Material derivative of the dot product

Question

I was wondering if it is the case that:

$$\frac{d}{dt} \left( \vec{a}\cdot\vec{b} \right)= \vec{a} \cdot \frac{d}{dt} \vec{b} + \vec{b} \cdot \frac{d}{dt} \vec{a}$$

Where $\frac{d}{dt}$ indicates the material derivative $\frac{\partial}{\partial t} + \vec{u} \cdot \nabla$

If so, why is this the case? I am having difficulty showing this to be true or false.

Answer 1

You really just have to expand out the material derivative. Consider the vectors $a = (a_1,\dots,a_n)$, $b = (b_1,\dots,b_n)$ and $u$ the velocity field. Then \begin{align*} \frac{d}{dt}\left(\sum_{j=1}^n a_jb_j\right) & = \frac{\partial}{\partial t}\left(\sum_{j=1}^n a_jb_j\right) + u\cdot\nabla\left(\sum_{j=1}^n a_jb_j\right) \\ & = \sum_{j=1}^n \left(\frac{\partial a_j}{\partial t}b_j + a_j\frac{\partial b_j}{\partial t}\right) + \sum_{k=1}^n u_k\frac{\partial}{\partial x_k}\left(\sum_{j=1}^n a_jb_j\right) \\ & = \sum_{j=1}^n \left(\frac{\partial a_j}{\partial t}b_j + a_j\frac{\partial b_j}{\partial t}\right) + \sum_{k=1}^n u_k\left(\sum_{j=1}^n \frac{\partial a_j}{\partial x_k}b_j + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial x_k}\right) \\ & = \sum_{j=1}^n \frac{\partial a_j}{\partial t}b_j + \sum_{k=1}^n u_k\left(\sum_{j=1}^n \frac{\partial a_j}{\partial x_k}b_j\right) + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial t} + \sum_{k=1}^n u_k\left(\sum_{j=1}^n a_j\frac{\partial b_j}{\partial x_k}\right) \end{align*} where we apply product rule. Interchanging sum for the second and fourth terms, we obtain \begin{align*} \frac{d}{dt}\left(\sum_{j=1}^n a_jb_j\right) & = \sum_{j=1}^n \frac{\partial a_j}{\partial t}b_j + \sum_{j=1}^n b_j\left(\sum_{k=1}^n u_k\frac{\partial a_j}{\partial x_k}\right) + \sum_{j=1}^n a_j\frac{\partial b_j}{\partial t} + \sum_{j=1}^n a_j\left(\sum_{k=1}^n u_k\frac{\partial b_j}{\partial x_k}\right) \\ & = \sum_{j=1}^n b_j\left(\frac{\partial a_j}{\partial t} + \sum_{k=1}^n u_k\frac{\partial a_j}{\partial x_k}\right) + \sum_{j=1}^n a_j\left(\frac{\partial b_j}{\partial t} + \sum_{k=1}^n u_k\frac{\partial b_j}{\partial x_k}\right) \\ & = b\cdot\frac{da}{dt} + a\cdot\frac{db}{dt}. \end{align*}

Answer 2

Consider $\vec a=[a_1\ \ a_2\cdots a_n]$ and $\vec b=[b_1 \ \ b_2\cdots b_n]$. Then it can simply be proved by applying scalar derivative property to their dot product $$\vec a \cdot \vec b= a_1b_1+a_2b_2+\cdots +a_nb_n.$$