Vector Calculus - Use of Identities

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The discussion revolves around using vector identities to prove that ∇ × (φ∇φ) = 0, where φ is a scalar field. Participants explore the identity ∇×(φA) = (∇φ)×A + φ(∇×A) and attempt to apply it to the problem. A key point is recognizing that the cross product of a vector with itself, such as (∇φ) × (∇φ), equals zero. Clarifications on notation and the importance of proper bracket placement are emphasized to avoid confusion. Ultimately, understanding these concepts leads to the conclusion that the expression evaluates to zero.
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Homework Statement



By using a suitable vector identity for ∇ × (φA), where φ(r) is a scalar field and A(r) is a vector field, show that
  1. ∇ × (φ∇φ) = 0,
    where φ(r) is any scalar field.

Homework Equations



∇×(φA) = (∇φ)×A+φ(∇×A)?

The Attempt at a Solution



I honestly have no idea how to even start this one. Any help would be hugely appreciated.
 
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You can apply the relevant equation to the expression you are asked to evaluate.
 
mfb said:
You can apply the relevant equation to the expression you are asked to evaluate.

Hi mfb, thank you for your reply. That is what I tried to do before but thought I must have been doing it wrong as i ended up with something that didn't cancel.

Okay so can I assume A = ∇φ
and the identity is ∇×(φA) = (∇φ)×A+φ(∇×A)

and follow that through like this:?

∇ x (φ∇φ) = ∇φ x ∇φ + φ∇ x ∇φ

I feel as if there is another identity that I must need to complete the question but have searched through my notes and online and can't find anything. Does ∇φ x ∇φ = -(φ∇ x ∇φ) in some way?
 
I've just had a thought. Are you able to say

A = ∇φ and therefore -A = φ∇

Allowing us to complete the question like this.

∇ x (φ∇φ) = (A x A) + (-A x A)
∇ x (φ∇φ) = (A x A) - (A x A)
= 0
 
12x4 said:
Hi mfb, thank you for your reply. That is what I tried to do before but thought I must have been doing it wrong as i ended up with something that didn't cancel.
So you had an idea where to start, and even a start. Why didn't you write that down? Then it would have been possible to see what went wrong.

∇ x (φ∇φ) = ∇φ x ∇φ + φ∇ x ∇φ
I would probably add brackets to make it more clear: (∇φ) x (∇φ) + φ(∇ x ∇φ)
What is the cross-product of something with itself?

12x4 said:
A = ∇φ and therefore -A = φ∇
No. The right equation is not even well-defined.
 
mfb said:
So you had an idea where to start, and even a start. Why didn't you write that down? Then it would have been possible to see what went wrong.

Sorry mfb, I thought my working was complete rubbish and irrelevant so didn't want to embarrass myself on here.

mfb said:
I would probably add brackets to make it more clear: (∇φ) x (∇φ) + φ(∇ x ∇φ)
What is the cross-product of something with itself?

No. The right equation is not even well-defined.

I think I understand now, I didn't realize that you can put a set of bracket in like that. And the cross product of something with itself is zero. Thanks!
 
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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