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!

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

1 Answer 1


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 $\vec b$.

  • $\begingroup$ Your second equality seems false. it is not divergence. $\endgroup$ Mar 13, 2021 at 19:27
  • $\begingroup$ The gradient of a vector should be a 2nd order tensor. Something like this (continuummechanics.org/velocitygradient.html) $\endgroup$
    – user853256
    Mar 13, 2021 at 19:29
  • $\begingroup$ @ManuelOliveira Is he talking about the gradient or divergence? Divergence of a vector is a scalar $\endgroup$
    – Andrei
    Mar 13, 2021 at 19:56

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