# Equality validation in vector calculus

I have getting myself into the area of fluid mechanics.

I would like to know if the following relationship holds.

$$\left (\vec{\mathbf{a}} \cdot \nabla \right ) \vec{\mathbf{b}} = \vec{\mathbf{a}} \cdot \nabla \vec{\mathbf{b}}$$

The left side of the equation is also known as the convective operator (https://mathworld.wolfram.com/ConvectiveOperator.html).

(vector $$\vec{\mathbf{a}}$$ dot product with the $$\nabla$$ operator) times vector $$\vec{\mathbf{b}}$$ is equal to the dot product of vector $$\vec{\mathbf{a}}$$ and the gradient of vector $$\vec{\mathbf{b}}$$.

Sorry if this a very basic question!

• No, it does not hold. The convective operator gives a vector, while the RHS is a scalar.
– 1__
Mar 13, 2021 at 19:13
• RHS is also a vector, along $\vec a$ Mar 13, 2021 at 19:14

It's not the same. Let's look at the $$\hat x$$ component for example in Cartesian coordinates. The left hand side (from the link that you have) is $$a_x\frac{\partial b_x}{\partial x}+a_y\frac{\partial b_x}{\partial y}+a_z\frac{\partial b_x}{\partial_z}$$ For the right hand side, $$\nabla\vec b=\frac{\partial b_x}{\partial x}+\frac{\partial b_y}{\partial y}+\frac{\partial b_z}{\partial_z}$$ This is a scalar. So when we multiply with vector $$\vec a$$, the $$\hat x$$ component will be $$a_x\frac{\partial b_x}{\partial x}+a_x\frac{\partial b_y}{\partial y}+a_x\frac{\partial b_z}{\partial_z}$$ Notice the difference: in the right hand side you have each term of the sum multiplied with a different component of $$\vec a$$, but each derivative acts on $$b_x$$. On the right hand side each term contains the same $$a_x$$, but the derivatives act on different components of $$\vec b$$.