Expression with two vectors and del operator

  • Thread starter BOYLANATOR
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  • #1
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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  • #2
SammyS
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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)?
Well, [itex]\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}\ .
[/itex]

So that [itex]\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}\ .
[/itex]
 
  • #3
BOYLANATOR
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Of course. It was the order (A.∇) rather than (∇.A) that confused me but now I realise that these are obviously equivalent. Thanks
 
  • #4
BOYLANATOR
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Of course. It was the order (A.∇) rather than (∇.A) that confused me but now I realise 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.
 
  • #5
dextercioby
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Yes, it's correct. The whole expression is called the derivative of B along A.
 

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