Two-dimensional vector representation

[Gribouille]

Hi,

Is there a method to represent a known two-dimensional vector field w of two coordinates x and y with zero divergence and non-zero curl as
$$ \vec{w}(x,y) = a \nabla b \, , \hspace{4mm} \nabla \cdot \vec{w} = 0 \, , \hspace{4mm} \nabla \times \vec{w} = f(x,y) \, ?$$
How would one proceed to calculate a and b?

 

[member 587159]

Hi,

Is there a method to represent a known two-dimensional vector field w of two coordinates x and y with zero divergence and non-zero curl as
$$ \vec{w}(x,y) = a \nabla b \, , \hspace{4mm} \nabla \cdot \vec{w} = 0 \, , \hspace{4mm} \nabla \times \vec{w} = f(x,y) \, ?$$
How would one proceed to calculate a and b?

No, I don't think so. Assuming a is constant, your vector field can be written as a gradient, which means it is conservative and therefore rotation-free, contradicting your assumption.

 

[Gribouille]

Thanks. a is not constant but depends on x and y, just as b.

 

[mfb]

Unless I made a mistake it is possible sometimes but not in general.

As negative example, consider $\vec w = \vec c \times \vec r$ where r is the position and c is a constant. Consider the unit circle. To get the direction right, $\nabla b$ has to be non-zero but going in a circle. It can't do that without having a rotation, contradiction.

As positive non-trivial example, use the w from above within the unit circle, then continue outside in a symmetric way with zero curl outside, and then add $d=(10,0)$ to it. Now our vector field doesn't have closed circles any more. We can introduce a suitable potential that gets the direction of the gradient right, and then fix the magnitude via a variable $a$.

 

[Hassan2]

if a is a vector and w =a×∇b, then yes. w then is the magnetic field, a is a unit vector normal to x-y plane, b is the magnetic vector potential, and f(x,y) is the electric current density that creates the magnetic field.