Trouble Understanding Vector Identity with curl and divergence

Question

I have trouble understanding the second line on the RHS. Why is a part of the property written as "∇·B + B·∇" and not as 2(∇·B) or 2(B·∇)? What is the difference between "∇·B" and "B·∇"? I thought the dot product was commutative.

Answer 1

The dot product of two vectors is commutative, but notice $\nabla\cdot\mathbf{B}$ is a scalar whereas $\mathbf{B}\cdot\nabla$ is an operator; they're not even the same type of thing, let alone equal. One then reinterprets the scalar $\nabla\cdot\mathbf{B}$ as the operator which multiplies by that scalar in order to understand the notation. The operator $\mathbf{B}\cdot\nabla$ is also problematic since it operates on scalar functions, so $(\mathbf{B}\cdot\nabla)\mathbf{A}$ wouldn't make sense, except we choose to interpret it as the operator $\mathbf{B}\cdot\nabla$ acting on each component of $\mathbf{A}$ separately (I mean, that's what you do when you multiply a vector by a scalar isn't it - multiply each component separately? although this isn't a scalar).

Notice these are not the same thing:

$$ (\nabla\cdot\mathbf{B})\mathbf{A} = \left(\frac{\partial B_1}{\partial x_1}+\frac{\partial B_2}{\partial x_2}+\frac{\partial B_3}{\partial x_3}\right)\mathbf{A} = \begin{bmatrix} \displaystyle\left(\frac{\partial B_1}{\partial x_1}+\frac{\partial B_2}{\partial x_2}+\frac{\partial B_3}{\partial x_3}\right) A_1 \\ \displaystyle\left(\frac{\partial B_1}{\partial x_1}+\frac{\partial B_2}{\partial x_2}+\frac{\partial B_3}{\partial x_3}\right) A_2 \\ \displaystyle\left(\frac{\partial B_1}{\partial x_1}+\frac{\partial B_2}{\partial x_2}+\frac{\partial B_3}{\partial x_3}\right) A_3 \end{bmatrix} $$

whereas

$$ (\mathbf{B}\cdot\nabla)\mathbf{A} = \left(B_1\frac{\partial}{\partial x_1}+B_2\frac{\partial}{\partial x_2}+B_3\frac{\partial}{\partial x_3}\right)\mathbf{A} = \begin{bmatrix} \displaystyle B_1\frac{\partial A_1}{\partial x_1}+B_2\frac{\partial A_1}{\partial x_2}+B_3\frac{\partial A_1}{\partial x_3} \\ \displaystyle B_1\frac{\partial A_2}{\partial x_1}+B_2\frac{\partial A_2}{\partial x_2}+B_3\frac{\partial A_2}{\partial x_3} \\ \displaystyle B_1\frac{\partial A_3}{\partial x_1}+B_2\frac{\partial A_3}{\partial x_2}+B_3\frac{\partial A_3}{\partial x_3} \end{bmatrix} $$

Expecting them to be equal is kind of like expecting $\left(x\frac{\mathrm{d}}{\mathrm{d}x}\right)y$ and $\left(\frac{\mathrm{d}}{\mathrm{d}x} x\right)y$ to be equal just because "multiplication is commutative"; the first is $xy'$ while the latter denotes $(xy)'$. In fact this is the 1D version of the this very issue, with $B(x)=x$ and $A(x)=y(x)$ and $\nabla=\frac{\mathrm{d}}{\mathrm{d}x}$. Multiplying scalars may be commutative, but multiplying operators is not.