How do I derive this vector calculus identity?

[Bright Liu]

Homework Statement

As the title says, how can I get this equation?

Relevant Equations

tensor analysis

$(\nabla\times\vec B) \times \vec B=\nabla \cdot (\vec B\vec B -\frac 1 2B^2\mathcal I)-(\nabla \cdot \vec B)\vec B$
$\mathcal I$ is the unit tensor

 

[archaic]

You need to know these.
$$\mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf c(\mathbf a\cdot\mathbf b)$$$$\mathbf a\cdot(\mathbf b\times\mathbf c)=\mathbf b\cdot(\mathbf c\times\mathbf a)=\mathbf c\cdot(\mathbf a\times\mathbf b)$$

 

[etotheipi]

Also note that since $\mathbf{u} \times \mathbf{v} = -\mathbf{v}\times \mathbf{u}$, you have$$\begin{align*}(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} &= - \mathbf{c} \times (\mathbf{a} \times \mathbf{b})\\ \\
&= \mathbf{a} (-\mathbf{c} \cdot \mathbf{b}) - \mathbf{b}(-\mathbf{c} \cdot \mathbf{a}) = -\mathbf{a}(\mathbf{c} \cdot \mathbf{b}) + \mathbf{b}(\mathbf{c} \cdot \mathbf{a})
\end{align*}$$

 

[haruspex]

Also note that since $\mathbf{u} \times \mathbf{v} = -\mathbf{v}\times \mathbf{u}$, you have$$\begin{align*}(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} &= - \mathbf{c} \times (\mathbf{a} \times \mathbf{b})\\ \\
&= \mathbf{a} (-\mathbf{c} \cdot \mathbf{b}) - \mathbf{b}(-\mathbf{c} \cdot \mathbf{a}) = -\mathbf{a}(\mathbf{c} \cdot \mathbf{b}) + \mathbf{b}(\mathbf{c} \cdot \mathbf{a})
\end{align*}$$

Yes, but you cannot do that with a vector operator. $\nabla\times\vec A$ is not $-\vec A\times\nabla$.
There’s a page of useful looking identities and their derivations at https://en.m.wikipedia.org/wiki/Vector_calculus_identities, but I do not see any that apply easily here.

 

[pasmith]

Homework Statement:: As the title says, how can I get this equation?
Relevant Equations:: tensor analysis

$(\nabla\times\vec B) \times \vec B=\nabla \cdot (\vec B\vec B -\frac 1 2B^2\mathcal I)-(\nabla \cdot \vec B)\vec B$
$\mathcal I$ is the unit tensor

Suffix notation is generally the best method (this being a more efficient form of "show that each cartesian component of the left hand side is equal to the corresponding cartesian component of the right hand side").

In particular, $$\mathcal{I}_{ij} = \delta_{ij} = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}$$ and
$$(\mathbf{a} \times \mathbf{b})_i = \epsilon_{ijk}a_jb_k$$ where $$\epsilon_{ijk} = \begin{cases} 1 & \mbox{(i,j,k) is an even permutation of (1,2,3)} \\ -1 & \mbox{(i,j,k) is an odd permutation of (1,2,3)} \\ 0 & \mbox{otherwise}\end{cases}$$ together with the identity $$\epsilon_{ijk} \epsilon_{kpq} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}$$

 

[etotheipi]

Yes, but you cannot do that with a vector operator. $\nabla\times\vec A$ is not $-\vec A\times\nabla$.
There’s a page of useful looking identities and their derivations at https://en.m.wikipedia.org/wiki/Vector_calculus_identities, but I do not see any that apply easily here.

Interesting, you're right! My previous, naïve, view was that so long as you work with Cartesian components then it's generally safe to treat $\nabla \equiv \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{pmatrix}^T$, however you are certainly correct that it doesn't work here, considering that the order of the $\frac{\partial}{\partial x_i}$ and the $A_j$ definitely matters.

In that case, I'll step back and see how yourself and @pasmith approach the problem, for fear of misleading the OP. Working with the components as mentioned in #5 seems like a good idea!