Gradient of the dot product of two vectors that are the same

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To find the gradient of the dot product of two identical vectors, specifically ∇(E · E), where E is expressed as E_{0}(exp(i(kx-ωt))), the discussion clarifies that E · E simplifies to ||E||². The gradient can be computed by treating the magnitude squared as a scalar function, allowing for the application of partial derivatives. The suggestion is to express the function without vectors to facilitate the calculation. The final expression for the gradient is indicated to be 2ikE_{0}². Understanding vector identities is essential for this computation.
wobblybird
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Hi,

I am trying find the simplified expression of this:
∇(E \cdot E)

Where E is the electric field that can written as E_{0}(exp(i(kx-ωt))

I know that since the two vectors are the same => E \cdot E = ||E||^{2}

Do I take the gradient of the magnitude then? It just doesn't feel right. Or should it be something like 2ikE_{0}^2?

Thank you so much!
 
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You can just take the gradient of the square of the magnitude - write it out as a function with no vectors involved, and calculate partial derivatives
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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