What Does (\vec{p} \cdot \nabla)\vec{E} Represent in Griffiths Eq. 4.5?

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[SOLVED] Griffiths eqn 4.5

Homework Statement


This question refers to Griffiths E and M book.
\vec{F} = (\vec{p} \cdot \nabla)\vec{E}
Can someone please explain what the expression in parenthesis is? I have never seen that before? Is it a vector or a scalar or an operator or what? And what is the operation between what is parenthesis and E? I think Griffiths really botched this. Obviously Griffiths wants to use this notation to write "more compactly" the equation before the previous equation, but he cannot just make up notation like that!

Homework Equations


The Attempt at a Solution

 
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ehrenfest said:

Homework Statement


This question refers to Griffiths E and M book.
\vec{F} = (\vec{p} \cdot \nabla)\vec{E}
Can someone please explain what the expression in parenthesis is? I have never seen that before? Is it a vector or a scalar or an operator or what? And what is the operation between what is parenthesis and E? I think Griffiths really botched this. Obviously Griffiths wants to use this notation to write "more compactly" the equation before the previous equation, but he cannot just make up notation like that!

Homework Equations





The Attempt at a Solution


I don't find anything ambiguous about it.

\vec{p} \cdot \nabla = p_x \partial_x + p_y \partial_y + p_z \partial_z
so this is a scalar operator. Then you apply it to each of the components of E to get each of the components of F.
 
I see what you mean.
 

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