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

[cristina89]

How to prove that [itex]\nabla[/itex] x ([itex]\phi[/itex][itex]\nabla[/itex][itex]\phi[/itex]) = 0?
([itex]\phi[/itex] 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 [itex]\Phi(x, y, z)[/itex] 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, [itex]\nabla \Phi[/itex] is the vector function
[tex]\frac{\partial\Phi}{\partial x}\vec{i}+ \frac{\partial\Phi}{\partial y}\vec{j}+ \frac{\partial\Phi}{\partial z}\vec{k}[/tex]

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

Finally,
[tex]\nabla\times(\Phi\nabla\Phi)[/tex]
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!

 

[CellCoree]

I'm assuming you mean a scalar field that is differentiable, in which case, the proof for \nabla x (\phi\nabla\phi) = 0 is quite straightforward.

First, let's define what \nabla x (\phi\nabla\phi) means. The cross product \nabla x \nabla is a vector operation that takes two vector fields and produces a third vector field. In this case, we are taking the cross product of the vector field \nabla\phi and the scalar field \phi, which produces a vector field.

Now, let's break down the components of this vector field. The components of \nabla\phi are the partial derivatives of \phi with respect to each variable (x, y, and z). The components of \phi are just the values of the scalar field at each point.

So, the cross product \nabla x (\phi\nabla\phi) can be written as:

(\frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z}) x (\phi, \phi, \phi)

Using the properties of the cross product, we can simplify this to:

(\frac{\partial\phi}{\partial y}\phi - \frac{\partial\phi}{\partial z}\phi, \frac{\partial\phi}{\partial z}\phi - \frac{\partial\phi}{\partial x}\phi, \frac{\partial\phi}{\partial x}\phi - \frac{\partial\phi}{\partial y}\phi)

Now, let's look at each component individually. We can see that each one is just the product of a partial derivative and the scalar field \phi. And since \phi is differentiable, its partial derivatives exist and are continuous. This means that as we take the limit of these components as the distance between two points approaches zero, the difference in the values of \phi at those two points also approaches zero. In other words, each component approaches zero as the distance between two points approaches zero.

Since this is true for all three components, we can conclude that the entire vector field \nabla x (\phi\nabla\phi) approaches the zero vector as the distance between two points approaches zero. And since this holds true for any point in the field, we can say that the vector