# Dot product with the gradient of a vector

I'm reading a physics book where the quantity $\mathbf B \cdot \nabla \mathbf B$ is often referred to, where $\nabla \mathbf B$ is the gradient of the magnetic field (a vector). I'm having trouble with this because the notation is sort of ambiguous. Let's say $\mathbf B$ is defined to be in the z-direction. If I split this into unit vectors:

$$\mathbf B \cdot \nabla \mathbf B = (B \hat z )\cdot \left ( \hat x \frac{\partial }{\partial x} + \hat y \frac{\partial}{\partial y} + \hat z \frac{\partial}{\partial z} \right )B \hat z \\ = (B \hat z )\cdot \left ( \hat x \hat z \frac{\partial B}{\partial x} + \hat y \hat z \frac{\partial B}{\partial y} + \hat z \hat z \frac{\partial B}{\partial z} \right ).$$

How do I do this dot product, do I use the first unit vector? So that

$$\mathbf B \cdot \nabla \mathbf B = B \frac{\partial B}{\partial z}\hat z?$$

Or should I use the second, in which

$$\mathbf B \cdot \nabla \mathbf B = B \left ( \hat x \frac{\partial B}{\partial x} + \hat y \frac{\partial B}{\partial y} + \hat z \frac{\partial B}{\partial z} \right ) = B \nabla B ?$$

• How do you take the gradient of a vector field? What you've written down (with $\hat x\hat z$, etc.) certainly doesn't make sense. – Ted Shifrin Apr 11 '18 at 20:24
• Well my book uses it all the time. – Mike Flynn Apr 11 '18 at 20:24
• So what's the definition? Uses what all the time? – Ted Shifrin Apr 11 '18 at 20:25
• It's possible to take the gradient of a vector, but the result is a tensor, not a vector. I'm not sure that's what the book really means, though. – Adrian Keister Apr 11 '18 at 20:26
• @TedShifrin the definition is never given. This is what I'm confused about. – Mike Flynn Apr 11 '18 at 20:28

The gradient of a vector $\nabla B$ is the 2nd order (2,0) tensor.
$$\nabla B = {\partial B_i\over\partial x_j}e_{ij}$$
$e_{i}$ is the unit (1,0) vector basis, and $e_{ij}=e_{i}\otimes e_{j}$ is the unit matrix (2,0) basis.
Hence the product $B\cdot\nabla B$ is: $$B\cdot \nabla B = B_i {\partial B_i\over\partial x_j}e_{ij}$$
For $B=B_3e_3$: $$B\cdot \nabla B = B_3{\partial B_3\over\partial x^{j}} e_{3j} = B_3{\partial B_3\over\partial x^{j}} e_{j} = B_3 \nabla B_3$$