Gradient of a dot product identity proof?

AI Thread Summary
The discussion focuses on proving the vector identity for the gradient of a dot product, specifically ∇(A·B) = A×(∇×B) + B×(∇×A) + (A·∇)B + (B·∇)A. The original poster struggles to simplify the equation and ends up with an incorrect result, indicating a misunderstanding of the gradient operator's properties. They express a desire to avoid complex methods involving Levi-Civita symbols and Kronecker Delta, seeking a more straightforward approach. However, other participants suggest that using these mathematical tools would simplify the proof significantly. Ultimately, the consensus is that while a component-wise approach may work, learning the advanced methods is beneficial for clarity and efficiency.
Libohove90
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Gradient of a dot product identity proof?

Homework Statement


I have been given a E&M homework assignment to prove all the vector identities in the front cover of Griffith's E&M textbook. I have trouble proving:

(1) ∇(A\bulletB) = A×(∇×B)+B×(∇×A)+(A\bullet∇)B+(B\bullet∇)A

Homework Equations


(2) A×(B×C) = B(A\bulletC)-C(A\bulletB)


The Attempt at a Solution


I applied the identity in equation (2) to the first two terms on the right hand side of equation (1), and that allowed me to cancel out 4 terms. Yet, I end up with:

∇(A\bulletB) = ∇(A\bulletB)+∇(B\bulletA)

...which I know cannot be correct. How can I prove this identity in a relatively straightforward way? I have seen other pages asking this yet they all involved the use of Levi-Cevita symbols and the Kronecker Delta, which I am trying not to use because we haven't learned them. I would gladly appreciate anyone's effort to help me out.
 
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The only sensible way I can see is to do it by hand for let's say the <x> component in both sides and show they are the same.
 


I was hoping I can get around the long calculations lol
 


You have to be careful with the ∇ operator in vector identities, as it is not commutative. I think this caused the problem here.
The long calculation will work, and it is sufficient to consider one component.
 


Libohove90 said:
I was hoping I can get around the long calculations lol
I know you wanted avoid them, but it's definitely worth learning about the Kronecker delta and Levi-Civita symbol. Using them makes verifying the identity much easier.
 
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