B ##\nabla## dot product

  • Thread starter Apashanka
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307
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From the vector identity ##\nabla •fA=f(\nabla • A)+A•\nabla f## where f is a scalar and A is a vector.
Now if f is an operator acting on A how does this formula change??
Like ##\nabla •[(v•\nabla)v]## where v is a vector
 
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Charles Link

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The most sure way of getting the answer would be to write out all of the terms in Cartesian coordinates... I will try to work on it a little and see what I get...## \\ ## Edit: With about 10 minutes of work on the above, I believe I get ## \nabla \cdot [(\vec{v} \cdot \nabla) \vec{v}]=\frac{1}{2} \nabla^2 (\vec{v}\cdot \vec{v}) ##. ## \\ ## I'll be happy to show more detail, but basically, I just worked with Cartesian coordinates.
 
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Charles Link

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A check on my above work shows## \nabla \cdot [(\vec{v} \cdot \nabla) \vec{v}]=\frac{1}{2}\nabla^2 (\vec{v} \cdot \vec{v})-\nabla \cdot (\vec{v} \times \nabla \times \vec{v}) ##. ## \\ ## I think the second expression on the right side is zero, but I haven't proven it yet. If the expression in post 2 is not correct, then post 3 contains your answer. I do think post 2 is correct. ## \\ ## Edit: I made an error in post 2. I believe post 3 is correct. ## \\ ## To see how I got the result of this post, begin with ## \nabla (\vec{v} \cdot \vec{v}) ## and use ## \nabla (\vec{a} \cdot \vec{b})=(\vec{a} \cdot \nabla ) \vec{b}+(\vec{b} \cdot \nabla )\vec{a}+\vec{a} \times \nabla \times \vec{b}+\vec{b} \times \nabla \times \vec{a} ##. ## \\ ## Next take ## \nabla \cdot ## on the expression, to give ## \nabla^2 (\vec{v} \cdot \vec{v}) ##.
 
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