Vector triple product with $\nabla$ operator

Question

I came across the following expression in several books (especially in plasma physics literature while deriving the magnetic pressure):

$$(\mathbf{\nabla} \times \mathbf{B})\times \mathbf{B} = \left(\mathbf{B} \cdot \mathbf{\nabla} \right)\mathbf{B}- \frac{1}{2}\nabla B^2 $$

But, if I use the BAC-CAB rule for this triple product operation I get: $$(\mathbf{\nabla} \times \mathbf{B})\times \mathbf{B} = \left(\mathbf{B} \cdot \mathbf{\nabla} \right)\mathbf{B}- \nabla B^2 $$

Which does not have the $1/2$ term on it. Do you know why is this?

Answer 1

The BAC-CAB identity does not apply here as $\nabla$ is an operator and not a vector. You cannot assume, in general, that vector identities hold for $\nabla$.

The correct product rule to use is $$\nabla(\mathbf{A}\cdot\mathbf{B}) = \mathbf{A}\times(\nabla\times\mathbf{B})+\mathbf{B}\times(\nabla\times\mathbf{A}) + (\mathbf{A}\cdot\nabla)\mathbf{B} + (\mathbf{B}\cdot\nabla)\mathbf{A}.$$ Setting $\mathbf{A}=\mathbf{B}$ yields $$\nabla\left(B^2\right) = 2[ (\mathbf{B}\cdot\nabla)\mathbf{B}-(\nabla\times\mathbf{B})\times\mathbf{B}]$$ from which the desired result is easily obtained.

Reference: Introduction to Electrodynamics by Griffiths, section 1.2.

Answer 2

Don't go from vector algebra to vector calculus identities. This is not a triple product of "regular" vectors, since $\nabla$ is used to write differential operators that can be written as formal vectors.

You should not try to extend vector algebra identities (involving regular vectors) to vector calculus identities (involving differential operators acting on vector fields), since this approach leads you to mistakes 99% of times, as you can realize comparing algebra vector identities with "similar" vector calculus identities in wikipedia pages.

Proof of the desired vector calculus identity. Anyway, you can easily prove vector calculus identities using index notations (thinking at Cartesian coordinates) and some properties of Levi-Civita and Dirac's delta symbols, like

$$\begin{aligned} \varepsilon_{ijk} & = \varepsilon_{jki} = \varepsilon_{kij} & \text{(even permutations)}\\ \varepsilon_{ijk} & = - \varepsilon_{ikj} & \text{(odd permutation)} \\ \varepsilon_{ijk} \varepsilon_{iab} & = \delta_{ja} \delta_{kb} - \delta_{jb} \delta_{ka} \ . \end{aligned}$$

Here,

$$\begin{aligned} \left\{ ( \nabla \times \mathbf{b} ) \times \mathbf{b} \right\}_i & = \varepsilon_{ijk} \varepsilon_{jlm} \partial_l b_m \, b_k = \\ & = (\delta_{kl} \delta_{im} - \delta_{km} \delta_{il}) \partial_l b_m \, b_k = \\ & = b_l \partial_l b_i - b_m \partial_i b_m = \\ & = b_m \partial_m b_i - \partial_i \left( \frac{b_m b_m}{2} \right) = \\ & = \left\{ ( \mathbf{b} \cdot \nabla ) \mathbf{b} - \frac{1}{2}\nabla |\mathbf{b}|^2 \right\}_i \ . \end{aligned}$$

Answer 3

Both answers here are very good, I only want to shed some light on how such a confusion had arisen for you. It seems it may have stemmed from a subtlety related to the following vector identity, to cite from the relevant Wikipedia entry:

The vector calculus identity of the cross product of a curl holds: $$\mathbf{v} \times \left( \nabla \times \mathbf{a} \right) = \nabla_a \left( \mathbf{v} \cdot \mathbf{a} \right) - \mathbf{v} \cdot \nabla \mathbf{a} $$ where the Feynman subscript notation $\nabla_\mathbf{a}$ is used, which means the subscripted gradient operates only on the factor $\mathbf{a}$.

Now, perhaps counter-intuitively, what happens when you put $\mathbf{v}=\mathbf{a}$ is that the operator in $\nabla_a \left( \mathbf{v} \cdot \mathbf{a} \right)$ is still only acting on one of the vectors, and not both! This is reflected quite clearly in @basics' derivation of the identity in his answer here, namely the term $ b_m\partial_i b_m $ shows you only one of the pair of vectors under the $\nabla_a$ is being differentiated.

That's why in order to put them both under the same regular $\nabla$, we must write:

$$\nabla_a \left( \mathbf{a} \cdot \mathbf{a} \right) = \frac{1}{2}\nabla(\mathbf{a}^2) $$

Which establishes the correctness of your first equation.

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The key takeaway as the other answers mention, is indeed not to try and apply regular vector identities to the $\nabla$ vector operator.

Not only triple vector identities like "BAC-CAB" fail when we try and do this, but even a basic identity that works for vectors like:

$$ \mathbf{A}\times\mathbf{B} = -\mathbf{B}\times\mathbf{A} $$

Makes completely no sense when we put $\mathbf{B}=\nabla$. When we do that, the LHS becomes $\mathbf{A}\times\nabla$, which is a vector operator, but the RHS, $-\nabla\times\mathbf{A}$ is still a vector!

Answer 4

Thanks everyone for your answers. Based on what you have answered I understood where I got it wrong!

I didn't use the Levi-Civita notation, but I have arrived to the conclusions and derivation that I have written in my document. See quoted part below.

Since the del operator, $\boldsymbol{\nabla}$, appears in the triple product, the normal $\text{BAC-CAB}$ vectorial rule for triple products cannot be applied directly. Considerino two vectors $\boldsymbol{A}$ and $\boldsymbol{B}$, applying a constrained version of the $\text{BAC-CAB}$ rule and knowing that $\boldsymbol{A} \times \boldsymbol{B} = - \boldsymbol{B} \times \boldsymbol{A}$, the following triple products result in: \begin{equation}\label{eq:311} (\boldsymbol{\nabla} \times \boldsymbol{A}) \times \boldsymbol{B} = (\boldsymbol{A} \cdot \boldsymbol{\nabla})\boldsymbol{B} - \boldsymbol{\nabla}_A (\boldsymbol{A} \cdot \boldsymbol{B}) \tag{3.11} \end{equation} \begin{equation}\label{eq:312} (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{A} = (\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{A} - \boldsymbol{\nabla}_B (\boldsymbol{A} \cdot \boldsymbol{B}). \tag{3.12} \end{equation} In the both left hand sides of the equations above the del operator acts only on vector $\boldsymbol{A}$ in equation \eqref{eq:311} and only on vector $\boldsymbol{B}$ in equation \eqref{eq:312}. Recognizing that $\boldsymbol{\nabla} = \boldsymbol{\nabla}_A + \boldsymbol{\nabla}_B$ yields that $\boldsymbol{\nabla}(\boldsymbol{A} \cdot \boldsymbol{B}) = \boldsymbol{\nabla}_A (\boldsymbol{A} \cdot \boldsymbol{B}) + \boldsymbol{\nabla}_B (\boldsymbol{A} \cdot \boldsymbol{B})$. Adding up both equations \eqref{eq:311} and \eqref{eq:312} leads to: \begin{equation}\label{eq:313} \boldsymbol{\nabla} (\boldsymbol{A} \cdot \boldsymbol{B}) = (\boldsymbol{A} \cdot \boldsymbol{\nabla})\boldsymbol{B} + (\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{A} - (\boldsymbol{\nabla} \times \boldsymbol{A}) \times \boldsymbol{B} - (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{A}. \tag{3.13} \end{equation} Making the substitution $\boldsymbol{A} = \boldsymbol{B}$ will finally provide the triple product: \begin{equation}\begin{aligned} \boldsymbol{\nabla} (\boldsymbol{B} \cdot \boldsymbol{B}) &= (\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{B} + (\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{B} - (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B} - (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B} \\ \boldsymbol{\nabla}B^2 &= 2[(\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{B} - (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B}] \\ (\boldsymbol{\nabla} \times \boldsymbol{B}) \times \boldsymbol{B} &= (\boldsymbol{B} \cdot \boldsymbol{\nabla})\boldsymbol{B}- \frac{1}{2}\boldsymbol{\nabla}B^2. \end{aligned} \tag{3.14} \end{equation}