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?

  • $\begingroup$ 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? $\endgroup$ – probably_someone Oct 20 '18 at 20:05
  • $\begingroup$ 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 $\endgroup$ – 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 $$ 0= \frac{1}{2} \nabla (|\mathbf{B}|^2) - (\mathbf{B}\cdot \nabla)~ \mathbf{B}. $$

The identity (z) above is arguably one of the most useful determinant identities in vector calculus. Check its antisymmetry in the respective pairs of indices.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.