Some complicated vector calculus..

**limddavid** · October 18th 2011, 10:21 PM

Not sure if the description fits the question I have, but here it goes:

"If a vector function f(x,y,z) is not irrotational but the product of f and a scalar function g(x,y,z) is irrotational, show that then:

$\mathbf{f}\cdot \boldsymbol{\triangledown} \times \mathbf{f}=0$ "

I've been working at this for a couple hours now... I was able to find that since:

$\boldsymbol{\triangledown} \times (g\mathbf{f})=0$
and
$\boldsymbol{\triangledown} \times (g\mathbf{f})=g(\triangledown \times \mathbf{f})+(\triangledown g)\times \mathbf{f}$ ,

then

$g(\triangledown \times \mathbf{f})=-(\triangledown g)\times \mathbf{f}$

I don't even know if this is in the right direction. Help!

**Danny** · October 19th 2011, 05:00 AM

Originally Posted by limddavid

Not sure if the description fits the question I have, but here it goes:

"If a vector function f(x,y,z) is not irrotational but the product of f and a scalar function g(x,y,z) is irrotational, show that then:

$\mathbf{f}\cdot \boldsymbol{\triangledown} \times \mathbf{f}=0$ "

I've been working at this for a couple hours now... I was able to find that since:

$\boldsymbol{\triangledown} \times (g\mathbf{f})=0$
and
$\boldsymbol{\triangledown} \times (g\mathbf{f})=g(\triangledown \times \mathbf{f})+(\triangledown g)\times \mathbf{f}$ ,

then

$g(\triangledown \times \mathbf{f})=-(\triangledown g)\times \mathbf{f}$

I don't even know if this is in the right direction. Help!

Ya, this looks good! I will use the identity that

$\mathbf{u} \cdot \left(\mathbf{v}\times \mathbf{w}) = \left(\mathbf{u} \times \mathbf{v}\right) \cdot \mathbf{w}$

From what you have

$\triangledown \times \mathbf{f}=-\frac{\triangledown g}{g} \times \mathbf{f} }= \mathbf{f} \times \frac{\triangledown g}{g}$ .

So, $\mathbf{f} \cdot \triangledown \times \mathbf{f} = \mathbf{f}\cdot \left( \mathbf{f} \times \frac{\triangledown g}{g}\right) = \left( \mathbf{f} \times \mathbf{f}\right) \cdot \frac{\triangledown g}{g}\right) = 0$ , since $\mathbf{f} \times \mathbf{f} = 0$ .

**limddavid** · October 19th 2011, 01:20 PM

Perfect. Now it's clear. thank you!

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