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We know from the tensor calculus that: $\vec\nabla (a\cdot b) = b\vec\nabla a + a \vec\nabla b$ , where $a$ and $b$ are two scalar functions.

But in the case where for example $a$ is a scalar function and $b$ is a vector how to develop that expression of gradient? $$\vec{\nabla}\left(a\cdot \vec v \right) = ?$$

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  • $\begingroup$ Two possible meanings. If there is no dot-product between $\vec{\nabla}$ and $a\vec{v}$ then you are taking the gradient of a vector-field. This is answered here. If there is a dot-product between $\vec{\nabla}$ and $a\vec{v}$ then you are taking the divergence of $a\vec{v}$ and you can find the relevant formula here. $\endgroup$ – Winther Aug 31 '15 at 13:41
  • $\begingroup$ thanks, I mean the gradient of a vector field, I know how to do that for a vector field, but it seems hard when that vector is multiplied by a scalar function $\endgroup$ – Navaro Aug 31 '15 at 14:48
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These sort of identities are usually proved in the component form and then transferred back to component-free form. In view of this, note that $\nabla(a\boldsymbol{v})$ is a second order tensor. Thus using the product rule,

$$\left(\nabla(a\boldsymbol{v})\right)_{ij} = \frac{\partial}{\partial x_j}\left(av_i\right)=\frac{\partial a}{\partial x_j}v_i+a\frac{\partial v_i}{\partial x_j}.$$

From the above component form, it is recognized that

$$\nabla(a\boldsymbol{v}) = \boldsymbol{v}\otimes\nabla a + a\nabla\boldsymbol{v}.$$

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  • $\begingroup$ Thank, I tried myself I find that result but I had the doubt about the dyadic product, so thank you again! $\endgroup$ – Navaro Sep 1 '15 at 19:59
  • $\begingroup$ Great. Glad I could help. $\endgroup$ – Lythia Sep 1 '15 at 20:29
  • $\begingroup$ I am a beginner in tensor calculus, So I am confused, for example how to write the following in the component form: $\mathrm{div} \left( \vec v\otimes \vec v) \right) ?$ $\endgroup$ – Navaro Sep 1 '15 at 23:15
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    $\begingroup$ Post another question and I'll be happy to answer it. That way the answer will be easier to find if others also have the same question. $\endgroup$ – Lythia Sep 1 '15 at 23:38
  • $\begingroup$ Hi, I cannot post another question, I need to wait one day to be able! (because when I try to ask a question, they say you reached the limit !!) $\endgroup$ – Navaro Sep 1 '15 at 23:59

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