Curl of Product

I

I'mLearning

New member
Joined
Jan 22, 2024
Messages
13
Do you know if k⋅∇×(a⋅∇a)\mathbf{k}\cdot\nabla\times(\mathbf{a}\cdot\nabla\mathbf{a}) can be rewritten?

If so, which rules are applied?

EDIT: The term seems to equal (a⋅∇k⋅∇×a)+(k⋅∇×a∇⋅a)(\mathbf{a}\cdot\nabla\mathbf{k}\cdot\nabla\times\mathbf{a}) + (\mathbf{k}\cdot\nabla\times\mathbf{a}\nabla\cdot\mathbf{a}) Can that be true?
 
I'mLearning said:
Do you know if k⋅∇×(a⋅∇a)\mathbf{k}\cdot\nabla\times(\mathbf{a}\cdot\nabla\mathbf{a}) can be rewritten?

If so, which rules are applied?

EDIT: The term seems to equal (a⋅∇k⋅∇×a)+(k⋅∇×a∇⋅a)(\mathbf{a}\cdot\nabla\mathbf{k}\cdot\nabla\times\mathbf{a}) + (\mathbf{k}\cdot\nabla\times\mathbf{a}\nabla\cdot\mathbf{a}) Can that be true?
Click to expand...
What is k and a? Are those related anyway?
 
I'mLearning said:
Do you know if k⋅∇×(a⋅∇a)\mathbf{k}\cdot\nabla\times(\mathbf{a}\cdot\nabla\mathbf{a}) can be rewritten?

If so, which rules are applied?

EDIT: The term seems to equal (a⋅∇k⋅∇×a)+(k⋅∇×a∇⋅a)(\mathbf{a}\cdot\nabla\mathbf{k}\cdot\nabla\times\mathbf{a}) + (\mathbf{k}\cdot\nabla\times\mathbf{a}\nabla\cdot\mathbf{a}) Can that be true?
Click to expand...
Hi I'mLearning,
not me I meant you! 😁


Let k=i+j+k\bold{k} = i + j + k and let a=2i+3j+4k\bold{a} = 2i + 3j + 4k

Solve the first form. Save your result for future comparison!

Solve the second form. Compare the two results. Are they equal?
 
You must log in or register to reply here.
Top