Vector Calculus - Use of Identities

12x4
Messages
28
Reaction score
0

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.
 
Physics news on Phys.org
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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top