# What is the difference between $\frac{(\vec B \cdot \nabla)\vec B}{ B}$ and $\nabla \cdot \vec B$?

I was doing a vector calculus problem, and in the hint I got to introduced with the first term, and I wondered, what is the difference between these two expression? My brain tells me that they are same, but I'm not sure.

Thank you.

• What is B, without the arrow? The norm of B?
– Paul
Apr 5, 2021 at 19:08
• @Paul yes, you got it right Apr 5, 2021 at 19:11

They are not the same. Notice that the divergence gives a number at each point (a scalar field) while the first expression in your answer is a vector. In fact, the $$j$$-th component of $$(\vec{B}\cdot \nabla)\vec{B}$$ is $$[(\vec{B}\cdot \nabla)\vec{B}]_j=\sum_{i=1}B_i\frac{\partial B_j}{\partial x_i},$$ where $$x_1=x$$, $$x_2=y$$ and $$x_3=z$$. Your first expression just divides each of these components by a number ($$B$$), and therefore still results in a vector field.
• @prAnjal Well, $(\vec{B}\cdot \nabla)$ is a vector operator that happens to appear in many vector calculus identities (specially some product rules, which might be why the hint mentioned it), but I never heard anyone give it a name or claim it has any special geometrical interpretation. $\frac{(\vec{B}\cdot \nabla)}{B}$ is a related vector operator that in your example is being applied to $\vec{B}$. Does the problem you're trying to solve ask anything about geometrical meaning? Like ordinary functions, not all vector operators have a straightforward/meaningful geometrical interpretation. Apr 6, 2021 at 19:57