Is \(\nabla \times (\phi \nabla \phi) = 0\) for a Differentiable Scalar Field?

[cristina89]

How to prove that $\nabla$ x ($\phi$$\nabla$$\phi$) = 0?
($\phi$ is a differentiable scalar field)

I'm a bit confused by this "differentiable scalar field" thing...

 

[HallsofIvy]

Are you saying that you do not know what a "differentiable scalar field" means? A "scalar" is simply a number, rather than a vector. This is just saying that $\Phi(x, y, z)$ is a (differentiable) function that returns a number for each point (x,y,z)- exactly the kind of function you are used to working with!

And, of course, $\nabla \Phi$ is the vector function
$$\frac{\partial\Phi}{\partial x}\vec{i}+ \frac{\partial\Phi}{\partial y}\vec{j}+ \frac{\partial\Phi}{\partial z}\vec{k}$$

so that $\Phi\nabla\Phi$ is that vector multiplied by the number $\Phi$:
$$\Phi\frac{\partial\Phi}{\partial x}\vec{i}+ \Phi\frac{\partial\Phi}{\partial y}\vec{j}+ \Phi\frac{\partial\Phi}{\partial z}\vec{k}$$
so is a "vector valued" function- it returns that vector at each point (x, y, z).

Finally,
$$\nabla\times(\Phi\nabla\Phi)$$
is the "curl" of that vector valued function.

 

[I like Serena]

Hi!

The way to proof such an expression is to write it out into x, y, and z components and simplify it (as HallsofIvy is suggesting).

As an alternative you can use the curl identities that you can find for instance on wiki:
http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities

 

[cristina89]

Hi!

The way to proof such an expression is to write it out into x, y, and z components and simplify it (as HallsofIvy is suggesting).

As an alternative you can use the curl identities that you can find for instance on wiki:
http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities

Thank you! Just figured out how to solve this!