Proving the Curl Identity for a Simple Curl Equation

[Hiero]

Homework Statement

Simplify $\nabla \times ( a\nabla b)$

Relevant Equations

$\nabla\times \vec V = \epsilon_{ijk}\frac{\partial V_k}{\partial x_j}\hat e_i$
$\nabla s = \frac{\partial s}{\partial x_i}\hat e_i$

Attempt:

$$\nabla \times ( a\nabla b) = \epsilon_{ijk}\frac{\partial}{\partial x_j}(a\frac{\partial b}{\partial x_k})\hat e_i$$ $$ = \epsilon_{ijk}\big(\frac{\partial a}{\partial x_j}\frac{\partial b}{\partial x_k}+a\frac{\partial b}{\partial x_j\partial x_k}\big)\hat e_i$$ $$= \nabla a \times \nabla b + \text{(final term)}$$

That “final term” (a triple sum) should be the zero vector, but I cannot see how. Maybe I messed up elsewhere.

Thanks.

 

[pasmith]

Well, \epsilon_{ijk} = -\epsilon_{ikj} but \partial_j\partial_kb = \partial_k\partial_jb so <br /> \epsilon_{ijk}\partial_j\partial_kb = -\epsilon_{ikj}\partial_k\partial_jb. But j and k are dummy indices, so we can relabel them on the right hand side ...

 

[Hiero]

Well, \epsilon_{ijk} = -\epsilon_{ikj} but \partial_j\partial_kb = \partial_k\partial_jb so <br /> \epsilon_{ijk}\partial_j\partial_kb = -\epsilon_{ikj}\partial_k\partial_jb. But j and k are dummy indices, so we can relabel them on the right hand side ...

That’s a very nice way to show it is zero. The last step (after relabeling) is that $[x=-x] \implies [x=0]$
I have to admit it seems magical to use this property ignoring the invisible nested summation.

Basically though, each component cancels out in pairs by virtue of the two properties you mentioned:

Well, \epsilon_{ijk} = -\epsilon_{ikj} but \partial_j\partial_kb = \partial_k\partial_jb

 

[pasmith]

Alternatively, \nabla \times (a\mathbf{v}) = (\nabla a) \times \mathbf{v} + a \nabla \times \mathbf{v} and if \mathbf{v} is a gradient then its curl is zero (which follows from the observation in my earlier post).