How can the following identity be proven?

$$\nabla(\vec{A} \cdot \vec{B}) = \vec{A} \times(\nabla \times \vec{B}) + \vec{B} \times(\nabla \times \vec{A}) + (\vec{A}\cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{A}$$

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    $\begingroup$ Hello, Ahsan, and welcome to math.SE. Since you are new, I want to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. $\endgroup$ – Zev Chonoles Jul 13 '12 at 6:14

Brian already gave a pretty clear answer. Here let me just remark upon the geometry of the expression. Clearly, by the product rule, you have that

$$ \nabla (\vec{A}\cdot \vec{B}) = (\nabla \vec{A}) \cdot \vec{B} + (\nabla \vec{B}) \cdot \vec{A} $$

Where $(\nabla \vec{A})$ denote the Jacobian matrix formed by taking the gradient of the vector $\vec{A}$. We can split any matrix into its symmetric and anti-symmetric parts. Here we do a slightly funny trick: write

$$ (\nabla \vec{A}) = \underbrace{(\nabla \vec{A}) - (\nabla\vec{A})^T}_{= (\mathrm{d}\vec{A})} + (\nabla\vec{A})^T $$

where $(M)^T$ denotes the matrix transpose. So

$$ (\nabla\vec{A})\cdot \vec{B} = (\mathrm{d}\vec{A})\cdot \vec{B} + (\nabla\vec{A})^T \cdot \vec{B} \tag{1} $$

The second term in the RHS of (1) can be written as

$$ (\nabla\vec{A})^T\cdot \vec{B} = \left( \vec{B}^T\cdot (\nabla\vec{A})\right)^T = (\vec{B}\cdot\nabla)\vec{A} $$

using the properties of the transpose operation. The first term on the RHS of (1) can be computed to be precisely

$$ \vec{B} \times (\nabla\times \vec{A}) $$

(This is more easily seen if you've learned the connection between the curl operator, the exterior derivatve, and the Hodge star operator.)


Let $\delta_{ij}$ and $\epsilon_{ijk}$ be the Kronecker delta and Levi-Civita symbol respectively. Then,

$$[\vec{A} \times(\nabla \times \vec{B}) + \vec{B} \times(\nabla \times \vec{A}) + (\vec{A}\cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{A}]_i = \epsilon_{ijk}\vec{A}_j[\epsilon_{klm}\partial_l\vec{B}_m] + \epsilon_{ijk}\vec{B}_j[\epsilon_{klm}\partial_l\vec{A}_m] + (\vec{A}_t\partial_t)\vec{B}_i + (\vec{B}_t\partial_t)\vec{A}_i$$

which simplifies as

$$(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})(\vec{A}_j\partial_l\vec{B}_m + \vec{B_j}\partial_l\vec{A_m}) + (\vec{A_t}\partial_t)\vec{B_i} + (\vec{B_t}\partial_t)\vec{A_i}$$

Replacing $j$ with $m$ and subsequently $l$ with $i$, we get

$$(1 - \delta_{im}\delta_{mi})(\vec{A}_m\partial_i\vec{B}_m + \vec{B_m}\partial_i\vec{A_m}) + (\vec{A_t}\partial_t)\vec{B_i} + (\vec{B_t}\partial_t)\vec{A_i}$$

Further simplification yields

$$\vec{A_m}\partial_i\vec{B_m} - \delta_{im}(\vec{A_m}\partial_m)\vec{B_m} + \vec{B_m}\partial_i\vec{A_m} - \delta_{im}(\vec{B_m}\partial_m)\vec{A_m} + (\vec{A_t}\partial_t)\vec{B_i} + (\vec{B_t}\partial_t)\vec{A_i}$$

Since $\delta_{im}\vec{B_m} = \vec{B_i}$ and $\delta_{im}\vec{A_m} = \vec{A_i}$, the above becomes

$$\vec{A_m}\partial_i\vec{B_m} - (\vec{A_m}\partial_m)\vec{B_i} + \vec{B_m}\partial_i\vec{A_m} - (\vec{B_m}\partial_m)\vec{A_i} + (\vec{A_t}\partial_t)\vec{B_i} + (\vec{B_t}\partial_t)\vec{A_i} = (\partial_i\vec{B_m})\vec{A_m} + (\partial_i\vec{A_m})\vec{B_m} = \partial_i(\vec{A} \cdot \vec{B})$$

which is precisely $[\nabla(\vec{A} \cdot \vec{B}]_i$


Indical notation is nice and all, but it's often not included in multivariable calculus syllabi, and sometimes it can be rewarding to just work through the algebra. It's nice to think of vector fields in terms of components; ie) $A(x,y,z) = (A_1,A_2,A_3).$ Let's look at the first component of $A \times (\textrm{curl }B).$ Well, just expanding according to the definition that should be $A_2 (\frac{\partial B_2}{\partial x} - \frac{\partial B_1}{\partial y}) - A_3(\frac{\partial B_1}{\partial z} - \frac{\partial B_3}{\partial x}).$ Now we just add to get the first component of $A \times (\textrm{curl }B) + B \times (\textrm{curl } A)$ :

$A_2 (\frac{\partial B_2}{\partial x} - \frac{\partial B_1}{\partial y}) - A_3(\frac{\partial B_1}{\partial z} - \frac{\partial B_3}{\partial x})+ B_2 (\frac{\partial A_2}{\partial x} - \frac{\partial A_1}{\partial y}) - B_3(\frac{\partial A_1}{\partial z} - \frac{\partial A_3}{\partial x}).$ With some algebra, this is seen to be equivalent to $(A_2 \frac{\partial B_2}{\partial x}+A_3 \frac{\partial B_3}{\partial x}) + (B_2 \frac{\partial A_2}{\partial x}+B_3 \frac{\partial A_3}{\partial x}) - (A_2 \frac{\partial B_1}{\partial y} +A_3 \frac{\partial B_1}{\partial z}) - (B_2 \frac{\partial A_1}{\partial y} +B_3 \frac{\partial A_1}{\partial z}).$

The first two terms almost look like the dot product rule for differentiation, and the last two terms almost look like a del expression. Thus we just have to simultaneously add and subtract a miracle quantity and rearrange some more. That miracle quantity is, as you can probably guess, $A_1 \frac{\partial B_1}{\partial x} +B_1 \frac{\partial A_1}{\partial x} ,$ giving the previous equation as

$(A_1 \frac{\partial B_1}{\partial x} +A_2 \frac{\partial B_2}{\partial x}+A_3 \frac{\partial B_3}{\partial x}) + (B_1 \frac{\partial A_1}{\partial x}+B_2 \frac{\partial A_2}{\partial x}+B_3 \frac{\partial A_3}{\partial x}) $

$- (A_1 \frac{\partial B_1}{\partial x} +A_2 \frac{\partial B_1}{\partial y} +A_3 \frac{\partial B_1}{\partial z}) - (B_1 \frac{\partial A_1}{\partial x}+B_2 \frac{\partial A_1}{\partial y} +B_3 \frac{\partial A_1}{\partial z}).$

This is the first component of $\nabla (A \cdot B) - ( A\cdot \nabla ) B - (B \cdot \nabla )A$ The rest of the work can be applied mutatis mutandis to the second and third components, completing the identity.


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