# Proof of vector identity $(\nabla \times B)\times B = -\frac{1}{2}\nabla B^2 + (B \cdot \nabla)B$

I'm working through the book:

Magnetohydrodynamics of the Earth's core (D. Gubbins, P.H. Roberts); J.A. Jacobs (Ed.), Geomagnetism, Vol. 2, Academic Press, London (1987)

... and I've come across the following identity I've had some trouble with.

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

I tried using Einstein notation, but I got a different RHS.

\begin{align} (\mathbf{\nabla} \times \textbf{B})\times \textbf{B} &=\varepsilon_{ijk}(\mathbf{\nabla} \times \textbf{B})_i B_j \\ &= \varepsilon_{ijk}\varepsilon_{abi}\partial_a B_b B_j\\ &= (\delta_{ja}\delta_{kb}-\delta_{jb}\delta_{ka})\partial_a B_b B_j\\ &= \partial_j B_k B_j - \partial_k B_j B_j\\ &= (\mathbf{B} \cdot \mathbf{\nabla})\mathbf{B}-\mathbf{\nabla}\mathbf{B}^2 \end{align}

The second term doesn't have a factor of $\frac{1}{2}$ needed. I tried using the regular vector calculations and got a similar answer. Where am I going wrong? Thank you

Edit

I've just realised that switching the order of one of the Levi-Civita symbols makes it work.

\begin{align} (\mathbf{\nabla} \times \textbf{B})\times \textbf{B} &=\varepsilon_{ijk}(\mathbf{\nabla} \times \textbf{B})_i B_j \\ &=-\varepsilon_{jik} B_j (\mathbf{\nabla} \times \textbf{B})_i \\ &=-\varepsilon_{jik} B_j \varepsilon_{abi}\partial_a B_b \\ &=-\varepsilon_{jik} \varepsilon_{abi} B_j \partial_a B_b \\ &=-\varepsilon_{ikj} \varepsilon_{iab} B_j \partial_a B_b \\ &=-(\delta_{ka}\delta_{jb}-\delta_{kb}\delta_{ja}) B_j \partial_a B_b\\ &= B_a \partial_a B_k - B_b \partial_k B_b\\ &= (\mathbf{B} \cdot \mathbf{\nabla}) B_k - \frac{1}{2}\partial_k (B_b B_b)\\ &= (\mathbf{B} \cdot \mathbf{\nabla})\mathbf{B}-\frac{1}{2}\mathbf{\nabla}\mathbf{B}^2 \end{align}

Using the product rule on $\partial_k (B_b B_b)=2B_b\partial_k B_b$.

Now I'm wondering why my initial approach didn't work? Can anyone help?

## 1 Answer

A word of caution, you should include the actual vectors in all your expressions, that accounts for the fact that the LHS has no free indices, while the RHS does. With this in mind consider

$$\frac{1}{2}\partial_k {\bf B}^2 = \frac{1}{2}\partial_k (B_j B_j) = \partial_k B_j B_j \tag{1}$$

So that

\begin{eqnarray} (\nabla \times {\bf B})\times{\bf B} &=& \varepsilon_{ijk} (\nabla \times {\bf B})_i B_j \color{blue}{e_k} \\ &=& \varepsilon_{ijk} \varepsilon_{abi}\partial_a B_b B_j \color{blue}{e_k} \\ &=& \varepsilon_{ijk} \varepsilon_{iab}\partial_a B_b B_j \color{blue}{e_k} \\ &=& (\delta_{ja}\delta_{kb} - \delta_{jb}\delta_{ka})\partial_a B_b B_j \color{blue}{e_k} \\ &=& \partial_j B_k B_j \color{blue}{e_k} - \partial_k B_j B_j \color{blue}{e_k} \\ &\stackrel{(1)}{=}& (B_j \partial_j)( B_k \color{blue}{e_k}) - \frac{1}{2}\color{blue}{e_k} \partial_k {\bf B}^2 \\ &=& ({\bf B}\cdot \nabla){\bf B} - \frac{1}{2}\nabla{\bf B}^2 \tag{2} \end{eqnarray}

• Another way is to write $[(\nabla \times {\bf B})\times{\bf B}]_k = \varepsilon_{ijk} (\nabla \times {\bf B})_i B_j = \cdots$ – md2perpe Jun 29 '18 at 19:37