# Show $2(B \cdot \nabla)B = \nabla |B|^2$ when the B-field is curl-less using summation notation

I was able to show for myself that $$2(\mathbf{B} \cdot \mathbf{\nabla})\mathbf{B} = \mathbf{\nabla} |\mathbf{B}|^2$$ when $$\mathbf{\nabla} \times \mathbf{B} = 0$$, but in order to do this, I had to actually write out all the components in regular (e.g. $$B_x\frac{\partial B_x}{B_y} + ...$$, etc.)

This is fine, but I would like to be able to present this in a more compact format (and I'm working on improving my familiarity with summation notation). Is there a way to write this proof in Einstein summation notation?

• If the magnetic field vanishes at infinity, then if your magnetic field has both zero divergence (from Gauss's Law for Magnetism) and zero curl, it's the zero field. I assume you're looking at conditions where nonzero field is allowed at infinity, then? – probably_someone Oct 20 '18 at 20:05
• Yes, @probably_someone -- The setup I have is a magnetic field generated by electromagnets several centimeters away from some nanoparticles, to which a (slowly-varying) magnetic force is being applied. See my previous question: link – Bunji Oct 20 '18 at 21:32

Observe the vanishing components i of $$\mathbf{B}\times (\nabla \times \mathbf{B})=0,$$ namely $$0= B_j \epsilon^{ijk}( \epsilon ^{klm}\partial_l B_m) = (\delta^{il}\delta^{jm}-\delta^{im}\delta^{jl}) B_j \partial_l B_m \tag{z}\\ = B_j \partial_i B_j - B_j \partial_j B_i = \frac{1}{2} \partial_i (B_j B_j)- B_j\partial_j B_i,$$ which but amount to the vanishing components of $$0= \frac{1}{2} \nabla (|\mathbf{B}|^2) - (\mathbf{B}\cdot \nabla)~ \mathbf{B}.$$