Proof: $(\nabla f \times \nabla g)$ is Solenoidal

[cristina89]

Be f and g two differentiable scalar field. Proof that ($\nabla$f) x ($\nabla$g) is solenoidal.

 

[Dick]

Be f and g two differentiable scalar field. Proof that ($\nabla$f) x ($\nabla$g) is solenoidal.

Show what you've done so far. What would you do to show a vector field is solenoidal?

 

[cristina89]

Show what you've done so far. What would you do to show a vector field is solenoidal?

Well, to be solenoidal, I know that $\nabla$ $\cdot$ ($\nabla$f x $\nabla$g) needs to be 0.

So,

$\nabla$ $\cdot$ ($\nabla$f x $\nabla$g) = $\nabla$g $\cdot$ ($\nabla$ x $\nabla$f) - $\nabla$f $\cdot$ ($\nabla$ x $\nabla$g)

Right? But why is this equal to zero?

 

[I like Serena]

Welcome to PF, cristina89!

Did you know that $\nabla \times \nabla f = 0$ for any differential scalar field f?

It is one of the non-trivial identities of the del operator.
See for instance: http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities

 

[cristina89]

Welcome to PF, cristina89!

Did you know that $\nabla \times \nabla f = 0$ for any differential scalar field f?

It is one of the non-trivial identities of the del operator.
See for instance: http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities

Thank you so much!