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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.

Anyone Please help.

Thank you.

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

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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.

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  • $\begingroup$ Ok got it, but what is its geometrical interpretation? $\endgroup$
    – prAnjal
    Apr 6, 2021 at 3:05
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    $\begingroup$ @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. $\endgroup$
    – Othin
    Apr 6, 2021 at 19:57
  • $\begingroup$ Ok got it, thank you $\endgroup$
    – prAnjal
    Apr 7, 2021 at 3:29

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