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Since the few (smooth) vector fields can denote themself in the unit vectors, $$\boldsymbol{\dbinom{B}{C}}=\dbinom{b_1\boldsymbol i+b_2\boldsymbol j+b_3\boldsymbol k}{c_1\boldsymbol i+c_2\boldsymbol j+c_3\boldsymbol k}$$ we have $$(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)+(b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3)\\=((\nabla b_1)c_1+b_1\nabla c_1)+((\nabla b_2)c_2+b_2\nabla c_2)+((\nabla b_3)c_3+b_3\nabla c_3)\\=\nabla(b_1c_1)+\nabla(b_2c_2)+\nabla(b_3c_3)\\=\nabla(b_1c_1+b_2c_2+b_3c_3)\\=\boldsymbol\nabla(\boldsymbol B\bullet\boldsymbol C)$$ I v found it once I was expanding $$\begin{cases}\boldsymbol{(\nabla\times B)\times C}&\equiv(\boldsymbol C\bullet\boldsymbol\nabla)\boldsymbol B-(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)\\\boldsymbol{(B\times\nabla)\times C}&\equiv(b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3)-\boldsymbol B(\boldsymbol\nabla\bullet\boldsymbol C)\end{cases}$$ seperately, then we have $$\boldsymbol{(\nabla\times B)\times\boldsymbol C}-\boldsymbol{(B\times\nabla)\times C}+\boldsymbol\nabla(\boldsymbol B\bullet\boldsymbol C)=(\boldsymbol C\bullet\boldsymbol\nabla)\boldsymbol B+\boldsymbol B(\boldsymbol\nabla\bullet\boldsymbol C)\ .\ :)$$

However I think an unique $(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)$ is sticky and v no ideas to his meanings and collapsing in the vector fields. Who can tutor me?? Ur all answers r welcome!! :D

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1 Answer 1

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The identity you have is correct:$\newcommand{\b}{\boldsymbol}$

$$ \nabla (\b{B}\cdot \b{C}) = (\b{B}\cdot \nabla) \b{C} + (\b{C}\cdot \nabla) \b{B} - (\nabla \times\b{B})\times\b{C} - (\nabla \times\b{C})\times\b{B} \tag{$\dagger$} $$ I rewrite it a bit to make it easier to understand and memorize.


$(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)$ has no specific meaning, but is sometimes denoted as the following in physics: $$ c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3 := \nabla_{\b{B}}(\b{B}\cdot \b{C}) $$ this means $\b{B}$ is differentiated while $\b{C}$ is held constant. It is defined as: $$ \nabla_{\b{B}}(\b{B}\cdot \b{C}) = (\b{C}\cdot \nabla) \b{B} - (\nabla \times\b{B})\times\b{C} \tag{1} $$ which is what you got up there.

Both terms in the right hand side of (1) have a geometric meaning: $(\b{C}\cdot \nabla) \b{B} $ is the directional derivative of $\b{B}$ in the direction of $\b{C}$, scaled by the magnitude of $\b{C}$; $(\nabla \times\b{B})\times\b{C}$ is the curl of $\b{B}$ projected to the plane which has $\b{C}$ as normal vector, then rotated a right angle such that it is perpendicular to both $\nabla \times \b{B}$ and $\b{C}$, also scaled by the magnitude of $\b{C}$.

Coincidentally, both terms' geometric meaning relies on a fixed $\b{C}$. We fixed a $\b{C}$ at a point, then we are literally decomposing the rate of change of the other vector $\b{B}$ into two parts, one part is the gradient along the direction of $\b{C}$, the other is the gradient perpendicular to $\b{C}$.

Similarly: $$ b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3 := \nabla_{\b{C}}(\b{B}\cdot \b{C}) = (\b{B}\cdot \nabla) \b{C} - (\nabla \times\b{C})\times\b{B} \tag{2} $$


Lastly, instead of your way of obtaining the identity, we can directly consider the following: $$ \nabla (\b{B}\cdot \b{C}) = \nabla_{\b{B}}(\b{B}\cdot \b{C}) + \nabla_{\b{C}}(\b{B}\cdot \b{C}) $$ which is the product rule for vector, then together with (1) and (2), we have the identity $(\dagger)$.

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