Vector Cross Product With Its Curl

  • #1

Homework Statement:

Prove A X (∇ X A) = ∇(A²/2) - A · ∇A

Relevant Equations:

εijk εlmk = δil δjm - δim δjl
Starting with LHS:

i εijk Aj (∇xA)k

i εijk εlmk Aj (d/dxl) Am

il δjm - δim δjl) Aj (d/dxl) Ami

δil δjm Aj (d/dxl) Ami - δim δjl Aj (d/dxl) Ami

Aj (d/dxi) Aji - Aj (d/dxj) Aii

At this point, the LHS should equal the RHS in the problem statement, but I have no clue where the 1/2 comes from...been trying to figure it out for the past few hours at this point. Rather, I get this and I'm not sure where my lapse in understanding is (I'm relatively new to index notation):

∇(A²) - A · ∇A
 
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Answers and Replies

  • #2
34,683
10,832
What is the derivative of f(x)=x2/2?
Your expression here is analogous.
 
  • #3
Notice that ##A^2 = A \cdot A##. When the gradient operator is applied to this term, you get two terms looking like this: ##A(\nabla A)##. In your initial work, you only have one of these terms, so to account for the duplicate, you must divide ##\nabla A^2## by two, which results in the term ##\nabla \frac{A^2}{2}## appearing in the identity you are attempting to prove.
 
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