the gradient of the product of a scalar by a vector

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

• 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. Commented Aug 31, 2015 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
– user265759
Commented Aug 31, 2015 at 14:48

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

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

• Great. Glad I could help. Commented Sep 1, 2015 at 20:29
• 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. Commented Sep 1, 2015 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, $$\text{div}(\vec{v}\otimes\vec{v}) = (\nabla\vec{v})\vec{v}+(\text{div}\,\vec{v})\vec{v}.$$ Commented Sep 2, 2015 at 0:41
• 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 $\vec v$
– user265759
Commented Sep 3, 2015 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. Commented Sep 3, 2015 at 11:58