Is the Curl of a Cross Product Affected by Directional Nabla?

[PhilDSP]

I have a number of books which give a vector identity equation for the curl of a cross product thus:

$$\nabla \times \left(a \times b \right) = a \left( \nabla \cdot b \right) + \left( b \cdot \nabla \right) a - b \left( \nabla \cdot a \right) - \left( a \cdot \nabla \right) b$$

But doesn't

$$b \left( \nabla \cdot a \right) = \left( a \cdot \nabla \right) b$$

If that is true then

$$\nabla \times \left(a \times b \right) = 2a \left( b \cdot \nabla \right) - 2b \left( a \cdot \nabla \right)$$

Or is there something I'm missing? (Since nabla is an operator the last equation as it's written might only make sense if it was multiplied by a vector)

 

[CompuChip]

It's easiest to see by writing it out in components:
$$[(\nabla \cdot a) b]_i = (\partial_x a_x + \partial_y a_y + \partial_z a_z) b_i = (\partial_x a_x) b_i + (\partial_y a_y) b_i + (\partial_z a_z) b_i$$
whereas
$$[(a \cdot \nabla) b]_i = (a_x \partial_x + a_y \partial_y + a_z \partial_z) b_i = (\partial_x b_i) a_x + (\partial_y b_i) a_y + (\partial_z b_i) a_z$$
and clearly these are not the same.

So while $a \cdot b = b \cdot a$ holds when a and b are really vectors, it is not necessarily true when one of them is a vector operator. This is one of the cases where the convenience of considering $\nabla$ as a vector satisfying all the rules for vectors does not apply.

 

[PhilDSP]

Thanks CompuChip. As I was mulling it over in afterthought I felt that I should have done just what you did. The notation seems deceptive so there is no substitute for doing the analysis as you have when any kind of operator is involved.

 

[lurflurf]

Nabla acts only to the right. Often it helps in these kind of manipulations to use a bidirectional nabla so that more vector identities are valid. What you have done is similar to the single variable equation
uDv=vDu
which is obviously does not hold in general, but would be true if D were bidirectional.