$ in the second term of the sum, and why are we allowed to compress the third term into a dot product without $\delta_ij$? Thank you very much!

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  • $\begingroup$ You can define the gradient of a vector to be a matrix. en.m.wikipedia.org/wiki/Matrix_calculus. From there, I don't know what to do because the last term seems to be a scalar. Where did you find this identity? It doesn't show up in a big list of vector calc identities. en.m.wikipedia.org/wiki/Vector_calculus_identities $\endgroup$ – Tyberius Apr 1 '18 at 0:24
  • $\begingroup$ Seems correct. Note that a) the result is a rank 2 tensor, or matrix and b) the second product is an "outer" product of essentially a column vector times a row vector, giving a matrix. Gradients of scalars are usually better seen as row vectors so it is slightly easier to understand your result by thinking of the result as $a_j$ in front and then a transpose of the whole thing as the last operation. $\endgroup$ – Jap88 Apr 1 '18 at 1:11
  • $\begingroup$ To echo the previous comment, this identity is incorrect. The second term should be a tensor product, not a dot product, i.e., it should be $(\nabla\phi)\mathbf a^T$ instead of $(\nabla\phi)^T\mathbf a$. $\endgroup$ – amd Apr 1 '18 at 1:59
  • $\begingroup$ Hello everyone, this was from one of the past exams which I am practicing. Why is the gradient of a scalar a row vector? Also, as a tensor product, how should I write it in index notation? $\endgroup$ – user107224 Apr 1 '18 at 13:46
  • $\begingroup$ Your notation $\phi\partial_i a_j + a_j \partial_i \phi = \phi \otimes \nabla a + a \otimes \nabla \phi$ is okay for the tensor product: simple juxtaposition with different indices. Now, why is the gradient of a scalar a row vector? well, because it is normally used to be dotted with a vector, and a vector is normally represented as a column, so for a vector $F$ and a scalar $f$ you can write ($\nabla f \cdot F = \nabla_i f F_i = [\nabla f][F]$ where the brackets mean matrix representatives. $\endgroup$ – Jackozee Hakkiuz May 19 '18 at 10:33

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