Vector calculus first derivative associative properties

Question

I am confused by the vector calculus first derivative associative properties found here on Wikipedia, in particular the notation.

For example, the directional derivative $$(\hat{n}\cdot \nabla) \vec{A}$$

by first derivative association properties shown in the wikipedia page, this is equivalent to

$$\hat{n} \cdot (\nabla \vec{A})$$

So my question is, whats with the different placement of the brackets? Is there really a difference between the two mathematically?

Answer 1

1. By $\nabla$ we mean an operator given by $*\rightarrow\frac{\partial * }{\partial x} e_x+\frac{\partial * }{\partial y} e_y+ \frac{\partial * }{\partial z} e_z$.

2. $\vec{n}$ is also an operator which to each point $p= p_1 e_1+p_2 e_2+ p_3 e_3$ in a surface takes it to the normal $\vec{n}(p)= \hat{n}_1(p) e_x+\hat{n}_2(p) e_y+\hat{n}_1(p) e_z$

What do we mean by $\nabla \vec{A}$?

Using the formula in 1 we have that if $A=A_1e_x+A_2e_y+A_3e_z$:

$$\nabla \vec{A}=\frac{\partial \vec{A} }{\partial x} e_x+\frac{\partial \vec{A} }{\partial y} e_y+ \frac{\partial \vec{A} }{\partial z} e_z=$$ $$\frac{\partial A_1 }{\partial x} e_x e_x+\frac{\partial A_2 }{\partial x} e_y e_x+\frac{\partial A_3 }{\partial x} e_z e_x+$$ $$\frac{\partial A_1 }{\partial y} e_x e_y+\frac{\partial A_2 }{\partial y} e_y e_y+\frac{\partial A_3 }{\partial y} e_z e_y+$$ $$\frac{\partial A_1 }{\partial z} e_x e_z+\frac{\partial A_2 }{\partial z} e_y e_z+\frac{\partial A_3 }{\partial z} e_z e_z$$

You can see $e_i e_j$ as a formal basis to a vector space. This is usually called the tensor product.

What do we mean by $\vec{n}\cdot \nabla$?

This is also an operator given by: $*\rightarrow\hat{n}_1\frac{\partial * }{\partial x} +\hat{n}_2\frac{\partial * }{\partial y} + \hat{n}_3\frac{\partial * }{\partial z} e_z$. You can see this as an inner product between $\hat{n}$ and $\nabla$ really.

Why is it that $(\vec{n}\cdot \nabla) \vec{A}= \vec{n}\cdot (\nabla \vec{A})$?

$$(\vec{n}\cdot \nabla) \vec{A}=\left(\hat{n}_1\frac{\partial * }{\partial x} +\hat{n}_2\frac{\partial * }{\partial y} + \hat{n}_3\frac{\partial * }{\partial z} e_z\right)\vec{A}=$$ $$ \hat{n}_1\frac{\partial A_1}{\partial x}e_x+\hat{n}_1\frac{\partial A_2}{\partial x}e_y+\hat{n}_1\frac{\partial A_3}{\partial x}e_z+$$ $$ \hat{n}_2\frac{\partial A_1}{\partial y}e_x+\hat{n}_2\frac{\partial A_2}{\partial y}e_y+\hat{n}_2\frac{\partial A_3}{\partial y}e_z+$$ $$ \hat{n}_3\frac{\partial A_1}{\partial z}e_x+\hat{n}_3\frac{\partial A_2}{\partial z}e_y+\hat{n}_3\frac{\partial A_3}{\partial z}e_z=$$ $$(\hat{n}_1 e_x )\cdot(\nabla \vec{A})+(\hat{n}_2 e_2 )\cdot(\nabla \vec{A})+(\hat{n}_3 e_z )\cdot(\nabla \vec{A}) $$

This last equality is seen by taking the inner product defined in tensor products $e_i\cdot (e_j e_k)= e_j \delta_{ik}$ Finally, using linearity of the inner product:

$$(\hat{n}_1 e_x )\cdot(\nabla \vec{A})+(\hat{n}_2 e_2 )\cdot(\nabla \vec{A})+(\hat{n}_3 e_z )\cdot(\nabla \vec{A}) =(\hat{n}_1 e_x+\hat{n}_2 e_y+ \hat{n}_3 e_z)\cdot (\nabla \vec{A})=\vec{n}\cdot \nabla \vec{A}$$