# Feynman's subscript notation

Consider this vector calculus identity: $$\mathbf{A} \times \left( \nabla \times \mathbf{B} \right) = \nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right) - \left( \mathbf{A} \cdot \nabla \right) \mathbf{B}$$

According to Wikipedia, the notation $\nabla_\mathbf{B}$ means that the subscripted gradient operates on only the factor $\mathbf{B}$. Can somebody explain the term $\nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right)$ in detail, give a concrete example, or an expression in components because I do not understand it at all. I encountered this identity in electromagnetism.

The notation $\vec \nabla_B$ means simply that the derivative are applying only on the vector $\vec B$.
$$(\vec \nabla_B)_i (\vec A\cdot \vec B) = \vec A\cdot\frac{\partial \vec B}{\partial x^i}\tag{1}$$
• What is the subscript $i$? And isn't the right side now a scalar, since it's a dot product? But the gradient is a vector, right? Feb 19 at 19:55
First of all, I always tell every student who asks me about such expressions to stop remembering all these infinite rules and learn to do that with Kronecker delta and Levi-Civita symbols. For example your initial expression looks like this: $$\mathbf{A}\times[\nabla\times\mathbf{B}]=\varepsilon_{ijk}A_j\varepsilon_{klm}\partial_lB_m = \varepsilon_{ijk}\varepsilon_{klm}A_j\partial_lB_m$$ As you can see, when you've written it like that, you don't need to care about non-commutativity of the cross-product any more. So you can take these Levi-Civitas and do the transformation: $$\varepsilon_{ijk}\varepsilon_{klm} = \varepsilon_{kij}\varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$ First I just made a cyclic permutation of the indexes, and the second equality is the only equality you need to remember (which is really easy: same indices go with "+", and swapped indices go with "-"). Inserting that, one gets: $$\varepsilon_{ijk}\varepsilon_{klm}A_j\partial_lB_m = \delta_{il}\delta_{jm}A_j\partial_lB_m - \delta_{im}\delta_{jl}A_j\partial_lB_m=A_j\partial_i B_j-A_j\partial_jB_i$$ Here I've used the property of Kronecker delta and actually came to the result as in your expression. Lets expand the sums to clarify what those terms actually mean:
$$A_j\partial_i B_j-A_j\partial_jB_i =\left(\begin{array}{c}A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_y}{\partial x}+A_z\frac{\partial B_z}{\partial x}\\A_x\frac{\partial B_x}{\partial y}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_z}{\partial y}\\A_x\frac{\partial B_x}{\partial z}+A_y\frac{\partial B_y}{\partial z}+A_z\frac{\partial B_z}{\partial z}\end{array}\right)-\left(\begin{array}{c}A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_x}{\partial y}+A_z\frac{\partial B_x}{\partial z}\\A_x\frac{\partial B_y}{\partial x}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_y}{\partial z}\\A_x\frac{\partial B_z}{\partial x}+A_y\frac{\partial B_z}{\partial y}+A_z\frac{\partial B_z}{\partial z}\end{array}\right)$$