vector calculus: distribution of dot product over outer product

Question

In continuum mechanics books usually I see the equation known as "vector identity":

$$\vec{v} \cdot \left( \vec{\nabla} \otimes\vec{v} \right)=\frac{1}{2}\vec{\nabla}\left(\vec{v} \cdot\vec{v} \right)-\vec{v} \times \left( \vec{\nabla} \times\vec{v} \right)\tag{1}$$

Where $\otimes$ is the outer product between vectors. I was wondering if this is valid for any arbitrary vectors and if I can write:

$$\vec{a} \cdot \left( \vec{b} \otimes\vec{a} \right)=\frac{1}{2}\vec{b}\left(\vec{a} \cdot\vec{a} \right)-\vec{a} \times \left( \vec{b} \times\vec{a} \right)\tag{2}$$

and if not where the equation 1 comes from?

Answer 1

In fact the following equation holds for $\vec{a},\vec{b}\in\mathbb{R}^3$:

$$\vec{a} \cdot \left( \vec{b} \otimes\vec{a} \right) = \vec{b}\left(\vec{a} \cdot\vec{a} \right)-\vec{a} \times \left( \vec{b} \times\vec{a} \right)\tag{3}$$

The reason is that in (1) $\vec{\nabla}\left(\vec{v} \cdot\vec{v} \right)$ means the differentiation, but in (2) $\vec{b}\left(\vec{a} \cdot\vec{a} \right)$ means the scalar multiplication.

Let me explain in detail.

(1) Set $\vec{v}=\langle v_x,v_y,v_z\rangle$. Then $\frac{1}{2}$ is cancelled by the differentiation as follows: $$ \begin{align*} \frac{1}{2} \vec{\nabla}\left(\vec{v} \cdot\vec{v} \right) &= \frac{1}{2} \vec{\nabla} (v_x^2+v_y^2+v_z^2) \\ &=\frac{1}{2} \left< \frac{\partial(v_x^2+v_y^2+v_z^2)}{\partial x},\frac{\partial(v_x^2+v_y^2+v_z^2)}{\partial y},\frac{\partial(v_x^2+v_y^2+v_z^2)}{\partial z} \right> \\ &=\left(\frac{1}{2}\cdot2\right) \left< v_x\frac{\partial v_x}{\partial x}+v_y\frac{\partial v_y}{\partial x}+v_z\frac{\partial v_z}{\partial x}, v_x\frac{\partial v_x}{\partial y}+v_y\frac{\partial v_y}{\partial y}+v_z\frac{\partial v_z}{\partial y}, v_x\frac{\partial v_x}{\partial z}+v_y\frac{\partial v_y}{\partial z}+v_z\frac{\partial v_z}{\partial z} \right> \end{align*} $$

(2) Set $\vec{a}=\langle a_x,a_y,a_z\rangle$ and $\vec{b}=\langle b_x,b_y,b_z\rangle$. Here we can't expect cancellation because of scalar multiplication. $$ \begin{align*} \vec{b}\left(\vec{a} \cdot\vec{a} \right) &= \langle b_x,b_y,b_z \rangle (a_x^2+a_y^2+a_z^2) \\ &= \langle (a_x^2+a_y^2+a_z^2)b_x,(a_x^2+a_y^2+a_z^2)b_y,(a_x^2+a_y^2+a_z^2)b_z \rangle \end{align*} $$

Answer 2

As mentioned by Chof equation (2) is not valid but a slightly different version of it is:

$$\vec{a} \cdot \left( \vec{b} \otimes\vec{a} \right) = \vec{b}\left(\vec{a} \cdot\vec{a} \right)-\vec{a} \times \left( \vec{b} \times\vec{a} \right)\tag{3}$$

the issue is that the del or nabla operator is not the same as other vectors.

Prove (3): from vector calculus we know:

$$\vec{a}\times\left(\vec{b}\times\vec{c}\right)=\vec{b}\left(\vec{a} \cdot\vec{c}\right)-\left(\vec{a} \cdot\vec{b}\right)\vec{c} \tag{4}$$

and we consider $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$ first rank tensors and representing them as row matrices we can use the dyadic product to simply write: $$\vec{a} \cdot \left( \vec{b} \otimes\vec{c} \right)\equiv \check{a}\check{b}^T \check{c}\equiv \left(\vec{a}\cdot\vec{b}\right)\vec{c} \tag{5}$$

Combining (4) and (5) leads to (3)

Prove (1): For nabla/del operator we have (Wikipedia entry):

$$\vec{\nabla}(\vec u \cdot \vec v) = \vec u \times (\vec{\nabla} \times \vec v) + \vec v \times (\vec{\nabla} \times \vec u) + ( \vec u \cdot \vec{\nabla}) \vec v + (\vec v \cdot \vec{\nabla} )\vec u \tag{6}$$

and from this wikipedia entry:

$$\vec{\nabla}\cdot\left(\vec{u}\otimes\vec{v}\right)\equiv\check{\nabla}\check{u}^T\check{v}=\left(\check{\nabla}\check{u}^T\right)\check{v}+\check{u}\check{\nabla}^T\check{v}\equiv \left(\vec{\nabla}\cdot\vec{u}\right)\vec{v}+\left( \vec{u}\cdot \vec{\nabla} \right)\vec{v} \tag{7}$$

also from (5) we can write:

$$\vec{u}\cdot\left(\vec{\nabla}\otimes\vec{v}\right)=\left( \vec{u}\cdot \vec{\nabla} \right)\vec{v}\tag{8}$$

combining (6) and (8) leads to equations (1) above.