Einstein Summation with Del Operator

Question

Can someone show explicitly me why $2B_k\nabla B_k = \nabla B^2$ ?

Is $B_k\nabla B_k$ just $B_x\nabla B_x+B_y\nabla B_y+B_z\nabla B_z$?

But then I end up with nine terms on the LHS and I can't match it with the RHS.

Answer 1

First of all you should notice that

$$2 B_k \nabla B_k = \sum_{k=1}^{3} 2 B_k \nabla B_k = 2 B_1 \nabla B_1 + 2 B_2 \nabla B_2 + 2 B_3 \nabla B_3$$

where ${\bf{B}}=B_k{\bf{e}}_k=B_1{\bf{e}}_1+B_2{\bf{e}}_3+B_3{\bf{e}}_3$ is a vector field. Also, ${\{\bf{e}}_1,{\bf{e}}_2,{\bf{e}}_3\}$ are the orthonormal basis in Cartesian coordinates. So you have three terms not nine! :)

Now, let us prove your identity

$$\begin{align} 2 B_k \nabla B_k &= 2 B_k \partial_i B_k {\bf{e}}_i \\ &= \partial_i (B_k B_k) {\bf{e}}_i \\ &= \partial_i ({\bf{B}} \cdot {\bf{B}}) {\bf{e}}_i \\ &= \nabla ({\bf{B}} \cdot {\bf{B}}) \\ &= \nabla B^2 \end{align}$$

That's all.