Proving the Irrotational Property of Vector Fields with an Example Solution

[neelakash]

Homework Statement

A vector field V is not irrotational.Show that it is always possible to find f such that fV is irrotational.

Homework Equations

The Attempt at a Solution

\nablax[fV]=f\nablaxV-Vx\nablaf
I have to equate the LHS to zero.But then,how can I extract f out of the resulting equation?
Please help

 

[neelakash]

I don't know why that \nabla didn't work.Please bear with this problem and help me.

 

[m2003]

me out



To prove the irrotational property of a vector field V, we need to show that the curl of V, denoted by \nabla \times V, is equal to zero. This means that the vector field has no rotational component and is therefore conservative.

Now, if we have a vector field V that is not irrotational, we can find a scalar function f such that fV is irrotational. This can be shown by using the vector identity \nabla \times (fV) = f\nabla \times V + V \times \nabla f.

Since we want fV to be irrotational, we can set the left-hand side of this equation to zero. This gives us f\nabla \times V = -V \times \nabla f.

To extract f from this equation, we can use the cross product property that states V \times \nabla f = -\nabla f \times V. This allows us to rewrite the equation as f\nabla \times V = \nabla f \times V.

Now, we can take the dot product of both sides with V to get f(\nabla \times V) \cdot V = (\nabla f \times V) \cdot V. Since the dot product of a vector with its own curl is always zero, we are left with f = 0.

Therefore, we have shown that it is always possible to find a scalar function f such that fV is irrotational. This proves the irrotational property of vector fields.