Expression with two vectors and del operator

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Homework Help Overview

The discussion revolves around the expression (A.∇)B, which involves vector A and the del operator applied to vector B. Participants are exploring the meaning and expansion of this expression using Cartesian components.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the implications of the expression (A.∇)B and how to expand it. There is a focus on the correct interpretation of the order of operations involving the vectors A and B.

Discussion Status

Some participants have provided insights into the mathematical representation of the expression, while others are questioning the equivalence of different orders of the vectors and the implications for the derivatives involved. There appears to be productive dialogue regarding the correct understanding of the operation.

Contextual Notes

There is mention of confusion regarding the order of the vectors in the expression, which may indicate a need for clarification on vector calculus concepts. Participants are also reflecting on the implications of the operation for the components of the vectors involved.

BOYLANATOR
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Homework Statement


(A.∇)B



What does this mean and how do I go about trying to expand this (using cartesian components)?
 
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BOYLANATOR said:

Homework Statement


(A.∇)B

What does this mean and how do I go about trying to expand this (using cartesian components)?
Well, \displaystyle \vec{\text{A}}\cdot\vec{\nabla}=\text{A}_x\,\frac{ \partial}{\partial x}+\text{A}_y\,\frac{ \partial}{\partial y}+\text{A}_z\,\frac{ \partial}{\partial z}\ .<br />

So that \displaystyle \left(\vec{\text{A}}\cdot\vec{\nabla}\right) \vec{\text{B}}=\text{A}_x\,\frac{ \partial\vec{\text{B}}}{\partial x}+\text{A}_y\,\frac{ \partial\vec{\text{B}}}{\partial y}+\text{A}_z\,\frac{ \partial\vec{\text{B}}}{\partial z}\ .<br />
 
Of course. It was the order (A.∇) rather than (∇.A) that confused me but now I realize that these are obviously equivalent. Thanks
 
BOYLANATOR said:
Of course. It was the order (A.∇) rather than (∇.A) that confused me but now I realize that these are obviously equivalent. Thanks

Actually that's wrong isn't it?
If the order is reversed you find the derivatives of each component of A. In the problem case though the derivation is carried out on the components of B. I hope this is right as it makes sense in my head.
 
Yes, it's correct. The whole expression is called the derivative of B along A.
 

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