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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? $$\vec{\nabla}\left(a\cdot \vec v \right) = ?$$

Najib Idrissi
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Samir Ouchene
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  • 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. – Winther Aug 31 '15 at 13:41
  • 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 –  Aug 31 '15 at 14:48

1 Answers1

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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

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

Lythia
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  • Great. Glad I could help. – Lythia Sep 01 '15 at 20:29
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    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. – Lythia Sep 01 '15 at 23:38
  • $(\text{div}(\vec{v}\otimes\vec{v}))_i=\tfrac{\partial}{\partial x_j}(v_iv_j)=\tfrac{\partial v_i}{\partial x_j}v_j+v_i\tfrac{\partial v_j}{\partial x_j}$. Thus, $$\text{div}(\vec{v}\otimes\vec{v}) = (\nabla\vec{v})\vec{v}+(\text{div},\vec{v})\vec{v}.$$ – Lythia Sep 02 '15 at 00:41
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    are you sure about this term: $\left( \nabla\vec v \right) \vec v$ I think there is a dot product between $\left( \nabla \vec v \right) $ and $\vec v$ –  Sep 03 '15 at 10:28
  • Some people put a dot between the two; it is a matter of notational preference. What is meant is the second order tensor $\nabla\vec{v}$ acting on the vector $\vec{v}$. This produces a vector. – Lythia Sep 03 '15 at 11:58
  • thank you for the reply. but, between the gradient of $\vec v$ (which a 2nd order tensor) and $\vec v$ is there a dot? –  Sep 05 '15 at 21:23
  • When a second order tensor $\boldsymbol{A}$ acts on a vector $\vec{x}$, I write $\boldsymbol{A}\vec{x}$. You can think of this like a matrix times a vector. In components, this is $\left(\boldsymbol{A}\vec{x}\right){ij} = A{ij}x_j$. This is what is going on in the term $(\nabla\vec{v})\vec{v}$. – Lythia Sep 06 '15 at 13:40
  • Would it be possible to show why not $\nabla a\otimes\vec v$ for the first term in the right? I cannot convince myself that my own derivation was something wrong, so I have asked a new question here https://math.stackexchange.com/questions/4213636 – MathArt Aug 01 '21 at 12:38